Question

Find the sum, invested at 10% compounded annually, on which the interest for the third year exceeds the interest of the first year by Rs. 252.

• A. $Rs.10,000$
• B. $Rs.12,000$
• C. $Rs.14,000$
• D. $Rs.16,000$

Answer 1

The correct option is B $Rs.12,000$

let the sum=Rs.P

rate=10%

C.I for 3 year=Amount at the end of 3rd year-Amount at the end of 2nd year

$=P{(1+\frac{10}{100})}^3-P{(1+\frac{10}{100})}^2$

$=P[1+{\left(\frac{11}{10}\right)}^3-{\left(\frac{11}{10}\right)}^2]$

$=P(\frac{11}{10}{)}^2-[\frac{11}{10}-1]$

$=P\times \frac{121}{100}\times \frac{1}{10}$

$=\frac{121}{1000}P$

C.P for 1 st year$=\frac{P\times 10\times 1}{100}=\frac{P}{10}$

Then according to the question

$\Rightarrow \frac{121}{1000}-\frac{P}{10}=252$

$\Rightarrow \frac{121P-100P}{1000}=252$

$\Rightarrow \frac{21P}{1000}=252$

$\Rightarrow P=\frac{252\times 1000}{21}=Rs.12000$