Question

Find the money, invested at 10% compounded annually, on which the sum of interest for the first year and the third year is Rs. 1,768.

• A. Rs. 4,000
• B. Rs. 6,000
• C. Rs. 7,000
• D. Rs. 8,000

Answer 1

Answer: (D)

Rs. 8,000

For the first year: $P=Rs.x,R=10\%,T=1year$

$\therefore$ Interest $\left(I\right)=\frac{P\times R\times T}{100}=\frac{x\times 10\times 1}{100}=Rs.\frac{x}{10}$

Then, Amount $=P+I=x+\frac{x}{10}=Rs.\frac{11x}{10}$

For the second year: $P=Rs.\frac{11x}{10},R=10\%,T=1year$

$\therefore$ Interest $\left(I\right)=\frac{P\times R\times T}{100}=\frac{\frac{11x}{10}\times 10\times 1}{100}=Rs.\frac{11x}{100}$

Then, Amount $=P+I=\frac{11x}{10}+\frac{11x}{100}=Rs.\frac{121x}{100}$

For the third year: $P=Rs.\frac{121x}{100},R=10\%,T=1year$

$\therefore$ Interest $\left(I\right)=\frac{P\times R\times T}{100}=\frac{\frac{121x}{100}\times 10\times 1}{100}=Rs.\frac{121x}{1000}$

Given: Interest for $1^{st}$ year $+$ Interest for $3^{rd}$ year $=1768$

$\Rightarrow \frac{x}{10}+\frac{121x}{1000}=1768$

$\Rightarrow \frac{100x+121x}{1000}=1768$

$\Rightarrow \frac{221x}{1000}=1768$

$\Rightarrow x=1768\times \frac{1000}{221}$

$\Rightarrow x=8000$

So, Money invested $=Rs.8000$.

Hence, Correct option is $\text{D}$.