6500+ Quantitative Aptitude Questions and Answers With Explanation

Q:

Sharon Stone deposits $500 at the beginning of each 3 months in an account earning 10% compounded quarterly. Determine how much money she has after 25 years

A) 22681.19	B) 23456
C) 44577	D) 25685

Answer & Explanation Answer: A) 22681.19

Explanation:

S=R[(1+i)^n-1)/i]

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0 3440

Q:

A retirement benefit of $12,000 is to be paid every 6 months for 25 years at interest rate of 7% compounded semi-annually. Find

(a) the present value to fund the end-of-period retirement benefit.

(b) the end-of-period semi-annual payment needed to accumulate the value in part (a) assuming regular investments for 30 years in an account yielding 8% compounded semi-annually

A) 245678	B) 234689
C) 281468.06	D) 234578

Answer & Explanation Answer: C) 281468.06

Explanation:

R=S[i/(1+i)^n-1]

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0 3431

Q:

A dealer sells two machines at Rs 12000 each. On one it gains 32% and on the other it looses 32%. What is its profit/loss percentage in the whole transaction?

A) No gain and no loss	B) 1% loss
C) 18% profit	D) 10.24% loss

Answer & Explanation Answer: D) 10.24% loss

Explanation:

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Filed Under: Profit and Loss
Exam Prep: Bank Exams

3 3430

Q:

$[\frac{3 \sqrt{2}}{\sqrt{6} - \sqrt{3}} - \frac{4 \sqrt{3}}{\sqrt{6} - \sqrt{2}} - \frac{6}{\sqrt{8} - \sqrt{12}}] = ?$

A) sqtr{3}-sqrt{2}	B) sqtr{3}+sqrt{2}
C) 5sqrt{3}	D) 1

Answer & Explanation Answer: C) 5sqrt{3}

Explanation:

Given Exp = $\frac{3 \sqrt{2}}{\sqrt{6} - \sqrt{3}} \times \frac{\sqrt{6} + \sqrt{3}}{\sqrt{6} + \sqrt{3}} - \frac{4 \sqrt{3}}{\sqrt{6} - \sqrt{2}} \times \frac{\sqrt{6} + \sqrt{2}}{\sqrt{6} + \sqrt{2}} - \frac{6}{2 (\sqrt{2} - \sqrt{3})}$

= $\frac{3 \sqrt{2} (\sqrt{6} + \sqrt{3})}{6 - 3} - \frac{4 \sqrt{3} (\sqrt{6} + \sqrt{2})}{6 - 2} + \frac{3}{(\sqrt{3} - \sqrt{2})} \times \frac{(\sqrt{3} + \sqrt{2})}{(\sqrt{3} + \sqrt{2})}$

= $\sqrt{2} (\sqrt{6} + \sqrt{3}) - \sqrt{3} (\sqrt{6} + \sqrt{2}) + 3 (\sqrt{3} + \sqrt{2})$

= $\sqrt{12} + \sqrt{6} - \sqrt{18} - \sqrt{6} + 3 \sqrt{3} + 3 \sqrt{2}$

= $5 \sqrt{3}$

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4 3423

Q:

In a joint family, there are 24 members whose age is decreased by 4 months, when one girl aged 24 years is replaced by a new boy. Find the age of the new boy ?

A) 18 years	B) 12 years
C) 15 years	D) 16 years

Answer & Explanation Answer: D) 16 years

Explanation:

Age of new boy = (Age of moved girl) – (Number of family members × Decreased in average)

= 24 – 24×4/12

= 24 – 8

= 16 years.

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7 3421

Q:

For a nominal interest rate of 8.4%, what is the compounding frequency if the periodic interest rate is:2.1

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

Answer & Explanation Answer: D) 4

Explanation:

i=j/m

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Q:

What amount must you invest now at 4% compounded monthly to accumulate $10,000 after 3 year

A) 8695	B) 7695
C) 3695	D) 4695

Answer & Explanation Answer: A) 8695

Explanation:

Given: j = 4%, m= 12, FV = $10,000, Term= 3.5 years
Then n =m *Term 12(3.5) 42

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Q:

A certain sum is invested for 2 years in scheme M at 20% p.a. compound interest (compounded annually), Same sum is also invested for the same period in scheme N at k% p.a. simple interest. The interest earned from scheme M is twice of that earned from scheme N. What is the value of k?

A) 7	B) 11
C) 9	D) 13

Answer & Explanation Answer: B) 11

Explanation:

Interest earned in scheme M =

$P [{(1 + \frac{20}{100})}^{2} - 1] = \frac{11 P}{25}$

Interest earned in scheme N =

$\frac{P x k x 2}{100} = \frac{P k}{50}$

Now, from the given data,

$\frac{11 P}{25} = 2 x \frac{P k}{50}$

k = 11

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20 3417

Quantitative Aptitude - Arithmetic Ability Questions

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