6500+ Quantitative Aptitude Questions and Answers With Explanation

Q:

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) 209	B) 290
C) 200	D) 208

Answer & Explanation Answer: A) 209

Explanation:

We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).

Required number of ways = $(6 C_{1} * 4 C_{3}) + (6 C_{2} * 4 C_{2}) + (6 C_{3} * 4 C_{1}) + 6 C_{4}$

= $(6 C_{1} * 4 C_{1}) + (6 C_{2} * 4 C_{2}) + (6 C_{3} * 4 C_{1}) + 6 C_{2}$ = 209.

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Filed Under: Permutations and Combinations

11 22697

Q:

A letter is takenout at random from 'ASSISTANT' and another is taken out from 'STATISTICS'. The probability that they are the same letter is :

A) 35/96	B) 19/90
C) 19/96	D) None of these

Answer & Explanation Answer: B) 19/90

Explanation:

$A S S I S T A N T \to A A I N S S S T T$

$S T A T I S T I C S \to A C I I S S S T T T$

Here N and C are not common and same letters can be A, I, S, T. Therefore

Probability of choosing A = $\frac{2_{C_{1}}}{9_{C_{1}}} \times \frac{1_{C_{1}}}{10_{C_{1}}}$ = 1/45

Probability of choosing I = $\frac{1}{9_{C_{1}}} \times \frac{2_{C_{1}}}{10_{C_{1}}}$ = 1/45

Probability of choosing S = $\frac{3_{C_{1}}}{9_{C_{1}}} \times \frac{3_{C_{1}}}{10_{C_{1}}}$ = 1/10

Probability of choosing T = $\frac{2_{C_{1}}}{9_{C_{1}}} \times \frac{3_{C_{1}}}{10_{C_{1}}}$ = 1/15

Hence, Required probability = $\frac{1}{45} + \frac{1}{45} + \frac{1}{10} + \frac{1}{15} = \frac{19}{90}$

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

97 22694

Q:

In a race of 1000 meters, A can beat B by 100 meters, in a race of 800 meters, B can beat C by 100 meters. By how many meters will A beat C in a race of 600 meters?

A) 125.5 meters	B) 126.5 meters
C) 127.5 meters	D) 128.5 meters

Answer & Explanation Answer: C) 127.5 meters

Explanation:

When A runs 1000 meters, B runs 900 meters and when B runs 800 meters, C runs 700 meters.

Therefore, when B runs 900 meters, the distance that C runs = (900 x 700)/800 = 6300/8 = 787.5 meters.

So, in a race of 1000 meters, A beats C by (1000 - 787.5) = 212.5 meters to C.

So, in a race of 600 meters, the number of meters by Which A beats C = (600 x 212.5)/1000 = 127.5 meters.

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Filed Under: Races and Games

30 22691

Q:

By mixing two qualities of pulses in the ratio 2: 3 and selling the mixture at the rate of Rs 22 per kilogram, a shopkeeper makes a profit of 10 %. If the cost of the smaller quantity be Rs 14 per kg, the cost per kg of the larger quantity is:

A) Rs 23	B) Rs 25
C) Rs 24	D) Nonof these

Answer & Explanation Answer: C) Rs 24

Explanation:

Cost Price of 5 kg = Rs.(14*2 + x*3) = (28 + 3x).

Sell price of 5 kg = Rs. (22x5) = Rs. 110.

[{110 - (28 + 3x)}/(28 + 3x) ]* 100 =10

[82-3x/28 + 3x]= 1 / 10

820 - 30x = 28 +3x

33x = 792

x = 24

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Filed Under: Profit and Loss

59 22677

Q:

In what time will Rs. 1000 become Rs. 1331 at 10% per annum compounded annually?

A) 1years	B) 2years
C) 3years	D) 4years

Answer & Explanation Answer: C) 3years

Explanation:

Principal = Rs. 1000; Amount = Rs. 1331; Rate = 10% p.a. Let the time be n years. Then,
[ 1000 (1+ (10/100))^n ] = 1331 or (11/10)^n = (1331/1000) = (11/10)^3
n = 3 years

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Filed Under: Compound Interest
Exam Prep: Bank Exams
Job Role: Bank PO

39 22646

Q:

What decimal of an hour is a second

A) .0028	B) .0027
C) .0026	D) .0025

Answer & Explanation Answer: B) .0027

Explanation:

1 / (60 * 60) = 1 / 3600 = .0027

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Filed Under: Decimal Fractions

53 22636

Q:

Which of the following numbers is composite?

A) 47	B) 57
C) 37	D) 67

Answer & Explanation Answer: A) 47

Explanation:

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Filed Under: Problems on Numbers
Exam Prep: Bank Exams

0 22620

Q:

The average age of 15 students of a class is 15 years. Out of these, the average age of 5 students is 14 years and that of the other 9 students is 16 years, The age of the 15th student is :

A) 11	B) 12
C) 13	D) 14

Quantitative Aptitude - Arithmetic Ability Questions

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