6500+ Quantitative Aptitude Questions and Answers With Explanation

Q:

$17 \frac{4}{7} + 13 \frac{2}{12} - ? = 11 \frac{2}{3}$

A) 16	B) 17
C) 18	D) 19

Answer & Explanation Answer: D) 19

Explanation:

$17 \frac{4}{7} + 13 \frac{2}{12} - ? = 11 \frac{2}{3}$

$\frac{123}{7} + \frac{158}{12} - ? = \frac{35}{3}$

=> $? = \frac{123}{7} + \frac{158}{12} - \frac{35}{3}$

=> ? = $\frac{1476 + 1106 - 980}{84}$

=> ? = 19.07

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Filed Under: Simplification
Exam Prep: AIEEE , Bank Exams , CAT
Job Role: Bank Clerk , Bank PO

13 9073

Q:

45 men working 9 hours per day dig 40 m deep. How many extra men should be put to dig to a depth of 56 m working 7 hours per day ?

A) 36	B) 61
C) 48	D) 54

Answer & Explanation Answer: A) 36

Explanation:

(45 x 9)/40 = (M x 7)/56 => M = 81

=> 81 – 45 = 36

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Filed Under: Time and Work
Exam Prep: AIEEE , Bank Exams , CAT , GATE
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6 9072

Q:

The value of a machine depreciates at the rate of 10% every year. It was purchased 3 years ago. If its present value is Rs. 8,748, its purchase price was ?

A) Rs.14,000	B) Rs.12,000
C) Rs. 12,800	D) Rs. 15,000

Answer & Explanation Answer: B) Rs.12,000

Explanation:

We have P = 8748, R = -10 and n = 3
Therefore, the purchase price of the machine
= $\frac{P}{{(1 + \frac{R}{100})}^{n}}$ = $\frac{8748}{{(1 - \frac{10}{100})}^{3}}$
= 8748 x (100/90) x (100/90) x (100/90)
= 12,000.

Therefore, the purchase price of the machine was Rs. 12,000.

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Filed Under: Percentage
Exam Prep: TOEFL , GRE , GATE , CAT , Bank Exams , AIEEE
Job Role: Bank PO , Bank Clerk , Analyst

7 9071

Q:

An Engine length 1000 m moving at 10 m/s. A bird is flying from engine to end with 'x' m/s and coming back at '2x' m/s. Take total time of bird travelling as 187.5 s. Find x and 2x in km/hr ?

A) 21.4 and 42.8	B) 25.2 and 50.4
C) 31.4208 and 62.8416	D) 33.12 and 66.24

Answer & Explanation Answer: C) 31.4208 and 62.8416

Explanation:

Birds speeds in mtrs/sec is 'x' and '2x' .
While flying from Engine to end, relative speed = (x+10) m/sec
from end to engine, flying speed = (2x - 10) mtr/sec
so
1000/(x+10) + 1000/(2x-10) = 187.5 secs
solving it, we get
so x = 8.728 m/sec and 2x= 17.456 m/sec

x = 31.4208 km/hr and 2x = 62.8416 km/hr.

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Filed Under: Problems on Trains
Exam Prep: AIEEE , Bank Exams , CAT , GATE , GRE
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11 9067

Q:

A man covered a certain distance at some speed. Had he moved 3 kmph faster, he would have taken 40 minutes less. If he had moved 2 kmph slower, he would have taken 40 minutes more. The distance (in km) is

A) 30	B) 36
C) 40	D) 42

Answer & Explanation Answer: C) 40

Explanation:

Let distance = x km and usual rate = y kmph.
Then, x/y - x/(y+3) = 40/60 --> 2y (y+3) = 9x ----- (i)
Also, x/(y-2) - x/y = 40/60 --> y(y-2) = 3x -------- (ii)
On dividing (i) by (ii), we get:

x = 40 km.

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Filed Under: Time and Distance
Exam Prep: AIEEE , Bank Exams , CAT , GATE
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12 9059

Q:

In a simultaneous throw of two dice, what is the probability of getting a total of 7 ?

A) 1/6	B) 1/4
C) 1/20	D) 3/4

Answer & Explanation Answer: A) 1/6

Explanation:

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Filed Under: Probability

16 9057

Q:

There are eight boxes of chocolates, each box containing distinct number of chocolates from 1 to 8. In how many ways four of these boxes can be given to four persons (one boxes to each) such that the first person gets more chocolates than each of the three, the second person gets more chocolates than the third as well as the fourth persons and the third person gets more chocolates than fourth person?

A) 70	B) 40
C) 72	D) 80

Answer & Explanation Answer: A) 70

Explanation:

All the boxes contain distinct number of chocolates.
For each combination of 4 out of 8 boxes, the box with the greatest number has to be given to the first person, the box with the second highest to the second person and so on.

The number of ways of giving 4 boxes to the 4 person is: 8 $C_{4}$ = 70

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Filed Under: Permutations and Combinations

5 9049

Q:

If 4/7 of P is equal to 12/14 of Q, then what is P : Q?

A) 15 : 8	B) 14 : 9
C) 7 : 8	D) 3 : 2

Quantitative Aptitude - Arithmetic Ability Questions

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