# Compound Interest Questions

Q:

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

60 12991
Q:

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.

28 9550
Q:

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)

21 8773
Q:

Find compound interest on Rs. 8000 at 15% per annum for 2 years 4 months, compounded annually

 A) 2109 B) 3109 C) 4109 D) 6109

Explanation:

Time = 2 years 4 months = 2(4/12) years = 2(1/3) years.
Amount = Rs'. [8000 X (1+(15/100))^2 X (1+((1/3)*15)/100)]
=Rs. [8000 * (23/20) * (23/20) * (21/20)]
= Rs. 11109. .
:. C.I. = Rs. (11109 - 8000) = Rs. 3109.

19 8093
Q:

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