6500+ Quantitative Aptitude Questions and Answers With Explanation

Q:

Select the correct option:

Convert decimal 101 to binary .

A) 1101001	B) 1100111
C) 1101011	D) 1100101

Answer & Explanation Answer: D) 1100101

Explanation:

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Filed Under: Numbers
Exam Prep: Bank Exams

0 22421

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 22392

Q:

If a=2b/3, b=2c/3, and c=2d/3, then find the ratio of b and d:

A) 8/9	B) 4/9
C) 4/3	D) 5/27

Answer & Explanation Answer: B) 4/9

Explanation:

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Filed Under: Ratios and Proportions
Exam Prep: Bank Exams

3 22373

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 22351

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 22346

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 22333

Q:

In a competitive examination in State A, 6% candidates got selected from the total appeared candidates. State B had an equal number of candidates appeared and 7% candidates got selected with 80 more candidates got selected than A. What was the number of candidates appeared from each State ?

A) 4000	B) 8000
C) 12000	D) 16000

Answer & Explanation Answer: B) 8000

Explanation:

Let the number of candidates appeared from each state be x.

In state A, 6% candidates got selected from the total appeared candidates

In state B, 7% candidates got selected from the total appeared candidates

But in State B, 80 more candidates got selected than State A

From these, it is clear that 1% of the total appeared candidates in State B = 80

=> total appeared candidates in State B = 80 x 100 = 8000

=> total appeared candidates in State A = total appeared candidates in State B = 8000

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Filed Under: Percentage

54 22317

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

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