# Compound Interest Question & Answers

## Compound Interest

Quantitative aptitude questions are asked in many competitive exams and placement exam. Compound Interest is a category in Quantitative Aptitude. Quantitative aptitude questions given here are extremely useful for all kind of competitive exams like Common Aptitude Test (CAT), MAT, GMAT, IBPS Exam, CSAT, CLAT, Bank Competitive Exams, ICET, UPSC Competitive Exams, CLAT, SSC Competitive Exams, SNAP Test, KPSC, XAT, GRE, Defense Competitive Exams, L.I.C/ G. I.C Competitive Exams, Railway Competitive Exam, TNPSC, University Grants Commission (UGC), Career Aptitude Test (IT Companies) and etc., Government Exams etc.

We have a large database of "Compound Interest" Problems answered with explanation. These will help students who are preparing for all types of competitive examinations.

A sum of money lent at compound interest for 2 years at 20% per annum would fetch Rs.482 more, if the interest was payable half yearly than if it was payable annually . The sum is

 A) 10000 B) 20000 C) 40000 D) 50000

Explanation:

Let sum=Rs.x

C.I. when compounded half yearly = $\inline \fn_jvn \left [ x(1+\frac{10}{100})^{4}-x \right ]=\frac{4641}{10000}x$

C.I. when compounded annually =$\inline \fn_jvn \left [ x(\frac{20}{100})^{2}-x \right ]=\frac{11}{25}x$

$\inline \fn_jvn \therefore \frac{4641}{10000}x-\frac{11}{25}x=482$

=> x=20000

Subject: Compound Interest - Quantitative Aptitude - Arithmetic Ability
Exam Prep: Bank Exams
Job Role: Bank PO

39

A sum of money amounts to Rs.6690 after 3 years and to Rs.10,035 after 6 years on compound interest.find the sum.

 A) 4360 B) 4460 C) 4560 D) 4660

Explanation:

Let the sum be Rs.P.then
P(1+R/100)^3=6690…(i) and P(1+R/100)^6=10035…(ii)
On dividing,we get (1+R/100)^3=10025/6690=3/2.
Substituting this value in (i),we get:
P*(3/2)=6690 or P=(6690*2/3)=4460
Hence,the sum is rs.4460.

Subject: Compound Interest - Quantitative Aptitude - Arithmetic Ability
Exam Prep: Bank Exams
Job Role: Bank PO

18

What is the rate of interest p.c.p.a.?

I. An amount doubles itself in 5 years on simple interest.

II. Difference between the compound interest and the simple interest earned on a certain amount in 2 years is Rs. 400.

III. Simple interest earned per annum is Rs. 2000

 A) I only B) II and III only C) All I, II and III D) I only or II and III only

Answer & Explanation Answer: D) I only or II and III only

Explanation:

$\inline \fn_jvn I. \frac{P\times R\times 5}{100}=P\Rightarrow R=20$

$\inline \fn_jvn II.P\left ( 1+\frac{R}{100} \right )^2-P-\frac{P\times R\times 2}{100}=400\Rightarrow pR^{2}=4000000$

$\inline \fn_jvn III.\frac{P\times R\times 1}{100}=2000\Rightarrow PR=200000$

$\inline \fn_jvn \therefore \frac{PR^{2}}{PR}=\frac{4000000}{200000}\Rightarrow R=20$

Thus I only or (II and III) give answer.

$\inline \fn_jvn \therefore$ Correct answer is (D)

Subject: Compound Interest - Quantitative Aptitude - Arithmetic Ability
Exam Prep: Bank Exams
Job Role: Bank PO

15

What annual payment will discharge a debt of Rs.7620 due in 3years at$\inline 16\frac{2}{3}$ % per annum interest?

 A) 5430 B) 4430 C) 3430 D) 2430

Explanation:

Let each installment be Rs.x. Then,

(P.W. of Rs.x due 1 year hence) + (P.W of Rs.x due 2 years hence) + (P.W of Rs. X due 3 years hence) = 7620.$\inline \left [ \frac{x}{1+(\frac{50}{3\times 100})} \right ]+\left [ \frac{x}{1+(\frac{50}{3\times 100})^2} \right ]+\left [ \frac{x}{1+(\frac{50}{3\times 100})^3} \right ]=7620$

$\inline \frac{6x}{7}+\frac{36x}{49}+\frac{216x}{343}=7620$

$\inline 294x+252x+216x=7620\times 343$

$\inline \fn_jvn \Rightarrow x=3430$

Amount of each installment = Rs.3430

Subject: Compound Interest - Quantitative Aptitude - Arithmetic Ability
Exam Prep: Bank Exams
Job Role: Bank PO

34

The population of a town was 3600 three years back. It is 4800 right now. What will be the population three years down the line, if the rate of growth of population has been constant over the years and has been compounding annually?

 A) Rs.600 B) Rs,6400 C) Rs.6500 D) Rs.6600

Explanation:

The population grew from 3600 to 4800 in 3 years. That is a growth of 1200 on 3600 during three year span.

Therefore, the rate of growth for three years has been constant.

The rate of growth during the next three years will also be the same.

Therefore, the population will grow from 4800 by $\inline \fn_jvn 4800\times \frac{1}{3}$= 1600

Hence, the population three years from now will be 4800 + 1600 = 6400