# 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

121. What would be the compound interest accrued on an amount of 10000 Rs. at the end of 2 years at the rate of 4 % per annum ?

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Correct Ans:816

Explanation:

Given principal = 10000 No. of years = 2 Rate of interest = 4
Amount = P x (1+r/100)^n, we get Amount = 10000 x (1+4/100)^2 = 10816

C.I = Amount - Principal = 10816 - 10000 = 816

C.I = Amount - Principal = 10816 - 10000 = 816

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122. What would be the compound interest accrued on an amount of 6250 Rs. at the end of 2 years at the rate of 8 % per annum?

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Correct Ans:1040

Explanation:

Given principal = 6250 No. of years = 2 Rate of interest = 8
Amount = P x (1+r/100)^n, we get Amount = 6250 x (1+8/100)^2 = 7290

C.I = Amount - Principal = 7290 - 6250 = 1040

C.I = Amount - Principal = 7290 - 6250 = 1040

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123. What would be the compound interest accrued on an amount of 6500 Rs. at the end of 2 years at the rate of 15 % per annum?

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Correct Ans:2096.25

Explanation:

Given principal = 6500 No. of years = 2 Rate of interest = 15
Amount = P x (1+r/100)^n, we get Amount = 6500 x (1+15/100)^2 = 8596.25

C.I = Amount - Principal = 8596.25 - 6500 = 2096.25

C.I = Amount - Principal = 8596.25 - 6500 = 2096.25

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124. What would be the compound interest accrued on an amount of 5000 Rs. at the end of 2 years at the rate of 9 % per annum?

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Correct Ans:940.5

Explanation:

Given

Principal = 5000 No. of years = 2 Rate of interest = 9

we get Amount

= 5000 x (1+9/100)^2

= 5940.5

= 5940.5 - 5000

=

Principal = 5000 No. of years = 2 Rate of interest = 9

**Amount = P x (1+r/100)^n,**we get Amount

= 5000 x (1+9/100)^2

= 5940.5

**C.I = Amount - Principal**= 5940.5 - 5000

=

**940.5.**
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125. What would be the compound interest obtained on an amount of Rs. 12,500 Rs. at the end of 3 years at the rate of 10% per annum?

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Correct Ans:4137.5

Explanation:

Given principal = 12500 No. of years = 3 Rate of interest = 10 Amount = P x (1+r/100)^n, we get Amount = 12500 x (1+10/100)^3 = 16637.5

C.I = Amount - Principal = 16637.5 - 12500 = 4137.5

C.I = Amount - Principal = 16637.5 - 12500 = 4137.5

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126. What would be the compound interest accrued on an amount of 14000 Rs. at the end of 3 years at the rate of 5 % per annum?

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Correct Ans:2206.75

Explanation:

Given

Principal = 14000 No. of years = 3 Rate of interest = 5

we get Amount = 14000 x (1+5/100)^3 = 16206.75

= 16206.75 - 14000 = 2206.75.

Principal = 14000 No. of years = 3 Rate of interest = 5

**Amount = P x (1+r/100)^n**,we get Amount = 14000 x (1+5/100)^3 = 16206.75

**C.I = Amount - Principal**= 16206.75 - 14000 = 2206.75.

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127. John invested an amount of Rs. 20000 for 2 years at compound interest at the rate of 6 % per annum. Find the amount he receives at the end of 2 years.

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Correct Ans:22472

Explanation:

Given

Principal : P = 20000 Rs. Rate of Interest : r = 6 % Number of years : n = 2

Amount = 20000 x (1+6/100)^2

= 20000 x (1+3/50)^2

= 20000 x (53/50) x (53/50)

=

Principal : P = 20000 Rs. Rate of Interest : r = 6 % Number of years : n = 2

**Amount = P x (1 + r/100) ^ n**Amount = 20000 x (1+6/100)^2

= 20000 x (1+3/50)^2

= 20000 x (53/50) x (53/50)

=

**22472.**
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128. Rs. 10000 is borrowed at compound interest at the rate of 4 % annum. What will be the amount to be paid after 2 years?

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Correct Ans:10816

Explanation:

Principal : P = 10000 Rs. Rate of Interest : r = 4 % Number of years : n = 2
Amount = P x (1 + r/100)^n Amount = 10000 x (1+4/100)^2 =10000 x (1+1/25)^2
=10000 x (26/25) x (26/25) =10816

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129. What would be the compound interest accrued on an amount of 5000 Rs. at the end of 2 years at the rate of 9 % per annum ?

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Correct Ans:940.5

Explanation:

Given principal = 5000
No. of years = 2
Rate of interest = 9

Amount = P x (1+r/100)^n, we get Amount = 5000 x (1+9/100)^2 = 5000 x (109/100) x (109/100) = 5940.5

Compound Interest, C. I = Amount - Principal = 5940.5 - 5000 = 940.5

Amount = P x (1+r/100)^n, we get Amount = 5000 x (1+9/100)^2 = 5000 x (109/100) x (109/100) = 5940.5

Compound Interest, C. I = Amount - Principal = 5940.5 - 5000 = 940.5

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130. What would be the compound interest accrued on an amount of 12500 Rs. at the end of 3 years at the rate of 10 % per annum ?

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Correct Ans:4137.5

Explanation:

Given principal = 12500
No. of years = 3
Rate of interest = 10

Amount = P x (1+r/100)^n, we get Amount = 12500 x (1+10/100)^3 = 12500 x (11/10)^3 = 12500 x (11/10)x (11/10)x (11/10) = 16637.5

Compound Interest, C. I = Amount - Principal = 16637.5 - 12500 = 4137.5

Amount = P x (1+r/100)^n, we get Amount = 12500 x (1+10/100)^3 = 12500 x (11/10)^3 = 12500 x (11/10)x (11/10)x (11/10) = 16637.5

Compound Interest, C. I = Amount - Principal = 16637.5 - 12500 = 4137.5

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131. What would be the compound interest accrued on an amount of 14000 Rs. at the end of 3 years at the rate of 5 % per annum ?

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Correct Ans:2206.75

Explanation:

Given principal = 14000
No. of years = 3
Rate of interest = 5 %

Amount = P x (1+r/100)^n, we get Amount = 14000 x (1+5/100)^3 = 14000 x (1 + 1/20)^3 = 14000 x (21/20)^3 = 14000 x (21/20) x (21/20) x (21/20) = 16206.75

Compound Interest = Amount - Principal = 16206.75 - 14000 = 2206.75

Amount = P x (1+r/100)^n, we get Amount = 14000 x (1+5/100)^3 = 14000 x (1 + 1/20)^3 = 14000 x (21/20)^3 = 14000 x (21/20) x (21/20) x (21/20) = 16206.75

Compound Interest = Amount - Principal = 16206.75 - 14000 = 2206.75

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132. John invested an amount of Rs. 20000 for 2 years at compound interest at the rate of 6 % per annum. Find the amount he receives at the end of 2 years.

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Correct Ans:22472

Explanation:

Principal : P = 20000 Rs. Rate of Interest : r = 6 % Number of years : n = 2
Amount = P x (1 + r/100)^n
=> Amount = 20000 x (1+6/100)^2
= 20000 x (1+3/50)^2
= 20000 x (53/50) x (53/50)
= 22472

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133. Rs. 10000 is borrowed at compound interest at the rate of 4 % annum. What will be the amount to be paid after 2 years ?

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Correct Ans:10816

Explanation:

Principal : P = 10000 Rs.
Rate of Interest : r = 4 %
Number of years : n = 2
Amount = P x (1 + r/100)^n
Amount = 10000 x (1+4/100)^2
=10000 x (1+1/25)^2
=10000 x (26/25) x (26/25) =10816

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134. Find the simple interest on Rs. 2000 at 7 % per annum for 4 years

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Correct Ans:560 Rs.

Explanation:

**Solution is :**

Given

Given

Principal : 2000

Rate of interest : 7

Number of years : 4

**Simple Interest = pnr / 100**

= ( 2000 x 4 x 7 ) / 100

=

**560 Rs**

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135. What would be the compound interest accrued on an amount of 10000 Rs. at the end of 2 years at the rate of 4 % per annum ?

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Correct Ans:816

Explanation:

**Solution is :**

Given principal = 10000

No. of years = 2

Rate of interest = 4

**Amount = P [ 1 + ( r / 100 )**

^{n}]= 10000 x [ 1 +( 4 / 100 )

^{2}]

= 10000 x ( 104 / 100 )

^{2 }= 10000 x ( 104 / 100 ) x ( 104 / 100 )

= 104 x 104

**= 10816**

Compound Interest = Amount - Principal

= 10816 - 10000

Compound Interest = Amount - Principal

=

**816 Rs**

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136. A person receives a sum of Rs. 2100 as interest for investing some amount at 10% p.a compounding annually for 2 years. Find the amount invested at the beginning

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Correct Ans:10000

Explanation:

**Solution is :**

Given Compound Interest = Rs.2100

Rate of Interest ( r ) = 10 % p.a

No.of years ( n ) = 2

To find , amount received at the beginning => principal

**Compound Interest = P [ 1 + ( r / 100 ) ^{n}- 1 ]**

=> 2100 = P[ 1 + ( 10 / 100 )

^{2}- 1 ]

=> 2100 = P[ 1 + ( 1 / 10 )

^{2}- 1 ]

=> 2100 = P[ ( 11 / 10 )

^{2}- 1 ]

=> 2100 = P[ ( 121 / 100 ) - 1 ]

=> 2100 = P[ 21 / 100 ]

=> 2100 x ( 100 / 21 ) = P

Principal = Rs. 10000

Amount invested at the beginning =

**Rs. 10000**

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137. What would be the compound interest accrued on an amount of 6500 Rs. at the end of 2 years at the rate of 15 % per annum ?

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Correct Ans:2096.25

Explanation:

**Solution is :**

Given principal = 6500

No. of years = 2

Rate of interest = 15

**Amount = P [ 1 + ( r / 100 )**

^{n}]= 6500 x [ 1 + ( 15 /100 )

^{2}]

= 6500 x [ 1 + ( 3 / 20 )

^{2}]

= 6500 x [ 23 / 20 ]

^{2}

= 6500 x [ 529 / 400 ]

**Amount =**

**8596.25**

Compound Interest = Amount - Principal

Compound Interest = Amount - Principal

= 8596.25 - 6500

**= 2096.25**

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138. What would be the compound interest accrued on an amount of 4500 Rs. at the end of 2 years at the rate of 10 % per annum ?

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Correct Ans:945

Explanation:

**Solution is :**

Given principal = 4500

No. of years = 2

Rate of interest = 10

**Amount = P [ 1 + ( r / 100 ) ]**

^{n}= 4500 x [ 1 + ( 10 / 100 ) ]

^{2 }= 4500 x [ 1 + ( 1 / 10 ) ]

^{2 }= 4500 x [ 11 / 10 ]

^{2}

= 4500 x [ 121 / 100 ]

**Amount = 5445**

Compound interest = Amount - principal

= 5445 - 4500

Compound interest = Amount - principal

= 945 Rs

**Ans is 945**

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139. Find the simple interest on Rs. 700 at 20 % per annum for 6 years

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Correct Ans:Rs. 840

Explanation:

**Solution is :**

Given

Given

Principal : 700

Rate of interest : 20

Number of years : 6

**Simple Interest = pnr / 100**

= ( 700 x 6 x 20 ) / 100

= 7 x 6 x 20

=

**840**

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140. What would be the compound interest accrued on an amount of 8000 Rs. at the end of 3 years at the rate of 10 % per annum ?

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Correct Ans:2648

Explanation:

**Solution is**

Given

principal = 8000

No. of years = 3

Rate of interest = 10

**Amount = P x ( 1 + ( r / 100 ) )**

^{n}= 8000 x ( 1 + ( 1 / 10 ) )

^{3}

= 8000 x ( 11 / 10 )

^{3}

= 8000 x ( 1331 / 1000 )

= 8 x 1331

**Amount = 10648**

**Compound Interest = Amount - Principal**

= 10648 - 8000

=

**2648 Rs**

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