Compound Interest - Solved Questions and Answers

Last Updated : 26 Jun, 2025

Question 1: Find the compound interest on Rs. 10,000 at 10% per annum for a time period of three and a half years.

Solution:

Time period of 3 years and 6 months means for 3 years, the interest is compounded yearly and for the remaining 6 months, the interest is compounded half-yearly.
This means that we have 3 cycles of interest compounded yearly and 1 cycle of interest compounded half yearly.
So, Amount = P [1 + (R / 100)]³ [1 + ( {R/2} / 100 )]
=> Amount = 10000 [1 + 0.1]³ [1 + 0.05]
=> Amount = 10000 (1.1)³ (1.05)
=> Amount = Rs. 13975.50
=> Compound Interest, CI = Amount - Principal = 13975.50 - 10000
Therefore, CI = Rs. 3975.50

Question 2: If Rs. 5000 amounts to Rs. 5832 in two years compounded annually, find the rate of interest per annum.

Solution:

Here, P = 5000, A = 5832, n = 2
A = P [1 + (R / 100)]ⁿ
5832 = 5000 [1 + (R / 100)]²
1 + (R / 100)]² = 5832 / 5000
[1 + (R / 100)]² = 11664 / 10000
[1 + (R / 100)] = 108 / 100 => R / 100 = 8 / 100
R = 8%
Thus, the required rate of interest per annum in 8%

Question 3: The difference between the SI and CI on a certain sum of money at a 10 % rate of annual interest for 2 years is Rs. 549. Find the sum.

Solution:

Let the sum be P
R = 10 %
n = 2 years
SI = P x R x n / 100
= P x 10 x 2 / 100
= 0.20 P
CI = A - P = P [1 + (R / 100)]n - P = 0.21 P
Now, it is given that CI - SI = 549
=> 0.21 P - 0.20 P = 549
=> 0.01 P = 549 => P = 54900
Therefore, the required sum of money is Rs. 54,900

Question 4: A sum of Rs. 1000 is to be divided among two brothers such that if the interest is compounded annually is 5 % per annum, then the money with the first brother after 4 years is equal to the money with the second brother after 6 years.

Solution:

Let the first brother be given Rs. P => Money with second brother = Rs. 1000 - P
Now, according to the question,
P [1 + (5 / 100)]⁴ = (1000 - P) [1 + (5 / 100)]⁶
=> P (1.05)⁴ = (1000 - P) (1.05)⁶
=> 0.9070 P = 1000 - P
=> 1.9070 P = 1000
=> P = 524.38
Therefore, share of first brother = Rs. 524.38
Share of second brother = Rs. 475.62

Question 5: A sum of money amounts to Rs. 669 after 3 years and to Rs. 1003.50 after 6 years on compound interest. Find the sum.

Solution:

Let the sum of money be Rs. P => P [1 + (R/100)]3= 669
P [1 + (R/100)]6= 1003.50
Dividing both equations, we get [1 + (R/100)]3 = 1003.50 / 669 = 1.50
Now, we put this value in the equation P [1 + (R/100)]3= 669
=> P x 1.50 = 669
=> P = 446
Thus, the required sum of money is Rs. 446

Question 6: An investment doubles itself in 15 years if the interest is compounded annually. How many years will it take to become 8 times?

Solution:

It is given that the investment doubles itself in 15 years.
Let the initial investment be Rs. P => At the end of 15 years, A = 2 P
Now, this 2 P will be invested. => Amount after 15 more years = 2 x 2 P = 4 P
Now, this 4 P will be invested. => Amount after 15 more years = 2 x 4 P = 8 P
Thus, the investment (P) will become 8 times (8 P) in 15 + 15 + 15 = 45 years

Question 7: If the difference between compound interest and simple interest on some principle amount is at the rate of interest of 20% per annum for 3 years in $48, then what is the principle amount?

Solution:

Here, T = 3 years, R = 20%, P = ?
Also, C.I - S.I = 48
P(1 + (R / 100))^T - P - (P × R × T) / 100 = 48
⇒ P(1 + (20/100))³- P - (P × 20 × 3) / 100 = 48
⇒ P(6/5)³ - P - P (3/5) = 48
⇒ P ((216/125)³ - 1 - (3/5)) = 48
⇒ P (0.128) = 48
⇒ P = (48/0.128)
⇒ P = 375
Therefore, the principle amount is $375.

Question 8: A sum of money placed at compound interest doubles itself in 3 years. In how many years will it amount to 8 times itself?

Solution:

Let in 3 years, P = y, C.I = 2y (since, it doubles itself)
Now, by formula,
C.I = P (1 + (R / 100))^T
⇒ 2y = y (1 + (R / 100))³
⇒ 2 = (1 + (R / 100))³
So, (1 + (R/100)) = 2^1/3 . . . (i)
We need to find, T = ? when C.I = 8y,
By formula,
C.I = P (1 + (R / 100))^T
⇒ 8y = y(2^1/3)^T [From equation (i)]
⇒ 2³y = y(2^T/3)
On Comparing, we get
T/3 = 3
⇒ T = 3(3)
⇒ T = 9 years.
Therefore, in 9 years the the sum of money will amount to 8times itself.

Question 9: In how many years will a sum of $800 at 10% per annum compound semi-annually become $926. ?

Solution:

Given that, P = $800, R = 10% per year, A = $1064.8, T = ?
By formula, Amount,
A = P (1 + ((R/2)/100))^2T
⇒ 926.1 = 800(1 + (5/100))^2T
⇒ (926.1 /800) = (21 /20)^2T
⇒ (9261 /8000) = (21/20)^2T
⇒ (21/20)³ = (21/10)^2T
On comparing, we get
2T = 3
⇒ T = 3/2 years
Therefore, in 3/2 years the semi-annually compounded sum will become $926.1

Question 10: A tree increases annually by (1/8)^th of its height. By how much will it increase after 2 years, if it stands today 64cm high?

Solution:

The increase% is given by, (1/8) × 100 % = (25/2) %
By formula,
C.I. = P (1 + (R / 100))^T
⇒ C.I.= 64(1 + (25/(2 × 100)))²
⇒ C.I.= 64(9/8)²
⇒ C.I.= 64(81/64)
⇒ C.I.= 81
Total Increase = Height after 2 years−Initial Height = 81−64 = 17 cm
Therefore, the tree will compound to height 17cms after 2 years.

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