# Compound Interest Questions and Answers updated daily – Aptitude

Compound Interest Questions: Solved 161 Compound Interest Questions and answers section with explanation for various online exam preparation, various interviews, Aptitude Category online test. Category Questions section with detailed description, explanation will help you to master the topic.

## Compound Interest Questions

101. The population of a town is 196000. It increases by 7% in the 1st year and decreases by 5% in the 2nd year. What is the population of the town at the end of 2 years?

Correct Ans:199234
Explanation:
Initial population is 196000
In First Year, population increasesby 7%
New population
= (107/100) x 196000
= 107 * 196000 / 100
= 107 * 1960
= 209720
Population after 1 year = 209720

In second year, Population decreases by 5%,
New population
= (100 - 5)/100 x 209720
= (95/100) x 209720
= 19 * 10486
= 199234
Population after 2 years will be 199234.
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102. A sum of money placed at compound interest doubles itself in 5 years .It will amount to eight times itself at the same rate of interest in:

Correct Ans:15 years
Explanation:
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103. A sum of money is borrowed and paid back in two annual instalments of Rs.882 each allowing 5% compound interest .The sum borrowed was:

Correct Ans:Rs.1620
Explanation:
Given
The sum borrowed
Present Worth of Rs.882 due 1 year + Present Worth of Rs.882 due 2 year
=> ( 882 ) / 1 + ( 5 / 100) ^ 1 + ( 882) / 1 + ( 5 / 100) ^ 2
=> (882 / 105 * 100 ) ^ 1 + (882 / 105 * 100 ) ^ 2
=> ( 882 /( 21 / 20 ) + ( 882 / (21 / 20) ^ 2
=> ( 882 * 20) / (21) + ( 882 * 20 * 20 / 21 * 21 )
=> 42 * 20 + 42 * 20 * 20 / 21
=> 840 + 2 * 20 * 20
=> 840 + 800
=> 1640
The sum borrowed = Rs.1640
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104. What annual payment will discharge a debt of Rs.1025 due in 2 years at the rate of 5% compound interest ?

Correct Ans:Rs.551.25
Explanation:
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105. The least number of complete years in which a sum of money put out at 20% compounded interest will be more than doubled is:

Correct Ans:4
Explanation:
Given,
R = 20%
Compound interest (C.I) = P [1 + (r/100)]^n
=> P [1 + (20/100)]^n > 2P
=> (120/100)^n > 2
=> (6/5)^n > 2
=> {(6/5) * (6/5) * (6/5) * (6/5)} > 2
=> (6/5)^4 = 2.0736 > 2
So, n = 4 years.
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106. A sum of money invested at compound interest amounts to Rs.800 in 3 years and to Rs.840 in 4 years. The rate of interest p.a. is:

Correct Ans:5%
Explanation:
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107. The effective annual rate of interest corresponding to a nominal rate at 6% p.a. payable half yearly is:

Correct Ans:6.09%
Explanation:
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108. The difference between the simple interest on a certain sum at the rate of 10% p.a. for 2 years and compound interest which is compounded every 6 months is Rs.124.05 .What is the principal sum?

Correct Ans:Rs.8000
Explanation:
Let the sum be P
Compound Interest on P at 10% for 2 years when interest is compounded half-yearly
=P(1+ (R / 2)100) 2T−P= P(1+(10 / 2)100) 2× 2− P
=P(1+120)4−P=P(21 / 20) 4− P
Simple Interest on P at 10% for 2 years
=P *R* T /100
=P×10×2100
= P / 5
Then P[(1+5 / 100)^4-1] - P x 10 x 2/100 = 124.05 (note: ^4 is the power of 4)
⇒ P[(21/20)^4 - 1 - 1/4] = 124.05
⇒ P[(194481/160000) - (6/5)] = 12405 /100
⇒ P[194481-192000 / 160000] = 12405 /100
⇒ P = [(12405/100) x (160000/2481)]
= 124.05 x 64.490
= 7999.9845
= 8000.
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109. The difference between compound interest and simple interest compounded annually on a certain sum of money for 2 years at 4% p.a. is Re.1 The sum (in Rs) is:

Correct Ans:625
Explanation:
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110. The difference between compound interest and simple interest on an amount of Rs.15,000 for 2 years in Rs.96 .What is the rate of interest per annum?

Correct Ans:8
Explanation:
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111. There is 60% increase in an amount in 6 years at simple interest. What will be the compound interest of Rs.12,000 after 3 years at the same rate?

Correct Ans:Rs.3972
Explanation:
Let the Principal be, P = Rs. 100

Given, S.I = 60% of 100 = Rs. 60, n = 6 years
Then, Rate of Interest, r = (S.I * 100 )/ (p * n)
=> r = (60 * 100) / (100 * 6)
=> r = 10 % p.a

Now, P = Rs. 12,000, n = 3 years, r = 10% p.a

C.I = P {[1 + (r/100)]n - 1}
= 12,000 * {[1 + (10/100)]3-1}
= 12,000 * [(11/10)3-1]
= 12,000 * [(1331/1000) - 1]
= 12,000 * (331/1000)
= 12 * 331
= 3972

Thus, Compound Interest = Rs. 3972
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112. The compound interest on a certain sum for 2 years at 10% p.a. is Rs.525.The simple interest on the same sum for double the time at the half the rate percent per annum is:

Correct Ans:Rs.500
Explanation:
Given, n = 2 years
r = 10 %
Compound interest (C.I) = Rs. 525
Compound interest (C.I) = P {[1 + (r/100)] ^n – 1}
=> P {[1 + (10/100)] ^2 – 1} = 525
=> P {[1 + (1/10)] ^2 – 1} = 525
=> P {[11/10] ^2 – 1} = 525
=> P {[121/100] – 1} = 525
=> 21* P / 100 = 525
=> P = 2500

Now for Simple Interest, S.I = p * n * r / 100
P = Rs. 2500
n = 4 years
r = 5%
S.I = (2500 * 4 * 5) / 100
=> S.I = 500 Rs.
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113. The compound interest on Rs.30,000 at 7% p.a. is Rs.4347.The period (in years) is :

Correct Ans:2
Explanation:
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114. What will be the difference between simple and compound interest @10% p.a. on a sum of Rs.1000
after 4 years?

Correct Ans:Rs.64.10
Explanation:
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115. The difference between the simple interest and compound interest on Rs.1200 for one year at @10% p.a. reckoned halfyearly is:

Correct Ans:Rs.3
Explanation:
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116. Find the compound interest on Rs.15,625 for 9 months at 16% per annum compounded quarterly ?

Correct Ans:Rs.1951
Explanation:
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117. What will be compounded interest on a sum of Rs.25,000 after 3 years at the rate of 12 p.c.p.a.?

Correct Ans:Rs.10123.20
Explanation:
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118. Albert invented an amount of Rs.8000 in a fixed deposit scheme for 2 years at compound interest rate 5% p.a. .How much amount will Albert get on maturity of the fixed deposit?

Correct Ans:Rs.8820
Explanation:
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119. Find the compound interest on Rs.10,000 in 2 years at 4% per annum,the interest being compounded half yearly.

Correct Ans:Rs.824.32
Explanation:
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120. What would be the compound interest accrued on an amount of 6250 Rs. at the end of 2 years at the rate of 12 % per annum ?

Correct Ans:1590
Explanation:
Given
Principal = 6250 No. of years = 2 Rate of interest = 12
Amount = P x (1+r/100)^n,
we get Amount = 6250 x (1+12/100)^2
= 6250 x 112 / 100
= 5 * 56 * 28
= 7840
C.I = Amount - Principal
= 7840 - 6250
= 1590.
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