299+ Compound Interest Questions and Answers

Q:

The difference between the simple interest on a certain sum at the rate of 10%per annum for 2 years and compound interest which is compounded every 6 months is Rs.124.05. what is the principal sum

A) Rs.6000	B) Rs.8000
C) Rs.12000	D) none of these

Answer & Explanation Answer: B) Rs.8000

Explanation:

Compound Interest on P at 10% for 2 years when interest is compounded half-yearly

= $P {(1 + \frac{\frac{R}{2}}{100})}^{2 T} - P = P {(1 + \frac{1}{20})}^{4} - P = P {(\frac{21}{20})}^{4} - P$

Simple Interest on P at 10% for 2 years = $\frac{P R T}{100} = \frac{P \times 10 \times 2}{100} = \frac{P}{5}$

Given that difference between compound interest and simple interest = 124.05

$P * {(\frac{21}{20})}^{4} - P - \frac{P}{5} = 124.05$

=> $P [{(\frac{21}{20})}^{4} - 1 - \frac{1}{5}] = 124.05$

P=8000

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Q:

The difference between the compound interest and the simple interest accrued on an amount of Rs. 18,000 in 2 years was Rs. 405. What was the rate of interest p.c.p.a. ?

A) 12%	B) 13%
C) 14%	D) 15%

Answer & Explanation Answer: D) 15%

Explanation:

Let the rate be R% p.a. Then,

$[1800 {(1 + \frac{R}{100})}^{2} - 1800] - (\frac{1800 * R * 2}{100}) = 405$

$1800 [\frac{{(100 + R)}^{2}}{10000} - 1 - \frac{2 R}{100}] = 405$

$1800 [\frac{{(100 + R)}^{2} - 10000 - 200 R}{10000}] = 405$

$\Leftrightarrow$

$\frac{9}{5} * R^{2} = 405 \Leftrightarrow R^{2} = (\frac{405 * 5}{9}) = 225 \Leftrightarrow R = 15$
Rate = 15%.

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Q:

The difference between the compound interest and the simple interest on a certain sum at 12% p.a. for two years is Rs.90. What will be the value of the amount at the end of 3 years?

A) 8560	B) 8673
C) 8746	D) 8780.80

Answer & Explanation Answer: D) 8780.80

Explanation:

when interest is reckoned using compound interest, interest being compounded annually. The difference in the simple interest and compound interest for two years is on account of the interest paid on the first year's interest Hence 12% of simple interest = 90 => simple interest =90/0.12 =750.

As the simple interest for a year = 750 @ 12% p.a., the principal =750/0.12 = Rs.6250.

If the principal is 6250, then the amount outstanding at the end of 3 years = 6250 + 3(simple interest on 6250) + 3 (interest on simple interest) + 1 (interest on interest on interest) = 6250 +3(750) + 3(90) + 1(10.80) = 8780.80.

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Q:

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

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

Answer & Explanation Answer: C) 3430

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.

$[\frac{x}{(1 + \frac{50}{3 x 100})}] + [\frac{x}{1 + {(\frac{50}{3 x 100})}^{2}}] + [\frac{x}{1 + {(\frac{50}{3 x 100})}^{3}}] = 7620$

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

$294 x + 252 x + 216 x = 7620 x 343$

=> x = 3430

Amount of each installment = Rs.3430

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Q:

Divide Rs. 1301 between A and B, so that the amount of A after 7 years is equal to the amount of B after 9 years, the interest being compounded at 4% per annum.

A) Rs.625	B) Rs.626
C) Rs.286	D) Rs.627

Answer & Explanation Answer: A) Rs.625

Explanation:

Let the two parts be Rs. x and Rs. (1301 - x).

$x {(1 + \frac{4}{100})}^{7} = (1301 - x) {(1 + \frac{4}{100})}^{9}$

$\frac{x}{(1301 - x)} = {(1 + \frac{4}{100})}^{2} = (\frac{26}{25} * \frac{25}{26})$

=> 625x=676(1301-x)

1301x=676 x 1301x=676.
So,the parts are rs.676 and rs.(1301-676)i.e rs.676 and rs.625

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Q:

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

Answer & Explanation Answer: B) Rs,6400

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 $4800 * \frac{1}{3}$ = 1600

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

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Q:

What will Rs.1500 amount to in three years if it is invested in 20% p.a. compound interest, interest being compounded annually?

A) Rs.2592	B) Rs.2492
C) Rs.2352	D) Rs.2352

Answer & Explanation Answer: A) Rs.2592

Explanation:

The usual way to find the compound interest is given by the formula A = .p(1+r/100)^n

In this formula,

A is the amount at the end of the period of investment

P is the principal that is invested

r is the rate of interest in % p.a

And n is the number of years for which the principal has been invested.

In this case, it would turn out to be A =1500(1+20/100)^3

= 2592.

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Q:

The difference between the compound interest and simple interest on a certain sum at 10% per annum for 2 years is Rs. 631. Find the sum.

A) 60100	B) 61100
C) 62100	D) 63100

Answer & Explanation Answer: D) 63100

Explanation:

Let the sum be Rs. x. Then,
C.I. = x ( 1 + ( 10 /100 ))^2 - x = 21x / 100 ,
S.I. = (( x * 10 * 2) / 100) = x / 5
(C.I) - (S.I) = ((21x / 100 ) - (x / 5 )) = x / 100
( x / 100 ) = 632 * x = 63100.
Hence, the sum is Rs.63,100.

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