Compound Interest Questions and Answers updated daily – Aptitude

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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?




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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:




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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:




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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 ?




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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:    




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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:




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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:




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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? 




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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:




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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?




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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? 




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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:  




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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 :




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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?




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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:




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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 ?




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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.?




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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?




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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. 




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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 ?   




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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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