Qunatitative Aptitude Questions

Q.- The L.C.M. of two numbers is 48. The numbers are in the ratio 2:3. Then sum of the number is:

    1. 1. 40
    2. 2. 50
    3. 3. 60
    4. 4. 35

1
Let the numbers be 2x and 3x Then their L.C.M. = 6x So 6x = 48 or x = 8 The numbers are 16 and 24 Hence required sum = (16 + 24) = 40

Q.- Which of the following has the most number of divisors?

    1. 1. 99
    2. 2. 101
    3. 3. 182
    4. 4. 176

4
99 = 1 x 3 x 3 x 11 101 = 1 x 101 176 = 1 x 2 x 2 x 2 x 2 x 11 182 = 1 x 2 x 7 x 13 So divisors of 99 are 1 3 9 11 33 .99 Divisors of 101 are 1 and 101 Divisors of 176 are 1 2 4 8 11 16 22 44 88 and 176 Divisors of 182 are 1 2 7 13 14 26 91 and 182. Hence 176 has the most number of divisors.

Q.- The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively is:

    1. 1. 271
    2. 2. 172
    3. 3. 721
    4. 4. 127

4
Required number = H.C.F. of (1657 - 6) and (2037 - 5) = H.C.F. of 1651 and 2032 = 127

Q.- The least number which when divided by 12 15 20 and 54 leaves in each case a remainder of 8 is:

    1. 1. 458
    2. 2. 548
    3. 3. 854
    4. 4. 584

2
Explanation : Required number = (L.C.M. of 12 15 20 54) + 8 = 540 + 8 = 548

Q.- Three numbers which are co-prime to each other are such that the product of the first two is 551 and that of the last two is 1073. The sum of the three numbers is:

    1. 1. 56
    2. 2. 65
    3. 3. 85
    4. 4. 58

3
Since the numbers are co-prime they contain only 1 as the common factor. Also the given two products have the middle number in common. So middle number = H.C.F. of 551 and 1073 = 29; First number = 551/29 = 19; Third number = 1073/29 = 37 Required sum = (19 + 29 + 37) = 85

Q.- The greatest possible length which can be used to measure exactly the lengths 7 m 3 m 85 cm 12 m 95 cm is:

    1. 1. 35 cm
    2. 2. 53 cm
    3. 3. 55 cm
    4. 4. 65 cm

1
Required length = H.C.F. of 700 cm 385 cm and 1295 cm = 35 cm

Q.- 252 can be expressed as a product of primes as:

    1. 1. 2x2x3x3x7
    2. 2. 3x3x2x2x7
    3. 3. 7x2x2x3x3
    4. 4. 2x7x3x2x3

1
Explanation : Clearly 252 = 2 x 2 x 3 x 3 x 7

Q.- The smallest number which when diminished by 7 is divisible 12 16 18 21 and 28 is:

    1. 1. 1501
    2. 2. 1015
    3. 3. 5110
    4. 4. 1510

2
Required number = (L.C.M. of 1216 18 21 28) + 7 = 1008 + 7 = 1015

Q.- What will be the least number which when doubled will be exactly divisible by 12 18 21 and 30 ?

    1. 1. 306
    2. 2. 360
    3. 3. 630
    4. 4. 260

3
L.C.M. of 12 18 21 30 = 2 x 3 x 2 x 3 x 7 x 5 = 1260 Required number = (1260 � 2) = 630

Q.- The least number which when divided by 5 6 7 and 8 leaves a remainder 3 but when divided by 9 leaves no remainder is:

    1. 1. 3168
    2. 2. 3138
    3. 3. 8163
    4. 4. 1683

4
L.C.M. of 5 6 7 8 = 840 Required number is of the form 840k + 3 Least value of k for which (840k + 3) is divisible by 9 is k = 2 Required number = (840 x 2 + 3) = 1683

Q.- Six bells commence tolling together and toll at intervals of 2 4 6 8 10 and 12 seconds respectively. In 30 minutes how many times do they toll together ?

    1. 1. 4
    2. 2. 10
    3. 3. 15
    4. 4. 16

4
L.C.M. of 2 4 6 8 10 12 is 120. So the bells will toll together after every 120 seconds(2 minutes). In 30 minutes they will toll together (30/2) + 1 = 16 times.

Q.- The greatest number of four digits which is divisible by 15 25 40 and 75 is:

    1. 1. 9000
    2. 2. 9400
    3. 3. 9600
    4. 4. 9800

3
Greatest number of 4-digits is 9999. L.C.M. of 15 25 40 and 75 is 600. On dividing 9999 by 600 the remainder is 399. Required number (9999 - 399) = 9600.

Q.- Three number are in the ratio of 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F. is:

    1. 1. 40
    2. 2. 80
    3. 3. 120
    4. 4. 200

1
Let the numbers be 3x 4x and 5x. Then their L.C.M. = 60x. So 60x = 2400 or x = 40. The numbers are (3 x 40) (4 x 40) and (5 x 40). Hence required H.C.F. = 40.

Q.- A B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds B in 308 seconds and c in 198 seconds all starting at the same point./n/n After what time will they again at the starting point ?

    1. 1. 2727 sec
    2. 2. 2772 sec
    3. 3. 7227 sec
    4. 4. 7272 sec

2
L.C.M. of 252 308 and 198 = 2772. So A B and C will again meet at the starting point in 2772 sec.

Q.- The H.C.F. of 9/10 12/25 18/35 and 21/40 is:

    1. 1. 3/140
    2. 2. 140/3
    3. 3. 3/1400
    4. 4. 3/1/1400

3
Required H.C.F. = (H.C.F. of 9 12 18 21)/(L.C.M. of 10 25 35 40) = 3/1400

Q.- If the product of two numbers is 84942 and their H.C.F. is 33 find their L.C.M.

    1. 1. 2574
    2. 2. 7425
    3. 3. 5274
    4. 4. 5247

1
HCF * LCM = 84942 because we know Product of two numbers = Product of HCF and LCM LCM = 84942/33 = 2574

Q.- Find the greatest number that will divide 400 435 and 541 leaving 9 10 and 14 as remainders respectively

    1. 1. 15
    2. 2. 17
    3. 3. 16
    4. 4. 18

2
Answer will be HCF of (400-9 435-10 541-14) HCF of (391 425 527) = 17

Q.- The H.C.F. and L.C.M. of two numbers are 12 and 5040 respectively. If one of the numbers is 144 find the other number

    1. 1. 240
    2. 2. 402
    3. 3. 204
    4. 4. 420

4
Solve this question by using below formula. Product of 2 numbers = product of their HCF and LCM 144 * x = 12 * 5040 x = (12*5040)/144 = 420

Q.- The ratio of two numbers is 3 : 4 and their H.C.F. is 4. Their L.C.M. is

    1. 1. 48
    2. 2. 50
    3. 3. 84
    4. 4. 52

1
Let the numbers be 3x and 4x. Then their H.C.F. = x. So x = 4. So the numbers 12 and 16. L.C.M. of 12 and 16 = 48.

Q.- An electronic device makes a beep after every 60 sec. Another device makes a beep after every 62 sec. They beeped together at 10 a.m. The time when they will next make a beep together at the earliest is

    1. 1. 0.43888888889
    2. 2. none
    3. 3. 0.43819444444
    4. 4. 0.4375

3
L.C.M. of 60 and 62 seconds is 1860 seconds 1860/60 = 31 minutes They will beep together at 10:31 a.m. Sometimes questions on red lights blinking comes in exam which can be solved in the same way

Q.- Which of the following fractions is the largest?

    1. 1. 42954
    2. 2. 13/16
    3. 3. 31/40
    4. 4. 63/80

1
L.C.M of 8 16 40 and 80 = 80. 7/8 = 70/80; 13/16 = 65/80; 31/40 = 62/80 Since 70/80 > 63/80 > 65/80 > 62/80 So 7/8 > 63/80 > 13/16 > 31/40 So 7/8 is the largest.

Q.- If the sum of two numbers is 55 and the H.C.F and L.C.M of these numbers are 5 and 120 respectively then the sum of the reciprocal of the numbers is equal to:

    1. 1. 55/601
    2. 2. 601/55
    3. 3. 11/120
    4. 4. 120/11

3
Let the numbers be a and b. Then a + b = 55 and ab = 5 * 120 = 600. Required sum = 1/a + 1/b = (a + b)/ab = 55/600 = 11/120.

Q.- Three numbers which are co-prime to each other are such that the product of the first two is 551 and that of the last two is 1073. The sum of the three numbers is:

    1. 1. 75
    2. 2. 81
    3. 3. 85
    4. 4. 89

3
Since the numbers are co-prime they contain only 1 as the common factor. Also the given two products have the middle number in common. So middle number = H.C.F of 551 and 1073 = 29; First number = 551/29 = 19 Third number = 1073/29 = 37. Required sum = 19 + 29 + 37 = 85.

Q.- The sum of two numbers is 528 and their H.C.F is 33. The number of pairs of numbers satisfying the above conditions is:

    1. 1. 4
    2. 2. 6
    3. 3. 8
    4. 4. 12

1
Let the required numbers be 33a and 33b. Then 33a + 33b = 528 => a + b = 16. Now co-primes with sum 16 are (1 15) (3 13) (5 11) and (7 9). Required numbers are (33 * 1 33 * 15) (33 * 3 33 * 13) (33 * 5 33 * 11) (33 * 7 33 * 9). The number of such pairs is 4.

Q.- A drink vendor has 80 liters of Maaza 144 liters of Pepsi and 368 liters of Sprite. He wants to pack them in cans so that each can contains the same number of liters of a drink and doesn't want to mix any two drinks in a can. What is the least number of c

    1. 1. 35
    2. 2. 37
    3. 3. 42
    4. 4. 30

2
The number of liters in each can = HCF of 80 144 and 368 = 16 liters. Number of cans of Maaza = 80/16 = 5 Number of cans of Pepsi = 144/16 = 9 Number of cans of Sprite = 368/16 = 23 The total number of cans required = 5 + 9 + 23 = 37 cans.

Q.- A room is 6 meters 24 centimeters in length and 4 meters 32 centimeters in width. Find the least number of square tiles of equal size required to cover the entire floor of the room?

    1. 1. 110
    2. 2. 124
    3. 3. 96
    4. 4. 117

4
Length = 6 m 24 cm = 624 cm Width = 4 m 32 cm = 432 cm HCF of 624 and 432 = 48 Number of square tiles required = (624 * 432)/(48 * 48) = 13 * 9 = 117

Q.- Find the lowest 4-digit number which when divided by 3 4 or 5 leaves a remainder of 2 in each case?

    1. 1. 1020
    2. 2. 1026
    3. 3. 1030
    4. 4. 1022

4
Lowest 4-digit number is 1000. LCM of 3 4 and 5 is 60. Dividing 1000 by 60 we get the remainder 40. Thus the lowest 4-digit number that exactly divisible by 3 4 and 5 is 1000 + (60 - 40) = 1020. Now add the remainder 2 that's required. Thus the answer is 1022.

Q.- The wheels revolve round a common horizontal axis. They make 15 20 and 48 revolutions in a minute respectively. Starting with a certain point on the circumference down wards. After what interval of time will they come together in the same position?

    1. 1. 1 min
    2. 2. 2 min
    3. 3. 3 min
    4. 4. None

1
Time for one revolution = 60/15 = 4 60/ 20 = 3 60/48 = 5/4 LCM of 4 3 5/4 LCM of Numerators/HCF of Denominators = 60/1 = 60

Q.- A B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds B in 308 seconds and C in 198 seconds all starting at the same point. After what time will they meet again at the starting point?

    1. 1. 26 min 18 sec
    2. 2. 42 min 36 sec
    3. 3. 45 min
    4. 4. 46 min 12 sec

4
L.C.M of 252 308 and 198 = 2772 So A B and C will again meet at the starting point in 2772 sec i.e. 46 min 12 sec.

Q.- Three numbers are in the ratio 3:4:5 and their L.C.M is 2400. Their H.C.F is:

    1. 1. 40
    2. 2. 80
    3. 3. 120
    4. 4. 200

1
Let the numbers be 3x 4x and 5x. Then their L.C.M = 60x. So 60x = 2400 or x = 40. The numbers are (3 * 40) (4 * 40) and (5 * 40). Hence required H.C.F = 40.

Q.- A goods train runs at the speed of 72 kmph and crosses a 250 m long platform in 26 seconds. What is the length of the goods train?

    1. 1. 230 m
    2. 2. 240 m
    3. 3. 260 m
    4. 4. 270 m

4
Speed = 72 x 5/18 m/sec = 20 m/sec Time = 26 sec Let the length of the train be x metres. Then (x + 250) / 26 = 20 x + 250 = 520 x = 270

Q.- A train 240 m long passes a pole in 24 seconds. How long will it take to pass a platform 650 m long?

    1. 1. 65sec
    2. 2. 89sec
    3. 3. 100sec
    4. 4. 150sec

2
Speed = 240/24 m/sec = 10 m/sec Required time = (240 + 650)/10 sec = 89 sec

Q.- A train 360 m long is running at a speed of 45 km/hr. In what time will it pass a bridge 140 m long?

    1. 1. 40 sec
    2. 2. 42 sec
    3. 3. 45 sec
    4. 4. 48 sec

1
Formula for converting from km/hr to m/s: X km/hr = X x 5 m/s. 18 Therefore Speed = 45 x 5 m/sec = 25 m/sec. 18 2 Total distance to be covered = (360 + 140) m = 500 m. Formula for finding Time = Distance Speed Required time = 500 x 2 sec = 40 sec.

Q.- Two trains are moving in opposite directions @ 60 km/hr and 90 km/hr. Their lengths are 1.10 km and 0.9 km respectively. The time taken by the slower train to cross the faster train in seconds is:

    1. 1. 36
    2. 2. 45
    3. 3. 48
    4. 4. 49

3
Relative speed = (60+ 90) km/hr = 150 x 5 m/sec 18 = 125 m/sec 3 Distance covered = (1.10 + 0.9) km = 2 km = 2000 m Required time = 2000 x 3 / 125sec =48 sec

Q.- A 270 metres long train running at the speed of 120 kmph crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds. What is the length of the other train?

    1. 1. 230m
    2. 2. 240m
    3. 3. 260m
    4. 4. 320m

1
Relative speed = (120 + 80) km/hr = 200 x 5/18 m/sec = 500/9 m/sec Let the length of the other train be x metres. Then( x + 270)/9 = 500/9 x + 270 = 500 x = 230

Q.- A train 110 metres long is running with a speed of 60 kmph. In what time will it pass a man who is running at 6 kmph in the direction opposite to that in which the train is going?

    1. 1. 5sec
    2. 2. 6sec
    3. 3. 7sec
    4. 4. 10sec

2
Speed of train relative to man = (60 + 6) km/hr = 66 km/hr = 66 x 5/18 m/sec = 55/3 m/sec Time taken to pass the man = 110 x 3/55 sec = 6 sec

Q.- A train 800 metres long is running at a speed of 78 km/hr. If it crosses a tunnel in 1 minute then the length of the tunnel (in meters) is:

    1. 1. 130
    2. 2. 360
    3. 3. 500
    4. 4. 540

3
Speed = ( 78 x 5/18) m/sec = ( 65/3 ) m/sec Time = 1 minute = 60 seconds Let the length of the tunnel be x metres Then ( 800 + x)/60 = 65/3 => 3(800 + x) = 3900 => x = 500

Q.- A 300 metre long train crosses a platform in 39 seconds while it crosses a signal pole in 18 seconds. What is the length of the platform?

    1. 1. 320m
    2. 2. 350m
    3. 3. 650m
    4. 4. none

2
Speed = ( 300 /18 m/sec = 50/3 m/sec Let the length of the platform be x metres Then ( x + 300)/39 = 50/3 => 3(x + 300) = 1950 => x = 350 m

Q.- How many seconds will a 500 metre long train take to cross a man walking with a speed of 3 km/hr in the direction of the moving train if the speed of the train is 63 km/hr?

    1. 1. 25
    2. 2. 30
    3. 3. 40
    4. 4. 45

2
Speed of the train relative to man = (63 - 3) km/hr+J11 = 60 km/hr = ( 60 x 5/18) m/sec = 50/3 m/sec Therefore Time taken to pass the man = (500 x 3)/50 sec = 30 sec

Q.- A train overtakes two persons who are walking in the same direction in which the train is going at the rate of 2 kmph and 4 kmph and passes them completely in 9 and 10 seconds respectively. The length of the train is:

    1. 1. 45m
    2. 2. 50m
    3. 3. 54m
    4. 4. 72m

2
Explanation : 2 kmph = ( 2 x 5/18) m/sec = 5/9 m/sec 4 kmph = ( 4 x 5/18 ) m/sec = 10/9 m/sec Let the length of the train be x metres and its speed by y m/sec Then ( x/(y-5/9) = 9 and x/(y-10/9) = 10 Therefore 9y - 5 = x and 10(9y - 10) = 9x => 9y - x = 5 and 90y - 9x = 100 On solving we get: x = 50 Therefore Length of the train is 50 m

Q.- A train of length 180 m crosses a man standing on a platform in 12 seconds and cross another train coming from opposite direction in 12 sec. If the second train running at 2/3 rd speed of the firstthen find the length of the second train?

    1. 1. 60m
    2. 2. 120m
    3. 3. 300m
    4. 4. 180m

2
Speed of the first train = 180/12 = 15 m/s Speed of the second train = 2/3 * 15 = 10 m/s Relative speed = 15+10 = 25 m/s Let length of the second train be L (180+L)/25 = 12 180+L = 300 L = 120 m length of the second train = 120 m

Q.- Two trains are approaching each other with the speed of 25km/hr. and 30km/hr. respectively from two different stations A and B. When two trains will meet each other the second train had covered 20 km more than the first train. What is the distance between

    1. 1. 220 km
    2. 2. 100 km
    3. 3. 120 km
    4. 4. 20 km

1
Let the distance traveled by the first train and second train be x and (x+20) respectively. Since time taken by both trains are same when they meet each other x/25 = (x+20)/30 x/5 = (x+20)/6 6x = 5x + 100 x = 100 Distance between A and B = x + (x+20) = 220 km

Q.- Two trains each 100 m long moving in opposite directions cross other in 8 sec. If one is moving twice as fast the other then the speed of the faster train is?

    1. 1. 30 km/hr
    2. 2. 45 km/hr
    3. 3. 60 km/hr
    4. 4. 75 km/hr

3
Let the speed of the slower train be x K101m/sec. Then speed of the train = 2x m/sec. Relative speed = ( x + 2x) = 3x m/sec. (100 + 100)/8 = 3x => x = 25/3. So speed of the faster train => 50/3 m/sec = 50/3 * 18/5 = 60 km/hr.

Q.- Two trains are running at 40 km/hr and 20 km/hr respectively in the same direction. Fast train completely passes a man sitting in the slower train in 5 sec. What is the length of the fast train?

    1. 1. 23 m
    2. 2. 23 2/9 m
    3. 3. 27 m
    4. 4. 27 7/9 m

4
Relative speed = (40 - 20) = 20 km/hr. = 20 * 5/ 18 = 50/9 m/sec. Length of faster train = 50/9 * 5 = 250/9 = 27 7/9 m.

Q.- Two goods trains each 500 m long are running in opposite directions on parallel tracks. Their speeds are 45 km/hr and 30 km/hr respectively. Find the time taken by the slower train to pass the driver of the faster one?

    1. 1. 12 sec
    2. 2. 24 sec
    3. 3. 48 sec
    4. 4. 60 sec

3
Relative speed = 45 + 30 = 75 km/hr. 75 * 5/18 = 125/6 m/sec. Distance covered = 500 + 500 = 1000 m. Required time = 1000 * 6/125 = 48 sec.

Q.- Two trains of equal lengths take 10 sec and 15 sec respectively to cross a telegraph post. If the length of each train be 120 m in what time will they cross other travelling in opposite direction?

    1. 1. 8 sec
    2. 2. 12 sec
    3. 3. 20 sec
    4. 4. 24 sec

2
Speed of the first train = 120/10 = 12 m/sec. Speed of the second train = 120/5 = 8 m/sec. Relative speed = 12 + 8 = 20 m/sec. Required time = (120 + 120)/20 = 12 sec.

Q.- Two trains are moving in opposite directions at 60 km/hr and 90 km/hr. Their lengths are 1.10 km and 0.9 km respectively. The time taken by the slower train to cross the faster train in seconds is?

    1. 1. 36
    2. 2. 45
    3. 3. 48
    4. 4. 49

3
Relative speed = 60 + 90 = 150 km/hr. = 150 * 5/18 = 125/3 m/sec. Distance covered = 1.10 + 0.9 = 2 km = 2000 m. Required time = 2000 * 3/125 = 48 sec

Q.- A train speeds past a pole in 15 sec and a platform 100 m long in 25 sec its length is?

    1. 1. 50m
    2. 2. 150m
    3. 3. 200m
    4. 4. None of these

2
Let the length of the train be x m and its speed be y m/sec. Then x/y = 15 => y = x/15 (x + 100)/25 = x/15 => x = 150 m.

Q.- A train running at the speed of 60 km/hr crosses a pole in 9 sec. What is the length of the train?

    1. 1. 120m
    2. 2. 200m
    3. 3. 180m
    4. 4. None of these

4
Speed = 60 * 5/18 = 50/3 m/sec Length of the train = speed * time = 50/3 * 9 = 150 m

Q.- A train 800 m long is running at a speed of 78 km/hr. If it crosses a tunnel in 1 min then the length of the tunnel is?

    1. 1. 130m
    2. 2. 360m
    3. 3. 500m
    4. 4. 540m

3
Speed = 78 * 5/18 = 65/3 m/sec. Time = 1 min = 60 sec. Let the length of the train be x meters. Then (800 + x)/60 = 65/3 x = 500 m.

Q.- A train travelling at a speed of 75 mph enters a tunnel 31/2 miles long. The train is 1/4 mile long. How long does it take for the train to pass through the tunnel from the moment the front enters to the moment the rear emerges?

    1. 1. 2.5 min
    2. 2. 3 min
    3. 3. 3.2 min
    4. 4. 3.5 min

2
Total distance covered = (7/2) + (1/4) miles = 15/4 miles. Therefore Time taken = ( 15/(4 x 75) hrs = 1/20 hrs = (1/20) x 60 min. = 3 min.

Q.- Two goods train each 500 m long are running in opposite directions on parallel tracks. Their speeds are 45 km/hr and 30 km/hr respectively. Find the time taken by the slower train to pass the driver of the faster one.

    1. 1. 12
    2. 2. 24
    3. 3. 48
    4. 4. 60

2
Relative speed = = (45 + 30) km/hr = ( 75 x 5/18) m/sec = 125/6 m/sec. We have to find the time taken by the slower train to pass the DRIVER of the faster train and not the complete train. So distance covered = Length of the slower train. Therefore Distance covered = 500 m. Therefore Required time = 500 x (6/125) = 24 sec.

Q.- Two trains are running in opposite directions with the same speed. If the length of each train is 120 metres and they cross each other in 12 seconds then the speed of each train (in km/hr) is:

    1. 1. 10
    2. 2. 18
    3. 3. 36
    4. 4. 72

3
Let the speed of each train be x m/sec. Then relative speed of the two trains = 2x m/sec. So 2x = (120 + 120)/12 => 2x = 20 => x = 10. Therefore Speed of each train = 10 m/sec = 10 x (18/5) km/hr = 36 km/hr.

Q.- Two trains of equal lengths take 10 seconds and 15 seconds respectively to cross a telegraph post. If the length of each train be 120 metres in what time (in seconds) will they cross each other travelling in opposite direction?

    1. 1. 10
    2. 2. 12
    3. 3. 15
    4. 4. 20

4
Speed of the first train = (120/10) m/sec = 12 m/sec. Speed of the second train = (120/15) m/sec = 8 m/sec. Relative speed = (12 + 8) = 20 m/sec. Therefore Required time = [ (120 + 120)/20 ] sec = 12 sec.

Q.- A train 108 m long moving at a speed of 50 km/hr crosses a train 112 m long coming from opposite direction in 6 seconds. The speed of the second train is:

    1. 1. 48 km/hr
    2. 2. 54 km/hr
    3. 3. 66 km/hr
    4. 4. 82 km/hr

4
Let the speed of the second train be x km/hr. Relative speed = (x + 50) km/hr = [ (x + 50) x (5/18) ] m/sec = (250 + 5x)/18 m/sec. Distance covered = (108 + 112) = 220 m. Therefore 220/ [(250 + 5x)/18] = 6 => 250 + 5x = 660 => x = 82 km/hr.

Q.- Two trains are running at 40 km/hr and 20 km/hr respectively in the same direction. Fast train completely passes a man sitting in the slower train in 5 seconds. What is the length of the fast train?

    1. 1. 23 m
    2. 2. 27.77 m
    3. 3. 24 m
    4. 4. 23.5 m

2
Relative speed = (40 - 20) km/hr = ( 20 x (5/18) m/sec = 50/9 m/sec. Therefore Length of faster train = (50/9) x 5 ( m = 250/9 m = 27.77 m.

Q.- A train overtakes two persons walking along a railway track. The first one walks at 4.5 km/hr. The other one walks at 5.4 km/hr. The train needs 8.4 and 8.5 seconds respectively to overtake them. What is the speed of the train if both the persons are walk

    1. 1. 66 km/hr
    2. 2. 72 km/hr
    3. 3. 78 km/hr
    4. 4. 81 km/hr

4
4.5 km/hr = 4.5 x (5/18) m/sec = 5/4 m/sec = 1.25 m/sec and 5.4 km/hr = 5.4 x (5/18) m/sec = 3/2 m/sec = 1.5 m/sec. Let the speed of the train be x m/sec. Then (x - 1.25) x 8.4 = (x - 1.5) x 8.5 => 8.4x - 10.5 = 8.5x - 12.75 => 0.1x = 2.25 => x = 22.5 Therefore Speed of the train = 22.5 x (18/5) km/hr = 81 km/hr.

Q.- A train travelling at 48 kmph completely crosses another train having half its length and travelling in opposite direction at 42 kmph in 12 seconds. It also passes a railway platform in 45 seconds. The length of the platform is

    1. 1. 400 m
    2. 2. 450 m
    3. 3. 560 m
    4. 4. 600 m

1
Let the length of the first train be x metres. Then the length of the second train is x/2 metres. Relative speed = (48 + 42) kmph = 90 x (5/18) m/sec = 25 m/sec. Therefore [x + (x/2)]/25 = 12 or 3x/2 = 300 or x = 200. Therefore Length of first train = 200 m. Let the length of platform be y metres. Speed of the first train = 48 x (5/18) m/sec = 40/3 m/sec. Therefore (200 + y) x (3/40) = 45 => 600 + 3y = 1800 => y = 400 m.

Q.- Two stations A and B are 110 km apart on a straight line. One train starts from A at 7 a.m. and travels towards B at 20 kmph. Another train starts from B at 8 a.m. and travels towards A at a speed of 25 kmph. At what time will they meet?

    1. 1. 9 a.m.
    2. 2. 10 a.m.
    3. 3. 10.30 a.m.
    4. 4. 11 a.m.

2
Suppose they meet x hours after 7 a.m. Distance covered by A in x hours = 20x km. Distance covered by B in (x - 1) hours = 25(x - 1) km. Therefore 20x + 25(x - 1) = 110 => 45x = 135 => x = 3. So they meet at 10 a.m.

Q.- Two trains one from Howrah to Patna and the other from Patna to Howrah start simultaneously. After they meet the trains reach their destinations after 9 hours and 16 hours respectively. The ratio of their speeds is:

    1. 1. 0.12777777778
    2. 2. 0.16875
    3. 3. 0.085416666667
    4. 4. 0.12638888889

2
Let us name the trains as A and B. Then (A's speed) : (B's speed) = sqrt. of b : sqrt. of a = sqrt. of 16 : sqrt. Of 9 = 4:3

Q.- A B and C can do a piece of work in 20 30 and 60 days respectively. In how many days can A do the work if he is assisted by B and C on every third day?

    1. 1. 12days
    2. 2. 15days
    3. 3. 16days
    4. 4. 18days

2
A's 2 day's work = ( 1 /20)x 2) = 1/10 (A + B + C)'s 1 day's work = ( 1/20 + 1/30 + 1/60) = 6/60 = 1/10 Work done in 3 days = ( 1/10 + 1/10) = 1/5 Now 1/5 work is done in 3 days. Whole work will be done in (3 x 5) = 15 days.

Q.- A alone can do a piece of work in 6 days and B alone in 8 days. A and B undertook to do it for Rs. 3200. With the help of C they completed the work in 3 days. How much is to be paid to C?

    1. 1. Rs 350
    2. 2. Rs 400
    3. 3. Rs 600
    4. 4. Rs 800

2
Explanation : C's 1 day's work = 1/3 - ( 1/6 + 1/8) = 1/3 - 7/24 = 1/24 A's wages : B's wages : C's wages = 1/6 : 1/8 : 1/24 = 4 : 3 : 1 C's share (for 3 days) = Rs.( 3 x (1/24) x 3200 = Rs. 400

Q.- If 6 men and 8 boys can do a piece of work in 10 days while 26 men and 48 boys can do the same in 2 days the time taken by 15 men and 20 boys in doing the same type of work will be:

    1. 1. 4 days
    2. 2. 5 days
    3. 3. 6 days
    4. 4. 7 days

1
Explanation : Let 1 man's 1 day's work = x and 1 boy's 1 day's work = y Then 6x + 8y = 1 /10 and 26x + 48y = 1/2 Solving these two equations we get : x = 1/100 and y = 1/200 (15 men + 20 boy)'s 1 day's work =( 15/100) +( 20 /200) = 1/4 15 men and 20 boys can do the work in 4 days.

Q.- A can do a certain work in the same time in which B and C together can do it. If A and B together could do it in 10 days and C alone in 50 days then B alone could do it in:

    1. 1. 15 days
    2. 2. 20 days
    3. 3. 25 days
    4. 4. 30 days

3
Explanation : (A + B)'s 1 day's work = 1/10 C's 1 day's work = 1/50 (A + B + C)'s 1 day's work = ( 1/10) +( 1/50) = 6/50 = 3/25 . .... (i) A's 1 day's work = (B + C)'s 1 day's work .... (ii) From (i) and (ii) we get: 2 x (A's 1 day's work) = 3/25 A's 1 day's work = 3/50 . B's 1 day's work (1/10) - (3/50) = 2/50 = 1/25 So B alone could do the work in 25 days.

Q.- 4 men and 6 women can complete a work in 8 days while 3 men and 7 women can complete it in 10 days. In how many days will 10 women complete it?

    1. 1. 35
    2. 2. 40
    3. 3. 45
    4. 4. 50

2
Let 1 man's 1 day's work = x and 1 woman's 1 day's work = y Then 4x + 6y = 1/5 and 3x + 7y = 1/10 Solving the two equations we get: x = 11/400 y = 1/400 1 woman's 1 day's work = 1/400 10 women's 1 day's work = (1/400) x 10 = 1/40 Hence 10 women will complete the work in 40 days.

Q.- A and B can together finish a work 30 days. They worked together for 20 days and then B left. After another 20 days A finished the remaining work. In how many days A alone can finish the work?

    1. 1. 40
    2. 2. 50
    3. 3. 54
    4. 4. 60

4
Explanation : (A + B)'s 20 day's work = ( 1/30) x 20 = 2/3 Remaining work = 1 -( 2/3 ) = 1/3 Now 1 work is done by A in 20 days. Therefore the whole work will be done by A in (20 x 3) = 60 days.

Q.- X and Y can do a piece of work in 20 days and 12 days respectively. X started the work alone and then after 4 days Y joined him till the completion of the work. How long did the work last?

    1. 1. 6 days
    2. 2. 10 days
    3. 3. 15 days
    4. 4. 20 days

2
Explanation : Work done by X in 4 days = (1/20) x 4 = 1/5 Remaining work = 1 - 1/5 = 4/5 (X + Y)'s 1 day's work = (1/20 )+ (1/12) = 8/60 = 2/15 Now 2/15 work is done by X and Y in 1 day So 4/5 work will be done by X and Y in (15/2) x (4/5) = 6 days Hence total time taken = (6 + 4) days = 10 days

Q.- A takes twice as much time as B or thrice as much time as C to finish a piece of work. Working together they can finish the work in 2 days. B can do the work alone in:

    1. 1. 4 days
    2. 2. 6 days
    3. 3. 8 days
    4. 4. 12 days

2
Explanation : Suppose A B and C take x x/2 and x/3 days respectively to finish the work. Then (1/x )+ (2/x )+ (3 /x) = 1/2 6/x = 1/2 x = 12 So B takes (12/2) = 6 days to finish the work

Q.- A works twice as fast as B. If B can complete a work in 12 days independently the number of days in which A and B can together finish the work in :

    1. 1. 4 days
    2. 2. 6 days
    3. 3. 8 days
    4. 4. 18 days

1
Explanation : Ratio of rates of working of A and B = 2 : 1 So ratio of times taken = 1 : 2 B's 1 day's work = 1 /12 Therefore A's 1 day's work = 1/6 ; (2 times of B's work) (A + B)'s 1 day's work = ( 1/6) +( 1/12) = 3/12 = 1/4 So A and B together can finish the work in 4 days.

Q.- A and B together can do a piece of work in 30 days. A having worked for 16 days B finishes the remaining work alone in 44 days. In how many days shall B finish the whole work alone?

    1. 1. 30 days
    2. 2. 40 days
    3. 3. 60 days
    4. 4. 70 days

3
Let A's 1 day's work = x and B's 1 day's work = y Then x + y = 1/30 and 16x + 44y = 1 Solving these two equations we get: x = 1/60 and y = 1/60 B's 1 day's work = 1/60 Hence B alone shall finish the whole work in 60 days.

Q.- A is 30% more efficient than B. How much time will they working together take to complete a job which A alone could have done in 23 days?

    1. 1. 11 days
    2. 2. 13 days
    3. 3. 20 3/7 days
    4. 4. None

2
Ratio of times taken by A and B = 100:130 = 10:13 Suppose B takes x days to do the work. x = (23 * 13)/10 = 299/10 A's 1 day work = 1/23; B's 1 day work = 10/299 (A + B)'s 1 day work = (1/23 + 10/299) = 1/13 A and B together can complete the job in 13 days.

Q.- A can do a piece of work in 4 hours; B and C together can do it in 3 hours which A and C together can do it in 2 hours. How long will B alone take to do it?

    1. 1. 8 h
    2. 2. 10 h
    3. 3. 12 h
    4. 4. 24 h

3
A's 1 hour work = 1/4; (B + C)'s 1 hour work = 1/3; (A + C)'s 1 hour work = 1/2 (A + B + C)'s 1 hour work = (1/4 + 1/3) = 7/12 B's 1 hour work = (7/12 + 1/2) = 1/12 B alone will take 12 hours to do the work.

Q.- Rohan and Elan are working on an assignment. Rohan takes 6 hrs to type 32 pages on a computer while Elan takes 5 hrs to type 40 pages. How much time will they take working together on two different computers to type an assignment of 110 pages?

    1. 1. 7 hr 30 min
    2. 2. 8 hr
    3. 3. 8 hr 15 min
    4. 4. 8 hr 25 min

3
Number of pages typed by Ronald in 1 hour = 32/6 = 16/3 Number of pages typed by Elan in 1 hour = 40/5 = 8 Number of pages typed by both in 1 hour = (16/3 + 8) = 40/3 Time taken by both to type 110 pages = (110 * 3/40) = 8 1/4 = 8 hrs 15 min

Q.- A alone can do a piece of work in 6 days and B alone in 8 days. A and B undertook to do it for Rs. 3200. With the help of C they completed the work in 3 days. How much is to be paid to C?

    1. 1. Rs. 375
    2. 2. Rs. 400
    3. 3. Rs. 600
    4. 4. Rs. 800

2
C's 1 day's work = 1/3 - (1/6 + 1/8) = 1/3 - 7/24 = 1/24 A's wages : B's wages : C's wages 1/6 : 1/8 : 1/24 = 4:3:1 C's share = 1/8 * 3200 = Rs. 400

Q.- A B and C can do a piece of work in 20 30 and 60 days respectively. In how many days A do the work if he is assisted by B and C on every third day?

    1. 1. 12 days
    2. 2. 15 days
    3. 3. 16 days
    4. 4. 18 days

2
A's 2 day's work = (1/20 * 2) = 1/10 (A + B + C)'s 1 day work = (1/20 + 1/30 + 1/60) = 1/10 Work done in 3 days = (1/10 + 1/10) = 1/5 Now 1/5 work is done in 3 days. Whole work will be done in (3 * 5) = 15 days.

Q.- 20 women can do a work in 9 days. After they have worked for 6 days. 6 more men join them. How many days will they take to complete the remaining work?

    1. 1. 0.12777777778
    2. 2. 0.16875
    3. 3. 0.21041666667
    4. 4. 0.12847222222

2
(20 * 16) women can complete the work in 1 day. 1 woman's 1 day work = 1/320 (16 * 15) men can complete the work in 1 day 1 man's 1 day work = 1/240 So required ratio = 1/240 : 1/320 = 4:3

Q.- 12 men can complete a piece of work in 4 days while 15 women can complete the same work in 4 days. 6 men start working on the job and after working for 2 days all of them stopped working. How many women should be put on the job to complete the remaining w

    1. 1. 15
    2. 2. 18
    3. 3. 20
    4. 4. 24

1
1 man's 1 day work = 1/48; 1 woman's 1 day work = 1/60. 6 men's 2 day's work = 6/48 * 2 = 1/4. Remaining work = (1 - 1/4) = 3/4 Now 1/60 work is done in 1 day by 1 woman. So 3/4 work will be done in 3 days by (60 * 3/4 * 1/3) = 15 women.

Q.- A man a woman and a boy can complete a job in 3 4 and 12 days respectively. How many boys must assist 1 man and 1 woman to complete the job in 1/4 of a day?

    1. 1. 1
    2. 2. 4
    3. 3. 41
    4. 4. 48

3
(1 man + 1 woman)'s 1 day work = (1/3 + 1/4) = 7/12 Work done by 1 man and 1 woman in 1/4 day = (7/12 * 1/4) = 7/48 Remaining work = (1 - 7/48) = 41/48 Work done by 1 boy in 1/4 day = ( 1/12 * 1/4) = 1/48 Number of boys required = 41/48 * 41 = 41.

Q.- Twelve men and six women together can complete a piece of work in four days. The work done by a women in one day is half the work done by a man in one day. If 12 men and six women started working and after two days six men left and six women joined then i

    1. 1. 1
    2. 2. 1(1/2)
    3. 3. 2
    4. 4. 2(1/2)

4
Work done by a women in one day = 1/2 (work done by a man/day) One women's capacity = 1/2(one man's capacity) One man = 2 women. 12 men = 24 women. 12 men + 6 women = 30 women 30 women can complete the work in four days. In the first 2 days they can complete 1/2 of the work. Remaining part of the work = 1/2. If 6 men leave and 6 new women join then new work force = 30 women-12women + 6 women = 24 women. Time taken by them to complete the remaining work = 1/2 (Time taken by 24 women to complete the work) = 1/2 * (30 * 4)/24 = 2 (1/2) days.

Q.- X men can do a work in 120 days. If there were 20 men less the work would have taken 60 days more. What is the value of X?

    1. 1. 40
    2. 2. 50
    3. 3. 60
    4. 4. 70

3
We have M1 D1 = M2 D2 120X = (X - 20)180 => 2X = (X - 20) 3 2X = 3X - 60 => X = 60

Q.- 5 men and 2 boys working together can do four times as much work as a man and a boy. Working capacity of man and boy is in the ratio.

    1. 1. 0.12777777778
    2. 2. 0.16875
    3. 3. 0.084027777778
    4. 4. 0.12638888889

3
Let 1 man 1 day work = x 1 boy 1 day work = y then 5x + 2y = 4(x+y) => x = 2y => x/y = 2/1 => x:y = 2:1

Q.- 10 women can complete a work in 7 days and 10 children take 14 days to complete the work. How many days will 5 women and 10 children take to complete the work?

    1. 1. 10 days
    2. 2. 7 days
    3. 3. 6 days
    4. 4. 9 days

2
1 woman's 1 day's work = 1/70 1 Child's 1 day's work = 1/140 5 Women and 10 children 1 day work = [(5/70)+(10/140)]=1/7 So 5 women and 10 children will finish the work in 7 days.

Q.- A piece of work can be done by 6 men and 5 women in 6 days or 3 men and 4 women in 10 days. It can be done by 9 men and 15 women in how many days ?

    1. 1. 3 days
    2. 2. 5 days
    3. 3. 4 days
    4. 4. 6 days

1
To calculate the answer we need to get 1 man per day work and 1 woman per day work. Let 1 man 1 day work =x and 1 woman 1 days work = y. => 6x+5y = 1/6 and 3x+4y = 1/10 On solving we get x = 1/54 and y = 1/90 (9 men + 15 women)'s 1 days work = (9/54) + (15/90) = 1/3 9 men and 15 women will finish the work in 3 days

Q.- 4 men and 6 women finish a job in 8 days while 3 men and 7 women finish it in 10 days. In how many days will 10 women working together finish it ?

    1. 1. 35 days
    2. 2. 40 days
    3. 3. 30 days
    4. 4. 45 days

2
Let 1 man's 1 day work = x and 1 woman's 1 days work = y. Then 4x + 6y = 1/8 and 3x+7y = 1/10 solving we get y = 1/400 [means work done by a woman in 1 day] 10 women 1 day work = 10/400 = 1/40 10 women will finish the work in 40 days

Q.- A alone can do a piece of work in 6 days and B alone in 8 days. A and B undertook to do it for Rs. 3200. With the help of C. they completed the work in 3 days. How much is to be paid to C

    1. 1. Rs. 350
    2. 2. Rs. 400
    3. 3. Rs. 450
    4. 4. Rs. 500

2
C's 1 day's work = (1/3)-[(1/6)+(1/8)]=(1/3)-(7/24)=1/24 A:B:C=1/6:1/8:1/24=4:3:1 C'sShare=(1/8)*3200=400 If you are confused how we multiplied 1/8 then please study ratio and proportion chapter for small information it is the C ratio divided by total ratio.

Q.- A can do a piece of work in 4 hours . A and C together can do it in just 2 hours while B and C together need 3 hours to finish the same work. In how many hours B can complete the work ?

    1. 1. 11 hrs.
    2. 2. 12 hrs.
    3. 3. 8 hrs.
    4. 4. 10 hrs.

2
Work done by A in 1 hour = 1/4 Work done by B and C in 1 hour = 1/3 Work done by A and C in 1 hour = 1/2 Work done by AB and C in 1 hour = (1/4)+(1/3) = 7/12 Work done by B in 1 hour = (7/12)–(1/2) = 1/12 => B alone can complete the work in 12 hour

Q.- A can do a piece of work in 15 days and B alone can do it in 10 days. B works at it for 5 days and then leaves. A alone can finish the remaining work in

    1. 1. 9 days
    2. 2. 8.5 days
    3. 3. 7.5 days
    4. 4. 8 days

3
B's 5 days work = (1/10)*5=1/2 Remaining work =1-(1/2)=1/2 A can finish work =15*(1/2)=7.5 days

Q.- A is thrice as good a workman as B and takes 10 days less to do a piece of work than B takes. B alone can do the whole work in

    1. 1. 15 days
    2. 2. 14 days
    3. 3. 12 days
    4. 4. 13 days

1
Ratio of times taken by A and B = 1:3 Means B will take 3 times which A will do in 1 time If difference of time is 2 days B takes 3 days If difference of time is 10 days B takes (3/2) * 10 =15 days

Q.- A does half as much work as B in three-fourth of the time. If together they take 18 days to complete the work how much time shall B take to do it

    1. 1. 20 days
    2. 2. 25 days
    3. 3. 30 days
    4. 4. 35 days

3
Suppose B takes x dáys to do the work. As per question A will take 2*(3/4)*x=3x/2 days (A+B)s 1 days work= 1/18 1/x + 2/3x = 1/18 or x = 30 days

Q.- A does a work in 10 days and B does the same work in 15 days. In how many days they together will do the same work ?

    1. 1. 5 days
    2. 2. 6 days
    3. 3. 7 days
    4. 4. 8 days

2
Firstly we will find 1 day work of both A and B then by adding we can get collective days for them So A's 1 day work = 1/10 B's 1 day work = 1/15 (A+B)'s 1 day work = (1/10)+(1/15)=(3+2)/30=1/6 So together they can complete work in 6 days.

Q.- A trader mixes 26 kg of rice at Rs. 20 per kg with 30 kg of rice of other variety at Rs. 36 per kg and sells the mixture at Rs. 30 per kg. His profit percent is:

    1. 1. None
    2. 2. 0.05
    3. 3. 0.08
    4. 4. 0.1

2
C.P. of 56 kg rice = Rs. (26 x 20 + 30 x 36) = Rs. (520 + 1080) = Rs. 1600 S.P. of 56 kg rice = Rs. (56 x 30) = Rs. 168n Gain = (80/1600) x 100 % = 5%

Q.- 100 oranges are bought at the rate of Rs. 350 and sold at the rate of Rs. 48 per dozen. The percentage of profit or loss is:

    1. 1. 100/7 %
    2. 2. 98/7 %
    3. 3. 102/7 %
    4. 4. 7/100 %

1
C.P. of 1 orange = Rs. 350 /100 = Rs. 3.50 S.P. of 1 orange = Rs. 48/12 = Rs. 4 Gain% = (0.50/3.50) x 100 % = 100/7 %

Q.- When a plot is sold for Rs. 18700 the owner loses 15%. At what price must that plot be sold in order to gain 15%?

    1. 1. Rs 23500
    2. 2. Rs 32500
    3. 3. Rs 25300
    4. 4. Rs 52300

3
85 : 18700 = 115 : x x = (18700 x 115)/85 = 25300 Hence S.P. = Rs. 25300

Q.- On selling 17 balls at Rs. 720 there is a loss equal to the cost price of 5 balls. The cost price of a ball is:

    1. 1. Rs 60
    2. 2. Rs 80
    3. 3. Rs 100
    4. 4. Rs 65

1
(C.P. of 17 balls) - (S.P. of 17 balls) = (C.P. of 5 balls) C.P. of 12 balls = S.P. of 17 balls = Rs.720 C.P. of 1 ball = Rs. 720/12 = Rs. 60

Q.- Some articles were bought at 6 articles for Rs. 5 and sold at 5 articles for Rs. 6. Gain percent is:

    1. 1. 0.36
    2. 2. 0.44
    3. 3. 0.45
    4. 4. 0.5

2
Suppose number of articles bought = L.C.M. of 6 and 5 = 30 C.P. of 30 articles = Rs. (5 /6)x 30 = Rs. 25 S.P. of 30 articles = Rs. (6/5) x 30 = Rs. 36 Gain % = (11/25) x 100 % = 44%

Q.- Sam purchased 20 dozens of toys at the rate of Rs. 375 per dozen. He sold each one of them at the rate of Rs. 33. What was his percentage profit?

    1. 1. 0.065
    2. 2. 0.056
    3. 3. 0.045
    4. 4. 0.054

2
Cost Price of 1 toy = Rs. 375/12 = Rs. 31.25 Selling Price of 1 toy = Rs. 33 So Gain = Rs. (33 - 31.25) = Rs. 1.75 Profit % = (1.75/31.25) x 100 % = 28/5 % = 5.6%

Q.- A man buys a cycle for Rs. 1400 and sells it at a loss of 15%. What is the selling price of the cycle?

    1. 1. Rs 1290
    2. 2. Rs 1910
    3. 3. Rs 9110
    4. 4. Rs 1190

4
S.P. = 85% of Rs. 1400 = Rs. ( 85/100) x 1400 = Rs. 1190

Q.- A shopkeeper expects a gain of 22.5% on his cost price. If in a week his sale was of Rs. 392 what was his profit?

    1. 1. Rs 72
    2. 2. Rs 27
    3. 3. Rs77
    4. 4. Rs 56

1
C.P. = Rs. (100/122.5) x 392 = Rs. (1000/1225) x 392 = Rs. 320 Profit = Rs. (392 - 320) = Rs. 72

Q.- A vendor bought toffees at 6 for a rupee. How many for a rupee must he sell to gain 20%?

    1. 1. 10
    2. 2. 5
    3. 3. 15
    4. 4. 4

2
C.P. of 6 toffees = Re. 1 S.P. of 6 toffees = 120% of Re. 1 = Rs. 6/5 For Rs. 6 toffees sold = 6/5 For Re. 1 toffees sold = 6 x 5/6 = 5

Q.- In a certain store the profit is 320% of the cost. If the cost increases by 25% but the selling price remains constant approximately what percentage of the selling price is the profit?

    1. 1. 0.6
    2. 2. 0.5
    3. 3. 0.7
    4. 4. 0.65

3
Let C.P.= Rs. 100. Then Profit = Rs. 320 S.P. = Rs. 420 New C.P. = 125% of Rs. 100 = Rs. 125 New S.P. = Rs. 420 Profit = Rs. (420 - 125) = Rs. 295 Required percentage = (295/420) x 100 % = 1475/21 % = 70% (approximately)

Q.- A shopkeeper sold an article offering a discount of 5% and earned a profit of 23.5%. What would have been the percentage of profit earned if no discount was offered?

    1. 1. 0.2
    2. 2. 0.24
    3. 3. 0.3
    4. 4. 0.47

3
Let C.P. be Rs. 100. Then S.P. = Rs. 123.50 Let marked price be Rs. x. Then 95/100 x = 123.50 x = 12350/95 = Rs. 130 Now S.P. = Rs. 130 C.P. = Rs. 100 Profit % = 30%.

Q.- The cost price of an article is 64% of the marked price. Calculate the gain percent after allowing a discount of 12%?

    1. 1. 0.375
    2. 2. 0.4
    3. 3. 0.505
    4. 4. 0.56

1
Let marked price = Rs. 100. Then C.P. = RS. 64 S.P. = Rs. 88 Gain % = 24/64 * 100 = 37.5%.

Q.- The cash difference between the selling prices of an article at a profit of 4% and 6% is Rs. 3. The ratio of the two selling prices is:

    1. 1. 2.1611111111
    2. 2. 2.2020833333
    3. 3. 2.2034722222
    4. 4. 2.1618055556

3
Let C.P. of the article be Rs. x. Then required ratio = 104% of x / 106% of x = 104/106 = 52/53 = 52:53

Q.- Albert buys 4 horses and 9 cows for Rs. 13400. If he sells the horses at 10% profit and the cows at 20% profit then he earns a total profit of Rs. 1880. The cost of a horse is:

    1. 1. Rs. 2000
    2. 2. Rs. 2500
    3. 3. Rs. 3000
    4. 4. Rs. 4000

1
Let C.P. of each horse be Rs. x and C.P. of each cow be Rs. y. Then 4x + 9y = 13400 -- (i) And 10% of 4x + 20% of 9y = 1880 2/5 x + 9/5 y = 1880 => 2x + 9y = 9400 -- (ii) Solving (i) and (ii) we get : x = 2000 and y = 600. Cost price of each horse = Rs. 2000.

Q.- A fair price shopkeeper takes 10% profit on his goods. He lost 20% goods during theft. His loss percent is

    1. 1. 8
    2. 2. 10
    3. 3. 11
    4. 4. 12

4
Suppose he has 100 items. Let C.P. of each item be Re. 1. Total cost = Rs. 100. Number of items left after theft = 80. S.P. of each item = Rs. 1.10 Total sale = 1.10 * 80 = Rs. 88 Hence loss % = 12/100 * 100 = 12%

Q.- A sells a bicycle to B at a profit of 20%. B sells it to C at a profit of 25%. If C pays Rs. 225 for it The cost price of the bicycle for A is:

    1. 1. Rs. 110
    2. 2. Rs. 120
    3. 3. Rs. 125
    4. 4. Rs. 150

4
125% of 120% of A = 225 125/100 * 120/100 * A = 225 A = 225 * 2/3 = 150.

Q.- A man buys an article for 10% less than its value and sells it for 10% more than its value. His gain or loss percent is:

    1. 1. no profit no loss
    2. 2. 20% profit
    3. 3. less than 20% profit
    4. 4. more than 20% profit

4
Let the article be worth Rs. x. C.P. 90% of Rs. x = Rs. 9x/10 S.P. = 110% of Rs. x = Rs. 11x/10 Gain = (11x/10 - 9x/10) = Rs. x/5 Gain % = x/5 * 10/9x * 100 = 22 2/9 % > 20%

Q.- On an order of 5 dozen boxes of a consumer product a retailer receives an extra dozen free. This is equivalent to allowing him a discount of:

    1. 1. 0.15
    2. 2. 0.1617
    3. 3. 0.1667
    4. 4. 0.2

3
Clearly the retailer gets 1 dozen out of 6 dozens free. Equivalent discount = 1/6 * 100 = 16 2/3%.

Q.- Arun purchased 30 kg of wheat at the rate of Rs. 11.50 per kg and 20 kg of wheat at the rate of 14.25 per kg. He mixed the two and sold the mixture. Approximately what price per kg should be sell the mixture to make 30% profit?

    1. 1. Rs. 14.8
    2. 2. Rs. 15.4
    3. 3. Rs. 15.6
    4. 4. Rs. 16.3

4
C.P. of 50 kg wheat = (30 * 11.50 + 20 * 14.25) = Rs. 630. S.P. of 50 kg wheat = 130% of Rs. 630 = 130/100 * 630 = Rs. 819. S.P. per kg = 819/50 = Rs. 16.38 = 16.30

Q.- The ratio between the sale price and the cost price of an article is 7:5. What is the ratio between the profit and the cost price of that article?

    1. 1. 0.088194444444
    2. 2. 0.29305555556
    3. 3. 0.086805555556
    4. 4. 0.20972222222

3
Let C.P. = Rs. 5x and S.P. = Rs. 7x. Then Gain = Rs. 2x Required ratio = 2x : 5x = 2:5

Q.- A pair of articles was bought for Rs. 37.40 at a discount of 15%. What must be the marked price of each of the articles ?

    1. 1. 23
    2. 2. 24
    3. 3. 22
    4. 4. 21

3
As question states that rate was of pair of articles So rate of One article = 37.40/2 = Rs. 18.70 Let Marked price = Rs X then 85% of X = 18.70 => X = 1870/85 = 22

Q.- A shopkeeper fixes the marked price of an item 35% above its cost price. The percentage of discount allowed to gain 8% is

    1. 1. 0.18
    2. 2. 0.2
    3. 3. 0.22
    4. 4. 0.25

2
Let the cost price = Rs 100 then Marked price = Rs 135 Required gain = 8% So Selling price = Rs 108 Discount = 135 - 108 = 27 Discount% = (27/135)*100 = 20%

Q.- A shopkeeper sold an article for Rs 2564.36. Approximately. what was his profit percent if the cost price of the article was Rs 2400

    1. 1. 0.075
    2. 2. 0.08
    3. 3. 0.06
    4. 4. 0.07

4
Gain % = (164.36*100/2400) = 6.84 % = 7% approx

Q.- A producer of tea blends two varieties of tea from two tea gardens one costing Rs 18 per kg and another Rs 20 per kg in the ratio 5 : 3. If he sells the blended variety at Rs 21 per kg then his gain percent is

    1. 1. 0.12
    2. 2. 0.14
    3. 3. 0.15
    4. 4. 0.125

1
Suppose he bought 5 kg and 3 kg of tea. Cost Price = Rs. (5 x 18 + 3 x 20) = Rs. 150. Selling price = Rs. (8 x 21) = Rs. 168. Profit = 168 - 150 = 18 So Profit % = (18/150) * 100 = 12%

Q.- A fruit seller sells mangoes at the rate of Rs.9 per kg and thereby loses 20%. At what price per kg he should have sold them to make a profit of 5%

    1. 1. Rs.18.12
    2. 2. Rs.18.11
    3. 3. Rs.11.18
    4. 4. Rs.11.81

4
85 : 9 = 105 : x x= (9×105/85) = Rs 11.81

Q.- A man bought an article and sold it at a gain of 5 %. If he had bought it at 5% less and sold it for Re 1 less he would have made a profit of 10%. The C.P. of the article was

    1. 1. Rs.225
    2. 2. Rs.300
    3. 3. Rs.200
    4. 4. Rs.250

3
Let original Cost price is x Its Selling price = 105/100 * x = 21x/20 New Cost price = 95/100 * x = 19x/20 New Selling price = 110/100 * 19x/20 = 209x/200 [(21x/20) - (209x/200)] = 1 => x = 200

Q.- If the cost price of 12 items is equal to the selling price of 16 items the loss percent is

    1. 1. 0.2
    2. 2. 0.25
    3. 3. 0.3
    4. 4. 0.28

2
Let the Cost Price of 1 item = Re. 1 Cost Price of 16 items = 16 Selling Price of 16 items = 12 Loss = 16 - 12 = Rs 4 Loss % = (4/16)* 100 = 25%

Q.- A man gains 20% by selling an article for a certain price. If he sells it at double the price the percentage of profit will be.

    1. 1. 1.35
    2. 2. 1.4
    3. 3. 1.5
    4. 4. 1.45

2
Let the C.P. = x Then S.P. = (120/100)x = 6x/5 New S.P. = 2(6x/5) = 12x/5 Profit = 12x/5 - x = 7x/5 Profit% = (Profit/C.P.) * 100 => (7x/5) * (1/x) * 100 = 140 %

Q.- A plot is sold for Rs. 18700 with a loss of 15%. At what price it should be sold to get profit of 15%.

    1. 1. Rs. 25300
    2. 2. Rs.25003
    3. 3. Rs.23500
    4. 4. Rs.23005

1
This type of question can be easily and quickly solved as following: Let at Rs x it can earn 15% profit 85:18700 = 115:x [as loss = 100 -15 Profit = 100 +15] x = (18700*115)/85 = Rs.25300

Q.- A man buys an item at Rs. 1200 and sells it at the loss of 20 percent. Then what is the selling price of that item

    1. 1. Rs.690
    2. 2. Rs.609
    3. 3. Rs.906
    4. 4. Rs.960

4
Here always remember when ever x% loss it means S.P. = (100 - x)% of C.P when ever x% profit it means S.P. = (100 + x)% of C.P So here will be (100 - x)% of C.P. = 80% of 1200 = 80/100 * 1200 = 960

Q.- The sum of ages of 5 children born at the intervals of 3 years each is 50 years. What is the age of the youngest child?

    1. 1. 4 years
    2. 2. 8 years
    3. 3. 10 years
    4. 4. None

1
Let the ages of children be x (x + 3) (x + 6) (x + 9) and (x + 12) years Then x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 50 5x = 20 x = 4 Age of the youngest child = x = 4 years

Q.- Present ages of Sameer and Anand are in the ratio of 5 : 4 respectively. Three years hence the ratio of their ages will become 11 : 9 respectively. What is Anand's present age in years?

    1. 1. 24
    2. 2. 27
    3. 3. 40
    4. 4. None

1
Let the present ages of Sameer and Anand be 5x years and 4x years respectively. Then (5x + 3)/4x+3 = 11/9 9(5x + 3) = 11(4x + 3) 45x + 27 = 44x + 33 45x - 44x = 33 - 27 x = 6 Anand's present age = 4x = 24 years

Q.- A is two years older than B who is twice as old as C. If the total of the ages of A B and C be 27 the how old is B?

    1. 1. 7
    2. 2. 8
    3. 3. 9
    4. 4. 10

4
Let C's age be x years. Then B's age = 2x years. A's age = (2x + 2) years (2x + 2) + 2x + x = 27 5x = 25 x = 5 Hence B's age = 2x = 10 years

Q.- The age of father 10 years ago was thrice the age of his son. Ten years hence father's age will be twice that of his son. The ratio of their present ages is:

    1. 1. 0.20972222222
    2. 2. 0.29375
    3. 3. 0.37638888889
    4. 4. 0.54444444444

2
Let the ages of father and son 10 years ago be 3x and x years respectively. Then (3x + 10) + 10 = 2[(x + 10) + 10] 3x + 20 = 2x + 40 x = 20 Required ratio = (3x + 10) : (x + 10) = 70 : 30 = 7 : 3.

Q.- A person's present age is two-fifth of the age of his mother. After 8 years he will be one-half of the age of his mother. How old is the mother at present?

    1. 1. 32
    2. 2. 36
    3. 3. 40
    4. 4. 48

3
Let the mother's present age be x years Then the person's present age = ( 2/5) x years ( 2/5) x + 8 = 1/2 (x + 8) 2(2x + 40) = 5(x + 8) x = 40

Q.- Ayesha's father was 38 years of age when she was born while her mother was 36 years old when her brother four years younger to her was born. What is the difference between the ages of her parents?

    1. 1. 2 years
    2. 2. 4 years
    3. 3. 6 years
    4. 4. 8 years

3
Mother's age when Ayesha's brother was born = 36 years Father's age when Ayesha's brother was born = (38 + 4) years = 42 years Required difference = (42 - 36) years = 6 years

Q.- The present ages of three persons in proportions 4 : 7 : 9. Eight years ago the sum of their ages was 56.Find their present ages (in years).

    1. 1. 82028
    2. 2. 162836
    3. 3. 203545
    4. 4. None

2
Let their present ages be 4x 7x and 9x years respectively Then (4x - 8) + (7x - 8) + (9x - 8) = 56 20x = 80 x = 4 Their present ages are 4x = 16 years 7x = 28 years and 9x = 36 years respectively

Q.- Sachin is younger than Rahul by 7 years. If their ages are in the respective ratio of 7 : 9 how old is Sachin?

    1. 1. 16 years
    2. 2. 18 years
    3. 3. 28 years
    4. 4. 24.5 years

4
Let Rahul's age be x years Then Sachin's age = (x - 7) years ( x - 7)/x = 7/9 9x - 63 = 7x 2x = 63 x = 31.5 Hence Sachin's age =(x - 7) = 24.5 years

Q.- At present the ratio between the ages of Arun and Deepak is 4 : 3. After 6 years Arun's age will be 26 years. What is the age of Deepak at present ?

    1. 1. 12 years
    2. 2. 15 years
    3. 3. 19 years
    4. 4. 21 years

2
Let the present ages of Arun and Deepak be 4x years and 3x years respectively. Then 4x + 6 = 26 4x = 20 x = 5 Deepak's age = 3x = 15 years

Q.- The sum of the present ages of a father and his son is 60 years. Six years ago father's age was five times the age of the son. After 6 years son's age will be:

    1. 1. 12 years
    2. 2. 14 years
    3. 3. 18 years
    4. 4. 20 years

4
Let the present ages of son and father be x and (60 -x) years respectively. Then (60 - x) - 6 = 5(x - 6) 54 - x = 5x - 30 6x = 84 x = 14 Son's age after 6 years = (x+ 6) = 20 years

Q.- Ayesha's father was 38 years of age when she was born while her mother was 36 years old when her brother four years younger to her was born. What is the difference between the ages of her parents?

    1. 1. 2 yrs
    2. 2. 4 yrs
    3. 3. 6 yrs
    4. 4. 8 yrs

3
Mother's age when Ayesha's brother was born = 36 years. Father's age when Ayesha's brother was born = (38 + 4) = 42 years. Required difference = (42 - 36) = 6 years.

Q.- The sum of ages of family members (both children and parents) is 360 years.The total ages of children and parents are in the ratio 2:1 and the ages of wife and husband are in the ratio 5:7. What will be the age of husband?

    1. 1. 60
    2. 2. 75
    3. 3. 77
    4. 4. 70

4
Sum of ages is 360 years.The ratio of children and parents ages is 2:1. Total age of parents = 360 x 1 / 3 = 120 years Ratio of wife and husband age is 5:7. Therefore the age of husband = 120 x 7 / 12 = 70 Hence the age of husband is 70 years.

Q.- Mr. Nagaraj got married 12 years ago to Parvathy and she become 7th member of the family. Today his mother died and Mrs. Parvathy gave birth to a girl baby. The average age of Nagaraj's family at the time of his marriage is the same as of today. What is

    1. 1. 54
    2. 2. 60
    3. 3. 72
    4. 4. 84

4
Let the average age of family members at the time of marriage be x Sum of the ages of the family members at that time is 7x --> eq 1 After 12 years (i.e. now) the ages of all the seven would have increased by 12 each i.e by 84 years (12 * 7) Hence total ages of all the members currently = 7x + 84 Now mother has died and new baby is born. New baby's age now is zero. Hence mother's age should be subtracted from the current total. Also the baby's age should be added to the total. Current age total all the members = 7x + 84 - mother's age + baby's age = 7x + 84 - mother's age (since baby's age is zero) --> eq 2 But the question says that the average now has become same as that of 12 years before. Hence 7x + 84 - mother's age = 7x Hence mother's age is 84 years.

Q.- If 6 years are subtracted from the present age of Arun and the remainder is divided by 18 then the present age of his grandson Gokul is obtained.If Gokul is 2 years younger to Madan whose age is 5 years then what is the age of Arun?

    1. 1. 48
    2. 2. 60
    3. 3. 84
    4. 4. 96

2
Let Arun's age be x. Gokul is 2 years younger than Madan so Gokul is 3 years (i.e 5 - 2 = 3) If Arun had born 6 years beforehis age would had been x - 6. As per the question x - 6 should be 18 times as that of Gokul's age. i.e. x - 6 = 3 x 18 x = 60 Hence the age of Arun is 60 years

Q.- A father said to his son "I was as old as you are at present at the time of your birth." If the father's age is 38 years now the son's age five years back was:

    1. 1. 14 yrs
    2. 2. 19 yrs
    3. 3. 33 yrs
    4. 4. 39 yrs

1
Let the son's present age be x years. Then (38 - x) = x 2x = 38 => x = 19 Son's age 5 years back = (19 - 5) = 14 years.

Q.- Eighteen years ago a father was three times as old as his son. Now the father is only twice as old his son. Then the sum of the present ages of the son and the father is:

    1. 1. 54
    2. 2. 72
    3. 3. 105
    4. 4. 108

4
Let the present ages of the father and son be 2x and x years respectively. Then (2x - 18) = 3(x - 18) => x = 36 Required sum = (2x + x) = 108 years.

Q.- Tanya's grandfather was 8 times older to her 16 years ago. He would be 3 times of her age 8 years from now. Eight years ago what was the ratio of Tanya's age to that of her grandfather?

    1. 1. 0.043055555556
    2. 2. 0.045138888889
    3. 3. 0.13055555556
    4. 4. None of these

4
16 years ago let T = x years and G = 8x years. After 8 years from now T = (x + 16 + 8) years and G = (8x + 16 + 8) years. 8x + 24 = 3(x + 24) => 5x = 48 8 years ago T/G = (x + 8)/(8x + 8) = (48/5 + 8) / (8 * 48/5 + 8) = 88/424 = 11/53

Q.- The age of father 10 years ago was thrice the age of his son. Ten years hence father's age was five times the age of the son. After 6 years son's age will be:

    1. 1. 0.20972222222
    2. 2. 0.29375
    3. 3. 0.37638888889
    4. 4. 0.37777777778

2
Let the age of father and son 10 years ago be 3x and x years respectively. Then (3x + 10) + 10 = 2[(x + 10) + 10] 3x + 20 = 2x + 40 => x = 20. Required ratio = (3x + 10):(x + 10) = 70:30 = 7:3

Q.- The ratio between the present ages of A and B is 5:3 respectively. The ratio between A's age 4 years ago and B's age 4 years hence is 1:1. What is the ratio between A's age 4 years hence and B's age 4 years ago?

    1. 1. 0.12569444444
    2. 2. 0.04375
    3. 3. 0.085416666667
    4. 4. 0.12638888889

1
Let the present ages of A and B be 5x and 3x years respectively. Then (5x - 4)/(3x + 4) = 1/1 2x = 8 => x = 4 Required ratio = (5x + 4):(3x - 4) = 24:8 = 3:1

Q.- Six years ago the ratio of ages of Kunal and Sagar was 6:5. Four years hence the ratio of their ages will be 11:10. What is Sagar's age at present?

    1. 1. 16 yrs
    2. 2. 18 yrs
    3. 3. 20 yrs
    4. 4. None of these

1
Let the ages of Kunal and Sagar 6 years ago be 6x and 5x years respectively. Then [(6x + 6) + 4] / [(5x + 6) + 4] = 11/10 10(6x + 10) = 11(5x + 10) => x = 2 Sagar's present age = (5x + 6) = 16 years.

Q.- The sum of the present ages of a father and his son is 60 years. Six years ago father's age was five times the age of the son. After 6 years son's age will be

    1. 1. 15 years
    2. 2. 18 years
    3. 3. 20 years
    4. 4. 25 years

3
Clue : (60 - x) - 6 = 5(x - 6)

Q.- A person's present age is two-fifth of the age of his mother. After 8 years he will be one-half of the age of his mother. How old is the mother at present

    1. 1. 35
    2. 2. 40
    3. 3. 45
    4. 4. 42

2
Let the mother's present age be x years. Then the person's present age = (2/5 x) years. => (2/5 x + 8/2) = 1 (x + 8) => 2(2x + 40) = 5(x + 8) => x = 40

Q.- A man is 24 years older than his son. In two years his age will be twice the age of his son. The present age of this son is

    1. 1. 20 years
    2. 2. 22 years
    3. 3. 18 years
    4. 4. 23 years

2
Let's Son age is x then Father age is x+24. => 2(x+2) = (x+24+2) => 2x+4 = x+26 => x = 22 years

Q.- Ratio of ages of three persons is 4:7:9 Eight years ago the sum of their ages was 56. Find their present ages.

    1. 1. 163536
    2. 2. 353616
    3. 3. 122836
    4. 4. 162836

4
Let the present ages are 4x 7x 9x. => (4x-8) + (7x-8) + (9x-8) = 56 => 20x = 80 => x = 4 So their present ages are: 162836

Q.- A man is 24 years older than his son. In two years his age will be twice the age of his son. The present age of his son is

    1. 1. 20 years
    2. 2. 21 years
    3. 3. 22 years
    4. 4. 23 years

3
Let the son's present age be x years. Then man's present age = (x + 24) years => (x + 24) + 2 = 2(x + 2) => x + 26 = 2x + 4 So x = 22

Q.- Total of the ages of A B ahd C at present is 90 years. Ten years ago the ratio of their ages was 1: 2: 3. What is the age of B at present

    1. 1. 28
    2. 2. 30
    3. 3. 29
    4. 4. 31

2
Let their ages 10 years ago is x 2x and 3x years. 10 + 2x + 10 + 3x + 10 = 90 hence x= 10 B's present age = (2x + 10) =30 years

Q.- Six years ago the ratio of the ages of Kunal and Sagar was 6:5 Four years hence the ratio of their ages will be 11:10. What is Sagar age at present

    1. 1. 17 years
    2. 2. 18 years
    3. 3. 20 years
    4. 4. 16 years

4
Let six years ago the age of Kunal and Sagar are 6x and 5x resp. then =>[(6x+6)+4]/(5x+6)+4=11/10 <=>10(6x+10)=11(5x+10) <=>5x=10=>x=2 So Sagar age is (5x+6) = 16

Q.- Sushil was thrice as old as Snehal 6 years back. Sushil will be times as old as Snehal 6 years hence. How old is Snehal today

    1. 1. 12
    2. 2. 14
    3. 3. 10
    4. 4. 16

1
Let Snehals age 6 years back = x. Then Sushils age 6 years back = 3x. (5/3) * (X + 6 + 6) = (3X + 6 + 6) So 5(x+ 12) = 3(3x+ 12) so x=6. Snehal Age = (x+ 6) years = 12 years

Q.- Ten years ago A was half of B in age. If the ratio of their present ages is 3 : 4 what will be the total of their present ages

    1. 1. 30
    2. 2. 35
    3. 3. 40
    4. 4. 45

2
Let A's age 10 years ago = x years. Then B's age 10 years ago = 2x years. (x + 10) / (2x+ lO) = 3/4 => x = 5. So the total of their present ages =(x + 10 + 2x + 10) = (3x + 20) = 35 years.

Q.- In 10 years A will be twice as old 5 as B was 10 years ago. If A is now 9 years older than B the present age of B is

    1. 1. 36
    2. 2. 38
    3. 3. 39
    4. 4. 40

3
Let B's age = x years. Then As age = (x+ 9) years. (x+9+10)=2(x—10) hence x=39. Present age of B = 39 years

Q.- The average weight of 8 person's increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg. What might be the weight of the new person?

    1. 1. 76 kg
    2. 2. 76.5 kg
    3. 3. 85 kg
    4. 4. None

3
Total weight increased = (8 x 2.5) kg = 20 kg. Weight of new person = (65 + 20) kg = 85 kg.

Q.- The average of 20 numbers is zero. Of them at the most how many may be greater than zero?

    1. 1. 0
    2. 2. 1
    3. 3. 10
    4. 4. 19

4
Average of 20 numbers = 0 Sum of 20 numbers (0 x 20) = 0 It is quite possible that 19 of these numbers may be positive and if their sum is a then 20th number is (-a)

Q.- A grocer has a sale of Rs. 6435 Rs. 6927 Rs. 6855 Rs. 7230 and Rs. 6562 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Rs. 6500?

    1. 1. 4991
    2. 2. 5991
    3. 3. 6001
    4. 4. 6991

1
Total sale for 5 months = Rs. (6435 + 6927 + 6855 + 7230 + 6562) = Rs. 34009 Required sale = Rs. [ (6500 x 6) - 34009 ] = Rs. (39000 - 34009) = Rs. 4991

Q.- A pupil's marks were wrongly entered as 83 instead of 63. Due to that the average marks for the class got increased by half (1/2). The number of pupils in the class is:

    1. 1. 10
    2. 2. 20
    3. 3. 40
    4. 4. 73

3
Let there be x pupils in the class Total increase in marks = x x 1/2 = x/2 x/2 = (83 - 63) x/2 = 20 x= 40

Q.- If the average marks of three batches of 55 60 and 45 students respectively is 50 55 60 then the average marks of all the students is:

    1. 1. 53.33
    2. 2. 54.68
    3. 3. 55
    4. 4. None

2
Required average = (55 x 50 + 60 x 55 + 45 x 60)/(55 + 60 + 45) = ( 2750 + 3300 + 2700)/160 = 8750/160 = 54.68

Q.- A library has an average of 510 visitors on Sundays and 240 on other days. The average number of visitors per day in a month of 30 days beginning with a Sunday is:

    1. 1. 250
    2. 2. 276
    3. 3. 280
    4. 4. 285

4
Since the month begins with a Sunday to there will be five Sundays in the month. Required average = (510 x 5 + 240 x 25)/30 = 8550/30 = 285

Q.- The average weight of A B and C is 45 kg. If the average weight of A and B be 40 kg and that of B and C be 43 kg then the weight of B is:

    1. 1. 17 kg
    2. 2. 20 kg
    3. 3. 26 kg
    4. 4. 31 kg

4
Let A B C represent their respective weights. Then we have : A + B + C = (45 x 3) = 135 .... (i) A + B = (40 x 2) = 80 .... (ii) B + C = (43 x 2) = 86 ....(iii) Adding (ii) and (iii) we get: A + 2B + C = 166 .... (iv) Subtracting (i) from (iv) we get : B = 31 B's weight = 31 kg.

Q.- The average monthly income of P and Q is Rs. 5050. The average monthly income of Q and R is Rs. 6250 and the average monthly income of P and R is Rs. 5200. The monthly income of P is:

    1. 1. 3500
    2. 2. 4000
    3. 3. 4050
    4. 4. 5000

2
Let P Q and R represent their respective monthly incomes. Then we have : P + Q = (5050 x 2) = 10100 .... (i) Q + R = (6250 x 2) = 12500 .... (ii) P + R = (5200 x 2) = 10400 .... (iii) Adding (i) (ii) and (iii) we get: 2(P + Q + R) = 33000 or P + Q + R = 16500 .... (iv) Subtracting (ii) from (iv) we get P = 4000 P's monthly income = Rs. 4000

Q.- The average age of husband wife and their child 3 years ago was 27 years and that of wife and the child 5 years ago was 20 years. The present age of the husband is:

    1. 1. 35
    2. 2. 40
    3. 3. 50
    4. 4. None

2
Sum of the present ages of husband wife and child = (27 x 3 + 3 x 3) years = 90 years Sum of the present ages of wife and child = (20 x 2 + 5 x 2) years = 50 years Husband's present age = (90 - 50) years = 40 years

Q.- A car owner buys petrol at Rs.7.50 Rs. 8 and Rs. 8.50 per litre for three successive years. What approximately is the average cost per litre of petrol if he spends Rs. 4000 each year?

    1. 1. Rs 7.98
    2. 2. Rs 8
    3. 3. Rs 8.50
    4. 4. Rs 9

1
Total quantity of petrol consumed in 3 years = 4000/7.50 + 4000/8 + 4000/8.5 litres = 4000( 2/15 + 1/8 + 2/17 ) litres = 76700/51 litres Total amount spent = Rs. (3 x 4000) = Rs. 12000 Average cost = Rs. ( 12000 x 51)/76700 = Rs. 6120/767 = Rs. 7.98

Q.- The average salary of all the workers in a workshop is Rs. 8000. The average salary of 7 technicians is Rs. 12000 and the average salary of the rest is Rs. 6000. The total number of workers in the workshop is:

    1. 1. 20
    2. 2. 21
    3. 3. 22
    4. 4. 23

2
Let the total number of workers be x. Then 8000x = (12000 * 7) + 6000(x - 7) 2000x = 42000 x = 21.

Q.- The arithmetic mean of the scores of a group of students in a test was 52. The brightest 20% of them secured a mean score of 80 and the dullest 25% a mean score of 31. The mean score of remaining 55% is:

    1. 1. 45
    2. 2. 50
    3. 3. 51.4 approx
    4. 4. 54.1 approx

3
Let the required means score be x. Then 20 * 80 + 25 * 31 + 55 * x = 52 * 100 1600 + 775 + 55x = 5200 55x = 2825 x = 565/11 = 51.4.

Q.- The average age of husband wife and their child 3 years ago was 27 years and that of wife and the child 5 years ago was 20 years. The present age of the husband is:

    1. 1. 35 yrs
    2. 2. 40 yrs
    3. 3. 45 yrs
    4. 4. 50 yrs

2
Sum of the present ages of husband wife and child = (27 * 3 + 3 * 3) = 90 years. Sum of the present age of wife and child = (20 * 2 + 5 * 2) = 50 years. Husband's present age = (90 - 50) = 40 years.

Q.- In an examination a pupil's average marks were 63 per paper. If he had obtained 20 more marks for his Geography paper and 2 more marks for his History paper his average per paper would have been 65. How many papers were there in the examination?

    1. 1. 8
    2. 2. 9
    3. 3. 10
    4. 4. 11

4
Let the number of papers be x. Then 63x + 20 + 2 = 65x 2x = 22 x = 11

Q.- The average age of students of a class is 15.8 years. The average age of boys in the class is 16.4 years and that of the girls is 15.4 years. The ration of the number of boys to the number of girls in the class is:

    1. 1. 0.043055555556
    2. 2. 0.085416666667
    3. 3. 0.12777777778
    4. 4. 0.17013888889

2
Let the ratio be k : 1. Then k * 16.4 + 1 * 15.4 = (k + 1) * 15.8 (16.4 - 15.8)k = (15.8 - 15.4) k = 0.4/0.6 = 2/3 Required ratio = 2/3 : 1 = 2:3.

Q.- The average monthly salary of 20 employees in an organisation is Rs. 1500. If the manager's salary is added then the average salary increases by Rs. 100. What is the manager's monthly salary?

    1. 1. Rs.2000
    2. 2. Rs.2400
    3. 3. Rs.3600
    4. 4. Rs.4800

3
Manager's monthly salary = Rs. (1600 * 21 - 1500 * 20) = Rs. 3600

Q.- The captain of a cricket team of 11 members is 26 years old and the wicket keeper is 3 years older. If the ages of these two are excluded the average age of the remaining players is one year less than the average age of the whole team. What is the averag

    1. 1. 23 yrs
    2. 2. 25 yrs
    3. 3. 35 yrs
    4. 4. 46 yrs

1
Let the average of the whole team be x years. 11x - (26 + 29) = 9(x - 1) 11x - 9x = 46 2x = 46 => x = 23 So average age of the team is 23 years.

Q.- The mean of 50 observations was 36. It was found later that an observation 48 was wrongly taken as 23. The corrected new mean is:

    1. 1. 35.2
    2. 2. 36.1
    3. 3. 36.5
    4. 4. 39.1

3
Correct sum = (36 * 50 + 48 - 23) = 1825. Correct mean = 1825/50 = 36.5

Q.- The average weight of A B and C is 45 kg. If the average weight of A and B be 40 kg and that of B and C be 43 kg then the weight of B is:

    1. 1. 17 kg
    2. 2. 20 kg
    3. 3. 26 kg
    4. 4. 31 kg

4
Let A B C represent their respective weights. Then we have: A + B + C = (45 * 3) = 135 --- (i) A + B = (40 * 2) = 80 --- (ii) B + C = (43 * 2) = 86 --- (iii) Adding (ii) and (iii) we get: A + 2B + C = 166 --- (iv) Subtracting (i) from (iv) we get: B = 31 B's weight = 31 kg

Q.- A total of 3000 chocolates were distributed among 120 boys and girls such that each boy received 2 chocolates and each girl received 3 chocolates. Find the respective number of boys and girls?

    1. 1. 7050
    2. 2. 6060
    3. 3. 5070
    4. 4. 8040

2
Let the number of boys be x. Number of girls is 120 - x. Total number of chocolates received by boys and girls = 2x + 3(120 - x) = 300 => 360 - x = 300 => x = 60. So the number of boys or girls is 60.

Q.- When a student weighing 45 kgs left a class the average weight of the remaining 59 students increased by 200g. What is the average weight of the remaining 59 students

    1. 1. 56
    2. 2. 59
    3. 3. 57
    4. 4. 58

3
Let the average weight of the 59 students be A. So the total weight of the 59 of them will be 59*A. The questions states that when the weight of this student who left is added the total weight of the class = 59A + 45 When this student is also included the average weight decreases by 0.2 kgs (59A+45)/60=A-0.2 => 59A + 45 = 60A - 12 => 45 + 12 = 60A - 59A => A = 57

Q.- A motorist travel to a place 150 km away at an avearge speed of 50 km/hr and returns ar 30 km/hr.His average speed for the whole jouney in km/hr is

    1. 1. 57.3 km/hr
    2. 2. 37.5 km/hr
    3. 3. 37.9 km/hr
    4. 4. 75.3 km/hr

2
Average speed will be 2xy/(x+y) km/hr = \frac{2(50)(30)}{50+30} km/hr = 37.5 km/hr

Q.- The average of six numbers is X and the average of three of these is Y.If the average of the remaining three is z then

    1. 1. x = y + z
    2. 2. 2x = y + z
    3. 3. x = 2y + z
    4. 4. x = y + 2z

2
X =((3y+3z)/6) or 2X= y + z

Q.- The average age of husband wife and their child 3 years ago was 27 years and that of wife and the child 5 years ago was 20 years. The present age of the husband is

    1. 1. 40
    2. 2. 45
    3. 3. 50
    4. 4. 35

1
Sum of the present ages of husband wife and child = (27 * 3 + 3 * 3) years = 90 years. Sum of the present ages of wife and child = (20 * 2 + 5 * 2) years = 50 years. Husband's present age = (90 - 50) years = 40 years

Q.- If the average marks of three batches of 55 60 and 45 students respectively is 50 55 60 then the average marks of all the students is

    1. 1. 54.86
    2. 2. 54.68
    3. 3. 45.68
    4. 4. 45.86

2
(55×50)+(60×55)+(45×60)55+60+45 =8750/160 =54.68

Q.- Average of 10 matches is 32. How many runs one should should score to increase his average by 4 runs.

    1. 1. 75
    2. 2. 76
    3. 3. 74
    4. 4. 73

2
Average after 11 innings should be 36 So Required score = (11 * 36) - (10 * 32) = 396 - 320 = 76

Q.- Average of first five multiples of 3 is

    1. 1. 9
    2. 2. 11
    3. 3. 13
    4. 4. 10

1
Average=3(1+2+3+4+5)/5=45/5=9

Q.- A library has an average of 510 visitors on Sundays and 240 on other day. The average number of visitors in a month of 30 days starting with sunday is

    1. 1. 280
    2. 2. 285
    3. 3. 290
    4. 4. 275

2
As the month begin with sunday so there will be five sundays in the month. So result will be: =[(510×5)+(240×25)]/30)=(8550/30)=285

Q.- A batsman makes a score of 87 runs in the 17th match and thus increases his average by 3. Find his average after 17th match

    1. 1. 40
    2. 2. 38
    3. 3. 37
    4. 4. 39

4
Let the average after 17th match is x then the average before 17th match is x-3 so 16(x-3) + 87 = 17x => x = 87 - 48 = 39

Q.- Average weight of 10 people increased by 1.5 kg when one person of 45 kg is replaced by a new man. Then weight of the new man is

    1. 1. 45
    2. 2. 50
    3. 3. 60
    4. 4. 55

3
Total weight increased is 1.5 * 10 = 15. So weight of new person is 45+15 = 60

Q.- In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels always come together?

    1. 1. 702
    2. 2. 207
    3. 3. 270
    4. 4. 720

4
The word 'OPTICAL' contains 7 different letters. When the vowels OIA are always together they can be supposed to form one letter. Then we have to arrange the letters PTCL (OIA). Now 5 letters can be arranged in 5! = 120 ways. The vowels (OIA) can be arranged among themselves in 3! = 6 ways. Required number of ways = (120 x 6) = 720.

Q.- How many 4-letter words with or without meaning can be formed out of the letters of the word 'LOGARITHMS' if repetition of letters is not allowed?

    1. 1. 5400
    2. 2. 5004
    3. 3. 5040
    4. 4. 4500

3
LOGARITHMS' contains 10 different letters. Required number of words = Number of arrangements of 10 letters taking 4 at a time. = 10P4 = (10 x 9 x 8 x 7) = 5040.

Q.- In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?

    1. 1. 36
    2. 2. 63
    3. 3. 76
    4. 4. 67

2
Required number of ways = (7C5 x 3C2) = (7C2 x 3C1) = [(7 x 6)/(2x1)] x3 = 63.

Q.- How many 3-digit numbers can be formed from the digits 2 3 5 6 7 and 9 which are divisible by 5 and none of the digits is repeated?

    1. 1. 20
    2. 2. 25
    3. 3. 15
    4. 4. 18

1
Since each desired number is divisible by 5 so we must have 5 at the unit place. So there is 1 way of doing it. The tens place can now be filled by any of the remaining 5 digits (2 3 6 7 9). So there are 5 ways of filling the tens place. The hundreds place can now be filled by any of the remaining 4 digits. So there are 4 ways of filling it. Required number of numbers = (1 x 5 x 4) = 20.

Q.- In how many ways a committee consisting of 5 men and 6 women can be formed from 8 men and 10 women?

    1. 1. 11670
    2. 2. 11760
    3. 3. 11067
    4. 4. 11076

2
Required number of ways = (8C5 x 10C6) = (8C3 x 10C4) = [ (8 x 7 x 6)/(3 x 2 x 1)] x[( 10 x 9 x 8 x 7)/ (4 x 3 x 2 x 1)] = 11760

Q.- In how many ways can the letters of the word 'LEADER' be arranged?

    1. 1. 630
    2. 2. 306
    3. 3. 360
    4. 4. 603

3
The word 'LEADER' contains 6 letters namely 1L 2E 1A 1D and 1R. Required number of ways = 6!/(1!)(2!)(1!)(1!)(1!) = 360.

Q.- In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together?

    1. 1. 270
    2. 2. 702
    3. 3. 207
    4. 4. 720

4
The word 'LEADING' has 7 different letters. When the vowels EAI are always together they can be supposed to form one letter. Then we have to arrange the letters LNDG (EAI). Now 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways. The vowels (EAI) can be arranged among themselves in 3! = 6 ways. Required number of ways = (120 x 6) = 720.

Q.- In how many different ways can the letters of the word 'DETAIL' be arranged in such a way that the vowels occupy only the odd positions?

    1. 1. 83
    2. 2. 38
    3. 3. 36
    4. 4. 63

3
There are 6 letters in the given word out of which there are 3 vowels and 3 consonants. Let us mark these positions as under: (1) (2) (3) (4) (5) (6) Now 3 vowels can be placed at any of the three places out 4 marked 1 3 5. Number of ways of arranging the vowels = 3P3 = 3! = 6. Also the 3 consonants can be arranged at the remaining 3 positions. Number of ways of these arrangements = 3P3 = 3! = 6. Total number of ways = (6 x 6) = 36.

Q.- A box contains 2 white balls 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box if at least one black ball is to be included in the draw?

    1. 1. 45
    2. 2. 50
    3. 3. 46
    4. 4. 64

4
We may have(1 black and 2 non-black) or (2 black and 1 non-black) or (3 black). Required number of ways = (3C1 x 6C2) + (3C2 x 6C1) + (3C3) = [(3 x 6 x 5)/(2x1)] + [(3 x 2 x 6)/(2x1)] + 1 = (45 + 18 + 1) = 64.

Q.- Out of 7 consonants and 4 vowels how many words of 3 consonants and 2 vowels can be formed?

    1. 1. 52200
    2. 2. 25002
    3. 3. 25200
    4. 4. 25202

3
Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4) = (7C3 x 4C2) = [(7 x 6 x 5)/(3x2x1)] x[( 4 x 3)/(2x1)] = 210. Number of groups each having 3 consonants and 2 vowels = 210. Each group contains 5 letters. Number of ways of arranging 5 letters among themselves = 5! = 5 x 4 x 3 x 2 x 1 = 120. Required number of ways = (210 x 120) = 25200.

Q.- Ten points are marked on a straight line and eleven points are marked on another straight line. How many triangles can be constructed with vertices from among the above points?

    1. 1. 495
    2. 2. 550
    3. 3. 1045
    4. 4. 2475

3
We can get the triangles in two different ways. Taking two points from the line having 10 points (in 10C2 ways i.e. 45 ways) and one point from the line consisting of 11 points (in 11 ways). So the number of triangles here is 45×11=495. Taking two points from the line having 11 points (in 11C2 i.e. 55 ways) and one point from the line consisting of 10 points (in 10 ways) the number of triangles here is 55×10=550 Total number of triangles =495+550 =1045 triangles

Q.- Suppose you have a currency named Miso in three denominations: 1 Miso 10 Misos and 50 Misos. In how many ways can you pay a bill of 107 Misos?

    1. 1. 16
    2. 2. 17
    3. 3. 18
    4. 4. 19

3
Let the number of currency 1 Miso 10 Misos and 50 Misos be xy and z respectively. x+10y+50z=107 Now the possible values of z could be 0 1 and 2. For z=0: x+10y=107 Number of integral pairs of values of x and y that satisfy the equation: x+10y=107 will be 11. These values of x and y in that order are: (710);(179);(278)…(1070) For z=1: x+10y=57 Number of integral pairs of values of x and y that satisfy the equation: x+10y=57 will be 6. These values of x and y in that order are: (75);(174);(273);(372);(471) and (570) For z=2: x+10y=7 There is only one integer value of x and y that satisfies the equation: x+10y=7 in that order is (70) Therefore total number of ways in which you can pay a bill of 107 Misos: =11+6+1 =18

Q.- Company BELIANCE hosted a party for 8 members of Company AXIAL. In the party no member of AXIAL had interacted with more than three members of BELIANCE. Out of all the members of BELIANCE three members - each interacted with four members of AXIAL and the

    1. 1. 9
    2. 2. 10
    3. 3. 11
    4. 4. 12

1
The important constraint here is that in the party no member of AXIAL had interacted with more than three members of BELIANCE. Given that there are 8 members of company AXIAL and three members of company BELIANCE interacted with four members of AXIAL Therefore the maximum possible number of members of company BELIANCE in the party will be 3+ 4C2 (Since each of the remaining members of the company BELIANCE have interacted with two members of AXIAL) =9

Q.- Each of the 11 letters AHIMOTUVWX and Z appears same when looked at in a mirror. They are called symmetric letters. Other letters in the alphabet are asymmetric letters. How many three letter computer passwords can be formed (no repetition allowed) with a

    1. 1. 12000
    2. 2. 12870
    3. 3. 13000
    4. 4. 15870

2
As there are 26 alphabets we can choose a 3 letter password from 11 symmetric letters and 26−11=15 from asymmetric letters. There are three possible cases that will satisfy the condition of forming three letter passwords with at least 1 symmetric letter. Case 1: 1 symmetric and 2 asymmetric =(11C1×15C2) =11×(15×4/1×2) =1155 Case 2: 2 symmetric and 1 asymmetric =(11C2×15C1) =(11×10/1×2)×15 =825 Case 3: all 3 sym+K145metric =11C3 =(11×10×9/1×2×3) =165 Hence total possible cases ={case (1)+case (2)+case (3)}×3! =[1155+825+165]×6 =2145×6 =12870

Q.- When four fair dice are rolled simultaneously in how many outcomes will at least one of the dice show 3?

    1. 1. 1296
    2. 2. 625
    3. 3. 671
    4. 4. 1921

3
When 4 dice are rolled simultaneously there will be a total of pow(64)=1296 outcomes. The number of outcomes in which none of the 4 dice show 3 will be pow(54)=625 outcomes. Therefore the number of outcomes in which at least one die will show 3: =1296–625 =671

Q.- In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels always come together?

    1. 1. 610
    2. 2. 720
    3. 3. 825
    4. 4. 120

2
The word 'OPTICAL' has 7 letters. It has the vowels 'O''I''A' in it and these 3 vowels should always come together. Hence these three vowels can be grouped and considered as a single letter. That is PTCL(OIA). Hence we can assume total letters as 5. and all these letters are different. Number of ways to arrange these letters = 5! = [5 x 4 x 3 x 2 x 1] = 120 All The 3 vowels (OIA) are different Number of ways to arrange these vowels among themselves = 3! = [3 x 2 x 1] = 6 Hence required number of ways = 120 x 6 = 720

Q.- Some boys are standing on a circle at distinct points. Each possible pair of persons who are not adjacent sing a 3 minute song one pair after another.The total time taken by all the pairs to sing is 1 hour. Find the number of boys?

    1. 1. 6
    2. 2. 7
    3. 3. 8
    4. 4. 9

3
Each boy would pair with n−3 other boys Number of possible pairs =n(n−3)/2 n(n−3)/2=603 n(n−3)/2=20 n(n−3)=40 n=8

Q.- In how many rearrangements of the word AMAZED is the letter E positioned in between the 2 As (Not necessarily flanked)?

    1. 1. 24
    2. 2. 72
    3. 3. 120
    4. 4. 360

3
In any rearrangement of the word consider only the positions of the letters AA and E. These can be as AAE AEA or EAA. So effectively one-third of all words will have E in between the two As. The total number of rearrangements are 6!/2!=360 One-third of 360 is =(1/3)×360 =120

Q.- abc are three distinct integers from 2 to 10 (both inclusive). Exactly one of abbc and ca is odd. abc is a multiple of 4. The arithmetic mean of a and b is an integer and so is the arithmetic mean of ab and c. How many such triplets are possible (unorder

    1. 1. 4
    2. 2. 5
    3. 3. 6
    4. 4. 7

1
Exactly one of abbc and ca is odd ⇒ Two are odd and one is even abc is a multiple of 4 ⇒ the even number is a multiple of 4 The arithmetic mean of a and b is an integer ⇒ a and b are odd and so is the arithmetic mean of ab and c. ⇒ a+b+c is a multiple of 3 c can be 4 or 8. c=4;ab can be 35 or 59 c=8;ab can be 37 or 79 Four triplets are possible.

Q.- If F(xn) be the number of ways of distributing x toys to n children so that each child receives at the most 2 toys then F(43)= _______?

    1. 1. 4
    2. 2. 5
    3. 3. 6
    4. 4. 7

3
We have to find the number of ways in which 4 toys can be distributed to 3 children so that each child receives at the most 2 toys. There are two possible cases: Case 1: Two of them receive 2 toys each and one of them doesn't get any toy. There are 3 possible ways to distribute the toys in this case i.e. the three possible ways of selecting the child who will not get any toy. Case 2: Two of them receive 1 toy each and one of them receives 2 toys. Again there are 3 possible ways to distribute the toys in this case i.e. the three possible ways of selecting the child who will get 2 toys. So there are a total of 6 possible ways.

Q.- In how many different ways can the letters of the word 'RUMOUR' be arranged?

    1. 1. 180
    2. 2. 108
    3. 3. 18
    4. 4. 200

1
The word 'RUMOUR' has 6 letters. But in these 6 letters 'R' occurs 2 times 'U' occurs 2 times and rest of the letters are different. Hence number of ways to arrange these letters = 6!/(2!)(2!)=(6×5×4×3×2×1)/(2×1)(2×1) = 180

Q.- How many 6 digit telephone numbers can be formed. if each number starts with 35 and no digit appears more than once?

    1. 1. 1680
    2. 2. 6180
    3. 3. 6108
    4. 4. 1608

1
The first two places can only be filled by 3 and 5 respectively and there is only 1 way of doing this Given that no digit appears more than once. Hence we have 8 digits remaining(01246789) So the next 4 places can be filled with the remaining 8 digits in 8P4 ways Total number of ways = 8P4 = 8 x 7 x 6 x 5 = 1680

Q.- An event manager has ten patterns of chairs and eight patterns of tables. In how many ways can he make a pair of table and chair?

    1. 1. 90
    2. 2. 80
    3. 3. 100
    4. 4. 70

2
He has has 10 patterns of chairs and 8 patterns of tables Hence A chair can be arranged in 10 ways and A table can be arranged in 8 ways Hence one chair and one table can be arranged in 10 x 8 ways = 80 ways

Q.- 25 buses are running between two places P and Q. In how many ways can a person go from P to Q and return by a different bus?

    1. 1. 620
    2. 2. 610
    3. 3. 600
    4. 4. 650

3
He can go in any bus out of the 25 buses. Hence He can go in 25 ways. Since he can not come back in the same bus that he used for travelling He can return in 24 ways. Total number of ways = 25 x 24 = 600

Q.- A box contains 4 red 3 white and 2 blue balls. Three balls are drawn at random. Find out the number of ways of selecting the balls of different colours?

    1. 1. 20
    2. 2. 21
    3. 3. 22
    4. 4. 24

4
1 red ball can be selected in 4C1 ways 1 white ball can be selected in 3C1 ways 1 blue ball can be selected in 2C1 ways Total number of ways = 4C1 x 3C1 x 2C1 =4 x 3 x 2 = 24

Q.- In how many different ways can 5 girls and 5 boys form a circle such that the boys and the girls alternate?

    1. 1. 2880
    2. 2. 2808
    3. 3. 8820
    4. 4. 8802

1
In a circle 5 boys can be arranged in 4! Ways Given that the boys and the girls alternate. Hence there are 5 places for girls which can be arranged in 5! Ways Total number of ways = 4! x 5! = 24 x 120 = 2880

Q.- Find out the number of ways in which 6 rings of different types can be worn in 3 fingers?

    1. 1. 729
    2. 2. 792
    3. 3. 297
    4. 4. 279

1
The first ring can be worn in any of the 3 fingers => There are 3 ways of wearing the first ring Similarly each of the remaining 5 rings also can be worn in 3 ways Hence total number of ways =3×3×3×3×3×6=pow(36)=729

Q.- In how many ways can 5 man draw water from 5 taps if no tap can be used more than once?

    1. 1. 210
    2. 2. 120
    3. 3. 320
    4. 4. 302

2
1st man can draw water from any of the 5 taps 2nd man can draw water from any of the remaining 4 taps 3rd man can draw water from any of the remaining 3 taps 4th man can draw water from any of the remaining 2 taps 5th man can draw water from remaining 1 tap 5 4 3 2 1 Hence total number of ways = 5 x 4 x 3 x 2 x 1 = 120

Q.- What is the value of 100P2 ?

    1. 1. 8008
    2. 2. 8800
    3. 3. 9900
    4. 4. 9009

3
100P2 = 100 x 99 = 9900

Q.- How many words with or without meaning can be formed by using all the letters of the word 'DELHI' using each letter exactly once?

    1. 1. 310
    2. 2. 210
    3. 3. 201
    4. 4. 120

4
The word 'DELHI' has 5 letters and all these letters are different. Total words (with or without meaning) formed by using all these 5 letters using each letter exactly once = Number of arrangements of 5 letters taken all at a time = 5P5 = 5! = 5 x 4 x 3 x 2 x 1 = 120

Q.- In a division sum the remainder is 0. As student mistook the divisor by 12 instead of 21 and obtained 35 as quotient. What is the correct quotient ?

    1. 1. 22
    2. 2. 15
    3. 3. 25
    4. 4. 20

4
Number = (12 x 35) Correct Quotient = 420 � 21 = 20

Q.- Which one of the following numbers is exactly divisible by 11?

    1. 1. 414256
    2. 2. 412456
    3. 3. 415624
    4. 4. 416524

3
(4 + 5 + 2) - (1 + 6 + 3) = 1 not divisible by 11. (2 + 6 + 4) - (4 + 5 + 2) = 1 not divisible by 11. (4 + 6 + 1) - (2 + 5 + 3) = 1 not divisible by 11. (4 + 6 + 1) - (2 + 5 + 4) = 0 So 415624 is divisible by 11.

Q.- If the number 517*324 is completely divisible by 3 then the smallest whole number in the place of * will be :

    1. 1. 1
    2. 2. 2
    3. 3. 3
    4. 4. 4

2
Sum of digits = (5 + 1 + 7 + x + 3 + 2 + 4) = (22 + x) which must be divisible by 3. x = 2.

Q.- 72519 x 9999 = ?

    1. 1. 725114718
    2. 2. 725114781
    3. 3. 725117481
    4. 4. 725117418

3
72519 x 9999 = 72519 x (10000 - 1) = 72519 x 10000 - 72519 x 1 = 725190000 - 72519 = 725117481.

Q.- (12345679 x 72) = ?

    1. 1. 88888888
    2. 2. 888888888
    3. 3. 8888888
    4. 4. 888888

2
12345679 x 72 = 12345679 x (70 +2) = 12345679 x 70 + 12345679 x 2 = 864197530 + 24691358 = 888888888

Q.- The smallest prime number is :

    1. 1. 1
    2. 2. 3
    3. 3. 2
    4. 4. 4

3
The smallest prime number is 2.

Q.- How many prime numbers are less than 50 ?

    1. 1. 16
    2. 2. 14
    3. 3. 13
    4. 4. 15

4
Prime numbers less than 50 are: 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 Their number is 15

Q.- A boy multiplied 987 by a certain number and obtained 559981 as his answer. If in the answer both 9 are wrong and the other digits are correct then the correct answer would be :

    1. 1. 555816
    2. 2. 555168
    3. 3. 555681
    4. 4. 555861

3
987 = 3 x 7 x 47 So the required number must be divisible by each one of 3 7 47 553681 (Sum of digits = 28 not divisible by 3) 555181 (Sum of digits = 25 not divisible by 3) 555681 is divisible by 3 7 47.

Q.- On dividing 2272 as well as 875 by 3-digit number N we get the same remainder. The sum of the digits of N is:

    1. 1. 11
    2. 2. 8
    3. 3. 9
    4. 4. 10

4
Clearly (2272 - 875) = 1397 is exactly divisible by N. Now 1397 = 11 x 127 The required 3-digit number is 127 the sum of whose digits is 10.

Q.- Which one of the following numbers is completely divisible by 45?

    1. 1. 208260
    2. 2. 202806
    3. 3. 202860
    4. 4. 208206

3
45 = 5 x 9 where 5 and 9 are co-primes. Unit digit must be 0 or 5 and sum of digits must be divisible by 9. Among given numbers such number is 202860.

Q.- On simplification [{(6/8)+(9/11)} -0.35]/(3/5) gives

    1. 1. 76/33
    2. 2. 67/33
    3. 3. 33/76
    4. 4. 33/67

2
[{(66+72)/88}-(35/100)]/(3/5) ={(138/88)-(35/100)}/(3/5) ={(69/44)-(7/20)}/(3/5) ={(690-154)/440}/(3/5) =(536/440)x(5/3) =67/33

Q.- The value of (1)+{1/1+[1/{1+(1/9)}]}

    1. 1. 29/19
    2. 2. 47392
    3. 3. 43037
    4. 4. 19/29

1
(1)+[1/{(1)+(9/10)}] =(1)+(10/19)=29/19

Q.- {1+3+5+...+3983}/1992 is equal to

    1. 1. 1988
    2. 2. 1989
    3. 3. 1990
    4. 4. 1992

4
Let the number of terms in 1+3+5+…+3983 be n then 1+ (n-1)x2=3983 means n=1992 hence 1+3+5…+3983={(1992)/2}x(1+3983)=1992x1992 hence1+3+5+…+3983)/1992 =(1992x1992)/1992 =1992

Q.- on dividing a number by 999 the quotient is377 and the reminder is 105. The number is:

    1. 1. 476727
    2. 2. 376538
    3. 3. 376728
    4. 4. 359738

3
Number=(Divisorxquotient)+ Remainder =(999x377)+105 ={(1000-1)x377}+105 =376728

Q.- The least number by which 72 must be multiplied in order to produce a multiple of 112 is :

    1. 1. 6
    2. 2. 12
    3. 3. 14
    4. 4. 18

3
A number which is divisible by both 72 and 112 is their lcm which is 1008. therefore Required number=1008/72=14

Q.- The unit digit in the product (784 x 618 x 917 x 463) is:

    1. 1. 2
    2. 2. 3
    3. 3. 4
    4. 4. 5

1
Unit digit in the given product = Unit digit in (4 x 8 x 7 x 3) = (672) = 2

Q.- If the number 653 xy is divisible by 90 then (x + y) = ?

    1. 1. 2
    2. 2. 3
    3. 3. 4
    4. 4. 6

3
90 = 10 x 9 Clearly 653xy is divisible by 10 so y = 0 Now 653x0 is divisible by 9. So (6 + 5 + 3 + x + 0) = (14 + x) is divisible by 9. So x = 4. Hence (x + y) = (4 + 0) = 4.

Q.- Which one of the following can't be the square of natural number ?

    1. 1. 32761
    2. 2. 81225
    3. 3. 42437
    4. 4. 20164

3
The square of a natural number never ends in 7. 42437 is not the square of a natural number.

Q.- The sum of first 45 natural numbers is:

    1. 1. 1035
    2. 2. 1280
    3. 3. 2070
    4. 4. 2140

1
Let Sn = (1 + 2 + 3 + ... + 45) This is an A.P. in which a = 1 d = 1 n = 45 and l = 45 Sn = (n/2) (a + l) = (45/2) x (1 + 45) = (45 x 23) = 1035 Required sum = 1035.

Q.- The sum of all two digit numbers divisible by 5 is:

    1. 1. 1035
    2. 2. 1245
    3. 3. 1230
    4. 4. 945

4
Required numbers are 10 15 20 25 ... 95 This is an A.P. in which a = 10 d = 5 and l = 95. tn = 95 a + (n - 1)d = 95 10 + (n - 1) x 5 = 95 (n - 1) x 5 = 85 (n - 1) = 17 n = 18 Requuired Sum = (n/2) (a + l) = (18/2) x (10 + 95) = (9 x 105) = 945.

Q.- The difference between a two-digit number and the number obtained by interchanging the digit is 36. What is the difference between the sum and the difference of the digits of the number if the ratio between the digits of the number is 1:2?

    1. 1. 6
    2. 2. 8
    3. 3. 16
    4. 4. 10

2
Since the number is greater than the number obtained on reversing the digits so the ten's digit is greater than the unit's digit. Let the ten's and unit's digits be 2x and x respectively. Then (10 * 2x + x) - (10x + 2x) = 36 9x = 36 x = 4 Required difference = (2x + x) - (2x - x) = 2x = 8.

Q.- Two-third of a positive number and 25/216 of its reciprocal are equal. The number is:

    1. 1. 43074
    2. 2. 42867
    3. 3. 25/144
    4. 4. 42867

1
Let the number be x. Then 2/3 x = 25/216 * 1/x pow(x2) = 25/216 * 3/2 = 25/144 x = 5/12

Q.- If the sum of one-half of a number exceeds one-third of that number by 7 1/3 the number is:

    1. 1. 16
    2. 2. 18
    3. 3. 20
    4. 4. 22

3
Let the number be x. Then (1/2 x + 1/5 x) - 1/3 x = 22/3 11/30 x = 22/3 x = 20

Q.- The difference of two numbers is 1365. On dividing the larger number by the smaller we get 6 as quotient and 15 as remainder. What is the smaller number?

    1. 1. 720
    2. 2. 270
    3. 3. 702
    4. 4. 207

2
Let the smallest number be x. Then larger number = (x + 1365) x + 1365 = 6x + 15 = 5x = 1350 x = 270 Smaller number = 270.

Q.- Which one of the following numbers is exactly divisible by 11?

    1. 1. 235641
    2. 2. 245642
    3. 3. 315624
    4. 4. 415624

4
(4 + 5 + 2) - (1 + 6 + 3) = 1 not divisible by 11. (2 + 6 + 4) - (4 + 5 + 2) = 1 not divisible by 11. (4 + 6 + 1) - (2 + 5 + 3) = 1 not divisible by 11. (4 + 6 + 1) - (2 + 5 + 4) = 0. So 415624 is divisible by 11.

Q.- Find the least number with which 16200 should be multiplied to make it a perfect cube.

    1. 1. 55
    2. 2. 50
    3. 3. 45
    4. 4. 54

3
16200 = 23 * 34 * 52 A perfect cube has a property of having the indices of all its prime factors divisible by 3. Required number = 32 * 5 = 45.

Q.- The sum of five consecutive odd numbers of set p is 435. What is the sum of five consecutive numbers of another set q. Whose largest number is 45 more than the largest number of set p?

    1. 1. 607
    2. 2. 670
    3. 3. 706
    4. 4. 760

2
Let the five consecutive odd numbers of set p be 2n - 3 2n - 1 2n + 1 2n + 3 2n + 5. Sum of these five numbers = 2n - 3 + 2n - 1 + 2n + 1 + 2n + 3 + 2n + 5 = 10n + 5 = 435 => n = 43 Largest number of set p = 2(43) + 5 = 91 The largest number of set q = 91 + 45 = 136 => The five numbers of set q are 132 133 134 135 136. Sum of above numbers = 132 + 133 + 134 + 135 + 136 = 670.

Q.- Three seventh of a number is 12 more than 40% of that number. What will be the 60% of that number?

    1. 1. 216
    2. 2. 162
    3. 3. 126
    4. 4. 261

3
3/7 x – 40/100 x = 12 x = 35 * 12 35 * 12 * 60/100 = 252/2 = 126

Q.- A two-digit number is such that the product of the digits is 8. When 18 is added to the number then the digits are reversed. The number is:

    1. 1. 42
    2. 2. 24
    3. 3. 23
    4. 4. 32

2
Let the ten's and unit's digit be x and 8/x respectively. Then (10x + 8/x) + 18 = 10 * 8/x + x 9x2 + 18x - 72 = 0 x2 + 2x - 8 = 0 (x + 4)(x - 2) = 0 x = 2 So ten's digit = 2 and unit's digit = 4. Hence required number = 24.

Q.- A number consists of 3 digit whose sum is 10. The middle digit is equal to the sum of the other two and the number will be increased by 99 if its digits are reversed. The number is:

    1. 1. 235
    2. 2. 253
    3. 3. 352
    4. 4. 325

2
Let the middle digit be x. Then 2x = 10 or x = 5. So the number is either 253 or 352. Since the number increases on reversing the digits so the hundred's digit is smaller than the unit's digit. Hence required number = 253.

Q.- Two pipes A and B can fill a cistern in 75/2 minutes and 45 minutes respectively. Both pipes are opened. The cistern will be filled in just half an hour if the B is turned off after :

    1. 1. 7 min
    2. 2. 6 min
    3. 3. 8 min
    4. 4. 9 min

4
Let B be turned off after x minutes. Then Part filled by (A + B) in x min. + Part filled by A in (30 -x) min. = 1. x [(2/75) + (1/45)] + (30 - x). 2/75 = 1 (11x/225) + (60 -2x)/75 = 1 11x + 180 - 6x = 225. x = 9.

Q.- A pump can fill a tank with water in 2 hours. Because of a leak it took 2 hours to fill the tank. The leak can drain all the water of the tank in :

    1. 1. 12 hrs.
    2. 2. 13 hrs.
    3. 3. 14 hrs.
    4. 4. 15 hrs.

3
Work done by the leak in 1 hour = (1/2) - (3/7) = 1/14 Leak will empty the tank in 14 hrs.

Q.- Pipes A and B can fill a tank in 5 and 6 hours respectively. Pipe C can empty it in 12 hours. If all the three pipes are opened together then the tank will be filled in :

    1. 1. 17/60
    2. 2. 60/17
    3. 3. 15/16
    4. 4. 42903

2
Net part filled in 1 hour ( 1/5) +( 1/6) - (1/12) = 17/60 The tank will be full in 60/17 hours

Q.- Three pipes A B and C can fill a tank from empty to full in 30 minutes 20 minutes and 10 minutes respectively. When the tank is empty all the three pipes are opened. A B and C discharge chemical solutions PQ and R respectively. What is the proportion of

    1. 1. 43046
    2. 2. 43045
    3. 3. 43044
    4. 4. 43043

2
Part filled by (A + B + C) in 3 minutes = 3[ (1/30) + (1/20) + (1/10)] = 3 x11/60 = 11/20 Part filled by C in 3 minutes/10 = 3/10 Required ratio = (3/10) x (20/11) = 6/11

Q.- Two pipes A and B can fill a tank in 20 and 30 minutes respectively. If both the pipes are used together then how long will it take to fill the tank ?

    1. 1. 15 min
    2. 2. 14 min
    3. 3. 12 min
    4. 4. 10 min

3
Part filled by A in 1 min = 1 /20 Part filled by B in 1 min = 1/30 Part filled by (A + B) in 1 min = (1/20) + (1/30) = 1/12 Both pipes can fill the tank in 12 minutes.

Q.- Two pipes A and B together can fill a cistern in 4 hours. Had they been opened separately then B would have taken 6 hours more than A to fill the cistern. How much time will be taken by A to fill the cistern separately ?

    1. 1. 7 hrs.
    2. 2. 4 hrs.
    3. 3. 5 hrs.
    4. 4. 6 hrs.

4
Let the cistern be filled by pipe A alone in x hours. Then pipe B will fill it in (x + 6) hours. (1/x) + 1/(x+6) = 1/4 (x + 6 + x)/x(x + 6) = 1/4 x2 - 2x - 24 = 0 (x -6)(x + 4) = 0 x = 6. [neglecting the negative value of x]

Q.- A tank is filled in 5 hours by three pipes A B and C. The pipe C is twice as fast as B and B is twice as fast as A. How much time will pipe A alone take to fill the tank ?

    1. 1. 45 hrs.
    2. 2. 40 hrs.
    3. 3. 53 hrs.
    4. 4. 35 hrs.

4
Suppose pipe A alone takes x hours to fill the tank. Then pipes B and C will take x/2 and x/4 hours respectively to fill the tank. (1/x) + (2/x) + (4/x) = 1/5 7/x = 1/5 x = 35 hrs.

Q.- Two pipes can fill a tank in 20 and 24 minutes respectively and a waste pipe can empty 3 gallons per minute. All the three pipes working together can fill the tank in 15 minutes. The capacity of the tank is :

    1. 1. 201 gallons
    2. 2. 102 gallons
    3. 3. 120 gallons
    4. 4. 210 gallons

3
Work done by the waste pipe in 1 minute = (1/15) - [ (1/20) + (1/24) ] = (1/15) - (11/120) = - 1/40 [-ve sign means emptying] Volume of 1/40 part = 3 gallons. Volume of whole = (3 x 40) gallons = 120 gallons.

Q.- One pipe can fill a tank three times as fast as another pipe. If together the two pipes can fill the tank in 36 minutes then the slower pipe alone will be able to fill the tank in :

    1. 1. 441 min
    2. 2. 144 min
    3. 3. 414 min
    4. 4. 405 min

2
Let the slower pipe alone fill the tank in x minutes. Then faster pipe will fill it in x/3 minutes. (1/x) + (3/x) = 1/36 4/x = 1/36 x = 144 min.

Q.- A large tanker can be filled by two pipes A and B in 60 minutes and 40 minutes respectively. How many minutes will it take to fill the tanker from empty state if B is used for half the time and A and B fill it together for the other half ?

    1. 1. 15 min
    2. 2. 20 min
    3. 3. 25 min
    4. 4. 30 min

4
Part filled by (A + B) in 1 minute = (1/60) + (1/40) = 1/24 . Suppose the tank is filled in x minutes. Then x/2[ (1/24) + (1/40) ] = 1 (x/2) x (1/15) = 1 x = 30 min.

Q.- Two pipes A and B can fill a tank in 6 hours and 4 hours respectively. If they are opened on alternate hours and if pipe A is opened first in how many hours the tank shall be full ?

    1. 1. 4
    2. 2. 5
    3. 3. 6
    4. 4. 7

2
(A+B)'s 2 hour's work when opened = (1/6)+(1/4)=5/12 (A+B)′s 4 hour's work =(5/12)∗2=5/6 Remaining work = 1−(5/6)=1/6 Now its A turn in 5 th hour1/6 work will be done by A in 1 hour Total time = 4+1=5 hours

Q.- 12 buckets of water fill a tank when the capacity of each tank is 13.5 litres. How many buckets will be needed to fill the same tank if the capacity of each bucket is 9 litres?

    1. 1. 15
    2. 2. 17
    3. 3. 18
    4. 4. 16

3
Capacity of the tank = (12*13.5) litres = 162 litres Capacity of each bucket = 9 litres. So we can get answer by dividing total capacity of tank by total capacity of bucket. Number of buckets needed = (162/9) = 18 buckets

Q.- One pipe can fill a tank three times as fast as another pipe. If together the two pipes can fill the tank in 36 minutes then the slower pipe alone will be able to fill the tank in

    1. 1. 144 min
    2. 2. 145 min
    3. 3. 414 min
    4. 4. 441 min

1
Let the slower pipe alone fill the tank in x minutes then faster will fill in x/3 minutes. Part filled by slower pipe in 1 minute = 1/x Part filled by faster pipe in 1 minute = 3/x Part filled by both in 1 minute = (1/x)+(3/x)=1/36 =>4/x=1/36 x=36∗4=144mins

Q.- Taps A and B can fill a bucket in 12 minutes and 15 minutes respectively. If both are opened and A is closed after 3 minutes how much further time would it take for B to fill the bucket?

    1. 1. 8.25 min
    2. 2. 8.52 min
    3. 3. 5.28 min
    4. 4. 5.82 min

1
Part filled in 3 minutes = 3∗[(1/12)+(1/15)]=3∗(9/60)=9/20 Remaining part =1−(9/20)=11/20 =>(1/15):(11/20)=1:X =>X=(11/20)∗(15/1) =>X=8.25mins

Q.- A cistern can be filled in 9 hours but due to a leak at its bottom it takes 10 hours. If the cistern is full then the time that the leak will take to make it empty will be ?

    1. 1. 70 hrs.
    2. 2. 85 hrs.
    3. 3. 90 hrs.
    4. 4. 80 hrs.

3
Part filled without leak in 1 hour = 1/9 Part filled with leak in 1 hour = 1/10 Work done by leak in 1 hour =19−110=190 We used subtraction as it is getting empty. So total time to empty the cistern is 90 hours

Q.- An electric pump can fill a tank in 3 hours. Because of a leak in the tank it took 3 hours 30 min to fill the tank. In what time the leak can drain out all the water of the tank and will make tank empty ?

    1. 1. 12 hrs.
    2. 2. 18 hrs.
    3. 3. 20 hrs.
    4. 4. 21 hrs.

4
We can get the answer by subtrating work done by leak in one hour by subtraction of filling for 1 hour without leak and with leak as Work done for 1 hour without leak = 1/3 Work done with leak =7/2 Work done with leak in 1 hr= 2/7 Work done by leak in 1 hr = (1/3)−(2/7) = 1/21 So tank will be empty by the leak in 21 hours.

Q.- A leak in the bottom of a tank can empty the full tank in 6 hours. An inlet pipe fills water at the rate of 4 litres a minute. When the tank is full the inlet is opened and due to the leak the tank is empty in 8 hours. The capacity of the tank (in litres)

    1. 1. 5670
    2. 2. 5607
    3. 3. 5760
    4. 4. 5706

3
Work done by the inlet in 1 hour =(1/6)−(1/8)=1/24 Work done by inlet in 1 min=(1/24)∗(1/60)=1/1440=>Volume of 1/1440 part = 4 liters Volume of whole = (1440 * 4) litres = 5760 litres.

Q.- Pipe A can fill a tank in 5 hours pipe B in 10 hours and pipe C in 30 hours. If all the pipes are open in how many hours will the tank be filled ?

    1. 1. 4.5 hrs.
    2. 2. 3.5 hrs.
    3. 3. 4 hrs.
    4. 4. 3 hrs.

4
Part filled by A in 1 hour = 1/5 Part filled by B in 1 hour = 1/10 Part filled by C in 1 hour = 1/30 Part filled by (A+B+C) in 1 hour = (1/5)+(1/10)+(1/30)=1/3 So all pipes will fill the tank in 3 hours.

Q.- A cistern can be filled by a tap in 4 hours while it can be emptied by another tap in 9 hours. If both the taps are opened simultaneously then after how much time cistern will get filled ?

    1. 1. 2.8 hrs.
    2. 2. 8.2 hrs.
    3. 3. 7.2 hrs.
    4. 4. 2.7 hrs.

3
When we have question like one is filling the tank and other is empting it then we subtraction as Filled in 1 hour = 1/4 Empties in 1 hour = 1/9 Net filled in 1 hour = 1/4 - 1/9 = 5/36 So cistern will be filled in 36/5 hours i.e. 7.2 hours

Q.- Two pipes A and B can fill a tank in 20 and 30 minutes respectively. If both the pipes are used together then how long it will take to fill the tank ?

    1. 1. 10 min
    2. 2. 12 min
    3. 3. 14 min
    4. 4. 15 min

2
In this type of questions we first get the filling in 1 minute for both pipes then we will add them to get the result as Part filled by A in 1 min = 1/20 Part filled by B in 1 min = 1/30 Part filled by (A+B) in 1 min = 1/20 + 1/30 = 1/12 So both pipes can fill the tank in 12 mins.

Q.- A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hour

    1. 1. 12 hrs.
    2. 2. 51 hrs.
    3. 3. 15 hrs.
    4. 4. 14 hrs.

3
Suppose first pipe alone takes x hours to fill the tank. Then second and third pipes will take (x - 5) and (x - 9) hours respectively to fill the tank. 1/x + 1/(x - 5) = 1/(x - 9) (2x - 5)(x - 9) = x(x - 5) x2 - 18x + 45 = 0 (x- 15)(x - 3) = 0 x = 15

Q.- Two pipes A and B can fill a tank in 15 min and 20 min respectively. Both the pipes are opened together but after 4 min pipe A is turned off. What is the total time required to fill the tank?

    1. 1. 10 min 20 sec
    2. 2. 11 min 45 sec
    3. 3. 12 min 30 sec
    4. 4. 14 min 40 sec

4
Part filled in 4 minutes = 4(1/15 + 1/20) = 7/15 Remaining part = 1 - 7/15 = 8/15 Part filled by B in 1 minute = 1/20 1/20 : 8/15 :: 1 ; x x = 8/15 * 1 * 20 = 10 2/3 min = 10 min 40 sec. The tank will be full in (4 min. + 10 min. 40 sec) = 14 min 40 sec.

Q.- Three pipes A B and C can fill a tank from empty to full in 30 minutes 20 minutes and 10 minutes respectively. When the tank is empty all the three pipes are opened. A B and C discharge chemical solutions P Q and R respectively. What is the proportion of

    1. 1. 42897
    2. 2. 43045
    3. 3. 43044
    4. 4. 42866

2
Part filled by (A + B + C) in 3 minutes = 3(1/30 + 1/20 + 1/10) = 11/20 Part filled by C in 3 minutes = 3/10 Required ratio = 3/10 * 20/11 = 6/11

Q.- A water tank is two-fifth full. Pipe A can fill a tank in 10 minutes and pipe B can empty it in 6 minutes. If both the pipes are open how long will it take to empty or fill the tank completely?

    1. 1. 6 min. to empty
    2. 2. 6 min. to full
    3. 3. 9 min. to empty
    4. 4. 9 min. to full

1
Clearly pipe B is faster than pipe A and so the tank will be emptied. Part to be emptied = 2/5 Part emptied by (A + B) in 1 minute = (1/6 - 1/10) = 1/15 1/15 : 2/5 :: 1 : x x = (2/5 * 1 * 15) = 6 min. So the tank will be emptied in 6 min.

Q.- A tank is filled in eight hours by three pipes A B and C. Pipe A is twice as fast as pipe B and B is twice as fast as C. How much time will pipe B alone take to fill the tank?

    1. 1. 82 hours
    2. 2. 28 hours
    3. 3. 29 hours
    4. 4. 27 hours

2
1/A + 1/B + 1/C = 1/8 (Given) Also given that A = 2B and B = 2C => 1/2B + 1/B + 2/B = 1/8 => (1 + 2 + 4)/2B = 1/8 => 2B/7 = 8 => B = 28 hours.

Q.- A cistern has three pipes A B and C. The pipes A and B can fill it in 4 and 5 hours respectively and C can empty it in 2 hours. If the pipes are opened in order at 1 2 and 3 A.M. When will the cistern be empty?

    1. 1. 0.66666666667
    2. 2. 5.30 PM
    3. 3. 0.75
    4. 4. 0.70833333333

4
1 to 2 = 1/4 2 to 3 = 1/4 + 1/5 = 9/20 After 3 AM = 1/4 + 1/5 - 1/2 = -1/20 1/4 + 9/20 = 14/20 1 h ---- 1/20 ? ----- 14/20 14 hours ==> 5 PM

Q.- A leak in the bottom of a tank can empty the full tank in 6 hours. An inlet pipe fills water at the rate of 4 liters per minute. When the tank is full in inlet is opened and due to the leak the tank is empties in 8 hours. The capacity of the tank is?

    1. 1. 5260 liters
    2. 2. 5760 liters
    3. 3. 5846 liters
    4. 4. 6970 liters

2
1/x - 1/6 = -1/8 x = 24 hrs 24 * 60 * 4 = 5760

Q.- Two pipes A and B can fill a cistern in 12 and 15 minutes respectively. Both are opened together but after 3 minutes A is turned off. After how much more time will the cistern be filled?

    1. 1. 3 1/4 min
    2. 2. 5 1/4 min
    3. 3. 8 1/4 min
    4. 4. 9 1/4 min

3
3/12 + (3 + x)/15 = 1 x = 8 1/4

Q.- Two pipes can fill a tank in 20 and 24 minutes respectively and a waste pipe can empty 3 gallons per minute. All the three pipes working together can fill the tank in 15 minutes. The capacity of the tank is?

    1. 1. 60 gallons
    2. 2. 100 gallons
    3. 3. 120 gallons
    4. 4. 180 gallons

3
Work done by the waste pipe in 1 minute = 1/15 - (1/20 + 1/24) = - 1/40 Volume of 1/40 part = 3 gallons Volume of whole = 3 * 40 = 120 gallons.

Q.- Three pipes A B and C can fill a tank in 6 hours. After working at it together for 2 hours. C is closed A and B can fill the remaining par in 7 hours. The number of hours taken by C alone to fill the tank is?

    1. 1. 15
    2. 2. 12
    3. 3. 14
    4. 4. 41

3
Part filled in 2 hours = 2/6 = 1/3. Remaining part = 1 - 1/3 = 2/3 (A + B)'s 1 hour work = 2/21 C's 1 hour work = [(A + B + C)'s 1 hour work - (A + B)'s 1 hour work] = (1/6 - 2/21) = 1/14 C alone can fill the tank in 14 hours.

Q.- If log a/b + log b/a = log (a + b) then:

    1. 1. a+b=1
    2. 2. a+b=2
    3. 3. a+b=3
    4. 4. a+b=7

1
log a/b + log b/a = log (a + b) log (a + b) = log [ (a/b) x (b/a) ] = log 1. so a + b = 1.

Q.- If log 27 = 1.431 then the value of log 9 is :

    1. 1. 0.549
    2. 2. 0.594
    3. 3. 0.945
    4. 4. 0.954

4
log 27 = 1.431 log (33 ) = 1.431 3 log 3 = 1.431 log 3 = 0.477 log 9 = log[pow(3 2)] = 2 log 3 = (2 x 0.477) = 0.954.

Q.- Which of the following statements is not correct ?

    1. 1. log2-log3
    2. 2. log(2+3)=log(2x3)
    3. 3. log2+log3
    4. 4. log5+log6

2
(a) Since loga a = 1 so log10 10 = 1. (b) log (2 + 3) = log 5 and log (2 x 3) = log 6 = log 2 + log 3 log (2 + 3) is not equal to log (2 x 3) (c) Since loga 1 = 0 so log10 1 = 0. (d) log (1 + 2 + 3) = log 6 = log (1 x 2 x 3) = log 1 + log 2 + log 3. So (b) is incorrect.

Q.- The value of [(1/(log3 60) + (1/(log4 60) + (1/ log5 60)] is :

    1. 1. 5
    2. 2. 4
    3. 3. 2
    4. 4. 1

4
Given expression = log60 3 + log60 4 + log60 5 =log60 (3 x 4 x 5) = log60 60 = 1.

Q.- If log 5 + log (5x + 1) = log(x + 5) + 1 then x is equal to :

    1. 1. 4
    2. 2. 5
    3. 3. 3
    4. 4. 2

3
log10 5 + log10 (5x + 1) = log10 (x + 5) + 1 log10 5 + log10 (5x + 1) = log10 (x + 5) + log10 10 log10 [5 (5x + 1)] = log10 [10(x + 5)] 5(5x + 1) = 10(x + 5) 5x + 1 = 2x + 10 3x = 9 x = 3.

Q.- If log 2 = 0.3010 the value of log10 80 is :

    1. 1. 1.93
    2. 2. 1.903
    3. 3. 1.309
    4. 4. 3.109

2
log10 80 = log10 (8 x 10) = log10 8 + log10 10 = log10 pow[23] + 1 = 3 log10 2 + 1 = (3 x 0.3010) + 1 = 1.9030.

Q.- If log10 2 = 0.3010 then log2 10 is equal to :

    1. 1. 1000/203
    2. 2. 1000/302
    3. 3. 1000/103
    4. 4. 1000/301

4
log2 10 = 1/log10 2 = 1/0.3010 = 10000/3010 = 1000/301

Q.- If log10 7 = a then log10 (1/70) is equal to :

    1. 1. a-2
    2. 2. a+2
    3. 3. (-)(a+1)
    4. 4. a+1

3
log10 (1/70) = log10 1 - log10 70 = - log10 (7 x 10) = - (log10 7 + log10 10) = - (a + 1).

Q.- The value of log2 16 is :

    1. 1. 5
    2. 2. 3
    3. 3. 2
    4. 4. 4

4
Let log2 16 = n. Then pow(2n) = 16 = pow(24) so n = 4. log2 16 = 4.

Q.- If log 2 = 0.30103 the number of digits in pow(264) is :

    1. 1. 20
    2. 2. 19
    3. 3. 18
    4. 4. 15

1
log[pow(264) = 64 x log 2 = (64 x 0.30103) = 19.26592 Its characteristic is 19. Hence then number of digits in pow(264) is 20.

Q.- log(-2)(-2) = ?

    1. 1. -1
    2. 2. 0
    3. 3. 1
    4. 4. none

4
logbx is undefined for x ≤ 0

Q.- If log2[log3(log2 x)] = 1 x is equal to:

    1. 1. 251
    2. 2. 215
    3. 3. 512
    4. 4. 521

3
log2[log3(log2 x)] = 1 log2[log3(log2 x)] = log2(2) log3(log2 x) = 2 log2 x =pow(32) = 9 x = pow(29) = 512

Q.- (log3 4) (log4 5) (log5 6) (log6 7) (log7 8) (log8 9) (log9 9) = ?

    1. 1. 2
    2. 2. 4
    3. 3. 3
    4. 4. 1

1
(log3 4)× (log4 5)× (log5 6)× (log6 7)× (log7 8)× (log8 9)× (log9 9) =(log4/log3)×(log5/log4)×(log6/log5)×(log7/log6)×(log8/log7)×(log9/log8)×1 =log9/log3 =logpow(32)/log3 =2log3/log3 = 2

Q.- if log2x = -6 x is equal to :

    1. 1. 64
    2. 2. 32
    3. 3. 6.4
    4. 4. 23377

4
log2x=−6 ⇒x=pow(2−6)=1/pow(26)=1/64

Q.- If log5 [pow(x2)+x] - log5 (x+1) = 3 find the value of x

    1. 1. 125
    2. 2. 120
    3. 3. 135
    4. 4. 145

1
log5[pow(x2)+x]−log5(x+1)=3 ⇒log5[pow(x2)+x]/(x+1)=3 ⇒log5[x(x+1)]/(x+1)=3 ⇒log5x=3 ⇒x=pow(53)=125

Q.- If log10 3 = 0.4771 then log3 10 is

    1. 1. 10000/4717
    2. 2. 10000/4771
    3. 3. 4717/10000
    4. 4. 4771/10000

2
log310=1/log103=1/0.4771=10000/4771

Q.- If log102 = a what is the value of log10(1/200)

    1. 1. (-)(a+1)
    2. 2. (a+1)
    3. 3. (-)(a+2)
    4. 4. (a+2)

3
log10(1/200)=log101−log10200=0−log10(2×100)=−[log102+log10100]=−(a+2)

Q.- If log10 5 + log10 (5x + 1) = log10 (x + 5) + 1 then x is equal to:

    1. 1. 3
    2. 2. 5
    3. 3. 2
    4. 4. 4

1
log10 5 + log10 (5x + 1) = log10 (x + 5) + 1 => log10 5 + log10 (5x + 1) = log10 (x + 5) + log10 10 => log10 [5(5x+1)] = log10 [10(x+5)] => 5(5x+1) = 10(x+5) => 5x+1 = 2(x+5) => 5x + 1 = 2x + 10 => 3x = 9 => x = 3

Q.- if log102 = 0.3010 what is the value of log101600 ?

    1. 1. 2.43
    2. 2. 2.403
    3. 3. 3.24
    4. 4. 3.204

4
log101600 = log10(16 × 100) = log10(16) + log10(100) = log10pow(24) + log10pow(102) = 4 log10(2) + 2 = (4 × 0.3010) + 2 = 1.204 + 2 = 3.204

Q.- log2 512 = ?

    1. 1. 27
    2. 2. 8
    3. 3. 9
    4. 4. 16

3
log2 512 = log2pow(29) = 9

Q.- Given: log8(5) = b. Express log4(10) in terms of b.

    1. 1. (1 + 3b) / 2
    2. 2. (1 - 3b) / 2
    3. 3. (1 + 2b) / 3
    4. 4. (1 - 2b) / 3

1
Use log rule of product: log4(10) = log4(2) + log4(5) log4(2) = log4(41/2) = 1/2 Use change of base formula to write: log4(5) = log8(5) / log8(4) = b / (2/3) since log8(4) = 2/3 log4(10) = log4(2) + log4(5) = (1 + 3b) / 2

Q.- log6(216) + [ log(42) - log(6) ] / log(49) =?

    1. 1. 7
    2. 2. 42376
    3. 3. 42553
    4. 4. 42407

3
log6(216) + [ log(42) - log(6) ] / log(49) = log6(6^3) + log(42/6) / log(72) = 3 + log(7) /2 log(7) = 3 + 1/2 = 7/2

Q.- Simplify: [(3^-1) - (9^-1) / 6]^1/3

    1. 1. 3
    2. 2. 42372
    3. 3. 42431
    4. 4. 42403

2
[(3^-1 - 9^-1) / 6]^1/3 = [(1/3 - 1/9) / 6]^1/3 = [(6 / 27) / 6]^1/3 = 1/3

Q.- Single logarithm to express (logxa)(logab) is:

    1. 1. logx b
    2. 2. logb a
    3. 3. logb x
    4. 4. None

1
Use change of base formula: (logxa)(logab) = logxa (logxb / logxa) = logx b

Q.- Find a so that the graph of y = loga x passes through the point (e 2).

    1. 1. e^1/2
    2. 2. e
    3. 3. e^2
    4. 4. None

1
2 = loga e a^2 = e ln(a^2) = ln e 2 ln a = 1 a = e^1/2

Q.- Find constant A such that log3 x = A log5x for all x > 0.

    1. 1. ln(3) / ln(5)
    2. 2. ln(5) / ln(3)
    3. 3. ln(5)
    4. 4. ln(3)

2
Use change of base formula using ln to rewrite the given equation as follows ln (x) / ln(3) = A ln(x) / ln(5) A = ln(5) / ln(3)

Q.- Solve for x the equation log [ log (2 + log2(x + 1)) ] = 0

    1. 1. 2^8
    2. 2. 2^8-1
    3. 3. 2^7
    4. 4. 2^7-1

2
Rewrite given equation as: log [ log (2 + log2 (x + 1)) ] = log (1) since log(1) = 0. log (2 + log2 (x + 1)) = 1 2 + log2 (x + 1) = 10 log2(x + 1) = 8 x + 1 = 2^8 x = 2^8 - 1

Q.- Find the x intercept of the graph of y = 2 log( sqrt(x - 1) - 2)

    1. 1. 8
    2. 2. 9
    3. 3. 10
    4. 4. 11

3
Solve: 0 = 2 log( sqrt(x - 1) - 2) Divide both sides by 2: log( sqrt(x - 1) - 2) = 0 Use the fact that log(1)= 0: sqrt(x - 1) - 2 = 1 Rewrite as: sqrt(x - 1) = 3 Raise both sides to the power 2: (x - 1) = 32 x - 1 = 9 x = 10

Q.- Solve for x the equation 4^(x - 2) = 3^(x + 4)

    1. 1. 6 ln(4) / ln(3)
    2. 2. 4 ln(6) / ln(4/3)
    3. 3. 3 ln(6) / ln(3/4)
    4. 4. 4 ln(6) / ln(3/4)

2
Given: 4x - 2 = 3x + 4 Take ln of both sides: ln ( 4x - 2 ) = ln ( 3x + 4 ) Simplify: (x - 2) ln 4 = (x + 4) ln 3 Expand: x ln 4 - 2 ln 4 = x ln 3 + 4 ln 3 Group like terms: x ln 4 - x ln 3 = 4 ln 3 + 2 ln 4 Solve for x: x = ( 4 ln 3 + 2 ln 4 ) / (ln 4 - ln 3) = ln (34 * 42) / ln (4/3) = ln (34 * 24) / ln (4/3) = 4 ln(6) / ln(4/3)

Q.- (log 8) + (log 3) - (log 6) = ?

    1. 1. log 1
    2. 2. log 4
    3. 3. log 5
    4. 4. log 17

2
(log 8)+(log 3)-(log 6) =(log 8*3)-(log 6) =log 24- log6 =log(24/6) =log 4

Q.- A bag contains 6 black and 8 white balls. One ball is drawn at random. What is the probability that the ball drawn is white ?

    1. 1. 42920
    2. 2. 42919
    3. 3. 42921
    4. 4. 42922

1
Let number of balls = (6 + 8) = 14. Number of white balls = 8. P (drawing a white ball) = 8/14 = 4/7

Q.- One card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is a face card (Jack Queen and King only) ?

    1. 1. 41395
    2. 2. 41334
    3. 3. 41365
    4. 4. 41426

2
Clearly there are 52 cards out of which there are 12 face cards. P (getting a face card) = 12/52 = 3/13

Q.- A card is drawn from a pack of 52 cards. The probability of getting a queen of club or a king of heart is :

    1. 1. 46023
    2. 2. 46054
    3. 3. 46082
    4. 4. 46143

1
Here n(S) = 52. Let E = event of getting a queen of club or a king of heart. Then n(E) = 2. P(E) = n�/n(s) = 2/52 = 1/26

Q.- Two dice are tossed. The probability that the total score is a prime number is :

    1. 1. 43072
    2. 2. 43074
    3. 3. 42867
    4. 4. 42806

2
Clearly n(S) = (6 x 6) = 36. Let E = Event that the sum is a prime number. Then E = { (1 1) (1 2) (1 4) (1 6) (2 1) (2 3) (2 5) (3 2) (3 4) (4 1) (4 3)(5 2) (5 6) (6 1) (6 5) } n(E) = 15. P(E) = n(E)/n(s) = 15/36 = 5/12

Q.- In a lottery there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize ?

    1. 1. 42773
    2. 2. 42918
    3. 3. 42919
    4. 4. 42920

2
P (getting a prize) = 10(10 + 25) = 10/35 = 2/7

Q.- Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even ?

    1. 1. 42858
    2. 2. 42799
    3. 3. 42798
    4. 4. 42828

4
In a simultaneous throw of two dice we have n(S) = (6 x 6) = 36. Then E = {(1 2) (1 4) (1 6) (2 1) (2 2) (2 3) (2 4) (2 5) (2 6) (3 2) (3 4)(3 6) (4 1) (4 2) (4 3) (4 4) (4 5) (4 6) (5 2) (5 4) (5 6) (6 1)(6 2) (6 3) (6 4) (6 5) (6 6)} n(E) = 27. P(E) = n(E)/n(s) = 27/36 = 3/4

Q.- Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5 ?

    1. 1. 42998
    2. 2. 44075
    3. 3. 44440
    4. 4. 44805

2
Here S = {1 2 3 4 .... 19 20}. Let E = event of getting a multiple of 3 or 5 = {3 6 9 12 15 18 5 10 20}. P(E) = n�/n(s) = 9/20 .

Q.- In a box there are 8 red 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green ?

    1. 1. 42795
    2. 2. 42738
    3. 3. 42826
    4. 4. 42796

1
Total number of balls = (8 + 7 + 6) = 21. Let E = event that the ball drawn is neither red nor green = event that the ball drawn is blue. n(E) = 7. P(E) = n(E)/n(s) = 7/21 = 1/3

Q.- What is the probability of getting a sum 9 from two throws of a dice ?

    1. 1. 42775
    2. 2. 42980
    3. 3. 42744
    4. 4. 42979

4
In two throws of a die n(S) = (6 x 6) = 36. Let E = event of getting a sum ={(3 6) (4 5) (5 4) (6 3)}. P(E) =n(E) /n(S) = 4/36 = 1/9 .

Q.- Three unbiased coins are tossed. What is the probability of getting at most two heads ?

    1. 1. 42893
    2. 2. 42922
    3. 3. 42954
    4. 4. 42924

3
Here S = {TTT TTH THT HTT THH HTH HHT HHH} Let E = event of getting at most two heads. Then E = {TTT TTH THT HTT THH HTH HHT}. P(E) = n(E)/n(s) = 7/8

Q.- What is the probability that a number selected from the numbers 1 2 3 4 5 ...16 is a prime number is :

    1. 1. 42370
    2. 2. 42952
    3. 3. 42950
    4. 4. 42552

3
S = [123...1516] and E = [23571113]. so P(E) = n(E)/n(S) = 6/16 = 3/8

Q.- One card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is a face card ?

    1. 1. 41365
    2. 2. 42826
    3. 3. 42744
    4. 4. 41275

1
Clearly n(S) = 52 and there are 16 face cards. so P(E) = 16/52 = 4/13

Q.- One card is drawn of random from a pack of 52 cards. What is the probability that the card drawn is a king ?

    1. 1. 41275
    2. 2. 18994
    3. 3. 41334
    4. 4. 42826

1
Cearly n(E) = 4/52 = 1/13

Q.- What is the probability that an ordinary year has 53 sundays ?

    1. 1. 53/365
    2. 2. 42917
    3. 3. 42918
    4. 4. 48/53

2
An ordinary year has 365 days i. e. 52 weaks and 1 day. so the probability that this day is a sunday is 1/7.

Q.- Tickets numbered from 1 to 20 are mixed up and a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 7 ?

    1. 1. 42005
    2. 2. 42767
    3. 3. 42857
    4. 4. 44013

3
Clearly N(S) = 20 and E = [369121518714] so P(E) = n(E)/n(S) = 8/20 = 2/5

Q.- A bag contains 6 black balls and 8 white balls. One ball is drawn at random.What is the probability that the ball drawn is white ?

    1. 1. 42920
    2. 2. 42828
    3. 3. 42798
    4. 4. 42948

1
Total number of balls = (6+8) =14. Number of white balls = 8 so P(drawing a white ball) = 8/14 = 4/7

Q.- The odds in favour of an event are 3:5.The probability of occurrence of the event is :

    1. 1. 42858
    2. 2. 42950
    3. 3. 42795
    4. 4. 42856

2
Number of cases favourable to E = 3. Total number of cases= (3+5) = 8 so P(E) = 3/8

Q.- The odds against the occurrence of an event are 5:4.The probability of its occurrence is :

    1. 1. 42859
    2. 2. 42982
    3. 3. 42856
    4. 4. 42826

2
Number of cases favourable to E = 4. Total number of cases = 5+4 = 9 so P(E) = 4/9

Q.- What is the probability of getting a king or queen in a single draw from a pack of 52 cards ?

    1. 1. 46023
    2. 2. 41275
    3. 3. 41306
    4. 4. none

3
Clearly n(S) = 52. There are 4 kings and 4 queens. so P(E) = n(E)/n(S) P(E)= 8/52 = 2/13

Q.- In a lottery there are 10 prizes and 25 blanks.What is the probability of getting prize ?

    1. 1. 43009
    2. 2. 42857
    3. 3. 42918
    4. 4. 42921

3
P(getting a prize) = 10/(10+25) = 10/35 = 2/7

Q.- 10 books are placed at random in a shelf. The probability that a pair of books will always be together is -.

    1. 1. 43009
    2. 2. 43017
    3. 3. 42856
    4. 4. 43011

3
10 books can be rearranged in 10! ways consider the two books taken as a pair then number of favourable ways of getting these two books together is 9! 2! Required probability = 1/5

Q.- What is the probability that a leap year has 53 Sundays and 52 Mondays?

    1. 1. 0
    2. 2. 42917
    3. 3. 42918
    4. 4. 42921

2
A leap year has 52 weeks and two days Total number of cases = 7 Number of favourable cases = 1 i.e. {Saturday Sunday} Required Probability = 1/7

Q.- A bag contains five white and four red balls. Two balls are picked at random from the bag. What is the probability that they both are different color?

    1. 1. 42982
    2. 2. 42983
    3. 3. 42985
    4. 4. 42986

2
Two balls can be picked from nine balls in ?C2 ways. We select one white ball and one red ball from five white balls and four red balls. This can be done 5C1 . 4C1 ways. The required probability = (5 * 4)/?C2 = 20/36 = 5/9

Q.- A box contains 3 blue marbles 4 red 6 green marbles and 2 yellow marbles. If three marbles are drawn what is the probability that one is yellow and two are red?

    1. 1. 24/455
    2. 2. 33298
    3. 3. 12/455
    4. 4. 15/91

3
Given that there are three blue marbles four red marbles six green marbles and two yellow marbles. When three marbles are drawn the probability that one is yellow and two are red = (1C1)(4C2)/15C3 = (2 * 4 * 3 * 3 * 2)/(1 * 2 * 15 * 14 * 13) = 12/455

Q.- A box contains 3 blue marbles 4 red 6 green marbles and 2 yellow marbles. If two marbles are picked at random what is the probability that they are either blue or yellow?

    1. 1. 44621
    2. 2. 44287
    3. 3. 44228
    4. 4. 41640

3
Given that there are three blue marbles four red marbles six green marbles and two yellow marbles. Probability that both marbles are blue = 3C2/15C2 = (3 * 2)/(15 * 14) = 1/35 Probability that both are yellow = 2C2/15C2 = (2 * 1)/(15 * 14) = 1/105 Probability that one blue and other is yellow = (3C1 * 2C1)/15C2 = (2 * 3 * 2)/(15 * 14) = 2/35 Required probability = 1/35 + 1/105 + 2/35 = 3/35 + 1/105 = 1/35(3 + 1/3) = 10/(3 * 35) = 2/21

Q.- Three 6 faced dice are thrown together. The probability that no two dice show the same number on them is -.

    1. 1. 43076
    2. 2. 42983
    3. 3. 13150
    4. 4. 43074

2
No two dice show same number would mean all the three faces should show different numbers. The first can fall in any one of the six ways. The second die can show a different number in five ways. The third should show a number that is different from the first and second. This can happen in four ways. Thus 6 * 5 * 4 = 120 favourable cases. The total cases are 6 * 6 * 6 = 216. The probability = 120/216 = 5/9.

Q.- A bag contains 7 green and 8 white balls. If two balls are drawn simultaneously the probability that both are of the same colour is -.

    1. 1. 42217
    2. 2. 42857
    3. 3. 42309
    4. 4. 42186

4
Drawing two balls of same color from seven green balls can be done in 7C2 ways. Similarly from eight white balls two can be drawn in 8C2 ways. P = 7C2/15C2 + 8C2/15C2 = 7/15

Q.- If two dice are thrown together the probability of getting an even number on one die and an odd number on the other is -.

    1. 1. 42826
    2. 2. 42767
    3. 3. 42828
    4. 4. 42858

2
The number of exhaustive outcomes is 36. Let E be the event of getting an even number on one die and an odd number on the other. Let the event of getting either both even or both odd then PE = 18/36 = 1/2 P(E) = 1 - 1/2 = 1/2.

Q.- From a pack of cards two cards are drawn one after the other with replacement. The probability that the first is a red card and the second is a king is -.

    1. 1. 46023
    2. 2. 19054
    3. 3. 15/26
    4. 4. 46327

1
Let E1 be the event of drawing a red card. Let E2 be the event of drawing a king . P(E1 n E2) = P(E1) . P(E2) (As E1 and E2 are independent) = 1/2 * 1/13 = 1/26

Q.- In a party there are 5 couples. Out of them 5 people are chosen at random. Find the probability that there are at the least two couples?

    1. 1. 44317
    2. 2. 41760
    3. 3. 41883
    4. 4. 16/21

1
Number of ways of (selecting at least two couples among five people selected) = (5C2 * 6C1) As remaining person can be any one among three couples left. Required probability = (5C2 * 6C1)/10C5 = (10 * 6)/252 = 5/21

Q.- From a point P on a level ground the angle of elevation of the top tower is 30 deg. If the tower is 100 m high the distance of point P from the foot of the tower is :

    1. 1. 173 m
    2. 2. 137 m
    3. 3. 713 m
    4. 4. 731 m

1
Let AB be the tower. Then APB = 30� and AB = 100 m. AB/AP = tan 30� = 1/sqrt.3 AP = (AB xsqrt. 3) m = 100xsqrt.3 m = (100 x 1.73) m = 173 m.

Q.- The angle of elevation of a ladder leaning against a wall is 60 deg and the foot of the ladder is 4.6 m away from the wall. The length of the ladder is :

    1. 1. 9.7 m
    2. 2. 7.9 m
    3. 3. 9.2 m
    4. 4. 2.9 m

3
Let AB be the wall and BC be the ladder. Then ACB = 60� and AC = 4.6 m. AC/BC = cos 60� = 1/2 BC = 2 x AC = (2 x 4.6) m = 9.2 m.

Q.- A man standing at a point P is watching the top of a tower which makes an angle of elevation of 30 deg with the man's eye. The man walks some distance towards the tower to watch its top and the angle of the elevation becomes 60 deg. What is the distance

    1. 1. 12
    2. 2. none
    3. 3. 10
    4. 4. 6

2
One of AB AD and CD must have given. So the data is inadequate.

Q.- Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30 deg and 45 deg respectively. If the lighthouse is 100 m high the distance between the two ships is :

    1. 1. 372 m
    2. 2. 327 m
    3. 3. 237 m
    4. 4. 273 m

4
Let AB be the lighthouse and C and D be the positions of the ships. Then AB = 100 m ACB = 30� and ADB = 45�. AB/AC = tan 30� = 1/sqrt 3 AC = AB xsqrt 3 = 100xsqrt 3 m. AB/AD= tan 45� = 1 AD = AB = 100 m. CD = (AC + AD) = (100sqrt 3 + 100) m = 100(sqrt 3 + 1) = (100 x 2.73) m = 273 m.

Q.- The angle of elevation of the sun when the length of the shadow of a tree 3 times the height of the tree is :

    1. 1. Cot20
    2. 2. Cot50
    3. 3. Cot40
    4. 4. Cot30

4
Let AB be the tree and AC be its shadow. Then AC/AB =sqrt(3) * cot30 / sqrt(3) = cot30

Q.- An observer 1.6 m tall is 20(sqrt 3) away from a tower. The angle of elevation from his eye to the top of the tower is 30 deg. The heights of the tower is :

    1. 1. 12.6 m
    2. 2. 21.6 m
    3. 3. 61.2 m
    4. 4. 16.2 m

2
Let AB be the observer and CD be the tower. Then CE = AB = 1.6 m BE = AC = 20xsqrt 3 m. DE/BE = tan 30� = 1/sqrt 3 DE = (20xsqrt 3)/(sqrt 3) m = 20 m. CD = CE + DE = (1.6 + 20) m = 21.6 m.

Q.- The shadow of a building is 20 m long when the angle of elevation of the sun is 60 degree. Find the height of the building.

    1. 1. 64.46
    2. 2. 64.34
    3. 3. 34.46
    4. 4. 34.64

4
Let AB be the building and AC be its shadow. Then AC=20 m and angle ACB=60 degree. Let AB=h then AB/AC= tan 60 degree=sqrt. 3= h/20=sqrt. 3. hence h=(20 xsqrt.3) m=(20x1.732) m=34.64m

Q.- If a vertical pole 6 m high has a shadow of length 2xsqrt.3 m find the angle of elevation of the sun.

    1. 1. 60 degree
    2. 2. 40 degree
    3. 3. 70 degree
    4. 4. 50 degree

1
Let AB be the vertical pole and AC be its shadow. Let the angle of elevation be x. then AB=6 m AC=2sqrt.3 m and angle ACB=x Now tan x=AB/Ac=6/2sqrt.3=sqrt.3=tan 60 degree. So x = 60 degree

Q.- A ladder leaning against a vertical wall makes an angle of 45 degree with the ground. The foot of the ladder is 3 m from the wall. Find the length of the ladder.

    1. 1. 4.32 m
    2. 2. 4.23 m
    3. 3. 32.6
    4. 4. 32.4

2
Let AB be the wall and CB be the ladder. Then AC = 3 m and Angle ACB = 45 degree Now CB/AC = sec 45 degree = sqrt.2 =CB/3 = sqrt. 2 Hence Length of the ladder = CB = 3 sqrt. 2 = (3x1.41) m = 4.23 m

Q.- A balloon is connected to a station by a cable of length 200 m inclined at 60 degree to the horizontal. Find the height of the balloon from the ground. Assume that there is no slack in the cable

    1. 1. 173.2 m
    2. 2. 172.3 m
    3. 3. 327.1 m
    4. 4. 371.2 m

1
Let be the balloon and AB be the vertical height. Let C be the cable. Then BC=200 m and angle ACB=60 degree. Then AB/BC=sin60=sqrt.3/2=AB/200=sqrt.3/2=AB=[(200xsqrt.3)/2] m=173.2 m

Q.- The shadow of a building is 20 m long when the angle of the elevation of the sun is 60.Find the height of the building.

    1. 1. 34.64 m
    2. 2. 34.46 m
    3. 3. 46.34 m
    4. 4. 46.64 m

1
Let AB be the building and AC be its shadow. Then AC = 20 m and angle ACB = 60. Let AB = h.Then AB/AC = tan60 = sqrt.3 so h/20 = sqrt.3. So h = (20*sqrt.3)m = (20*1.732)m = 34.64 m

Q.- If a vertical pole 6 m high has a shadow of length 2(sqrt.3) m.Find the angle of elevation of the sun.

    1. 1. 60
    2. 2. 70
    3. 3. 45
    4. 4. 65

1
Let AB be the vertical pole and AC be its shadow. Let the angle of elevation be x. Then AB = 6 m AC = 2(sqrt.3)m and angle ACB = x. Now tanx = AB/AC = 6/(2sqrt.3) = sqrt.3 = tan60. so x = 60

Q.- A ladder leaning against a vertical wall makes an angle of 45 with the ground. The foot of the ladder is 3 m from the wall.Find the length of the ladder

    1. 1. 4.23 m
    2. 2. 4.32 m
    3. 3. 3.42 m
    4. 4. 3.24 m

1
Let AB be the wall and CB be the ladder. Then AC = 3 m and angle ACB = 45. Now CB/AC = sec45 = sqrt.2 = CB/3 = sqrt.2. So length of the ladder = CB = 3(sqrt.2) = (3*1.41)m = 4.23 m

Q.- The ratio of the length of a rod and its shadow is 1:sqrt.3.The angle of elevation of the sun is :

    1. 1. 45
    2. 2. 60
    3. 3. 90
    4. 4. 30

4
Let AB be the rod and AC be its shadow. Let angle ACB = p. Let AB = x. Then AC = sqrt.3x. tanp = AB/AC = x/(sqrt.3*x) = 1/sqrt.3 So angle p = 30

Q.- The angle of elevation of a moon when the length of the shadow of a pole is equal to its heights is :

    1. 1. 30
    2. 2. 45
    3. 3. 90
    4. 4. 70

2
Let AB = x then AC = x So tany =AB/AC = x/x = 1 so angle(y) = 45 degree

Q.- If the length of shadow of a pole on a level ground is twice the length of that pole the angle of elevation of the sun is :

    1. 1. 30
    2. 2. 40
    3. 3. 90
    4. 4. none

4
Let AB = x then AC = 2x. So tany = AB/AC = x/2x = 1/2 = y = Inv.tan(1/2). So Ans. is none

Q.- The angle of elevation of a tower from a distance 100 m. from its foot is 30 degree.The height of the tower is :

    1. 1. 100*(sqrt.3)
    2. 2. 50*(sqrt.3)
    3. 3. 100/(sqrt.3)
    4. 4. 50/(sqrt.3)

3
Let AB be the tower AC = 100 m.and y = 30 degree. Then AB/Ac = tan30 so AB = 100*(1/sqrt.3)m

Q.- The attitude of the sun at any instant is 60 degree.The height of the vertical pole that will cast a shadow of 30 m. is :

    1. 1. 40
    2. 2. 30
    3. 3. 30*(sqrt.3)
    4. 4. 40*(sqrt.3)

3
Let AB be the pole and AC be its shadow. Then angle(y) = 60 and AC = 30 m. So AB/AC = tan60 = AB/30 = sqrt.3 so Ab = 30*(sqrt.3) m.

Q.- When the sun is 30 degree above the horizontal the length of shadow cast by a building 50 m. high is :

    1. 1. sqrt.2
    2. 2. 2
    3. 3. 3
    4. 4. sqrt.3

4
Let AB be the building and AC be its shadow. Then AB = 50 m. and angle(y) = 30 degree. So AC/AB = cot30 = sqrt.3 so (AC/50)*(sqrt.3) so AC = 50 = sqrt.3 m.

Q.- If the elevation of the sun changed from 30 degree to 60 degree then the difference between the lengths of shadows of a pole 15 m. high made at these two positions is :

    1. 1. 10
    2. 2. 10*(sqrt.3)
    3. 3. 12
    4. 4. 11

2
When AB = 15 m Angle(y) = 30 Then AC/AB = tan30 so AC = 15/(sqrt.3)m. When AB = 15 m y = 60 then AC/AB = tan60 so AC = 15*(sqrt.3)m. So Diff. in lengths of shadows = 15*(sqrt.3)-15/(sqrt.3) = 30/(sqrt.3) = 10*(sqrt.3) m.

Q.- A ladder 10 m long just reaches the top of a wall and makes an angle of 60 deg with the wall.Find the distance of the foot of the ladder from the wall (sqrt.3=1.73)

    1. 1. 4.32 m
    2. 2. 17.3 m
    3. 3. 5 m
    4. 4. 8.65 m

4
Let BA be the ladder and AC be the wall as shown above. Then the distance of the foot of the ladder from the wall = BC Given that BA = 10 m angleBAC = 60° sin 60°= BC/BA (sqrt.3)/2 = BC/10 BC = 10×(sqrt.3)/2 = 5×1.73=8.65 m

Q.- From a tower of 80 m high the angle of depression of a bus is 30 deg. How far is the bus from the tower?

    1. 1. 40 m
    2. 2. 138.4 m
    3. 3. 46.24 m
    4. 4. 160 m

2
Let AC be the tower and B be the position of the bus. Then BC = the distance of the bus from the foot of the tower. Given that height of the tower AC = 80 m and the angle of depression angleDAB = 30° angleABC = angleDAB = 30° ( Because DA || BC) tan 30°=AC/BC=>tan 30°=80/BC BC = 80/tan30° =80/(1/sqrt.3)=80×1.73=138.4 m i.e. Distance of the bus from the foot of the tower = 138.4 m

Q.- The angle of elevation of the top of a lighthouse 60 m high from two points on the ground on its opposite sides are 45 deg and 60 deg. What is the distance between these two points?

    1. 1. 45 m
    2. 2. 30 m
    3. 3. 103.8 m
    4. 4. 94.6 m

4
Let BD be the lighthouse and A and C be the two points on ground. Then BD the height of the lighthouse = 60 m angle BAD = 45° angle BCD = 60° tan 45° = BD/BA?1=60/BA?BA=60 m------ (1) tan 60° = BD/BC sqrt.3=60/BC BC=60/sqrt.3=(60×sqrt.3)/(sqrt.3×sqrt.3) =60sqrt.3/3=20sqrt.3=20×1.73=34.6 m------ (2) Distance between the two points A and C = AC = BA + BC = 60 + 34.6 [? Substituted the value of BA and BC from (1) and (2)] = 94.6 m

Q.- Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30 deg and 45 deg respectively. If the lighthouse is 100 m high the distance between the two ships is:

    1. 1. 300 m
    2. 2. 173 m
    3. 3. 273 m
    4. 4. 200 m

3
Let BD be the lighthouse and A and C be the positions of the ships. Then BD = 100 m angle BAD = 30° angle BCD = 45° tan 30° = BD/BA?1/sqrt.3=100/BA?BA=100sqrt.3 tan 45° = BD/BC?1=100/BC?BC=100 Distance between the two ships = AC = BA + BC =100sqrt.3+100=100(sqrt.3+1)=100(1.73+1)=100×2.73=273 m

Q.- A man standing at a point P is watching the top of a tower which makes an angle of elevation of 30 deg with the man's eye. The man walks some distance towards the tower to watch its top and the angle of the elevation becomes 45 deg. What is the distance b

    1. 1. 9 units
    2. 2. 3sqrt.3 units
    3. 3. Data inadequate
    4. 4. 12 units

3
tan 45° = SR/QR tan 30° = SR/PR = SR/(PQ + QR) two equations and 3 variables. Hence we can not find the required value with the given data. (Note that if one of SR PQ QR is known this becomes two equations and two variables and if that was the case we could have found out the required value)

Q.- From a point P on a level ground the angle of elevation of the top tower is 30 deg. If the tower is 200 m high the distance of point P from the foot of the tower is:

    1. 1. 346 m
    2. 2. 400 m
    3. 3. 312 m
    4. 4. 298 m

1
tan 30° = RQ/PQ 1/sqrt.3 = 200/PQ PQ = 200sqrt.3 = 200×1.73 = 346 m

Q.- The angle of elevation of the sun when the length of the shadow of a tree is equal to the height of the tree is:

    1. 1. None of these
    2. 2. 60 deg
    3. 3. 45 deg
    4. 4. 30 deg

3
Consider the diagram shown above where QR represents the tree and PQ represents its shadow We have QR = PQ Let angle QPR = ? tan ? = QR/PQ =1 (since QR = PQ) => ? = 45° i.e. required angle of elevation = 45°

Q.- An observer 2 m tall is 10sqrt.3 m away from a tower. The angle of elevation from his eye to the top of the tower is 30 deg. The height of the tower is:

    1. 1. None of these
    2. 2. 12 m
    3. 3. 14 m
    4. 4. 10 m

2
SR = PQ = 2 m PS = QR = 10sqrt.3 m tan 30°=TS/PS 1/sqrt.3 = TS/10sqrt.3 TS = 10sqrt.3/sqrt.3 = 10 m TR = TS + SR = 10 + 2 = 12 m

Q.- The angle of elevation of a ladder leaning against a wall is 60 deg and the foot of the ladder is 12.4 m away from the wall. The length of the ladder is:

    1. 1. 14.8 m
    2. 2. 6.2 m
    3. 3. 12.4 m
    4. 4. 24.8 m

4
Consider the diagram shown above where PR represents the ladder and RQ represents the wall. cos 60° = PQ/PR 1/2 = 12.4/PR PR = 2×12.4 = 24.8 m

Q.- The top of a 15 metre high tower makes an angle of elevation of 60 deg with the bottom of an electronic pole and angle of elevation of 30 deg with the top of the pole. What is the height of the electric pole?

    1. 1. 5 metres
    2. 2. 8 metres
    3. 3. 10 metres
    4. 4. 12 metres

3
Consider the diagram shown above. AC represents the tower and DE represents the pole Given that AC = 15 m angleADB = 30° angleAEC = 60° Let DE = h Then BC = DE = h AB = (15-h) (? AC=15 and BC = h) BD = CE tan 60° = AC/CE => sqrt.3 = 15/CE => CE = 15/sqrt.3--- (1) tan 30° = AB/BD => 1/sqrt.3 = 15-h/BD => 1/sqrt.3 = 15-h/(15/sqrt.3) (? BD = CE and Substituted the value of CE from equation 1 ) =>(15-h) = 1/sqrt.3×15/sqrt.3 = 15/3 = 5 => h = 15-5 = 10 m i.e. height of the electric pole = 10 m

Q.- The population of a town increased from 175000 to 262500 in a decade. The average percent increase of population per year is :

    1. 1. 0.05
    2. 2. 0.06
    3. 3. 0.07
    4. 4. 0.005

1
Increase in 10 years = (262500 - 175000) = 87500 Increase% = (87500/175000)x 100 % = 50%. Required average = 50/10 % = 5%

Q.- Rajeev buys good worth Rs. 6650. He gets a rebate of 6% on it. After getting the rebate he pays sales tax @ 10%. Find the amount he will have to pay for the goods.

    1. 1. Rs. 6867.1
    2. 2. Rs. 6876.1
    3. 3. Rs. 7668.1
    4. 4. Rs. 7686.1

2
Rebate = 6% of Rs. 6650 = Rs. (6/100) x 6650 = Rs. 399 Sales tax = 10% of Rs. (6650 - 399) = Rs. (10/100) x 6251 = Rs. 625.10 Final amount = Rs. (6251 + 625.10) = Rs. 6876.10

Q.- Gauri went to the stationers and bought things worth Rs. 25 out of which 30 paise went on sales tax on taxable purchases. If the tax rate was 6% then what was the cost of the tax free items ?

    1. 1. Rs. 91.07
    2. 2. Rs. 91.7
    3. 3. Rs. 19.70
    4. 4. Rs. 19.07

3
Let the amount taxable purchases be Rs. x Then 6% of x = 30/100 x = (30/100) x (100/6) = 5 Cost of tax free items = Rs. [25 - (5 + 0.30)] = Rs. 19.70

Q.- Two tailors X and Y are paid a total of Rs. 550 per week by their employer. If X is paid 120 percent of the sum paid to Y how much is Y paid per week ?

    1. 1. Rs. 502
    2. 2. Rs. 205
    3. 3. Rs.520
    4. 4. Rs. 250

4
Let the sum paid to Y per week be Rs. z. Then z + 120% of z = 550. z + (120/100) z = 550 11 z /5= 550 z = 550 x 5/11 = 250

Q.- Three candidates contested an election and received 1136 7636 and 11628 votes respectively. What percentage of the total votes did the winning candidate get ?

    1. 1. 0.45
    2. 2. 0.55
    3. 3. 0.57
    4. 4. 0.75

3
Total number of votes polled = (1136 + 7636 + 11628) = 20400 Required percentage = (11628/20400) x 100 % = 57%

Q.- In an election between two candidates one got 55% of the total valid votes 20% of the votes were invalid. If the total number of votes was 7500 the number of valid votes that the other candidate got was :

    1. 1. 7200
    2. 2. 2700
    3. 3. 2007
    4. 4. 7002

2
Number of valid votes = 80% of 7500 = 6000. Valid votes polled by other candidate = 45% of 6000 = (45/100) x 6000 = 2700.

Q.- A student multiplied a number by 3/5 instead of 5/3 What is the percentage error in the calculation?

    1. 1. 0.55
    2. 2. 0.5
    3. 3. 0.64
    4. 4. 0.46

3
Let the number be x Then error = (5/3 x) - (3/5 x) = (16/15) x Error% = (16x/15) x (3/5x) x 100 % = 64%

Q.- In a certain school 20% of students are below 8 years of age. The number of students above 8 years of age is 2/3 of the number of students of 8 years of age which is 48. What is the total number of students in the school ?

    1. 1. 80
    2. 2. 90
    3. 3. 120
    4. 4. 100

4
Let the number of students be x. Then Number of students above 8 years of age = (100 - 20)% of x = 80% of x 80% of x = 48 + 2/3 of 48 (80/100) x = 80 x = 100

Q.- What percentage of numbers from 1 to 70 have 1 or 9 in the unit's digit ?

    1. 1. 0.2
    2. 2. 0.3
    3. 3. 0.4
    4. 4. 0.25

1
Clearly the numbers which have 1 or 9 in the unit's digit have squares that end in the digit 1. Such numbers from 1 to 70 are 1 9 11 19 21 29 31 39 41 49 51 59 61 69. Number of such number =14 Required percentage = (14/70) x 100 % = 20%.

Q.- Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56% of the sum of their marks. The marks obtained by them are :

    1. 1. 33 42
    2. 2. 42 33
    3. 3. 24 42
    4. 4. 42 36

2
Let their marks be (x + 9) and x Then x + 9 = (56/100) (x + 9 + x) 25(x + 9) = 14(2x + 9) 3x = 99 x = 33 So their marks are 42 and 33.

Q.- In a hotel 60% had vegetarian lunch while 30% had non-vegetarian lunch and 15% had both type of lunch. If 96 people were present how many did not eat either type of lunch ?

    1. 1. 25
    2. 2. 26
    3. 3. 42
    4. 4. 24

4
n(A)=[(60/100)∗96]=288/5 n(B)=[(30/100)∗96]=144/5 n(A∩B)=[(15/100)∗96]=725 People who have either or both lunchn(A∪B)=(288/5)+(144/5)−(72/5)=360/5=72 So People who do no have either lunch were = 96 -72 = 24

Q.- Out of 450 students of a school 325 play football 175 play cricket and 50 neither play football nor cricket. How many students play both football and cricket ?

    1. 1. 90
    2. 2. 100
    3. 3. 120
    4. 4. 110

2
Students who play cricket n(A) = 325 Students who play football n(B) = 175 Total students who play either or both games =n(A∪B)=450−50=400 Requirednumber n(A∩B)=n(A)+n(B)−n(A∪B) =325+175−400 =100

Q.- In an examination 34% of the students failed in mathematics and 42% failed in English. If 20% of the students failed in both the subjects then find the percentage of students who passed in both the subjects.

    1. 1. 0.43
    2. 2. 0.45
    3. 3. 0.44
    4. 4. 0.42

3
Failed in mathematics n(A) = 34 Failed in English n(B) = 42 n(A∪B)=n(A)+n(B)−n(A∩B) =34+42−20 =56 Failed in either or both subjects are 56. Percentage passed = (100−56)%=44%

Q.- Two numbers are less than third number by 30% and 37% respectively. How much percent is the second number less than by the first

    1. 1. 0.12
    2. 2. 0.08
    3. 3. 0.1
    4. 4. 0.11

3
Let the third number is x. then first number = (100-30)% of x = 70% of x = 7x/10 Second number is (63x/100) Difference = 7x/10 - 63x/100 = 7x/10 So required percentage is difference is what percent of first number => (7x/100 * 10/7x * 100 )% = 10%

Q.- If x% of y is 100 and y% of z is 200 then find the relation between x and z.

    1. 1. z=4x
    2. 2. 2z=x
    3. 3. z=2x
    4. 4. z=3x

3
It is y% of z = 2(x% of y) => yz/100 = 2xy/100 => z = 2x

Q.- Raman's salary was decreased by 50% and subsequently increased by 50%. How much percent does he loss.

    1. 1. 24
    2. 2. 28
    3. 3. 26
    4. 4. 25

4
Let the origianl salary = Rs. 100 It will be 150% of (50% of 100) = (150/100) * (50/100) * 100 = 75 So New salary is 75 It means his loss is 25%

Q.- How many litres of pure acid are there in 8 litres of a 20% solution

    1. 1. 1.5
    2. 2. 1.6
    3. 3. 6.1
    4. 4. 1.4

2
Question of this type looks a bit typical but it is too simple as below… It will be 8 * 20/100 = 1.6

Q.- 10% of inhabitants of a village having died of cholera a panic set in during which 25% of the remaining inhabitants let the village. The population is then reduced to 4050. Find the original inhabitants

    1. 1. 5000
    2. 2. 6000
    3. 3. 4000
    4. 4. 6500

2
Let the total number is x then (100-25)% of (100 - 10)% x = 4050 => 75% of 90% of x = 4050 => 75/100 * 90/100 * x = 4050 => x = (4050*50)/27 = 6000

Q.- One fourth of one third of two fifth of a number is 15. What will be40% of that number

    1. 1. 175
    2. 2. 170
    3. 3. 180
    4. 4. 185

3
(1/4) * (1/3) * (2/5) * x = 15 then x = 15 * 30 = 450 40% of 450 = 180

Q.- Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56% of the sum of their marks. The marks obtained by them are

    1. 1. 3344
    2. 2. 3341
    3. 3. 4234
    4. 4. 4233

4
Let their marks be (x+9) and x. Then x+9 = 56/100(x + 9 +x) => 25(x+9) => 14 (2x + 9) => 3x = 99 => x = 33. So their marks are 42 and 33

Q.- If 20% of a number then 120% of that number will be?

    1. 1. 270
    2. 2. 207
    3. 3. 702
    4. 4. 720

4
Let the number x. Then 20% of x = 120 x = (120 * 100)/20 = 600 120% of x = (120/100 * 600) = 720.

Q.- In an election between two candidates A and B the number of valid votes received by A exceeds those received by B by 15% of the total number of votes polled. If 20% of the votes polled were invalid and a total of 8720 votes were polled then how many valid

    1. 1. 2843
    2. 2. 8243
    3. 3. 2834
    4. 4. 8234

3
Let the total number of votes polled in the election be 100k. Number of valid votes = 100k - 20% (100k) = 80k Let the number of votes polled in favour of A and B be a and b respectively. a - b = 15% (100k) => a = b + 15k => a + b = b + 15k + b Now 2b + 15k = 80k and hence b = 32.5k It is given that 100k = 8720 32.5k = 32.5k/100k * 8720 = 2834 The number of valid votes polled in favour of B is 2834.

Q.- Two tests had the same maximum mark. The pass percentages in the first and the second test were 40% and 45% respectively. A candidate scored 216 marks in the second test and failed by 36 marks in that test. Find the pass mark in the first test?

    1. 1. 236
    2. 2. 228
    3. 3. 252
    4. 4. None of these

4
Let the maximum mark in each test be M. The candidate failed by 36 marks in the second test. pass mark in the second test = 216 + 36 = 252 45/100 M = 252 Pass mark in the first test = 40/100 M = 40/45 * 252 = 224.

Q.- Ten percent of Ram's monthly salary is equal to eight percent of Shyam's monthly salary. Shyam's monthly salary is twice Abhinav's monthly salary. If Abhinav's annual salary is Rs. 1.92 lakhs find Ram's monthly salary?

    1. 1. Rs.26500
    2. 2. Rs.26005
    3. 3. Rs. 25600
    4. 4. Rs.26050

3
Let the monthly salaries of Ram and Shyam be Rs. r and Rs. s respectively. 10/100 r = 8/100 s r = 4/5 s Monthly salary of Abhinav = (1.92 lakhs)/12 = Rs. 0.16 lakhs s = 2(0.16 lakhs) = 0.32 lakhs r = 4/5(0.32 lakhs) = Rs. 25600

Q.- Anil spends 40% of his income on rent 30% of the remaining on medicines and 20% of the remaining on education. If he saves Rs. 840 every month then find his monthly salary?

    1. 1. Rs. 1800
    2. 2. Rs. 2000
    3. 3. Rs. 2200
    4. 4. Rs. 2500

4
Let's Anil's salary be Rs. 100. Money spent on Rent = 40% of 100 = Rs. 40. Money spent on medical grounds = 30% of (100 - 40) = 3/10 * 60 = Rs. 18. Money spent on education = 20% of (60 - 18) = 1/5 * 42 = Rs. 8.40 Anil saves 100 - (40 + 18 + 8.40) i.e. Rs. 33.60 for 33.6 ---> 100 ; 840 ---> ? Required salary = 840/33.6 * 100 = Rs. 2500

Q.- There is a 30% increase in the price of an article in the first year a 20% decrease in the second year and a 10% increase in the next year. If the final price of the article is Rs. 2288 then what was the price of the article initially?

    1. 1. Rs. 1500
    2. 2. Rs. 1800
    3. 3. Rs. 2000
    4. 4. Rs. 2400

3
Let the price of the article four years age be Rs. 100 in the 1st year price of the article = 100 + 30 = Rs. 130. In the 2nd year price = 130 - 20% of 130 = 130 - 26 = Rs. 104. In the 3rd year price = 104 + 10% of 104 = 104 + 10.4 = Rs. 114.40. But present price of the article is Rs. 2288 for 114.4 ---> 100 ; 2288 ---> ? Required price = (2288 * 100)/114.4 = 20 * 100 = Rs. 2000.

Q.- Ajay spends 45% of his monthly income on household items 25% of his monthly income on buying cloths 7.5% of his monthly income on medicines and saves the remaining amount which is Rs. 9000. Find his monthly income.

    1. 1. Rs. 40000
    2. 2. Rs. 36000
    3. 3. Rs. 50000
    4. 4. Rs. 45000

1
Let the monthly income of Ajay be Rs. x Savings of Ajay = x - (45 + 25 + 7.5)/100 * x = 22.5/100 x 22.5/100 x = 9000 x = 40000.

Q.- A person spends 1/5th of his income on the education of his children and 20% of the remaining on food. If he is left with Rs.576 find his income?

    1. 1. Rs.900
    2. 2. Rs.800
    3. 3. Rs.500
    4. 4. Rs.1000

1
X * 4/5 * 80/100 = 576 X = 14400/16 X = 900

Q.- The population of a town is 45000; 5/9th of them are males and the rest females 40% of the males are married. What is the percentage of married females?

    1. 1. 0.6
    2. 2. 0.5
    3. 3. 0.45
    4. 4. 0.4

2
Male = 45000* 5/9 = 25000 Female = 45000* 4/9 = 20000 Married Male = 25000* 40/100 = 10000 Married Female = 10000 20000 ------------ 10000 100 ------------- ? => 50%

Q.- 5% people of a village in Sri Lanka died by bombardment 15% of the remainder left the village on account of fear. If now the population is reduced to 3553 how much was it in the beginning?

    1. 1. 4004
    2. 2. 4040
    3. 3. 4400
    4. 4. 4500

3
X * (95/100) * (85/100) = 3553 X = 4400

Q.- Find the ratio in which rice at Rs. 7.20 a kg be mixed with rice at Rs. 5.70 a kg to produce a mixture worth Rs. 6.30 a kg.

    1. 1. 0.085416666667
    2. 2. 0.12638888889
    3. 3. 0.17013888889
    4. 4. 0.21111111111

1
By the rule of alligation : Cost of 1 kg of 1st kind 720 p Cost of 1 kg of 2nd kind 570 p Mean Price 630 p Required ratio = 60 : 90 = 2 : 3.

Q.- A merchant has 1000 kg of sugar part of which he sells at 8% profit and the rest at 18% profit. He gains 14% on the whole. The quantity sold at 18% profit is :

    1. 1. 506 kg
    2. 2. 650 kg
    3. 3. 500 kg
    4. 4. 600 kg

4
By the rule of alligation we have: Profit on 1st part -- 8% Profit on 2nd part -- 18% Mean Profit --14% Ration of 1st and 2nd parts = 4 : 6 = 2 : 3 Quantity of 2nd kind = (3/5) x 1000 kg = 600 kg

Q.- The cost of Type 1 rice is Rs. 15 per kg and Type 2 rice is Rs. 20 per kg. If both Type 1 and Type 2 are mixed in the ratio of 2 : 3 then the price per kg of the mixed variety of rice is :

    1. 1. Rs 18.5
    2. 2. Rs 20
    3. 3. Rs 81
    4. 4. Rs 18

4
Let the price of the mixed variety be Rs. x per kg. By rule of alligation we have : Cost of 1 kg of Type 1 rice Rs 15 Cost of 1 kg of Type 2 rice Rs 20 Mean Price--Rs x (20 - x)/ (x - 15) = 2/3 60 - 3x = 2x - 30 5x = 90 x = 18

Q.- In what ratio must a grocer mix two varieties of tea worth Rs. 60 a kg and Rs. 65 a kg so that by selling the mixture at Rs. 68.20 a kg he may gain 10%?

    1. 1. 0.084027777778
    2. 2. 0.043055555556
    3. 3. 0.12638888889
    4. 4. 0.085416666667

3
S.P. of 1 kg of the mixture = Rs. 68.20 Gain = 10% C.P. of 1 kg of the mixture = Rs. 100 x 68.20 = Rs. 62 110 By the rule of alligation we have : Cost of 1 kg tea of 1st kind.Cost of 1 kg tea of 2nd kind Rs. 60 Mean Price Rs. 62 Rs. 65 3 2 Required ratio = 3 : 2

Q.- A jar full of whisky contains 40% alcohol. A part of this whisky is replaced by another containing 19% alcohol and now the percentage of alcohol was found to be 26%. The quantity of whisky replaced is:

    1. 1. 44/100
    2. 2. 66/100
    3. 3. 55/100
    4. 4. 65/100

2
By the rule of alligation we have : Strength of first jar--40% Strength of 2nd jar--19% Mean Strength--26% So ratio of 1st and 2nd quantities = 7 : 14 = 1 : 2 Required quantity replaced = 2/3

Q.- In what ratio must water be mixed with milk to gain 16% on selling the mixture at cost price?

    1. 1. 0.045833333333
    2. 2. 0.25069444444
    3. 3. 0.085416666667
    4. 4. 0.12638888889

1
Let C.P. of 1 litre milk be Re. 1 S.P. of 1 litre of mixture = Re.1 Gain = 50/3 % C.P. of 1 litre of mixture = (100 x 3/350) x 1 = 6/7 By the rule of alligation we have : C.P. of 1 litre of water--Rs 0 C.P. of 1 litre of milk--Rs 01 Mean Price--Rs 6/7 Ratio of water and milk = 1/7 : 6/7 = 1 : 6

Q.- A container contains 40 litres of milk. From this container 4 litres of milk was taken out and replaced by water. This process was repeated further two times. How much milk is now contained by the container?

    1. 1. 21.96
    2. 2. 92.16
    3. 3. 16.29
    4. 4. 29.16

4
Amount of milk left after 3 operations = 40 {(1 -0.1)(1-.01)(1-0.1)} = 40 x (9/10) x (9/10) x (9/10) = 29.16 litres.

Q.- How many kilogram of sugar costing Rs. 9 per kg must be mixed with 27 kg of sugar costing Rs. 7 per kg so that there may be a gain of 10% by selling the mixture at Rs. 9.24 per kg?

    1. 1. 63 kg
    2. 2. 36 kg
    3. 3. 56 kg
    4. 4. 65 kg

1
S.P. of 1 kg of mixture = Rs. 9.24 Gain 10% C.P. of 1 kg of mixture = Rs. (100/110) x 9.24 = Rs. 8.40 By the rule of allilation we have : C.P of 1 kg sugar of 1st kind -Rs 9 Cost of 1 kg sugar of 2nd kind Rs 7 Mean Price--Rs. 8.40 Ratio of quantities of 1st and 2nd kind = 14 : 6 = 7 : 3 Let x kg of sugar of 1st be mixed with 27 kg of 2nd kind Then 7 : 3 = x : 27 x = (7 x 27/3) = 63 kg.

Q.- A container contains 40 litres of milk. From this container 4 litres of milk was taken out and replaced by water. This process was repeated further two times. How much milk is now contained by the container?

    1. 1. 61.29
    2. 2. 92.16
    3. 3. 16.29
    4. 4. 29.16

4
Amount of milk left after 3 operations = 40[{(1-0.1)(1-0.1)(1-0.1)}] = 40 x( 9/10) x (9/10) x (9/10) = 29.16 litres.

Q.- A dishonest milkman professes to sell his milk at cost price but he mixes it with water and thereby gains 25%. The percentage of water in the mixture is:

    1. 1. 0.2
    2. 2. 0.3
    3. 3. 0.4
    4. 4. 0.25

1
Let C.P. of 1 litre milk be Re. 1 Then S.P. of 1 litre of mixture = Re. 1 Gain = 25% C.P. of 1 litre mixture = Re. (100/125) x 1 = 4/5 By the rule of alligation we have : C.P. of 1 litre of milk--Rs 01 C.P. of 1 litre of water--Rs 0 Mean Price--4/5 Ratio of milk to water = 4/5 : 1/5 = 4 : 1 Hence percentage of water in the mixture = (1/5) x 100 % = 20%

Q.- In a mixture of 60 litres the ratio of milk and water is 2:1.If the ratio of the milk and water is to be 1:2 then the amount of water to be further added is :

    1. 1. 20 litres
    2. 2. 30 litres
    3. 3. 40 litres
    4. 4. 60 litres

4
Ratio of milk and water in mixture of 60 litres = 2:1. Quantity of milk = 40 litres. Quantity of water = 20 litres. If ratio of milk and water is to be 1:2 then in 40 litres of milk water should be 80 litres So Quantity of water to be added = 60 litres.

Q.- A sum of Rs. 41 was divided among 50 boys and girls. Each boy gets 90 paise and a girl 65 paise.The number of boys is :

    1. 1. 16
    2. 2. 34
    3. 3. 14
    4. 4. 36

2
Average money received by each = Rs.41/50 = 82 paise. Ratio of boys and girls = 17:8. So number of boys = 50*(17/25) = 34

Q.- 729 ml of a mixture contains milk and water in the ratio 7:2.How much more water is to be added to get a new mixture containing milk and water in ratio 7:3 ?

    1. 1. 600 ml
    2. 2. 710 ml
    3. 3. 520 ml
    4. 4. none of these

4
Milk = 729*(7/9) = 567 ml. Water = 729-567 = 162 ml. Now 567/(162+x) =7/3 so x = 81

Q.- Some amount out of Rs.7000 was lent at 6% p.a. and the remaining at 4% p.a. If the total simple interest from both the fractions in 5 years was Rs.1600.The sum lent at 6% p.a. was :

    1. 1. Rs.2000
    2. 2. Rs.5000
    3. 3. Rs.3500
    4. 4. Rs.2500

1
Average annual rate = (1600/7000)*(1000/5) = 32/7% So amount at 6% : amount at 4% = 4/7 : 10/7 = 2 : 5. Hence sum lent at 6% = Rs.7000*(2/7) = Rs.2000

Q.- Two vessels A and B contain milk and water mixed in the ratio 5:3 and 2:3. When the mixture was mixed to form a new mixture containing half milk and half water.They must be taken in the ratio :

    1. 1. 0.086805555556
    2. 2. 0.12847222222
    3. 3. 0.17013888889
    4. 4. 0.29375

3
Milk in A = 5/8 of whole. Milk in B = 2/5 of whole. Milk in mixture of A and B = 1/2. So By alligation rule (Mix. in A) : (Mix in B) = 1/10 :1/8 = 4 : 5

Q.- A jar full of whisky contains 40% of alcohol. A part of this whisky is replaced by another containing 19% alcohol. The percentage of alcohol was found to be 26. The quantity of whisky replaced is :

    1. 1. 42405
    2. 2. 42372
    3. 3. 42403
    4. 4. 42434

3
Using the method of alligation Required ratio = 7:14 = 1:2. So Required quantity = 2/3

Q.- Kantilal mixes 80 kg. of sugar worth of Rs.6.75 per kg. with 120 kg. worth of Rs.8 per kg.At what rate shall he sell the mixture to gain 20% ?

    1. 1. Rs.7.50
    2. 2. Rs.9
    3. 3. Rs.8.20
    4. 4. Rs.8.85

2
Total CP of 200 kg. of mixture = Rs.(80*6.75)+(120*8) = Rs.1500. Average rate = Rs.7.50 per kg. Rquired rate = 120% of Rs.7.50 = Rs. 9 per kg.

Q.- Rs.1000 is lent in two parts one at 6% simple interest and the other at 8% simple interest. The yearly income is Rs.75.The sum lent at 8% is :

    1. 1. Rs.250
    2. 2. Rs.500
    3. 3. Rs.750
    4. 4. Rs.600

3
Total interest = Rs.75. Average rate = (100*75)/(1000*1)% = 15/2% So (sum at 6%) : (sum at 8%) = 1/2 : 3/2 = 1 : 3. Hence sum at 8% = Rs.1000*(3/4) = Rs.750

Q.- A grocer buys two kind of rice at Rs.1.80 and Rs.1.20 per kg. respectively.In what proportion should these be mixed so that by selling the mixture at Rs.1.75 per kg. 25% may be gained ?

    1. 1. 0.084027777778
    2. 2. 0.12638888889
    3. 3. 0.12777777778
    4. 4. 0.043055555556

4
SP of 1 kg. mixture = Rs.1.75 Gain = 25% So mean price = Rs.(1.75*100)/125 = Rs.1.40 So dearer rice : cheaper rice = 20:40 = 1:2

Q.- 15 litres of a mixture contains 20% alcohol and the rest water.If 3 litres of water be mixed in it the percentage of alcohol in the new mixture will be :

    1. 1. 17
    2. 2. 50/3
    3. 3. 56/3
    4. 4. 63/5

2
Initially the mixture contains 3 litres of alcohol and 12 litres of water. Afterwards the mixture contains 3 litres of alcohol and 15 litres of water. So percentage of alcohol = (3/18)*100 = 50/3%

Q.- How many litres of water should be added to a 30 litre mixture of milk and water containing milk and water in the ratio of 7 : 3 such that the resultant mixture has 40% water in it?

    1. 1. 5 litres
    2. 2. 7 litres
    3. 3. 10 litres
    4. 4. None of these

1
30 litres of the mixture has milk and water in the ratio 7 : 3. i.e. the solution has 21 litres of milk and 9 litres of water. When you add more water the amount of milk in the mixture remains constant at 21 litres. In the first case before addition of further water 21 litres of milk accounts for 70% by volume. After water is added the new mixture contains 60% milk and 40% water. Therefore the 21 litres of milk accounts for 60% by volume. Hence 100% volume=frac{21}{0.6} =35 litres. We started with 30 litres and ended up with 35 litres. Therefore 5 litres of water was added.

Q.- In what ratio must a person mix three kinds of tea costing Rs.60/kg Rs.75/kg and Rs.100 /kg so that the resultant mixture when sold at Rs.96/kg yields a profit of 20%?

    1. 1. 0.12993055556
    2. 2. 0.043101851852
    3. 3. 0.044467592593
    4. 4. 0.084074074074

3
The resultant mixture is sold at a profit of 20% at Rs.96/kg i.e. 1.2 (cost) = Rs.96 ? Cost =96/1.2= Rs.80 / kg. Let the three varieties be A B and C costing Rs.60 Rs.75 and Rs.100 respectively. The mean price falls between B and C. Hence the following method should be used to find the ratio in which they should be mixed. Step 1: Find out the ratio of Qa:Qc using alligation rule: QaQc=100-80/80-60=11 Step 2: Find out the ratio of Qb:Qc using alligation rule: QbQc=100-80/80-75=41 Step 3: Qc the resultant ratio of variety c can be found by adding the value of Qc in step 1 and step 2=1+1=2. However in CAT if you try and solve the problem using the above method you will end up spending more than 2 and may be 3 minutes on this problem which is a criminal mismanagement of time. The best way to solve a problem of this kind in CAT is to go from the answer choices as shown below The resultant ratio Qa:Qb:Qc::1:4:2 1 kg of variety A at Rs.60 is mixed with 4 kgs of variety B at Rs.75 and 2 kgs of variety C at Rs.100. The total cost for the 7kgs: =60+(4×75)+(2×100) =60+300+200=560 Cost per kg. of the mixture =560/7=80 kgs.

Q.- Two varieties of wheat - A and B costing Rs. 9 per kg and Rs. 15 per kg were mixed in the ratio 3 : 7. If 5 kg of the mixture is sold at 25% profit find the profit made?

    1. 1. Rs.13.50
    2. 2. Rs.14.50
    3. 3. Rs.15.50
    4. 4. Rs.16.50

4
Let the quantities of A and B mixed be 3x kg and 7x kg. Cost of 3x kg of A = 9(3x) = Rs. 27x Cost of 7x kg of B = 15(7x) = Rs. 105x Cost of 10x kg of the mixture = 27x + 105x = Rs. 132x Cost of 5 kg of the mixture = 132x/10x (5) = Rs. 66 Profit made in selling 5 kg of the mixture = 25/100 (cost of 5 kg of the mixture) = 25/100 * 66 = Rs. 16.50

Q.- In a mixture of milk and water the proportion of milk by weight was 80%. If in a 180 gm mixture 36 gms of pure milk is added what would be the percentage of milk in the mixture formed?

    1. 1. 0.8333
    2. 2. 0.85
    3. 3. 0.8733
    4. 4. 0.9033

1
Percentage of milk in the mixture formed = [80/100 (180) + 36] / (180 + 36) * 100% = (144 + 36)/216 * 100% = 5/6 * 100% = 83.33%.

Q.- In a can there is a mixture of milk and water in the ratio 4 : 5. If it is filled with an additional 8 litres of milk the can would be full and ratio of milk and water would become 6 : 5. Find the capacity of the can?

    1. 1. 40
    2. 2. 44
    3. 3. 48
    4. 4. 52

2
Let the capacity of the can be T litres. Quantity of milk in the mixture before adding milk = 4/9 (T - 8) After adding milk quantity of milk in the mixture = 6/11 T. 6T/11 - 8 = 4/9(T - 8) 10T = 792 - 352 T = 44

Q.- A zookeeper counted the heads of the animals in a zoo and found it to be 80. When he counted the legs of the animals he found it to be 260. If the zoo had either pigeons or horses how many horses were there in the zoo?

    1. 1. 30
    2. 2. 40
    3. 3. 50
    4. 4. 60

3
Let the number of horses =x Then the number of pigeons =80-x. Each pigeon has 2 legs and each horse has 4 legs. Therefore total number of legs =4x+2(80-x)=260 4x+160-2x=260 2x=100 x=50

Q.- How many litres of a 12 litre mixture containing milk and water in the ratio of 2 : 3 be replaced with pure milk so that the resultant mixture contains milk and water in equal proportion?

    1. 1. 1 litre
    2. 2. 2 litre
    3. 3. 3 litre
    4. 4. 6 litre

2
The mixture contains 40% milk and 60% water in it. That is 4.8 litres of milk and 7.2 litres of water. Now we are replacing the mixture w+K162ith pure milk so that the amount of milk and water in the mixture is 50% and 50%. That is we will end up with 6 litres of milk and 6 litres of water. Water gets reduced by 1.2 litres. To remove 1.2 litres of water from the original mixture containing 60% water we need to remove 1.2/0.6 litres of the mixture = 2 litres.

Q.- A mixture of 70 litres of milk and water contains 10% water. How many litres of water should be added to the mixture so that the mixture contains 12 1/2% water?

    1. 1. 2
    2. 2. 8
    3. 3. 4
    4. 4. 5

1
Quantity of milk in the mixture = 90/100 (70) = 63 litres. After adding water milk would form 87 1/2% of the mixture. Hence if quantity of mixture after adding x liters of water (87 1/2) / 100 x = 63 x = 72 Hence 72 - 70 = 2 litres of water must be added.

Q.- Two vessels P and Q contain 62.5% and 87.5% of alcohol respectively. If 2 litres from vessel P is mixed with 4 litres from vessel Q the ratio of alcohol and water in the resulting mixture is?

    1. 1. 0.67013888889
    2. 2. 0.67152777778
    3. 3. 0.58680555556
    4. 4. 0.79513888889

4
Quantity of alcohol in vessel P = 62.5/100 * 2 = 5/4 litres Quantity of alcohol in vessel Q = 87.5/100 * 4 = 7/2 litres Quantity of alcohol in the mixture formed = 5/4 + 7/2 = 19/4 = 4.75 litres As 6 litres of mixture is formed ratio of alcohol and water in the mixture formed = 4.75 : 1.25 = 19 : 5.

Q.- An alloy contains zinc copper and tin in the ratio 2:3:1 and another contains copper tin and lead in the ratio 5:4:3. If equal weights of both alloys are melted together to form a third alloy then the weight of lead per kg in new alloy will be:

    1. 1. 42767
    2. 2. 42826
    3. 3. 42948
    4. 4. 42370

3
In the first alloy ratio of Zinc Copper and Tin is given as Z:C:T=2:3:1 Similarly In the second alloy ratio of Copper Tin and Lead is given as C:T:L=5:4:3 The trick here is to arrive at a quantity where calculation becomes easy. To do that we take LCM of 6(=2+3+1) taken as 2 kg Zinc 3 kg Copper and 1 Kg Lead) and 12(=5+4+3) taken as 5 kg Copper 4 kg Tin and 3 Kg Lead which is 12. Thus we assume that both the alloys are being mixed at 12 Kgs each. Alloys are mixed together to form third alloy. Then the ratio of content in it Z:C:T:L=4:(6+5):(2+4):3 Weight of the third alloy T=12+12=24 Kg. So weight of the Lead L=3/24 =1/8 kg

Q.- Sakshi invests a part of Rs. 12000 in 12% stock at Rs. 120 and the remainder in 15% stock at Rs. 125. If his total dividend per annum is Rs. 1360 how much does he invest in 12% stock at Rs. 120?

    1. 1. 4100
    2. 2. 4050
    3. 3. 3000
    4. 4. 4000

4
Let investment in 12% stock be Rs. x Then investment in 15% stock = Rs. (12000 - x) (12/120)x x + (15/125) x (12000 - x) = 1360 x/10 + 3/25 (12000 - x) = 1360 5x + 72000 - 6x = 1360 x 50 x = 4000

Q.- The cost price of a Rs. 100 stock at 4 discount when brokerage is 1/4 % is :

    1. 1. Rs 96.25
    2. 2. Rs 96.52
    3. 3. Rs 69.52
    4. 4. Rs 69.25

1
C.P. = Rs. 100 - 4 + 1 /4 = Rs. 96.25

Q.- A 12% stock yielding 10% is quoted at :

    1. 1. Rs 102
    2. 2. Rs 120
    3. 3. Rs 201
    4. 4. Rs 69.25

2
To earn Rs. 10 money invested = Rs. 100 To earn Rs. 12 money invested = Rs. 100 x 12 / 10 = Rs. 120 Market value of Rs. 100 stock = Rs. 120

Q.- A man invested Rs. 1552 in a stock at 97 to obtain an income of Rs. 128. The dividend from the stock is :

    1. 1. 0.1
    2. 2. 0.09
    3. 3. 0.08
    4. 4. 0.07

3
By investing Rs. 1552 income = Rs. 128 By investing Rs. 97 income = Rs. 128 x 97/1552 = Rs. 8 Dividend = 8%

Q.- By investing Rs. 1620 in 8% stock Michael earns Rs. 135. The stock is then quoted at :

    1. 1. Rs 96
    2. 2. Rs 69
    3. 3. Rs 100
    4. 4. Rs 69.25

1
To earn Rs. 135 investment = Rs. 1620 To earn Rs. 8 investment = Rs. (1620/135) x 8 = Rs. 96 Market value of Rs. 100 stock = Rs. 96

Q.- A man invests some money partly in 9% stock at 96 and partly in 12% stock at 120. To obtain equal dividends from both he must invest the money in the ratio :

    1. 1. 0.13472222222
    2. 2. 0.09375
    3. 3. 0.13611111111
    4. 4. 0.17708333333

4
For an income of Re. 1 in 9% stock at 96 investment = Rs. 96/9 = Rs. 32/3 For an income Re. 1 in 12% stock at 120 investment = Rs. 120/12 = Rs.10 Ratio of investments = 32/3 : 10 = 32 : 30 = 16 : 15

Q.- A man invested Rs. 4455 in Rs. 10 shares quoted at Rs. 8.25. If the rate of dividend be 12% his annual income is :

    1. 1. Rs 846
    2. 2. Rs 486
    3. 3. Rs 648
    4. 4. Rs 69.25

3
Number of shares = 4455/8.25 = 540 Face value = Rs. (540 x 10) = Rs. 5400 Annual income = Rs. (12/100) x 5400 = Rs. 648

Q.- A 6% stock yields 8%. The market value of the stock is :

    1. 1. Rs 60
    2. 2. Rs 70
    3. 3. Rs 57
    4. 4. Rs 75

4
For an income of Rs. 8 investment = Rs. 100 For an income of Rs. 6 investment = Rs. (100/8)x 6 = Rs. 75 Market value of Rs. 100 stock = Rs. 75

Q.- In order to obtain an income of Rs. 650 from 10% stock at Rs. 96 one must make an investment of :

    1. 1. Rs 6240
    2. 2. Rs 2406
    3. 3. 4062
    4. 4. 2460

1
To obtain Rs. 10 investment = Rs. 96 To obtain Rs. 650 investment = Rs. (96/10) x 650 = Rs. 6240

Q.- A man buys Rs. 20 shares paying 9% dividend. The man wants to have an interest of 12% on his money. The market value of each share is :

    1. 1. Rs 20
    2. 2. Rs 15
    3. 3. Rs 51
    4. 4. Rs 55

2
Dividend on Rs. 20 = Rs. (9/100) x 20 = Rs. 9/5 Rs. 12 is an income on Rs. 100 Rs. 9/5 is an income on Rs. (100/12) x 9/5 = Rs. 15

Q.- A man invested Rs.4455 in Rs.10 shares quoted at Rs.8.25.If the rate of dividend be 6% his annual income is :

    1. 1. Rs.423
    2. 2. Rs.432
    3. 3. Rs.324
    4. 4. Rs.234

3
Number of shares = 4455/8.25 = 540. Face value = Rs.540*10 = Rs.5400. Income = Rs(6/100)*5400 = Rs.324

Q.- A man buys Rs.20 shares paying 9% dividend.The man wants to have an interest of 12% on his money.The market value of each share must be :

    1. 1. Rs.12
    2. 2. Rs.15
    3. 3. Rs.18
    4. 4. Rs.21

2
Dividend on Rs. 20 = Rs.(9/100)*20 = Rs.9/5. So Rs.12 is an income on Rs.100. So Rs.9/5 is an income on Rs.(100/12)*(9/5) = Rs.15

Q.- By investing in 15/4% stock at 96 one ears Rs.100.The investment made is:

    1. 1. Rs.5260
    2. 2. Rs.5206
    3. 3. Rs.2560
    4. 4. Rs.2506

3
For earning Rs.15/4 investment = Rs.96. For earning Rs. 100 investment = Rs.(96*4/15)*100 = Rs.2560

Q.- A man invested Rs.388 in a stock at 97 to obtain an income of Rs.22.The dividend from the stock is :

    1. 1. 0.03
    2. 2. 0.12
    3. 3. 11/2%
    4. 4. 2/11%

3
When investment is Rs.388 income = Rs.22. When investment is Rs.97 income = Rs.(22/388)*97 = Rs.5.50. Dividend on Rs.100 stock = 11/2%

Q.- A man invests some money partly in 3% stock at 96 and partly in 4% stock at 120.To get equal dividends from both he must invest the money in the ratio ?

    1. 1. 0.67708333333
    2. 2. 0.63611111111
    3. 3. 0.425
    4. 4. 0.50694444444

1
For an income of Rs.1 in 3% stock investment = Rs.96/3 = Rs.32. For an income of Rs.1 in 4% stock investment = Rs.120/4 = Rs.30. So Ratio of investment = 32:30 = 16:15

Q.- I want to purchase a 6% stock which must yield 5% on my capital.At what price must I buy the stock ?

    1. 1. Rs.111
    2. 2. Rs.102
    3. 3. Rs. 210
    4. 4. Rs.120

4
For an investment of Rs.5 investment = Rs.100. For an income of Rs.6 investment = Rs.(100/5)*6 = Rs.120

Q.- A invested some money in 4% stock at 96. Now B wants to invest in an equally good 5% stock. B must purchase a stock worth of :

    1. 1. Rs.120
    2. 2. Rs.220
    3. 3. Rs.320
    4. 4. Rs.202

1
For an income of Rs.4 investment = Rs.96. For an income of Rs.5 investment = Rs.(96/4)*5 = Rs.120

Q.- A 4% stock yields 5%.The market value of the stock is :

    1. 1. Rs.125
    2. 2. Rs.80
    3. 3. Rs.99
    4. 4. Rs.110

2
For an income of Rs.5 investment = Rs.100. For an income of Rs.4 investment = Rs.(100/5)*4 = Rs.80

Q.- By investing in a 6% stock at 96 an income of Rs.100 is obtained by making an investment of :

    1. 1. Rs.1600
    2. 2. Rs.1006
    3. 3. Rs.1060
    4. 4. Rs.6100

1
For an income of Rs.6 investment = Rs.96. For an income of Rs.100 investment = Rs.(96/6)*100 = Rs.1600

Q.- The cash realized by selling a 11/2% stock at 425/4 brokerage being 1/4% is :

    1. 1. Rs.610
    2. 2. Rs.160
    3. 3. Rs.106
    4. 4. none

3
Cash realized = Rs.(425/4)-(1/4) = Rs.106

Q.- Find the annual income derived from Rs. 2500 8% stock at 106?

    1. 1. 100
    2. 2. 150
    3. 3. 200
    4. 4. 250

3
Income from Rs. 100 stock = Rs. 8. Income from Rs. 2500 stock = (8/100)*2500 = Rs. 200.

Q.- A 6% stock yields 8%. The market value of the stock is:

    1. 1. 70
    2. 2. 75
    3. 3. 80
    4. 4. 85

2
For an income of Rs. 8 investment = Rs. 100. For an income of Rs. 6 investment =(100/8)*6 = Rs. 75. Market value of Rs. 100 stock = Rs. 75.

Q.- A man invested Rs. 4455 in Rs. 10 shares quoted at Rs. 8.25. If the rate of dividend be 12% his annual income is:

    1. 1. 107.04
    2. 2. 648
    3. 3. 500
    4. 4. 648.6

2
Number of shares =4455/8.25 = 540. Face value = Rs. (540 x 10) = Rs. 5400. Annual income = (12/100)*5400 = Rs. 648.

Q.- Find the cost of Rs. 6400 10% stock at 15 discount?

    1. 1. 2000
    2. 2. 5440
    3. 3. 5140
    4. 4. 8500

2
Cost of Rs. 100 stock = Rs. (100-15) Cost of Rs. 6400 stock = (85/100)*6400 = Rs. 5440.

Q.- A man invests some money partly in 9% stock at 96 and partly in 12% stock at 120. To obtain equal dividends from both he must invest the money in the ratio:

    1. 1. 0.12847222222
    2. 2. 0.043055555556
    3. 3. 0.67708333333
    4. 4. 0.17013888889

3
For an income of Re. 1 in 9% stock at 96 investment = Rs. 96/9 = Rs.32/3 For an income Re. 1 in 12% stock at 120 investment = Rs. 120/12 = Rs. 10. Ratio of investments =(32/3) : 10 = 32 : 30 = 16 : 15.

Q.- By investing Rs. 1620 in 8% stock Michael earns Rs. 135. The stock is then quoted at:

    1. 1. 80
    2. 2. 96
    3. 3. 106
    4. 4. 108

2
To earn Rs. 135 investment = Rs. 1620. To earn Rs. 8 investment =(1620/135)*8= Rs. 96. Market value of Rs. 100 stock = Rs. 96.

Q.- In order to obtain an income of Rs. 650 from 10% stock at Rs. 96 one must make an investment of:

    1. 1. 3100
    2. 2. 6240
    3. 3. 6400
    4. 4. 9600

2
To obtain Rs. 10 investment = Rs. 96. To obtain Rs. 650 investment = 650*96/10 = Rs. 6240.

Q.- A invested some money in 10% stock at 96. If B wants to invest in an equally good 12% stock he must purchase a stock worth of :

    1. 1. Rs.80
    2. 2. Rs.120.25
    3. 3. Rs.115.20
    4. 4. Rs.125.40

3
For an income of Rs. 10 investment = Rs. 96. For an income of Rs. 12 investment = 96*12/10= Rs. 115.20

Q.- By investing in 16 2/3% stock at 64 one earns Rs. 1500. The investment made is:

    1. 1. 5460
    2. 2. 5600
    3. 3. 5760
    4. 4. 5840

3
To earn Rs.50/3 investment = Rs. 64. To earn Rs. 1500 investment = 64*(3/50)*1500 = Rs. 5760.

Q.- A man buys Rs. 20 shares paying 9% dividend. The man wants to have an interest of 12% on his money. The market value of each share is:

    1. 1. 12
    2. 2. 15
    3. 3. 18
    4. 4. 20

2
Dividend on Rs. 20 = Rs. (9/100)x 20 = Rs.9/5. Rs. 12 is an income on Rs. 100. Rs.9/5 is an income on Rs.[ (100/12) x (9/5)] = Rs. 15.

Q.- The cash realised on selling a 14% stock is Rs.106.25 brokerage being 1/4% is

    1. 1. 100
    2. 2. 106
    3. 3. 123
    4. 4. 120.25

2
Cash realised= Rs. (106.25 - 0.25)= Rs. 106.

Q.- The true discount on a bill of Rs. 540 is Rs. 90. The banker's discount is:

    1. 1. Rs 801
    2. 2. Rs. 108
    3. 3. Rs 810
    4. 4. Rs 180

2
P.W. = Rs. (540 - 90) = Rs. 450 S.I. on Rs. 450 = Rs. 90 S.I. on Rs. 540 = Rs. (90/450)x 540 = Rs. 108 B.D. = Rs. 108

Q.- The banker's gain on a bill due 1 year hence at 12% per annum is Rs. 6. The true discount is:

    1. 1. Rs 45
    2. 2. Rs 65
    3. 3. Rs 60
    4. 4. Rs 50

4
T.D. =( B.G. x 100 )/RxT= Rs. ( 6 x 100) /12x1 = Rs. 50

Q.- The certain worth of a certain sum due sometime hence is Rs. 1600 and the true discount is Rs. 160. The banker's gain is:

    1. 1. Rs 16
    2. 2. Rs 61
    3. 3. Rs 38
    4. 4. Rs 18

1
B.G. = (T.D.xT.D.)/P.W. = Rs. ( 160 x 160 )/1600 = Rs. 16.

Q.- The banker's discount of a certain sum of money is Rs. 72 and the true discount on the same sum for the same time is Rs. 60. The sum due is:

    1. 1. Rs 630
    2. 2. Rs 360
    3. 3. Rs 306
    4. 4. Rs 612

2
Explanation : Sum = B.D. x T.D./B.D.-T.D. = Rs. (72 x 60)/(72-60) = Rs. ( 72 x 60 )/12 = Rs. 360

Q.- The banker's discount on Rs. 1600 at 15% per annum is the same as true discount on Rs. 1680 for the same time and at the same rate. The time is:

    1. 1. 6 months
    2. 2. 8 months
    3. 3. 4 months
    4. 4. 2 months

3
S.I. on Rs. 1600 = T.D. on Rs. 1680 Rs. 1600 is the P.W. of Rs. 1680 i.e. Rs. 80 is on Rs. 1600 at 15% Time = (100 x 80)/(1600x15) year = 1/3 year = 4 months

Q.- The banker's gain of a certain sum due 2 years hence at 10% per annum is Rs. 24. The present worth is:

    1. 1. Rs 500
    2. 2. Rs 600
    3. 3. Rs 400
    4. 4. Rs 650

2
T.D. = B.G. x 100 = Rs. 24 x 100 = Rs. 120 P.W. = 100 x T.D. = Rs. 100 x 120 = Rs. 600

Q.- The banker's gain on a sum due 3 years hence at 12% per annum is Rs. 270. The banker's discount is:

    1. 1. Rs 1020
    2. 2. Rs 2010
    3. 3. Rs 2001
    4. 4. Rs 1002

1
T.D. = (B.G. x 100)/RxT = Rs. ( 270 x 100)/12x3 = Rs. 750 B.D. = Rs.(750 + 270) = Rs. 1020

Q.- The banker's discount on a certain sum due 2 years hence is 11/10 of the true discount. The rate percent is:

    1. 1. 0.06
    2. 2. 0.05
    3. 3. 0.07
    4. 4. 0.08

2
Let T.D. be Re. 1 Then B.D. = Rs. 11/10 = Rs. 1.10 Sum = Rs. ( 1.10 x 1)/(1.10-1.10) = Rs. 110/10 = Rs. 11 S.I. on Rs. 11 for 2 years is Rs. 1.10 Rate = ( 100 x 1.10)/(11x2) % = 5%

Q.- The present worth of a certain bill due sometime hence is Rs. 800 and the true discount is Rs. 36. The banker's discount is :

    1. 1. Rs 37.52
    2. 2. Rs 37.5
    3. 3. Rs 37
    4. 4. Rs 37.62

4
B.G. = (T.D.xT.D.)/P.W. = Rs. (36 x 36)/800 = Rs. 1.62 B.D. = (T.D. + B.G.) = Rs. (36 + 1.62) = Rs. 37.62

Q.- The banker's discount on a bill due 4 months hence at 15% is Rs. 420. The true discount is :

    1. 1. Rs350
    2. 2. Rs 300
    3. 3. Rs 400
    4. 4. Rs 450

3
T.D. = (B.D. x 100)/(100 + (R x T)) = Rs. ( 420 x 100 )/100 + ( 15 x 1/3) = Rs.( 420 x 100 )/105 = Rs. 400

Q.- The banker's gain on sum due 3 years. hence at 5% is Rs.90. The banker`s discount is :

    1. 1. Rs.690
    2. 2. Rs.720
    3. 3. Rs.810
    4. 4. Rs.150

1
T.D. = (B.G.*100)/(R*T) = Rs.(90*100)/(5*3) = Rs.600 so B.D. = Rs.(600+90) = Rs.690

Q.- The present worth of a sum due sometimes hence is Rs.576 and the banker`s gain is Rs.1.The true discount is :

    1. 1. Rs.16
    2. 2. Rs.18
    3. 3. Rs.24
    4. 4. Rs.32

3
T.D. = sqrt.[(P.W.)*(B.G.)] = Rs.sqrt(576*1) = Rs. 24

Q.- The banker discount on a bill due 6 months hence at 6% is Rs.37.08.The true discount is :

    1. 1. Rs.6.18
    2. 2. Rs.12.36
    3. 3. Rs.48
    4. 4. Rs.36

4
TD = (BD*100)/[100+(R*T)] = Rs. [(37.08*100)/100+(6*1/2)] = Rs.36

Q.- The banker gain on a bill due 1 year hence at 5% is Rs.1.The true discount is :

    1. 1. Rs.15
    2. 2. Rs.20
    3. 3. Rs.25
    4. 4. Rs.5

2
TD = (BG*100)/(R*T) = Rs.(1*100)/(5*1) = Rs.20

Q.- The banker discount on Rs.1600 at 6% is the same as the true discount on Rs.1624 for the same time and at the same rate.Then the time is :

    1. 1. 3 months
    2. 2. 4 months
    3. 3. 6 months
    4. 4. 8 months

1
SI on Rs.1600 = TD on Rs.1624 so Rs.1600 is PW of Rs.1624. i.e. Rs.24 is The SI on Rs.1600 at 6%. so Time = (100*24)/(1600*6)year = 1/4 year = 3 months

Q.- The banker gain of a certain sum of money is Rs.36 and the true discount on the same sum for the same time and at the same rate is Rs.30.The sum is :

    1. 1. Rs.1080
    2. 2. Rs.180
    3. 3. Rs.500
    4. 4. Rs.300

2
Sum = (BD*TD)/(BD-TD) = Rs.(36*30)/6 = Rs.180

Q.- The banker gain of a certain sum due 2 years hence at 5% per annum is Rs.8.The present worth is :

    1. 1. Rs.800
    2. 2. Rs.1600
    3. 3. Rs.1200
    4. 4. Rs.880

1
TD = (BG*100)/(Rate*Time) = Rs.(8*100)/(5*2) = Rs.800

Q.- The present worth of a certain sum due sometime hence is Rs.1600 and the true discount is Rs.160.The banker gain is :

    1. 1. Rs.10
    2. 2. Rs.16
    3. 3. Rs.20
    4. 4. Rs.24

2
BG = (TD*TD)/PW = Rs.(160*160)/1600 = Rs.16

Q.- The present worth of a certain bill due sometime hence is Rs.800 and the true discount is Rs.36.Then the banker discount is :

    1. 1. Rs.37
    2. 2. Rs.34.38
    3. 3. Rs.37.62
    4. 4. Rs.38.98

3
BG = (TD*TD)/PW = Rs.(36*36)/800 = Rs.1.62 so BD = TD+BG = Rs.(36+1.62) = Rs.37.62

Q.- The true discount on a bill of Rs.540 is Rs.90.The banker discount is :

    1. 1. Rs.60
    2. 2. Rs.150
    3. 3. Rs.108
    4. 4. Rs.110

3
PW = Rs.(540-90) = Rs.450 SI on Rs.450 = Rs.90 BD = SI on Rs.540 = Rs.(90/450)*540 = Rs.108

Q.- The banker's discount on a bill due 4 months hence at 15% is Rs. 420. What is the true discount?

    1. 1. Rs.350
    2. 2. Rs.400
    3. 3. Rs.450
    4. 4. Rs.430

2
TD = (BD ×100)/(100+ TR) =(420×100)/[100+{(4/12)×15}] =(420×100)/[100+{(1/3)×15}] =(420×100)/(100+5) =(420×100)/105 =(84×100)/21 =4×100=400

Q.- The banker's discount on a certain amount due 2 years hence is 11/10 of the true discount. What is the rate percent?

    1. 1. 0.01
    2. 2. 0.05
    3. 3. 0.06
    4. 4. 0.04

2
Let TD = Rs. 1 Then BD = (11/10)×1= Rs. 11/10 T = 2 R = ? F = (BD ×TD)/(BD – TD)=[(11/10)×1] / [(11/10)-1]=(11/10)/(1/10)= Rs. 11 BD = FTR/100 ?(11/10)=(11×2×R)/100 ?110=22R ?R=110/22=5%

Q.- The present worth of a sum due sometimes hence is Rs.5760 and the banker's gain is Rs.10. What is the true discount?

    1. 1. Rs. 480
    2. 2. Rs. 420
    3. 3. Rs. 120
    4. 4. Rs. 240

4
TD = sqrt.of (PW × BG) = sqrt.of (5760×10) = sqrt.of (57600) = Rs. 240

Q.- What is the banker's discount if the true discount on a bill of Rs.540 is Rs.90 ?

    1. 1. Rs.108
    2. 2. Rs.180
    3. 3. Rs.210
    4. 4. Rs.120

1
Present Worth PW = F - TD = 540 - 90 = Rs. 450 Simple Interest on the Present Worth = True Discount Hence Simple Interest on 450 = 90 ------(Equation 1) Simple Interest on the face value = Bankers Discount => Simple Interest on 540 = Bankers Discount From Equation 1 Simple Interest on 450 = 90 Hence Simple Interest on 540 = (90/450)×540=540/5 = Rs. 108 => Bankers Discount = Rs. 108

Q.- A bill for Rs. 3000 is drawn on 14th July at 5 months. It is discounted on 5th October at 10%. What is the Banker's Discount?

    1. 1. Rs. 60
    2. 2. Rs. 82
    3. 3. Rs. 90
    4. 4. Rs. 120

1
F = Rs. 3000 R = 10% Date on which the bill is drawn = 14th July at 5 months Nominally Due Date = 14th December Legally Due Date = 14th December + 3 days = 17th December Date on which the bill is discounted = 5th October Unexpired Time = [6th to 31st of October] + [30 Days in November] + [1st to 17th of December] = 26 + 30 + 17 = 73 Days =(73/365) year=1/5 year BD = Simple Interest on the face value of the bill for unexpired time = FTR/100 = [3000×(1/5)×10]/100 = [30×(1/5)×10] = Rs. 60

Q.- The bankers discount and the true discount of a sum at 10% per annum simple interest for the same time are Rs.100 and Rs.80 respectively. What is the sum and the time?

    1. 1. Sum = Rs.400 and Time = 5 years
    2. 2. Sum = Rs.200 and Time = 2.5 years
    3. 3. Sum = Rs.400 and Time = 2.5 years
    4. 4. Sum = Rs.200 and Time = 5 years

3
BD = Rs.100 TD = Rs.80 R = 10% F = (BD ×TD)/(BD – TD) = (100×80)/(100–80) = (100×80)/20 = Rs. 400 BD = Simple Interest on the face value of the bill for unexpired time =FTR/100 ?100 = (400×T×10)/100 ?100 = 4×T×10 ?10 = 4×T ?T = 10/4 = 2.5 years

Q.- The banker's gain on a bill due 1 year hence at 10% per annum is Rs. 20. What is the true discount?

    1. 1. Rs. 200
    2. 2. Rs. 100
    3. 3. Rs. 150
    4. 4. Rs. 250

1
TD = (BG ×100)/TR = (20×100)/(1×10) = Rs. 200

Q.- The present worth of a certain bill due sometime hence is Rs. 1296 and the true discount is Rs. 72. What is the banker's discount?

    1. 1. Rs.76
    2. 2. Rs.74
    3. 3. Rs.70
    4. 4. Rs.78

1
BG = pow(TD2)/PW = pow(722)/1296 = (72×72)/1296 = (12×12)/36 = 12/3 = Rs. 4 BG = BD – TD => 4 = BD - 72 => BD = 72 + 4 = Rs. 76

Q.- The present worth of a certain sum due sometime hence is Rs. 3400 and the true discount is Rs. 340. The banker's gain is:

    1. 1. Rs. 21
    2. 2. Rs.44
    3. 3. Rs.54
    4. 4. Rs.34

4
BG = pow(TD2)/PW = pow(3402)/3400 = (340×340)/3400 = 340/10 = Rs. 34

Q.- A banker paid Rs.5767.20 for a bill of Rs.5840 drawn of Apr 4 at 6 months. If the rate of interest was 7% what was the day on which the bill was discounted?

    1. 1. 3rd March
    2. 2. 3rd September
    3. 3. 3rd October
    4. 4. 3rd August

4
F = Rs.5840 R = 7% BD = 5840 - 5767.20 = Rs.72.8 BD = FTR/100?72.8=(5840×T×7)/100?T=(72.8×100)/(7×5840)=(10.4×100)/5840=1040/5840=104/584=13/73 years=(13×365)/73 Days=65 Days => Unexpired Time = 65 Days Given that Date of Draw of the bill = 4th April at 6 months => Nominally Due Date = 4th October => Legally Due Date = (4th October + 3 days) = 7th October Hence The date on which the bill was discounted = (7th October - 65 days) = (7th October - 7 days in October - 30 days in September - 28 days in August) = 3rd August

Q.- Excluding stoppages the speed of a bus is 54 kmph and including stoppages it is 45 kmph. For how many minutes does the bus stop per hour?

    1. 1. 14
    2. 2. 12
    3. 3. 9
    4. 4. 10

4
Due to stoppages it covers 9 km less. Time taken to cover 9 km = (9/54) x 60 min = 10 min.

Q.- If a person walks at 14 km/hr instead of 10 km/hr he would have walked 20 km more. The actual distance travelled by him is :

    1. 1. 55 km
    2. 2. 40 km
    3. 3. 50 km
    4. 4. 60 km

3
Let the actual distance travelled be x km. Then 14x = 10x + 200 4x = 200 x = 50 km.

Q.- An aeroplane covers a certain distance at a speed of 240 kmph in 5 hours. To cover the same distance in 5/3 hours it must travel at a speed of :

    1. 1. 270 km/hr
    2. 2. 720 km/hr
    3. 3. 207 km/hr
    4. 4. 702 km/hr

2
Distance = (240 x 5) = 1200 km. Speed = Distance/Time Speed = 1200/(5/3) km/hr.[We can write as 5/3 hours] Required speed = 1200 x (3/5) km/hr = 720 km/hr.

Q.- A person crosses a 600 m long street in 5 minutes. What is his speed in km per hour?

    1. 1. 7.2 km/hr
    2. 2. 2.7 km/hr
    3. 3. 6.4 km/hr
    4. 4. 4.6 km/hr

1
Speed = 600/(5x60) m/sec. = 2 m/sec. Converting m/sec to km/hr (see important formulas section) = 2 x (18/5) km/hr = 7.2 km/hr.

Q.- A man on tour travels first 160 km at 64 km/hr and the next 160 km at 80 km/hr. The average speed for the first 320 km of the tour is :

    1. 1. 17.11 km/hr
    2. 2. 71.11 km/hr
    3. 3. 11.17 km/hr
    4. 4. 11.71 km/hr

2
Total time taken = (160/64) + (160/80) hrs. = 9/2 hrs. Average speed = 320 x (2/9) km/hr = 71.11 km/hr.

Q.- The ratio between the speeds of two trains is 7 : 8. If the second train runs 400 km in 4 hours then the speed of the first train is :

    1. 1. 87.95 km/hr
    2. 2. 57.8 km/hr
    3. 3. 87.5 km/hr
    4. 4. 78.5 km/hr

3
Let the speed of two trains be 7x and 8x km/hr. Then 8x = 400/4 = 100 x = 100/8 = 12.5 Speed of first train = (7 x 12.5) km/hr = 87.5 km/hr.

Q.- A man covered a certain distance at some speed. Had he moved 3 kmph faster he would have taken 40 minutes less. If he had moved 2 kmph slower he would have taken 40 minutes more. The distance (in km) is :

    1. 1. 45
    2. 2. 60
    3. 3. 50
    4. 4. 40

4
Let distance = x km and usual rate = y kmph. Then (x/y) - x/(y+3) = 40/60 2y(y + 3) = 9x ....(i) And x/(y-2) - (x/y) = 40/60 y(y - 2) = 3x ....(ii) On dividing (i) by (ii) we get: x = 40.

Q.- A farmer travelled a distance of 61 km in 9 hours. He travelled partly on foot @ 4 km/hr and partly on bicycle @ 9 km/hr. The distance travelled on foot is :

    1. 1. 16 km
    2. 2. 18 km
    3. 3. 20 km
    4. 4. 15 km

1
Let the distance travelled on foot be x km. Then distance travelled on bicycle = (61 -x) km. So (x/4) + (61 -x)/9 = 9 9x + 4(61 -x) = 9 x 36 5x = 80 x = 16 km.

Q.- In covering a distance of 30 km Abhay takes 2 hours more than Sameer. If Abhay doubles his speed then he would take 1 hour less than Sameer. Abhay's speed is :

    1. 1. 10 km/hr
    2. 2. 5 km/hr
    3. 3. 6 km/hr
    4. 4. 7 km/hr

2
Let Abhay's speed be x km/hr. Then (30/x) - (30/2x) = 3 6x = 30 x = 5 km/hr.

Q.- In a flight of 600 km an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km/hr and the time of flight increased by 30 minutes. The duration of the flight is:

    1. 1. 3 hr
    2. 2. 1.5 hr
    3. 3. 1 hr
    4. 4. 2 hr

3
Let the duration of the flight be x hours. (2x + 3)(x - 1) = 0 x = 1 hr.[neglecting the -ve value of x]

Q.- The ratio between the speeds of two trains is 7: 8. If the second train runs 400 kms in 4 hours then the speed of the first train is ?

    1. 1. 58.7 km/hr
    2. 2. 57.8 km/hr
    3. 3. 78.5 km/hr
    4. 4. 87.5 km/hr

4
Let the speeds of two trains be 7X and 8X km/hr. 8X= 400/4 =>X=12.5Km/hr So speed of first train is 12.5*7 = 87.5 km/hr

Q.- A thief is noticed by a policeman from a distance of 200 m. The thief starts running and the policeman chases him. The thief and the policeman run at the rate of 10 km and 11 km per hour respectively. What is the distance between them after 6 minutes ?

    1. 1. 90 m
    2. 2. 100 m
    3. 3. 110 m
    4. 4. 105 m

2
Relative speed of the thief and policeman = (11 – 10) km/hr = 1 km/hr Distance covered in 6 minutes = (1/60)∗6=1/10=100meters So distance between them after 6 minutes = 200 - 100 = 100 meters

Q.- A walks around a circular field at the rate of one round per hour while B runs around it at the rate of six rounds per hour. They start at same point at 7:30 am. They shall first cross each other at ?

    1. 1. 4.27 am
    2. 2. 4.72 am
    3. 3. 7.42 am
    4. 4. 7.24 am

3
Relative speed between two = 6-1 = 5 round per hour They will cross when one round will complete with relative speed which is 1/5 hour = 12 mins. So 7:30 + 12 mins = 7:42

Q.- A person travels equal distances with speed of 3 km/hr 4 km/hr and 5 km/hr and takes a total of 47 minutes. Find the total distance.

    1. 1. 3 km
    2. 2. 3.5 km
    3. 3. 4 km
    4. 4. 2.5 km

1
Let the distance be 3x km then (x/3)+(x/4)+(x/5)=47/60 47x/60=47/60 x=1 So total distance = 3*1 = 3 Km

Q.- The distance between two cities A and B is 330 Km. A train starts from A at 8 a.m. and travel towards B at 60 km/hr. Another train starts from B at 9 a.m and travels towards A at 75 Km/hr. At what time do they meet?

    1. 1. 11.45 a.m.
    2. 2. 11 a.m.
    3. 3. 10 a.m.
    4. 4. 11.30 a.m.

2
Suppose they meet x hrs after 8 a.m. then (Distance moved by first in x hrs) + [Distance moved by second in (x-1) hrs] = 330 So 60x+75(x-1) = 330 x=3. Sothey meet at (8+3). i.e 11a.m.

Q.- A man in a train notices that he can count 41 telephone posts in one minute. If they are known to be 50 metres apart then at what speed is the train travelling?

    1. 1. 110 km/hr
    2. 2. 90 km/hr
    3. 3. 100 km/hr
    4. 4. 120 km/hr

4
Number of gaps between 41 poles = 40 So total distance between 41 poles = 40*50 = 2000 meter = 2 km In 1 minute train is moving 2 km/minute. Speed in hour = 2*60 = 120 km/hour

Q.- A man on tour travels first 160 km at 64 km/hr and the next 160 km at 80 km/hr. Find the average speed for first 320 km of tour.

    1. 1. 11.17 km/hr
    2. 2. 71.11 km/hr
    3. 3. 71.15 km/hr
    4. 4. 17.11 km/hr

2
We know Time = Distance/speed + So total time taken = (160/64)+(160/80)=9/2 hours Time taken for 320 Km = 320∗(2/9)=71.11km/hr

Q.- A Man travelled a distance of 61 km in 9 hours. He travelled partly on foot at 4 km/hr and partly on bicycle at 9 km/hr. What is the distance travelled on foot?

    1. 1. 16 km
    2. 2. 15 km
    3. 3. 16.5 km
    4. 4. 14 km

1
Let the time in which he travelled on foot = x hour Time for travelling on bicycle = (9 - x) hr Distance = Speed * Time and Total distance = 61 km So 4x + 9(9-x) = 61 => 5x = 20 => x = 4 So distance traveled on foot = 4(4) = 16 km

Q.- 2 trains starting at the same time from 2 stations 200 km apart and going in opposite direction cross each other at a distance of 110 km from one of the stations. What is the ratio of their speeds ?

    1. 1. 0.38263888889
    2. 2. 99:111
    3. 3. 111:99
    4. 4. 0.46458333333

4
We know total distance is 200 Km If both trains crossed each other at a distance of 110 km then one train covered 110 km and other 90 km [110+90=200km] So ratio of their speed = 110:90 = 11:9

Q.- Excluding stoppages the speed of a bus is 54 kmph and including stoppages it is 45 kmph. For how many minutes does the bus stop per hour ?

    1. 1. 12 min
    2. 2. 10 min
    3. 3. 15 min
    4. 4. 11 min

2
Due to stoppages it covers 9 km less. Time taken to cover 9 km = (9/54) hour = (1/6)*60 minutes = 10 minutes

Q.- A man driving his bike at 24 kmph reaches his office 5 minutes late. Had he driven 25 faster on an average he would have reached 4 minutes earlier than the scheduled time. How far is his office?

    1. 1. 24 km
    2. 2. 72 km
    3. 3. 18 km
    4. 4. 36 km

3
Let x km be the distance between his house and office. While travelling at 24 kmph he would take x/24 hours While travelling at 25% faster speed i.e. 24+25%of24=24×(1/4) = 30 kmph he would take x30 hours Now as per the problem time difference = 5 min late + 4 min early = 9 min (x/24)-(x/30)=9 min x = 18 km

Q.- Two motorists Anil and Sunil are practicing with two different sports car; Ferrari and Maclarun on the circular racing track for the car racing tournament to be held next month. Both Anil and Sunil start from the same point on the circular track. Anil com

    1. 1. 382
    2. 2. 284
    3. 3. 255
    4. 4. 420

1
Time taken by Sunil for 1st round = 2 min 2nd round = 4min 3rd round = 8 min 4th round = 16 min 5th round = 32 min 6th round = 64 min 7th round = 128 min 8th round = 256 min ? Anil tales one minute for every round. He meets 127 times in 7th and 255 times in 8th round Total meet =127+255= 382

Q.- In a 3600 m race around a circular track of length 400m the faster runner and the slowest runner meet at the end of the fourth minute for the first time after the start of the race. All the runners maintain uniform speed throughout the race. If the faster

    1. 1. 12 min
    2. 2. 16 min
    3. 3. 24 min
    4. 4. 36 min

3
As the faster runner is thrice as fast as the slowest runner the faster runner would have completed three rounds by the time the slowest runner completes one round. And that is their second meeting. Their first meeting takes place after the fastest runner takes 4 min to complete one and the half round. 400×3/2=600 m He takes (3600/600)×4=24minutes to finish the race.

Q.- P and Q travels from D to A and break journey at M in between. Somewhere between D and M P asks how far have we travelled? Q replies Half as far as the distance from here to M. Somewhere between M and A exactly 300 km from the point where P asks the first

    1. 1. 250 km
    2. 2. 300 km
    3. 3. 450 km
    4. 4. 500 km

3
Let's say it was point Z when P asked the first question and point Y when P asked the second question. It can be shown as: D----Z---->----M-------->--------Y--------A Let DM=x km MA=y km Therefore distance is DA=DM+MA=x+y As per the information given in the question: ZY=300 km Also ZM=2x/3 and MY=2y/3 ZY=ZM+MY =(2x/3)+(2y/3) =300 x+y=450 km So distance between D and A is 450 km

Q.- From a point P on the surface of radius 3 cm two cockroaches A and B started moving along two different circular paths each having the maximum possible radius on the surface of the sphere that lie in the two different planes which are inclined at an angle

    1. 1. 9 s
    2. 2. 12 s
    3. 3. 18 s
    4. 4. 27 s

1
Both the circular paths have the maximum possible radius hence both have a radius of 3 cm each. Irrespective of the angle between the planes of their circular paths the two cockroaches will meet again at the point Q only which is diametrically opposite end of P. A will takes 9 seconds to reach point Q completing half a revolution. On the other hand B would have completed 3/2 of his revolution and it will also reach point Q simultaneously.

Q.- Two trains start from P and Q respectively and travel towards each other at a speed of 50 km/hr and 40 km/hr respectively. By the time they meet the first train has traveled 100 km more than the second. The distance between P and Q is?

    1. 1. 500 km
    2. 2. 600 km
    3. 3. 630 km
    4. 4. 900 km

4
At the time of meeting let the distance traveled by the second train be x km. Then distance covered by the first train is (x + 100) km. x/40 = (x + 100)/50 50x = 40x + 4000 => x = 400 So distance between P and Q = (x + x + 100)km = 900 km.

Q.- A B and C start simultaneously from X to Y. A reaches Y turns back and meet B at a distance of 11 km from Y. B reached Y turns back and meet C at a distance of 9 km from Y. If the ratio of the speeds of A and C is 3:2 what is the distance between X and Y?

    1. 1. 99 km
    2. 2. 100 km
    3. 3. 129 km
    4. 4. 142 km

1
Let the distance between X and Y be d km In the first instance distance travelled by A=d+11 In the first instance distance travelled by B=d-11 The time taken by both is same (d+11/A)=(d-11/B) AB=d+11/d-11-------(i) In the second instance distance travelled by B=d+9 While distance travelled by C=d-9 BC=d+9/d-9------(ii) From (i) and (ii) AC=(A/B)×(B/C)=32 Thus By solving we get d=1 or 99

Q.- A and B walk around a circular track. They start at 8 a.m. from the same point in the opposite directions. A and B walk at a speed of 2 rounds per hour and 3 rounds per hour respectively. How many times shall they cross each other before 9.30 a.m. ?

    1. 1. 5
    2. 2. 6
    3. 3. 7
    4. 4. 8

3
Relative speed = 2 + 3 = 5 rounds per hour. So they cross each other 5 times in an hour and 2 times in half an hour. Hence they cross each other 7 times before 9.30 a.m.

Q.- Two cars P and Q start at the same time from A and B which are 120 km apart. If the two cars travel in opposite directions they meet after one hour and if they travel in same direction (from A towards B) then P meets Q after 6 hours. What is the speed of

    1. 1. 50 km/h
    2. 2. 60 km/h
    3. 3. 70 km/h
    4. 4. 120 km/h

3
Let their speed be x km/hr and y km/he respectively. Then 120/(x + y) = 1 => x + y = 120 --- (i) Now when they move in same direction: (Distance traveled by P in 6 hrs) - (Distance traveled by Q in 6 hrs) = 120 km 6x - 6y = 120 => x - y = 20 --- (ii) Sloving (i) and (ii) we get x = 70 y = 50 P's speed = 70 km/hr.

Q.- In a 1000 m race A beats B by 50 m and B beats C by 100 m. In the same race by how many meters does A beat C?

    1. 1. 145
    2. 2. 150
    3. 3. 155
    4. 4. 160

1
By the time A covers 1000 m B covers (1000 - 50) = 950 m. By the time B covers 1000 m C covers (1000 - 100) = 900 m. So the ratio of speeds of A and C = 1000/950 * 1000/900 = 1000/855 So by the time A covers 1000 m C covers 855 m. So in 1000 m race A beats C by 1000 - 855 = 145 m.

Q.- A person borrows Rs. 5000 for 2 years at 4% p.a. simple interest. He immediately lends it to another person at 25/4 p.a for 2 years. Find his gain in the transaction per year.

    1. 1. Rs. 112.50
    2. 2. Rs. 112.05
    3. 3. Rs. 11.205
    4. 4. Rs. 112.75

1
Gain in 2 years = Rs. 5000 x (25/4) x (2/100) - [(5000 x 4 x 2)/100] = Rs. (625 - 400) = Rs. 225 Gain in 1 year = Rs. 225/2 = Rs. 112.50

Q.- A certain amount earns simple interest of Rs. 1750 after 7 years. Had the interest been 2% more how much more interest would it have earned ?

    1. 1. Rs. 35
    2. 2. none
    3. 3. Rs. 125
    4. 4. Rs. 50

2
We need to know the S.I. principal and time to find the rate. Since the principal is not given so data is inadequate.

Q.- What will be the ratio of simple interest earned by certain amount at the same rate of interest for 6 years and that for 9 years ?

    1. 1. 0.086111111111
    2. 2. 0.16805555556
    3. 3. 0.085416666667
    4. 4. 0.12638888889

3
Let the principal be P and rate of interest be R%. Required ratio = [(P x R x 6) /100] / [(P x R x 9)/100 = 6PR/9PR = 6/9 = 2 : 3

Q.- A sum of money amounts to Rs. 9800 after 5 years and Rs. 12005 after 8 years at the same rate of simple interest. The Rate of Interest per annum is :

    1. 1. 0.18
    2. 2. 0.2
    3. 3. 0.21
    4. 4. 0.12

4
S.I. for 3 years = Rs. (12005 - 9800) = Rs. 2205. S.I. for 5 years = Rs. (2205/3) x 5 = Rs. 3675 Principal = Rs. (9800 - 3675) = Rs. 6125. Hence rate = (100 x 3675)/(6125 x 5) % = 12%

Q.- A man took loan from a bank at the rate of 12% p.a. Simple Interest. After 3 years he had to pay Rs. 5400 interest only for the period. The principal amount borrowed by him was :

    1. 1. Rs. 15000
    2. 2. Rs. 51000
    3. 3. Rs. 10500
    4. 4. Rs. 14000

1
Principal = Rs. (100 x 5400)/(12x3) = Rs. 15000

Q.- A lent Rs. 5000 to B for 2 years and Rs. 3000 to C for 4 years on simple interest at the same rate of interest and received Rs. 2200 in all from both of them as interest. The rate of interest per annum is :

    1. 1. 0.11
    2. 2. 0.1
    3. 3. 0.12
    4. 4. 0.14

2
Let the rate be R% p.a. Then [(5000 x R x 2)/100] + [(3000 x R x 4)/100] = 2200 100R + 120R = 2200 R = 2200/220 = 10 Rate = 10%

Q.- An automobile financier claims to be lending money at simple interest but he includes the interest every six months for calculating the principal. If he is charging an interest of 10% the effective rate of interest becomes :

    1. 1. 0.2501
    2. 2. 0.251
    3. 3. 0.1025
    4. 4. 0.1052

3
Let the sum be Rs. 100. Then S.I. for first 6 months = Rs. (100 x 10 x 1)/(100x2) = Rs. 5 S.I. for last 6 months = Rs. (105 x 10 x 1)/(100x2) = Rs. 5.25 So amount at the end of 1 year = Rs. (100 + 5 + 5.25) = Rs. 110.25 Effective rate = (110.25 - 100) = 10.25%

Q.- A sum of Rs. 12500 amounts to Rs. 15500 in 4 years at the rate of simple interest. What is the rate of interest ?

    1. 1. 0.09
    2. 2. 0.08
    3. 3. 0.07
    4. 4. 0.06

4
S.I. = Rs. (15500 - 12500) = Rs. 3000 Rate = (100 x 3000)/(12500x4) % = 6%

Q.- A sum of Rs. 725 is lent in the beginning of a year at a certain rate of interest. After 8 months a sum of Rs. 362.50 more is lent but at the rate twice the former. At the end of the year Rs. 33.50 is earned as interest from both the loans. What was the

    1. 1. 0.0436
    2. 2. 0.0634
    3. 3. 0.0346
    4. 4. 0.0364

3
Let the original rate be R%. Then new rate = (2R)% Note: Here original rate is for 1 year(s); the new rate is for only 4 months i.e. 1/3 year(s). (725 x R x 1 /100) + (362.50 x 2R x 1)/(100x3) = 33.50 (2175 + 725) R = 33.50 x 100 x 3 (2175 + 725) R = 10050 (2900)R = 10050 R = 10050 / 2900 = 3.46 Original rate = 3.46%

Q.- How much time will it take for an amount of Rs. 450 to yield Rs. 81 as interest at 4.5% per annum of simple interest ?

    1. 1. 5 years
    2. 2. 2 years
    3. 3. 3 years
    4. 4. 4 years

4
Time = (100 x 81 )/(450 x 4.5) years = 4 years.

Q.- A sum of money amounts to Rs 9800 after 5 years and Rs 12005 after 8 years at the same rate of simple interest. The rate of interest per annum is

    1. 1. 0.125
    2. 2. 0.1
    3. 3. 0.11
    4. 4. 0.12

4
We can get SI of 3 years = 12005 - 9800 = 2205 SI for 5 years = (2205/3)*5 = 3675 [so that we can get principal amount after deducting SI] Principal = 12005 - 3675 = 6125 So Rate = (100*3675)/(6125*5) = 12%

Q.- A financier claims to be lending money at simple interest But he includes the interest every six months for calculating the principal. If he is charging an interest of 10% the effective rate of interest becomes.

    1. 1. 0.2501
    2. 2. 0.251
    3. 3. 0.1025
    4. 4. 0.1052

3
Let the sum is 100. As financier includes interest every six months. then we will calculate SI for 6 months then again for six months as below: SI for first Six Months = (100*10*1)/(100*2) = Rs. 5 Important: now sum will become 100+5 = 105 SI for last Six Months = (105*10*1)/(100*2) = Rs. 5.25 So amount at the end of year will be (100+5+5.25) = 110.25 Effective rate = 110.25 - 100 = 10.25

Q.- In how many years Rs 150 will produce the same interest at 8% as Rs. 800 produce in 3 years at 9/2%

    1. 1. 9
    2. 2. 10
    3. 3. 12
    4. 4. 14

1
Firstly we need to calculate the SI with prinical 800Time 3 years and Rate 9/2% it will be Rs. 108 Then we can get the Time as Time = (100*108)/(150*8) = 9

Q.- At what rate percent per annum will the simple interest on a sum of money be 2/5 of the amount in 10 years

    1. 1. 0.03
    2. 2. 0.04
    3. 3. 0.045
    4. 4. 0.05

2
Let sum = x Time = 10 years. S.I = 2x /5 [as per question] Rate =( (100 * 2x) / (x*5*10))% => Rate = 4%

Q.- A sum of money at simple interest amounts to Rs. 2240 in 2 years and to Rs. 2600 in 5 years. What is the principal amount

    1. 1. 2200
    2. 2. 1800
    3. 3. 1500
    4. 4. 2000

4
SI for 3 year = 2600-2240 = 360 SI for 2 year 360/3 * 2 = 240 principal = 2240 - 240 = 2000

Q.- What is the present worth of Rs. 132 due in 2 years at 5% simple interest per annum

    1. 1. 102
    2. 2. 110
    3. 3. 120
    4. 4. 210

3
Let the present worth be Rs.x ThenS.I.= Rs.(132 - x) =› (x*5*2/100) = 132 - x =› 10x = 13200 - 100x =› 110x = 13200 x= 120

Q.- If a sum of money doubles itself in 8 years at simple interest the ratepercent per annum is

    1. 1. 12.5
    2. 2. 12
    3. 3. 14
    4. 4. 13.5

1
Let sum = x then Simple Interest = x Rate = (100 * x) / (x * 8) = 12.5

Q.- At 5% per annum simple interest Rahul borrowed Rs. 500. What amount will he pay to clear the debt after 4 years

    1. 1. 550
    2. 2. 400
    3. 3. 500
    4. 4. 600

4
We need to calculate the total amount to be paid by him after 4 years So it will be Principal + simple interest. So =>500+(500∗5∗4)/100=>Rs.600

Q.- Find the rate at Simple interest at which a sum becomes four times of itself in 15 years.

    1. 1. 0.2
    2. 2. 0.15
    3. 3. 0.25
    4. 4. 0.205

1
Let sum be x and rate be r% then (x*r*15)/100 = 3x [important to note here is that simple interest will be 3x not 4x beause 3x+x = 4x] => r = 20%

Q.- If A lends Rs. 3500 to B at 10% p.a. and B lends the same sum to C at 11.5% p.a. then the gain of B (in Rs.) in a period of 3 years is

    1. 1. 571.5
    2. 2. 157.5
    3. 3. 715.5
    4. 4. 155.7

2
We need to calculate the profit of B. It will be SI on the rate B lends - SI on the rate B gets Gain of B=[(3500×11.5×3)/100]−[(3500×10×3)/100]=157.50

Q.- Mr. Thomas invested an amount of Rs. 13900 divided in two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be Rs. 3508 what was the amount invested in sch

    1. 1. Rs.6400
    2. 2. Rs.6500
    3. 3. Rs.7200
    4. 4. Rs.7500

1
Let the sum invested in scheme A be Rs. x and that in scheme B be Rs. (13900 - x). Then (x * 14 * 2)/100 + [(13900 - x) * 11 * 2]/100 = 3508 28x - 22x = 350800 - (13900 * 22) 6x = 45000 => x = 7500 So sum invested in scheme B = (13900 - 7500) = Rs. 6400.

Q.- An amount of Rs. 100000 is invested in two types of shares. The first yields an interest of 9% p.a and the second 11% p.a. If the total interest at the end of one year is 9 3/4 % then the amount invested in each share was?

    1. 1. Rs. 52500; Rs. 47500
    2. 2. Rs. 62500; Rs. 37500
    3. 3. Rs. 72500; Rs. 27500
    4. 4. Rs. 82500; Rs. 17500

2
Let the sum invested at 9% be Rs. x and that invested at 11% be Rs. (100000 - x). Then (x * 9 * 1)/100 + [(100000 - x) * 11 * 1]/100 = (100000 * 39/4 * 1/100) (9x + 1100000 - 11x)/100 = 39000/4 = 9750 x = 62500 Sum invested at 9% = Rs. 62500 Sum invested at 11% = Rs. (100000 - 62500) = Rs. 37500.

Q.- David invested certain amount in three different schemes. A B and C with the rate of interest 10% p.a. 12% p.a. and 15% p.a. respectively. If the total interest accrued in one year was Rs. 3200 and the amount invested in scheme C was 150% of the amount in

    1. 1. Rs.5000
    2. 2. Rs.6000
    3. 3. Rs.7000
    4. 4. Rs.8000

1
Let x y and z be the amount invested in schemes A B and C respectively. Then (x * 10 * 1)/100 + (y * 12 * 1)/100 + (z * 15 * 1)/100 = 3200 10x + 12y + 15z = 320000 Now z = 240% of y = 12/5 y And z = 150% of x = 3/2 x x = 2/3 z = ( 2/3 * 12/5) y = 8/5 y 16y + 12y + 36y = 320000 y = 5000 Sum invested in scheme B = Rs. 5000.

Q.- A person invested in all Rs. 2600 at 4% 6% and 8% per annul simple interest. At the end of the year he got the same interest in all the three cases. The money invested at 4% is?

    1. 1. Rs.200
    2. 2. Rs.400
    3. 3. Rs.800
    4. 4. Rs.1200

4
Let the parts be x y and [2600 - (x + y)]. Then (x * 4 * 1)/100 = (y * 6 * 1)/100 = {[2600 - (x + y)] * 8 * 1}/100 y/x = 4/6 = 2/3 or y = 2/3 x So (x * 4 * 1)/100 = [(2600 - 5/3 x) * 80/100 52x = (7800 * 8) => x = 1200 Money invested at 4% = Rs. 1200.

Q.- At what rate percent per annum will the simple interest on a sum of money be 2/5 of the amount in 10 years?

    1. 1. 0.04
    2. 2. 0.0567
    3. 3. 0.06
    4. 4. 0.0867

1
Let sum = x. Then S.I. = 2x/5 Time = 10 years. Rate = (100 * 2x) / (x * 5 * 10) = 4%

Q.- A lent Rs. 5000 to B for 2 years and Rs. 3000 to C for 4 years on simple interest at the same rate of interest and received Rs. 2200 in all from both of them as interest. The rate of interest per annum is?

    1. 1. 0.05
    2. 2. 0.07
    3. 3. 0.1
    4. 4. 0.12

3
Let the rate be R% p.a. Then (5000 * R * 2)/100 + (3000 * R * 4)/100 = 2200 100 R + 120 R = 2200 R = 10%

Q.- A lends Rs. 2500 to B and a certain to C at the same time at 7% p.a. simple interest. If after 4 years A altogether receives Rs. 1120 as interest from B and C then the sum lent to C is?

    1. 1. Rs.750
    2. 2. Rs.1500
    3. 3. Rs.4500
    4. 4. Rs.6000

2
Let the sum lent to C be Rs. x. Then (2500 * 7 * 4) / 100 + (x * 7 * 4) / 100 = 1120 7/25 x = (1120 - 700) => x = 1500

Q.- In how much time would the simple interest on a certain sum be 0.125 times the principal at 10% per annum?

    1. 1. 1 1/4 years
    2. 2. 1 3/4 years
    3. 3. 2 1/4 years
    4. 4. 2 3/4 years

1
Let sum = x. Then S.I. = 0.125x = 1/8 x R = 10% Time = (100 * x) / (x * 8 * 10) = 5/4 = 1 1/4 years.

Q.- A sum of money becomes 7/6 of itself in 3 years at a certain rate of simple interest. The rate per annum is?

    1. 1. 0.0556
    2. 2. 0.0655
    3. 3. 0.18
    4. 4. 0.255

1
Let sum = x. Then amount = 7x/6 S.I. = 7x/6 - x = x/6; Time = 3 years. Rate = (100 * x) / (x * 6 * 3) = 5 5/9 %.

Q.- If the annual rate of simple interest increases from 10% to 12 1/2 % a man's yearly income increases by Rs. 1250. His principal in Rs. is?

    1. 1. 45000
    2. 2. 50000
    3. 3. 60000
    4. 4. 65000

2
Let the sum be Rs. x. Then (x * 25/2 * 1/100) - (x * 10 * 1)/100 = 1250 25x - 20x = 250000 x = 50000

Q.- The compound interest on Rs. 30000 at 7% per annum is Rs. 4347. The period (in years) is :

    1. 1. 4
    2. 2. 3
    3. 3. 1
    4. 4. 2

4
Amount = Rs. (30000 + 4347) = Rs. 34347. Let the time be n years. Then 30000[pow{1 + (7)/100)} n] n = 34347 pow(107 n) = 11449 = pow(107 2) n = 2 years.

Q.- The effective annual rate of interest corresponding to a nominal rate of 6% per annum payable half-yearly is :

    1. 1. 0.0906
    2. 2. 0.096
    3. 3. 0.0609
    4. 4. 0.069

3
Amount of Rs. 100 for 1 year when compounded half-yearly = Rs. 100 xpow[{1 + (3/100)} 2] = Rs. 106.09 Effective rate = (106.09 - 100)% = 6.09%

Q.- Albert invested an amount of Rs. 8000 in a fixed deposit scheme for 2 years at compound interest rate 5 p.c.p.a. How much amount will Albert get on maturity of the fixed deposit ?

    1. 1. Rs. 2088
    2. 2. Rs.8820
    3. 3. Rs. 8280
    4. 4. Rs. 8202

2
Amount= Rs. 8000 x pow{1 + (5/100)} 2] = Rs. 8000 x (21/20) x (21/20) = Rs. 8820.

Q.- The least number of complete years in which a sum of money put out at 20% compound interest will be more than doubled is :

    1. 1. 4
    2. 2. 6
    3. 3. 8
    4. 4. 5

1
Px pow{1 + (20/100) n} > 2P pow(6/5 n) > 2. Now (6/5) x (6/5) x (6/5) x (6/5) > 2. So n = 4 years.

Q.- What will be the compound interest on a sum of Rs. 25000 after 3 years at the rate of 12 p.c.p.a.?

    1. 1. Rs. 10312.20
    2. 2. Rs. 10123.20
    3. 3. Rs. 10123.02
    4. 4. 10132.2

2
Amount = Rs. 25000 xpow[ 1 + (12/100) 3] = Rs. 25000 x (28/25) x (28/25) x (28/25) = Rs. 35123.20 C.I. = Rs. (35123.20 - 25000) = Rs. 10123.20

Q.- Simple interest on a certain sum of money for 3 years at 8% per annum is half the compound interest on Rs. 4000 for 2 years at 10% per annum. The sum placed on simple interest is :

    1. 1. Rs. 5071
    2. 2. Rs. 5017
    3. 3. Rs. 1750
    4. 4. Rs. 1705

3
C.I. = Rs. 4000 xpow[ 1 + (10/100) 2] - 4000 = Rs. 4000 x (11/10) x (11/10) - 4000 = Rs. 840. Sum = Rs. 420 x 100/(3*8) = Rs. 1750.

Q.- If the simple interest on a sum of money for 2 years at 5% per annum is Rs. 50 what is the compound interest on the same at the same rate and for the same time ?

    1. 1. Rs. 25.15
    2. 2. Rs. 25.51
    3. 3. Rs. 51.52
    4. 4. Rs. 51.25

4
Sum = Rs. 50 x 100/(2x5) = Rs. 500. Amount = Rs. 500 x [pow( 1 + (5/100) 2] = Rs. 500 x (21/20) x (21/20) = Rs. 551.25 C.I. = Rs. (551.25 - 500) = Rs. 51.25

Q.- The difference between simple interest and compound on Rs. 1200 for one year at 10% per annum reckoned half-yearly is :

    1. 1. Rs. 3
    2. 2. Rs. 4
    3. 3. Rs. 3.5
    4. 4. Rs. 6

1
S.I. = Rs (1200 x 10 x 1)/100 = Rs. 120. C.I. = Rs. 1200 xpow[ 1 + (5/100) 2] - 1200 = Rs. 123. Difference = Rs. (123 - 120) = Rs. 3.

Q.- The difference between simple and compound interests compounded annually on a certain sum of money for 2 years at 4% per annum is Re. 1. The sum (in Rs.) is :

    1. 1. 652
    2. 2. 625
    3. 3. 562
    4. 4. 526

2
Let the sum be Rs. x. Then C.I. =[ x pow {1 + (4/100) 2} - x] = [ (676/625) x - x] = (51/625) x. S.I. = (x x 4 x 2)/100 = 2x/25 . (51x/625) - (2x/25) = 1 x = 625.

Q.- There is 60% increase in an amount in 6 years at simple interest. What will be the compound interest of Rs. 12000 after 3 years at the same rate ?

    1. 1. 7293
    2. 2. 7239
    3. 3. 3972
    4. 4. 3927

3
Let P = Rs. 100. Then S.I. Rs. 60 and T = 6 years. R = (100 x 60)/(100x6) = 10% p.a. Now P = Rs. 12000. T = 3 years and R = 10% p.a. C.I. = Rs. 12000 x[pow{1 + (10/100) 3}] - 1 = Rs. 12000 x (331/1000) = 3972.

Q.- A sum of money invested at compound interest to Rs. 800 in 3 years and to Rs 840 in 4 years. The rate on interest per annum is.

    1. 1. 0.05
    2. 2. 0.06
    3. 3. 0.07
    4. 4. 0.055

1
S.I. on Rs 800 for 1 year = 40 Rate = (100*40)/(800*1) = 5%

Q.- On a sum of money simple interest for 2 years is Rs 660 and compound interest is Rs 696.30 the rate of interest being the same in both cases.

    1. 1. 0.1175
    2. 2. 0.115
    3. 3. 0.11
    4. 4. 0.12

3
Difference between C.I and S.I for 2 years = 36.30 S.I. for one year = 330. S.I. on Rs 330 for one year = 36.30 So R% = \frac{100*36.30}{330*1} = 11%

Q.- Effective annual rate of interest corresponding to nominal rate of 6% per annum compounded half yearly will be

    1. 1. 0.069
    2. 2. 0.096
    3. 3. 0.0906
    4. 4. 0.0609

4
Let the amount Rs 100 for 1 year when compounded half yearly n = 2 Rate = 6/2 = 3% Amount=100[1+(3/100)][1+(3/100)]=106.09 Effective rate = (106.09 - 100)% = 6.09%

Q.- Find the compound interest on Rs. 7500 at 4% per annum for 2 years compounded annually.

    1. 1. Rs.612
    2. 2. Rs.621
    3. 3. Rs.216
    4. 4. Rs.261

1
Amount=7500×[1+(4/100)][1+(4/100)]=7500×(26/25)×(26/25)=8112 So compound interest = (8112 - 7500) = 612

Q.- Albert invested amount of 8000 in a fixed deposit for 2 years at compound interest rate of 5 % per annum. How much Albert will get on the maturity of the fixed deposit.

    1. 1. 2880
    2. 2. 2088
    3. 3. 8802
    4. 4. 8820

4
8000×[(1+(5/100)][1+(5/100)] =>8000×(21/20)×(21/20) =>8820

Q.- What will be the compound interest on Rs. 25000 after 3 years at the rate of 12 % per annum

    1. 1. 10123.2
    2. 2. 10123.02
    3. 3. 10132.2
    4. 4. 10132.02

1
25000×[1+(12/100)][1+(12/100)][1+(12/100)] =>25000×(28/25)×(28/25)×(28/25) =>35123.20 So Compound interest will be 35123.20 - 25000 = Rs 10123.20

Q.- A man saves Rs 200 at the end of each year and lends the money at 5% compound interest. How much will it become at the end of 3 years.

    1. 1. 662.3
    2. 2. 626.02
    3. 3. 662.2
    4. 4. 662.02

4
[200(21/20)×(21/20)×(21/20)]+[200(21/20)×(21/20)]+[200(2120)]=662.02

Q.- The present worth of Rs.169 due in 2 years at 4% per annum compound interest is

    1. 1. 165.52
    2. 2. 165.25
    3. 3. 156.25
    4. 4. 156.52

3
Amount=[169/{1+(4/100)}{1+(4/100)}] Amount=(169∗25∗25)/26∗26 Amount=156.25

Q.- Simple interest on a certain sum of money for 3 years at 8% per annum is half the compound interest on Rs. 4000 for 2 years at 10% per annum. The sum placed on simple interest is

    1. 1. 1705
    2. 2. 1750
    3. 3. 1507
    4. 4. 1570

2
C.I.={4000×square of[1+(10/100)]−4000} =[4000∗(11/10)∗(11/10)−4000] =840 So S.I. = 840/2=420 So Sum = (S.I.∗100)/R∗T =(420∗100)/3∗8 =Rs1750

Q.- If the simple interest on a sum of money for 2 years at 5% per annum is Rs.50. what will be the compound interest on same values

    1. 1. 55.21
    2. 2. 55.12
    3. 3. 51.52
    4. 4. 51.25

4
S.I.=(P∗R∗T)/100 P=(50∗100)/5∗2=500 Amount=500xsquare of [1+(5/100)]=[500(21/20)∗(21/20)] =551.25 C.I.=551.25−500=51.25

Q.- A sum of money is borrowed and paid back in two annual installments of Rs. 882 each allowing 5% C.I. The sum borrowed was?

    1. 1. Rs.1620
    2. 2. Rs.1640
    3. 3. Rs.1680
    4. 4. Rs.1700

2
Principal = (P.W. of Rs. 882 due 1 year hence) + (P.W. of Rs. 882 due 2 years hence) = [882/(1 + 5/100) + 882/pow(1 + 5/1002)] = (882 * 20)/21 + (882 * 400)/441 = Rs. 1640.

Q.- A sum of money placed at C.I. interest doubles itself in 5 years. It will amount to eight times itself at the same rate of interest in?

    1. 1. 7 yrs
    2. 2. 10 yrs
    3. 3. 15 yrs
    4. 4. 20 yrs

3
P*pow(1 + R/1005) = 2P => pow(1 + R/1005) = 2 Let P*pow(1 + R/100n) = 8P => pow(1 + R/100n) = 8 = pow(23) = pow{ pow(1 + R/100 5 } 3 } => pow(1 + R/100n) = pow(1 + R/10015) => n = 15 Required time = 15 years.

Q.- The difference between C.I. and S.I. on an amount of Rs. 15000 for 2 years is Rs. 96. What is the rate of interest per annum?

    1. 1. 3
    2. 2. 8
    3. 3. 10
    4. 4. 12

2
[15000 * pow(1 + R/1002) - 15000] - (15000 * R * 2)/100 = 96 15000[pow(1 + R/1002) - 1 - 2R/100] = 96 15000[pow(100 + R2) - 10000 - 200R]/10000 = 96 pow(R2) = (96 * 2)/3 = 64 => R = 8 Rate = 8%

Q.- The least number of complete years in which a sum of money put out at 20% C.I. will be more than doubled is?

    1. 1. 3
    2. 2. 4
    3. 3. 5
    4. 4. 6

2
P*pow(1 + 20/100n) > 2P or pow(6/5n)> 2 Now (6/5 * 6/5 * 6/5 * 6/5) > 2. So n = 4 years.

Q.- What annual payment will discharge a debt of Rs. 1025 due in 2 years at the rate of 5% compound interest?

    1. 1. Rs.550.50
    2. 2. Rs.551.25
    3. 3. Rs. 560.50
    4. 4. Rs. 560.75

2
Let each installment be Rs. x. Then x/(1 + 5/100) + x/pow(1 + 5/1002) = 1025 820x + 1025 * 441 x = 551.25 So value of each installment = Rs. 551.25

Q.- The compound interest on Rs. 30000 at 7% per annum is Rs. 4347. The period(in years) is?

    1. 1. 2
    2. 2. 3
    3. 3. 4
    4. 4. 5

1
Amount = (30000 + 4347) = Rs. 34347 Let the time be n years. Then 30000*pow(1 + 7/100n) = 34347 = pow(107/100n) = 34347/30000 = pow(107/1002) n = 2 years.

Q.- If the C.I. on a sum for 2 years at 12 1/2 % per annum is Rs. 510 the S.I. on the same sum at the same rate for the same period of time is?

    1. 1. Rs.400
    2. 2. Rs.450
    3. 3. Rs.460
    4. 4. Rs.480

4
Let the sum be Rs. P. Then [P*pow{1 + 25/(2 * 100)2} - P] = 510 P[pow{9/82} - 1] = 510. Sum = Rs. 1920 So S.I. = (1920 * 25 * 2) / (2 * 100) = Rs. 480

Q.- Find the C.I. on a sum of Rs.1600 for 9 months at 20% per annum interest being compounded quarterly?

    1. 1. Rs.17684
    2. 2. Rs.1684
    3. 3. Rs.2522
    4. 4. Rs.3408

3
A = 1600*pow{(21/20)3} = 2522

Q.- The C.I. on a certain sum for 2 years Rs.41 and the simple interest is Rs.40. What is the rate percent?

    1. 1. 0.04
    2. 2. 0.05
    3. 3. 0.06
    4. 4. 0.08

2
SI = 20 + 20 CI = 20 + 21 20 ------ 1 100 ---- ? => 5%

Q.- Simran started a software business by investing Rs. 50000. After six months Nanda joined her with a capital of Rs. 80000. After 3 years they earned a profit of Rs. 24500. What was Simran's share in the profit ?

    1. 1. Rs. 10500
    2. 2. Rs. 15000
    3. 3. Rs. 50001
    4. 4. Rs. 11000

1
Simran : Nanda = (50000 x 36) : (80000 x 30) = 3 : 4 Simran's share = Rs. 24500 x (3/7) = Rs. 10500.

Q.- Arun Kamal and Vinay invested Rs. 8000 Rs. 4000 and Rs. 8000 respectively in a business. Arun left after six months. If after eight months there was a gain of Rs. 4005 then what will be the share of Kamal ?

    1. 1. Rs.980
    2. 2. Rs. 890
    3. 3. Rs. 908
    4. 4. Rs. 809

2
Arun : Kamal : Vinay = (8000 x 6) : (4000 x 8) : (8000 x 8) = 48 : 32 : 64 = 3 : 2 : 4 Kamal's share = Rs. 4005 x (2/9) = Rs. 890

Q.- Aman started a business investing Rs. 70000. Rakhi joined him after six months with an amount of Rs.. 105000 and Sagar joined them with Rs. 1.4 lakhs after another six months. The amount of profit earned should be distributed in what ratio among Aman Rakh

    1. 1. 0.63625
    2. 2. 0.51128472222
    3. 3. 0.51060185185
    4. 4. 0.63351851852

3
Aman : Rakhi : Sagar = (70000 x 36) : (105000 x 30) : (140000 x 24) = 12 : 15 : 16.

Q.- A began a business with Rs. 85000. He was joined afterwards by B with Rs. 42500. For how much period does B join if the profits at the end of the year are divided in the ratio of 3 : 1 ?

    1. 1. 9 months
    2. 2. 6 months
    3. 3. 7 months
    4. 4. 8 months

4
Suppose B joined for x months.Then (85000 x 12)/(42500 x X) = 3/1 X = (85000 x 12)/(42500 x 3) = 8. So B joined for 8 months.

Q.- A and B started a business in partnership investing Rs. 20000 and Rs. 15000 respectively. After six months C joined them with Rs. 20000. What will be B's share in total profit of Rs. 25000 earned at the end of 2 years from the starting of the business ?

    1. 1. Rs. 7500
    2. 2. Rs. 5700
    3. 3. Rs 7005
    4. 4. Rs. 5007

1
A : B : C = (20000 x 24) : (15000 x 24) : (20000 x 18) = 4 : 3 : 3. B's share = Rs. 25000 x (3/10) = Rs. 7500.

Q.- A B C rent a pasture. A puts 10 oxen for 7 months B puts 12 oxen for 5 months and C puts 15 oxen for 3 months for grazing. If the rent of the pasture is Rs. 175 how much must C pay as his share of rent ?

    1. 1. Rs. 50
    2. 2. Rs. 45
    3. 3. Rs. 55
    4. 4. Rs. 35

2
A : B : C = (10 x 7) : (12 x 5) : (15 x 3) = 70 : 60 : 45 = 14 : 12 : 9. C's rent = Rs. 175 x (9/35) = Rs. 45

Q.- A and B started a partnership business investing some amount in the ratio of 3 : 5. C joined then after six months with an amount equal to that of B. In what proportion should the profit at the end of one year be distributed among A B and C ?

    1. 1. 0.25358796296
    2. 2. 0.4208912037
    3. 3. 0.25700231481
    4. 4. 0.4208912037

3
Let the initial investments of A and B be 3x and 5x. A : B : C = (3x x 12) : (5x x 12) : (5x x 6) = 36 : 60 : 30 = 6 : 10 : 5.

Q.- A starts business with Rs. 3500 and after 5 months B joins with A as his partner. After a year the profit is divided in the ratio 2 : 3. What is B's contribution in the capital ?

    1. 1. 6000
    2. 2. 7000
    3. 3. 8000
    4. 4. 9000

4
Let B's capital be Rs. x. Then (3500 x 12)/7x = 2/3 14x = 126000 x = 9000.

Q.- A B C subscribe Rs. 50000 for a business. A subscribes Rs. 4000 more than B and B Rs. 5000 more than C. Out of a total profit of Rs. 35000 A receives :

    1. 1. Rs. 14700
    2. 2. Rs. 71400
    3. 3. Rs. 14007
    4. 4. Rs. 74100

1
Let C = x. Then B = x + 5000 and A = x + 5000 + 4000 = x + 9000 So x + x + 5000 + x + 9000 = 50000 3x = 36000 x = 12000 A : B : C = 21000 : 17000 : 12000 = 21 : 17 : 12. A's share = Rs. 35000 x (21/50) = Rs. 14700.

Q.- A B and C jointly thought of engaging themselves in a business venture. It was agreed that A would invest Rs. 6500 for 6 months B Rs. 8400 for 5 months and C Rs. 10000 for 3 months. A wants to be the working member for which he was to receive 5% of the pr

    1. 1. Rs. 6206
    2. 2. Rs.6260
    3. 3. Rs. 2606
    4. 4. Rs. 2660

4
For managing A received = 5% of Rs. 7400 = Rs. 370 Balance = Rs. (7400 - 370) = Rs. 7030. Ratio of their investments = (6500 x 6) : (8400 x 5) : (10000 x 3) = 39000 : 42000 : 30000 = 13 : 14 : 10 B's share = Rs. 7030 x (14/37) = Rs. 2660.

Q.- A and B entered into a partnership investing Rs. 16000 and Rs. 12000 respectively. After 3 months A withdrew Rs. 5000 while B invested Rs. 5000 more. After 3 more months C joins the business with a capital of Rs. 21000. The share of B exceeds that of C ou

    1. 1. Rs.6003
    2. 2. Rs.6300
    3. 3. Rs.3600
    4. 4. Rs.3006

3
A:B: C = 16000*3 + 11000*9:12000*3 + 17000*9:21000*6 = 147:189:126 = 7:9:6 Difference of B and C's shares = Rs. [26400 * (9/22) — 26400 * (6/22)) = Rs. 3600

Q.- Arun Kamal and Vinay invested Rs. 8000 Rs. 4000 and Rs. 8000 respectively in a business. Arun left after six months. If after eight months there was a gain of Rs. 4005 then what will be the share of Kamal?

    1. 1. Rs.980
    2. 2. Rs.908
    3. 3. Rs.890
    4. 4. Rs.809

3
Arun : Kamal : Vinay = (8000 x 6) : (4000 x 8) : (8000 x 8) = 48 : 32 : 64 = 3 : 2 : 4 Kamal's share=4005∗(2/9)=Rs 890

Q.- P and Q started a business investing Rs 85000 and Rs 15000 resp. In what ratio the profit earned after 2 years be divided between P and Q respectively.

    1. 1. 0.13680555556
    2. 2. 0.71041666667
    3. 3. 0.17847222222
    4. 4. 0.71111111111

2
In this type of question as time frame for both investors is equal then just get the ratio of their investments. P:Q = 85000:15000 = 85:15 = 17:3

Q.- In business A and C invested amounts in the ratio 2:1 whereas the ratio between amounts invested by A and B was 3:2. If Rs 157300 was their profit how much amount did B receive.

    1. 1. Rs.44008
    2. 2. Rs.44800
    3. 3. Rs.48004
    4. 4. Rs.48400

4
A:B = 3:2 = 6:4 => A:C = 2:1 = 6:3 => A:B:C = 6:4:3 B share = (4/13)*157300 = 48400

Q.- Yogesh started a business investing Rs. 45000. After 3 months Pranab joined him with a capital of Rs. 60000. After another 6 months Atul joined them with a capital of Rs. 90000. At the end of the year they made a profit of Rs. 20000. What would be Atuls s

    1. 1. Rs.3000
    2. 2. Rs.6000
    3. 3. Rs.5000
    4. 4. Rs.4000

4
Just take care of the months of investment rest all will be simple. Yogesh:Pranab:Atul = 45000*12:60000*9:90000*3 = 2:2:1 Atul's share = Rs. 20000 * (1/5) = Rs. 4000

Q.- Manoj received Rs. 6000 as his share out of the total profit of Rs. 9000 which he and Ramesh earned at the end of one year. If Manoj invested Rs.120000 for 6 months whereas Ramesh invested his amount for the whole year what was the amount invested by Rame

    1. 1. Rs.4500
    2. 2. Rs.3500
    3. 3. Rs.6000
    4. 4. Rs.5000

4
Suppose Ramesh invested Rs. x. Then Manoj : Ramesh = 20000 * 6 : x * 12. 120000/12x : 6000/3000 => x = 120000/24 = 5000

Q.- A B and C invested Rs. 8000 Rs. 4000 and Rs. 8000 respectively. in a business. A left after six months. If after eight months there was a gain of Rs. 4005 then what will be the share of B?

    1. 1. Rs.980
    2. 2. Rs.908
    3. 3. Rs.890
    4. 4. Rs.809

3
A:B:C = (8000*6):(4000*8):(8000*8) = 48:32:64 = 3:2:4 So B share = (2/9)*4005 = Rs 890

Q.- A started a business with Rs.21000 and is joined afterwards by B with Rs.36000. After how many months did B join if the profits at the end of the year are divided equally?

    1. 1. 4
    2. 2. 5
    3. 3. 8
    4. 4. 6

2
Suppose B joined after x months then 21000*12=36000*(12-x) => 36x = 180 => x = 5

Q.- Anand and Deepak started a business investing Rs.22500 and Rs.35000 respectively. Out of a total profit of Rs. 13800. Deepak's share is

    1. 1. Rs.4080
    2. 2. Rs.4008
    3. 3. Rs.8004
    4. 4. Rs.8400

4
Ratio of their shares = 22500 : 35000 =9 : 14 Deepak's share = Rs.(13800×14/23) = Rs. 8400

Q.- Rs. 700 is divided among A B C so that A receives half as much as B and B half as much as C. Then C's share is

    1. 1. 350
    2. 2. 300
    3. 3. 400
    4. 4. 350

3
Let C = x. Then B = x/2 and A = x/4 A:B:C = 1:2:4. C's share Rs.[(4/7)*700) = 400

Q.- A and B started business in partnership investing Rs. 20000 and Rs. 15000 respectively. After six months C joined them with Rs. 20000. What will be B's share in the total profit of Rs. 25000 earned at the end of 2 years from the starting of the business?

    1. 1. Rs.7500
    2. 2. Rs.8000
    3. 3. Rs.9000
    4. 4. None of these

1
A:B:C = (20000 * 24) : (15000 * 24) : (20000 * 18) = 4:3:3 B's share = 25000 * 3/10 = Rs. 7500.

Q.- Aman started a business investing Rs. 70000. Rakhi joined him after six months with an amount of Rs. 105000 and Sagar joined them with Rs. 1.4 lakhs after another six months. The amount of profit earned should be distributed in what ratio among Aman Rakhi

    1. 1. 0.21258101852
    2. 2. 0.51060185185
    3. 3. 0.28189814815
    4. 4. None of these

2
Aman : Rakhi : Sagar = (70000 * 36) : (105000 * 30) : (140000 * 24) = 12:15:16

Q.- X and Y started a business with capitals Rs. 20000 and Rs. 25000. After few months Z joined them with a capital of Rs. 30000. If the share of Z in the annual profit of Rs. 50000 is Rs. 14000 then after how many months from the beginning did Z join?

    1. 1. 3
    2. 2. 4
    3. 3. 5
    4. 4. 6

3
Investments of X Y and Z respectively are Rs. 20000 Rs. 25000 and Rs. 30000 Let investment period of Z be x months. Ratio of annual investments of X Y and Z is (20000 * 12) : (25000 * 12) : (30000 * x) = 240 : 300 : 30x = 8 : 10 : x The share of Z in the annual profit of Rs. 50000 is Rs. 14000. => [x/ (18 + x)] 50000 = 14000 => [x/ (18 + x)] 25 = 7 => 25x = 7x + (18 * 7) => x = 7 months. Z joined the business after (12 - 7) months. i.e. 5 months.

Q.- A B C enter into a partnership investing Rs. 35000 Rs. 45000 and Rs. 55000 respectively. The respective shares of A B C in annual profit of Rs. 40500 are:

    1. 1. Rs. 10500 Rs. 13500 Rs. 16500
    2. 2. Rs. 11500 Rs. 13000 Rs. 16000
    3. 3. Rs. 11000 Rs. 14000 Rs. 15500
    4. 4. Rs. 11500 Rs. 12500 Rs. 16500

1
A:B:C = 35000 : 45000 : 55000 = 7:9:11 A's share = 40500 * 7/27 = Rs. 10500 B's share = 40500 * 9/27 = Rs. 13500 C's share = 40500 * 11/27 = Rs. 16500

Q.- A B and C started a business with capitals of Rs. 8000 Rs. 10000 and Rs. 12000 respectively. At the end of the year the profit share of B is Rs. 1500. The difference between the profit shares of A and C is?

    1. 1. Rs.300
    2. 2. Rs.400
    3. 3. Rs.500
    4. 4. Rs.600

4
Ratio of investments of A B and C is 8000 : 10000 : 12000 = 4 : 5 : 6 And also given that profit share of B is Rs. 1500 => 5 parts out of 15 parts is Rs. 1500 Now required difference is 6 - 4 = 2 parts Required difference = 2/5 (1500) = Rs. 600

Q.- P and Q started a business with respective investments of Rs. 4 lakhs and Rs. 10 lakhs. As P runs the business his salary is Rs. 5000 per month. If they earned a profit of Rs. 2 lakhs at the end of the year then find the ratio of their earnings?

    1. 1. 0.042361111111
    2. 2. 0.086805555556
    3. 3. 0.12638888889
    4. 4. 0.20972222222

1
Ratio of investments of P and Q is 2 : 5 Total salary claimed by P = 12 * 5000 = Rs. 60000 Total profit = Rs. 2 lakhs. Profit is to be shared = Rs. 140000 Share of P = (2/7) * 140000 = Rs. 400000 Share of Q = Rs. 100000 Total earnings of P = (60000 + 40000) = Rs. 100000 Ratio of their earnings = 1 : 1

Q.- A and B invests Rs.10000 each A investing for 8 months and B investing for all the 12 months in the year. If the total profit at the end of the year is Rs.25000 find their shares?

    1. 1. Rs.8000 Rs.17000
    2. 2. Rs.9000 Rs.16000
    3. 3. Rs.18000 Rs.7000
    4. 4. Rs.10000 Rs.15000

4
The ratio of their profits A:B = 8:12 = 2:3 Share of A in the total profit = 2/5 * 25000 = Rs.10000 Share of B in the total profit = 3/5 * 25000 = Rs.15000

Q.- The ratio of investments of two partners P and Q is 7:5 and the ratio of their profits is 7:10. If P invested the money for 5 months find for how much time did Q invest the money?

    1. 1. 7 months
    2. 2. 9 months
    3. 3. 10 months
    4. 4. 11 months

3
7*5: 5*x = 7:10 x = 10

Q.- A and B rent a pasture for 10 months. A put in 80 cows for 7 months. How many can B put in for the remaining 3 months if he pays half as much again as A?

    1. 1. 120
    2. 2. 180
    3. 3. 200
    4. 4. 280

4
80* 7: x* 3 = 1:1 1/2 560: 3x = 2: 3 x = 280

Q.- A and B invests Rs.3000 and Rs.4000 respectively in a business. If A doubles his capital after 6 months. In what ratio should A and B divide that year's profit?

    1. 1. 0.38055555556
    2. 2. 0.38194444444
    3. 3. 0.33958333333
    4. 4. 0.42291666667

1
(3*6 + 6*6): (4*12) 54:48 => 9:8

Q.- January 1 2007 was Monday. What day of the week lies on Jan. 1 2008?

    1. 1. Tueday
    2. 2. Sunday
    3. 3. Monday
    4. 4. Friday

1
The year 2007 is an ordinary year. So it has 1 odd day 1st day of the year 2007 was Monday 1st day of the year 2008 will be 1 day beyond Monday Hence it will be Tuesday

Q.- January 1 2008 is Tuesday. What day of the week lies on Jan 1 2009?

    1. 1. Monday
    2. 2. Thursday
    3. 3. Sunday
    4. 4. Saturday

2
The year 2008 is a leap year. So it has 2 odd days. 1st day of the year 2008 is Tuesday (Given) So 1st day of the year 2009 is 2 days beyond Tuesday. Hence it will be Thursday.

Q.- On 8th Dec 2007 Saturday falls. What day of the week was it on 8th Dec 2006?

    1. 1. Friday
    2. 2. Monday
    3. 3. Tuesday
    4. 4. Sunday

1
The year 2006 is an ordinary year. So it has 1 odd day. So the day on 8th Dec 2007 will be 1 day beyond the day on 8th Dec 2006. But 8th Dec 2007 is Saturday. 8th Dec 2006 is Friday.

Q.- The calendar for the year 2007 will be the same for the year :

    1. 1. 2015
    2. 2. 2018
    3. 3. 2016
    4. 4. 2017

2
Count the number of odd days from the year 2007 onwards to get the sum equal to 0 odd day. Year : 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 2017 Odd day : 1 2 1 1 1 2 1 1 1 2 1 Sum = 14 odd days 0 odd days. Calendar for the year 2018 will be the same as for the year 2007.

Q.- Which of the following is not a leap year?

    1. 1. 800
    2. 2. 900
    3. 3. 600
    4. 4. 700

4
The century divisible by 400 is a leap year. The year 700 is not a leap year.

Q.- On 8th Feb 2005 it was Tuesday. What was the day of the week on 8th Feb 2004?

    1. 1. Monday
    2. 2. Thursday
    3. 3. Sunday
    4. 4. Friday

3
The year 2004 is a leap year. It has 2 odd days. The day on 8th Feb 2004 is 2 days before the day on 8th Feb 2005. Hence this day is Sunday.

Q.- How many days are there in x weeks x days?

    1. 1. 6x
    2. 2. 8x
    3. 3. 7x
    4. 4. 5x

2
x weeks x days = (7x + x) days = 8x days.

Q.- If 6th March 2005 is Monday what was the day of the week on 6th March 2004?

    1. 1. Monday
    2. 2. Thursday
    3. 3. Sunday
    4. 4. Friday

3
The year 2004 is a leap year. So it has 2 odd days. But Feb 2004 not included because we are calculating from March 2004 to March 2005. So it has 1 odd day only. The day on 6th March 2005 will be 1 day beyond the day on 6th March 2004. Given that 6th March 2005 is Monday. 6th March 2004 is Sunday (1 day before to 6th March 2005).

Q.- Today is Monday. After 61 days it will be :

    1. 1. Saturday
    2. 2. Sundat
    3. 3. Monday
    4. 4. Thursday

1
Each day of the week is repeated after 7 days. So after 63 days it will be Monday. After 61 days it will be Saturday.

Q.- It was Sunday on Jan 1 2006. What was the day of the week Jan 1 2010?

    1. 1. Thursday
    2. 2. Monday
    3. 3. Sunday
    4. 4. Friday

4
On 31st December 2005 it was Saturday. Number of odd days from the year 2006 to the year 2009 = (1 + 1 + 2 + 1) = 5 days. On 31st December 2009 it was Thursday. Thus on 1st Jan 2010 it is Friday.

Q.- If the seventh day of a month is three days earlier than Friday. What day will it be on the nineteenth day of the month?

    1. 1. Thursday
    2. 2. Friday
    3. 3. Monday
    4. 4. Sunday

4
Given that seventh day of a month is three days earlier than Friday => Seventh day is Tuesday => 14th is Tuesday => 19th is Sunday

Q.- If 25th of August in a year is Thursday the number of Mondays in that month is

    1. 1. 7
    2. 2. 4
    3. 3. 5
    4. 4. 6

3
Given that 25th August = Thursday Hence 29th August = Monday So 22nd15th and 8th and 1st of August also will be Mondays Number of Mondays in August = 5

Q.- How many days will there be from 26th January1996 to 15th May1996(both days included)?

    1. 1. 115
    2. 2. 111
    3. 3. 121
    4. 4. 110

2
Number of days from 26-Jan-1996 to 15-May-1996 (both days included) = 6(Jan) + 29(Feb) + 31 (Mar) + 30(Apr)+ 15(May) = 111

Q.- The second day of a month is Friday. What will be the last day of the next month which has 31 days?

    1. 1. Data inadequate
    2. 2. Friday
    3. 3. Monday
    4. 4. Sunday

1
We cannot find out the answer because the number of days of the current month is not given.

Q.- Arun went for a movie nine days ago. He goes to watch movies only on Thursdays. What day of the week is today?

    1. 1. Sunday
    2. 2. Monday
    3. 3. Tuesday
    4. 4. Saturday

4
Clearly it can be understood from the question that 9 days ago was a Thursday Number of odd days in 9 days = 2 (As 9-7 = 2 reduced perfect multiple of 7 from total days) Hence today = (Thursday + 2 odd days) = Saturday

Q.- If 1st October is Sunday then 1st November will be

    1. 1. Sunday
    2. 2. Wednesday
    3. 3. Monday
    4. 4. Tuesday

2
Given that 1st October is Sunday Number of days in October = 31 31 days = 3 odd days (As we can reduce multiples of 7 from odd days which will not change anything) Hence 1st November = (Sunday + 3 odd days) = Wednesday

Q.- If the first day of a year (other than leap year) was Friday then which was the last day of that year?

    1. 1. Thursday
    2. 2. Monday
    3. 3. Friday
    4. 4. Sunday

3
Given that first day of a normal year was Friday Odd days of the mentioned year = 1 (Since it is an ordinary year) Hence First day of the next year = (Friday + 1 Odd day) = Saturday Therefore last day of the mentioned year = Friday

Q.- Today is Thursday. The day after 59 days will be?

    1. 1. Sunday
    2. 2. Monday
    3. 3. Tuesday
    4. 4. Friday

1
59 days = 8 weeks 3 days = 3 odd days Hence if today is Thursday After 59 days it will be = (Thursday + 3 odd days) = Sunday

Q.- 8th Dec 2007 was Saturday.what day of the week was it on 8th Dec 2006?

    1. 1. Thursday
    2. 2. Sunday
    3. 3. Friday
    4. 4. Monday

3
Given that 8th Dec 2007 was Saturday Number of days from 8th Dec 2006 to 7th Dec 2007 = 365 days 365 days = 1 odd day Hence 8th Dec 2006 was = (Saturday - 1 odd day) = Friday

Q.- Which of the following is not a leap year?

    1. 1. 900
    2. 2. 1200
    3. 3. 700
    4. 4. 800

3
Remember the leap year rule (Given in the formulas) 1. Every year divisible by 4 is a leap year if it is not a century. 2. Every 4th century is a leap year but no other century is a leap year. 8001200 and 2000 comes in the category of 4th century (such as 400800120016002000 etc). Hence 8001200 and 2000 are leap years 700 is not a 4th century but it is a century. Hence it is not a leap year

Q.- What day of the week does May 28 2006 fall on

    1. 1. Saturday
    2. 2. Monday
    3. 3. Sunday
    4. 4. Thursday

3
28th May 2006 = (2005 years + period from 1-Jan-2006 to 28-May-2006) We know that number of odd days in 400 years = 0 Hence the number of odd days in 2000 years = 0 (Since 2000 is a perfect multiple of 400) Number of odd days in the period 2001-2005 = 4 normal years + 1 leap year = 4 x 1 + 1 x 2 = 6 Days from 1-Jan-2006 to 28-May-2006 = 31 (Jan) + 28 (Feb) + 31 (Mar) + 30 (Apr) + 28(may) = 148 148 days = 21 weeks + 1 day = 1 odd day Total number of odd days = (0 + 6 + 1) = 7 odd days = 0 odd day 0 odd day = Sunday Hence May 28 2006 is Sunday.

Q.- On what dates of April 2001 did Wednesday fall?

    1. 1. 2nd 9th 16th 23rd
    2. 2. 4th 11th 18th 25th
    3. 3. 3rd 10th 17th 24th
    4. 4. 1st 8th 15th 22nd 29th

2
We need to find out the day of 01-Apr-2001 01-Apr-2001 = (2000 years + period from 1-Jan-2001 to 01-Apr-2001) We know that number of odd days in 400 years = 0 H+K234ence the number of odd days in 2000 years = 0 (Since 2000 is a perfect multiple of 400) Days from 1-Jan-2001 to 01-Apr-2001 = 31 (Jan) + 28 (Feb) + 31 (Mar) + 1(Apr) = 91 91 days = 13 weeks = 0 odd day Total number of odd days = (0 + 0) = 0 odd days 0 odd day = Sunday. Hence 01-Apr-2001 is Sunday. Hence first Wednesday of Apr 2011 comes in 04th and successive Wednesdays come in 11th 18th and 25th

Q.- The calendar for the year 2007 will be the same for the year

    1. 1. 2015
    2. 2. 2018
    3. 3. 2016
    4. 4. 2014

2
For a year to have the same calendar with 2007 the total odd days from 2007 should be 0. Year2 0072008200920102011201220132014201520162017 Odd Days:12111211121 respectively. Take the year 2014 given in the choice. Total odd days in the period 2007-2013 = 5 normal years + 2 leap year = 5 x 1 + 2 x 2 = 9 odd days = 2 odd day (As we can reduce multiples of 7 from odd days which will not change anything) Take the year 2016 given in the choice. Number of odd days in the period 2007-2015 = 7 normal years + 2 leap year = 7 x 1 + 2 x 2 = 11 odd days = 4 odd days (Even if the odd days were 0 calendar of 2007 will not be same as the calendar of 2016 because 2007 is not a leap year whereas 2016 is a leap year. In fact you can straight away ignore this choice due to this fact without even bothering to check the odd days) Take the year 2017 given in the choice. Number of odd days in the period 2007-2016 = 7 normal years + 3 leap year = 7 x 1 + 3 x 2 = 13 odd days = 6 odd days Take the year 2018 given in the choice. Number of odd days in the period 2007-2017 = 8 normal years + 3 leap year = 8 x 1 + 3 x 2 = 14 odd days = 0 odd day (As we can reduce multiples of 7 from odd days which will not change anything) Also both 2007 and 2018 are not leap years. Since total odd days in the period 2007-2017 = 0 and both 2007 and 2018 are of same type 2018 will have the same calendar as that of 2007

Q.- The last day of a century cannot be

    1. 1. Monday
    2. 2. Wednesday
    3. 3. Tuesday
    4. 4. Friday

3
We know that number of odd days in 100 years = 5 Hence last day of the first century is Friday Number of odd days in 200 years = 5 x 2 = 10 = 3 (As we can reduce multiples of 7 from odd days which will not change anything) Hence last day of the 2nd century is Wednesday Number of odd days in 300 years = 5 x 3 = 15 = 1 Hence last day of the 3rd century is Monday We know that umber of odd days in 400 years = 0. (? 5 x 4 + 1 = 21 = 0) Hence last day of the 4th century is Sunday Now this cycle will be repeated. Hence last day of a century will not be Tuesday or Thursday or Saturday. its better to learn this by heart which will be helpful to save time in objective type exams

Q.- What day of the week was 1 January 1901

    1. 1. Monday
    2. 2. Tuesday
    3. 3. Saturday
    4. 4. Friday

2
1 Jan 1901 = (1900 years + 1st Jan 1901) We know that number of odd days in 400 years = 0 Hence the number of odd days in 1600 years = 0 (Since 1600 is a perfect multiple of 400) Number of odd days in the period 1601-1900 = Number of odd days in 300 years = 5 x 3 = 15 = 1 (As we can reduce perfect multiples of 7 from odd days without affecting anything) 1st Jan 1901 = 1 odd day Total number of odd days = (0 + 1 + 1) = 2 2 odd days = Tuesday Hence 1 January 1901 is Tuesday.

Q.- What day of the week will 22 Apr 2222 be?

    1. 1. Monday
    2. 2. Tuesday
    3. 3. Sunday
    4. 4. Thursday

1
22 Apr 2222 = (2221 years + period from 1-Jan-2222 to 22-Apr-2222) We know that number of odd days in 400 years = 0 Hence the number of odd days in 2000 years = 0 (Since 2000 is a perfect multiple of 400) Number of odd days in the period 2001-2200 = Number of odd days in 200 years = 5 x 2 = 10 = 3 (As we can reduce perfect multiples of 7 from odd days without affecting anything) Number of odd days in the period 2201-2221 = 16 normal years + 5 leap years = 16 x 1 + 5 x 2 = 16 + 10 = 26 = 5 odd days Number of days from 1-Jan-2222 to 22 Apr 2222 = 31 (Jan) + 28 (Feb) + 31 (Mar) + 22(Apr) = 112 112 days = 0 odd day Total number of odd days = (0 + 3 + 5 + 0) = 8 = 1 odd day 1 odd days = Monday Hence 22 Apr 2222 is Monday.

Q.- What is the year next to 1990 which will have the same calendar as that of the year 1990?

    1. 1. 1992
    2. 2. 2001
    3. 3. 1995
    4. 4. 1996

2
For a year to have the same calendar with 1990 total odd days from 1990 should be 0. Take the year 1992 from the given choices. Total odd days in the period 1990-1991= 2 normal years = 2 x 1 = 2 odd days Take the year 1995 from the given choices. Number of odd days in the period 1990-1994 = 4 normal years + 1 leap year = 4 x 1 + 1 x 2 = 6 odd days Take the year 1996 from the given choices. Number of odd days in the period 1990-1995= 5 normal years + 1 leap year = 5 x 1 + 1 x 2 = 7 odd days = 0 odd days (As we can reduce multiples of 7 from odd days which will not change anything) Though number of odd days in the period 1990-1995 is 0 there is a catch here. 1990 is not a leap year whereas 1996 is a leap year. Hence calendar for 1990 and 1996 will never be the same. Take the year 2001 from the given choices. Number of odd days in the period 1990-2000= 8 normal years + 3 leap years = 8 x 1 + 3 x 2 = 14 odd days = 0 odd days Also both 1990 and 2001 are normal years. Hence 1990 will have the same calendar as that of 2001

Q.- Arun went for a movie nine days ago. He goes to watch movies only on Thursdays. What day of the week is today?

    1. 1. Wednesday
    2. 2. Friday
    3. 3. Sunday
    4. 4. Saturday

4
Clearly it can be understood from the question that 9 days ago was a Thursday Number of odd days in 9 days = 2 (As 9-7 = 2 reduced perfect multiple of 7 from total days) Hence today = (Thursday + 2 odd days) = Saturday

Q.- What will be the day of the week 15th August 2010?

    1. 1. Thursday
    2. 2. Sunday
    3. 3. Monday
    4. 4. Saturday

2
15th Aug 2010 = (2009 years + period from 1-Jan-2010 to 15-Aug-2010) We know that number of odd days in 400 years = 0 Hence the number of odd days in 2000 years = 0 (Since 2000 is a perfect multiple of 400) Number of odd days in the period 2001-2009 = 7 normal years + 2 leap year = 7 x 1 + 2 x 2 = 11 = (11 - 7x1) odd day = 4 odd day Days from 1-Jan-2010 to 15-Aug-2010 = 31 (Jan) + 28 (Feb) + 31 (Mar) + 30 (Apr) + 31(may) + 30(Jun) + 31(Jul) + 15(Aug) = 227 227 days = 32 weeks + 3 day = 3 odd day Total number of odd days = (0 + 4 + 3) = 7 odd days = 0 odd day 0 odd day = Sunday Hence 15th August 2010 is Sunday.

Q.- How many days are there in x weeks x days

    1. 1. 14x
    2. 2. 7x
    3. 3. 7
    4. 4. 8x

4
x weeks x days = (7*x)+x=7x+x= 8x days

Q.- A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. If the area of the lawn is 2109 sq. m then what is the width of the road ?

    1. 1. 6 m
    2. 2. 4 m
    3. 3. 2 m
    4. 4. 3 m

4
Area of the park = (60 x 40) m2 = 2400 m2. Area of the lawn = 2109 m2. Area of the crossroads = (2400 - 2109) m2 = 291 m2. Let the width of the road be x metres. Then 60x + 40x - pow(x2) = 291 x2 - 100x + 291 = 0 (x - 97)(x - 3) = 0 x = 3.

Q.- The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes then the area of the park (in sq. m) is :

    1. 1. 156300
    2. 2. 153060
    3. 3. 153600
    4. 4. 153006

3
Perimeter = Distance covered in 8 min. = (12000/60) x 8 m = 1600 m. Let length = 3x metres and breadth = 2x metres. Then 2(3x + 2x) = 1600 or x = 160. Length = 480 m and Breadth = 320 m. Area = (480 x 320) m2 = 153600 m2.

Q.- An error 2% in excess is made while measuring the side of a square. The percentage of error in the calculated area of the square is :

    1. 1. 0.0404
    2. 2. 0.044
    3. 3. 0.0435
    4. 4. 0.0453

1
100 cm is read as 102 cm. A1 = (100 x 100) cm2 and A2 (102 x 102) cm2. (A2 - A1) = [pow(1022) -pow(1002)] = (102 + 100) x (102 - 100) = 404 cm2. Percentage error = 404/(100x100) x 100 % = 4.04%

Q.- A towel when bleached was found to have lost 20% of its length and 10% of its breadth. The percentage of decrease in area is :

    1. 1. 0.24
    2. 2. 0.3
    3. 3. 0.82
    4. 4. 0.28

4
Let original length = x and original breadth = y. Decrease in area = xy - [(80/100) x] x (90/100) y = xy - (18/25) xy = (7/25)xy Decrease %=[(7/25) xy] x [1/(xy)] x 100 % = 28%.

Q.- What is the least number of squares tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad ?

    1. 1. 148
    2. 2. 814
    3. 3. 841
    4. 4. 418

2
Length of largest tile = H.C.F. of 1517 cm and 902 cm = 41 cm. Area of each tile = (41 x 41) cm2. Required number of tiles = (1517 x 902)/(41x41) = 814.

Q.- The difference between the length and breadth of a rectangle is 23 m. If its perimeter is 206 m then its area is :

    1. 1. 5220
    2. 2. 5202
    3. 3. 2502
    4. 4. 2520

4
We have: (l - b) = 23 and 2(l + b) = 206 or (l + b) = 103. Solving the two equations we get: l = 63 and b = 40. Area = (l x b) = (63 x 40) m2 = 2520 m2.

Q.- The length of a rectangle is halved while its breadth is tripled. What is the percentage change in area ?

    1. 1. 0.5
    2. 2. 0.4
    3. 3. 0.45
    4. 4. 0.35

1
Let original length = x and original breadth = y. Original area = xy. New length = x/2 New breadth = 3y. New area = (x/2) x 3y = (3/2) xy. Increase % = [(1/2( xy)] x (1/xy) x 100 % = 50%.

Q.- The length of a rectangular plot is 20 metres more than its breadth. If the cost of fencing the plot @ 26.50 per metre is Rs. 5300 what is the length of the plot in metres ?

    1. 1. 50 m
    2. 2. 60 m
    3. 3. 45 m
    4. 4. 75m

2
Let breadth = x metres. Then length = (x + 20) metres. Perimeter = (5300/26.50) m = 200 m. 2[(x + 20) + x] = 200 2x + 20 = 100 2x = 80 x = 40. Hence length = x + 20 = 60 m.

Q.- A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq. feet how many feet of fencing will be required ?

    1. 1. 88 ft
    2. 2. 77 ft
    3. 3. 88.5 ft
    4. 4. 76 ft

1
We have: l = 20 ft and lb = 680 sq. ft. So b = 34 ft. Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft.

Q.- A tank is 25 m long 12 m wide and 6 m deep. The cost of plastering its walls and bottom at 75 paise per sq. m is :

    1. 1. Rs. 555
    2. 2. Rs. 855
    3. 3. Rs. 585
    4. 4. Rs. 558

4
Area to be plastered = [2(l + b) x h] + (l x b) = {[2(25 + 12) x 6] + (25 x 12)} m2 = (444 + 300) m2n = 744m2 cost of plastering = Rs. 744 x (75/100)= Rs. 558.

Q.- A circular path of 13 m radius has marginal walk 2 m wide all round it. Find the cost of leveling the walk at 25p per m2?

    1. 1. Rs.45
    2. 2. Rs.78
    3. 3. Rs.44
    4. 4. Rs.40

3
Let the original radius be R cm. New radius = 2R Area=πR2 New Area =π2R2=4πR2 Increase in area =(4πR2−πR2)=3πR2 Increase percent = (3πR2/πR2)*100=300%

Q.- The radius of the two circular fields is in the ratio 3: 5 the area of the first field is what percent less than the area of the second?

    1. 1. 0.5
    2. 2. 0.6
    3. 3. 0.64
    4. 4. 0.84

3
Area of Square =π∗r2 =>π∗r2=24.64 =>r2=(24.64/22)∗7 =>r2=7.84 =>r=sqrt.(7.84) =>r=2.8 Circumference =2π∗r=2∗(22/7)∗2.8=17.60m

Q.- A rope of which a calf is tied is increased from 12 m to 23 m how much additional grassy ground shall it graze(in meter sq)?

    1. 1. 1120
    2. 2. 1220
    3. 3. 1210
    4. 4. 1250

3
We know the product of diagonals is 1/2*(product of diagonals) Let one diagonal be d1 and d2 So as per question (1/2)∗d1∗d2=150 (1/2)∗10∗d2=150 d2=150/5=30

Q.- What is the ratio between perimeters of two squares one having 3 times the diagonal then the other?

    1. 1. 0.16736111111
    2. 2. 0.044444444444
    3. 3. 0.12569444444
    4. 4. 0.04375

3
Let the triangle and parallelogram have common base b let the Altitude of triangle is h1 and of parallelogram is h2(which is equal to 100 m) then Area of triangle =(1/2)∗b∗h1 Area of rectangle =b∗h2 As per question (1/2)∗b∗h1=b∗h2 (1/2)∗b∗h1=b∗100 h1=100∗2=200m

Q.- The radius of a semi circle is 6.3 cm then its perimeter is?

    1. 1. 35.4
    2. 2. 32.4
    3. 3. 30.4
    4. 4. 28.4

2
Lets solve this Area of rectangle =l∗b Area of triangle =(1/2)l∗b Ratio =l∗b:(1/2)l∗b=1:1/2=2:1

Q.- The perimeter of a semi circle is 144 cm then the radius is?

    1. 1. 25cm
    2. 2. 28cm
    3. 3. 30 cm
    4. 4. 35cm

2
We know area of square =a2 Area of triangle =(1/2)∗a∗h=>(1/2)∗a∗h=a2=>h=2a

Q.- The radius of a circular wheel is 1.75 m how many revolutions will it make in traveling 1 km?

    1. 1. 1000
    2. 2. 2000
    3. 3. 3000
    4. 4. 4000

1
Area of triangle A1 = (1/2)∗base∗height=(1/2)∗15∗12=90cm2 Area of second triangle =2∗A1=180cm2 (1/2)∗20∗height=180 =>height=18cm

Q.- If the wheel is 14 cm then the number of revolutions to cover a distance of 1056 cm is?

    1. 1. 10
    2. 2. 12
    3. 3. 14
    4. 4. 16

2
Let the lengths of the line segments be x and x+2 cm then (x+2)2−x2=32 x2+4x+4−x2=32 4x=28 x=7cm

Q.- The ratio of the radius of two circles is 1: 3 and then the ratio of their areas is?

    1. 1. 0.04375
    2. 2. 0.045138888889
    3. 3. 0.045833333333
    4. 4. 0.047916666667

4
So lets solve this Length of largest tile = HCF of (1517 cm and 902 cm) = 41 cm Required number of tiles = Area of floor/Area of tile = (1517×902)/(41×41)=814

Q.- Find the area of the quadrilateral of one of its diagonals is 20 cm and its off sets 9 cm and 6 cm?

    1. 1. 120
    2. 2. 150
    3. 3. 180
    4. 4. 310

2
Number of bricks =Courtyard area/1 brick area = (2500×1600)/(20×10)=20000

Q.- If the radius of a circle is diminished by 10% then the area is diminished by:

    1. 1. 4
    2. 2. 2
    3. 3. 3.5
    4. 4. 3

4
p (pow(152) - pow(132)) = 176 176 * 1/4 = Rs.44

Q.- Find the circumference of a circle whose area is 24.64 meter sqaure

    1. 1. 17.60 m
    2. 2. 17.06 m
    3. 3. 60.17 m
    4. 4. 60.71 m

1
r = 3 p*r*r = 9 r = 5 p*r*r = 25 25 p – 16 p 100 ---- ? => 64%

Q.- The area of rhombus is 150 cm square. The length of one of the its diagonals is 10 cm. The length of the other diagonal is:

    1. 1. 25 cm
    2. 2. 30 cm
    3. 3. 35 cm
    4. 4. 40 cm

2
p (pow{232} – pow{122}) = 1210

Q.- A triangle and a parallelogram are constructed on the same base such that their areas are equal. If the altitude of the parallelogram is 100 m then the altitude of the triangle is.

    1. 1. 350 m
    2. 2. 300 m
    3. 3. 200 m
    4. 4. 250 m

3
d1 = 3d d2 = d d1:d2=3d:d=3:1

Q.- What will be the ratio between the area of a rectangle and the area of a triangle with one of the sides of the rectangle as base and a vertex on the opposite side of the rectangle ?

    1. 1. 0.043055555556
    2. 2. 0.084027777778
    3. 3. 0.044444444444
    4. 4. 0.16736111111

2
2*r+3.14*r=32.38

Q.- If the area of a square with the side a is equal to the area of a triangle with base a then the altitude of the triangle is.

    1. 1. 2a
    2. 2. 3a
    3. 3. 4a
    4. 4. 5a

1
3.14*r+2*r=144 r=28

Q.- The base of a triangle is 15 cm and height is 12 cm. The height of another triangle of double the area having the base 20 cm is :

    1. 1. 20 cm
    2. 2. 18 cm
    3. 3. 19 cm
    4. 4. 16 cm

2
2 * 22/7 * 1.75 * x = 11000 x = 1000

Q.- The difference of the areas of two squares drawn on two line segments in 32 sq. cm. Find the length of the greater line segment if one is longer than the other by 2 cm.

    1. 1. 9 cm
    2. 2. 6 cm
    3. 3. 7 cm
    4. 4. 8 cm

3
2 * 22/7 * 14 * x = 1056 => x = 12

Q.- What are the least number of square tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad ?

    1. 1. 841
    2. 2. 418
    3. 3. 481
    4. 4. 814

4
r1: r2 = 1: 3 ?*pow(r12): ?*pow(r22) pow(r12): pow(r22) = 1: 9

Q.- A courtyard is 25 meter long and 16 meter board is to be paved with bricks of dimensions 20 cm by 10 cm. The total number of bricks required is :

    1. 1. 20000
    2. 2. 30000
    3. 3. 25000
    4. 4. 22000

1
1/2 * 20(9 + 6) = 150 cm2

Q.- How many times in a day the hands of a clock are straight ?

    1. 1. 55
    2. 2. 22
    3. 3. 33
    4. 4. 44

4
In 12 hours the hands coincide or are in opposite direction 22 times. In 24 hours the hands coincide or are in opposite direction 44 times a day.

Q.- At what time in minutes between 3 o'clock and 4 o'clock both the needles will coincide each other ?

    1. 1. 61.63 min
    2. 2. 61.36 min
    3. 3. 16.36 min
    4. 4. 16.63 min

3
At 3 o'clock the minute hand is 15 min. spaces apart from the hour hand. To be coincident it must gain 15 min. spaces. 55 min. are gained in 60 min. 15 min. are gained in (60/55) x 15 min = (16+ 4/11) min. The hands are coincident at 16.36 min. past 3.

Q.- At what time between 9 and 10 o'clock will the hands of a watch be together ?

    1. 1. 45 min past 9
    2. 2. 49.09 min past 9
    3. 3. 40 min past 9
    4. 4. 30 min past9

2
To be together between 9 and 10 o'clock the minute hand has to gain 45 min. spaces. 55 min. spaces gained in 60 min. 45 min. spaces are gained in (60/55) x 45 min or 49+ (1/11) min. The hands are together at 49.09 min. past 9.

Q.- How many times in a day are the hands of a clock in straight line but opposite in direction ?

    1. 1. 24
    2. 2. 33
    3. 3. 22
    4. 4. 26

3
The hands of a clock point in opposite directions (in the same straight line) 11 times in every 12 hours. (Because between 5 and 7 they point in opposite directions at 6 o'clcok only). So in a day the hands point in the opposite directions 22 times.

Q.- How many times are the hands of a clock at right angle in a day ?

    1. 1. 40
    2. 2. 36
    3. 3. 24
    4. 4. 44

4
In 12 hours they are at right angles 22 times. In 24 hours they are at right angles 44 times.

Q.- At what time between 5.30 and 6 will the hands of a clock be at right angles ?

    1. 1. 43.63 min past 5
    2. 2. 40.63 min past 5
    3. 3. 43.60 min past 5
    4. 4. 43.03 min past 5

1
At 5 o'clock the hands are 25 min. spaces apart. To be at right angles and that too between 5.30 and 6 the minute hand has to gain (25 + 15) = 40 min. spaces. 55 min. spaces are gained in 60 min. 40 min. spaces are gained in (60/55) x 40 min = 43.63 min. Required time = 43.63 min. past 5.

Q.- At what time between 7 and 8 o'clock will the hands of a clock be in the same straight line but not together ?

    1. 1. 6.45 min past 7
    2. 2. 5.35 min past 7
    3. 3. 5.40 min past 7
    4. 4. 5.45 min past 7

4
When the hands of the clock are in the same straight line but not together they are 30 minute spaces apart. At 7 o'clock they are 25 min. spaces apart. Minute hand will have to gain only 5 min. spaces. 55 min. spaces are gained in 60 min. 5 min. spaces are gained in (60/55) x 5 min = 5.45 min. Required time = 5.45 min. past 7.

Q.- How much does a watch lose per day if its hands coincide every 64 minutes ?

    1. 1. 32.27 min
    2. 2. 72.32 min
    3. 3. 32.72 min
    4. 4. 32.27 min

3
55 min. spaces are covered in 60 min. 60 min. spaces are covered in (60/55) x 60 min. = 65.45 min. Loss in 64 min. = 65.45 - 64 = 16/11 min. Loss in 24 hrs = (16/11) x (1/64) x 24 x 60 min. = 32.72 min.

Q.- An accurate clock shows 8 o'clock in the morning. Through how may degrees will the hour hand rotate when the clock shows 2 o'clock in the afternoon ?

    1. 1. 150 deg
    2. 2. 180 deg
    3. 3. 144 deg
    4. 4. 168 deg

2
Angle traced by the hour hand in 6 hours = [(360/12) x 6]� = 180�.

Q.- A clock is started at noon. By 10 minutes past 5 the hour hand has turned through :

    1. 1. 155 deg
    2. 2. 145 deg
    3. 3. 150 deg
    4. 4. 160 deg

1
Angle traced by hour hand in 12 hrs = 360�. Angle traced by hour hand in 5 hrs 10 min. i.e. (31/6) hrs = [ (360/12) x 31/6]� = 155�.

Q.- A clock strikes 4 taking 9 seconds. In order to strike 12 at the same rate. the time taken is

    1. 1. 53 sec
    2. 2. 43 sec.
    3. 3. 23 sec.
    4. 4. 33 sec.

4
There are 3 intervals when the clock strikes 4 Time taken at 3 intervals = 9 seconds Time taken for 1 interval = 9/3=3 seconds In order to strike 12 there are 11 intervals. Hence time needed =3×11=33 seconds

Q.- An accurate clock shows 8 o'clock in the morning. Through how may degrees will the hour hand rotate when the clock shows 2 o'clock in the afternoon?

    1. 1. 154 deg
    2. 2. 180 deg
    3. 3. 170 deg
    4. 4. 160 deg

2
We know that Angle traced by hour hand in 12 hrs = 360° From 8 to 2 there are 6 hours The angle traced by the hour hand in 6 hours = 6×36012=180°

Q.- A clock is started at noon. By 10 minutes past 5 the hour hand has turned through

    1. 1. 155 deg
    2. 2. 145 deg
    3. 3. 70 deg
    4. 4. 140 deg

1
We know that Angle traced by hour hand in 12 hrs = 360° Time duration from noon to 10 minutes past 5 = 5 hours 10 minutes=5(10/60)hour=31/6 hours. Hence the angle traced by hour hand from noon to 10 minutes past 5 = (31/6)×(360/12)=(31/6)x30=31x5=155 degree

Q.- At what angle the hands of a clock are inclined at 15 minutes past 5?

    1. 1. 70 deg
    2. 2. 155 deg
    3. 3. 145deg
    4. 4. 67.5deg

4
When the minute hand is behind the hour hand the anglebetween the two hands at M minutes past H'o clock=30[H−(M/5)]+M/2 degree. When the minute hand is ahead of the hour hand the angle between the two hands at M minutes past H'o clock=30[(M/5)−H]−M/2 degree Here H = 5 M = 15 and the minute hand is behind the hour hand. Hence the angle =30(H−M/5)+M/2=30(5−15/5)+15/2=30(5−3)+7.5=30×2+7.5=67.5°

Q.- How many times in a day are the hands of a clock in straight line but opposite in direction?

    1. 1. 33
    2. 2. 22
    3. 3. 25
    4. 4. 44

2
The hands of a clock point in opposite directions (in the same straight line making an angle 180° between them) 11 times in every 12 hours because between 5 and 7 they point in opposite directions at 6 'o clock only. Hence the hands point in the opposite directions 22 times in a day However this is already given as a formula and its is better to by heart the answer as 22 which can save time in competitive exams.(However if you should know the theory behind)

Q.- How many times will the hands of a clock coincide in a day?

    1. 1. 26
    2. 2. 33
    3. 3. 22
    4. 4. 24

3
The hands of a clock coincide 11 times in every 12 hours (Between 11 and 1 they coincide only once at 12 o'clock). 12:00 am 1:05 am 2:11 am 3:16 am 4:22 am 5:27 am 6:33 am 7:38 am 8:44 am 9:49 am 10:55 am 12:00 pm 1:05 pm 2:11 pm 3:16 pm 4:22 pm 5:27 pm 6:33 pm 7:38 pm 8:44 pm 9:49 pm 10:55 pm Hence the hands coincide 22 times in a day. However this is already given as a formula and its is better to by heart the answer as 22 which can save time in competitive exams.(However if you should know the theory behind)

Q.- How many times in a day the hands of a clock are straight

    1. 1. 46
    2. 2. 40
    3. 3. 42
    4. 4. 44

4
The hands of a clock point in opposite directions (in the same straight line making an angle 180° between them) 11 times in every 12 hours because between 5 and 7 they point in opposite directions at 6 'o clock only. Hence the hands point in the opposite directions 22 times in a day. The hands of a clock coincide(0 ° between them) 11 times in every 12 hours (Between 11 and 1 they coincide only once at 12 o'clock). Hence the hands coincide 22 times in a day. So In 24 hours the hands come in opposite direction or coincide 44 times . However this is already given as a formula and its is better to by heart the answer as 44 which can save time in competitive exams.(However if you should know the theory behind)

Q.- How many times are the hands of a clock at right angle in a day?

    1. 1. 44
    2. 2. 43
    3. 3. 42
    4. 4. 41

1
In 12 hours hands of a clock are at right angles at 22 times. In 24 hours hands of a clock are at right angles at 44 times.

Q.- An accurate clock shows 8 o'clock in the morning. Through how may degrees will the hour hand rotate when the clock shows 2 o'clock in the afternoon?

    1. 1. 168 deg
    2. 2. 180 deg
    3. 3. 150 deg
    4. 4. 144 deg

2
Angle traced by the hour hand in 6 hours = (360/12) x 6º = 180º.

Q.- A clock is started at noon. By 10 minutes past 5 the hour hand has turned through:

    1. 1. 145 deg
    2. 2. 150 deg
    3. 3. 155 deg
    4. 4. 160 deg

3
Angle traced by hour hand in 12 hrs = 360º. Angle traced by hour hand in 5 hrs 10 min. i.e. 31/6 hrs = (360/12) x (31/6)º = 155º.

Q.- What is the angle between the hands at 4.40?

    1. 1. 95 deg
    2. 2. 100 deg
    3. 3. 120 deg
    4. 4. 110

2
4 hour 40 minutes = 14/3 hour Angle traced by hour hand in 12 hrs = 360° Angle traced by hour hand in 14/3 hour = (360/12)×(14/3)=30×(14/3)=10×14=140° Angle traced by minute hand in 60 min. = 360°. Angle traced by minute hand in 40 min. = (360/60)×40=6×40=240°. Angle between the hands of the clock when the time is 4.40 = 240° - 140° =100°.

Q.- What is the angle between the hour and the minute hand of a clock when the time is 3.25?

    1. 1. 46 deg
    2. 2. 47 deg
    3. 3. 150 deg
    4. 4. 47.5 deg

4
3 hour 25 minutes = 41/12 hour Angle traced by hour hand in 12 hrs = 360° Angle traced by hour hand in 41/12 hour = (360/12)×(41/12)=30×(41/12) = 10×41/4=10×10.25=102.5° Angle traced by minute hand in 60 min. = 360°. Angle traced by minute hand in 25 min. = (360/60)×25=6×25=150°. Angle between the hands of the clock when the time is 10.25 = 150° - 102.5° = 47.5 °.

Q.- A watch which gains uniformly is 2 minutes low at noon on Monday and is 4 min. 48 sec fast at 2 p.m. on the following Monday. When was it correct?

    1. 1. 2 p.m. on Tuesday
    2. 2. 2 p.m. on Wednesday
    3. 3. 3 p.m. on Thursday
    4. 4. 1 p.m. on Friday

2
Time from 12 p.m. on Monday to 2 p.m. on the following Monday = 7 days 2 hours = 170 hours. The watch gains(2 + 24/5) min.or34/5 min. in 170 hrs. Now34/5 min. are gained in 170 hrs. 2 min. are gained in[170 x(5/34)x 2]hrs= 50 hrs. Watch is correct 2 days 2 hrs. after 12 p.m. on Monday i.e. it will be correct at 2 p.m. on Wednesday.

Q.- At what time between 4 and 5 o'clock will the hands of a watch point in opposite directions?

    1. 1. 45 min. past 4
    2. 2. 40 min. past 4
    3. 3. 504/11 min. past 4
    4. 4. 54 6/11 min. past 4

4
At 4 o'clock the hands of the watch are 20 min. spaces apart. To be in opposite directions they must be 30 min. spaces apart. Minute hand will have to gain 50 min. spaces. 55 min. spaces are gained in 60 min. 50 min. spaces are gained in (60/55)x 50 min. or 54 6/11 min. Required time = 54 6/11 min. past 4.

Q.- At what time between 9 and 10 o'clock will the hands of a watch be together?

    1. 1. 45 min. past 9
    2. 2. 50 min. past 9
    3. 3. 49 1/11min. past 9
    4. 4. 48 2/11min. past 9

3
To be together between 9 and 10 o'clock the minute hand has to gain 45 min. spaces. 55 min. spaces gained in 60 min. 45 min. spaces are gained in(60/55)x 45 min or 491/11 min. The hands are together at 49 1/11 min. past 9.

Q.- At what time between 5 and 6 will the hands of the clock coincide?

    1. 1. 26 2/11 minutes past 5
    2. 2. 26 3/11 minutes past 5
    3. 3. 28 3/11 minutes past 5
    4. 4. 27 3/11 minutes past 5

4
At 5 o' clock the hands are 25 minute spaces apart Hence minute hand needs to gain 25 more minute spaces so that the hands will be in opposite direction We know that 55 min spaces are gained by minute hand (with respect to hour hand) in 60 min Hence time taken for gaining 25 minute spaces by minute hand =(60/55)×25 minute=(12/11)×25 minute=300/11 minute = 27 3/11 minute Hence hands of the clock will be together at 27 3/11 minute past 5

Q.- The minute hand of a clock overtakes the hour hand at intervals of 65 minutes. How much a day does the clock gain or loss?

    1. 1. 10 9/143 minutes
    2. 2. 11 9/143 minutes
    3. 3. 11 10/143 minutes
    4. 4. 10 10/143 minutes

4
The minute hand of a clock overtakes the hour hand at intervals of M minutes of correct time. The clock gains or loses in a day by=[(720/11)-M] [(60×24)/M] minutes Here M = 65. The clock gains or losses in a day by =[(720/11)-M][(60×24)/M] = [(720/11)-65][(60×24)/65]) = 5/11[(12×24)/13] = 1440/143 = 10 10/143 minutes

Q.- A clock is set at 5 am. If the clock loses 16 minutes in 24 hours what will be the true time when the clock indicates 10 pm on 4th day?

    1. 1. 9.30 pm
    2. 2. 0.91666666667
    3. 3. 10.30 pm
    4. 4. 0.95833333333

4
Time from 5 am to 10 pm on the 4th day = 3 days 17 hours = 3×24+17=89 hours Given that clock loses 16 minutes in 24 hours ?> 23 hour 44 minutes of the given clock = 24 hours in a normal clock ? 23 44/60 hours of the given clock = 24 hours in a normal clock ? 23 11/15 hours of the given clock = 24 hours in a normal clock ? 356/15 hours of the given clock = 24 hours in a normal clock ? 89 hours of the given clock = 24×(15/356)×89 hours in a normal clock=24×(15/4)=6×15=90 hours So the correct time is 90 hours after 5 am = 3 days 18 hours after 5 am = 11 pm on the 4th day

Q.- A watch which gains 5 seconds in 3 minutes was set right at 7 a.m. In the afternoon of the same day when the watch indicated quarter past 4 o'clock the true time is

    1. 1. 0.625
    2. 2. 3.45 pm
    3. 3. 3.30 pm
    4. 4. 0.66666666667

4
Time from 7 am to 4.15 pm = 9 hours 15 minutes = 9 1/4 hours = 37/4 hours 3 minute 5 seconds of the given clock = 3 minutes of a normal clock 3 1/12 minutes of the given clock = 3 minutes of a normal clock 37/12 minutes of the given clock = 3 minutes of a normal clock 37/720 hours of the given clock = 3/60 hours of a normal clock 37/720 hours of the given clock = 1/20 hours of a normal clock 37/4 hours of the given clock = (1/20)×(720/37)×(37/4) hours of the given clock= 9 hours of the given clock Hence the correct time = 9 hours after 7 am = 4 pm

Q.- The angle between the minute hand and the hour hand of a clock when the time is 8.30 is

    1. 1. 75 deg
    2. 2. 85 deg
    3. 3. 80 deg
    4. 4. 70deg

1
8.30 = 8 hour 30 minutes = 8 1/2 hour=17/2 hour Angle traced by hour hand in 12 hours = 360° Hence Angle traced by hour hand in 17/7 hour = (360/12)×(17/2) = 30×(17/2)=15×17=255° Angle traced by minute hand in 60 minutes = 360° Angle traced by minute hand in 30 minutes = (360/60)×30=180° Required angle = 255- 180= 75°

Q.- A hollow iron pipe is 21 cm long and its external diameter is 8 cm. If the thickness of the pipe is 1 cm and iron weighs 8 g/cm3 then the weight of the pipe is :

    1. 1. 3.669 kg
    2. 2. 3.966 kg
    3. 3. 3.600 kg
    4. 4. 3.696 kg

4
External radius = 4 cm Internal radius = 3 cm. Volume of iron = (22/7) x [pow(42) -pow(32)] x 21 cm3 7 = (22/7) x 7 x 1 x 21 cm3 = 462 cm3. Weight of iron = (462 x 8) gm = 3696 gm = 3.696 kg.

Q.- 66 cubic centimetres of silver is drawn into a wire 1 mm in diameter. The length of the wire in metres will be :

    1. 1. 84 m
    2. 2. 48 m
    3. 3. 52 m
    4. 4. 25 m

1
Let the length of the wire be h. Radius = 1/2 mm = 1/20 cm. Then (22/7) x (1/20) x (1/20) x h = 66. h = (66 x 20 x 20 x 7)/22 = 8400 cm = 84 m.

Q.- A hall is 15 m long and 12 m broad. If the sum of the areas of the floor and the ceiling is equal to the sum of the areas of four walls the volume of the hall is :

    1. 1. 1002
    2. 2. 1200
    3. 3. 1203
    4. 4. 1205

2
2(15 + 12) x h = 2(15 x 12) h = 180/27 m = 20/3 m. Volume = 15 x 12 x (20/3) m3 = 1200 m3.

Q.- In a shower 5 cm of rain falls. The volume of water that falls on 1.5 hectares of ground is :

    1. 1. 760
    2. 2. 705
    3. 3. 750
    4. 4. 507

3
1 hectare = 10000 m2 So Area = (1.5 x 10000) m2 = 15000 m2. Depth = 5/100 m = 1/20 m. Volume = (Area x Depth) = 15000 x (1/20) m3 = 750 m3.

Q.- A metallic sheet is of rectangular shape with dimensions 48 m x 36 m. From each of its corners a square is cut off so as to make an open box. If the length of the square is 8 m the volume of the box (in m3) is :

    1. 1. 1520
    2. 2. 1502
    3. 3. 5102
    4. 4. 5120

4
Clearly l = (48 - 16)m = 32 m b = (36 -16)m = 20 m h = 8 m. Volume of the box = (32 x 20 x 8) m3 = 5120 m3.

Q.- A cistern 6m long and 4 m wide contains water up to a depth of 1 m 25 cm. The total area of the wet surface is :

    1. 1. 49
    2. 2. 48
    3. 3. 49.5
    4. 4. 94

1
Area of the wet surface = [2(lb + bh + lh) - lb] = 2(bh + lh) + lb = [2 (4 x 1.25 + 6 x 1.25) + 6 x 4] m2 = 49 m2.

Q.- 50 men took a dip in a water tank 40 m long and 20 m broad on a religious day. If the average displacement of water by a man is 4 meter cube then the rise in the water level in the tank will be :

    1. 1. 52 cm
    2. 2. 25 cm
    3. 3. 35 cm
    4. 4. 53 cm

2
Total volume of water displaced = (4 x 50) m3 = 200 m3. Rise in water level = 200/(40x20) m 0.25 m = 25 cm.

Q.- A boat having a length 3 m and breadth 2 m is floating on a lake. The boat sinks by 1 cm when a man gets on it. The mass of the man is :

    1. 1. 50 kg
    2. 2. 40 kg
    3. 3. 06 kg
    4. 4. 60 kg

4
Volume of water displaced = (3 x 2 x 0.01) m3 = 0.06 m3. Mass of man = Volume of water displaced x Density of water = (0.06 x 1000) kg = 60 kg.

Q.- A cistern of capacity 8000 litres measures externally 3.3 m by 2.6 m by 1.1 m and its walls are 5 cm thick. The thickness of the bottom is :

    1. 1. 10 cm
    2. 2. 12 cm
    3. 3. 14 cm
    4. 4. 9 cm

1
Let the thickness of the bottom be x cm. Then [(330 - 10) x (260 - 10) x (110 - x)] = 8000 x 1000 320 x 250 x (110 - x) = 8000 x 1000 (110 - x) = 8000 x 1000 = 100 320 x 250 x = 10 cm

Q.- How many bricks each measuring 25 cm x 11.25 cm x 6 cm will be needed to build a wall of 8 m x 6 m x 22.5 cm ?

    1. 1. 6400
    2. 2. 7400
    3. 3. 4600
    4. 4. 4700

1
Number of bricks = (Volume of the wall)/( Volume of 1 brick)= ( 800 x 600 x 22.5) /(25 x 11.25 x 6) = 6400.

Q.- If the radius of the base of a right circular cone is 3r and its height is equal to the radius of the base then its volume is :

    1. 1. 1/3*22/7*pow(r3)
    2. 2. 2/3*22/7*pow(r3)
    3. 3. 3*22/7*pow(r3)
    4. 4. 9*22/7*pow(r3)

4
Volume=1/3*22/7*pow(3r2)*3r =9*22/7*pow(r3)

Q.- If a hemi-spherical dome has an inner diameter of 28 m then its volume is :

    1. 1. 6186.6
    2. 2. 5749.33
    3. 3. 7099.33
    4. 4. 7459.33

2
Volume=[(2/3)*(22/7)*14*14*14] =5749.33 pow(m3)

Q.- The total surface area of a solid hemispere whose diameter is 14 cm is :

    1. 1. 588*22/7
    2. 2. 392*22/7
    3. 3. 147*22/7
    4. 4. 98*22/7

3
Total surface area = 3*22/7*pow(R2) = 3*22/7*pow(72) = 147*22/7 pow(cm3)

Q.- The number of solid spheres each of diameter 6 cm that could be moulded to form a solid metal cylinder of height 45 cm and diameter 4 cm is :

    1. 1. 3
    2. 2. 4
    3. 3. 5
    4. 4. 6

3
Let the number of spheres be n. then n*4/3*22/7*3*3*3 = 22/7*2*2*45 so n = 5

Q.- A conical vessel whose internal radius is 10 cm and height 48 cm is full of water. If this water is poured in to a cylindrical vessel with internal radius 20 cm the height to which water rises in it is :

    1. 1. 3 cm
    2. 2. 4 cm
    3. 3. 5 cm
    4. 4. 6 cm

2
1/3*22/7*pow(102)*48 = 22/7*pow(202)*H therefore H = 4 cm

Q.- A sphere of radius 6 cm is dropped into a cylindrical vessel partly filled with water. The radius of the vessel is 8 cm. If the sphere is submerged completely then the surface of the water rises by :

    1. 1. 4.5 cm
    2. 2. 3 cm
    3. 3. 4 cm
    4. 4. 2 cm

1
22/7*8*8*H = 4/3*22/7*6*6*6 Hence H = 4.5 cm

Q.- If the volume and the surface area of a sphere are numerically the same then its radius is :

    1. 1. 1 units
    2. 2. 2 units
    3. 3. 3 units
    4. 4. 4 units

3
4*22/7*pow(R2) = 4/3*22/7*pow(R3) Hence R = [4*3/4] = 3

Q.- A sphere of diameter 12.6 cm is melted and cast into a right circular cone of height 25.2 cm. The radius of the base of the cone is :

    1. 1. 6.3 cm
    2. 2. 2.1 cm
    3. 3. 2 cm
    4. 4. 3 cm

1
4/3*22/7*pow(6.33) = 1/3*22/7*pow(R2)*25.2 Hence R = 6.3 cm

Q.- A cylindrical tub of radius 12 cm contains water up to a depth of 20 cm. A spherical iron ball is dropped into the tub and thus the of water is raised by 6.75 cm. The radius of the ball is :

    1. 1. 4.5 cm
    2. 2. 6 cm
    3. 3. 9 cm
    4. 4. 7.25 cm

3
4/3*22/7*pow(R3) = 22/7*12*12*6.75 Hence pow(R3) = 9*9*9 so R = 9

Q.- The volume of a pyramid of base area 16 sq.cm and height 9 cm is :

    1. 1. 36
    2. 2. 48
    3. 3. 72
    4. 4. 144

4
Volume = (Area of the base)*Height = (16*9) = 144

Q.- A rectangular field 30 m long and 20 m broad. How much deep it should be dug so that from the earth taken out a platform can be formed which is 8 m long 5.5 m broad and 1.5 m high where as the earth taken out is increase by 10/5?

    1. 1. 12 cm
    2. 2. 10 cm
    3. 3. 14 cm
    4. 4. 11 cm

2
30 * 20 * x = (8 * 5.5 * 1.5)/2

Q.- Calculate the number of bricks each measuring 25 cm * 15 cm * 8 cm required to construct a wall of dimensions 10 m * 4 m * 5 m when 10% of its volume is occupied by mortar?

    1. 1. 4000
    2. 2. 5000
    3. 3. 6000
    4. 4. 7000

3
10 * 4/100 * 5 * 90/100 = 25/100 * 15/100 * 8/100 * x 10 * 20 * 90 = 15 * 2 * x => x = 6000

Q.- If in a box of dimensions 6 m * 5 m * 4 m smaller boxes of dimensions 60 cm * 50 cm * 40 cm are kept in it then what will be the maximum number of the small boxes that can be kept in it?

    1. 1. 500
    2. 2. 1000
    3. 3. 900
    4. 4. 600

2
6 * 5 * 4 = 60/100 * 50/100 * 40/100 * x 1 = 1/10 * 1/10 * 1/10 * x => x = 1000

Q.- A brick measures 20 cm * 10 cm * 7.5 cm how many bricks will be required for a wall 25 m * 2 m * 0.75 m?

    1. 1. 24000
    2. 2. 23000
    3. 3. 22000
    4. 4. 25000

4
25 * 2 * 0.75 = 20/100 * 10/100 * 7.5/100 * x 25 = 1/100 * x => x = 25000

Q.- he dimensions of an open box are 52 40 and 29 cms. Its thickness is 2 cms. If 1 cm3 of metal used in the box weighs 0.5 gms the weight of the box is?

    1. 1. 8.56 kg
    2. 2. 7.76 kg
    3. 3. 7.756 kg
    4. 4. 6.832 kg

4
52 * 40 * 29 = 60320 48 * 36 * 27 = 46656 ------------(subtract) 13664 13664 * 1/2 = 6832 => 6.832 kg

Q.- How many meters of carpet 50cm wide will be required to cover the floor of a room 30m * 20m?

    1. 1. 1000
    2. 2. 600
    3. 3. 2400
    4. 4. 1200

4
50/100 * x = 30 * 20 x = 1200

Q.- How many shots of 1cm radius can be prepared from a sphere of 3cm radius?

    1. 1. 26
    2. 2. 25
    3. 3. 27
    4. 4. 28

3
4/3 Pi * 3 * 3 * 3 = 4/3 p * 1 * 1 * 1 * x x = 27

Q.- Find the expenditure on digging a well 14m deep and of 3m diameter at Rs.15 per cubic meter?

    1. 1. Rs.1185
    2. 2. Rs.1285
    3. 3. Rs.1385
    4. 4. Rs.1485

4
22/7 * 14 * 3/2 * 3/2 = 99 m2 99 * 15 = 1485

Q.- The diameter of a cylindrical tin is 6cm and height is 5 cm. Find the volume of the cylinder?

    1. 1. 30 p cc
    2. 2. 45 p cc
    3. 3. 150 p cc
    4. 4. 180 p cc

2
r = 3 h = 5 p * 3 * 3 * 5 = 45 p cc

Q.- How many cubes of edge 2 dm can be cut out of a meter cube?

    1. 1. 251
    2. 2. 215
    3. 3. 152
    4. 4. 125

4
1 * 1* 1 = 2/10 * 2/10 * 2/10 * x x = 125

Q.- In a mixture 60 litres the ratio of milk and water 2 : 1. If this ratio is to be 1 : 2 then the quanity of water to be further added is :

    1. 1. 35 litres
    2. 2. 40 litres
    3. 3. 50 litres
    4. 4. 60 litres

4
Quantity of milk = (60 x 2/3) litres = 40 litres. Quantity of water in it = (60- 40) litres = 20 litres. New ratio = 1 : 2 Let quantity of water to be added further be x litres. Then milk : water = 40/(20+x) Now 40/(20+x) = 1/2 20 + x = 80 x = 60. Quantity of water to be added = 60 litres.

Q.- A sum of money is to be distributed among A B C D in the proportion of 5 : 2 : 4 : 3. If C gets Rs. 1000 more than D what is B's share ?

    1. 1. Rs.1500
    2. 2. Rs. 2000
    3. 3. Rs. 1000
    4. 4. Rs. 2500

2
Let the shares of A B C and D be Rs. 5x Rs. 2x Rs. 4x and Rs. 3x respectively. Then 4x - 3x = 1000 x = 1000. B's share = Rs. 2x = Rs. (2 x 1000) = Rs. 2000.

Q.- Two numbers are respectively 20% and 50% more than a third number. The ratio of the two numbers is :

    1. 1. 0.12916666667
    2. 2. 0.25208333333
    3. 3. 0.17013888889
    4. 4. 0.21111111111

3
Let the third number be x. Then first number = 120% of x = 120x/100 = 6x/5 Second number = 150% of x = 150x/100 = 3x/2 Ratio of first two numbers = (6x/5) : (3x/2) = 12x : 15x = 4 : 5.

Q.- If Rs. 782 be divided into three parts proportional to 1/2 :2/3 :3/4 then the first part is :

    1. 1. Rs.402
    2. 2. Rs.204
    3. 3. Rs. 420
    4. 4. Rs. 240

2
Given ratio =1/2 :2/3 :3/4 = 6 : 8 : 9. 1st part = Rs. 782 x (6/23) = Rs. 204

Q.- If 0.75 : x :: 5 : 8 then x is equal to :

    1. 1. 1.3
    2. 2. 3.1
    3. 3. 2.1
    4. 4. 1.2

4
(x x 5) = (0.75 x 8) x = 6/5 = 1.20

Q.- Salaries of Ravi and Sumit are in the ratio 2 : 3. If the salary of each is increased by Rs. 4000 the new ratio becomes 40 : 57. What is Sumit's salary ?

    1. 1. Rs. 50000
    2. 2. Rs. 40000
    3. 3. Rs. 83000
    4. 4. Rs. 38000

4
Let the original salaries of Ravi and Sumit be Rs. 2x and Rs. 3x respectively. Then (2x + 4000)/57 = 403x + 4000 57(2x + 4000) = 40(3x + 4000) 6x = 68000 3x = 34000 Sumit's present salary = (3x + 4000) = Rs.(34000 + 4000) = Rs. 38000.

Q.- The ratio of the number of boys and girls in a college is 7 : 8. If the percentage increase in the number of boys and girls be 20% and 10% respectively what will be the new ratio ?

    1. 1. 1.0555555556
    2. 2. 0.85069444444
    3. 3. 0.93125
    4. 4. 0.89027777778

4
Originally let the number of boys and girls in the college be 7x and 8x respectively. Their increased number is (120% of 7x) and (110% of 8x). (120/100) x 7x and (110/100) x 8x 42x/5 and 44x/5 The required ratio = 42x/5 : 44x/5 = 21 : 22.

Q.- Two number are in the ratio 3 : 5. If 9 is subtracted from each the new numbers are in the ratio 12 : 23. The smaller number is :

    1. 1. 44
    2. 2. 33
    3. 3. 55
    4. 4. 22

2
Let the numbers be 3x and 5x. Then (3x - 9)/(5x-9) = 12/23 23(3x - 9) = 12(5x - 9) 9x = 99 x = 11. The smaller number = (3 x 11) = 33.

Q.- The fourth proportional to 5 8 15 is :

    1. 1. 24
    2. 2. 42
    3. 3. 40
    4. 4. 45

1
Let the fourth proportional to 5 8 15 be x. Then 5 : 8 : 15 : x 5x = (8 x 15) x = (8 x 15)/5 = 24.

Q.- If 40% of a number is equal to two-third of another number what is the ratio of first number to the second number ?

    1. 1. 0.12777777778
    2. 2. 0.16875
    3. 3. 0.21041666667
    4. 4. 0.12847222222

3
Let 40% of A = 2/3 B Then 40A/100 = 2B/3 2A /5= 2B/3 A/B = (2/3) x 5/2 = 5/3 A : B = 5 : 3.

Q.- If 40% of a number is equal to two-third of another number. what is the ratio of first number to the second number.

    1. 1. 0.12847222222
    2. 2. 0.12777777778
    3. 3. 0.16875
    4. 4. 0.21041666667

4
Let the first number is A and second number is B. As per question 40/100 A=2/3 B A/B=(2/3)∗(100/40) A/B=5/3=>A:B=5:3

Q.- The least whole number which when subtracted from both the terms of the ratio 6 : 7 to give a ratio less than 16 : 21 is

    1. 1. 3
    2. 2. 4
    3. 3. 5
    4. 4. 8

1
Let x is subtracted. Then (6−x)/(7−x)<16/21 21(6—x)<16(7—x) =>5x>14 x>2.8 Least such number is 3

Q.- The sum of three numbers is 98. If the ratio of the first to the second is 2:3 and that of the second to the third is 5:8 then the second number is

    1. 1. 35
    2. 2. 20
    3. 3. 25
    4. 4. 30

4
a:b=2:3 b:c=5:8 =5∗(3/5):8∗(3/5) = 3:24/5 a:b:c=2:3:24/5=10:15:24 b=98∗(15/49)=30

Q.- A dog takes 3 leaps for every 5 leaps of a hare. If one leap of the dog is equal to 3 leaps of the hare the ratio of the speed of the dog to that of the hare is

    1. 1. 0.21458333333
    2. 2. 0.37847222222
    3. 3. 0.37777777778
    4. 4. 0.17291666667

2
Dog : Hare = (3*3) leaps of hare : 5 leaps of hare = 9:5

Q.- If three numbers in the ratio 3 : 2: 5 be such that the sum of their squares is 1862 the middle number will be

    1. 1. 15
    2. 2. 14
    3. 3. 16
    4. 4. 18

2
Let the numbers be 3x 2x and 5x. Then 9x + 4x + 25x =1862 ⇒ 38x = 1862 ⇒ x = 49 ⇒ x = 7. middle number = 2x = 14

Q.- In a mixture 60 litres the ratio of milk and water 2 : 1. If the this ratio is to be 1 : 2 then the quanity of water to be further added is

    1. 1. 50
    2. 2. 55
    3. 3. 40
    4. 4. 60

4
Quantity of Milk = 60*(2/3) = 40 liters Quantity of water = 60-40 = 20 liters As per question we need to add water to get quantity 2:1 => 40/(20+x) = 1/2 => 20 + x = 80 => x = 60 liters

Q.- Two numbers are respectively 20% and 50% more than a third number. The ratio of the two numbers is

    1. 1. 0.21111111111
    2. 2. 0.12777777778
    3. 3. 0.17013888889
    4. 4. 0.16875

3
Let the third number be x. First Number (120/100)*x = 6x/5 Second Number (150/100)*x = 3x/2 Ratio = 6x/5:3x/2 => 4:5

Q.- In a college the ratio of the number of boys to girls is 8 : 5. If there are 200 girls the total number of students in the college is

    1. 1. 205
    2. 2. 520
    3. 3. 502
    4. 4. 250

2
Let the boy are 8x and Girls are 5x => 5x = 200 => x = 40 Total students = 8x+5x = 13x = 13(40) = 520

Q.- Rs. 120 are divided among A B C such that A's share is Rs. 20 more than B's and Rs. 20 less than C's. What is B's share

    1. 1. Rs.25
    2. 2. Rs.20
    3. 3. Rs.18
    4. 4. Rs.30

2
Let C = x. Then A = (x—20) and B = (x—40). x + x - 20 + x - 40 = 120 Or x=60. A:B:C = 40:20:60 = 2:1 :3. B's share = Rs. 120*(1/6) = Rs. 20

Q.- If 2 : 9 :: x : 18 then find the value of x

    1. 1. 7
    2. 2. 6
    3. 3. 4
    4. 4. 5

3
Treat 2:9 as 2/9 and x:18 as x/18 treat :: as = So we get 2/9 = x/18 => 9x = 36 => x = 4

Q.- Ratio and Proportion - Aptitude. If 10% of x = 20% of y then x:y is equal to:

    1. 1. 0.043055555556
    2. 2. 0.084027777778
    3. 3. 0.20902777778
    4. 4. 0.41736111111

2
10% of x = 20% of y 10x/100 = 20y/100 x/10 = y/5 x/y = 10/5 = 2/1 x:y = 2:1.

Q.- The ratio of the number of boys and girls in a college is 7:8. If the percentage increase in the number of boys and girls be 20% and 10% respectively. What will be the new ratio?

    1. 1. 0.33958333333
    2. 2. 0.22083333333
    3. 3. 0.39027777778
    4. 4. Cannot be determined

3
Originally let the number of boys and girls in the college be 7x and 8x respectively. Their increased number is (120% of 7x) and (110% of 8x). i.e. (120/100 * 7x) and (110/100 * 8x) i.e. 42x/5 and 44x/5 Required ratio = 42x/5 : 44x/5 = 21:22

Q.- Seats for Mathematics Physics and Biology in a school are in the ratio 5:7:8. There is a proposal to increase these seats by 40% 50% and 75% respectively. What will be the ratio of increased seats?

    1. 1. 0.085462962963
    2. 2. 0.2549537037
    3. 3. 0.25565972222
    4. 4. None of these

1
Originally let the number of seats for Mathematics Physics and Biology be 5x 7x and 8x respectively. Number of increased sears are (140% of 5x) (150% of 7x) and (175% of 8x) i.e. (140/100 * 5x) (150/100 * 7x) and (175/100 * 8x) i.e. 7x 21x/2 and 14x Required ratio = 7x:21x/2:14x = 14x : 21x : 28x = 2:3:4

Q.- In a bag there are coins of 25 p 10 p and 5 p in the ratio of 1:2:3. If there are Rs. 30 in all how many 5 p coins are there?

    1. 1. 50
    2. 2. 100
    3. 3. 150
    4. 4. 200

3
Let the number of 25 p 10 p and 5 p coins be x 2x and 3x respectively. Then sum of their values = [25x/100 + (10 * 2x)/100 + (5 * 3x)/100] = Rs. 60x/100 60x/100 = 30 => x = 50. Hence the number of 5 p coins = 3 * 50 = 150.

Q.- If A:B = 2:3 B:C = 4:5 and C:D = 6:7 then A:B:C:D is:

    1. 1. 16:22:30:35
    2. 2. 16:24:15:35
    3. 3. 16:24:30:35
    4. 4. 18:24:30:35

3
A:B = 2:3 B:C = 4:5 = (4 * 3/4):(5 * 3/4) = 3:15/4 and C:D = 6:7 = (6 * 15/24):(7 * 15/24) = 15/4:35/8 A:B:C:D = 2:3:15/4:35/8 = 16:24:30:35

Q.- In a fort there are 1200 soldiers. If each soldier consumes 3 kg per day the provisions available in the fort will last for 30 days. If some more soldiers join the provisions available will last for 25 days given each soldier consumes 2.5 kg per day. Find

    1. 1. 582
    2. 2. 528
    3. 3. 258
    4. 4. 285

2
Assume x soldiers join the fort. 1200 soldiers have provision for 1200 (days for which provisions last them)(rate of consumption of each soldier) = (1200)(30)(3) kg. Also provisions available for (1200 + x) soldiers is (1200 + x)(25)(2.5) k As the same provisions are available => (1200)(30)(3) = (1200 + x)(25)(2.5) x = [(1200)(30)(3)] / (25)(2.5) - 1200 x = 528.

Q.- Two men Amar and Bhuvan have the ratio of their monthly incomes as 6 : 5. The ratio of their monthly expenditures is 3 : 2. If Bhuvan saves one-fourth of his income find the ratio of their monthly savings?

    1. 1. 0.12847222222
    2. 2. 0.13194444444
    3. 3. 0.13055555556
    4. 4. 0.043055555556

2
Let the monthly incomes of Amar and Bhuvan be 6x and 5x respectively. Let the monthly expenditure of Amar and Bhuvan be 3y and 2y respectively. Savings of Bhuvan every month = 1/4(5x) = (His income) - (His expenditure) = 5x - 2y. => 5x = 20x - 8y => y = 15x/8. Ratio of savings of Amar and Bhuvan = 6x - 3y : 1/4(5x) = 6x - 3(15x/8) : 5x/4 = 3x/8 : 5x/4 = 3 : 10.

Q.- Find the numbers which are in the ratio 3 : 2 : 4 such that the sum of the first and the second added to the difference of the third and the second is 21 ?

    1. 1. 12 8 16
    2. 2. 6 4 8
    3. 3. 9 6 24
    4. 4. 9 6 12

4
Let the numbers be a b and c. Given that a b and c are in the ratio 3 : 2 : 4. let a = 3x b = 2x and c = 4x Given (a+b) + (c - b) = 21 a + b + c - b = 21 a + c = 21 3x + 4x = 21 7x = 21 x = 3 a b c are 3x 2x 4x. a b c are 9 6 12.

Q.- Amar Bhavan and Chetan divide an amount of Rs. 5600 among themselves in the ratio 3 : 6 : 5. If an amount of Rs. 400 is deducted from each of their shares what will be the new ratio of their shares of the amount?

    1. 1. 0.17159722222
    2. 2. 0.044479166667
    3. 3. 0.086851851852
    4. 4. 0.21607638889

3
Let the shares of Amar Bhavan and Chetan be Rs. 3x Rs. 6x and Rs. 5x respectively. 3x + 6x + 5x = 5600 => 14x = 5600 => x = 400. Required ratio = 3x - 400 : 6x - 400 : 5x - 400 = 3x - x : 6x - x : 5x - x = 2x : 5x : 4x => 2 : 5 : 4

Q.- Three persons A B and C divide a certain amount of money such that A's share is Rs.4 less than half of the total amount. B's share is Rs.8 more than half of what is left and finally C takes the which is Rs.14. Find the total amount they initially had with

    1. 1. Rs.60
    2. 2. Rs.70
    3. 3. Rs.80
    4. 4. Rs.75

3
Let the total amount be Rs.P Let shares of A and B be Rs.X and Rs.Y respectively. Given C's share was Rs.14. X + Y + 14 = P --- (1) From the given data X = P/2 - 4 --- (2) Remaining amount = P - (P/2 - 4) = P/2 + 4 Y = 1/2(P/2 + 4) + 8 P/4 + 10 --- (3) From (1) (2) and (3) P/2 - 4 + P/4 + 10 + 14 = P 3P/4 + 20 = P P - 3P/4 = 20 P/4 = 20 P = Rs.80

Q.- In one hour a boat goes 11 km/hr along the stream and 5 km/hr against the stream. The speed of the boat in still water (in km/hr) is :

    1. 1. 8 kmph
    2. 2. 6 kmph
    3. 3. 7 kmph
    4. 4. 9 kmph

1
Speed in still water = 1/2 (11 + 5) kmph = 8 kmph.

Q.- A man's speed with the current is 15 km/hr and the speed of the current is 2.5 km/hr. The man's speed against the current is :

    1. 1. 8 kmph
    2. 2. 10kmph
    3. 3. 7 kmph
    4. 4. 6kmph

2
Man's rate in still water = (15 - 2.5) km/hr = 12.5 km/hr. Man's rate against the current = (12.5 - 2.5) km/hr = 10 km/hr.

Q.- A boat can travel with a speed of 13 km/hr in still water. If the speed of the stream is 4 km/hr find the time taken by the boat to go 68 km downstream.

    1. 1. 6 hrs
    2. 2. 5 hrs
    3. 3. 4 hrs
    4. 4. 3 hrs

3
Speed downstream = (13 + 4) km/hr = 17 km/hr. Time taken to travel 68 km downstream = 68 /17 hrs = 4 hrs.

Q.- A boat covers a certain distance downstream in 1 hour while it comes back in 3/2 hours. If the speed of the stream be 3 kmph what is the speed of the boat in still water ?

    1. 1. 10 kmph
    2. 2. 15 kmph
    3. 3. 12 kmph
    4. 4. 14 kmph

2
Let the speed of the boat in still water be x kmph. Then Speed downstream = (x + 3) kmph Speed upstream = (x - 3) kmph. (x + 3) x 1 = (x - 3) x 3/2 2x + 6 = 3x - 9 x = 15 kmph.

Q.- A man can row at 5 kmph in still water. If the velocity of current is 1 kmph and it takes him 1 hour to row to a place and come back how far is the place ?

    1. 1. 3.6 km
    2. 2. 3.2 km
    3. 3. 4.2 km
    4. 4. 2.4 km

4
Speed downstream = (5 + 1) kmph = 6 kmph. Speed upstream = (5 - 1) kmph = 4 kmph. Let the required distance be x km. Then x + x = 1 2x + 3x = 12 5x = 12 x = 2.4 km.

Q.- A boat takes 90 minutes less to travel 36 miles downstream than to travel the same distance upstream. If the speed of the boat in still water is 10 mph the speed of the stream is :

    1. 1. 5 mph
    2. 2. 4 mph
    3. 3. 2 mph
    4. 4. 3 mph

3
Let the speed of the stream x mph. Then Speed downstream = (10 + x) mph Speed upstream = (10 - x) mph. 36 +36/(10 - x) (10 + x) = 90/60 72x x 60 = 90 (100 - x2) x2 + 48x - 100 = 0 (x+ 50)(x - 2) = 0 x = 2 mph.

Q.- The speed of a boat in still water in 15 km/hr and the rate of current is 3 km/hr. The distance travelled downstream in 12 minutes is :

    1. 1. 2.5 km
    2. 2. 5.2 km
    3. 3. 6.3 km
    4. 4. 3.6 km

4
Speed downstream = (15 + 3) kmph = 18 kmph. Distance travelled = 18 x (12/60) km = 3.6 km.

Q.- A boat running downstream covers a distance of 16 km in 2 hours while for covering the same distance upstream it takes 4 hours. What is the speed of the boat in still water ?

    1. 1. 6 kmph
    2. 2. 7 kmph
    3. 3. 8 kmph
    4. 4. 5 kmph

1
Rate downstream = 16/2 kmph = 8 kmph. Rate upstream = 16/4 kmph = 4 kmph. Speed in still water = 1/2 (8 + 4) kmph = 6 kmph.

Q.- A boatman goes 2 km against the current of the stream in 1 hour and goes 1 km along the current in 10 minutes. How long will it take to go 5 km in stationary water ?

    1. 1. 1 hr 15 min
    2. 2. 2 hr 15 min
    3. 3. 3 hr 15 min
    4. 4. 4 hr 15 min

1
Rate downstream = (1/10) x 60 km/hr = 6 km/hr. Rate upstream = 2 km/hr. Speed in still water = 1/2 (6 + 2) km/hr = 4 km/hr. Required time = 5/4 hrs = 5/4 hrs = 1 hr 15 min.

Q.- Speed of a boat in standing water is 9 kmph and the speed of the stream is 1.5 kmph. A man rows to a place at a distance of 105 km and comes back to the starting point. The total time taken by him is :

    1. 1. 30 hrs
    2. 2. 40 hrs
    3. 3. 42 hrs
    4. 4. 24 hrs

4
Speed upstream = 7.5 kmph. Speed downstream = 10.5 kmph. Total time taken = (105/7.5) + (105/10.5) hours = 24 hours.

Q.- A boatman can row 3 km against the stream in 20 minutes and return in 18 minutes. Find the rate of current

    1. 1. 3/1 km/hr
    2. 2. 2/3 km/hr
    3. 3. 2/1 km/hr
    4. 4. 1/2 km/hr

4
Speed upstream = 3/(20/60) = 9 km/hr Speed downstream = 3/(18/60) = 10 km/hr Rate of current = (10−9)/2=1/2 km/hr

Q.- A boatman can row 96 km downstream in 8 hr. If the speed of the current is 4 km/hr then find in what time will be able to cover 8 km upstream?

    1. 1. 2 hrs.
    2. 2. 3 hrs.
    3. 3. 2.5 hrs.
    4. 4. 3.5 hrs.

1
Speed downstream = 96⁄8 = 12 kmph Speed of current = 4 km/hr Speed of the boatman in still water = 12-4 = 8 kmph Speed upstream = 8-4 = 4 kmph Time taken to cover 8 km upstream = 8⁄4 = 2 hours

Q.- A man can row 40 kmph in still water and the river is running at 10 kmph. If the man takes 1 hr to row to a place and back how far is the place?

    1. 1. 18.57 km
    2. 2. 18.75 km
    3. 3. 51.87 km
    4. 4. 51.78 km

2
Let the distance be x Speed upstream = (40 - 10) = 30 kmph Speed downstream = (40 + 10) = 50 kmph Total time taken = 1 hr (x/50)+(x/30)=1 ⇒8x/150=1 ⇒x=150/8 x= 18.75 km

Q.- A man can row 4 kmph is still water. If the river is running at 2 kmph it takes 90 min to row to a place and back. How far is the place?

    1. 1. 3.22 km
    2. 2. 5.22 km
    3. 3. 2.25 km
    4. 4. 2.52 km

3
Speed in still water = 4 kmph Speed of the stream = 2 kmph Speed upstream = (4-2)= 2 kmph Speed downstream = (4+2)= 6 kmph Total time = 90 minutes = 90⁄60 hour = 3⁄2 hour Let L be the distance. Then L6+L2=32 => L + 3L = 9 => 4L = 9 => L = 9⁄4= 2.25 km

Q.- If a man rows at the rate of 5 kmph in still water and his rate against the current is 3 kmph then the man's rate along the current is:

    1. 1. 5 kmph
    2. 2. 7.5 kmph
    3. 3. 6 kmph
    4. 4. 7 kmph

4
Let the rate along with the current is x km/hr (x+3)/2=5 => x + 3 = 10 => x = 7 kmph

Q.- Two pipes A and B can fill a tank in 10 hrs and 40 hrs respectively. If both the pipes are opened simultaneously how much time will be taken to fill the tank?

    1. 1. 8.5 hrs.
    2. 2. 7 hrs.
    3. 3. 8 hrs.
    4. 4. 9 hrs.

3
Pipe A can fill 1⁄10 of the tank in 1 hr Pipe B can fill 1⁄40 of the tank in 1 hr Pipe A and B together can fill 1⁄10 + 1⁄40 = 1⁄8 of the tank in 1 hr i.e. Pipe A and B together can fill the tank in 8 hours

Q.- In a river flowing at 2 km/hr a boat travels 32 km upstream and then returns downstream to the starting point. If its speed in still water be 6 km/hr find the total journey time.

    1. 1. 13 hr
    2. 2. 12.5 hr
    3. 3. 11 hr
    4. 4. 12 hr

4
speed of the boat = 6 km/hr Speed downstream = (6+2) = 8 km/hr Speed upstream = (6-2) = 4 km/hr Distance travelled downstream = Distance travelled upstream = 32 km Total time taken = Time taken downstream + Time taken upstream =(32/8)+(32/4)=(32/8)+(64/8)=96/8 = 12 hr

Q.- A man can row at 5 kmph in still water. If the velocity of current is 1 kmph and it takes him 1 hour to row to a place and come back how far is the place?

    1. 1. 2.4 km
    2. 2. 2.3 km
    3. 3. 3.2 km
    4. 4. 4.2 km

1
Speed in still water = 5 kmph Speed of the current = 1 kmph Speed downstream = (5+1) = 6 kmph Speed upstream = (5-1) = 4 kmph Let the requited distance be x km Total time taken = 1 hour =>x6+x4=1 => 2x + 3x = 12 => 5x = 12 => x = 2.4 km

Q.- A man takes twice as long to row a distance against the stream as to row the same distance in favour of the stream. The ratio of the speed of the boat (in still water) and the stream is:

    1. 1. 0.04375
    2. 2. 0.12569444444
    3. 3. 0.085416666667
    4. 4. 0.12638888889

2
Let speed upstream = x Then speed downstream = 2x Speed in still water = (2x+x)/2=3x/2 Speed of the stream = (2x−x)/2=x/2 Speed in still water : Speed of the stream = 3x/2:x/2 = 3:1

Q.- A boat can travel with a speed of 22 km/hr in still water. If the speed of the stream is 5 km/hr find the time taken by the boat to go 54 km downstream

    1. 1. 2.5 hrs.
    2. 2. 4 hrs.
    3. 3. 3 hrs.
    4. 4. 2 hrs.

4
Speed of the boat in still water = 22 km/hr speed of the stream = 5 km/hr Speed downstream = (22+5) = 27 km/hr Distance travelled downstream = 54 km Time taken = distance/speed=54/27 = 2 hours

Q.- The speed at which a man can row a boat in still water is 15 kmph. If he rows downstream where the speed of current is 3 kmph what time will he take to cover 60 metres?

    1. 1. 10 seconds
    2. 2. 15 seconds
    3. 3. 20 seconds
    4. 4. 12 seconds

4
Speed of the boat downstream = 15 + 3 = 18 kmph = 18 * 5/18 = 5 m/s Hence time taken to cover 60 m = 60/5 = 12 seconds.

Q.- A person can row at 9 kmph and still water. He takes 4 1/2 hours to row from A to B and back. What is the distance between A and B if the speed of the stream is 1 kmph?

    1. 1. 32 km
    2. 2. 25 km
    3. 3. 28 km
    4. 4. None of these

4
Let the distance between A and B be x km. Total time = x/(9 + 1) + x/(9 - 1) = 4.5 => x/10 + x/8 = 9/2 => (4x + 5x)/40 = 9/2 => x = 20 km.

Q.- A man can row downstream at 18 kmph and upstream at 10 kmph. Find the speed of the man in still water and the speed of stream respectively?

    1. 1. 13 3
    2. 2. 12 6
    3. 3. 14 4
    4. 4. 15 3

3
Let the speed of the man in still water and speed of stream be x kmph and y kmph respectively. Given x + y = 18 --- (1) and x - y = 10 --- (2) From (1) & (2) 2x = 28 => x = 14 y = 4.

Q.- A man can row 30 km downstream and 20 km upstream in 4 hours. He can row 45 km downstream and 40 km upstream in 7 hours. Find the speed of man in still water?

    1. 1. 15 kmph
    2. 2. 10 kmph
    3. 3. 12 kmph
    4. 4. 12.5 kmph

4
Let the speed of the man in still water be a kmph and let the speed of the stream be b kmph. Now 30/(a + b) + 20/(a - b) = 4 and 45/(a + b) + 40/(a - b) = 7 Solving the equation The speed of man in still water is 12.5 kmph.

Q.- The time taken by a man to row his boat upstream is twice the time taken by him to row the same distance downstream. If the speed of the boat in still water is 42 kmph find the speed of the stream?

    1. 1. 12 kmph
    2. 2. 13 kmph
    3. 3. 14 kmph
    4. 4. 15 kmph

3
The ratio of the times taken is 2:1. The ratio of the speed of the boat in still water to the speed of the stream = (2+1)/(2-1) = 3/1 = 3:1 Speed of the stream = 42/3 = 14 kmph.

Q.- The speed of a boat in upstream is 60 kmph and the speed of the boat downstream is 80 kmph. Find the speed of the boat in still water and the speed of the stream?

    1. 1. 70 10 kmph
    2. 2. 35 27 kmph
    3. 3. 50 60 kmph
    4. 4. 45 55 kmph

1
Speed of the boat in still water = (60+80)/2 = 70 kmph. Speed of the stream = (80-60)/2 = 10 kmph.

Q.- A man whose speed is 4.5 kmph in still water rows to a certain upstream point and back to the starting point in a river which flows at 1.5 kmph find his average speed for the total journey?

    1. 1. 8 kmph
    2. 2. 4 kmph
    3. 3. 2 kmph
    4. 4. 10 kmph

2
M = 45 S = 1.5 DS = 6 US = 3 AS = (2 * 6 * 3) /9 = 4

Q.- The speed of a boat in still water is 60kmph and the speed of the current is 20kmph. Find the speed downstream and upstream?

    1. 1. 35 25 kmph
    2. 2. 80 40 kmph
    3. 3. 40 60 kmph
    4. 4. 50 55 kmph

2
Speed downstream = 60 + 20 = 80 kmph Speed upstream = 60 - 20 = 40 kmph

Q.- A man swims downstream 30 km and upstream 18 km taking 3 hours each time what is the speed of the man in still water?

    1. 1. 2 kmph
    2. 2. 8 kmph
    3. 3. 16 kmph
    4. 4. 4 kmph

2
30 --- 3 DS = 10 ? ---- 1 18 ---- 3 US = 6 ? ---- 1 M = ? M = (10 + 6)/2 = 8

Q.- A man can row with a speed of 15 kmph in still water. If the stream flows at 5 kmph then the speed in downstream is?

    1. 1. 10 kmph
    2. 2. 5 kmph
    3. 3. 20 kmph
    4. 4. 22 kmph

3
M = 15 S = 5 DS = 15 + 5 = 20

Q.- In a 100 m race A can beat B by 25 m and B can beat C by 4 m. In the same race A can beat C by :

    1. 1. 28 m
    2. 2. 38 m
    3. 3. 30 m
    4. 4. 26 m

1
A : B = 100 : 75 B : C = 100 : 96. A : C = (A/B) x (B/C) = (100/75) x (100/96) = 100/72A = 100 : 72. A beats C by (100 - 72) m = 28 m.

Q.- A runs 5/3 times as fast as B. If A gives B a start of 80 m how far must the winning post be so that A and B might reach it at the same time ?

    1. 1. 250 m
    2. 2. 200 m
    3. 3. 350 m
    4. 4. 300 m

2
Ratio of the speeds of A and B = 5 : 3. Thus in race of 5 m A gains 2 m over B. 2 m are gained by A in a race of 5 m. 80 m will be gained by A in race of (5/2) x 80 m = 200 m. Winning post is 200 m away from the starting point.

Q.- In a 300 m race A beats B by 22.5 m or 6 seconds. B's time over the course is :

    1. 1. 70 sec
    2. 2. 60 sec
    3. 3. 80 sec
    4. 4. 08 sec

3
B runs 45/2 m in 6 sec. B covers 300 m in 6 x (2/45) x 300 sec = 80 sec.

Q.- A can run 22.5 m while B runs 25 m. In a kilometre race B beats A by :

    1. 1. 950 m
    2. 2. 100 m
    3. 3. 850 m
    4. 4. 700 m

2
When B runs 25 m A runs 45/2 m. When B runs 1000 m A runs (45/2) x (1/25) x 1000 m = 900 m. B beats A by 100 m.

Q.- In a 200 metres race A beats B by 35 m or 7 seconds. A's time over the course is :

    1. 1. 30 sec
    2. 2. 23 sec
    3. 3. 33 sec
    4. 4. 40 sec

3
B runs 35 m in 7 sec. B covers 200 m in (7 /35)x 200 = 40 sec. B's time over the course = 40 sec. A's time over the course (40 - 7) sec = 33 sec.

Q.- In a game of 100 points A can give B 20 points and C 28 points. Then B can give C :

    1. 1. 6 points
    2. 2. 9 points
    3. 3. 8 points
    4. 4. 10 points

4
A : B = 100 : 80. A : C = 100 : 72. B/C = (B/A) x (A/C) = (80/100) x( 100/72) = 10/9 = 100/90 = 100 : 90. B can give C 10 points.

Q.- In 100 m race A covers the distance in 36 seconds and B in 45 seconds. In this race A beats B by :

    1. 1. 35 m
    2. 2. 30 m
    3. 3. 20 m
    4. 4. 25 m

3
Distance covered by B in 9 sec. = (100/45) x 9 m = 20 m. A beats B by 20 metres.

Q.- At a game of billiards A can give B 15 points in 60 and A can give C to 20 points in 60. How many points can B give C in a game of 90 ?

    1. 1. 8 points
    2. 2. 10 points
    3. 3. 7 points
    4. 4. 9 points

2
A : B = 60 : 45. A : C = 60 : 40. B/C = (B/A) x (A/C) = (45/60) x (60/40) = 45/40 = 90/80 = 90 : 80. B can give C 10 points in a game of 90.

Q.- In a 500 m race the ratio of the speeds of two contestants A and B is 3 : 4. A has a start of 140 m. Then A wins by :

    1. 1. 20 m
    2. 2. 15 m
    3. 3. 16 m
    4. 4. 18 m

1
To reach the winning post A will have to cover a distance of (500 - 140)m i.e. 360 m. While A covers 3 m B covers 4 m. While A covers 360 m B covers 4 x 360 m = 480 m. Thus when A reaches the winning post B covers 480 m and therefore remains 20 m behind. A wins by 20 m.

Q.- In a 100 m race A can give B 10 m and C 28 m. In the same race B can give C :

    1. 1. 50 m
    2. 2. 20 m
    3. 3. 30 m
    4. 4. 40 m

2
A : B = 100 : 90. A : C = 100 : 72. B : C = (B/A) x (A/C) = (90/100) x (100/72) = 90/72 When B runs 90 m C runs 72 m. When B runs 100 m C runs (72/90) x 100 m = 80 m. B can give C 20 m.

Q.- In one km race A beats B by 4 seconds or 40 metres. How long does B take to run the kilometer?

    1. 1. 80
    2. 2. 90
    3. 3. 110
    4. 4. 100

4
This means B takes 4 seconds to run 40 metres => B takes 4/40=110 seconds to run 1 metre => B takes (1/10)×1000=100 seconds to run 1000 metre

Q.- P runs 1 km in 3 minutes and Q in 4 minutes 10 secs. How many metres start can P give Q in 1 kilometre race so that the race may end in a dead heat?

    1. 1. 280 m.
    2. 2. 208 m.
    3. 3. 802 m.
    4. 4. 820 m.

1
P run 1 km in 3 minutes Q run 1 km in 4 minutes 10 secs => Q runs 1 km in 25/6 minutes=> Q runs (1×(6/25)×3)=18/25=0.72 km in 3 minutes Hence in a 1 km race P can give Q (1 - 0.72)=0.28 km = 280 metre

Q.- In a 100 metres race. A runs at a speed of 2 metres per seconds. If A gives B a start of 4 metres and still beats him by 10 seconds. find the speed of B.

    1. 1. 3.2
    2. 2. 2.3
    3. 3. 6.1
    4. 4. 1.6

4
Speed of A = 2 m/s Time taken by A to run 100 m distance/speed=100/2 = 50 seconds A gives B a start of 4 metres and still A beats him by 10 seconds => B runs (100-4)=96 m in (50+10)=60 seconds Speed of B = distance/time=96/60 = 1.6 m/s

Q.- In a game of 90 points A can give B 15 points and C 30 points. How many points can B give C in a game of 100 points?

    1. 1. 10
    2. 2. 20
    3. 3. 30
    4. 4. 25

2
While A scores 90 points B scores (90-15)=75 points and C scores (90-30)= 60 points i.e. when B scores 75 points C scores 60 points => When B scores 100 points C scores (60/75)×100 = 80 points i.e. in a game of 100 points B can give C (100-80)=20 points

Q.- In a 200 metres race A beats B by 35 m or 7 seconds. A's time over the course is:

    1. 1. 35
    2. 2. 22
    3. 3. 33
    4. 4. 36

3
B runs 35 m in 7 sec. => B runs 200 m in (7/35)×200 = 40 seconds Since A beats B by 7 seconds A runs 200 m in (40-7) = 33 seconds Hence A's time over the course = 33 seconds

Q.- A B and C are the three contestants in one km race. If A can give B a start of 40 metres and A can give C a start of 64 metres. How many metres start can B give C?

    1. 1. 30 m
    2. 2. 25 m
    3. 3. 20 m
    4. 4. 22 m

2
While A covers 1000 m B covers (1000-40)=960 m and C covers (1000-64)=936 m i.e. when B covers 960 m C covers 936 m When B covers 1000 m C covers (936/960)×1000 = 975 m i.e. B can give C a start of (1000-975) = 25 m

Q.- In a 100 m race A beats B by 10 m and C by 13 m. In a race of 180 m B will beat C by:

    1. 1. 6 m
    2. 2. 7 m
    3. 3. 8 m
    4. 4. 5 m

1
While A runs 100 m B runs (100-10)=90 m and C runs (100-13)=87 m i.e. when B runs 90 m C runs 87 m => when B runs 180 m C runs (87/90)×180= 174 m Hence in a 180 m race B will beat C by (180-174)=6 m

Q.- In a game of 100 points A can give B 20 points and C 28 points. Then B can give C:

    1. 1. 12 points
    2. 2. 20 points
    3. 3. 15 points
    4. 4. 10 points

4
In a game of 100 points A scores 100 points B scores (100-20)=80 points and C scores (100-28)=72 points i.e. when B scores 80 points C scores 72 points => When B scores 100 points C scores (72/80)×100 = 90 points i.e. In a game of 100 points B can give C (100-90)=10 points

Q.- In 100 m race A covers the distance in 36 seconds and B in 45 seconds. In this race A beats B by:

    1. 1. 25 m
    2. 2. 20 m
    3. 3. 23 m
    4. 4. 24 m

2
In 100 m race A covers the distance in 36 seconds and B in 45 seconds. Clearly A beats B by (45-36)=9 seconds Speed of B = Distance/Time=100/45 m/s Distance Covered by B in 9 seconds = Speed × Time = (100/45)×9 = 20 metre i.e. A beats B by 20 metre

Q.- In a 300 m race A beats B by 22.5 m or 6 seconds. B's time over the course is:

    1. 1. 80 sec.
    2. 2. 85 sec.
    3. 3. 70 sec.
    4. 4. 75 sec.

1
B runs 22.5 m in 6 seconds => B runs 300 m in (6/22.5)×300 = 80 seconds i.e. B's time over the course = 80 seconds

Q.- In a 100 m race A runs at 6 km/hr. A gives B a start of 4 m and still beats him by 4 sec. Find the speed of B.

    1. 1. 4.6 km/hr
    2. 2. 5.6 km/hr
    3. 3. 4.4 km/hr
    4. 4. 5.4 km/hr

4
Speed of A = 6 km/hr = 6×5/18 = 5/3 m/s Time taken by A = distance/speed = 100/(5/3) = 60 sec A gives B a start of 4 m i.e. B has to cover (100-4)=96 metre A beats B by 4 sec Time taken by B to run the race = 60 + 4 = 64 sec Speed of B = Distance/time = 96/64 m/s = 5.4 km/hr

Q.- A and B run a 5 km race on a round course of 400 m. If their speeds be in the ratio 5 : 4 how often does the winner pass the other?

    1. 1. 5 times
    2. 2. 6 times
    3. 3. 5/2 times
    4. 4. 4 times

3
speeds of A and B are in the ratio 5 : 4 when A covers 5 metre B covers 4 metre i.e. when A covers 5 metre A gains 1 metre over B when A covers 5 ×400 = 2000 metre A gains 400 metre over B. i.e. when A covers 2000 metre the winner pass the other 1 time. when A covers 5000 metre the winner pass the other 5000/2000 = 5/2 times

Q.- A and B run a km race. If A gives B a start of 50 m A wins by 14 sec and if A gives B a start of 22 sec B wins by 20 m. The time taken by A to run a km is

    1. 1. 120
    2. 2. 110
    3. 3. 130
    4. 4. 100

4
Let time taken by A to run 1000 m = a and time taken by B to run 1000 m = b If A gives B a start of 50 m A wins by 14 sec distance travelled by A = 1000 distance travelled by B = (1000-50)=950 Time taken by B to run 950 m - Time taken by A to run 1000 m = 14 sec (950/1000)(b-a) = 14 95/100 (b-a) = 14---(Equation 1) If A gives B a start of 22 sec B wins by 20 m i.e. A starts 22 seconds after B starts from the same starting point. i.e. here B runs for b seconds where A runs for (b-22) seconds => Distance Covered by B in b seconds - Distance Covered by A in (b-22) seconds = 20 =>1000-(1000/a)×(b-22)=20 (1000/a)×(b-22)=980 1000b-22000=980a 100b-2200=98a 50b-1100=49a 50b-49a = 1100---(Equation 2) (Equation 1)×49?[(95×49)/100]b-49a = 14×49 ? 46.55b-49a = 686---(Equation 3) (Equation 2) - (Equation 3) ? 3.45b = 414 b = 414/3.45 = 120 seconds Substituting this value of b in Equation 1 [(95×120)/100]-a = 14 114-a = 14 a =114-14 = 100 i.e. The time taken by A to run a km = 100 seconds

Q.- A can run a kilometre in 4 min 54 sec and B in 5 min. How many metres start can A give B in a km race so that the race may end in a dead heat?

    1. 1. 22 m
    2. 2. 20 m
    3. 3. 18 m
    4. 4. 24 m

2
i.e. in this race A will win by (5 min - 4 min 54 sec) = (300 sec - 294 sec) = 6 sec B covers 1000 m in 5 min => B covers 1000 m in 300 sec => B covers 1000/300 = 10/3 m in 1 sec=> B covers (10/3)×6 = 20 m in 6 sec i.e. A can give B a start of 20 metre so that the race will end in a dead heat.

Q.- A runs 4/3 as fast as B. If A gives B a start of 30 metres. How far must be the wining post so that the race ends in a dead heat?

    1. 1. 100 m
    2. 2. 110 m
    3. 3. 210 m
    4. 4. 120 m

4
Let the distance to the winning point = x Speed of A : Speed of B = 4/3 : 1 Let Speed of A = 4k/3 Speed of B = k A gives B a start of 30 metres => A run x metres and B run (x-30)metres Time taken by A = Distance run by A / Speed of A = x/(4k /3) Time taken by B= Distance run by B / Speed of B = (x-30)/k For the race to be a a dead-heat race Time taken by A = Time taken by B ?+K258 x/(4k/3) = (x-30)/k ? x/(4/3) = x-30 ? 3x = 4x-120 ? x = 120 i.e. the distance to the winning point = 120 metre

Q.- In a game of billiards A can give B 20 points in 60 and he can give C 30 points in 60. How many points can B give C in a game of 100?

    1. 1. 40
    2. 2. 30
    3. 3. 25
    4. 4. 20

3
A scores 60 while B score 40 and C scores 30. The number of points that C scores when B scores 100 = (100 * 30)/40 = 25 * 3 = 75. In a game of 100 points B gives (100 - 75) = 25 points to C.

Q.- In a race of 1000 m A can beat by 100 m in a race of 800m B can beat C by 100m. By how many meters will A beat C in a race of 600 m?

    1. 1. 57.5 m
    2. 2. 127.5 m
    3. 3. 150.7 m
    4. 4. 98.6 m

2
When A runs 1000 m B runs 900 m and when B runs 800 m C runs 700 m. When B runs 900 m distance that C runs = (900 * 700)/800 = 6300/8 = 787.5 m. In a race of 1000 m A beats C by (1000 - 787.5) = 212.5 m to C. In a race of 600 m the number of meters by which A beats C = (600 * 212.5)/1000 = 127.5 m.

Q.- A can run a kilometer race in 4 1/2 min while B can run same race in 5 min. How many meters start can A give B in a kilometer race so that the race mat end in a dead heat?

    1. 1. 150 m
    2. 2. 125 m
    3. 3. 130 m
    4. 4. 100 m

4
A can give B (5 min - 4 1/2 min) = 30 sec start. The distance covered by B in 5 min = 1000 m. Distance covered in 30 sec = (1000 * 30)/300 = 100 m. A can give B 100m start.

Q.- A can give B 100 meters start and C 200 meters start in a kilometer race. How much start can B give C in a kilometer race?

    1. 1. 111.12 m
    2. 2. 888.88 m
    3. 3. 777.52 m
    4. 4. 756.34 m

1
A runs 1000 m while B runs 900 m and C runs 800 m. The number of meters that C runs when B runs 1000 m = (1000 * 800)/900 = 8000/9 = 888.88 m. B can give C = 1000 - 888.88 = 111.12 m.

Q.- In a kilometer race A beats B by 50 meters or 10 seconds. What time does A take to complete the race?

    1. 1. 200 sec
    2. 2. 190 sec
    3. 3. 210 sec
    4. 4. 150 sec

2
Time taken by B run 1000 meters = (1000 * 10)/50 = 200 sec. Time taken by A = 200 - 10 = 190 sec.

Q.- If the true discount on s sum due 2 years hence at 14% per annum be Rs. 168 the sum due is :

    1. 1. Rs. 768
    2. 2. Rs. 678
    3. 3. Rs. 876
    4. 4. Rs. 786

1
P.W. = 100 x T.D./(RxT) = 100 x 168/(14x2) = 600. Sum = (P.W. + T.D.) = Rs. (600 + 168) = Rs. 768.

Q.- The simple interest and the true discount on a certain sum for a given time and at a given rate are Rs. 85 and Rs. 80 respectively. The sum is :

    1. 1. Rs. 1306
    2. 2. Rs.1360
    3. 3. Rs. 6013
    4. 4. Rs. 6031

2
Sum = S.I. x T.D./(S.I.) - (T.D.) = (85 x 80)/ (85 - 80) = Rs. 1360.

Q.- The interest on Rs. 750 for 2 years is the same as the true discount on Rs. 960 due 2 years hence. If the rate of interest is the same in both cases it is :

    1. 1. 0.11
    2. 2. 0.12
    3. 3. 0.14
    4. 4. 0.15

3
S.I. on Rs. 750 = T.D. on Rs. 960. This means P.W. of Rs. 960 due 2 years hence is Rs. 750. T.D. = Rs. (960 - 750) = Rs. 210. Thus S.I. on R.s 750 for 2 years is Rs. 210. Rate = (100 x 210)/(750 x 2) % = 14%

Q.- Rs. 20 is the true discount on Rs. 260 due after a certain time. What will be the true discount on the same sum due after half of the former time the rate of interest being the same ?

    1. 1. 40.4
    2. 2. 40.04
    3. 3. 10.04
    4. 4. 10.4

4
S.I. on Rs. (260 - 20) for a given time = Rs. 20. S.I. on Rs. 240 for half the time = Rs. 10. T.D. on Rs. 250 = Rs. 10. T.D. on Rs. 260 = Rs. (10/250) x 260 = Rs. 10.40

Q.- The present worth of Rs. 2310 due 2 years hence the rate of interest being 15% per annum is :

    1. 1. Rs.1680
    2. 2. Rs. 1608
    3. 3. Rs. 8016
    4. 4. Rs.8061

1
P.W. = Rs. (100 x 2310)/[100+(15x5/2)] = Rs. 1680.

Q.- The true discount on Rs. 1760 due after a certain time at 12% per annum is Rs. 160. The time after which it is due is :

    1. 1. 8 months
    2. 2. 10 months
    3. 3. 7 months
    4. 4. 6 months

2
P.W. = Rs. (1760 -160) = Rs. 1600. S.I. on Rs. 1600 at 12% is Rs. 160. Time = 100 x 160 = 5 years = (5 x 12)/(1600x12) months =5/6 years=(5/6x12) months= 10 months.

Q.- The true discount on a bill due 9 months hence at 16% per annum is Rs. 189. The amount of the bill is :

    1. 1. Rs. 1764
    2. 2. Rs. 1746
    3. 3. Rs. 4617
    4. 4. Rs. 4671

1
Let P.W. be Rs. x. Then S.I. on Rs. x at 16% for 9 months = Rs. 189. x x 16 x 9 x 1/(12*100) = 189 or x = 1575. P.W. = Rs. 1575. Sum due = P.W. + T.D. = Rs. (1575 + 189) = Rs. 1764.

Q.- A man buys a watch for Rs. 1950 in cash and sells it for Rs. 2200 at a credit of 1 year. If the rate of interest is 10% per annum the man :

    1. 1. Rs. 60
    2. 2. Rs.50
    3. 3. Rs.65
    4. 4. Rs. 70

2
S.P. = P.W. of Rs. 2200 due 1 year hence Rs. (2200 x 100)/[100 + (10 x 1)] = Rs. 2000. Gain = Rs. (2000 - 1950) = Rs. 50.

Q.- If Rs. 10 be allowed as true discount on a bill of Rs. 110 due at the end of a certain time then the discount allowed on the same sum due at the end of double the time is :

    1. 1. Rs. 18.22
    2. 2. Rs. 18.45
    3. 3. Rs. 18.33
    4. 4. Rs. 18.40

3
S.I. on Rs. (110 - 10) for a certain time = Rs. 10. S.I. on Rs. 100 for double the time = Rs. 20. T.D. on Rs. 120 = Rs. (120 - 100) = Rs. 20. T.D. on Rs. 110 = Rs. (20/120) x 110 = Rs. 18.33

Q.- A man purchased a cow for Rs. 3000 and sold it the same day for Rs. 3600 allowing the buyer a credit of 2 years. If the rate of interest be 10% per annum then the man has a gain of :

    1. 1. 0.01
    2. 2. 0
    3. 3. 0.02
    4. 4. 0.03

2
C.P. = Rs. 3000. S.P. = Rs. (3600 x 100)/[100 + (10 x 2)] = Rs. 3000. Gain = 0%.

Q.- The true dicount on a bill due 10 months hence at 6% per annum is Rs.26.25.The amount of the bill is :

    1. 1. Rs.1575
    2. 2. Rs.500
    3. 3. Rs.650.25
    4. 4. Rs.551.25

4
Amount = TD*[{100+(R*T)}/(R*T)] = Rs.(26.25*105)/5 = Rs.551.25

Q.- The true discount on Rs.2575 due 4 months hence is Rs.75.The rate present is :

    1. 1. 0.06
    2. 2. 0.08
    3. 3. 0.09
    4. 4. 0.05

3
PW = Rs.(2575-75) = Rs.2500 so Rate = (100*75*3)/2500*1 = 9%

Q.- The true discount on Rs.1860 due after a certain time at 5% is Rs.60.The time after which it is due is :

    1. 1. 6 months
    2. 2. 8 months
    3. 3. 9 months
    4. 4. 10 months

2
PW = (sum due)-TD = Rs.1860-60 = Rs.1800. Thus Rs.60 is SI on Rs.1800 at 5% per annum. so Time = (100*60)/(1800*5)years = 2/3 years = 8 months

Q.- A man buys a watch for Rs.195 in cash and sells it for Rs.220 at a credit of 1 year.If the rate of interest is 10% the man gains :

    1. 1. Rs.15
    2. 2. Rs.3
    3. 3. Rs.5
    4. 4. Rs.10

3
PW of Rs.220 due 1 year hence = Rs.(100*220)/(100+10) = Rs.200 so the man gains Rs.5

Q.- If Rs.10 be allowed as true discount on a bill of Rs.110 due at the end of a certain time.Then the discount allowed on the same sum due at the end of double the time is :

    1. 1. Rs.20
    2. 2. Rs.21.81
    3. 3. Rs.22
    4. 4. Rs.18.33

4
SI on Rs.(110-10) for a given time = Rs.10 SI on Rs.100 for double time= Rs.20. Sum = Rs.100+20 = Rs.120 TD on Rs.110 = Rs.(20/120)*110 = Rs.18.33

Q.- The simple interest and the true discount on a certain sum for a given time and at a given rate are Rs.25 and Rs.20 respectively.The sum is :

    1. 1. Rs.500
    2. 2. Rs.200
    3. 3. Rs.250
    4. 4. Rs.100

4
Sum = (SI*TD)/(SI-TD) = Rs.(25*20)/(25-20) = Rs.100

Q.- The interest on Rs.750 for two years is equal to the true discount on Rs.810 for the same time and at the same rate.The rate percent is :

    1. 1. 0.06
    2. 2. 0.07
    3. 3. 0.04
    4. 4. 0.05

3
Since TD is SI on PW. We have Rs(810-750) or Rs 60 as SI on Rs 750 for two years. So Rate = (100*60)/750*2 = 4%

Q.- If the true discount on a sum due 2 years hence at 5% per annum be Rs 75.Then the sum due is :

    1. 1. Rs 750
    2. 2. Rs 825
    3. 3. Rs 875
    4. 4. Rs 800

2
PW = (100*TD)/R*T = Rs.(100*75)/5*2 = Rs.750. So sum due = Rs.750+75 = Rs.825

Q.- The true discount on Rs.2575 due 4 months hence is Rs.75.Find the rate percent of interest.

    1. 1. 0.1
    2. 2. 0.11
    3. 3. 0.095
    4. 4. 0.09

4
PW = Rs.(2575-75) = Rs.2500 so SI on Rs.2500 for 4 months is Rs.75. Hence rate = (100*75*3)/2500*1 = 9%

Q.- The true discount on a bill due 9 months hence at 6% per annum is Rs.180.Find the amount of the bill and its present worth.

    1. 1. Rs.4180
    2. 2. Rs.1480
    3. 3. Rs4108
    4. 4. Rs.1408

1
PW = (100*TD)/(R*T) = Rs.(100*180)/6*(3/4) = Rs.4000. Amount = (PW+TD) = Rs.(4000+180) = Rs.4180

Q.- The difference between the simple interest and true discount on a certain sum of money for 6 months at 12% per annum is Rs. 25. Find the sum.

    1. 1. 8600
    2. 2. 6800
    3. 3. 6008
    4. 4. 6080

2
Let the sum be Rs. x. T.D. = (x*25/2*1/2)/(100+(25/2*1/2)) = x*25/4*4/425 = x/17 S.I=x*25/2*1/2*1/100=x/16 x/16-x/17=25 =>17x-16x=25*16*17 =>x=6800 Hence sum due = Rs. 6800.

Q.- Find the present worth of Rs. 930 due 3 years hence at 8% per annum. Also find the discount.

    1. 1. 650 280
    2. 2. 750 180
    3. 3. 800 130
    4. 4. 850 80

2
P.W=100 x Amount /[100 + (R x T)] =Rs.100 x 930/(100 + (8x3)) = (100x930)/124 = Rs. 750 T.D. = (Amount) - (P.W.) = Rs. (930 - 750) = Rs. 180.

Q.- The true discount on a bill due 9 months hence at 12% per annum is Rs.540. Find the amount of the bill and its present worth.

    1. 1. Rs.6000
    2. 2. Rs.6500
    3. 3. Rs.5500
    4. 4. Rs.5000

1
Let amount be Rs.X then X x R x T/100+(R x T)=T.D = X x 12 x 3/4/100+(12 x 3/4) = 540 X=(540 x 109/9) = Rs. 6540 Amount= Rs. 6540. P.W=Rs.(6540-540) = Rs. 6000.

Q.- The true discount on a certain sum of money due 3 years hence is Rs. 250 and the simple interest on the same sum for the same time and at the same rate is Rs. 375. Find the sum and the rate percent.

    1. 1. 666.67%
    2. 2. 773.33%
    3. 3. 766.67%
    4. 4. 573.33%

3
T.D. = Rs. 250 and S.I. = Rs. 375. Sum due =S.I. xT.D./ S.I. -T.D. =375x250/375- 250 =Rs.750. Rate=[100*375/750*3]%= 50/3%

Q.- A trader owens a merchant Rs.10028 due 1 year hence. The trader wants to settle the account after 3 months.If the rate of interest is 12% per annum how much cash should he pay?

    1. 1. Rs. 9025
    2. 2. Rs. 9200
    3. 3. Rs. 9600
    4. 4. Rs. 9560

2
Required money P.W of Rs.10028 due 9 months Rs.[10028x100/100+(12x9/12)] =Rs. 9200

Q.- A man purchased a cow for Rs. 3000 and sold it the same day for Rs. 3600 allowing the buyer a credit of 2 years. If the rate of interest be 10% per annum then the man has a gain of

    1. 1. 0
    2. 2. 0.05
    3. 3. 0.04
    4. 4. 0.1

1
C.P= Rs. 3000. S.P=Rs[3600x100/100+(10x2)] =3000 SoGain is 0 %.

Q.- A man buys a watch for Rs.1950 in cash and sells it for Rs.2200 at a credit of 1 year. If the rate of interests is 10% per annum the man

    1. 1. gain Rs.55
    2. 2. gain Rs.30
    3. 3. loses Rs.30
    4. 4. gains Rs.50

4
S.P=P.W. of Rs.2200 due 1 year = Rs.[2200x100/100+(10x1)] = Rs.2000. Gain=Rs.(2000-1950) = Rs.50.

Q.- Rs. 20 is the true discount on Rs. 260 due after a certain time. What will be the true discount on the same sum due after half of the former time the rate of interest being the same?

    1. 1. Rs.10.04
    2. 2. Rs.10.10
    3. 3. Rs.10.40
    4. 4. Rs.10.20

3
S.I.on Rs.(260-20) for a given time=Rs. 20 S.I. on Rs. 240 for half the time=Rs. 10. T.D. on Rs. 250=Rs. 10. T.D. on Rs. 260=Rs.(10/250x260) =Rs. 10.40

Q.- The simple interest and the true discount on a certain sum for a given time and at a given rate are Rs. 85 and Rs. 80 respectively. The sum is:

    1. 1. Rs.1350
    2. 2. Rs.1306
    3. 3. Rs.1360
    4. 4. Rs.1300

3
Sum =(S.I. x T.D.)/(S.I..) =(85 x 80)/(85 - 80) = Rs. 1360.

Q.- If the true discount on s sum due 2 years hence at 14% per annum be Rs. 168 the sum due is:

    1. 1. Rs.768
    2. 2. Rs.786
    3. 3. Rs.678
    4. 4. Rs.687

1
P.W. =(100 x T.D.)/(R x T) =(100 x 168)/(14 x 2) = 600. Sum = (P.W. + T.D.) = Rs. (600 + 168) = Rs. 768.