Question

17. A man borrows Rs 12,000 at a certain rate of compound interest per annum. His loan becomes
Rs 12,730.80 at the end of 2
years.
(i) Find the principal in the beginning of the second year.
(ii) Find the compound interest (to the nearest rupee) on the sum for 3 years.
​

Answer 1

Answer:

Solution:

Let the rate of interest = R%

Principal (P) = Rs.12,000

Time period (n) = 2 yrs

\text{We have amount (A)} = P\left (1+\frac{R}{100} \right )^n

Where P is principal, R is rate of interest and n is time period

It is given that amount after two years = Rs. 12730.80

Therefore, Rate of interest = 3 % 1)

Interest on 1st year \small = \frac{PTR}{100} = \frac{12000\times 1 \times3 }{100}= Rs.360 Now,

Principal in the beginning of second year = Principal + interest = 12,000 + 360 = Rs. 12360

2) Amount after 3 years Now, Compound interest on the sum of 3 years = Amount after 3 years - Principal = 13112.70 - 12000 = Rs. 1112.7