299+ Compound Interest Questions and Answers

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:

$I . \frac{P * R * 5}{100} = P \Leftrightarrow R = 20$

$I I . P {(1 + \frac{R}{100})}^{2} - P - \frac{P * R * 2}{100} = 400 = > p R^{2} = 4000000$

$I I I . \frac{P * R * 1}{100} = 2000 = > P R = 200000$

$\frac{P R^{2}}{P R} = \frac{4000000}{200000} \Leftrightarrow R = 20$

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

Correct answer is (D)

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Q:

The compound interest on rs.30000 at 7% per annum is Rs.4347. The period is

A) 2 years	B) 2.5 years
C) 3 years	D) 4 years

Answer & Explanation Answer: A) 2 years

Explanation:

Amount = Rs.(30000+4347) = Rs.34347

let the time be n years

Then,30000(1+7/100)^n = 34347

(107/100)^n = 34347/30000 = 11449/10000 = (107/100)^2

$\therefore$ n = 2years

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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

Answer & Explanation Answer: B) 4460

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.

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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

Answer & Explanation Answer: B) 3109

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.

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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

Answer & Explanation Answer: B) 20000

Explanation:

Let sum=Rs.x

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

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

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

=> x=20000

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Q:

Find the compound interest on Rs. 10,000 in 2 years at 4% per annum, the interest being compounded half-yearly.

A) 524.32	B) 624.32
C) 724.32	D) 824.32

Answer & Explanation Answer: D) 824.32

Explanation:

Principal = Rs. 10000; Rate = 2% per half-year; Time = 2 years = 4 half-years.
Amount == Rs. 10824.32.

$R s . [10000 * {(1 + \frac{2}{100})}^{4}] = R s . [10000 * \frac{51}{50} * \frac{51}{50} * \frac{51}{50} * \frac{51}{50}]$

$\therefore$ C.I. = Rs. (10824.32 - 10000) = Rs. 824.32.

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Q:

Find the compound interest on Rs. 16,000 at 20% per annum for 9 months, compounded quarterly

A) 2422	B) 2522
C) 2622	D) 2722

Answer & Explanation Answer: B) 2522

Explanation:

Principal = Rs. 16000; Time = 9 months =3 quarters;
Rate = 20% per annum = 5% per quarter.
Amount = Rs. [16000 x (1+(5/100))3] = Rs. 18522.

CJ. = Rs. (18522 - 16000) = Rs. 2522

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Q:

Reena took a loan of Rs. 1200 with simple interest for as many years as the rate of interest. If she paid Rs. 432 as interest at the end of the loan period, what was the rate of interest?

A) 3.6	B) 6
C) 18	D) Cannot be determined

Answer & Explanation Answer: B) 6

Explanation:

Explanation:

Let rate = R% and time = R years.

Then, $(\frac{1200 * R * R}{100}) = 432$

$12 R^{2} = 432$

=> $R^{2} = 36 \Leftrightarrow R = 6$

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