Compound Interest - Solved Questions and Answers

Compound Interest is interest calculated on the initial principal and the accumulated interest from previous periods. It grows exponentially over time because you earn "interest on interest."

Formula:

A =P(1+ \frac{R}{n})^{nT}

Where:

• A = Final amount (Principal + Interest)
• P = Principal amount (initial investment/loan)
• R = Annual interest rate
• n = Number of times interest is compounded per year
• T = Time in years

Compound Interest questions and answers are provided below for you to learn and practice.

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 = $926, 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.