299+ Compound Interest Questions and Answers

Q:

Rs 1000 is being charged at 50% per annum. what is the interest for 3rd year at compound interest?

A) 1122	B) 1025
C) 1125	D) 625

Answer & Explanation Answer: C) 1125

Explanation:

Total amount for two years = $1000 \times {(1.5)}^{2} = 2250$

Now , interest for third year = $\frac{2250 \times 50 \times 1}{100} = 1125$

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16 9953

Q:

At a certain rate of interest the compound interest of 3 years and simple interest of 5 years for certain sum of money is respectively Rs. 1513.2 and Rs. 2400. Find the common rate of interest per annum ?

A) 5%	B) 6%
C) 4%	D) 3%

Answer & Explanation Answer: A) 5%

Explanation:

Given compound interest for 3 years = Rs. 1513.2

and simple interest for 5 years = Rs. 2400

Now, we know that C.I = $[P {(1 + \frac{R}{100})}^{n} - 1]$

=> 1513.2 = $[P {(1 + \frac{R}{100})}^{3} - 1]$ ...........(A)

And S.I = PTR/100

=> 2400 = P5R/100 ..................(B)

By solving (A) & (B), we get

R = 5%.

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17 9566

Q:

Mr. Gupta borrowed a sum of money on compound interest. What will be the amount to
be repaid if he is repaying the entire amount at the end of 2 years?
I. The rate of interest is 5 p.c.p.a.
II. Simple interest fetched on the same amount in one year is Rs. 600.
III. The amount borrowed is 10 times the simple interest in 2 years.

A) .I only	B) III only
C) I or II	D) II and Either I or III only

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

Explanation:

I. gives, Rate = 5% p.a.

II. gives, S.I. for 1 year = Rs. 600.

III. gives, sum = 10 x (S.I. for 2 years).

Now I, and II give the sum.

For this sum, C.I. and hence amount can be obtained.

Thus, III is redundant.

Again, II gives S.I. for 2 years = Rs. (600 x 2) = Rs. 1200.

Now, from III, Sum = Rs. (10 x 1200) = Rs . 12000.

Thus,Rate = $\frac{100 * 1200}{2 * 12000}$ =5%

Thus, C.I. for 2 years and therefore, amount can be obtained.

Thus, I is redundant.

Hence, I or III redundan

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5 9385

Q:

Sonika invested an amount of Rs.5800 for 2 years. At what rate of compound interest will she get an amount of Rs.594.50 at the end of 2 years ?

A) 6.5%	B) 6%
C) 5%	D) 4.5%

Answer & Explanation Answer: C) 5%

Explanation:

C.I = $P [{(1 + \frac{r}{100})}^{t} - 1]$

594.5 = $5800 [{(1 + \frac{r}{100})}^{2} - 1]$

$\frac{594.5}{5800} + 1 = {(1 + \frac{r}{100})}^{2}$

$\frac{r}{100} = 1.05 - 1 = 0.05 = > r = 5$ % .

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13 9088

Q:

The minimum number of half years in which a sum of money kept at 60% compound interest will be quadrupled is ?

A) 2	B) 4
C) 3	D) 6

Answer & Explanation Answer: D) 6

Explanation:

Let us assume Amount be 100 Rs and we calculate in CI

First year 60% of 100 = 60 amount (100+60) is 160

Second year 60% of 160 = 96 amount (160+96) is 256

Third year 60% of 256 =153.6 amount (256+153.6) is 409.6

Here the Amount of 100 Rs is quadrapled in 3 years.

One year contains 2 half years

Three year has 6 half years.

Therefore, It takes 6 half years.

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14 8697

Q:

There is 50% increase in an amount in 5 years at simple interest. What will be the compound interest of Rs. 12,000 after 3 years at the same rate ?

A) Rs. 5422	B) Rs. 5489
C) Rs. 3972	D) Rs. 6145

Answer & Explanation Answer: C) Rs. 3972

Explanation:

Let p = Rs. 100.

Then, S.I is Rs. 50 and time = 5 years.

$∴ R = \frac{100 \times 50}{100 \times 5}$ = 10% p.a.

Now, p = Rs. 12,000 , T = 3 years and R = 10% p.a.

C.I. = Rs. $12000 \times \{{(1 + \frac{10}{100})}^{3} - 1\}$ = Rs. 3972

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6 8484

Q:

If you deposit $8000 into an account paying 7% annual interest compounded quarterly, how long until there is $12400 in the account?

A) 2.3	B) 3.3
C) 4.3	D) 6.3

Answer & Explanation Answer: D) 6.3

Explanation:

FV=P(1+r/n)^nt

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6 8458

Q:

Find the principal if the interest compounded at the rate of 10% per annum for two years is Rs. 420 ?

A) Rs. 2000	B) Rs. 2100
C) Rs. 1900	D) Rs. 2150

Answer & Explanation Answer: A) Rs. 2000

Explanation:

Given,
Compound rate, R = 10% per annum
Time = 2 years
C.I = Rs. 420
Let P be the required principal.
A = (P+C.I)
Amount, A = $P {(1 + (\frac{r}{100}))}^{n}$

(P+C.I) = $P {[1 + (\frac{10}{100})]}^{2}$

(P+420) = P[11/10][11/10]

P-1.21P = -420

0.21P = 420

Hence, P = 420/0.21 = Rs. 2000

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Compound Interest Questions

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