6500+ Quantitative Aptitude Questions and Answers With Explanation

Q:

The ratio of the present ages of P and Q is 3 : 4. Five years ago, the ratio of their ages was 5 : 7. Find their present ages.

Answer

As the ratio of their present ages is 3 : 4 , let their present ages be 3X and 4X.

So, 5 years ago, as the ratio of their ages was 5 : 7, we can write (3x-5) : (4x-5) = 5 : 7. Solving, we get X = 10. Hence, their present ages are 3X = 30 and 4X = 40

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Subject: Problems on Ages

209 46776

Q:

If $\tan^{4} θ + \tan^{2} θ = 2$ , then $\sec^{4} θ - \sec^{2} θ = ?$

A) 0	B) 3
C) 2	D) 1

Answer & Explanation Answer: C) 2

Explanation:

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Filed Under: Simplification
Exam Prep: Bank Exams

0 46698

Q:

∆DEF is right angled at E. If tanD = 12/5, then what is the value of secF ?

A) 5/12	B) 13/5
C) 5/13	D) 13/12

Answer & Explanation Answer: D) 13/12

Explanation:

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Filed Under: Simplification
Exam Prep: Bank Exams

0 46661

Q:

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) 135	B) 63
C) 125	D) 64

Answer & Explanation Answer: B) 63

Explanation:

Required number of ways = $(7 C_{5} * 3 C_{2}) = (7 C_{2} * 3 C_{1}) = 63$

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Filed Under: Permutations and Combinations

52 46545

Q:

Find the Odd One Out?

3 24 186 1008 5040 20160

A) 24	B) 20160
C) 186	D) 5040

Answer & Explanation Answer: C) 186

Explanation:

The given number series is 3 24 186 1008 5040 20160

Here the given series follows a pattern that,

3

3 x 8 = 24

24 x 7 = 168 not 186

168 x 6 = 1008

1008 x 5 = 5040

5040 x 4 = 20160

Hence the Odd One in the number series is 186.

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Filed Under: Odd Man Out
Exam Prep: AIEEE , Bank Exams , CAT , GATE
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3 46318

Q:

ΔABC is right angled at B. If m∠C = 45 deg, then find the value of (cosecA - √3).

A) (4-√3)/2	B) √2-√3
C) -√3/2	D) (√6-1)/√3

Answer & Explanation Answer: B) √2-√3

Explanation:

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Filed Under: Simplification
Exam Prep: Bank Exams

0 46285

Q:

Three unbiased coins are tossed.What is the probability of getting at least 2 heads?

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

Answer & Explanation Answer: B) 1/2

Explanation:

Here S= {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}.

Let E = event of getting at least two heads = {THH, HTH, HHT, HHH}.

P(E) = n(E) / n(S)

= 4/8= 1/2

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Filed Under: Probability

92 46224

Q:

Solve the following :

$60 \div 5 \times (16 - 8 \div 2) \div 3 = ?$

A) 3	B) 48
C) 1	D) 1/3

Quantitative Aptitude - Arithmetic Ability Questions

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