299+ Compound Interest Questions and Answers

Q:

Determine the nominal rate of interest if:

The periodic rate is 0.83% per month

A) 7%	B) 8%
C) 9%	D) 10%

Answer & Explanation Answer: D) 10%

Explanation:

j=mi

Report Error

View Answer Report Error Discuss

Filed Under: Compound Interest
Exam Prep: Bank Exams
Job Role: Bank PO

2 28610

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.

Report Error

View Answer Report Error Discuss

Filed Under: Compound Interest
Exam Prep: Bank Exams
Job Role: Bank PO

20 26865

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%.

Report Error

View Answer Report Error Discuss

Filed Under: Compound Interest
Exam Prep: Bank Exams
Job Role: Bank PO

22 26577

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

Report Error

View Answer Report Error Discuss

Filed Under: Compound Interest
Exam Prep: Bank Exams
Job Role: Bank PO

78 26480

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

Report Error

View Answer Report Error Discuss

Filed Under: Compound Interest
Exam Prep: Bank Exams
Job Role: Bank PO

33 25609

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

Report Error

View Answer Report Error Discuss

Filed Under: Compound Interest
Exam Prep: Bank Exams
Job Role: Bank PO

30 24526

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.

Report Error

View Answer Report Error Discuss

Filed Under: Compound Interest
Exam Prep: Bank Exams
Job Role: Bank PO

27 24492

Q:

The principal that amounts to Rs.4913 in 3 years at $4 \frac{1}{3}$ % per annum compound interest compounded annually is

A) Rs.3096	B) Rs.4076
C) Rs.4085	D) Rs.4096

Answer & Explanation Answer: D) Rs.4096

Explanation:

principal = $A = 4913, T = 3, R = 4 \frac{1}{3}$

$P = {\frac{A}{(\frac{1 + R}{100})}}^{T}$

= Rs.4096

Report Error

View Answer Report Error Discuss

Filed Under: Compound Interest
Exam Prep: GRE

30 23930

Compound Interest Questions

Quick Links