Quantitative Aptitude Archives - JumpWhere

HCF and LCM

MS — Mon, 08 Aug 2022 05:45:21 +0000

Highest Common Factor (H.C.F) or Greatest Common Divisor (G.C.D.):

The H.C.F. of two or more than two numbers is the greatest number that divides each of them exactly. There are two methods of finding the H.C.F. of a given set of numbers:

Factorization Method: Express the each one of the given numbers as the product of prime factors. The product of least powers of common prime factors gives H.C.F.

Division Method: Suppose we have to find the H.C.F. of two given numbers, divide the larger by the smaller one. Now, divide the divisor by the remainder. Repeat the process of dividing the preceding number by the remainder last obtained till zero is obtained as remainder. The last divisor is required H.C.F.

Finding the H.C.F. of more than two numbers: Suppose we have to find the H.C.F. of three numbers then, H.C.F. of [(H.C.F. of any two) and (the third number)] gives the H.C.F. of three given number. Similarly, the H.C.F. of more than three numbers may be obtained.

Least Common Multiple (L.C.M.):

The least number which is exactly divisible by each one of the given numbers is called their L.C.M. There are two methods of finding the L.C.M. of a given set of numbers:

Factorization Method: Resolve each one of the given numbers into a product of prime factors. Then, L.C.M. is the product of highest powers of all the factors.

Division Method (short-cut): Arrange the given numbers in a rwo in any order. Divide by a number which divided exactly at least two of the given numbers and carry forward the numbers which are not divisible. Repeat the above process till no two of the numbers are divisible by the same number except 1. The product of the divisors and the undivided numbers is the required L.C.M. of the given numbers.

Product of two numbers = Product of their H.C.F. and L.C.M.

Co-primes: Two numbers are said to be co-primes if their H.C.F. is 1.

H.C.F. and L.C.M. of Fractions:

H.C.F = H.C.F of Numerators / L.C.M of Denominators
L.C.M = L.C.M of Numerators / H.C.F of Denominators

Instruction to take HCF Aptitude Test

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Hcf and Lcm – Question and Answers

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Question 1

A teacher can divide her class into groups of 5,13, and 17. What is the smallest possible strength of the class?

A	835
B	940
C	1105
D	1220

Question 1 Explanation:

Smallest possible strength of the class will be L.C.M of 5,13,17 ans:1105

Question 2

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

A	544
B	548
C	504
D	536

Question 2 Explanation:

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

Question 3

Find the highest common factor of 36 and 84.

A	4
B	6
C	12
D	18

Question 3 Explanation:

36 = 22 x 32
84 = 22 x 3 x 7
H.C.F. = 22 x 3 = 12.

Question 4

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:

A	85
B	75
C	65
D	60

Question 4 Explanation:

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.

Question 5

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

A	15 cm
B	35 cm
C	25 cm
D	52 cm

Question 5 Explanation:

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

Question 6

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

A	127
B	305
C	235
D	123

Question 6 Explanation:

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

Question 7

Which of the following has the most number of divisors?

A	176
B	182
C	99
D	101

Question 7 Explanation:

Option A, 176 = 1 x 2 x 2 x 2 x 2 x 11.
Option B, 182 = 1 x 2 x 7 x 13.
Option C, 99 = 1 x 3 x 3 x 11.
Option D, 101 = 1 x 101.
Divisors of 99 are 1, 3, 9, 11, 33, 99.
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 have the most number of divisors

Question 8

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

A	30
B	40
C	38
D	36

Question 8 Explanation:

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.

Question 9

The H.C.F. of two numbers is 11 and their L.C.M. is 7700. If one of the numbers is 275, then the other is:

A	267
B	318
C	308
D	279

Question 9 Explanation:

Other number = (11 x 7700/275) = 308.

Question 10

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

A	1260
B	630
C	2524
D	None of these

Question 10 Explanation:

L.C.M. of 12, 18, 21 30 2 | 12 – 18 – 21 – 30
—————————-
= 2 x 3 x 2 x 3 x 7 x 5 = 1260. 3 | 6 – 9 – 21 – 15
—————————- Required number = (1260 ÷ 2) | 2 – 3 – 7 – 5
= 630.

Question 11

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

A	12
B	68
C	48
D	78

Question 11 Explanation:

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.

Question 12

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

A	1008
B	1032
C	1015
D	1022

Question 12 Explanation:

Required number = (L.C.M. of 12,16, 18, 21, 28) + 7
= 1008 + 7
= 1015

Question 13

252 can be expressed as a product of primes as:

A	2 x 3 x 3 x 3 x 7
B	3 x 3 x 3 x 3 x 7
C	2 x 2 x 2 x 3 x 7
D	2 x 2 x 3 x 3 x 7

Question 13 Explanation:

Clearly, 252 = 2 x 2 x 3 x 3 x 7.

Question 14

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 again at the starting point ?

A	26 minutes and 18 seconds
B	45 minutes
C	46 minutes and 12 seconds
D	42 minutes and 36 second

Question 14 Explanation:

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.

Question 15

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

A	1677
B	2523
C	1683
D	3363

Question 15 Explanation:

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.

Question 16

The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is:

A	13
B	33
C	23
D	43

Question 16 Explanation:

L.C.M. of 5, 6, 4 and 3 = 60.
On dividing 2497 by 60, the remainder is 37.
Number to be added = (60 – 37) = 23.

Question 17

The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:

A	114
B	294
C	364
D	364

Question 17 Explanation:

L.C.M. of 6, 9, 15 and 18 is 90.
Let required number be 90k + 4, which is multiple of 7.
Least value of k for which (90k + 4) is divisible by 7 is k = 4.
Required number = (90 x 4) + 4 = 364.

Question 18

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

A	123
B	127
C	235
D	305

Question 18 Explanation:

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

Question 19

Which of the following has the most number of divisors?

A	99
B	101
C	176
D	182

Question 19 Explanation:

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.

Question 20

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

A	28
B	32
C	40
D	64

Question 20 Explanation:

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.

Question 21

The H.C.F.

9	,	12	,	18	and	21	is:
10		25		35		40

A	3/5
B	252/5
C	3/1400
D	63/700

Question 21 Explanation:

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

Question 22

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 reciprocals of the numbers is equal to:

A	55 601
B	601 55
C	11 120
D	120 11

Question 22 Explanation:

Let the numbers be a and b.
Then, a + b = 55 and ab = 5 x 120 = 600.
The required sum = 1/a+1/b = a+b/ab == 55/600 = 11/200

Question 23

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

A	15 cm
B	25 cm
C	35 cm
D	42 cm

Question 23 Explanation:

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

Question 24

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:

A	75
B	81
C	85
D	89

Question 24 Explanation:

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.

Question 25

Find the highest common factor of 36 and 84.

A	4
B	6
C	12
D	18

Question 25 Explanation:

36 = 22 x 32
84 = 22 x 3 x 7
H.C.F. = 22 x 3 = 12.

Question 26

Which of the following fraction is the largest ?

A	7/8
B	13/16
C	31/40
D	63/80

Question 26 Explanation:

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>65/80>63/80>62/80>, so 7/8,>13/16>63/80>31/40
So, 7/8 is the largest.

Question 27

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

A	504
B	536
C	544
D	548

Question 27 Explanation:

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

Question 28

The H.C.F. of two numbers is 11 and their L.C.M. is 7700. If one of the numbers is 275, then the other is:

A	279
B	283
C	308
D	318

Question 28 Explanation:

Other number = (11 x 7700/275) = 308.

Question 29

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

A	196
B	630
C	1260
D	2520

Question 29 Explanation:

L.C.M. of 12, 18, 21 30 2 | 12 – 18 – 21 – 30
= 2 x 3 x 2 x 3 x 7 x 5 = 1260. 3 | 6 – 9 – 21 – 15
—————————- Required number = (1260 ÷ 2) | 2 – 3 – 7 – 5
= 630.

Question 30

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

A	12
B	16
C	24
D	48

Question 30 Explanation:

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.

Question 31

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

A	1008
B	1015
C	1022
D	1032

Question 31 Explanation:

Required number = (L.C.M. of 12,16, 18, 21, 28) + 7
= 1008 + 7
= 1015

Question 32

252 can be expressed as a product of primes as:

A	2 x 2 x 3 x 3 x 7
B	2 x 2 x 2 x 3 x 7
C	3 x 3 x 3 x 3 x 7
D	2 x 3 x 3 x 3 x 7

Question 32 Explanation:

Clearly, 252 = 2 x 2 x 3 x 3 x 7.

Question 33

Find the lowest common multiple of 24, 36 and 40.

A	120
B	240
C	360
D	480

Question 33 Explanation:

2 | 24 – 36 – 40
——————–
2 | 12 – 18 – 20
——————–
2 | 6 – 9 – 10
——————-
3 | 3 – 9 – 5
——————-
| 1 – 3 – 5
L.C.M. = 2 x 2 x 2 x 3 x 3 x 5 = 360.

Question 34

The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is:

A	3
B	13
C	23
D	33

Question 34 Explanation:

L.C.M. of 5, 6, 4 and 3 = 60.
On dividing 2497 by 60, the remainder is 37.
Number to be added = (60 – 37) = 23.

Question 35

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:

A	1677
B	1683
C	2523
D	3363

Question 35 Explanation:

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.

Question 36

A	26 minutes and 18 seconds
B	42 minutes and 36 seconds
C	45 minutes
D	46 minutes and 12 seconds

Question 36 Explanation:

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.

Question 37

The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is:

A	101
B	107
C	111
D	185

Question 37 Explanation:

Let the numbers be 37a and 37b.
Then, 37a x 37b = 4107
ab = 3.
Now, co-primes with product 3 are (1, 3).
So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111).
Greater number = 111.

Question 38

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

A	40
B	80
C	120
D	200

Question 38 Explanation:

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.

Question 39

The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:

A	74
B	94
C	184
D	364

Question 39 Explanation:

L.C.M. of 6, 9, 15 and 18 is 90.
Let required number be 90k + 4, which is multiple of 7.
Least value of k for which (90k + 4) is divisible by 7 is k = 4.
Required number = (90 x 4) + 4 = 364.

Question 40

The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is:

A	276
B	299
C	322
D	345

Question 40 Explanation:

Clearly, the numbers are (23 x 13) and (23 x 14).
Larger number = (23 x 14) = 322.

Question 41

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?

A	4
B	10
C	15
D	16

Question 41 Explanation:

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.

Question 42

Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is:

A	4
B	5
C	6
D	8

Question 42 Explanation:

N = H.C.F. of (4665 – 1305), (6905 – 4665) and (6905 – 1305)
= H.C.F. of 3360, 2240 and 5600 = 1120.
Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4

Question 43

The least number which when divided by 16, 18 and 21, leave the remainder 3, 5 and 8 respectively is:

A	982
B	893
C	1024
D	995

Question 43 Explanation:

16———-3
18———-5 CD = 13
21———-8
LCM od 16,18,21 = 1008
1008-13 = 995

Question 44

Four bells begin to toll together respectively at the intervals of 8, 10, 12 and 16 seconds. After how many seconds will they toll together again?

A	246 seconds
B	242 seconds
C	240 seconds
D	243 seconds

Question 44 Explanation:

LCM = 240

Question 45

A heap of coconuts is divided into groups of 2, 3 and 5 and each time one coconut is left over. The least number of Coconuts in the heap is?

A	31
B	41
C	51
D	61

Question 45 Explanation:

LCM = 30 => 30 + 1 = 31

Question 46

Five bells first begin to toll together and then at intervals of 5, 10, 15, 20 and 25 seconds respectively. After what interval of time will they toll again together? s?

A	5 min
B	5.5 min
C	5.2 min
D	61

Question 46 Explanation:

LCM = 300/60 = 5 min

Question 47

HCF and LCM two numbers are 12 and 396 respectively. If one of the numbers is 36, then the other number is?

A	36
B	66
C	132
D	264

Question 47 Explanation:

12 * 396 = 36 * x
x = 132

Question 48

The L.C.M of two numbers is 495 and their H.C.F is 5. If the sum of the numbers is 10, then their difference is:

A	10
B	46
C	70
D	90

Question 48 Explanation:

Let the numbers be x and (100 – x).
Then, x(100 – x) = 5 * 495
x2 – 100x + 2475 = 0
(x – 55)(x – 45) = 0
x = 55 or 45
The numbers are 45 and 55.
Required difference = 55 – 45 = 10.

Question 49

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:

A	55/601
B	601/55
C	11/120
D	120/11

Question 49 Explanation:

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.

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Time, Work and Distance

MS — Tue, 22 Aug 2017 05:15:28 +0000

Explanation / Important formulas

Important formulas:

km/hr to m/s conversion:
a km/hr = a x 5/18 m/s.

m/s to km/hr conversion:
a m/s = a x 18/5 km/hr.

Formulas for finding Speed, Time and Distance:

Time taken by a train of length l metres to pass a pole or standing man or a signal post is equal to the time taken by the train to cover l metres.

Time taken by a train of length l metres to pass a stationery object of length b metres is the time taken by the train to cover (l + b) metres.

Suppose two trains or two objects bodies are moving in the same direction at u m/s and v m/s, where u > v, then their relative speed is = (u – v) m/s.

Suppose two trains or two objects bodies are moving in opposite directions at u m/s and v m/s, then their relative speed is = (u + v) m/s.

If two trains of length a metres and b metres are moving in opposite directions at u m/s and v m/s, then:
The time taken by the trains to cross each other = (a + b) / (u + v) sec.

If two trains of length a metres and b metres are moving in the same direction at u m/s and v m/s, then:
The time taken by the faster train to cross the slower train = (a + b) / (u -v) sec.

If two trains (or bodies) start at the same time from points A and B towards each other and after crossing they take a and b sec in reaching B and A respectively, then:
(A’s speed) : (B’s speed) = (b : a)

Instructions to take Aptitude Test

Click on start to start taking the test.
Click on the option (A, B, C or D) to figure out the right answer.
You can answer multiple times till you get the right answer.
Once you get the right answer, explanations (if any) for the same will be showcased down.
On click of list, you get to see total no of questions, no of questions you answered and no of questions pending to answer.
On click of question number you will go to that particular question.
On click of END your test will end.
On click of ‘Get Results’ you will get to see correct answer for each questions.

Time, Work and Distance – Question and Answers

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Question 1

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

A	120 meters
B	180 meters
C	324 meters
D	150 meters

Question 1 Explanation:

Speed = (60 * 5/18) m/sec = (50/3) m/sec
Length of the train = (Speed * Time)
Length of the train = ((50/3) * 9) m = 150 m

Question 2

A train 125 m long passes a man, running at 5 km/hr in the same direction in which the train is going in 10 seconds. The speed of the train is?

A	45 km/hr
B	54 km/hr
C	50 km/hr
D	55 km/hr

Question 2 Explanation:

Speed of the train relative to man = 125/10 m/sec
=> 25/2 m/sec
=> 25/2 * 18/5 km/hr
=> 45 km/hr
Let the speed of the train be x km/hr
Then, relative speed = (x – 5) km/hr
=> x – 5 = 45
=> x = 50 km/hr

Question 3

P is able to do a piece of work in 15 days and Q can do the same work in 20 days. If they can work together for 4 days, what is the fraction of work left?

A	8/15
B	7/15
C	11/15
D	2/11

Question 3 Explanation:

Amount of work P can do in 1 day = 1/15
Amount of work Q can do in 1 day = 1/20
Amount of work P and Q can do in 1 day = 1/15 + 1/20 = 7/60
Amount of work P and Q can together do in 4 days = 4 * (7/60) = 7/15
Fraction of work left = 1 – 7/15 = 8/15

Question 4

P can lay railway track between two stations in 16 days. Q can do the same job in 12 days. With the help of R, they completes the job in 4 days. How much days does it take for R alone to complete the work?

A	9 (3/5) days
B	9 (1/5) days
C	9 (2/5) days
D	10 days

Question 4 Explanation:

Amount of work P can do in 1 day = 1/16
Amount of work Q can do in 1 day = 1/12
Amount of work P, Q and R can together do in 1 day = 1/4
Amount of work R can do in 1 day = 1/4 – (1/16 + 1/12) = 3/16 – 1/12 = 5/48
=> Hence R can do the job on 48/5 days = 9 (3/5) days

Question 5

P, Q and R can do a work in 20, 30 and 60 days respectively. How many days does it. Need to complete the work if P does the work and he is assisted by Q and R on every third day?

A	10 days
B	14 days
C	15 days
D	9 days

Question 5 Explanation:

Amount of work P can do in 1 day = 1/20
Amount of work Q can do in 1 day = 1/30
Amount of work R can do in 1 day = 1/60
P is working alone and every third day Q and R is helping him
Work completed in every three days = 2 * (1/20) + (1/20 + 1/30 + 1/60) = 1/5
So work completed in 15 days = 5 * 1/5 = 1
I.e, the work will be done in 15 days

Question 6

A is thrice as good as B in work. A is able to finish a job in 60 days less than B. They can finish the work in how many days if they work together?

A	18 days
B	22 1/2 days
C	24 days
D	26 days

Question 6 Explanation:

If A completes a work in 1 day, B completes the same work in 3 days
Hence, if the difference is 2 days, B can complete the work in 3 days
=> if the difference is 60 days, B can complete the work in 90 days
=> Amount of work B can do in 1 day = 1/90
Amount of work A can do in 1 day = 3 * (1/90) = 1/30
Amount of work A and B can together do in 1 day = 1/90 + 1/30 = 4/90 = 2/45
=> A and B together can do the work in 45/2 days = 22 ½ days

Question 7

A can do a particular work in 6 days. B can do the same work in 8 days. A and B signed to do it for Rs. 3200. They completed the work in 3 days with the help of C. How much is to be paid to C?

A	380
B	600
C	420
D	400

Question 7 Explanation:

Amount of work A can do in 1 day = 1/6
Amount of work B can do in 1 day = 1/8
Amount of work A + B can do in 1 day = 1/6 + 1/8 = 7/24
Amount of work A + B + C can do = 1/3
Amount of work C can do in 1 day = 1/3 – 7/24 = 1/24
Work A can do in 1 day : work B can do in 1 day : work C can do in 1 day
= 1/6 : 1/8 : 1/24 = 4 : 3 : 1
Amount to be paid to C = 3200 * (1/8) = 400

Question 8

6 men and 8 women can complete a work in 10 days. 26 men and 48 women can finish the same work in 2 days. 15 men and 20 women can do the same work in how many days?

A	4 days
B	6 days
C	2 days
D	8 days

Question 8 Explanation:

Let work done by 1 man in 1 day = m and work done by 1 woman in 1 day = b
Work done by 6 men and 8 women in 1 day = 1/10
=> 6m + 8b = 1/10
=> 60m + 80b = 1 — (1)
Work done by 26 men and 48 women in 1 day = 1/2
=> 26m + 48b = ½
=> 52m + 96b = 1— (2)
Solving equation 1 and equation 2. We get m = 1/100 and b = 1/200
Work done by 15 men and 20 women in 1 day
=> 15/100 + 20/200 = 1/4
=> Time taken by 15 men and 20 women in doing the work = 4 days

Question 9

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. B alone can complete the work in how many hours?

A	12 hours
B	6 hours
C	8 hours
D	10 hours

Question 9 Explanation:

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 A,B 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 hours

Question 10

P can do a work in the same time in which Q and R together can do it. If P and Q work together, the work can be completed in 10 days. R alone needs 50 days to complete the same work. then Q alone can do it in

A	30 days
B	25 days
C	20 days
D	15 days

Question 10 Explanation:

Work done by P and Q in 1 day = 1/10
Work done by R in 1 day = 1/50
Work done by P, Q and R in 1 day = 1/10 + 1/50 = 6/50
But Work done by P in 1 day = Work done by Q and R in 1 day
Hence the above equation can be written as
Work done by P in 1 day * 2 = 6/50
=> Work done by P in 1 day = 3/50
=> Work done by Q and R in 1 day = 3/50
Hence work done by Q in 1 day = 3/50 – 1/50 = 2/50 = 1/25
So Q alone can do the work in 25 days

Question 11

A completes 80% of a work in 20 days. Then B also joins and A and B together finish the remaining work in 3 days. How long does it need for B if he alone completes the work?

A	37 1/2 days
B	22 days
C	31 days
D	27 days

Question 11 Explanation:

Work done by A in 20 days = 80/100 = 8/10 = 4/5
Work done by A in 1 day = (4/5)/20 = 4/100 = 1/25 — (1)
Work done by A and B in 3 days = 20/100 = 1/5
(Because remaining 20% is done in 3 days by A and B)
Work done by A and B in 1 day = 1/15 —(2)
Work done by B in 1 day = 1/15 – 1/25 = 2/75
=> B can complete the work in 75/2 days = 37 ½ days

Question 12

Machine P can print one lakh books in 8 hours. Machine Q can print the same number of books in 10 hours while machine R can print the same in 12 hours. All the machines started printing at 9 A.M. Machine P is stopped at 11 A.M. and the remaining two machines complete work. Approximately at what time will the printing of one lakh books be completed?

A	3 PM
B	1PM
C	11 AM
D	2 PM

Question 12 Explanation:

Work done by P in 1 hour = 1/8
Work done by Q in 1 hour = 1/10
Work done by R in 1 hour = 1/12
Work done by P,Q and R in 1 hour = 1/8 + 1/10 + 1/12 = 37/120
Work done by Q and R in 1 hour = 1/10 + 1/12 = 22/120 = 11/60
From 9 am to 11 am, all the machines were operating
i.e, they all operated for 2 hours and work completed = 2 * (37/120) = 37/60
Pending work = 1 – 37/60 = 23/60
Hours taken by Q an R to complete the pending work = (23/60) / (11/60) = 23/11
which is approximately equal to 2
Hence the work will be completed approximately 2 hours after 11 am ; i.e. around 1 pm

Question 13

P can finish a work in 18 days. Q can finish the same work in 15 days. Q worked for 10 days and left the job. how many days does P alone need to finish the remaining work?

A	8
B	5
C	4
D	6

Question 13 Explanation:

Work done by P in 1 day = 1/18
Work done by Q in 1 day = 1/15 Work done by Q in 10 days = 10/15 = 2/3 Remaining work = 1 – 2/3 = 1/3 Number of days in which P can finish the remaining work = (1/3) / (1/18) = 6

Question 14

3 men and 7 women can complete a work in 10 days. But 4 men and 6 women need 8 days to complete the same work. In how many days will 10 women complete the same work?

A	50
B	40
C	30
D	20

Question 14 Explanation:

Work done by 4 men and 6 women in 1 day = 1/8
Work done by 3 men and 7 women in 1 day = 1/10
Let 1 man does m work in 1 day and 1 woman does w work in 1 day
The above equations can be written as
4m + 6w = 1/8 —(1)
3m + 7w = 1/10 —(2)
Solving equation (1) and (2) , we get m=11/400 and w=1/400
Amount of work 10 women can do in a day = 10 * (1/400) = 1/40
i.e, 10 women can complete the work in 40 days

Question 15

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

A	60
B	50
C	40
D	30

Question 15 Explanation:

Amount of work done by A and B in 1 day = 1/30
Amount of work done by A and B in 20 days = 20 *(1/30) = 20/30 = 2/3
Remaining work – 1– 2/3 = 1/3
A completes 1/3 work in 20 days
Amount of work A can do in 1 day = (1/3)/20 = 1/60
=> A can complete the work in 60 days

Question 16

A can complete a work in 12 days with a working of 8 hours per day. B can complete the same work in 8 days when working 10 hours a day. If A and B work together, working 8 hours a day, the work can be completed in how many days?

A	5 5/11
B	4 5/11
C	6 4/11
D	6 5/11

Question 16 Explanation:

A can complete the work in 12 days working 8 hours a day
=> Number of hours A can complete the work = 12*8 = 96 hours
=> Work done by A in 1 hour = 1/96
B can complete the work in 8 days working 10 hours a day
=> Number of hours B can complete the work = 8*10 = 80 hours
=> Work done by B in 1 hour = 1/80
Work done by A and B in 1 hour = 1/96 + 1/80 = 11/480
=> A and B can complete the work in 480/11 hours
A and B works 8 hours a day
Hence total days to complete the work with A and B working together
= (480/11)/ (8) = 60/11 days = 5 5 ⁄ 11 days

Question 17

P is 30% more efficient than Q. P can complete a work in 23 days. If P and Q work together, how much time will it take to complete the same work?

A	9
B	11
C	13
D	15

Question 17 Explanation:

Work done by P in 1 day = 1/23
Let work done by Q in 1 day = q
q * (130/100) = 1/23
=> q = 100 / (23*130) = 10 / (23*13)
Work done by P and Q in 1 day = 1/23 + 10 / (23*13) = 23 / (23*13) = 1/13
=> P and Q together can do the work in 13 days

Question 18

P, Q and R can complete a work in 24, 6 and 12 days respectively. The work will be completed in how many days if all of them are working together?

A	2
B	3 3/7
C	4 1/4
D	5

Question 18 Explanation:

Work done by P in 1 day = 1/24
Work done by Q in 1 day = 1/6
Work done by R in 1 day = 1/12
Work done by P,Q and R in 1 day = 1/24 + 1/6 + 1/12 = 7/24
=> Working together, they will complete the work in 24/7 days = 3 3⁄7 days

Question 19

10 men can complete a work in 7 days. But 10 women need 14 days to complete the same work. How many days will 5 men and 10 women need to complete the work?

A	5
B	6
C	7
D	8

Question 19 Explanation:

Work done by 10 men in 1 day = 1/7
Work done by 1 man in 1 day = (1/7)/10 = 1/70
Work done by 10 women in 1 day = 1/14 Work done by 1 woman in 1 day = 1/140 Work done by 5 men and 10 women in 1 day = 5 × (1/70) + 10 × (1/140) = 5/70 + 10/140 = 1/7 => 5 men and 10 women can complete the work in 7 days

Question 20

Kamal will complete work in 20 days. If Suresh is 25% more efficient than Kamal, he can complete the work in how many days?

A	14
B	16
C	18
D	20

Question 20 Explanation:

Work done by Kamal in 1 day = 1/20
Work done by Suresh in 1 day = (1/20) × (125/100) = 5/80 = 1/16
=> Suresh can complete the work in 16 days

Question 21

Anil and Suresh are working on a special assignment. Anil needs 6 hours to type 32 pages on a computer and Suresh needs 5 hours to type 40 pages. If both of them work together on two different computers, how much time is needed to type an assignment of 110 pages?

A	7 hour 15 minutes
B	7 hour 30 minutes
C	8 hour 15 minutes
D	8 hour 30 minutes

Question 21 Explanation:

Pages typed by Anil in 1 hour = 32/6 = 16/3
Pages typed by Suresh in 1 hour = 40/5 = 8
Pages typed by Anil and Suresh in 1 hour = 16/3 + 8 = 40/3
Time taken to type 110 pages when Anil and Suresh work together = 110 * 3 /40 = 33/4
= 8 ¼ hours = 8 hour 15 minutes

Question 22

P and Q can complete a work in 20 days and 12 days respectively. P alone started the work and Q joined him after 4 days till the completion of the work. How long did the work last?

A	5 days
B	10 days
C	14 days
D	22 days

Question 22 Explanation:

Work done by P in 1 day = 1/20
Work done by Q in 1 day = 1/12
Work done by P in 4 days = 4 × (1/20) = 1/5
Remaining work = 1 – 1/5 = 4/5
Work done by P and Q in 1 day = 1/20 + 1/12 = 8/60 = 2/15
Number of days P and Q take to complete the remaining work = (4/5) / (2/15) = 6
Total days = 4 + 6 = 10

Question 23

P takes twice as much time as Q or thrice as much time as R to finish a piece of work. They can finish the work in 2 days if work together. How much time will Q take to do the work alone?

A	4
B	5
C	6
D	7

Question 23 Explanation:

Let P takes x days to complete the work
Then Q takes x/2 days and R takes x/3 days to finish the work
Amount of work P does in 1 day = 1/x
Amount of work Q does in 1 day = 2/x
Amount of work R does in 1 day = 3/x
Amount of work P,Q and R do in 1 day = 1/x + 2/x + 3/x = 1/x (1 + 2 + 3) = 6/x
6/x = 2
=> x = 12
=> Q takes 12/2 days = 6 days to complete the work

Question 24

P and Q can complete a work in 15 days and 10 days respectively. They started the work together and then Q left after 2 days. P alone completed the remaining work. The work was finished in how many days?

A	12
B	16
C	20
D	24

Question 24 Explanation:

Work done by P in 1 day = 1/15
Work done by Q in 1 day = 1/10
Work done by P and Q in 1 day = 1/15 + 1/10 = 1/6
Work done by P and Q in 2 days = 2 * (1/6) = 1/3
Remaining work = 1 – 1/3 = 2/3
Time taken by P to complete the remaining work 2/3 = (2/3) / (1/15) = 10 days
Total time taken = 2 + 10 = 12 days

Question 25

P and Q can do a work in 30 days. Q and R can do the same work in 24 days and R and P in 20 days. They started the work together, but Q and R left after 10 days. How many days more will P take to finish the work?

A	10
B	15
C	18
D	22

Question 25 Explanation:

Let work done by P in 1 day = p
Work done by Q in 1 day = q
Work done by R in 1 day = r
p + q = 1/30
q + r = 1/24
r + p = 1/20
Adding all the above, 2p + 2q + 2r = 1/30 + 1/24+ 1/20 = 15/120 = 1/8
=> p + q + r = 1/16
=> Work done by P,Q and R in 1 day = 1/16
Work done by P, Q and R in 10 days = 10 × (1/16) = 10/16 = 5/8
Remaining work = 1 = 5/8 = 3/8
Work done by P in 1 day = Work done by P,Q and R in 1 day – Work done by Q and R in 1 day
= 1/16 – 1/24 = 1/48
Number of days P needs to work to complete the remaining work = (3/8) / (1/48) = 18

Question 26

P works twice as fast as Q. If Q alone can complete a work in 12 days, P and Q can finish the work in how many days?

A	1
B	2
C	3
D	4

Question 26 Explanation:

Work done by Q in 1 day = 1/12
Work done by P in 1 day = 2 * (1/12) = 1/6 Work done by P and Q in 1 day = 1/12 + 1/6 = ¼ => P and Q can finish the work in 4 days

Question 27

A work can be finished in 16 days by twenty women. The same work can be finished in fifteen days by sixteen men. The ratio between the capacity of a man and a woman is?

A	1:3
B	4:3
C	2:3
D	2:1

Question 27 Explanation:

Work done by 20 women in 1 day = 1/16
Work done by 1 woman in 1 day = 1/(16*20)
Work done by 16 men in 1 day = 1/15
Work done by 1 man in 1 day = 1/(15*16)
Ratio of the capacity of a man and woman =1/(15*16) : 1/(16*20) = 1/15 : 1/20
= 1/3 :1/4 = 4:3

Question 28

P and Q need 8 days to complete a work. Q and R need 12 days to complete the same work. But P, Q and R together can finish it in 6 days. How many days will be needed if P and R together do it?

A	3
B	8
C	12
D	4

Question 28 Explanation:

Let work done by P in 1 day = p
work done by Q in 1 day = q
Work done by R in 1 day = r
p + q = 1/8 —(1)
q + r = 1/12 —(2)
p + q + r = 1/6 —(3)
(3) – (2) => p = 1/6 – 1/12 = 1/12
(3) – (1) => r = 1/6 – 1/8 = 1/24
p + r = 1/12 + 1/24 = 3/24 = 1/8
=> P and R will finish the work in 8 days

Question 29

P can do a work in 24 days. Q can do the same work in 9 days and R can do the same in 12 days. Q and R start the work and leave after 3 days. P finishes the remaining work in how many days?

A	7
B	8
C	9
D	10

Question 29 Explanation:

Work done by P in 1 day = 1/24
Work done by Q in 1 day = 1/9
Work done by R in 1 day = 1/12
Work done by Q and R in 1 day = 1/9 + 1/12 = 7/36
Work done by Q and R in 3 days = 3*7/36 = 7/12
Remaining work = 1 – 7/12 = 5/12
Number of days in which P can finish the remaining work = (5/12) / (1/24) = 10

Question 30

If daily wages of a man is double to that of a woman, how many men should work for 25 days to earn Rs.14400? Given that wages for 40 women for 30 days are Rs.21600?

A	12
B	14
C	16
D	18

Question 30 Explanation:

Wages of 1 woman for 1 day = 2160040*30
Wages of 1 man for 1 day = 21600×240*30
Wages of 1 man for 25 days = 21600*2*2540*30
Number of men = 14400(21600*2*2540*30) = 144(216*5040*30) = 1449 = 16

Question 31

P,Q and R together earn Rs.1620 in 9 days. P and R can earn Rs.600 in 5 days. Q and R in 7 days can earn Rs.910. How much amount does R can earn per day?

A	Rs. 40
B	Rs. 70
C	Rs. 90
D	Rs. 100

Question 31 Explanation:

Amount Earned by P,Q and R in 1 day = 1620/9 = 180 —(1)
Amount Earned by P and R in 1 day = 600/5 = 120 —(2)
Amount Earned by Q and R in 1 day = 910/7 = 130 —(3)
(2)+(3)-(1) => Amount Earned by P , Q and 2R in 1 day
Amount Earned by P,Q and R in 1 day = 120+130-180 = 70
=> Amount Earned by R in 1 day = 70

Question 32

Assume that 20 cows and 40 goats can be kept for 10 days for Rs.460. If the cost of keeping 5 goats is the same as the cost of keeping 1 cow, what will be the cost for keeping 50 cows and 30 goats for 12 days?

A	Rs. 1104
B	Rs. 1000
C	Rs. 934
D	Rs. 1210

Question 32 Explanation:

Assume that cost of keeping a cow for 1 day = c , cost of keeping a goat for 1 day = g
Cost of keeping 20 cows and 40 goats for 10 days = 460
Cost of keeping 20 cows and 40 goats for 1 day = 460/10 = 46
=> 20c + 40g = 46
=> 10c + 20g = 23 —(1)
Given that 5g = c
Hence equation (1) can be written as 10c + 4c = 23 => 14c =23
=> c = 23/14
cost of keeping 50 cows and 30 goats for 1 day
= 50c + 30g
= 50c + 6c (substituted 5g = c)
= 56 c = 56*23/14
= 92
Cost of keeping 50 cows and 30 goats for 12 days = 12*92 = 1104

Question 33

There is a group of people, each of whom can complete a piece of work in 16 days, when they are working individually. On the first day one person works, on the second day another person joins him, on the third day one more person joins them and this process continues till the work is completed. How many days are needed to complete the work?

A	3 1⁄4 days
B	4 1⁄3 days
C	5 1⁄6 days
D	6 1⁄5 days

Question 33 Explanation:

Work completed in 1st day = 1/16
Work completed in 2nd day = (1/16) + (1/16) = 2/16
Work completed in 3rd day = (1/16) + (1/16) + (1/16) = 3/16
An easy way to attack such problems is from the choices. You can see the choices are
very close to each other. So just see one by one
For instance, The first choice given in 3 1⁄4
The work done in 3 days = 1/16 + 2/16 + 3/16 = (1+2+3)/16 = 6/16
The work done in 4 days = (1+2+3+4)/16 = 10/16
The work done in 5 days = (1+2+3+4+5)/16 = 15/16, almost close, isn’t it?
The work done in 6 days = (1+2+3+4+5+6)/16 > 1
Hence the answer is less than 6, but greater than 5
Hence the answer is 5 1⁄6 days
(Just for your reference, work done in 5 days = 15/16
Pending work in 6th day = 1 – 15/16 = 1/16
In 6th day, 6 people are working and work done = 6/16
To complete the work 1/16, time required = (1/16) / (6/16) = 1/6 days
Hence total time required = 5 + 1/6 = 5 1⁄6 days

Question 34

Riya and priya set on a journey. Riya moves eastward at a speed of 20km/h and priya moves westward at a speed of 30km/h. How far will be priya from riya after 30 min?

A	25 km
B	10 km
C	50 km
D	30 km

Question 34 Explanation:

Distance = relative speed * time
= (20+15) * 1/2
=25

Question 35

A man takes 5 hours 45 min in walking to a certain place and riding back. He would have gained 2 hours by riding both ways. The time he would take to walk both ways is?

A	11 hrs
B	8 hrs 45 min
C	7 hrs 45 min
D	9 hrs 20 min

Question 35 Explanation:

Given that time taken for riding both ways will be 2 hours lesser than the time needed for waking one way and riding back
From this, we can understand that time needed for riding one way = time needed for waking one way – 2 hours
Given that time taken in walking one way and riding back = 5 hours 45 min
Hence The time he would take to walk both ways = 5 hours 45 min + 2 hours = 7 hours 45 min

Question 36

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?

A	12
B	11
C	10
D	9

Question 36 Explanation:

Due to stoppages, it covers 9 km less
Time taken to cover 9 km = (9/54 * 60) min = 10 min

Question 37

A man complete a journey in 10 hours. He travels first half of the journey at the rate of 21 km/hr and second half at the rate of 24 km/hr. Find the total journey in km?

A	121
B	242
C	224
D	112

Question 37 Explanation:

(1/2)x /21 + (1/2)x /24 = 10
x/21 + x/24 = 20
15x = 168 * 20
x = (168 * 20 / 15) = 224 km

Question 38

A car traveling with 5/7 of its actual speed covers 42 km in 1 hr 40 min 48 sec. What is the actual speed of the car?

A	30 km/hr
B	35 km/hr
C	25 km/hr
D	40 km/hr

Question 38 Explanation:

Time taken = 1 hr 40 min 48 sec = 1 hr 40 4/5 min = 1 51/75 hrs = 126/75 hrs
Let the actual speed be x km/hr
Then, 5/7x * 126/75 = 42
x = (42 * 7 * 75/5 * 126) = 35 km/hr

Question 39

A man covered a certain distance at some speed. If he had 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. What is the the distance in km?

A	36
B	38
C	40
D	42

Question 39 Explanation:

Let distance = x km and usual rate = y kmph
Then, x/y – x /(y+3) = 40/60 => 2y(y+3) = 9x ….(1)
And, x/(y-2) – x/y = 40/60 => y(y-2) = 3x ………(2)
On dividing (1) by (2), we get x = 40

Question 40

A and B walk around a circular track. A and B walk at a speed of 2 rounds per hour and 3 rounds per hour respectively. If they start at 8 a.m. from the same point in opposite directions, how many times shall they cross each other before 9.30 a.m.?

A	5
B	6
C	7
D	8

Question 40 Explanation:

Relative speed = Speed of A + Speed of B (∴ they walk in opposite directions)
= 2 + 3 = 5 rounds per hour
=> They cross each other 5 times in 1 hour and 2 times in 1/2 hour
Time duration from 8 am to 9.30 am = 1.5 hour
Hence they cross each other 7 times before 9.30 am

Question 41

Two boys starts from the same place walking at the rate of 5 kmph and 5.5 kmph respectively in the same direction. What time will they take to be 8.5 km apart?

A	17 hr
B	14 hr
C	12 hr
D	19 hr

Question 41 Explanation:

Relative speed = 5.5 – 5 = 0.5 kmph (because they walk in the same direction)
Distance = 8.5 km
Time = Distance * Speed = 8.5 * 0.5 = 17 hr

Question 42

In covering a distance of 30 km, Arun takes 2 hours more than Anil. If Arun doubles his speed, then he would take 1 hour less than Anil. What is Arun’s speed?

A	8 kmph
B	5 kmph
C	4 kmph
D	7 kmph

Question 42 Explanation:

Let Abhay’s speed be x km/hr
Then, 30/x – 30/2x = 3
6x = 30
x = 5 km/hr

Question 43

A car travels first 160 km at 64 km/hr and the next 160 km at 80 km/hr. What is the average speed for the first 320 km of the tour?

A	70.24 km/hr
B	74. 24 km/hr
C	71.11 km/hr
D	72.21 km/hr

Question 43 Explanation:

We know time = Distance/speed
So, total time taken = (160/64 + 160/80) = 9/2 hours
Time taken for 320 km = 320 * 2/9 = 71.11
= 71.11 km/hr

Question 44

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?

A	12 km
B	14 km
C	16 km
D	18 km

Question 44 Explanation:

let the distance he travelled on foot = x km
Then the distance he travelled on bicycle = (61-x) km
So, x/4 + (61−x)/9 = 9
9x + 4(61−x) = 9 * 36
5x= 80
x = 16 km

Question 45

Walking 6/7^th of his usual speed, a man is 12 minutes too late. What is the usual time taken by him to cover that distance?

A	1 hr 42 min
B	1 hr
C	2 hr
D	1 hr 12 min

Question 45 Explanation:

New speed = 6/7 of usual speed
Speed and time are inversely proportional
Hence new time = 7/6 of usual time
Hence, 7/6 of usual time – usual time = 12 minutes
=> 1/6 of usual time = 12 minutes
=> usual time = 12 * 6 = 72 minutes
=> 1 hour 12 minutes

Question 46

A man goes to his office from his house at a speed of 3 km/hr and returns at a speed of 2 km/hr. If he takes 5 hours in going and coming, what is the distance between his house and office?

A	3 km
B	4 km
C	5 km
D	6 km

Question 46 Explanation:

If a car covers a certain distance at x kmph and An Equal distance at y kmph, the average speed of the whole journey = 2xy/x+y kmph
Hence, average speed = (2*3*2) /2+3 = 12/5 km/hr
Total time taken = 5 hours
⇒ Distance travelled = 12/5*5 = 12 km
⇒ Distance between his house and office = 12/2 = 6 km

Question 47

A man rides his bicycle 10 km at an average speed of 12 km/hr and again travels 12 km at an average speed of 10 km/hr. What is his average speed for the entire trip approximately?

A	11.2 kmph
B	10 kmph
C	10.2 kmph
D	10.8 kmph

Question 47 Explanation:

Total distance travelled = (10 + 12)km = 22 km
Time taken to travel 10 km at an average speed of 12 km/hr = distance * speed = 10/12 hr
Time taken to travel 12 km at an average speed of 10 km/hr = distance * speed = 12/10 hr
Total time taken = [10/12 + 12/10] hr = 61/30 hrs
Average speed = [22*30/61] km/hr = 10.8 km/hr

Question 48

An airplane covers a certain distance at a speed of 240 kmph in 5 hours. To cover the same distance in 1 2/3 hours, it must travel at a speed of?

A	660 km/hr
B	680 km/hr
C	700 km/hr
D	720 km/hr

Question 48 Explanation:

Distance = (240 * 5) = 1200 km
Speed = Distance/Time
Speed = 1200 / (5/3) km/hr
Therefore required speed = (1200 * 3/5) km/hr
= 720 km/hr

Question 49

A train can travel 50% faster than a car. Both start from point A at the same time and reach point B 75 kms away from A at the same time. On the way, however, the train lost about 12.5 minutes while stopping at the stations. What is the speed of the car?

A	80 kmph
B	102 kmph
C	120 kmph
D	140 kmph

Question 49 Explanation:

Let speed of the car be x kmph
Then, speed of the train = 150/100x = (3/2x) kmph
75/x – 75/(3/2)x = 125/10 * 60
75/x – 50/x = 5/24
x = (25 * 24 / 5) = 120 kmph

Question 50

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. What is the duration of the flight?

A	2 hour
B	112 hour
C	12 hour
D	1 hour

Question 50 Explanation:

Let the duration of the flight be x hours
Then, 600/x – 600/x + (1/2) = 200
600/x – 1200/2x + 1 = 200
x(2x + 1) = 3
2*2 + x – 3 = 0
(2x + 3)(x – 1) = 0
x = 1 hr [neglecting the -ve value of x]

Question 51

If a person walks at 14 km/hr instead of 10 km/hr, he would have walked 20 km more. What is the actual distance travelled by him?

A	80 km
B	70 km
C	60 km
D	50 km

Question 51 Explanation:

Let the actual distance travelled be x km
Then, x / 10 = x+20 / 14
14x = 10x + 200
4x = 200
x = 50 km

Question 52

The ratio between the speeds of two trains is 7:8. If the second train runs 400 km in 4 hours, What is the the speed of the first train?

A	85 km/hr
B	87.5 km/hr
C	90 km/hr
D	92.5 km/hr

Question 52 Explanation:

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 * 12.5) km/hr = 87.5 km/hr

Question 53

A car travels at an average of 50 miles per hour for 212 hours and then travels at a speed of 70 miles per hour for 112 hours. How far did the car travel in the entire 4 hours?

A	210 miles
B	230 miles
C	250 miles
D	260 miles

Question 53 Explanation:

Speed1 = 50 miles/hour
Time1 = 212 hour = 52 Hour
Distance1 = Speed1 × Time1
=> 50 × 52 = 25 × 5 = 125 Miles
Speed2 = 70 miles/hour
Time2 = 112 hour = 32 hour
Distance2 = Speed2 × Time2 = 70*32 = 35*3 = 105 miles
Total Distance = Distance1 + Distance2
= 125 + 105 = 230 miles

Question 54

The speed of a bus increases by 2 km after every one hour. If the distance travelling in the first one hour was 35 km. what was the total distance travelled in 12 hours?

A	422 km
B	552 km
C	502 km
D	492 km

Question 54 Explanation:

Given that distance travelled in 1st hour = 35 km
And speed of the bus increases by 2 km after every one hour
Hence distance travelled in 2nd hour = 37 km
Hence distance travelled in 3rd hour = 39 km
Total Distance Travelled = [35 + 37 + 39 + … (12 terms)]
This is an Arithmetic Progression(AP) with first term, a = 35, number of terms ,n = 12 and common difference, d = 2
The sequence a , (a + d), (a + 2d), (a + 3d), (a + 4d), . . . is called an Arithmetic Progression(AP)where a is the first term and d is the common difference of the AP
Sum of the first n terms of an Arithmetic Progression(AP),
Sn=n2[2a+(n−1)d]where n = number of terms
Hence, [35+37+39+… (12 terms)] = S12 = 122[2*35+(12−1)2] = 6[70+22] = 6 * 92 = 552
Hence the total distance travelled = 552 km

Question 55

Sound is said to travel in air at about 1100 feet per second. A man hears the axe striking the tree, ¹¹/₅seconds after he sees it strike the tree.How far is the man from the wood chopper?

A	1800 ft
B	2810 ft
C	3020 ft
D	2420 ft

Question 55 Explanation:

Speed of the sound = 1100 ft/s
Time = 11/5 second
Distance = Speed * Time = 1100 *115 = 220*11 = 2420 ft

Question 56

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 will they meet?

A	10.30 a.m
B	10 a.m.
C	12 noon
D	11 a.m.

Question 56 Explanation:

Assume that they meet x hours after 8 a.m
Then, train1, starting from A , travelling towards B, travels x hours till the trains meet
⇒ Distance travelled by train1 in x hours = Speed *Time = 60x
Then, train2, starting from B , travelling towards A, travels (x-1) hours till the trains meet
⇒ Distance travelled by train2 in (x-1) hours = Speed *Time = 75(x-1)
Total distance travelled = Distance travelled by train1 + Distance travelled by train2
=> 330 = 60x + 75(x-1)
=> 12x + 15(x-1) = 66
=> 12x + 15x – 15 = 66
=> 27x = 66 + 15 = 81
=> 3x = 9
=> x = 3
Hence the trains meet 3 hours after 8 a.m., i.e. at 11 a.m

Question 57

A man walking at the rate of 5 km/hr crosses a bridge in 15 minutes. What is the length of the bridge (in metres)?

A	1250
B	1280
C	1320
D	1340

Question 57 Explanation:

We need to get the answer in meters
So we will first of change distance from km/hour to meter/sec by multiplying it with 5/18 and also change 15 minutes to seconds by multiplying it with 60
Speed = 5*5/18 = 25/18 m/sec
Time = 15 * 60 seconds = 900 seconds
Distance = Time * Speed
Distance = (25 / 18) * 900 = 1250 meter

Question 58

A train travelled at an average speed of 100 km/hr, stopping for 3 minutes after every 75 km. How long did it take to reach its destination 600 km from the starting point?

A	6 hrs 21 min
B	7 hrs 14 min
C	7 hrs 22 min
D	6 hrs

Question 58 Explanation:

Average Speed = 100 km/hr
With this speed, train can cover 600 km in 6 hours but train used to stop for 3 minutes after every 75 km
So, Train will stop = 600/75 = 8 times in the whole journey
But, you need to understand that after 7th stopping train will be at the destiny in next stop. Therefore, train will stop 7 times in journey
So, extra time = 7 *3 = 21 minutes
Total time taken in the journey = 6 hour 21 minutes

Question 59

A person travels from A to B at a speed of 40 km/hr and returns by increasing his speed by 50%. What is his average speed for both the trips?

A	60 km/hr
B	56 km/hr
C	52 km/hr
D	48 km/hr

Question 59 Explanation:

Speed while going = 40 km/hr
Speed while returning = 150% 0f 40 = 60 km/hr
Average speed = 2xy/x+y
= (2*40*60/40+60)
= 4800/100 = 48 km/hr

Question 60

A person crosses a 600 m long street in 5 minutes. What is his speed in km per hour?

A	3.6
B	7.2
C	8.4
D	10

Question 60 Explanation:

Speed = [600 / (5*60)] m/sec = 2 m / sec
(Converting m/sec to km/hr (see important formulas section)
= [2* (18 / 5)] km/hr
= 7.2 km/hr

Question 61

If a person walks at 14 km/hr instead of 10 km/hr, he would have walked 20 km more. The actual distance traveled by him is?

A	50 km
B	80 km
C	70 km
D	90 km

Question 61 Explanation:

Let the actual distance travelled be x km
x / 10 = x + 20/14
14x = 10x + 200
4x = 200
x = 50 km

Question 62

It takes eight hours for a 600 km journey, if 120 km is done by train and the rest by car. It takes 20 minutes more, if 200 km is done by train and the rest by car. What is the ratio of the speed of the train to that of the car?

A	3 : 4
B	2 : 3
C	1 : 2
D	1 : 3

Question 62 Explanation:

Let the speed of the train be x km/hr and that of the car be y km/hr
Then, 120/x + 480/y = 8 => 1/x + 4/y = 1/15 …..(1)
And, 200/x + 400/y = 25/3 => 1/x +2/y =1/24 …..(2)
Solving (1) and (2), we get x = 60 and y = 80
Ratio of speeds = 60 : 80 = 3 : 4

Question 63

Arun is traveling on his cycle and has calculated to reach point A at 2 pm if he travels at 10 kmph, he will reach there at 12 noon if he travels at 15 kmph. At what speed must he travel to reach A at 1 pm?

A	8 kmph
B	10 kmph
C	12 kmph
D	14 kmph

Question 63 Explanation:

Let the distance travelled be x km
Then, x/10 – x/15 = 2
= 3x – 2x = 60 => x = 60 km
Time taken to travel 60 km at 10 km/hr = (60/10)hrs = 6 hrs
So, Arun started 6 hours before 2 P.M i.e.., at 8 A.M
Therefore Required speed = (60/5)kmph = 12 kmph

Question 64

An athlete runs 200 meters race in 24 seconds. What is his speed?

A	20 km/hr
B	25 km/hr
C	27.5 km/hr
D	30 km/hr

Question 64 Explanation:

Speed = Distance/Time = 200/24 m/s
=> 200/24*185 km/hr
=> 40*34 km/hr = 10*3 km/hr = 30 km/hr

Question 65

A train is moving at the speed of 80 km/hr. What is its speed in meters per second?

A	2229 m/s
B	22 m/s
C	2119 m/sec
D	21 m/s

Question 65 Explanation:

Speed = 80 km/hr = 80*5/18 m/s
=> 40×59 m/s = 2009 m/s
=> 2229 m/s

Question 66

A man in a train notices that he can count 21 telephone posts in one minute. If they are known to be 50 metres apart, at what speed is the train travelling?

A	61 km/hr
B	56 km/hr
C	63 km/hr
D	60 km/hr

Question 66 Explanation:

Number of poles = 21
Distance between them = 50m
Total distance = 20*50m = 1000m = 1 km
Time taken is = 1min = 1/60 hr
Speed = Distance / time
Speed = 1/(1/60) = 1*60 = 60 kmph

Question 67

A truck covers a distance of 550 metres in 1 minute whereas a train covers a distance of 33 kms in 45 minutes. What is the ratio of their speed?

A	2 : 1
B	1 : 2
C	4 : 3
D	3 : 4

Question 67 Explanation:

Ratio of speeds = (550/60 * 18/5) : (33/45 * 60)
= 33 : 44 = 3 : 4

Question 68

A person has to cover a distance of 6 km in 45 minutes. If he covers one-half of the distance in two-thirds of the total time; to cover the remaining distance in the remaining time, what should be his speed in km/hr?

A	14 km/hr
B	12 km/hr
C	10 km/hr
D	8 km/hr

Question 68 Explanation:

The person needs to cover 6 km in 45 minutes
Given that he covers one-half of the distance in two-thirds of the total time
He covers half of 6 km in two-thirds of 45 minutes
He covers 3 km in 30 minutes
Hence, now he need to cover the remaining 3 km in the remaining 15 minutes
Distance = 3 km
Time = 15 minutes = 1/4 hour
Required Speed = Distance/Time = 3/(1/4) = 12 km/hr

Question 69

An aeroplane covers a certain distance at a speed of 240 kmph in 5 hours. To cover the same distance in 1 2/3 hours, it must travel at a speed of?

A	300 kmph
B	360 kmph
C	600 kmph
D	720 kmph

Question 69 Explanation:

Distance = (240 x 5) = 1200 km
Speed = Distance/Time
Speed = 1200 / (5/3) km/hr [We can write 1 2/3 hours as 5/3 hours]
Therefore required speed = (1200 * 3/5) km/hr = 720 km/hr

Question 70

Robert is travelling on his cycle and has calculated to reach point A at 2 P.M. If he travels at 10 kmph, he will reach there at 12 noon if he travels at 15 kmph. At what speed must he travel to reach A at 1 P.M?

A	8 kmph
B	11 kmph
C	12 kmph
D	14 kmph

Question 70 Explanation:

Let the distance travelled by x km
Then, x/10 – x/15 = 2
3x – 2x = 60
x = 60 km
Time taken to travel 60 km at 10 km/hr = (60/10)hrs = 6 hrs
So, Robert started 6 hours before 2 P.M
i.e., at 8 A.M
Therefore, Required speed = (60/5) kmph = 12 kmph

Question 71

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?

A	14 km
B	15 km
C	16 km
D	17 km

Question 71 Explanation:

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 * 36
5x = 80
x = 16 km

Question 72

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?

A	35.55 km/hr
B	36 km/hr
C	71.11 km/hr
D	71 km/hr

Question 72 Explanation:

Total time taken = (160/64 + 160/80) hrs = 9/2 hrs
Average speed = (320 * 2/9) km/hr = 71.11 km/hr

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Stocks and Shares

MS — Mon, 21 Aug 2017 05:43:41 +0000

Explanation / Important formulas

Shares or stocks: The whole capital is divided into small units called shares or stock.

Example: For each investment, the company issues a ‘share-certificate’, showing the value of each share and the number of shares held by a person. The person who subscribes in shares or stock is called a share holder or stock holder.

Stock Capital: The total amount of money needed to run the company is called the stock capital.

Dividend: The annual profit distributed among share holders is called dividend. Dividend is paid annually as per share or as a percentage.

Market Value: The stock of different companies are sold and bought in the open market through brokers at stock-exchanges. A share or stock is said to be:

At premium or Above par, if its market value is more than its face value.
At par, if its market value is the same as its face value.
At discount or Below par, if its market value is less than its face value.
Thus, if a Rs. 100 stock is quoted at premium of 16, then market value of the stock = Rs.(100 + 16) = Rs. 116.
Likewise, if a Rs. 100 stock is quoted at a discount of 7, then market value of the stock = Rs. (100 – 7) = 93.

Face Value: The value of a share or stock printed on the share-certificate is called its Face Value or Nominal Value or Par Value.

Brokerage: The broker’s charge is called brokerage.

When stock is purchased, brokerage is added to the cost price.
When stock is sold, brokerage is subtracted from the selling price.

Note:

The face value of a share always remains the same.
The market value of a share changes from time to time.
Dividend is always paid on the face value of a share.
Number of shares held by a person = Total Investment / Investment in 1 share = Total Income / Income from 1 share = Total face Value / Face of 1 share

Thus, by a Rs. 100, 9% stock at 120, we mean that:

Face Value of stock = Rs. 100.
Market Value (M.V) of stock = Rs. 120.
Annual dividend on 1 share = 9% of face value = 9% of Rs. 100 = Rs. 9.
An investment of Rs. 120 gives an annual income of Rs. 9.
Rate of interest p.a = Annual income from an investment of Rs. 100 = (9/120 x 100)% = 7 ½%

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Stocks and shares – Question and Answers

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Question 1

To produce an annual income of Rs. 1200 from a 12% stock at 90, the amount of stock needed is

A	Rs. 10800
B	Rs. 10000
C	Rs. 14400
D	Rs. 16000

Question 1 Explanation:

For an income of Rs.12, stock needed = 100 For an income of Rs. 1200, stock needed = 100 * 1200/12 = 10000

Question 2

The cost price of a Rs. 100 stock at 4 discount, when brokerage is ¼ % is

A	Rs. 96
B	Rs. 95.75
C	Rs. 96.25
D	Rs. 104.25

Question 2 Explanation:

C.P = Rs. 100 – 4 + ¼ = Rs. 96.25

Question 3

Y investing Rs. 1620 in 8% stock, sandeep earns Rs. 135. The stock is then quoted at:

A	Rs. 106
B	Rs. 80
C	Rs. 96
D	Rs. 108

Question 3 Explanation:

To earn Rs. 135, investment = Rs. 1620 To earn Rs. 8, investment = Rs. (1620/135) * 8 = Rs. 96 Market value of Rs. 100 stock = Rs. 96

Question 4

A 6% stock yields 8%. The market value of the stock is

A	Rs. 48
B	Rs. 75
C	Rs. 96
D	Rs. 133.33

Question 4 Explanation:

For an income of Rs. 8, Investment = Rs. 100 For an income of Rs. 6, investment = Rs. 100/8 * 6 = Rs. 75 Market value of Rs. 100 stock = Rs. 75

Question 5

Rs. 9800 are invested partly in 9% stock at 75 and 10 % stock at 80 to have equal amount of incomes. The investment in 9% stock is

A	Rs. 4800
B	Rs. 5400
C	Rs. 5000
D	Rs. 5600

Question 5 Explanation:

Let the investment in 9% stock is x Investment in 10% stock = (9800 – x) 9/75 * x = 10/80(9800- x) hence x = 5000

Question 6

A 9 % stock yields 8%. The market value of the stock is

A	Rs. 72
B	Rs. 116.50
C	Rs. 90
D	Rs. 112.50

Question 6 Explanation:

To get Rs.8, investment = Rs. 100 To get Rs.9, investment = Rs. 100/8 * 9 = Rs. 112.50 Market value of Rs. 100 stock = Rs. 112.50

Question 7

Rajinder invests a part of Rs. 12000 in 12% stock at Rs. 120 and the remainder in 15% stock at Rs. 125. If his taotal dividend per annum is Rs. 1360, how much does he invest in 12% stock at Rs. 120?

A	Rs. 4500
B	Rs. 6000
C	Rs. 5500
D	Rs 4000

Question 7 Explanation:

Let investment in 12% stock is Rs. X Investment in 15% stock Rs. (12000-x) (12/120)* x +15/125(12000-x) = 1360 5x + 72000- 6x hence x = 4000

Question 8

A man buys Rs. 20 shares paying 9% divided. The man wants to have an interest of 12% on his money. The market value of each share is

A	12
B	15
C	18
D	20

Question 8 Explanation:

Dividend on Rs. 20 = Rs. (90/100) 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

Question 9

Find the cost of Rs. 6400, 10% stock at 15 discount.

A	5440
B	5480
C	6440
D	6480

Question 9 Explanation:

Cost of Rs. 100 stock = Rs. (100 – 15) Cost of Rs. 6400 stock = Rs. (85/100 * 6400) = Rs. 5440

Question 10

Find the cash required to purchase Rs. 3200, 7(1/2)% stock at 107(brokerage (1/2)%).

A	3440
B	4440
C	5440
D	6440

Question 10 Explanation:

Cash required to purchase Rs. 100 stock = Rs. (107 + (1/2)) = Rs. (215/2) Cash required to purchase Rs. 100 stock = Rs [(215/2)*(1/100)*3200] = Rs. 3440

Question 11

Find the annual income derived from Rs. 2500, 8% stock at 106.

A	100
B	150
C	200
D	250

Question 11 Explanation:

Income from Rs. 100 stock = Rs. 8 Income from Rs. 2500 = Rs. [(8/1000*2500)] = Rs. 200

Question 12

F ind the annual income derived by investing Rs. 6800 in 10% stock at 136.

A	200
B	300
C	400
D	500x

Question 12 Explanation:

By investing Rs. 136, income obtained = Rs. 10 By investing Rs. 6800, income obtained = Rs. [(10/136)*6800] = Rs. 500

Question 13

Find the cost of 96 shares of Rs. 10 each at (3/4)discount, brokerage being(1/4) per share.

A	812
B	912
C	1012
D	1112

Question 13 Explanation:

Cost of 1 share = Rs. [(10 –(3/4)) +(1/4)] = Rs. (19/2). Cost of 96 shares = Rs. [(19/2)*96] = Rs. 912

Question 14

Find the income derived from 88 shares of Rs.25 each at 5 premium, brokerage being (1/4) per share and the rate of dividend being 7(1/2)% per annum.

A	150
B	165
C	180
D	201

Question 14 Explanation:

Cost of 1 share = Rs. [25+5+(1/4)] = Rs. (121/4) Cost of 88 shares = Rs. [(121/4)*88] = Rs. 2662 Investment made = Rs. 2662 Face value of 88 shares = Rs. (88*25) = Rs. 2200 Dividend on Rs. 2200 = Rs. [(15/20)*(1/100)*2200] = Rs. 165 Income derived = Rs. 165

Question 15

A man sells Rs. 5000, 12 % stock at 156 and invests the proceeds parity in 8% stock at 90 and 9% stock at 108. He hereby increases his income by Rs. 70. How much of the proceeds were invested in each stock?

A	3600(8%), 4200(9%)
B	4000(8%), 4200(9%)
C	3600(8%), 4000(9%)
D	4000(8%),4000(9%)4000(8%),4000(9%)

Question 15 Explanation:

S.P of Rs. 5000 stock = Rs. [(156/100)*5000] = Rs. 7800 Income from this stock = Rs. [(12/100)*5000] = Rs. 600 Let investment in * % stock be x and that in 9% stock = (7800 – x) [x*(8/90)] + (7800-x) * (9/108) = (600 +7) (4x/45) + [(7800 – x)*(9/108) = (600+7)] (4x/45) + [(7800 – x) / 12] = 670 If 3 16x + 117000 – 15x = (670*180) X= 3600 Money invested in 9% at 108 = Rs. (7800 – 3600) = Rs. 4200

Question 16

A man invested Rs. 1552 in a stock at 97 to obtain an income of Rs. 128.The dividend from the stock is?

A	7.5%
B	8%
C	9.7%
D	none of these

Question 16 Explanation:

By investing Rs. 1552, income = Rs. 128 By investing Rs. 97, income = Rs (128/1552 x 97) = Rs. 8 Therefore, Dividend = 8%

Question 17

In order to obtain an income of Rs. 650 from 10% stock at Rs. 96, one must make an investment of

A	Rs. 6240
B	Rs. 3100
C	Rs. 6500
D	Rs. 9600

Question 17 Explanation:

Answer: Option A To obtain Rs. 10, investment = Rs. 96 To obtain Rs. 650, Investment Rs.(96/10 x 650) = Rs. 6240

Question 18

A man buys Rs. 25 shares in a company which pays 9% divided. The money invested is such that it gives 10% on investment. At what price did he buy the shares?

A	Rs. 22
B	Rs. 22.50
C	Rs. 25
D	Rs. 22.50

Question 18 Explanation:

Suppose he buys each share be Rs. X Then, (25 x 9/100) = (X x 10/100) X= 22.50 Cost of each share is Rs. 22.50

Question 19

A man invested Rs. 14,400 in Rs. 100 shares of a company at 20% premium. If the company declared 5% divided at the end of the year, then how much does he get?

A	Rs. 500
B	Rs. 600
C	Rs. 650
D	Rs. 720

Question 19 Explanation:

Number of shares = (14400/120) = 120 Face value = Rs. (100×120) = Rs. 12000 Annual income = Rs. (5/100×12000) = Rs. 600

Question 20

A investment some money in 105 stock at 96. If B wants to invest in an equality good 12% stock, he must purchase a stock worth of?

A	Rs. 50
B	Rs. 105
C	Rs. 115.20
D	Rs. 125.40

Question 20 Explanation:

For an income of Rs. 10, investments = Rs. 96 For an income of Rs. 12, Investments = Rs. (96/10 x 12) = Rs. 115.20

Question 21

A man buys Rs. 50 shares in a company which pays 10% divided. If the man gets 12.5% on his investments, at what price did he buys the shares?

A	Rs.40
B	Rs. 37.50
C	Rs. 48
D	Rs. 52

Question 21 Explanation:

Dividend on 1 share = Rs. (10/100 x 50) = Rs. 5 Rs. 12.50 is an income on an investment of Rs. 100 Rs. 5 is an income on investments of Rs. (100 x 2/25 x5) = Rs. 40 Cost of 1 share = Rs. 40

Question 22

A man invests in a 16% stock at 128. The interest obtained by him is?

A	16
B	12.5
C	8
D	12

Question 22 Explanation:

Investing Rs. 128, income is = Rs. 16 Investing Rs. 100, income is Rs.(16*100/128) = 12.5

Question 23

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?

A	3:4
B	3:5
C	4:5
D	16: 15

Question 23 Explanation:

For an income of Re. 1 in 9% Stock at 96, investment = Rs. (96/9). = Rs.(32/3) For an income of Re.1in 12% stock at 120, investments = Rs.(120/12) = Rs. 10 Ratio of investments = 32/3:10 = 32: 30 = 16: 25

Question 24

Sakshi invests a part of Rs. 12,000 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?

A	4100
B	4050
C	3000
D	4000

Question 24 Explanation:

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

Question 25

The cost price of a Rs. 100 stock at 4 discount, when brokerage is ¼ % is?

A	Rs. 96.25
B	Rs. 96.52
C	Rs. 69.52
D	Rs. 69.25

Question 25 Explanation:

C.P = Rs. 100 – 4 + ¼ = Rs. 96.25

Question 26

A 12% stock yielding 10% is quoted at

A	Rs. 102
B	Rs. 120
C	Rs. 201
D	Rs. 69.25

Question 26 Explanation:

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

Question 27

A man invested Rs. 1552 in a stock at 97 to obtain an income of Rs. 128. The dividend from the stock is

A	10%
B	9%
C	8%
D	7%

Question 27 Explanation:

By investing Rs. 1552, income = Rs. 128 B y investing Rs. 97, income = Rs.128 x 97/1552 = Rs. 8 Dividend = 8%

Question 28

By investing Rs. 1620 in 8% stock, Michael earns Rs. 135. The stock is then quoted at

A	Rs. 96
B	Rs. 69
C	Rs. 100
D	Rs. 69.25

Question 28 Explanation:

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

Question 29

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

A	15:14
B	14:15
C	15:16
D	16: 15

Question 29 Explanation:

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

Question 30

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

A	Rs.846
B	Rs. 486
C	Rs. 648
D	Rs. 69.25

Question 30 Explanation:

Number of shares = 4455/8.25 = 540 Face value = Rs. (540 x 10) = Rs. 5400 Annual income = Rs. (12/100) x 5400 = Rs. 648

Question 31

A 6% stock yields 8%. The market value of the stock is

A	Rs. 60
B	Rs. 70
C	Rs. 57
D	Rs. 75

Question 31 Explanation:

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

Question 32

In order to obtain an income of Rs. 650 from 10% stock at Rs. 96, one must make an investment of

A	Rs. 6240
B	Rs. 2406
C	Rs. 4062
D	Rs. 2460

Question 32 Explanation:

To obtain Rs. 10, investment = Rs. 96 To obtain Rs. 650, investment = Rs. (96/10) x 650 = Rs. 6240

Question 33

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

A	Rs. 20
B	Rs. 15
C	Rs. 51
D	Rs. 55

Question 33 Explanation:

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

Question 34

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?

A	Rs. 423
B	Rs. 432
C	Rs. 432
D	Rs. 234

Question 34 Explanation:

Number of shares = 4455/8.25 = 540 Face value = Rs. 540*10 = Rs. 5400 Income = Rs(6/100)*5400 = Rs. 324

Question 35

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:

A	Rs. 12
B	Rs. 15
C	Rs. 18
D	Rs. 21

Question 35 Explanation:

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

Question 36

By investing in 15/4% stock at 96, one ears Rs.100. The investment made is?

A	Rs. 5260
B	Rs. 5206
C	Rs. 2560
D	Rs. 2506

Question 36 Explanation:

For earning Rs. 15/4, investment = Rs. 96 For earning Rs. 100, investment = Rs. (96*4/15)*100 = Rs. 2560

Question 37

A man invested Rs. 388 in a stock at 97 to obtain an income of Rs. 22. The dividend from the stock is?

A	3%
B	12%
C	11/2%
D	2/11%

Question 37 Explanation:

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%

Question 38

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?

A	16: 15
B	15:16
C	10:12
D	12: 10

Question 38 Explanation:

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

Question 39

I want to purchase a 6% stock which must yield 5% on my capital. At what price must i buy the stock?

A	Rs. 111
B	Rs. 102
C	Rs. 210
D	Rs. 120

Question 39 Explanation:

For an investment of Rs. 5, investment = Rs. 100 For an income of Rs. 6, investment = Rs. (100/5)*6 = Rs. 120

Question 40

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?

A	Rs. 120
B	Rs. 220
C	Rs. 320
D	Rs. 202

Question 40 Explanation:

For an income of Rs.4, investment = Rs.96 For an income of Rs. 5, Investment = Rs. (96/4)*5 = Rs. 120

Question 41

A 4% stock yields 5%. The market value of the stock is?

A	Rs. 125
B	Rs. 80
C	Rs. 80
D	Rs. 110

Question 41 Explanation:

For an income of Rs.5, investment = Rs.100 For an income of Rs. 4, investment = Rs.(100/5)*4 = Rs.80

Question 42

By investing in a 6% stock at 96, an income of Rs. 100 is obtained by making an investment of:

A	Rs. 1600
B	Rs. 1006
C	Rs. 1060
D	Rs. 6100

Question 42 Explanation:

For an income of Rs. 6, investment = Rs. 96 For an income of Rs.100, investment = Rs. (96/6)*100 = Rs. 1600

Question 43

The cash realized by selling a 11/2% stock at 425/4, brokerage being ¼% is

A	Rs. 610
B	Rs. 160
C	Rs. 106
D	none

Question 43 Explanation:

Cash realized = Rs. (425/4) – (1/4) = Rs. 106

Question 44

Find the annual income derived from Rs. 2500, 8% stock at 106?

A	100
B	150
C	200
D	250

Question 44 Explanation:

Income from Rs. 100 stock = Rs. 8 Income from Rs. 2500 stock = (8/100)*2500 = Rs. 200

Question 45

A 6% stock yields 8%. The market value of the stock is?

A	70
B	75
C	80
D	85

Question 45 Explanation:

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

Question 46

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

A	107.04
B	648
C	500
D	648.6

Question 46 Explanation:

Number of shares = 4455/8.25 = 540 Face value = Rs. (540 x 10) = Rs. 5400 Annual income = (12/100)*5400 = Rs. 648

Question 47

Find the cost of Rs. 6400, 10% stock at 15 discount?

A	2000
B	5440
C	5140
D	8500

Question 47 Explanation:

Cost of Rs. 100 stock = Rs. (100 – 15) Cost of Rs. 6400 stock = (85/100)*6400 = Rs. 5440

Question 48

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?

A	3:5
B	1:2
C	16:15
D	4:5

Question 48 Explanation:

For an income of Re. 1 in 9% stock atr 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

Question 49

By investing Rs. 1620 in 8% stock, Michael earns Rs. 135. The stock is then quoted at?

A	80
B	94
C	106
D	108

Question 49 Explanation:

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

Question 50

In order to obtain an income of Rs. 650 from 10% stock at Rs. 96, one must make an investment of?

A	3100
B	6240
C	6400
D	9600

Question 50 Explanation:

To obtain Rs. 10, investment = Rs. 96 To obtain Rs. 650, investment = 650*96/10 = Rs. 6240

Question 51

A investment 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?

A	Rs. 80
B	Rs. 120.25
C	Rs. 115.20
D	Rs. 125.40

Question 51 Explanation:

For an income of Rs. 10, investment = Rs. 96. For an income of Rs. 12, investment = 96*12/10 = Rs. 115.20

Question 52

By investing in 16 2/3% stocks at 64, one earns Rs.1500, The investment made is?

A	5460
B	5600
C	5760
D	5840

Question 52 Explanation:

To earn Rs. 50/3, investment = Rs. 64 To earn Rs. 1500, investment = 64*(3/50)*1500 = Rs. 5760

Question 53

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?

A	12
B	15
C	18
D	20

Question 53 Explanation:

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

Question 54

The cash realised on selling a 14% stock is Rs. 106.25, Brokerage being ¼% is

A	100
B	106
C	123
D	120.25

Question 54 Explanation:

Cash realised = Rs. (106.25 – 0.25) = Rs. 106

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Simple and Compound Interest

MS — Sun, 20 Aug 2017 04:43:23 +0000

Explanation / Important formulas

SIMPLE INTEREST

Principal: The money borrowed or lent out for a certain period is called the principal or the sum.

Interest: Extra money paid for using other’s money is called interest.

Simple Interest(S.I.): If the interest on a sum borrowed for certain period is reckoned uniformly, then it is called simple interest.

Let Principal = P, Rate = R% per annum (p.a.) and Time = T years. Then

Simple Interest = (P x R x T / 100)
P = (100 x S.I / R x T); R = (100 x S.l / P x T) and T= (100 x S.I / P x R)

COMPOUND INTEREST

To find compound interest Let Principal = P, Rate = R% per annum, Time = n years.

When interest is compounded Quarterly: Amount = P [1+(R/4) /100]⁴ⁿ
When interest is compounded Half-yearly: Amount = P [1+(R/2) / 100]²ⁿ
When interest is compound Annually: Amount = P (1 + R/100)ⁿ
When interest is compounded annually but time is in fraction, say 3(2/5) years: Amount = P (1+R/100)³ x (1+ (2/5R) / 100)
When Rates are different for different years, say R₁%, R₂%, R₃% for 1^st, 2^ndand 3^rd year respectively. Then the amount = P (1+R₁/100) (1+R₂/100) (1+R₃/100)
Present worth of Rs.x due n years hence is given by: Present worth = [X / (1+ R/100)].

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Simple and compound interest – Question and Answers

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Question 1

A money lender lent Rs. 1000 at 3% per year and Rs. 1400 at 5% per year. The amount should be returned to him when the total interest comes to Rs. 350. Find the number of years?

A	3.5
B	3.75
C	4
D	4.5

Question 1 Explanation:

(1000 * t * 3 / 100) + (1400 * t * 5 / 100)
= 350 => t =3.5

Question 2

How long will it take a certain amount to increase by 30% at the rate of 15% simple interest?

A	2%
B	4%
C	8%
D	2%

Question 2 Explanation:

Let the principal be Rs. x
Simple interest = x * 30 / 100 = 3x / 10
T = 100 * SI / PR = 100 * (3x / 10) / x * 15 = 2%
Alternatively, this can be solved by considering
principal amount to be Rs. 100
Then simple interest becomes Rs. 30
Then, T = 100 * 30 / 100 * 15 = 2%

Question 3

A certain amount becomes Rs. 5760 in 2 years and Rs. 6912 in 3 years. What is the principal amount and the rate of interest?

A	Rs. 2500
B	Rs. 7000
C	Rs. 4000
D	Rs. 3000

Question 3 Explanation:

S.I on Rs. 5760 for 1 year = 6912 – 5760 = Rs. 1152
Therefore, Rate of interest for 1 year = 100 * 1152/5760 * 1 = 20%
Let the principal be p
Then, Principal = p[1 + 20 / 100]²= 5760
Solving which gives Principal = Rs. 4000

Question 4

Find the simple interest on Rs. 5000 at a certain rate if the compound interest on the same amount for 2 years is Rs. 253.125?

A	Rs. 250
B	Rs. 400
C	Rs. 500
D	Rs.100

Question 4 Explanation:

Let the rate of interest be r
5000[1+ r /100]² = 5000 + 253.125
[1+ r /100]² = 5253.125 / 5000
Solving which gives
[1+ r/100]² = 1681 / 1600
1+ r/100 = 41 / 40
r = 2.5
Therefore, SI = 5000 * 2 * 2.5 / 100 = Rs. 250

Question 5

The difference between SI and CI compounded annually on a certain sum of money for 2 years at 8% per annum is Rs. 12.80. Find the principal?

A	Rs. 1500
B	Rs. 2000
C	Rs. 1000
D	Rs. 5000

Question 5 Explanation:

Let the principal amount be x
SI = x * 2 * 8 / 100 = 4x / 25
CI = x[1+ 8 / 100]² – x = 104x / 625
Therefore, 104x / 625 – 4x / 25 = 12.80
Solving which gives x, Principal = Rs. 2000

Question 6

Find the compound interest on Rs. 10000 at 12% rate of interest for 1 year, compounded half-yearly?

A	Rs. 7654
B	Rs. 1236
C	Rs. 1543
D	Rs. 2765

Question 6 Explanation:

Amount with CI = 10000 [1+ (12 / 2 * 100)]²= Rs. 11236
Therefore, CI = 11236 – 10000 = Rs. 1236

Question 7

Find the compound interest on Rs. 3000 at 5% for 2 years, compounded annually?

A	Rs. 301.2
B	Rs. 307.5
C	Rs. 387.2
D	Rs. 309.3

Question 7 Explanation:

Amount with CI = 3000 (1+ 5 / 100)² = Rs. 3307.5
Therefore, CI = 3307.5 – 3000 = Rs. 307.5

Question 8

A certain principal amounts to Rs. 15000 in 2.5 years and to Rs. 16500 in 4 years at the same rate of interest. Find the rate of interest?

A	8%
B	5%
C	3%
D	9%

Question 8 Explanation:

Amount becomes 15000 in 2.5 years and 16500 in 4 years
Simple interest for (4 – 2.5) years = 16500 – 15000
Therefore, SI for 1.5 years = Rs. 1500
SI for 2.5 years = 1500 / 1.5 * 2.5 = 2500
Principal amount = 15000 – 2500 = Rs. 12500
Rate of Interest = 2500 * 100 / 12500 * 2.5
R = 8%

Question 9

Find the present worth of Rs. 78000 due in 4 years at 5% interest per year?

A	Rs. 65000
B	Rs. 63000
C	Rs. 67000
D	Rs. 61000

Question 9 Explanation:

Amount with interest after 4 years = Rs. 78000
Therefore, simple interest = 78000 – Principal
Let the principal amount be p
78000 – p = p * 4 * 5 / 100
p = 13000
Principal = 78000 – 13000 = Rs. 65000

Question 10

A sum of Rs. 25000 becomes Rs. 27250 at the end of 3 years when calculated at simple interest. Find the rate of interest?

A	9%
B	3%
C	7%
D	2%

Question 10 Explanation:

Simple interest = 27250 – 25000 = 2250
Time = 3 years
SI = PTR / 100
R = SI * 100 / PT
R = 2250 * 100 / 25000 * 3
R = 3%

Question 11

Simple interest on a sum at 5% per annum for 2 years is Rs. 60. The compound interest on the same sum?

A	Rs. 62
B	Rs. 62.4
C	Rs. 61.5
D	Rs. 60.5

Question 11 Explanation:

The difference between compound interest and simple interest on Rs. P for 2 years at R% per annum = (R × SI) / (2 × 100)
Difference between compound interest and simple interest = (R × SI) / (2 × 100) = (5 × 60) / (2 × 100) = 1.5
Hence, compound interest = Simple Interest + 1.5 = 60 + 1.5 = Rs. 61.5

Question 12

What will be the amount if a sum of Rs. 10000 is placed at compound interest for 3 years while rate of interest for the first, second and third years is 2, 5 and 10 percent, respectively?

A	Rs. 11231
B	Rs. 11781
C	Rs. 11658
D	Rs. 11244

Question 12 Explanation:

Amount after 3 years = 10000 (1 + 2 / 100) (1 + 5 / 100) (1 + 10 / 100)
= 10000 (102 / 100) (105 / 100) (110 / 100) = (102 × 105 × 11) / 10
= Rs. 11781

Question 13

If in a certain number of years Rs. 10000 amount to Rs. 160000 at compound interest, in half that time Rs. 10000 will amount to?

A	Rs. 80000
B	Rs. 40000
C	Rs. 50000
D	Rs. 60000

Question 13 Explanation:

Let the rate of interest be R% per annum
Assume that Rs. 10000 amount to Rs. 160000 in T years
10000(1+R/100)^T= 160000
=>(1+R/100)^T= 160000 / 10000 = 16
=>(1+R / 100)^T/2= 16 = 4 — (1)
In T/2 years, Rs.10000 amounts to 10000(1+R/100)^T/2
= 10000×4 [∵ from (1)]
= 40000

Question 14

A sum of Rs. 6600 was taken as a loan. This is to be repaid in two equal annual instalments. If the rate of interest be 20% compounded annually then the value of each instalment is?

A	Rs. 4320
B	Rs. 2220
C	Rs. 4400
D	Rs. 4420

Question 14 Explanation:

Present worth of Rs. x due T years hence is given by
Present Worth (PW) = x / (1+R/100)^T
Let x be the annual payment
Then, present worth of x due 1 year hence + present worth of x due 2 year hence = 6600
x / (1 + (20/100))¹ + x(1+(20/100))² = 6600
X / (6 / 5)+ x (6 / 5)
² = 6600
5x / 6 + 25x/ 36 = 6600
55x /36 = 6600
x = (6600×36) / 55 = 4320

Question 15

The difference between the simple interest on a certain sum at the rate off 10% per annum for 2 years and compound interest which is compounded every 6 months is Rs. 124.05. What is the principal sum?

A	Rs. 10000
B	Rs. 12000
C	Rs. 6000
D	Rs. 8000

Question 15 Explanation:

Let the sum be P
Compound Interest on P at 10% for 2 years when interest is compounded half-yearly
=P(1+(R/2) /100)^2T– P = P(1+(10/2) /100)^{2 x 2} –P = P(1+1/20)⁴ – P =P(21/20)⁴ –P
Simple Interest on P at 10% for 2 years = PRT/100=(P×10×2) /100=P / 5
Given that difference between compound interest and simple interest = 124.05
=>P(21 / 20)⁴ – P –P/5 = 124.05
=>P[(21/20)⁴−1 −1/5] = 124.05
=>P[194481 / 160000 – 1 −1/5] = 124.05
=>P[(194481−160000−32000) / 160000]=124.05
=>P[2481 / 160000] = 124.05
=>P = (124.05 × 160000) / 2481 = 160000 / 20 = 8000

Question 16

A sum is invested for 3 years compounded at 5%, 10% and 20 % respectively. In three years, if the sum amounts to Rs. 1386, then find the sum?

A	Rs. 1500
B	Rs. 1400
C	Rs. 1200
D	Rs. 1000

Question 16 Explanation:

1386 = P(1 + (5 / 100)) (1 + (10 / 100)) (1+(20 / 100))
1386 = P(21 / 20) (11 / 10) (6 / 5)
P = (1386 × 20 × 10 × 5) / (21 × 11 × 6)
= (66 × 20 × 10 × 5) / (11 × 6) = 20 × 10 × 5 = Rs. 1000
i.e., the sum is Rs.1000

Question 17

Divide Rs. 3364 between A and B, so that A’s Share at the end of 5 years may equal to B’s share at the end of 7 years, compound interest being at 5 percent?

A	Rs. 1764 and Rs. 1600
B	Rs. 1756 and Rs. 1608
C	Rs. 1722 and Rs. 1642
D	None of these

Question 17 Explanation:

A’s share after 5 years = B’s share after 7 years
(A’s present share)(1 + (5 / 100))⁵ = (B’s present share)(1 + (5 / 100))⁷
=>(A’s present share) / (B’s present share)
=> (1 + (5 / 100))⁷/ (1 + (5 / 100))⁵ = (1 + (5 / 100))^(7-5)
=> (1 + (5 / 100))² = (21 /20)² = 441 / 400
i.e, A’s present share : B’s present share = 441 : 400
Since the total present amount is Rs.3364, A’s share = 3364 × [441 (441 + 400)]
= 3364 × (441 / 841) = 4 × 441 = Rs. 1764
B’s present share = 3364 – 1764 = Rs.1600

Question 18

If a sum on compound interest becomes three times in 4 years, then with the same interest rate, the sum will become 81 times in?

A	12 years
B	18 years
C	16 years
D	14 years

Question 18 Explanation:

Let the sum be P
The sum P becomes 3P in 4 years on compound interest
3P = P(1 + R / 100)⁴
⇒ 3 = (1 + R / 100)⁴
Let the sum P becomes 81P in n years
=> 81P = P(1 + R / 100)ⁿ
=> 81=(1 + R / 100)ⁿ
=> (3)⁴= (1 + R / 100)ⁿ
=> ((1 + R / 100)⁴)⁴ = (1 + R / 100)ⁿ
=> (1 + R / 100)¹⁶
=> (1 + R / 100)ⁿ = 16
i.e, the sum will become 81 times in 16 years

Question 19

Arun borrowed a certain sum from Manish at a certain rate of simple interest for 2 years. He lent this sum to Sunil at the same rate of interest compounded annually for the same period. At the end of two years, he received Rs. 2400 as compound interest but paid Rs. 2000 only as simple interest. Find the rate of interest?

A	40%
B	30%
C	20%
D	10%

Question 19 Explanation:

Let the sum be x
Simple interest on x for 2 years = Rs.2000
Simple interest = PRT / 100
2000 = (x * R * 2) / 100
⇒ xR = 100000— (1)
Compound Interest on x for 2 years = 2400
P(1 + R / 100)^T– P = 2400
x(1 + R / 100)² – x = 2400
x[1 + (2R / 100) + (R²/ 10000) – x] =2400
x (2R / 100) + (R² / 10000) = 2400
(2xR /100) + (xR² / 10000 = 2400— (2)
Substituting the value of xR from (1) in (2) ,we get
(2 × 100000) / 100 + (100000 × R) / 10000 = 2400
2000 + 10R = 2400
10R = 400
R = 40%

Question 20

A	40%
B	30%
C	20%
D	10%

Question 20 Explanation:

Question 21

A sum is invested at compounded interest payable annually. The interest in the first two successive years was Rs. 400 and Rs. 420. The sum is?

A	Rs. 8000
B	Rs.7500
C	Rs. 8500
D	Rs. 8200

Question 21 Explanation:

This means that, simple Interest on Rs.400 for 1 year = 420 – 400 = 20
Rate = 100×SI / PT = (100×20) / 400×1 = 5%
Rs.400 is the interest on the sum for 1st year
Hence, sum = (100×SI) / RT = (100×400) / (5×1)
= Rs. 8000

Question 22

The population of a town is 40,000. It decreases by 20 per thousand per year. Find out the population after 2 years?

A	38484
B	38266
C	38416
D	38226

Question 22 Explanation:

This problem is similar to the problems we saw in compound interest
We can use the formulas of compound interest here as well
In compound interest, interest (a certain percentage of the principal) will be added to the principal after every year
Similarly, in this problem, a certain count(a certain percentage of the population) will be decreased from the total population after every year
i.e., the formula becomes, A = P(1 − R / 100)T
where Initial population = P, Rate = R% per annum,
Time = T years and A = the population after T years
Please note that we have to use the -ve sign here instead of the + sign as the population gets decreased
R = (20 × 100) / 1000 = 2%(∵ percentage is calculated for twenty per thousand)
Population after 2 years = P(1 − R / 100)T
= 40000(1 − 2 / 100)2
= 40000(1 − 1 / 50)2 = 40000(49 / 50)2 = (40000 × 49 × 49) / (50 × 50) = 16 × 49 × 49 = 38416

Question 23

Andrews earns an interest of Rs. 1596 for the third year and Rs. 1400 for the second year on the same sum. Find the rate of interest if it is lent at compound interest?

A	12%
B	13%
C	14%
D	15%

Question 23 Explanation:

Interest earned in 3rd year = Rs. 1596
Interest earned in 2nd year = Rs. 1400
i.e, in 3rd year, Andrews gets additional interest of (Rs. 1596 – Rs. 1400) = Rs.196
This means, Rs.196 is the interest obtained for Rs.1400 for 1 year
R = (100×SI) / PT = (100×196) / (1400×1) = 196 / 14 = 14%

Question 24

If the difference between the simple interest and compound interests on some principal amount at 20% for 3 years is Rs. 48, then the principal amount is?

A	Rs. 365
B	Rs. 325
C	Rs. 395
D	Rs. 375

Question 24 Explanation:

Let the sum be Rs. x
Amount after 3 years on Rs.x at 20% per annum when interest is compounded annually
=P(1 + R / 100)^T = x(1 + 20 / 100)³ = x(120 / 100)³= x(6 / 5)3
Compound Interest = x(6 / 5)³ – x = x[(6 / 5)³ − 1]
= x [216 / 125 − 1] = 91x / 125
Simple Interest = PRT / 100 = (x * 20 * 3) / 100 = 3x / 5
Given that difference between compound interest and simple interest is Rs. 48
91x / 125 − 3x / 5 = 48
(91x − 75x) / 125 = 48
16x / 125 = 48
x = (48 × 125) / 16 = 3 × 125 = Rs. 375
i.e, the sum is Rs.375

Question 25

What sum invested for 2 years at 14% compounded annually will grow to Rs. 5458.32?

A	4120
B	3300
C	4200
D	4420

Question 25 Explanation:

Let sum be P
P(1 + R / 100)^T = 5458.32
P(1 + 14 / 100)² = 5458.32
P(114 / 100)²= 5458.32
P = (5458.32 × 100 × 100) / (114 × 114)
= (47.88 × 100 × 100) / 114 = 0.42 × 100 × 100 = 4200

Question 26

A sum of money is borrowed and paid back in two annual instalments of Rs. 882 each allowing 5% compound interest. The sum borrowed was?

A	Rs.1820
B	Rs.1640
C	Rs.1260
D	Rs.1440

Question 26 Explanation:

Present worth of Rs. x due T years hence is given by
Present Worth (PW) = x / (1+R / 100)^T
The sum borrowed = Present Worth of Rs.882 due 1 year hence + Present Worth of Rs.882 due 2 year hence
=> 882 / (1 + (5 / 100))¹+ 882(1 + (5 / 100))²
=> 882 / (105 /100) + 882 / (105 / 100)²
=> [882(21 / 20) + 882 / (21 / 20)²
=> (882 × 20) / 21+(882 × 20 × 20) / (21 × 21)
=> 42 × 20 + (42 × 20 × 20) / 21 = 840 + 2 × 20 × 20
=> 840 + 800 = 1640
i.e., The sum borrowed = Rs.1640

Question 27

A sum put out at 4% compound interest payable half-yearly amounts to Rs. 13265.10 in 1¹⁄₂ years. The sum is?

A	Rs. 12500
B	Rs. 11200
C	Rs. 8840
D	Rs. 12600

Question 27 Explanation:

Let the sum be P
Time, T = 1 1/2 year = 3 / 2 year
Amount after 1 1/2 years = P(1 + (R / 2) / 100)^2T = P(1 + (4 / 2) / 100)^{2 x 3/2}
= P(1 + 2 / 100)³ = P(102 / 100) ³ = P(51 / 50)³
Given that amount after 1¹⁄₂ years = 13265.10
=> P(51 / 50)³= 13265.10
=> P = 13265.10(50 / 51)³ = (13265.10 × 50 × 50 × 50) / (51 × 51 × 51) = (260.1 × 50 × 50 × 50) / (51 × 51)
=> (5.1 × 50 × 50 × 50) / 51 = 0.1 × 50 × 50 × 50 = Rs. 12500

Question 28

A sum amounts to Rs. 882 in 2 years at 5% compound interest. The sum is?

A	Rs. 800
B	Rs. 822
C	Rs. 840
D	Rs. 816

Question 28 Explanation:

Let the sum be P
Amount After 2 years = P(1 + R / 100)^T= P(1 + 5 / 100)²=P(105 / 100)²= P(21 / 20)²
Given that amount After 2 years = 882
=> P(21 / 20)²= 882
=> P=(882 × 20 × 20) / (21 × 21) = 2 × 20 × 20 = Rs. 800

Question 29

The Simple interest on a certain sum for 2 years at 20% per annum is Rs. 80. The corresponding compound interest is?

A	Rs. 66
B	Rs. 82
C	Rs. 86
D	Rs. 88

Question 29 Explanation:

Principal, P = (100 × SI) / RT = (100 × 80) / (20 × 2) = Rs. 200
Amount after 2 year on Rs.200 at 20% per annum when interest is compounded annually
=> P(1 + R100)^T = 200(1 + 20 / 100)² = 200(120 / 100)²
=> (200 × 120 × 120) / (100 × 100) = (2 × 120 × 120) / 100
=> 2 × 12 × 12 = Rs. 288
Compound Interest = 288 – 200 = Rs.88

Question 30

There is 80% increase in an amount in 8 years at simple interest. What will be the compound interest of Rs. 14,000 after 3 years at the same rate?

A	Rs.3794
B	Rs.3714
C	Rs.4612
D	Rs.4634

Question 30 Explanation:

Let P = Rs.100
Simple Interest = Rs. 80 ( ∵ 80% increase is due to the simple interest)
Rate of interest = (100 × SI) / PT = (100 × 80) / (100 × 8) = 10% per annum
Now let’s find out the compound interest of Rs. 14,000 after 3 years at 10%
P = Rs. 14000
T = 3 years
R = 10%
Amount after 3 years = P(1 + R / 100)^T = 14000(1 + 10 / 100)³ = 14000 (110 / 100)³
= 14000(11 / 10)³ = 14 × 11³ = 18634
Compound Interest = Rs.18634 – Rs.14000 = Rs.4634

Question 31

The compound interest on Rs. 30,000 at 7% per annum is Rs. 4347. The period (in years) is?

A	2
B	3
C	8
D	6

Question 31 Explanation:

Amount = Rs. (30000 + 4347) = Rs. 34347
Let the time be n years
Then, 30000(1 + 7 / 100)ⁿ= 34347
=> (107 / 100) ⁿ= (34347 / 30000) = 11449 / 10000 = (107 / 100)
n = 2 years

Question 32

What is the difference between the compound interests on Rs. 5000 for 1 1/2 years at 4% per annum compounded yearly and half-yearly?

A	A. Rs. 2.04
B	Rs. 3.06
C	Rs. 4.80
D	Rs. 8.30

Question 32 Explanation:

Rs.(5000 x (26 / 25) x (51 / 50)) = Rs 5304
C.I when interest is compounded half- yearly = Rs [5000 x (1 + 2 / 100)³]
= Rs.(5000 x (51 / 50) x (51 / 50) x (51 / 50)) = Rs 5306.04
Difference = Rs. (5306.04 – 5304) = Rs. 2.04

Question 33

There is 60% increase in an amount in 6 years at simple interest. What will be the compound interest of Rs. 12,000 after 3 years at the same rate?

A	Rs. 2160
B	Rs. 3120
C	Rs. 3972
D	Rs. 6240

Question 33 Explanation:

Let P = Rs. 100. Then, S.I. Rs. 60 and T = 6 years
R =(100 x 60 / 100 x 6) = 10% p.a
Now, P = Rs. 12000. T = 3 years and R = 10% p.a
C.I = Rs [12000 x (1+10/100)3-1]
= Rs (12000 x 331 / 1000) = 3972

Question 34

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?

A	A. 625
B	630
C	640
D	650

Question 34 Explanation:

Let the sum be Rs. x. Then,
C.I = [x (1 + (4 / 100) 2 – x)] = (676 / 625 x -x) = 51 / 625 x
S.I = (X * 4 * 2) / 100
= 2x / 25
(51x / 625) – (2x / 25) = 1
x = 625

Question 35

A bank offers 5% compound interest calculated on half-yearly basis. A customer deposits Rs. 1600 each on 1^st January and 1^st July of a year. At the end of the year, the amount he would have gained by way of interest is?

A	Rs. 120
B	Rs. 121
C	Rs. 122
D	Rs. 123

Question 35 Explanation:

Amount = Rs.[1600 x (1 + (5 / 2x 100)²) + 1600 x (1+ (5 / 2x 100))]
Rs.[1600 x 41/40 x 41/40 + 1600 x 41/40]
Rs [1600 x 41/40(41/40 + 1)]
Rs [(1600 x 41 x 81)/ (40 x 40)]
= Rs. 3321
C.I. = Rs. (3321 – 3200) = Rs. 121

Question 36

The effective annual rate of interest corresponding to a nominal rate of 6% per annum payable half-yearly is?

A	6.06%
B	6.07%
C	6.08%
D	6.09%

Question 36 Explanation:

Amount of Rs.100 for 1 year when compounded half-yearly = Rs [ 100 x (1 + 3 / 100)2] Rs.106.09
Effective rate = (106.09 – 100)% = 6.09%

Question 37

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?

A	Rs. 8620
B	A. Rs. 8600
C	Rs. 8820
D	None of these

Question 37 Explanation:

Amount = Rs. [8000 x (1 + 5 / 100)²]
= Rs. (8000 x 21 / 20 x 21 / 20)
= Rs. 8820

Question 38

The least number of complete years in which a sum of money put out at 20% compound interest will be more than doubled is?

A	3
B	4
C	5
D	6

Question 38 Explanation:

P(1 + 20 / 100)n > 2P = (6 / 5)n > 2
Now, (6 / 5 x 6 / 5 x 6 / 5 x 6 / 5) >2
So, n = 4 years

Question 39

At what rate of compound interest per annum will a sum of Rs. 1200 become Rs. 1348.32 in 2 years?

A	6.5%
B	6%
C	7%
D	8%

Question 39 Explanation:

Let the rate be R% p.a
Then, 1200 x (1 + R / 100)² = 1348.32
(1 + R/100)² = 134832 / 120000 = 11236 / 10000
(1+ R / 100)² = (106 / 100)²
1 + (R /100) =106 / 100 R = 6%

Question 40

What will be the compound interest on a sum of Rs. 25,000 after 3 years at the rate of 12 p.c.p.a.?

A	Rs. 9000.30
B	Rs. 9720
C	Rs. 10123.20
D	Rs. 10483.20

Question 40 Explanation:

Amount = Rs. [25000 x (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

Question 41

The difference between compound interest and simple interest on an amount of Rs. 15,000 for 2 years is Rs. 96. What is the rate of interest per annum?

A	8
B	10
C	12
D	Cannot be determined

Question 41 Explanation:

[15000 x (1+R/100)2 – 15000] – (15000 x R x 2) / 100 = 96
15000[(1+R/100)2 – 1 – 2R / 100] = 96
15000[(100+R)² – (10000 / 10000)- (200 x R)] = 96
R² = (96 x 2) / 3 = 64
R = 8
Rate = 8%

Question 42

The difference between simple interest and compound on Rs. 1200 for one year at 10% per annum reckoned half-yearly is?

A	Rs. 2.50
B	Rs. 3
C	Rs. 3.75
D	Rs 4

Question 42 Explanation:

S.I = Rs (1200 x 10 x 1) / 100 = Rs. 120
C.I = Rs (1200 x (1 + (5/100)2 – 1200)) = Rs 123
Difference = Rs. (123 – 120) = Rs. 3

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Ratio and Proportion

MS — Sat, 19 Aug 2017 06:45:01 +0000

Explanation / Important formulas

Ratio: The ratio of two quantities a and b in the same units is the fraction a/b and we write it as a : b.

In the ratio a : b, we call a as the first term or antecedent and b, the second term or consequent.
Example: The ratio 5 : 9 represents 5/9 With antecedent = 5, consequent = 9.

[Rule: The multiplication or division of each term of a ratio by the same non-zero number does not affect the ratio.]

Example: 4 : 5 = 8 : 10 = 12 : 15. Also, 4 : 6 = 2 : 3.

Proportion: The equality of two ratios is called proportion.

If a : b = c : d, we write a : b :: c : d and we say that a, b, c, d are in proportion.
Here a and d are called extremes, while b and c are called mean terms.
Product of means = Product of extremes.
Thus, a : b :: c : d (b x c) = (a x d).

Fourth Proportional: If a : b = c : d, then d is called the fourth proportional to a, b, c.

Third Proportional: a : b = c : d, then c is called the third proportion to a and b.

Mean Proportional: Mean proportional between a and b is ab.

Comparison of Ratios: We say that (a : b) > (c : d) <=> a / b > c / d

Compounded Ratio: The compounded ratio of the ratios: (a : b), (c : d), (e : f) is (ace : bdf).

Duplicate Ratios: Duplicate ratio of (a : b) is (a² : b²).

Sub-duplicate ratio of (a : b) is (a^1/2 : b^1/2).
Triplicate ratio of (a : b) is (a³ : b³).
Sub-triplicate ratio of (a : b) is (a^1/3 : b^1/3).
If a/b = c/d, then a + b / a – b =c + d / c – d [Componendo and dividendo]

Variations

We say that x is directly proportional to y, if x = ky for some constant k and we write, x ∝ y.
We say that x is inversely proportional to y, if xy = k for some constant k and we write, x ∝ 1/ y

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Ratio and proportion – Question and Answers

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Question 1

If the area of two circles are in the ratio 169:196 then the ratio of their radii is?

A	10:11
B	11:12
C	12:13
D	13:14

Question 1 Explanation:

If r and R are radii if two circles, then
pi * r^2 /pi * R^2 = 169 / 196
r^2 / R^2 = 169 / 196 = (13/14)^2
so r/R = 13/14

Question 2

The salaries of A, B, C are in the ratio 2 : 3 : 5. If the increments of 15%, 10% and 20% are allowed respectively in their salaries, then what will be new ratio of their salaries?

A	3 : 3 : 10
B	10 : 11 : 20
C	23 : 33 : 60
D	Cannot be determined

Question 2 Explanation:

Let A = 2k, B = 3k and C = 5k
A’s new salary = (115 / 100) of 2k = (115 / 100) x 2k = 23k / 10
B’s new salary = (110 / 100) of 3k = (110 / 100) x 3k = 33k / 10
C’s new salary = (120 / 100) of 5k = (120 / 100) x 5k = 6k
New ratio = [23k / 10 : 33k / 10 : 6k] = 23 : 33 : 60

Question 3

If 40% of a number is equal to two-third of another number, what is the ratio of first number to the second number?

A	2:5
B	3:7
C	5:3
D	7:3

Question 3 Explanation:

Let 40% of A = 2 / 3 B
Then, 40A / 100 = 2B / 3
2A / 5 = 2B / 3
A / B = (2 / 3 * 5 / 2) = 5 / 3
A : B = 5 : 3

Question 4

The fourth proportional to 5, 8, 15 is?

A	18
B	24
C	19
D	20

Question 4 Explanation:

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

Question 5

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?

A	27
B	33
C	49
D	55

Question 5 Explanation:

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

Question 6

In a bag, there are coins of 25 p, 10 p and 5 p in the ratio of 1 : 2 : 3. If there is Rs. 30 in all, how many 5 p coins are there?

A	50
B	100
C	150
D	200

Question 6 Explanation:

Then, sum of their values = Rs. {(25x / 100) +[(10 * 2x) / 100)]+ (5 * 3x /100)} = Rs. 60x / 100
(60x / 100) = 30
x = (30 x 100) / 60 = 50
Hence, the number of 5 p coins = (3 x 50) = 150

Question 7

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?

A	8:9
B	17:18
C	21:22
D	Cannot be determined

Question 7 Explanation:

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 * 7x) And (110 / 100 * 8x)
42x / 5 and 44x / 5
The required ratio = 42x / 5 : 44x / 5 = 21 : 22

Question 8

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?

A	17,000
B	20,000
C	25,500
D	38000

Question 8 Explanation:

Let the original salaries of Ravi and Sumit be Rs. 2x and Rs. 3x respectively
Then, (2x + 4000) / (3x + 4000) = 40 / 57
57(2x + 4000) = 40(3x + 4000)
6x = 68,000
3x = 34,000
Sumit’s present salary = (3x + 4000) = Rs.(34000 + 4000) = Rs. 38,000

Question 9

If 0.75 : x :: 5 : 8, then x is equal to?

A	1.12
B	1.20
C	1.25
D	1.30

Question 9 Explanation:

(x * 5) = (0.75 * 8) x = 6 / 5 = 1.20

Question 10

The sum of three numbers is 98. If the ratio of the first to second is 2 : 3 and that of the second to the third is 5 : 8, then the second number is?

A	20
B	30
C	48
D	58

Question 10 Explanation:

Let the three parts be A, B, C. Then,
A : B = 2 : 3 and B : C = 5 : 8 = (5 x 3 / 5 : 8 x 3 / 5) = 3 : (24 / 5)
A : B : C = 2 : 3 : (24 / 5) = 10 : 15 : 24
B = 98 x (15 / 49) = 30

Question 11

A and B together have Rs. 1210. If 4/15 of A’s amount is equal to 2/5 of B’s amount, how much amount does B have?

A	460
B	484
C	550
D	664

Question 11 Explanation:

4 /15A = 2 /5B
A = (2/5 * 15/4)B
A = 3 / 2B
A / B = 3 / 2
A : B = 3 : 2
B’s share = Rs. (1210 * (2 / 5)) = Rs. 484

Question 12

Two numbers are respectively 20% and 50% more than a third number. The ratio of the two numbers is?

A	2 : 5
B	3 : 5
C	4 : 5
D	6 : 7

Question 12 Explanation:

Let the third number be x
Then the 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

Question 13

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?

A	500
B	1500
C	2000
D	None of these

Question 13 Explanation:

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

Question 14

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?

A	2 : 3: 4
B	6 : 7 : 8
C	6 : 8 : 9
D	None of these

Question 14 Explanation:

Originally, let the number of seats for Mathematics, Physics and Biology be 5x, 7x and 8x respectively
Number of increased seats are (140% of 5x), (150% of 7x) and (175% of 8x)
= (140/100 * 5x) ,(150/100 * 7x) and (175/100 * 8x)
= 7x, 21x /2 and 14x
The required ratio = 7x : 21x /2 : 14x
= 14x : 21x : 28x
= 2 : 3 : 4

Question 15

In a mixture 60 litres, the ratio of milk and water 2 : 1. If this ratio is to be 1 : 2, then the quantity of water to be further added is?

A	20
B	30
C	40
D	60

Question 15 Explanation:

Quantity of milk = (60 * 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

Question 16

A sum of Rs.312 was divided among 100 boys and girls in such a way that the boy gets Rs.3.60 and each girl Rs.2.40 the number of girls is?

A	35
B	40
C	45
D	50

Question 16 Explanation:

Let x be the number of boys and y be the number of girls
Given total number of boys and girls = 100
x + y = 100
A boy gets Rs. 3.60 and a girl gets Rs. 2.40
The amount given to 100 boys and girls = Rs. 312
3.6x + 2.4y = 312
By Solving
3.6x + 3.6y = 360 multiplying by 3.6
3.6x + 2.4y1.20yy = 312 = 48 = 40
The number of girls is 40

Question 17

In a mixture 60 litres, the ratio of milk and water 2:1. If the this ratio is to be 1:2, then the quantity of water to be further added is?

A	20
B	30
C	40
D	60

Question 17 Explanation:

Quantity of milk = 60 × 23 = 40 litres
Quantity of water in it = (60 − 40) litres = 20 litres
New ratio = 1 : 2
Let, the quantity of water to be added further be x litres
Then milk : water = 4020 + x
Now, 4020 + x20 + x ⇒x =12 = 80 = 60

Question 18

Two numbers are respectively 20% and 50% more than a third number. The ratio of the two numbers is?

A	2 : 5
B	3 : 5
C	4 : 5
D	5 : 4

Question 18 Explanation:

Let the third number be x
Then, first number = 120% of x = 120 x 100=6 x 5
Second number = 150% of x = 150 x 100 = 3 x 2
Ratio of first two numbers = 6 x 5 : 3 x 2 = 4 : 5

Question 19

A sum of Rs. 36.90 is made up of 180 coins which are either 10 paise coins or 25 p coins. The number of 10 p coins is?

A	48
B	54
C	56
D	60

Question 19 Explanation:

Total number of coins = 180
Let x be number of 10p coins and y be number of 25p coins
x + y = 180 ——–(i)
Step (ii) Given 10p coins and 25p coins make the sum = Rs. 36.90
10x 100 + 25y 100 ⇒ 10x + 25y = 36.9 = 3690—–(ii)
Step (iii)
Solving (i) and (ii),
10x + 10y10x + 25y − 15yy = 1800 = 3690 = 1890 = 126
Substitute y value in equation (i)
x = 180 – 126 = 54
Number of 10p coins = 54

Question 20

The salaries A, B, C are in the ratio 2:3:5. If the increments of 15%, 10% and 20% are allowed respectively in their salaries, then what will be new ratio of their salaries?

A	3 : 3 : 10
B	10 : 11 : 20
C	23 : 33 : 60
D	Cannot be determined

Question 20 Explanation:

Let A = 2k, B = 3k and C = 5k
A’s new salary = (115100 × 2k) = 23k10 B’s new salary = (110100 × 3k) = 33k10 C’s new salary = (120100 × 5k) = 6k ⇒New ratio = (23k10 : 33k10 : 6k) = 23 : 33 : 60

Question 21

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?

A	Rs. 17,000
B	Rs. 20,000
C	Rs. 34,000
D	Rs. 38,000

Question 21 Explanation:

Let the original salaries of Ravi and Sumit be Rs. 2x and Rs. 3x respectively
then, 2x + 40003x + 4000 ⇒57 × (2x + 4000) 6x3x = 4057 = 40 × (3x + 4000) = 68,000 = 34,000
Sumit’s present salary = (3x + 4000) = Rs.(34000 + 4000) = Rs. 38,000

Question 22

The sum of three numbers is 98. If the ratio of the first to second is 2:3 and that of the second to the third is 5:8, then the second number is?

A	20
B	30
C	48
D	58

Question 22 Explanation:

Let the three parts be A, B, C. Then,
A:B=2:3 and B:C=5:8
=> 5 × 35 : 8 × 35 ⇒ A : B : C = 2 : 3 : 245 = 10 : 15 : 24
=> B = 98 × 1549 = 30 = 3 : 245

Question 23

Rs.432 is divided amongst three workers A, B and C such that 8 times A’s share is equal to 12 times B’s share which is equal to 6 times C’s share. How much did A get?

A	Rs.192
B	Rs. 133
C	Rs. 144
D	Rs. 128

Question 23 Explanation:

8 times A’s share = 12 times B’s share = 6 times C’s share
Note that this is not the same as the ratio of their wages being 8 : 12 : 6
In this case, find out the L.C.M of 8, 12 and 6 and divide the L.C.M by each of the above numbers to get the ratio of their respective shares
The L.C.M of 8, 12 and 6 is 24
Therefore, the ratio A : B : C is
248 : 2412 : 246 ⇒ A : B : C :: 3 : 2 : 4
The sum of the total wages =3x + 2x + 4x = 432
9x = 432 or x = 48
Hence A gets 3 × 48 = Rs 144

Question 24

If 20 men or 24 women or 40 boys can do a job in 12 days working for 8 hours a day, how many men working with 6 women and 2 boys take to do a job four times as big working for 5 hours a day for 12 days?

A	120 men
B	122 men
C	128 men
D	134 men

Question 24 Explanation:

Amount of work done by 20 men = 24 women = 40 boys or 1 man = 1.2 woman = 2 boys
Let us therefore, find out the amount of men required
If only men were working on the job, to complete the new job under the new conditions and then make adjustments for the women and children working with the men
The man hours required to complete the new job = 4 times the man hours required to complete the old job
(As the new job is 4 times as big as the old job)
Let n be the number of men required
n × 5 × 12 = 20 × 8 × 12 × 4 = or n = 128
i.e. 128 men working on the job will be able to complete the given job
However, the problem states that 6 women and 2 boys are working on the job
6 women $ = \dfrac{6}{1.2} = 5 men and 2 boys = 1 man
∴ The equivalent of 5 + 1 = 6 men are already working
Thus, final number of men working
= 128 − 6 = 122 men

Question 25

P,Q and R enter into a partnership with capitals in the ratio 3:2:1. After 4 months, P leaves the business and after 4 more months Q also leaves the business and R continues till the end of the year. If R takes 10 of the profit for managing the business, then what part of the profit does R get?

A	37%
B	36%
C	27%
D	30%

Question 25 Explanation:

Let Rs 100 be the profit
Rs 90 is to be divided in the ratio 12 : 16 : 12. i.e 3 : 4 : 3
R gets 310 × 90 = 27 and 10 for managing
Thus 27 + 10 = 37%

Question 26

An outgoing batch of students wants to gift a PA system worth Rs 4,200 to their school. If the teachers, offer to pay 50% more than the students and an external benefactor gives three times the teacher’s contribution, then how much should the teachers donate?

A	Rs 600
B	Rs 840
C	Rs 900
D	Rs 1,200

Question 26 Explanation:

The ratio of the share students : teacher : benefactor = 1 : 1.5 : 4.5
So the proportion to teacher’s share = 1.57
Hence, the teachers would donate 1.57 × 4200 = Rs 900

Question 27

IBM and KTC quote for a tender. On the tender opening day, IBM realizes that their quotations are in the ratio 7:4 and hence decreases its price during negotiations to make it Rs 1 Lakh lower than KTC’s quoted price. KTC realizes that the final quotes of the two were in the ratio 3:4. What was the price at which IBM won the bid?

A	Rs 7 Lakh
B	Rs 4 Lakh
C	Rs 3 Lakh
D	Rs 1.5 Lakh

Question 27 Explanation:

IBM initially quoted Rs 7x lakh. KTC quoted 4x lakh
IBM’s final quote = (4×1) Lakh
Thus, 4x − 14x ⇒ x = 34 = 1
IBM’s bid winning price = Rs 3. Lakh
So IBM wins the bid at 4×1= Rs. 3 lakh

Question 28

The proportion of milk and water in 3 samples is 2:1, 3:2 and 5:3. A mixture comprising of equal quantities of all 3 samples is made. The proportion of milk and water in the mixture is?

A	2 : 1
B	5 : 1
C	99 : 61
D	227 : 133

Question 28 Explanation:

Proportion of milk in 3 samples is 23,35,58
Proportion of water in 3 samples is 13,25,38
Since equal quantities are taken,
Total proportion of milk is 23 + 35 + 58 = 227120
Total proportion of water is 13 + 25 + 38 = 133120
Proportion of milk and water in the solution is = 227 : 133

Question 29

Rs 4830 is divided among Abhishek, Dishant and Prashant such that if Abhishek’s share diminishes by Rs 5, Dishant’s share diminishes by Rs 10 and Prashant’s share diminishes by Rs 15, their shares will be in the ratio 5:4:3. Find the Dishant’s original share?

A	1610
B	2010
C	2410
D	1590

Question 29 Explanation:

Let actual share of Abhishek, Dishant and Prashant be A, D , P respectively
A + D + P = 4830
Hence, A’s, D’s and P’s share are diminished by Rs 5, Rs 10 and Rs 15, their net share will be Rs.4800
Dishant’s diminished share = 412 × 4800 = Rs 1600
Hence, Dishant actual share = Rs 1600 + Rs 10 = Rs 1610

Question 30

The ratio of squares of first n natural numbers to square of sum of first n natural numbers is 17:325. The value of n is?

A	15
B	25
C	35
D	40

Question 30 Explanation:

Sum of squares of first n natural numbers
= ∑n2 = n(n + 1)(2n + 1)6
Squares of sum of first n natural numbers
= (∑n)2 = n(n + 1)2 × n (n + 1)2
Now the ratio is
= n(n +1)(2n + 1) 6 : n (n+1) 2 × n (n+1) 2= 17 : 325
⇒ (2n + 13)(n (n + 1) 2) = 17325
⇒2(2n + 1)3n(n + 1) = 17325
⇒ 650 (2n + 1) = 51n (n + 1)
⇒ 1300n + 650 = 51n2 + 51n
⇒ 51n2 − 1300n + 51n − 650 = 0
⇒ 51n2 − 1249n − 650 = 0
Upon solving the above equation, we get,
n = 25, − 0.5098
Out of these two,
n = 25 is there in the solution

Question 31

Two men, sitting on the dining table. One man has 7 eggs and other had 5 eggs. A third man passing by requested them to share their food in return for money. The three of them shared the eggs equally and the third traveler paid the other two a total of Rs 24. Find the difference between the amounts received by first two men?

A	18
B	12
C	16
D	15

Question 31 Explanation:

As the two men had a total of 12 = ( 7 + 5) eggs, also they agreed to share eggs in such a manner that all three of them got an equal number of eggs
So each of them must have got 4 = (123) eggs
The first and the second man must have given 3 = (7 − 4) eggs and 1= (5 − 4) egg to the third man
Hence, the ratio of their share is 3 : 1
Now, money has to be distributed in the ratio of their contribution. As total money paid by the stranger is Rs. 24
Hence the first and the second man get Rs 18 = (24 × 33 + 1) and Rs 6 = (24 × 13 + 1) respectively
The 1st man gets Rs 12 = (18 − 6) more than the second

Question 32

A, B and C play cricket. A‘s runs are to B‘s runs and B‘s runs are to C‘s as 3:2. They get altogether 342 runs. How many runs did A make?

A	162
B	108
C	72
D	None

Question 32 Explanation:

A : B = 3 : 2 = 9 : 6
B : C = 3 : 2 = 6 : 4 (making B equal)
Therefore, A : B : C = 9 : 6 : 4
Therefore, the runs made by A = 919 × 342 = 162

Question 33

The most economical speed for this journey is?

A	20 km/hr
B	30 km/hr
C	35 km/hr
D	40 km/hr

Question 33 Explanation:

Cost = K²×speed²[ Where k is constant]
64 = k × 16 × 16
k = 1/4
Cost = speed²/4
Total cost = (speed²/ 4 + 400 ) x 100 / speed
using options, putting different values of speed, we find speed = 40km/hr to be most economical

Question 34

The total cost for the journey at this most economical speed is?

A	Rs. 8000
B	Rs. 6000
C	Rs. 10,000
D	Rs. 11,000

Question 34 Explanation:

E=ks²
⇒ E= 64 and S = 16
K=1/4
Total cost =1/4 × s²t + 400t
=14 × s ²t + 400t
Most economical speed, checking options we get most economical speed at 40km/hr
Total cost at 40km/hr
= ¼ x 40 ² + 400 ( 400 / 40)
= Rs. 8000

Question 35

Points A and B are both in the line segment PQ and on the same side of its midpoint. A divides PQ in the ratio 2:3, and B divides PQ in the ratio 3:4. If AB=2, then the length of PQ is?

A	70
B	75
C	80
D	85

Question 35 Explanation:

Let PA = 2x and AB = 3x
and PB = 3y and BQ = 4y
PB : BQ = 3 : 4
PA : QA = 2 : 3
PQ = 5x = 7y
x = 7 / 5 …..(i)
From equation (i) and (ii),
Now, AB = PQ − PA − BQ =7y − 4y −2x
⇒ 3y − 2x = 2 ————– (ii)
y = 10 and Hence, PQ = 70

Question 36

Three beakers have capacity of 250ml, 650ml and 200ml. 682 ml of Juice is poured into them so that the same fraction of each is filled. The volume filled in the largest beaker will be?

A	415 ml
B	403 ml
C	400 ml
D	424 ml

Question 36 Explanation:

As the beaker filled by the same fractions,
p250 = q650
= r200
= p + q + r250 + 650 + 200
Thus p + q + r = 682
⇒q = 403 ml

Question 37

The total surface area of a solid iron cube and a solid aluminium cuboid are the same. The length, breadth and height of the cuboid are in the ratio 1:2:4. Both are melted together in a vessel. What is the ratio of iron and aluminium in the resultant mixture?

A	(14/3)3/2:8
B	8:(14/3)3/2
C	(3/14)3/2:8
D	8:(3/14)3/2

Question 37 Explanation:

lbh = 2 (lb + bh + hl)=1 : 2 : 4 = 2l , h = 4l = 2×(2l2 + 8l2 + 4l2)=28l2 = (14/3)3/2l3 = (14/3)3/2:8

Question 38

The monthly salaries of two persons are in the ratio of 4:7. If each receives an increase of Rs.25 in the salary, the ratio is altered to 3:5. Find their respective salaries?

A	120 and 210
B	80 and 140
C	180 and 300
D	200 and 350

Question 38 Explanation:

Let the salaries be 4x and 7x
Therefore, 4x + 257x + 255(4x + 25)20x + 125x = 35 = 3(7x + 25) = 21x + 75 = 50
Therefore, their salaries are 4 × 50 and 7 × 50 i.e., 200 and 350

Question 39

On a certain day, the ratio of the passenger in the 1st class and the second class travelling by train is 1:3. The ratio of the fares collected from each first class and second class passengers is 30:1. If the total amount collected from all the passengers is Rs 1,320. Find the amount in Rs, collected from the second class passengers?

A	240
B	360
C	480
D	120

Question 39 Explanation:

Let the number of passengers travelling by first class and second class be x and 3x respectively
Lets the fares collected from each of the first class and second class passengers be 30y and y respectively
Hence x(30y) + 3x(y) = 30xy + 3xy = 33xy = 1320
xy = 40
Total amount collected from the second class
= 3xy = 3 × 40 = Rs 120

Question 40

By mistake, instead of dividing Rs 117 among three persons P, Q and R in the ratio (12,13,14), it was divided in the ratio 2:3:4. Who gains the most and how much?

A	28
B	35
C	25
D	27

Question 40 Explanation:

Ratio 12,13,14 is equivalent to 6 : 4 : 3
So, in case of correct distribution, P would have got,
=> 117 × 66 + 4 + 3 = Rs. 54
Q would have got, = 117 × 46 + 4 + 3 = Rs. 36
and R would have got, = 117 × 36 + 4 + 3 = Rs. 27
But actually the money was divided in the ratio 2 : 3 : 4 and shares of P, Q and R in this case would be
For P, = 117 × 22 + 3 + 4 = Rs. 26
For Q, = 117 × 32 + 3 + 4 = Rs. 39,
and for R, = 117 × 42 + 3 + 4 = Rs. 52
Thus P gains, = 54 − 26 = 28,
Q gains, = 36 − 39 = −3 (loss)
and R gains, = 27 − 52 = −25 (loss)
Thus, P gains the most Rs. 28

Question 41

20 of the students in a class failed in an examination. Out of the students who failed, 75 were males. Male students who failed constitute 90 of the economically poor students in the class. What is the ratio of the number of economically poor students to the number of students in the class?

A	1 : 4
B	1 : 5
C	5 : 6
D	1 : 6

Question 41 Explanation:

Let the total number of students in the class be ‘t’
Let total number of students in the class who are economically poor be ‘p’
Total number of students who failed = t5
Total number of male students who failed = 34 × t5 = 9p10
⇒ pt = 16
⇒ Number of economically poor studentsNumber of Students = 16 = 1 : 6

Question 42

The present ages of A and B are as 6:4. Five years ago their ages were in the ratio 5:3. Find their present ages?

A	42, 28
B	36, 24
C	30, 20
D	25, 15

Question 42 Explanation:

Go from the choices
Choice (C) 30 and 20 are in the ratio of 6 : 4
Five years ago their ages would be 25 and 15 which are in the ratio 5 : 3
Hence choice (C) is the right answer

Question 43

A, B and C enter into a partnership by investing Rs.3600, Rs.4400 and Rs.2800. A is a working partner and gets a fourth of the profit for his services and the remaining profit is divided amongst the three in the rate of their investments. What is the amount of profit that B gets if A gets a total of Rs. 8000?

A	4888.88
B	9333.33
C	4000
D	3666.66

Question 43 Explanation:

Let x be the profit
Their investment ratio = 3600 : 4400 : 2800 = 9 : 11 : 7
A’s profit of Rs. 8000 = 14 × x + 13 × 34 × x =12 × x x = Rs 16,000 Therefore B’s profit = 1127 × 34 × 16,000 = Rs 4888.88

Question 44

A noodles merchant buys two varieties of noodles the price of the first being twice that of the second. He sells the mixture at Rs 17.50 per kilogram thereby making a profit of 25. If the ratio of the amounts of the first noodles and the second noodles in the mixture is 2:3, then the respective costs of each noodles are?

A	Rs 20, Rs 10
B	Rs 24, Rs 12
C	Rs 16, Rs 8
D	Rs 26, Rs 13

Question 44 Explanation:

Let the price of one noodles = k
⇒ the price of other noodle = k2
Price of 1 kg = 2k5 + 35 × k2 = 7k10
But CP = 17.50 × 100125 = 14
⇒ 7k10 = 14
⇒ k = 20
So price of the noodles’s are 20 and 10

Question 45

In a house, there are dogs, cats and parrot in the ratio 3:7:5. If the number of cats was more than the number of dogs by a multiple of both 9 and 7, what is the minimum of pets in the house?

A	945
B	630
C	252
D	238

Question 45 Explanation:

If three kinds of pets are taken be 3k,7k and 5k respectively,
Then 7k − 3k = 63p (where p is any positive integer)
As the number is a multiple of both 9 and 7, it has to be multiple of 63
⇒ k = 63p4
Minimum value of p for which k is a natural number is 4
Thus, k=63
Hence, the number of pets = 15k = 945

Question 46

There are N numbers of apples in the house, in which four people are lived. If the first men woke up and divided the apples into 5 equal piles and found one extra apple. He took one of those piles along with the extra apple and hid them. He then gathered the 4 remaining piles into a big pile, woke up the second person and went to sleep. Each of the other 3 persons did the same one by one i.e. divided the big pile into 5 equal piles and found one extra apple. Each hid one of the piles along with the extra apple and gathered the remaining 4 piles into a big pile. If N>1000, what could be the least value of N?

A	1249
B	1023
C	1202
D	1246

Question 46 Explanation:

Suppose N=5x + 1
A took (x + 1) biscuit
Now 4x is of the form 5y + 1 then x must be in the form 5z + 4
⇒4(5z + 4) = 5y + 1
⇒ y = 4z + 3 and x = 5z + 4
The ratio of number of biscuits that A and B took is [(5z + 4) +1] : [(4z + 3) + 1] = 5 : 4
So, we can say that any two successive persons A, B, C and D take coins in the ratio of 5 : 4
Let the number of biscuits that A, B, C and D took be a, b, c and d respectively
a : b = b : c = c : d = 5 : 4
a : b : c : d = 125 : 100 : 80 : 64
⇒ a = 125k
⇒ x = 125k − 1 and N = 5x + 1 = 625k − 4
As, N>1000, the least value of N is when k = 2
N = 1246

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Problems on trains

MS — Thu, 17 Aug 2017 06:47:01 +0000

Explanation / Important formulas

km/hr to m/s conversion: a km/hr = (a X 5/18) m/s
m/s to km/hr conversion: a km/hr = (a X 18/5) km/hr

Time taken by a train of length metres to pass a pole or standing man or a signal post is equal to the time taken by the train to cover l metres
Time taken by a train of length metres to pass a stationery object of length b metres is the time taken by the train to cover (l + b) metres
Suppose two trains or two objects bodies are moving in the same direction at u m/s and v m/s, where u > v, then their relative speed is = (u – v) m/s
Suppose two trains or two objects bodies are moving in opposite directions at u m/s and v m/s, then their relative speed is = (u + v) m/s
If two trains of length metres and b metres are moving in opposite directions at u m/s and v m/s, then: their relative speed is = (u – v) m/s
Suppose two trains or two objects bodies are moving in opposite directions at u m/s and v m/s, then their relative speed is = (u + v) m/s
If two trains of length metres and b metres are moving in opposite directions at u m/s and v m/s, then: The time taken by the trains to cross each other = (a+b) / (u+v) sec
The time taken by the faster train to cross the slower train = (a+b) / (u-v) sec

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Problems on trains – Question and Answers

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Question 1

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?

A	48 km/hr
B	54 km/hr
C	66 km/hr
D	82 km/hr

Question 1 Explanation:

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

Question 2

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?

A	130
B	360
C	500
D	540

Question 2 Explanation:

Speed = (78 * 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

Question 3

Two trains of equal length are running on parallel lines in the same direction at 46 km/hr and 36 km/hr. The faster train passes the slower train in 36 seconds. The length of each train is?

A	50 m
B	72 m
C	80 m
D	82 m

Question 3 Explanation:

Let the length of each train be x metres
Then, distance covered = 2xmetres
Relative speed = (46 – 36) km/hr
= (10 * 5 / 18) m/sec
= (25 / 9) m/sec
Therefore,( 2x / 36) = (25 / 9)
2x = 100 x = 50

Question 4

A train 125 m long passes a man, running at 5 km/hr in the same direction in which the train is going, in 10 seconds. The speed of the train is?

A	45 km/hr
B	50 km/hr
C	54 km/hr
D	55 km/hr

Question 4 Explanation:

Speed of the train relative to man = (125 / 10)
= (25/2) m/sec
= (25 / 2 * 18 / 5) km/hr
= 45 km/hr
Let the speed of the train be x km/hr
Then, relative speed = (x – 5) km/hr
x – 5 = 45 => x = 50 km/hr

Question 5

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?

A	36
B	45
C	48
D	49

Question 5 Explanation:

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

Question 6

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?

A	230 m
B	240 m
C	260 m
D	270 m

Question 6 Explanation:

Speed = (72 * 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

Question 7

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?

A	9 a.m
B	10 a.m
C	10.30 a.m
D	11 a.m

Question 7 Explanation:

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

Question 8

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?

A	10
B	18
C	36
D	72

Question 8 Explanation:

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

Question 9

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?

A	40 sec
B	42 sec
C	45 sec
D	48 sec

Question 9 Explanation:

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

Question 10

The length of the bridge, which a train 130 metres long and travelling at 45 km/hr can cross in 30 seconds, is?

A	200 m
B	225 m
C	245 m
D	250 m

Question 10 Explanation:

Speed = (45 x 5 / 18) m/sec = (25 / 2) m/sec
Time = 30 sec
Let the length of bidge be x metres
Then, (130 + x / 30) = 25 / 2
2(130 + x) = 750 x = 245 m

Question 11

The length of the bridge, which a train 130 metres long and travelling at 45 km/hr can cross in 30 seconds, is?

A	200 m
B	225 m
C	245 m
D	250 m

Question 11 Explanation:

Speed = (45 x 5 / 18) m/sec = (25 / 2) m/sec
Time = 30 sec
Let the length of bidge be x metres
Then, (130 + x / 30) = 25 / 2
2(130 + x) = 750 x = 245 m

Question 12

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

A	65 sec
B	89 sec
C	100 sec
D	150 sec

Question 12 Explanation:

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

Question 13

A jogger running at 9 kmph alongside a railway track in 240 metres ahead of the engine of a 120 metres long train running at 45 kmph in the same direction. In how much time will the train pass the jogger?

A	39 sec
B	18 sec
C	36 sec
D	72 sec

Question 13 Explanation:

Speed of train relative to jogger = (45 – 9) km/hr = 36 km/hr
= (36 x (5 / 18)) m/sec
= 10 m/sec
Distance to be covered = (240 + 120) m = 360 m
Therefore, time taken = 360 /10 sec = 36 sec

Question 14

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?

A	25
B	30
C	40
D	45

Question 14 Explanation:

Speed of the train relative to man = (63 – 3) km/hr
= 60 km/hr
= (60 * 5 / 18) m/sec
= (50 / 3) m/sec
Therefore, time taken to pass the man = (500 * 3/50 ) sec
= 30 sec

Question 15

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

A	30 km/hr
B	45 km/hr
C	60 km/hr
D	75 km/hr

Question 15 Explanation:

Let the speed of the slower train be xm/sec
Then, speed of the faster train = 2xm/sec
Relative speed = (x + 2x) m/sec = 3xm/sec
Hence (100 + 100) / 8 = 3x
24x = 200
x = 25 / 3
So, speed of the faster train = 50/3 m/sec
= (50 / 3) * (18 / 5) km/hr
= 60 km/hr

Question 16

A train passes a station platform in 36 seconds and a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr, what is the length of the platform?

A	120 m
B	240 m
C	300 m
D	None of these

Question 16 Explanation:

Speed = (54 x 5/18)m/sec = 15 m/ sec Length of the train = (15 x 20)m = 300 m
Let the length of the platform be x metres
Then, (x + 300) / 36 = 15
x + 300 = 540 x = 240 m

Question 17

Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 23 seconds. The ratio of their speeds is?

A	1 : 3
B	3 : 2
C	3 : 4
D	None of these

Question 17 Explanation:

Let the speeds of the two trains be x m/sec and y m/sec respectively
Then, length of the first train = 27x metres, and length of the second train = 17y metres
Therefore, (27x + 17y) / (x + y) = 23
27x + 17y = 23x + 23y 4x = 6y
x/y = 3/2

Question 18

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?

A	23 m
B	23 2/9 m
C	27 m
D	27 7/9 m

Question 18 Explanation:

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

Question 19

How many seconds will a 500 meter long train moving with a speed of 63 km/hr take to cross a man walking with a speed of 3 km/hr in the direction of the train?

A	42
B	50
C	30
D	28

Question 19 Explanation:

Distance = 500 m
Speed of the train relative to man = ( 63 – 3 ) km/hr = 60 km/hr
= (60 x 5/18) m/s = 50/3 m/s
Time taken to pass the man = distance/speed = (500 x 3/50) sec = 30 sec

Question 20

Two trains leaving from two station 50 miles away from each other with costant speed of 60 miles per hour, approaches towards each other on diffrent tracks. if length of each train is 1/6 mile. When they meet how much time they need to pass each other totally?

A	1/6 minutes
B	1/4 minutes
C	1/8 minutes
D	1/2 minutes

Question 20 Explanation:

As we know ,Both are approaching each other relative speed = 60+60 = 120m/h
Distance = 50m(Given)
Time = 50/120 = 5/12 hrs = 25 minutes to meet Now,
Relative length of train = 1/6 + 1/6 = 1/3m
Relative speed = 120m/h
Rime to cross both the trains = ( 1/3 )/120 = 1/360 Hours
Convert it into minutes = 1/6 minutes

Question 21

The time a passenger train takes to cross another freight train is twice when the passenger train crosses the freight train running in opposite directions. What is the ratio of their speeds?

A	2:8
B	3:1
C	3:2
D	2:6

Question 21 Explanation:

Speed of freight train = x
Speed of passenger train = y
Sum of their length = s
So time = s/(x – y)when both are in same direction
Time = s/(x + y) when opposite direction
From question s/(x – y) = 2. s/(x + y) so solving this we get, x/y = 3/1
So x : y = 3 : 1

Question 22

A	9 a.m
B	10 a.m
C	10.30 a.m
D	11 a.m

Question 22 Explanation:

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
20x + 25(x – 1) = 110
45x = 135
x = 3
So, they meet at 10 a.m

Question 23

Two trains move in the same direction at speeds 50 kmph and 32 kmph respectively. A man in the slower train observes that 15 seconds elapse before the faster train completely passes by him. What is the length of faster train?

A	75m
B	80m
C	70m
D	86m

Question 23 Explanation:

Relative speed = (50-32)
t = 15/(60 * 60) = time taken in hours
length = t*rs = 75m

Question 24

Two trains of length 400 meters and 300 meters starts at 6.00 pm and 7.00 pm and runs at the speed of 80 km/hr and 90 km/hr respectively. If they are running on parallel tracks in the same direction then the time taken by them to cross each other is?

A	219 seconds
B	252 seconds
C	198 seconds
D	202 seconds

Question 24 Explanation:

Length of the two trains are 400 meters and 300 meters and their respective speeds are 80 km/hr and 90 km/hr
Since the trains are moving in the same direction,the time required to cross each other = (a + b)/(u – v) sec
Here, a = 300 m, b = 400 m and u = 90 km/hr, v = 80 km/hr
a + b = 300 + 400 = 700 m
And, u – v = (90 – 80)km/hr = 10 km/hr
Converting the unit of speed into m/Sec:
10 km/hr = 10 x 5/18 m/sec = 50/18 m/s
Now, the required time = 700 x 18/50 sec = 252 sec

Question 25

Two trains of length 0.55 km and 0.45 km run at the rate of 30 km/hr and 45 km/hr. If they are travelling on parallel tracks in the opposite direction then how long will the slower train takes to cross the faster train?

A	48 seconds
B	52 seconds
C	39 seconds
D	98 seconds

Question 25 Explanation:

Length of the two trains are 0.55 km and 0.45 km and their respective speeds are 30 km/hr and 45 km/hr
Since the trains are moving in the opposite direction, the time required to cross each other = (a + b)/(u + v) sec
Here, a = 0.55 km = 550 meters, b = 0.45 km = 450 meters and u = 30 km/hr, v = 45 km/hr
a + b = 550 + 450 = 1000 m
And, u + v = (30 + 45)km/hr = 75 km/hr
Converting the unit of speed into m/Sec:
75 km/hr = 75 x 5/18 m/sec = 375/18 m/s
Now, the required time = 1000 x 18/375 sec = 48 sec

Question 26

The speed ratio of two trains each of length 250 meters is same and are running on parallel tracks in the opposite direction. If they take 25 seconds to cross each other then their speed is?

A	10 km/hr
B	36 km/hr
C	48 km/hr
D	19 km/hr

Question 26 Explanation:

Given that, the speed and length of two trains are equal
And, they are moving in the opposite direction, they take 25 seconds to cross each other
Let the required speed of the trains be X m/sec
Now, from the formula (a + b)/(u + v), we have a = b = 250 m and u = v = X m/sec
Then, a + b = 250 + 250 = 500 m and u + v = 2X m/sec, 25 = 500/2X sec
X = 10 m/sec = 10 x 18/5 km/hr = 36 km/hr
Hence, the speed of two trains is 36 km/hr

Question 27

There is a 200 miles long tunnel. One train enters the tunnel at a speed of 200mph at the same time another train enters the tunnel in opposite direction at a speed of 1000 mph. A bee travels at a speed of 1500mph enters the tunnel goes to and fro until it reaches a train. What is the distance covered by the bee when the two trains collide?

A	280 miles
B	350 miles
C	250 miles
D	140 miles

Question 27 Explanation:

Relative speed of two trains = 1000+200 = 1200mph
Time taken to collide = 200/1200 = 1/6 hours
The bee travels for 1/6 hours. Hence distance covered = 1500 * 1/6 = 250 miles

Question 28

A train crosses a platform of 120m in 15sec, same train crosses another platform of length 180m in 18sec. then find the length of the train?

A	175m
B	180m
C	185m
D	170m

Question 28 Explanation:

Length of the train be ‘X’
X + 120/15 = X + 180/18
6X + 720 = 5X + 900
X = 180m

Question 29

A 1200m long train crosses a tree in 120sec, how much time will I take to pass a platform 700m long?

A	180 sec
B	190 sec
C	170 sec
D	175 sec

Question 29 Explanation:

L = S*T
S= 1200/120
S= 10M/Sec
Total length(D)= 1900m
T = D/S
T = 1900/10
T = 190Sec

Question 30

The length of the bridge, which a train 130 metres long and travelling at 45 km/hr can cross in 30 seconds, is?

A	150 m
B	160 m
C	240 m
D	245 m

Question 30 Explanation:

Speed = 45 x 5/18 m/sec = 25/2 m/sec
Time = 30 sec
Let the length of bridge be x metres
Then, (130 + x)/30 = 25/2
2(130 + x) = 750
x = 245 m

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Pipes and Cisterns

MS — Wed, 16 Aug 2017 05:45:58 +0000

[vc_row][/vc_row][vc_column][/vc_column][vc_toggle title=”Explanation / Important formulas:” el_id=”1481454282462-60885454-0285″]

Inlet: A pipe connected with a tank or a cistern or a reservoir that fills it, is known as an inlet.

Outlet: A pipe connected with a tank or cistern or reservoir emptying it, is known as an outlet.

If a pipe can fill a tank in x hours, then:part filled in 1 hour = 1 / x
If a pipe can empty a tank in y hours, then:part emptied in 1 hour = 1
If a pipe can fill a tank in y/x hours and another pipe can empty the full tank in y hours (where y > x), then on opening both the pipes, then the net part filled in 1 hour = (1/x – 1/y)
If a pipe can fill a tank in x hours and another pipe can empty the full tank in y hours (where x > y), then on opening both the pipes, then the net part emptied in 1 hour = (1/y – 1/x)

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Pipes and cisterns - Question and Answers

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Question 1

Pipe A and Pipe B can be fill a tank in 20 minutes and 30 minutes respectively. Pipe C can empty the tank in 60 minutes. The tank is initially empty. Both the pipes A and B are opened. Pipe C is opened after 6 minutes. How much time does it take to fill the tank?

A	12minutes
B	13.5 minutes
C	15minutes
D	14.5 minutes

Question 1 Explanation:

1/30th of the tank is filled by pipe B in 1 minutes
1/60th of the tank is emptied by pipe C in 1 minutes
Part of tank filled in the first 6 minutes = 6*[(1/20)+(1/30)] = ½
When all pipes are opened, part of tank filled in 1 minutes = (1/12)-(1/60) = 1/15
Time required to fill the remaining ½ of the tank = (1/2) / (1/15) = 7.5
Total time = 6+7.5 = 13.5 minutes

Question 2

Pipe A and Pipe B can fill a tank in 20 minutes and 10 minutes respectively. Pipe C can empty the tank in 15 minutes. If the three pipes are opened simultaneously, find the time taken to fill the tank/

A	12
B	15
C	20
D	18

Question 2 Explanation:

When all pipes are opened simultaneously, part of tank filled in 1minutes
= (1/20) + (1/10) – (1/15) = 1/12
Time taken to fill the tank = 12 minutes

Question 3

Pipes P and Q can empty a tank in 1 hr and 1.5 hrs respectively. Pipes R and S can fill a tank in 2 hrs and 2.5 hrs respectively. The tank is filled to its capacity of 15 litres and all four pipes are opened simultaneously. How much water remains in the tank after an hour?

A	7 liters
B	5 liters
C	3.5 liters
D	2.5 liters

Question 3 Explanation:

Part of tank emptied by P in an hour = 1
Part of tank emptied by Q in an hour = 1/1.5 = 2/3
Part of tank filled by R in an hour = ½
Part of tank filled by S in an hour = 1/ 2.5 = 2/5
Part of tank emptied in an hour = (1+ (2/3)) - (1/2 + (2/5)) = 23/30
Water remaining in the tank = 15*7/30 = 3.5 liters

Question 4

Three pipes A,B, C are attached to a tank. A and B can fill it in 20 and 30 minutes respectively while C can empty it in 15 minutes. If A,B and C are kept open successively for 1 minutes each, how soon will the tank be filled?

A	180 minutes
B	167 minutes
C	200 minutes
D	178 minutes

Question 4 Explanation:

Part of the tank filled in 3 minutes = (1/20) + (1/30) – (1/15) = 1/60
Part filled in 165 minutes = (165/3) * (1/60) = 55/60
Part filled in 167 minutes = (55/60) + (1/20) + (1/30) = 60/60 = 1
The tank is filled in 167 minutes

Question 5

Pipe A and B can fill a tank in 1 hour and1 ¼ hours respectively. Both pipes are opened in the beginning and after some time, pipe B is closed. The tank is filled in 40 minutes. Pipe B was closed after?

A	30 minutes
B	45 minutes
C	40 minutes
D	25 minutes

Question 5 Explanation:

Part of tank filled by A in a minute = 1/60
Part of tank filled by B in a minutes =1/75
Let pipe B be closed after x minutes
Part of tank filled in x minutes= x*[(1/60)+(1/75)] = 9x / 300 Part of tank filled in (40-x) minutes = [(40-x)*(1/60)] = (40-x) / 60
(9x / 300) + ((40 - x) / 60) = 1 4x = 100
X= 25

Question 6

Two pipes A and B can fill a tank in 20 min, when opened simultaneously. If pipe A alone takes 60 minutes to fill the tank, how much time will pipe B alone take to fill the tank?

A	60 minutes
B	40 minutes
C	30 minutes
D	20 minutes

Question 6 Explanation:

Let B take x minutes to fill the tank
Part of the tank filled by A and B = (1/60) + (1/x) = (1/20)
Solving this, x= 30

Question 7

A pipe can fill a tank in an hour. Because of a leak, it took 1 hour and 10 minutes to fill the tank. Find the time taken by the leak to drain all the water in the tank?

A	10 minutes
B	7/6 hours
C	6/7 hours
D	7hours

Question 7 Explanation:

Let the leak take x to drain the tank 1hr 10 min = 1+(1/6) = 7/6 hrs
1- (1/x) = 1/ (7/6) = 6/7
Part of tank leaked in an hour = 1-6/7 = 1/7
The leak takes7 hours to drain all the water

Question 8

Two taps A and B can fill a bathtub in 10 minutes and 15 minutes respectively. If A was open for the first two minutes, B for the next two minutes and so on, the bathtub is filled in ?

A	6 minutes
B	12 minutes
C	12.5 minutes
D	24 minutes

Question 8 Explanation:

Part of the tub filled in the first 2 minutes = 2*1/10 =1/5 Part of the tub filled in the next 2 minutes = 2*1/15 = 2/15
Part of the filled in 4 minutes = (1/5) +(2/15) =1/3
Time required to fill the tub = 3*4 = 12 minutes

Question 9

A tank has three inlets A,B and C. C takes twice the time taken by A to fill the tank and B takes half the time taken by A to fill the tank. If they can fill the tank together in 4 minutes, find the time taken by B to fill the tank?

A	7 minutes
B	14 minutes
C	3.5 minutes
D	28 minutes

Question 9 Explanation:

Let A take x minutes to fill the tank (1/x) + (1/ (x/2)) + (1/2x) = 1/4
(7/2x) = ¼
x=14
x /2 = 7

Question 10

An inlet pipe fills a tank at the rate of 5 liters of water a minutes. An outlet connected to the tank can empty a full tank i 5 hours. Both the piped are opened together for 30 minutes and then , the outlet is closed. It took another 36 minutes to fill the tank. Find the capacity of the tank

A	480 liters
B	360 liters
C	300 litres
D	240 liters

Question 10 Explanation:

Part of the tank filled by the inlet in a minute = (1/x)
Part of the tank emptied by the outlet in a minute = 1 / (5*60) = 1/300
Part of the tank filled in the first 30 minutes = 30*[(1/x) - (1/300)]
= (300 - x) / 10x Part of the tank filled in the next 36minutes = 36 * (1/x)
1 – [(300 - x) / 10 x] = 36 / x
Solving, x= 60
Capacity of tank = 60 * 5 = 300 liters

Question 11

Pipe A fills a tank at the rate of 5 liters every hour. Pipe B fills another tank at a rate of 1 litre in the first hour, 2litres in the second hour, 3 in the third and so on. After how many hours will the volume of water contained in the two tanks be equal?

A	10hours
B	5 hours
C	9 hours
D	6 hours

Question 11 Explanation:

Let the water contained in the tanks be equal after x hours
5x= 1+2+3+ .......+x
5x = x(x+1) / 2
X= 9 hours

Question 12

Pipe A can fill a tank thrice as faster as pipe B. If pipe A alone can fill the tank i n 20 minutes, how long does it take to fill tank if both the pipes are opened?

A	15 minutes
B	30 minutes
C	12 minutes
D	10 minutes

Question 12 Explanation:

Part of tank filled by pipe A in 1 minute= 1/20
Part of tank filled by B in 1 minute = (1/20) / 3 = 1/60
Part of tank filled by them together in 1 min = (1/20) + (1/60) = 1/15
They take 15 min to fill the tank

Question 13

A tap can fill a bucket in 10 min. Due to a leak in the bucket, it takes 12min to fill. After the tap is closed, the leak empties the bucket in?

A	20 minutes
B	1 hour
C	30 minutes
D	60 minutes

Question 13 Explanation:

Let the leak take x minutes to empty the bucket
Part of bucket filled by the tap in 1 minute =1/10
(1/10) – (1/x) = (1/12)
X = 60 minutes

Question 14

An empty drum can be filled with 18 buckets of water, if the capacity of each bucket is 16 liters. How many buckets will be need to fill the drum if the capacity of each bucket is 12 liters?

A	28
B	20
C	15
D	24

Question 14 Explanation:

Capacity of the drum = 18 * 16 = 288 liters
No. of buckets needed = 288 / 12 = 24

Question 15

Pipe A can fill a tank in 4 hours and pipe B that can fill a tank in 2 hours. Both are opened when the tank is empty. Pipe A is closed 30 minutes before the tank overflows. How long did it take for the tank to overflow?

A	100 minutes
B	75 minutes
C	90 minutes
D	80 minutes

Question 15 Explanation:

Part of tank filled by both pipes in 1 hour = (1/4) + (1/2) = ¾
Pipe A is closed before 30 minutes of overflow
Only pipe B was opened
Part of tank filled by pipe B in 30 minutes = ¼
The remaining 3/4th of the tank was filled by pipes A and B
To fill 3/4th of the tank together, 1 hour is taken
Time for overflow = 1 hours + 30 minutes = 90 minutes

Question 16

Three pipes can fill a tank in 6, 10 and 15 hours respectively. If they are opened together, how long does it take to fill the tank?

A	3 hours
B	30 hours
C	10 hours
D	33 hours

Question 16 Explanation:

Part of tank filled by the 3 pipes in 1 hour
= (1/6) +(1/10) + (1/15)
= 10/30
= 1/3
They take 3 hours

Question 17

There are 10 pipes connected to a tank, some are inlet pipes and the others are outlets. Each of the inlet pipes can fill the tank in 6 hours and each of the outlet pipes can drain the entire tank in 8 hours. If all the pipes are opened when the tank is full, the tank is drained in half a day. How many of them are outlet pipes?

A	4
B	5
C	6
D	7

Question 17 Explanation:

Let there be x inlet pipes and (10-x) outlet pipes Part of tank filled by an inlet pipe in an hour = 1/6 Part of tank drained by an outlet pipe in an hour =1/8 If all the pipes are opened, the tank is drained in half a day (12 hours)
Part of tank drained in an hour when all are open = 1/12
Part of tank drained by (10-x) outlet pipes – Part of tank filled by x inlet pipes = 1/12
[(10-x) / 8] – [x/6] = 1/12
Solving,
3(10-x)-4x = 2
30 – 7x = 2
7x=28
X=4
There are 4 inlet pipes and 6 outlet pipes

Question 18

Water flowing from a tap fills a 5 liters can inn 2 minutes. How many such cans can be filled in 1680 seconds?

A	70 cans
B	14 cans
C	28 cans
D	84 cans

Question 18 Explanation:

1680 seconds = 28 minutes
1 can is filled in 2 minutes
14 cans can be filled in 28 minutes

Question 19

Pipes A and B can fill a tank in 8 hours and 12 hours respectively. Both of them were opened simultaneously when the tank was empty. It took 6 hours to fill the tank due to a leak in the bottom. After the pipes were closed, the leak will drain the full tank in

A	12 hours
B	18 hours
C	1day
D	2 days

Question 19 Explanation:

Let the leak take x hours to empty the tank Part of tank emptied by the leak in an hour = 1/x Part of tank filled in an hour = (1/8) + (1/12) – (1-x) =(5/24) – 1/x = 1/6
1/x = (5/24) –(1/6)
= 1/24
X= 24 hours

Question 20

Pipes A fills a cistern twice as faster as pipe B. Pipe A was opened when the tank was empty and after an hour, pipe B was also Opened. Find the time taken to fill the cistern if pipe B alone can fill it in 8 hours?

A	2 hours
B	4 hours
C	4.5 hours
D	3 hours

Question 20 Explanation:

Part of tank filled by ppe B in an hour =1/8
Part of tank filled by pipe A in an hour = ¼
In the first hour, 1/4th of the tank is filled.
The remaining 3/4th of the tank is filled by A and B together
Part of tank filled by A and B in an hour = (1/4) +(1/8) = 3/8
Time taken to fill ¾ of the tank = (3/4) / (3/8) = 2 hours
The cistern is filled in 1+ 2 = 3 hours

Question 21

Pipe X can fill a tank in 5 hour. If pipe Y is also opened, the tank is filled in 3 hours. Find the time taken by pipe Y alone to fill the tank?

A	4.5 hours
B	6 hours
C	7.5 hours
D	) 7.5 hours

Question 21 Explanation:

(1/x) + (1/y) = (1/3)
(1/y) = (1/3) – (1/5)
= 2/ 15
Y takes 15/2 = 7.5 hours to fill the tank

Question 22

Water flows from a tap at the rate of 1.2 liters per minutes. If the tap is opened for 3 hours, find the amount of water collected in the tank?

A	360 liters
B	216 liters
C	180 liters
D	144 liters

Question 22 Explanation:

Amount of water collected in 180 mins
= 180 * 1.2
= 216 liters

Question 23

Pipes P and Q can fill a tank in 40 minutes and 1 hour respectively. If they are opened together, find the time taken to fill the tank?

A	1 hr 40 min
B	36 mins
C	20 mins
D	24 mins

Question 23 Explanation:

Part of tank filled P and Q in a minute =(1/40) + (1/60)
=5/120
= 1/24
It takes 24 minutes to fill the tank

Question 24

Three pipes A,B and C take 3 hours to fill a tank. If pipes A and B alone take 6 hours and 9 hours to fill he tank, find the time taken by pipe C lone to fill the tank?

A	12 hours
B	145 hours
C	18 hours
D	24 hours

Question 24 Explanation:

(1/c) = (1/3) – (1/6) – (1/9) = (1/18)

Question 25

Two pipes A and B can fill a tank in half an hour and 1(1/2)hour respectively. An outlet at the bottom of the tank can empty a full tank in 1 hour. If all the three pipes are opened, find the time taken to fill the tank?

A	1 hour
B	36 min
C	45 min
D	54 min

Question 25 Explanation:

Part of tank filled by pipe A in 1 minute = 1/30
Part of tank filled by pipe B in 1 minute = 1/90
Part of tank emptied by the outlet in 1 minute = 1/60
Part of tank filled when all the pipes are opened = part of tank filled by inlet-part of tank emptied by outlet = (1/30) + (1/90) – (1/60)
= (6+2-3) / 180
= 1/ 36
The tank is filled in 36 mins

Question 26

An outlet pipe can empty (5/7) of the cistern in half an hour. Find the time taken to empty the full cistern.

A	70mins
B	(7/5)hrs
C	21(3/7)mins
D	42 mins

Question 26 Explanation:

(5/7) of the cistern is emptied in 30 minutes
Full cistern is emptied in 30*7/5 = 42 minutes

Question 27

Tap A can fill a cistern in 1 hour and Tap B can fill it in 30 minutes. Tap A is opened first and tap B is also opened after 15 minutes. Find the time taken to fill the cistern?

A	35 mins
B	30 mins
C	25 mins
D	20 mins

Question 27 Explanation:

Part of cistern filled by A in 1minute = (1/60)
Part of cistern filled by A in 15 mins = 15/60 = ¼
Part of cistern filled by A and B in 1 minute = (1/60)+(1/30) = 1/20
¾ th of the cistern is filled by A and B
Time taken to fill ¾ of the cistern = (3/4) /(1/20) = 15 minutes
The cistern is filled in 30 minutes

Question 28

A tank is filled by two taps in half an hour. If the first tap alone takes 2 hours to fill the tank, find the time taken by the second tap alone to fill the tank?

A	40 mins
B	1 hour
C	75 mins
D	90 mins

Question 28 Explanation:

Part of tank filled by the two taps in 1 minute = 1/30
Part of tank filled by taps A in 1 minute = 1/120
(1/A) + (1/B) = (1/30)
(1/B) = (1/30) –(1/120) = (1/40)
B alone takes 40 mins to fill the tank

Question 29

A tank can be filled by 2 pipes X,Y and Z, when turned on separately, in 33 minutes, 11 minutes and 22 minutes respectively. If all are turned on for 2 mins and then Y and Z are turned off, find the time taken by X alone to fill the cistern.

A	22 minutes
B	15 minutes
C	11 minutes
D	6 minutes

Question 29 Explanation:

Part of tank filled X, Y and Z in 1 minute = (1/33) + (1/11) + (1/22) = 1/6
Part of tank filled in the first 2 minutes = 2*1/6 = 1/3
2/3 of the tank is filled by X alone
Time taken by X to fill 2/3 of the tank = (2/3) / (1/33) = 22 minutes

Question 30

Two pipes A and B can fill a tank in 36 hours and 45 hour respectively. If both the pipes are opened simultaneously, how much time will be taken to fill the tank?

A	20 hours
B	18 hours
C	24 hours
D	22.5 hours

Question 30 Explanation:

Part of tank filled by A in 1 hour = 1/36
Part of tank filled by B in 1 hour = 1/45
Part of tank filled by both in 1 hour = (1/36) + (1/45) = 81 / (36*45)
= 1/20
They can fill the tank in 20 hours

Question 31

15 buckets of water fill a tank when the capacity of each bucket in 12 liters. How many buckets will be needed to fill the tank in the same time, if the capacity of each bucket is 18 liters?

A	18
B	10
C	15
D	8

Question 31 Explanation:

No. of buckets needed = 15 * 12/18 = 10

Question 32

Two taps A and B can fill a tank in 5 hours and 20 hours respectively. If both the taps are open then due to a leakage, it took 30 minutes more to fill the tank. If the tank is full, how long will it take for the leakage alone to empty the tank?

A	18 hrs
B	36 hrs
C	35 hrs
D	9 hrs

Question 32 Explanation:

Part of tank filled by A and B in 1 hour = (1/5) + (1/20) = ¼
They take 4 hours to fill the tank. But due to the leak, they take 4.5 hours to fill the tank
Part of tank emptied by leak in 1 hours = ¼ - 1/ 4.5 = ¼ - 2/9
= 1/36
The leak can empty the tank in 36 hours

Question 33

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

A	192 mins
B	144 mins
C	108 mins
D	81 mins

Question 33 Explanation:

Let the tome taken by the faster pipe to fill tank be x mins and the time taken by the slower pipe be 3xmins
1/x + 1/3x = 1/36
4/3x = 1/36
X= 48
3x = 144

Question 34

Pipe 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, in how many hours will the tank be filled?

A	30/17
B	30/11
C	9/2
D	60/17

Question 34 Explanation:

(1/5) + (1/6) – (1/12)
= 17/60
The tank will be filled in (60/17 ) hours

Question 35

A tap can fill a tank in 6 hour. After half the tank is filled, three more similar taps are opened. What is the total time a taken to fill the tank completely?

A	4 hrs
B	3hrs 45 mins
C	4 hrs 15 mins
D	3 hrs 15 mins

Question 35 Explanation:

Half the tank is filled in 3 hours Now, 4 taps are filling the tank
Time taken by 1 tap to fill half the tank = 3
Time taken by 4 taps to fill half the tank = ¾
Total time taken = 3 hrs 45mins

Question 36

A cistern can be filled by a atap 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 will the cistern get filled?

A	6.5 hrs
B	4.5 hrs
C	5 hrs
D	7.2 hrs

Question 36 Explanation:

(1/4) – (1/9) = 5/36
The cistern gets filled in 36/5 = 7.2 hours

Question 37

A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time 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 hours slower than the third pipe. The time required by the first pipe alone to fill the tank is

A	10 hrs
B	15 hrs
C	30 hrs
D	6 hrs

Question 37 Explanation:

1/A + 1/B = 1/C
B = A -5
B = C + 4
C =A – 9
1/A + 1/(A-5) = 1/(A-9)
1/(A - 9) – 1/(A-5) = 1/A
Substitute the options given in the place of A
If A = 10,
1 – 1/5 not equal to 1/10
If A = 15,
1/6 – 1/10 = 1/15
So, A = 15

Question 38

Three pipes A,B and C can fill a tank from empty to fill in 30 min, 20 mins and 10 min 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 solution R in the liquid in the tank after 3 minutes?

A	1/11
B	5/11
C	6/11
D	8/11

Question 38 Explanation:

Part of tank filled in 3 minutes
= 3 * [(1/30)+(1/20)+(1/10)]
= 11/20
Part of tank filled by C in 3 minutes
= 3/10
Ratio = 3/10 : 11/20
= 6: 11

Question 39

Two taps can fill a cistern in 30 and 40 minutes respectively. If both the taps are opened simultaneously then the approximate time taken to fill the cistern is

A	17 minutes
B	12 minutes
C	19 minutes
D	21 minutes

Question 39 Explanation:

Let the time taken to fill the cistern be x minutes
1/x = 1/30 = 1/40 = 7/120
X= 120/7 = 17. 14 minutes

Question 40

A pipe P can fill a tank in 16 mins and the other pipe Q can empty the whole tank in 32 mins. If both P and Q are opened simultaneously, then the time taken to fill the tank is

A	16 mins
B	32 mins
C	48 mins
D	40 mins

Question 40 Explanation:

Let the time taken to fill the tank be x minutes
1/x = 1/P – 1/Q
= 1/16 – 1/32
= 1/32
X= 32 mins

Question 41

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

A	8 hrs
B	6 hrs
C	2 hrs
D	1 hrs

Question 41 Explanation:

1/A + 1/B – ¼
B= A+6
We can check the options to find the answer
If A=8,
1/8 + 1/1 4 not equal to ¼
If A = 6,
1/6 + 1/12 = ¼
So, A = 6

Question 42

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?

A	3.5
B	2
C	3
D	2.5

Question 42 Explanation:

Part of tank filled in 1 hour when all the pipes are open
= (1/A) + (1/B) + (1/C)
=(1/5) + (1/10) + (1/30)
= 1/3
The tank will be filled in 3 hours

Question 43

A Pump can fill a tank with water in 2 hours because of a leak, it took 7/3 hours to fill the tank. The leak can drain out the tank in

A	8 hrs
B	14 hrs
C	7 hrs
D	13/3

Question 43 Explanation:

Let the leak take x hours to drain out the tank
½ - 1/x = 3/7
1/x = ½ - 3/7 = 1/14
X= 14

Question 44

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

A	6mins to fill
B	9 mins to fill
C	6 mins to empty
D	9 mins to empty

Question 44 Explanation:

Part of tank filled by A in 1minute = 1/10
Part of tank emptied by B in 1 minute = 1/6
The rate at which pipe B empties is higher than the rate at which the pipe A fills the tank
Part of tank emptied in 1minute when A and B are opened = 1/6 – 1/10
= 1/ 15
Time taken to empty 2/5th of the tank
= 2/5 / 1/15
= 6 minutes

Question 45

A tank can be filled by a tap in 20 mins and by another tap in 60 mins. Both the taps are kept open for 10 mins and then the first tap is shut off. After this, the tank will be completely filled in?

A	10 mins
B	15 mins
C	12 mins
D	20 mins

Question 45 Explanation:

Part of tank filled by both the taps in 1 minute = (1/20) + (1/60)
= 1/15
Part of tank filled by both the taps in 10 minutes = 2/3
The remaining 1/3 of the tank is filled by the second tap alone.
Time taken to fill 1/3 of the tank = (1/3) /(1/60) = 20 minutes

Question 46

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

A	8
B	15
C	16
D	18

Question 46 Explanation:

Capacity of the tank = (12*13.5) liters = 162 litres
Number of 9 liters buckets needed = (162/9) = 18

Question 47

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

A	60 hrs
B	80 hrs
C	70 hrs
D	90 hrs

Question 47 Explanation:

Part of the cistern emptied by the leak in 1 hour = (1/9 - 1/10) = 1/90
The leak will empty the cistern in 90 hrs

Question 48

To fill a cistern, pipes A,B and C take 20 minutes, 15 minutes and 12 minutes respectively. The time taken by the three pipes together to fill the cistern is

A	5mins
B	12 mins
C	10 mins
D	15(2/3) mins

Question 48 Explanation:

Part of tank filled by A,B and C in 1 minute
= 1/20 + 1/15 + 1/12
= 1/5
The cistern will be filled in 5 mins

Question 49

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?

A	7mins 45 sec
B	8 mins 5 sec
C	7 mins 15 sec
D	8 mins 15 sec

Question 49 Explanation:

Part of the bucket filled by A and B in 1 minute
= (1/A) + (1/B)
= 3/20
Part of the bucket filled by A and B in 3 minutes = 9/20
The remaining 11/20 of the bucket is filled by B alone Time taken = (11/20) / (1/15)
= 33/4 minutes
= 8 mins 15 seconds

Question 50

Two pipes can fill a tank in 10 hours and 12 hours respectively while a third pipe empties the full tank in 20 hours. If all the three pipes operate simultaneously, in how much time the tank will be filled?

A	7 hrs
B	7hrs 30 mins
C	8 hrs
D	8 hrs 30 mins

Question 50 Explanation:

Part of tank filled in 1 hour = (1/10) + (1/12) + (1/20)
= 2/15
The tank will be filled in 15/2 hrs
= 7 hrs 30 minutes

Question 51

Two taps A and B can fill a tank in 10 hours and 15 hours respectively. If both the taps are opened together, the tank will be full in

A	5 hrs
B	12.5 hrs
C	6 hrs
D	7.5 hrs

Question 51 Explanation:

Part of tank filled by A and B in 1 hour = (1/10) + (1/15)
= 1/6
The tank will be filled in 6 hours

Question 52

A tap can fill a cistern in 2 hours and another tap can empty the cistern in 3 hours. How long will it take to fill the cistern if both the taps are opened?

A	5 hrs
B	7 hrs
C	6 hrs
D	8 hrs

Question 52 Explanation:

Part of the cistern filled by the first tap in 1 hour = ½
Part of the cistern filled by the second tap in 1 hour = 1/3
Part of the cistern filled in 1 hour when both the taps are opened = (1/2) – (1/3)
= 1/6
The cistern will be filled in 6 hours

Question 53

A cistern has two taps which fill it in 12 minutes and 15 minutes respectively. There is also a waste pipe in the cistern. When all the three are opened, the empty cistern is full in 20 minutes. How long will the waste pipe take to empty the full cistern?

A	8 minutes
B	12 minutes
C	10 minutes
D	16 minutes

Question 53 Explanation:

Part of tank emptied by the waste pipe in 1 minute = (1/12) + (1/15) – (1/20)
= 1/10
The waste pipe takes 10 minutes to empty the cistern

Question 54

An electric pump can fill a tank in 3 hours. Because of a leak in the tank, it took 3 hours 30 minutes to fill the tank. The leak can drain out all the water of the tank in

A	10 hrs 30 mins
B	21 hours
C	12 hours
D	24 hours

Question 54 Explanation:

3hrs 30 minutes
= 3.5 hours
= 7/2 hours
Work done by the leak in 1 hour
= (1/3) – (2/7)
= 1/21
The leak will empty the tank in 21 hours

Question 55

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?

A	20 hrs
B	20 hrs
C	35 hrs
D	cannot be determined

Question 55 Explanation:

1/A + 1/B + 1/C= 1/5
Let the time taken by A to fill the tank be X hours
B is twice as fast as A
Time taken by B = x/2 minutes
C is twice as fast as B
Time taken by C = x/4 minutes
1/x + 2/x + 4/x = 1/5
X= 35

Question 56

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

A	10 mins 20sec
B	11mins 45 sec
C	12mins 30
D	15 mins

Question 56 Explanation:

B is used for x minutes. A and B are used together for the next x minutes
Part of tank filled by B in x minutes = x/40
Part of tank filled by A and B in 1 minute = 1/60 + 1/40
= 1/24
Part of tank filled by A and B in x minutes= x/24
x/40 + x/24 = 1
8x = 120
X = 15
The total time taken is 2x = 30 minutes

Question 57

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?

A	15 mins
B	20 mins
C	27.5 mins
D	30 mins

Question 57 Explanation:

Let the time taken to fill the tanker be 2x minutes
Part of tank filled by B in x minutes = x/40
Part of tank filled by A and B in x minutes = x/40 + x/60
x/40 + x/40 +x/60 =1
x= 15
2x = 30

Question 58

Three pipes A, B and C can fill a tank in 6 hours. After working at it together for 2 hour, C is closed and A and B can fill the remaining part in 7 hours. The number of hours taken by C alone to fill the tank is

A	10
B	12
C	14
D	16

Question 58 Explanation:

Part of the tank filled by A, B and C in 2 hours = 2/6 = 1/3
The remaining 2/3 of the tank is filled by A and B in 7 hours
Part of tank filled by A and B in 1 hour = (2/3) / 7 = 2/21
1/C = 1/6 – 2/21
= 3/42
= 1/14
C alone can fill the tank in 14 hours

Question 59

A pipe can fill a tank in 6 hours. After three fourth the tank is filled, two more similar taps are opened. What is the total time taken to fill the tank completely?

A	4hrs 30 mins
B	5hrs
C	5 hrs 15 mins
D	5 hrs 30 mins

Question 59 Explanation:

Part of tank filled by 3 pipes in 1 hour = 3/6 = ½
1/4th of the tank is filled by 3 pipes in ½ hours
Time taken by the pipe to fill ¾ of the tank = 6*3/4 = 9/2
Total time taken = ½ + 9/2 = 5 hours

Question 60

If two pipes C and D can together fill a tank in 36 minutes and pipe C alone takes 1 hour to fill the tank, find the time taken by pipe D alone to fill the tank?

A	72 mins
B	80 mins
C	100 mins
D	90 mins

Question 60 Explanation:

Answer: Option D
1/D = 1/36 – 1/60 = 1/90
D alone takes 90 minutes to fill the tank

Question 61

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 pair in 7 hours. The number of hours taken by C alone to fill the tank is?

A	15
B	12
C	14
D	41

Question 61 Explanation:

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)’s1 hour work]
= (1/6 – 2/21) = 1/14
C alone can fill the tank in 14 hours

Question 62

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?

A	60 gallons
B	100 gallons
C	120 gallons
D	180 gallons

Question 62 Explanation:

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

Question 63

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?

A	3 ¼ min
B	5 ¼ min
C	8 ¼ min
D	9 ¼ min

Question 63 Explanation:

3/12 +(3+X)/15 =1
X= 8 ¼

Question 64

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?

A	5260 liters
B	5760 liters
C	5846 liters
D	6970 liters

Question 64 Explanation:

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

Question 65

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

A	4PM
B	5.30 PM
C	6PM
D	5 PM

Question 65 Explanation:

1 to 2 = ¼
2 to 3 = ¼ + 1/5 = 9/20
After 3AM = ¼ + 1/5 – ½ = - 1/20
¼ + 9/20 = 14/20
1 hr -----1/20
? --------14/20
14 hours  5 PM

Question 66

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 the solution R in the liquid in the tank after 3 minutes?

A	5/11
B	6/11
C	7/11
D	8/11

Question 66 Explanation:

Question 67

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:

A	1 13/17 hours
B	2 8/11
C	3 9/17
D	4 1/2 1

Question 67 Explanation:

Net part filled in 1 hour 1 + 1 - 1 = 17 . 5 6 12 60 The tank will be full in 60 hours i.e., 3 9 hours. 17 17

Question 68

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:

A	4 1/3 Hours
B	7 hours
C	8 hours
D	14 hours

Question 68 Explanation:

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

Question 69

Two pipes A and B can fill a cistern in 37 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:

A	5 min.
B	9 min.
C	10 min.
D	15 min.

Question 69 Explanation:

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 + 1 + (30 - x). 2 = 1 75 45 75 11x + (60 -2x) = 1 225 75 11x + 180 - 6x = 225. x = 9.

Question 70

A	6 hours
B	10 hours
C	15 hours
D	30 hours

Question 70 Explanation:

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 + 1 = 1 x (x - 5) (x - 9) x - 5 + x = 1 x(x - 5) (x - 9) (2x - 5)(x - 9) = x(x - 5) x2 - 18x + 45 = 0 (x - 15)(x - 3) = 0 x = 15. [neglecting x = 3]

Question 71

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:

A	81 min.
B	108 min.
C	144 min.
D	192 min.

Question 71 Explanation:

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

Question 72

A	15 min
B	20 min
C	27.5 min
D	30 min

Question 72 Explanation:

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

Question 73

A tap can fill a tank in 6 hours. After half the tank is filled, three more similar taps are opened. What is the total time taken to fill the tank completely?

A	3 hrs 15 min
B	3 hrs 45 min
C	4 hrs
D	4 hrs 15 min

Question 73 Explanation:

Time taken by one tap to fill half of the tank = 3 hrs. Part filled by the four taps in 1 hour = 4 x 1 = 2 . 6 3 Remaining part = 1 - 1 = 1 . 2 2 2 : 1 :: 1 : x 3 2 x = 1 x 1 x 3 = 3 hours i.e., 45 mins. 2 2 4 So, total time taken = 3 hrs. 45 mins.

Question 74

Three taps A, B and C can fill a tank in 12, 15 and 20 hours respectively. If A is open all the time and B and C are open for one hour each alternately, the tank will be full in:

A	6 hours
B	6 2/3 hours
C	7 hours
D	7 1/2 houurs

Question 74 Explanation:

c) 7 hours d) 7 1 hours 2 Answer: Option C Explanation: (A + B)'s 1 hour's work = 1 + 1 = 9 = 3 . 12 15 60 20 (A + C)'s hour's work = 1 + 1 = 8 = 2 . 12 20 60 15 Part filled in 2 hrs = 3 + 2 = 17 . 20 15 60 Part filled in 6 hrs = 3 x 17 = 17 . 60 20 Remaining part = 1 - 17 = 3 . 20 20 Now, it is the turn of A and B and 3 part is filled by A and B in 1 hour. 20 Total time taken to fill the tank = (6 + 1) hrs = 7 hrs.

Question 75

Three pipes A, B and C can fill a tank in 6 hours. After working at it together for 2 hours, C is closed and A and B can fill the remaining part in 7 hours. The number of hours taken by C alone to fill the tank is:

A	10
B	12
C	14
D	16

Question 75 Explanation:

Part filled in 2 hours = 2 = 1 6 3 Remaining part = 1 - 1 = 2 . 3 3 (A + B)'s 7 hour's work = 2 3 (A + B)'s 1 hour's work = 2 21 C's 1 hour's work = { (A + B + C)'s 1 hour's work } - { (A + B)'s 1 hour's work } = 1 - 2 = 1 6 21 14 C alone can fill the tank in 14 hours.

Question 76

A	60 gallons
B	100 gallons
C	120 gallons
D	180 gallons

Question 76 Explanation:

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

Question 77

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?

A	20 hours
B	25 hours
C	35 hours
D	Cannot be determined
E	None of these

Question 77 Explanation:

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

Question 78

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?

A	1 hour
B	2 hours
C	6 hours
D	8 hours

Question 78 Explanation:

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

Question 79

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?

A	12 min
B	15 min
C	15 min
D	50 min

Question 79 Explanation:

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 + 1 = 1 . 20 30 12 Both pipes can fill the tank in 12 minutes.

Question 80

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

A	10 min. 20 sec
B	11 min. 45 sec.
C	12 min. 30 sec.
D	14 min. 40 sec.

Question 80 Explanation:

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

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Probability

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Random Experiment: An experiment in which all possible outcomes are know and the exact output cannot be predicted in advance, is called a random experiment.

Examples:

Rolling an unbiased dice.
Tossing a fair coin.
Drawing a card from a pack of well-shuffled cards.
Picking up a ball of certain colour from a bag containing balls of different colours.

Details:

When we throw a coin, then either a Head (H) or a Tail (T) appears.
A dice is a solid cube, having 6 faces, marked 1, 2, 3, 4, 5, 6 respectively. When we throw a die, the outcome is the number that appears on its upper face.
A pack of cards has 52 cards.
It has 13 cards of each suit, name Spades, Clubs, Hearts and Diamonds.
Cards of spades and clubs are black cards.
Cards of hearts and diamonds are red cards.
There are 4 honours of each unit.
There are Kings, Queens and Jacks. These are all called face cards.

Sample Space: When we perform an experiment, then the set S of all possible outcomes is called the sample space.

Examples:

In tossing a coin, S = {H, T}
If two coins are tossed, the S = {HH, HT, TH, TT}.
In rolling a dice, we have, S = {1, 2, 3, 4, 5, 6}.

Event: Any subset of a sample space is called an event.

Probability of Occurrence of an Event: Let S be the sample and let E be an event. Then, P(E) = n(E) / n(S)

Results on Probability:

For any events A and B we have: P(A∪B) = P(A) + P(B) – P(A∩B)

If A denotes (not-A), then P(A) = 1 – P(A)

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Probability - Question and Answers

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Question 1

A bag contains 12 white and 18 black balls. Two balls are drawn in succession without replacement. What is the probability that first is white and second is black?

A	18/145
B	18/29
C	36/135
D	36/145

Question 1 Explanation:

The probability that first ball is white:
= ¹²C₁ / ³⁰C₁ = 12 / 30 = 2 / 5
Since, the ball is not replaced; hence the number of balls left in bag is 29
Hence, the probability the second ball is black:
= ¹⁸C₁ / ²⁹C₁= 18 / 29
Required probability,
= (2 / 5) × (18 / 29)
= 36 / 145

Question 2

One card is drawn at random from a pack of 52cards. What is the probability that the card drawn is a face card (jack, queen, king only)?

A	1 / 13
B	3 / 13
C	1 / 4
D	9 / 52

Question 2 Explanation:

Total no. of face cards (jack, queen and king only) =12
Total no .of cards = 52
So probability = 12 / 52 = 3 / 13

Question 3

A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?

A	10 / 21
B	11 / 21
C	2 / 7
D	5 / 7

Question 3 Explanation:

Total number of balls = (2 + 3 + 2) = 7
Let S be the sample space
Then, n(S) = Number of ways of drawing 2 balls out of 7
= 7C₂ = (7 x 6) / (2 x 1) = 21
Let E = Event of drawing 2 balls, none of which is blue
n(E) = Number of ways of drawing 2 balls out of (2 + 3) balls
= ⁵C₂
= (5 x 4) / (2 x 1)
= 10
Therefore P(E) = n(E) / n(S) = 10 / 21

Question 4

What is the probability of getting a sum 9 from two throws of a dice?

A	1 / 6
B	1 / 8
C	1 / 9
D	1 / 12

Question 4 Explanation:

In two throws of a dice, n(S) = (6 x 6) = 36
Let E = event of getting a sum ={(3, 6), (4, 5), (5, 4), (6, 3)}
Therefore, n(E) / n(S) = 4 / 36 = 1/ 9

Question 5

From a pack of 52 cards, two cards are drawn together at random. What is the probability of both the cards being kings?

A	1 / 15
B	25 / 57
C	35 / 256
D	1 / 221

Question 5 Explanation:

Let S be the sample space
Then, n(S) = ⁵²C₂ = (52 x 51) / (2 x1) = 1326
Let E = event of getting 2 kings out of 4
n(E) = ⁴C₂= (4 x 3) / (2 x1) = 6
P(E) = n(E) / n(S) = 6 / 1326 = 1 / 221

Question 6

Four dice are thrown simultaneously. Find the probability that all of them show the same face?

A	1 / 216
B	1 / 36
C	4 / 216
D	3 / 216

Question 6 Explanation:

The total number of elementary events associated to the random experiments of throwing four dice
simultaneously is:
= 6×6×6×6 = 6 ⁴
n(S) = 6 ⁴
Let X be the event that all dice show the same face
X={(1,1,1,1,),(2,2,2,2),(3,3,3,3),(4,4,4,4),(5,5,5,5),(6,6,6,6)}
n(X) = 6
Hence required probability,
= n(X) / n(S)
= 6 / 6 ⁴ = n(X) / n(S) = 6 / 6⁴ = 1 / 216

Question 7

In a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize?

A	1 / 10
B	2 / 5
C	2 / 7
D	5 / 7

Question 7 Explanation:

P (getting a prize) = 10 / (10 + 25) = 10 / 35 = 2 / 7

Question 8

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?

A	1 / 3
B	3 / 4
C	7 / 19
D	8 / 21

Question 8 Explanation:

Total number of balls = (8 + 7 + 6) = 21
Let E = event that the ball drawn is neither red nor green
n(E) = 7
Therefore P(E) = n(E) / n(S) = 7 / 21 = 1 / 3

Question 9

A group of 50 students were required to clear 2 tasks, one in rock-climbing and the other in bridge crossing during an adventure sports expedition. 30 students cleared both the tasks. 37 cleared bridge crossing, 38 student cleared rock-climbing. How many students could not clear any task?

A	4
B	5
C	9
D	10

Question 9 Explanation:

No. of student those completed at least one task: 37 + 38 - 30 = 45
No. of students those did not complete any task: 50 - 45 = 5

Question 10

200 pupil total, out of 125 like pizza, 115 like burger, then many like both?

A	30
B	35
C	40
D	60

Question 10 Explanation:

n(A U B) = n(A) + n(B) – n(A n B)
200 = 125 + 115 – Both
Both = 240 - 200 = 40

Question 11

Three unbiased coins are tossed. What is the probability of getting at most two heads?

A	3 / 4
B	1 / 4
C	3 / 8
D	7 / 8

Question 11 Explanation:

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}
Therefore, P(E) = n(E) / n(S) = 7 / 8

Question 12

A bag contains 4 white, 5 red and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red, is?

A	1 / 22
B	3 / 22
C	2 / 91
D	2 / 77

Question 12 Explanation:

Let S be the sample space
Then, n(S)= number of ways of drawing 3 balls out of 15
= ¹⁵C₃
(15 x 14 x 13) / (3 x 2 x 1) = 455
Let E = event of getting all the 3 red balls
Therefore, n(E) = ⁵C₃ = ⁵C₂ = (5 x 4) / (2 x 1) = 10
Therefore, P(E) = n(E) / n(S) = 10 / 455 = 2 / 91

Question 13

In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is?

A	21 / 46
B	25 / 117
C	1 / 50
D	3 / 25

Question 13 Explanation:

Let S be the sample space and E be the event of selecting 1 girl and 2 boys
Then, n(S)= Number ways of selecting 3 students out of 25
= ²⁵C₃
=> (25 x 24 x 23 ) / (3 x 2 x1) = 2300
n(E) = (¹⁰C₁ x ¹⁵C₂)
=> [10 x [(15 x 14) / (2 x1)]] = 1050
Therefore, P(E) = n(E) / n(S) = 1050 / 2300 = 21/ 46(3 x 2 x1)

Question 14

A bag contains 6 red, 5 blue and 2 green marbles. If 2 marbles are picked at random, What is the probability that both are red?

A	6 / 13
B	5 / 26
C	5 / 13
D	7 / 26

Question 14 Explanation:

The required probability i.e P(Both are red),
=> ⁶C₂ / ¹³C₂
=> 5 / 26

Question 15

Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?

A	1 / 2
B	3 / 4
C	3 / 8
D	5 / 16

Question 15 Explanation:

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
Therefore, P(E) = n(E) / n(S) = 27 / 36 = 3 / 4

Question 16

A card is drawn from a pack of 52 cards. The probability of getting a queen of club or a king of heart is?

A	1 / 13
B	2 / 13
C	1 / 26
D	1 / 52

Question 16 Explanation:

Here, n(S) = 52
Let E = event of getting a queen of club or a king of heart
Then, n(E) = 2
Therefore, P(E) = n(E) / n(S) = 2/ 52 = 1/ 26

Question 17

Two cards are drawn together from a pack of 52 cards. The probability that one is a spade and one is a heart, is?

A	3 / 20
B	29 / 34
C	47 100
D	13 / 102

Question 17 Explanation:

Let S be the sample space
Then, n(S) = ⁵²C₂ = (52 x 51) / (2 x 1) = 1326
Let E = event of getting 1 spade and 1 heart
n(E) = number of ways of choosing 1 spade out of 13 and 1 heart out of 13
=(¹³C₁ x ¹³C₁)
=> (13 x 13)= 169
P(E) = n(E) / n(S) = 169 / 1326 = 13 / 102

Question 18

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)?

A	1 / 13
B	3 / 13
C	1 / 4
D	9 / 52

Question 18 Explanation:

Clearly, there are 52 cards, out of which there are 12 face cards
Therefore, P(Getting a face card) = 12 / 52 = 3 / 13

Question 19

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?

A	1 / 2
B	2 / 5
C	8 / 15
D	9 / 20

Question 19 Explanation:

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}
Therefore P(E) = n(E) / n(S) = 9 / 20

Question 20

There are 76 people residing in a colony, 53 can read hindu, 46 can read times,39 can read deccan, 15 can read all.if 22 can read hindu and deccan, 23 can read deccan and times, then what is the number of persons who can read only times and hindu?

A	22
B	32
C	33
D	44

Question 20 Explanation:

Let x = the number of persons read only times and Hindu
(P(AUBUC) = P(A) + P(B) + P(C) - P(A n B) – P(A n C) – P(B n C) + P(A n B n C))
76 = 53 + 46 + 39 - 22 - 23 - x + 15
x = 108 - 76
x = 32

Question 21

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?

A	3 / 4
B	4 / 7
C	1 / 8
D	3 / 7

Question 21 Explanation:

Let number of balls = (6 + 8) = 14
Number of white balls = 8
P(drawing a white ball) = 8 / 14 = 4 / 7

Question 22

10 friends meet foe movie, 20 for picnic and 5 for games, 4 for movies and picnic, 2 for movies and games, 0 for picnic and games 2 for all, how many students are there totally?

A	31
B	15
C	21
D	25

Question 22 Explanation:

n(A U B U C) = n(A) + n(B) + n(C) – n(A n B) – n(A n C) – n(C n B) + n(A n B n C)
Total no.of students = 10 + 20 + 5 - 4 - 4 - 2 + 2 = 31

Question 23

There are two boxes namely A and B. A contains 20 green and 15 blue balls and B contains 7green and 8 blue balls. You can move the balls between the two boxes. If you are allowed tochoose a box at random then what will be the maximum probability of getting a green ball from the chosen box?

A	55/58
B	40/58
C	41/58
D	39/58

Question 23 Explanation:

Total number of green balls = 20 + 7 = 27 and total number of blue balls = 15 + 8 = 23
Given that, we can move the balls between the boxes
From the above arrangement, Probability to choose a green ball =Probability of choosing box A x Probability of choosing green ball from box A + Probability of choosing box B x Probability of choosing green ball from box B ...(1)
Since there are only two boxes, the probability of choosing box A = probability of choosing box B = ½
Probability of choosing a green ball from box A = number of green balls in box A / total number of ballsin box A = 1/1 = 1
Probability of choosing a green ball from box B = number of green balls in box B / total number of ballsin box B = 26/(26+23) = 26/49 = 26/49
Required probability (from eqn(1)) = 1/2 x 1 + 1/2 x 26/29 = 1/2 (1 + 26/29) = (1/2) x (55/29) = 55/58
Hence the required maximum probability = 55/58

Question 24

Four boys and three girls stand in queue for an interview. The probability that they stand in alternate positions is?

A	1/35
B	1/34
C	1/17
D	1/68

Question 24 Explanation:

Total number of possible arrangements for 4 boys and 3 girls in a queue = 7!
When they occupy alternate position the arrangement would be like: B G B G B G B
Thus, total number of possible arrangements for boys = (4×3×2)
Total number of possible arrangements for girls =(3×2)
Required probability = (4×3×2×3×2) /7! = 1/35

Question 25

A committee of 4 members is to be selected from a group of 4 women and 3 men. What is the probability that the committee has at least one man?

A	1/35
B	3/7
C	34/35
D	4/7

Question 25 Explanation:

Probability of there being all women in the committee = 4C4 divided by 7C4 = 1/35
But, Probability of there being all women in the committee + Probability of there being at least one man = Probability of formation of a committee with no restriction of gender = 1
Probability of there being at least one man = 1 - 1/35 = 34/35

Question 26

An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles. If four marbles are picked at random, what is the probability that at least one is blue?

A	4/15
B	69/91
C	11/15
D	22/91

Question 26 Explanation:

Total possible outcomes = n(S)
=> Selection of 4 marbles out of 15 marbles
=> 15C₄= (15 x 14 x 13 x 12) divided by 1 x 2 x 3 x 4 = 1365
When no marble is blue, favourable number of cases n(E)
=> Selection of 4 marbles out of 11 marbles
=> 11C₄ = 11 x 10 x 9 x 8 divided by 1 x 2 x 3 x 4 = 330
Probability that atleast one ball is blue + Probability that no ball is blue = 1
Therefore, Probability that atleast one ball is blue = 1 - Probability that no ball is blue
Therefore, Probability that atleast one ball is blue = 1 - n(E) / n(S)
= 1 - 330/1365 = 1 - 22/91 = 69 /91

Question 27

A leather box contains 8 black balls and 6 white balls. Two draws of three balls each are made, the balls being replaced after the first draw. What is the probability that the balls were black in the first draw and white in the second draw?

A	70/8281
B	140/20449
C	25/5445
D	35/5448

Question 27 Explanation:

Total number of balls = 8 + 6 = 14
Total ways of drawing 3 balls, N(S) = 14C3 = 364
No of ways to draw 3 black balls = N(E1) for black balls = 8C3 = 56
Probability of all balls being black = P(E1) = N(E1) / N(S) = 56/ 364 = 14/91
No of ways to draw 3 white balls = N(E2) for white balls = 6C3 = 20
Probability of all balls being white = P(E2) = 20 / 364 = 5/91
Probability of drawing 3 black balls in first draw and drawing 3 white balls in the second draw = P(E1) x P(E2)
Therefore P(E) = 14/91 x 5/91 = 70 /8281

Question 28

Fourteen persons are sitting around a circular table facing the centre. What is the probability that three particular persons sit together?

A	2/9
B	1/13
C	2/13
D	1/26

Question 28 Explanation:

n a circle of n different persons, the total number of arrangements possible = (n - 1)!
Total number of arrangements = n(S) = (14 – 1)! = 13 !
Taking three persons as a unit, total persons = 12 (in 4 units)
Therefore no. of ways for these 12 persons to around the circular table = (12 - 1)! = 11!
In any given unit, 3 particular person can sit in 3!
Hence total number of ways that any three person can sit = n(E) = 11! X 3!
Therefore P (E) = probability of three persons sitting together = n(E) / n(S) = 11! X 3 ! divided by 13!
3 x 2 divided by 13 x 12 = 1/26

Question 29

In how many different ways can the letter of the word “ECHRONICLEE” be arranged?

A	1663200
B	8316000
C	3326400
D	4158000

Question 29 Explanation:

ECHRONICLEE has 11 letters
Total number of rearrangements = (Number of letters)! / (Number of 1st repetitive letter)! x (Number of 2nd repetitive letter)!
In the word ECHRONICLEE, E occurs thrice. C occurs twice
Hence denominator of the above formula will become 3! x 2!
Therefore Total number of rearrangements = 11!/ 3! 2!
= 3326400

Question 30

In a game there are 70 people in which 40 are boys and 30 are girls, out of which 10 people are selected at random. One from the total group, thus selected is selected as a leader at random. What is the probability that the person, chosen as the leader is a boy?

A	4/7
B	4/9
C	5/7
D	2/7

Question 30 Explanation:

The total groups contains boys and girls in the ratio 4 : 3
If some person are selected at random from the group, the expected value of the ratio of boys and girls will be 4 : 3
If the leader is chosen at random from the selection, the probability of him being a boy = 4/7

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Permutation and Combination

MS — Mon, 14 Aug 2017 08:45:32 +0000

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Permutation implies arrangement where order of things is important and includes word formation number formation, circular permutation etc.

All permutations (or arrangements) made with the letters a, b, c by taking two at a time are (ab, ba, ac, ca, bc, cb).
All permutations made with the letters a, b, c taking all at a time are (abc, acb, bac, bca, cab, cba).

Number of Permutations:

Number of all permutations of n things, taken r at a time, is given by:

ⁿP_r = n(n – 1)(n – 2) … (n – r + 1) = n! / (n – r)!

Examples:

⁶P₂ = (6 x 5) = 30.
⁷P₃ = (7 x 6 x 5) = 210.

More information: https://en.wikipedia.org/wiki/Permutation

Combination means selection where order is not important and it involves selection of team forming geometrical figures, distribution of things etc.

Examples:

Suppose we want to select two out of three boys A, B, C. Then, possible selections are AB, BC and CA. [Note: AB and BA represent the same selection]
All the combinations formed by a, b, c taking ab, bc, ca.
The only combination that can be formed of three letters a, b, c taken all at a time is abc.
Various groups of 2 out of four persons A, B, C, D are: AB, AC, AD, BC, BD, CD.
Note that ab ba are two different permutations but they represent the same combination.

Number of Combinations:

The number of all combinations of n things, taken r at a time is:

ⁿC_r = n ! / [(r!)(n-r)]! = [n(n-1)(n-2)…. To r factors] / r!

Note: ⁿC_n = 1 and ⁿC₀ = 1 , ⁿC_r = ⁿC_{(n – r)}

Examples:

¹¹C₄ = (11 x 10 x 9 x 8) / (4 x 3 x 2 x 1) = 330
¹⁶C₁₃ = ¹⁶C_{(16 – 13)} = ¹⁶C₃ = 16 x 15 x 14 / 3!= (16 x 15 x 14) / (3 x 2 x 1) = 560

Factorial Notation

Let n be a positive integer. Then, factorial n, denoted n! is defined as:

n! = n(n – 1)(n – 2) … 3.2.1.

Examples:

We define 0! = 1.
4! = (4 x 3 x 2 x 1) = 24.
5! = (5 x 4 x 3 x 2 x 1) = 120.

More information: https://en.wikipedia.org/wiki/Combination

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Permutation and combination - Question and Answers

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Question 1

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

A	266
B	5040
C	11760
D	86400

Question 1 Explanation:

Required number of ways:
= (8C5 X 10C6)
= {(8*7*6/3*2*1)*(10*9*8*7/4*3*2*1)}
= 11760

Question 2

How many ways to draw 3balls from 16 balls pool.(balls are numbered as 1 to 16 and are distinct)

A	16P3
B	16C3
C	16!
D	72

Question 2 Explanation:

No. of ways to draw 3 balls from 16 balls: 16C3

Question 3

From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?

A	564
B	645
C	735
D	756

Question 3 Explanation:

We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only). Required number of ways = (⁷C₃ x ⁶C₂) + (⁷C₄ x ⁶C₁) + (⁷C₅)
(7 x 6 x 5)/(3x2x1) x (6 x 5/2x1) + (⁷C₃ x ⁶C₁) + (⁷C₂)
=525 +(7x6x5/3x2x1 x 6) + (7x6/2x1)
= (525 + 210 + 21)
=756

Question 4

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

A	360
B	480
C	720
D	5040

Question 4 Explanation:

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.

Question 5

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

A	810
B	1440
C	2880
D	50400

Question 5 Explanation:

In the word 'CORPORATION', we treat the vowels OOAIO as one letter.
Thus, we have CRPRTN (OOAIO).
This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different.
Number of ways arranging these letters = 7!/2! = 2520.
Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged
in 5!/3! = 20 ways.
Required number of ways = (2520 x 20) = 50400.

Question 6

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

A	210
B	1050
C	25200
D	21400

Question 6 Explanation:

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)
(⁷C₃ x ⁴C₂)
(7*6*5/3*2*1) * (4*3/2*1)
=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.

Question 7

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

A	72
B	144
C	360
D	720

Question 7 Explanation:

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

Question 8

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?

A	63
B	90
C	126
D	45

Question 8 Explanation:

Required number of ways = (⁷C₅ x ³C₂) =
(⁷C₂ x ³C₁) =
(7*6/2*3) = 63

Question 9

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?

A	40
B	400
C	5040
D	2520

Question 9 Explanation:

'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.

Question 10

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

A	10080
B	4989600
C	120960
D	None of these

Question 10 Explanation:

In the word 'MATHEMATICS', we treat the vowels AEAI as one letter.
Thus, we have MTHMTCS (AEAI).
Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different.
Number of ways of arranging these letters = 8!/(2!)(2!) = 10080.
Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.
Number of ways of arranging these letters = 4!/2! = 12..
Required number of words = (10080 x 12) = 120960.

Question 11

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

A	120
B	720
C	4320
D	2160

Question 11 Explanation:

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

Question 12

In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?

A	159
B	194
C	205
D	209

Question 12 Explanation:

We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).
number of ways = (⁶C₁ x ⁴C₃) + (⁶C₂ x ⁴C₂) + (⁶C₃ x ⁴C₁) + (⁶C₄)
= (⁶C₁ x ⁴C₁) + (⁶C₂ x ⁴C₂) + (⁶C₃ x ⁴C₁) + (⁶C₂)
=(6*4) +((6*5/2*1) *(4*3/2*1)) + ((6*5*4/3*2*1)*4)+(6*5/2*1)
= (24 + 90 + 80 + 15)
= 209.

Question 13

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?

A	5
B	10
C	15
D	20

Question 13 Explanation:

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.

Question 14

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

A	266
B	5040
C	11760
D	86400

Question 14 Explanation:

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

Question 15

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?

A	32
B	48
C	64
D	96

Question 15 Explanation:

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* 6x5/2x1) + (3x2/2x1 * 6) + 1
= (45 + 18 + 1)
= 64.

Question 16

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?

A	32
B	48
C	36
D	60

Question 16 Explanation:

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

Question 17

How many multiples of 5 are there from 10 to 95?

A	20
B	18
C	17
D	25

Question 17 Explanation:

As you know, multiples of 5 are integers having 0 or 5 in the digit to the extreme right (i.e. the unit's place). The first digit from the right can be chosen in 2 ways.
The second digit can be any one of 1,2,3,4,5,6,7,8,91,2,3,4,5,6,7,8,9
i.e. There are 9 choices for the second digit.
Thus, there are 2×9=2×9= 18 multiples of 5 from 10 to 95.

Question 18

In a city, the bus route numbers consist of a natural number less than 100, followed by one of the letters A, B, C, D, E and F. How many different bus routes are possible?

A	456
B	522
C	594
D	500

Question 18 Explanation:

The number can be any one of the natural numbers from 1 to 99.
There are 99 choices for the number.
The letter can be chosen in 6 ways.
Number of possible bus routes are 99×6=99×6= 594

Question 19

There are 3 questions in a question paper. If the questions have 4, 3 and 2 solutions respectively, find the total number of solutions.

A	20
B	15
C	45
D	24

Question 19 Explanation:

Here question 1 has 4 solutions, question 2 has 3 solutions and question 3 has 2 solutions.
=> By the multiplication (counting) rule,
Total number of solutions=4×3×2==4×3×2= 24

Question 20

Consider the word ROTOR. Whichever way you read it, from left to right or from right to left, you get the same word. Such a word is known as palindrome. Find the maximum possible number of 5-letter palindromes.

A	17576
B	17654
C	14789
D	16872

Question 20 Explanation:

The first letter from the right can be chosen in 26 ways because there are 26 alphabets.
Having chosen this, the second letter can be chosen in 26 ways.
=> The first two letters can be chosen in 26×26=67626×26=676 ways
Having chosen the first two letters, the third letter can be chosen in 26 ways.
=> All the three letters can be chosen in 676×26=17576676×26=17576 ways.
It implies that the maximum possible number of five letter palindromes is 17576 because the fourth letter is the same as the second letter and the fifth letter is the same as the first letter.

Question 21

How many 3-digit numbers can be formed with the digits 1,4,7,8 and 9 if the digits are not repeated?

A	30
B	40
C	60
D	59

Question 21 Explanation:

Three digit numbers will have unit's, ten's and hundred's place.
Out of 5 given digits any one can take the unit's place. This can be done in 5 ways ------- (i).
After filling the unit's place, any of the four remaining digits can take the ten's place..
This can be done in 4 ways ------- (ii).
After filling in ten's place, hundred's place can be filled from any of the three remaining digits..
This can be done in 3 ways ------- (iii).
=> By counting principle, the number of 3 digit numbers =5×4×3==5×4×3= 60

Question 22

Suppose you can travel from a place A to a place B by 3 buses, from place B to place C by 4 buses, from place C to place D by 2 buses and from place D to place E by 3 buses. In how many ways can you travel from A to E?

A	63
B	45
C	72
D	85

Question 22 Explanation:

The bus from A to B can be selected in 3 ways.
The bus from B to C can be selected in 4 ways.
The bus from C to D can be selected in 2 ways.
The bus from D to E can be selected in 3 ways.
So, by the General Counting Principle, one can travel from A to E in 3×4×2×3=3×4×2×3= 72 ways

Question 23

Suppose you want to arrange your English, Hindi, Mathematics, History, Geography and Science books on a shelf. In how many ways can you do it?

A	630
B	569
C	654
D	720

Question 23 Explanation:

We have to arrange 6 books.
The number of permutations of n objects isn!=n.(n−1).(n−2)...2.1n!=n.(n−1).(n−2)...2.1.
Here n=6n=6 and therefore, number of permutations is 6.5.4.3.2.1=6.5.4.3.2.1= 720.

Question 24

If you have 6 New Year greeting cards and you want to send them to 4 of your friends, in how many ways can this are done?

A	260
B	392
C	723
D	360

Question 24 Explanation:

We have to find number of permutations of 4 objects out of 6 objects.
This number is 6P4=6(6−1)(6−2)
(6−3)=6×5×4×3=3606P4=6(6−1)(6−2)
(6−3)=6×5×4×3=360

Question 25

Suppose 7 students are staying in a hall in a hostel and they are allotted 7 beds. Among them, Parvin does not want a bed next to Anju because Anju snores. Then, in how many ways can you allot the beds?

A	200ways
B	5400ways
C	3600ways
D	3505ways

Question 25 Explanation:

Let the beds be numbered 1 to 7.
Case 1:
Suppose Anju is allotted bed number 1.
Then, Parvin cannot be allotted bed number 2.
So Parvin can be allotted a bed in 5 ways.
After allotting a bed to Parvin, the remaining 5 students can be allotted beds in 5! ways.
So, in this case the beds can be allotted in 5×5!=6005×5!=600 ways.
Case 2:
Anju is allotted bed number 7.
Then, Parvin cannot be allotted bed number 6
As in Case 1, the beds can be allotted in 600 ways.
Case 3:
Anju is allotted one of the beds numbered 2,3,4,52,3,4,5 or 66
Parvin cannot be allotted the beds on the right hand side and left hand side of Anju's bed.
For example, if Anju is allotted bed number 2, beds numbered 1 or 3 cannot be allotted to Parvin.
Therefore, Parvin can be allotted a bed in 4 ways in all these cases.
After allotting a bed to Parvin, the other 5 can be allotted a bed in 5! ways.
Therefore, in each of these cases, the beds can be allotted 4×5!=4804×5!=480 ways.
=> The beds can be allotted in:
2×600+5×480=1200+2400=2×600+5×480=1200+2400= 3600 ways

Question 26

There are 4 books on fairy tales, 5 novels and 3 plays. In how many ways can you arrange these so that books on fairy tales are together, novels are together and plays are together and in the order, books on fairy tales, novels and plays.

A	28756
B	28756
C	17280
D	16547

Question 26 Explanation:

There are 4 books on fairy tales and they have to be put together.
They can be arranged in 4! ways.
Similarly, there are 5 novels.
They can be arranged in 5! ways.
And there are 3 plays.
They can be arranged in 3! ways.
So, by the counting principle all of them together can be arranged in 4!×5!×3!=4!×5!×3!= 17280 ways

Question 27

Suppose there are 4 books on fairy tales, 5 novels and 3 plays as in Example 5.3. They have to be arranged so that the books on fairy tales are together, novels are together and plays are together, but we no longer require that they should be in a specific order. In how many ways can this be done?

A	465765
B	678954
C	103680
D	876524

Question 27 Explanation:

First, we consider the books on fairy tales, novels and plays as single objects.
These three objects can be arranged in 3!=63!=6 ways.
Let us fix one of these 6 arrangements.
This may give us a specific order, say, novels -> fairy tales -> plays.
Given this order, the books on the same subject can be arranged as follows.
The 4 books on fairy tales can be arranged among themselves in 4!=244!=24 ways.
The 5 novels can be arranged in 5!=1205!=120 ways.
The 3 plays can be arranged in 3!=63!=6 ways.
For a given order, the books can be arranged in 24×120×6=1728024×120×6=17280 ways.
Therefore, for all the 6 possible orders the books can be arranged in 6×17280=6×17280= 103680 ways.

Question 28

In how many ways can 4 girls and 5 boys be arranged in a row so that all the four girls are together?

A	17654
B	17280
C	56783
D	78345

Question 28 Explanation:

Let 4 girls be one unit and now there are 6 units in all. They can be arranged in 6! ways.
In each of these arrangements 4 girls can be arranged in 4! ways.
=> Total number of arrangements in which girls are always together
=6!×4!=720×24==6!×4!=720×24= 17280

Question 29

Find the number of subsets of the set {1,2,3,4,5,6,7,8,9,10,11}{1,2,3,4,5,6,7,8,9,10,11} having 4 elements.

A	330 ways
B	340 ways
C	333 ways
D	348 ways

Question 29 Explanation:

Here the order of choosing the elements doesn't matter and this is a problem in combinations.
We have to find the number of ways of choosing 4 elements of this set which has 11 elements.
This can be done in:
11C4=11×10×9×81×2×3×4=11C4=11×10×9×81×2×3×4= 330 ways

Question 30

In a box, there are 5 black pens, 3 white pens and 4 red pens. In how many ways can 2 black pens, 2 white pens and 2 red pens can be chosen?

A	240ways
B	450ways
C	180ways
D	220ways

Question 30 Explanation:

Number of ways of choosing 2 black pens from 5 black pens
=5C2=5P22!=5×41×2=10=5C2=5P22!=5×41×2=10
Number of ways of choosing 2 white pens from 3 white pens
=3C2=3P22!=3×21×2=3=3C2=3P22!=3×21×2=3
Number of ways of choosing 2 red pens from 4 red pens
=4C2=4P22!=4×31×2=6=4C2=4P22!=4×31×2=6
=> By the Counting Principle, 2 black pens, 2 white pens, and 2 red pens can be chosen in
10×3×6=10×3×6= 180ways.

Question 31

There are 5 novels and 4 biographies. In how many ways can 4 novels and 2 biographies can be arranged on a shelf?

A	11234
B	21600
C	37432
D	37432

Question 31 Explanation:

4 novels can be selected out of 5 in 5C45C4 ways.
2 biographies can be selected out of4 in 4C24C2 ways.
Number of ways of arranging novels and biographies: =5C4×4C2=30=5C4×4C2=30
After selecting any 6 books (4 novels and 2 biographies) in one of the 30 ways, they can be arranged on the shelf in 6!=7206!=720 ways.
By the Counting Principle, the total number of arrangements =30×720==30×720= 21600

Question 32

Evaluate 7P2. 4P3

A	1456
B	1008
C	1080
D	1000

Question 32 Explanation:

(7!/ 5!). (4!/ 1!)
⇒(7.6). (4.3.2)
⇒1008

Question 33

Evaluate 5C2. 3C2

A	20
B	30
C	40
D	60

Question 33 Explanation:

(5!/3!2!). (3!/2!1!)
⇒(5.4/2). (3)
⇒30

Question 34

How many ways are there in selecting 5 members from 6 males and 5 females, consisting 3 males and 2 females?

A	120
B	240
C	100
D	200

Question 34 Explanation:

This is a case of combination i.e. selecting 3 males from 6 males and 2 females from 5 females.
⇒Required number of ways = (6C3 *5C2)
⇒(6.5.4/3.2)*(5.4/2)
⇒200.

Question 35

How many words can be formed by using letters of the word "DAUGHTER" so that the vowels come together?

A	4320
B	4563
C	4632
D	4700

Question 35 Explanation:

This is a case of permutation. In a word "DAUGHTER", there are 8 letters including 3 vowels (AUE)
According to the question, vowels should always come together. Therefore, in this case we will treat all the vowels as one entity or one alphabet. This implies, in total there are 6 words (one word which is a group of vowels)
These 6 words can be arranged in 6P6 ways ⇒6P6 = 6!/1! = 6! = 720 WAYS
Also, three vowels in a group may be arranged in 3! ways
⇒3! = 6 ways
Therefore, required number of words = (720*6) = 4320

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Percentage and True Discount

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PERCENTAGE: A percentage is defined as a number represented as a fraction of 100. Percentages are used to express numbers between zero and one. It is used to compare things and use it in ratios. It is denoted by the symbol %. Percentage = (Value / Totalvalue) * 100, thus 1/ 4 = (1/4 x 100)% = 25%
Percentage Increase/Decrease:
- If the price of a commodity increases by R%, then the reduction in consumption so as not to increase the expenditure is: [R / (100 + R) x 100]%
- If the price of a commodity decreases by R%, then the increase in consumption so as not to decrease the expenditure is: [R / (100 – R) x 100]%
Results on Population: Let the population of a town be P now and suppose it increases at the rate of R% per annum, then:
- Population after n years = P(1+ R/100)ⁿ
- Population n years ago = P / (1+ R/100)ⁿ
Results on Depreciation: Let the present value of a machine be P. Suppose it depreciates at the rate of R% per annum. Then:
- Value of the machine after n years = P (1 – R / 100)ⁿ
- Value of the machine n years ago = P /(1- R /100)ⁿ
- If A is R% more than B, then B is less than A by [R /(100 +R) x 100]%
- If A is R% less than B, then B is more than A by [R /(100 – R) x 100]%
TRUE DISCOUNT: The “True discount” is a term used when you are promised a given amount of money in the future. Since money collect interest, the actual value of that promise is less than the face value

Example:

Consider a man has to pay Rs. 156 after 4 years and the rate of interest is 14% per annum. Clearly, Rs. 100 at 14% will amount to R. 156 in 4 years. So, the payment of Rs. now will clear off the debt of Rs. 156 due 4 years hence. We say that:
- Sum due = Rs. 156 due 4 years hence;
- Present Worth (P.W.) = Rs. 100;
- True Discount (T.D.) = Rs. (156 – 100) = Rs. 56 = (Sum due) – (P.W.).
- We define: T.D. = Interest on P.W..; Amount = (P.W.) + (T.D.)
- Interest is reckoned on P.W. and true discount is reckoned on the amount.
IMPORTANT FORMULAS

Let rate = R% per annum and Time = T years. Then,
- P.W. = 100 x Amount / 100 + (R x T) = 100 x T.D / R x T
- T.D. = (P.W) x R x T / 100 = Amount x R x T / 100+(R x T)
- Sum = (S.I) x (T.D) / (S.I) – (T.D)
- (S.I.) – (T.D.) = S.I. on T.D.
- When the sum is put at compound interest, then P.W. = Amount / (1+R / 100)^T

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Percentage and true discount - Question and Answers

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Question 1

Which of the following is the greatest?

A	40% of 30
B	3/5 of 25
C	6.5% of 200
D	five more than the square of 3

Question 1 Explanation:

3/5 of 25b=15
40% of 30 = 12
6.5% of 200 = 13
5+9=14

Question 2

A new apartment complex purchased 60 toilets and 20 shower heads. If the price of a toilet is three times the price of a shower head, what percent of the total cost was the cost of all the shower heads?

A	9%
B	10%
C	11%
D	13%

Question 2 Explanation:

If the cost of toilet = x & shower head = y
Total cost = 60x+20y, given x=3y
So total cost = 60*3y+20y=200y% of shower head cost to total cost = 20y/200y = 10

Question 3

A batsman scored 110 runs which included 3 boundaries and 8 sixes. What percent of his total score did he make by running between the wickets?

A	45%
B	45 5/11 %
C	54 6/11 %
D	55%

Question 3 Explanation:

Number of runs made by running = 110 - (3 x 4 + 8 x 6)
= 110 - (60)
= 50.
Required percentage = ( (50 x 100)/100)% = 5/4511 %

Question 4

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:

A	39, 30
B	41, 32
C	42, 33
D	43, 34

Question 4 Explanation:

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.

Question 5

A fruit seller had some apples. He sells 40% apples and still has 420 apples. Originally, he had:

A	588 apples
B	600 apples
C	672 apples
D	700 apples

Question 5 Explanation:

Suppose originally he had x apples.
Then, (100 - 40)% of x = 420.
60/100 x x = 420
x = (420 x 100)/60 = 700.

Question 6

What percentage of numbers from 1 to 70 has 1 or 9 in the unit's digit?

A	1
B	14
C	20
D	21

Question 6 Explanation:

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. < br/> Number of such number =14 < br/> Required percentage = (14/70 x 100)% = 20%

Question 7

If A = x% of y and B = y% of x, then which of the following is true?

A	A is smaller than B.
B	A is greater than B
C	Relationship between A and B cannot be determined.
D	If x is smaller than y, then A is greater than B.
E	A = B

Question 7 Explanation:

x% of y = x/100 x y = y/100 x x = y% of x A = B.

Question 8

Three candidates contested an election and received 1136, 7636 and 11628 votes respectively. What percentage of the total votes did the winning candidate get?

A	57%
B	60%
C	65%
D	90%

Question 8 Explanation:

Total number of votes polled = (1136 + 7636 + 11628) = 20400.
Required percentage = (11628/20400 x 100) % = 57%.

Question 9

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?

A	Rs. 200
B	Rs. 250
C	Rs. 300
D	None of these

Question 9 Explanation:

Let the sum paid to Y per week be Rs. z.
Then, z + 120% of z = 550.
z + (120/100 )z = 550
(11/5)z = 550>br/> z = (550 x 5) 11 = 250.

Question 10

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?

A	Rs. 15
B	Rs. 15.70
C	Rs. 19.70
D	Rs. 20

Question 10 Explanation:

Let the amount taxable purchases be Rs. x.
Then, 6% of x = 30
x = (30/100 * 100/6) = 5

Question 11

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.

A	Rs. 6876.10
B	Rs. 6999.20
C	Rs. 6654
D	Rs. 7000

Question 11 Explanation:

Rebate = 6% of Rs. 6650 = Rs.(6 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

Question 12

The population of a town increased from 1,75,000 to 2,62,500 in a decade. The average percent increase of population per year is:

A	4.37%
B	5%
C	6%
D	8.75%

Question 12 Explanation:

Increase in 10 years = (262500 - 175000) = 87500.
Increase% = ((87500/175000) x 100 ) % = 50%.
Required average = (50/10) % = 5%. 10

Question 13

If 20% of a = b, then b% of 20 is the same as:

A	4% of a
B	5% of a
C	20% of a
D	None of these

Question 13 Explanation:

20% of a = b
20/100 a = b.
b% of 20 = (b/100) * 20 = (20/100)a * 1/100*20 = 4/100*a = 4% of a

Question 14

In a certain school, 20% of students are below 8 years of age. The number of students above 8 years of age is of the number of students of 8 years of age which is 48. What is the total number of students in the school?

A	72
B	80
C	120
D	150
E	100

Question 14 Explanation:

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.

Question 15

Two numbers A and B are such that the sum of 5% of A and 4% of B is two-third of the sum of 6% of A and 8% of B. Find the ratio of A : B.

A	2 : 3
B	1 : 1
C	3:4
D	4 : 3

Question 15 Explanation:

5% of A + 4% of B = 2/3 (6% of A + 8% of B)
5/100 A + 4/100 B = 2/3(6/100 A + 8/100B)
1/20A + 1/25B = 1/25A + 4/75B
(1/20-1/25)A = (4/75-1/25)B
1/100A = 1/75B
A/B = 100/75 = 4/3
Required ratio = 4:3

Question 16

A student multiplied a number by 3/5 and 5/3 What is the percentage error in the calculation?

A	34%
B	44%
C	54%
D	64%

Question 16 Explanation:

Let the number be x.
Then, error = 5/3 * 3/5 = 16/15
Error% =(16/15x * 3/5x * 100)% = 64%

Question 17

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:

A	2700
B	2900
C	3000
D	3100

Question 17 Explanation:

Number of valid votes = 80% of 7500 = 6000.
Valid votes polled by other candidate = 45% of 6000
= (45/100 x 6000) = 2700.

Question 18

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. What are the marks obtained by them?

A	42, 33
B	42, 36
C	44, 33
D	44, 36

Question 18 Explanation:

Let the marks secured by them be xx and (x+9)(x+9)
sum of their marks =x+(x+9)=2x+9=x+(x+9)=2x+9
Given that (x+9)(x+9) was 56% of the sum of their marks.
⇒(x+9)=56100(2x+9)⇒(x+9)=1425(2x+9)⇒25x+225=28x+126⇒3x=99⇒x=33⇒(x+9)=56100(2x+9)⇒(x+9)=1425(2x+9)⇒25x+225=28x+126⇒3x=99⇒x=33
Then (x+9)=33+9=42(x+9)=33+9=42
Hence their marks are 33 and 42

Question 19

A true discount on an amount of Rs.480 due 3 years hence is equal as simple interest on Rs.350 for 3 years at the same interest rate on true discount. Then what is the interest rate?

A	11.34%
B	12.38%
C	10.92%
D	13.12%

Question 19 Explanation:

S.I on Rs.350 = T.D on Rs.480
This means P.W of Rs.480 due 3 years hence is Rs.350
Therefore, T.D = Rs.(480 – 350) = Rs.130
Thus S.I of Rs.350 for 3 years is Rs.130
Rate of Interest = 100 x S.I / amount x time % = 100 x 130 / 350 x 3% = 12.38%
Hence the required rate of interest is 12.38%

Question 20

If Rs.160 and Rs.170 are true discount and simple interest of a certain amount respectively and if the time period and interest rate are equal for both simple interest and true discount then the amount is?

A	Rs.2120
B	Rs.2140
C	Rs.2720
D	Rs.2180

Question 20 Explanation:

Let rate = R% per annum and Time = T years
Sum = [(S.I.) x (T.D.)]/[(S.I.) – (T.D.)] Here, S.I = 170 and T.D = 160
Then required amount = 170 x 160 / 170 – 160 = 170 x 160/10 = 2720
Hence the required sum is Rs.2720

Question 21

The true discount on Rs. 2562 due 4 months hence is Rs. 122. The rate percent is?

A	12%
B	13%
C	14%
D	15%

Question 21 Explanation:

P.W = 2562-122 = 2440
Rate = 100 * 122 / 2440 * 1/3 = 15%

Question 22

The true discount on a certain sum due 6 months hence at 15% is Rs. 240. What is the banker's discount on the same sum for the same time at the same rate?

A	None of these
B	278
C	Rs. 228
D	Rs. 258

Question 22 Explanation:

TD = Rs. 240
T = 6 months = 1/2 year
R = 15%
TD = BG ×100/TR
240 = BG × 100 / [(1/2 × 15)]
BG = (240 × 15) / (100 × 2) = (120 × 15) / 100 = Rs. 18
BG = BD – TD
18 = BD - 240
BD = 18 + 240 = Rs. 258

Question 23

The present worth of a certain bill due sometime hence is Rs. 400 and the true discount is Rs. 20. What is the banker's discount?

A	Rs. 19
B	Rs. 22
C	Rs. 20
D	Rs. 21

Question 23 Explanation:

BG = (TD)² / PW = 20² / 400 = Rs.1
BG = BD – TD
=> 1 = BD - 20
=> BD = 1 + 20 = Rs. 21

Question 24

The banker's discount on a bill due 6 months hence at 6% is Rs. 18.54. What is the true discount?

A	Rs. 24
B	Rs. 12
C	Rs. 36
D	Rs. 18

Question 24 Explanation:

T = 6 months = 1/2 year
R = 6%
TD = (BD ×100)/(100+TR) = (18.54×100)/[100+(1/2 x 6)]
= (18.54×100)/103 = 1854/103 = Rs. 18

Question 25

The B.D. and T.D. on a certain sum is Rs.200 and Rs.100 respectively. Find out the sum?

A	Rs. 400
B	Rs. 300
C	Rs. 100
D	Rs. 200

Question 25 Explanation:

F = (BD×TD)/(BD–TD) = (200×100)/(200-100) = (200x100)/100 = Rs. 200

Question 26

If the true discount on a certain sum due 6 months hence at 15% is Rs. 120, what is the banker's discount on the same sum for the same time and at the same rate?

A	50
B	129
C	100
D	160

Question 26 Explanation:

B.G. = S.I. on T.D
=> Rs.(120 x 15 x 1/2 x 1/100)
=> Rs. 9
(B.D.) - (T.D.) = Rs. 9
B.D. = Rs. (120 + 9) = Rs. 129

Question 27

The true discount on a bill of Rs. 2160 is Rs. 360. What is the banker's discount?

A	Rs. 432
B	Rs. 422
C	Rs. 412
D	Rs. 442

Question 27 Explanation:

F = Rs. 2160
TD = Rs. 360
PW = F - TD = 2160 - 360 = Rs. 1800
True Discount is the Simple Interest on the present value for unexpired time
=> Simple Interest on Rs. 1800 for unexpired time = Rs. 360
Banker's Discount is the Simple Interest on the face value of the bill for unexpired time
= Simple Interest on Rs. 2160 for unexpired time
= (360/1800) x 2160 = (1/5) x 2160 = Rs. 432

Question 28

The present worth of a bill due sometime hence is Rs. 1100 and the true discount on the bill is Rs. 110. Find the banker's discount?

A	111
B	121
C	131
D	141

Question 28 Explanation:

T.D. = (P.W.*B.G)^(1/2)
B.G. =(T.D.)2/ P.W. = Rs.[(110x110)/ 1100] = Rs. 11
B.D.= (T.D. + B.G.) = Rs. (110 + 11) = Rs. 121

Question 29

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?

A	Rs. 21
B	Rs. 17
C	Rs. 18
D	Rs. 34

Question 29 Explanation:

BG = (TD)² / PW = (340)² /3400 = 340 x 340 / 3400 = 340/10 = Rs. 34

Question 30

The banker's discount on Rs. 1800 at 12% per annum is equal to the true discount on Rs. 1872 for the same time at the same rate. Find the time?

A	3 months
B	4 months
C	5 months
D	6 months

Question 30 Explanation:

S.I. on Rs. 1800 = T. D. on Rs. 1872
P.W. of Rs. 1872 is Rs. 1800
Rs. 72 is S.I. on Rs. 1800 at 12%
Time = [(100x72) / (12x1800)] years = 1/3 years = 4 months

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