6500+ Quantitative Aptitude Questions and Answers With Explanation

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 22552

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 22539

Q:

In a class of 78 students 41 are taking French, 22 are taking German. Of the students taking French or German, 9 are taking both courses. How many students are not enrolled in either course?

A) 6	B) 12
C) 24	D) 18

Answer & Explanation Answer: C) 24

Explanation:

You could solve this by drawing a Venn diagram. A simpler way is to realize that you can subtract the number of students taking both languages from the numbers taking French to find the number taking only French. Likewise find those taking only German. Then we have:Total = only French + only German + both + neither

78 = (41-9) + (22-9) + 9 + neither.

Not enrolled students = 24

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Filed Under: Average
Exam Prep: GRE

20 22535

Q:

If Mar 18th,1994 falls on Friday then Feb 25th,1995 falls on which day?

A) Wednesday	B) Monday
C) Saturday	D) Sunday

Answer & Explanation Answer: C) Saturday

Explanation:

First,we count the number of odd days for the left over days in the given period.Here,given period is 18.3.1994 to 25.2.1995

Month	Mar	Apr	May	Jun	Jul	Aug	Sep	Oct	Nov	Dec	Jan	Feb
Days	13	30	31	30	31	31	30	31	30	31	31	25
Odd Days	6	2	3	2	3	3	2	3	2	3	3	4

Therefore, No. of Odd Days = 6 + 2 + 3 + 2 + 3 + 3 + 2 + 3 + 2 + 3 + 3 + 4 = 36 = 1 odd day

So, given day Friday + 1 = Saturday is the required result.

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Filed Under: Calendar

103 22520

Q:

At what time, between 3 o’clock and 4 o’clock, both the hour hand and minute hand coincide each other ?

A) 3:16 7/11	B) 3:16 11/4
C) 3:30	D) 3:16 4/11

Answer & Explanation Answer: D) 3:16 4/11

Explanation:

Coincide means 00 angle.

This can be calculated using the formulafor time A to B means [11m/2 - 30 (A)]

Here m gives minutes after A the both hands coincides.

Here A = 3, B = 4

0 =11m/2 –30 × 3
11m = 90 × 2 = 180
m= 180/11 = 16 ⁴/₁₁

So time = 3 : 16 ⁴/₁₁

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

33 22517

Q:

What value should come in place of the question mark (?) in the following question?

${(?)}^{2} % of 650 = {(20)}^{2} + 4^{2}$

A) 8	B) 64
C) 8	D) 642

Answer & Explanation Answer: A) 8

Explanation:

$?^{2} = \frac{400 + 16}{650} x 100 ? = 8$

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Filed Under: Quantitative Aptitude - Arithmetic Ability
Exam Prep: Bank Exams

0 22514

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 22499

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 22493

Quantitative Aptitude - Arithmetic Ability Questions

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