Question
Mohit borrowed a certain sum at $5\%$ per annum compound interest and cleared this loan by paying $Rs. 12,600$ at the end of the first year and $Rs. 17,640$ at the end of the second year. Find the sum borrowed.
✓
Answer
For the payment of $Rs. 12,600$ at the end of first year : $A= Rs. 12,600 ; n=1$ year and $r=5 \%$
Now, $\mathrm{A}=\mathrm{P}\left(1+\frac{r}{100}\right)^n$
$\Rightarrow 12,600=\mathrm{P}\left(1+\frac{5}{100}\right)^1$
$ \Rightarrow 12,600=\mathrm{P}\left(\frac{21}{20}\right)$
$ \Rightarrow \mathrm{P}=\frac{20}{21} \times 12,600=\text { Rs. } 12,000$
For the payment of $Rs. 17,640$ at the end of second year : $A=Rs. 17,640, n=2$ years and $r=5 \%$
Now, $\mathrm{A}=\mathrm{P}\left(1+\frac{r}{100}\right)^n$
$\Rightarrow 17,640=\mathrm{P}\left(1+\frac{5}{100}\right)^2$
$ \Rightarrow 17,640=\mathrm{P}\left(\frac{21}{20}\right)^2$
$ \Rightarrow \mathrm{P}=\frac{20}{21} \times \frac{20}{21} \times 17,640=\text { Rs. } 16,000$
$\therefore$ Sum borrowed $= Rs. (12,000+16,000)= Rs. 28,000$.
Now, $\mathrm{A}=\mathrm{P}\left(1+\frac{r}{100}\right)^n$
$\Rightarrow 12,600=\mathrm{P}\left(1+\frac{5}{100}\right)^1$
$ \Rightarrow 12,600=\mathrm{P}\left(\frac{21}{20}\right)$
$ \Rightarrow \mathrm{P}=\frac{20}{21} \times 12,600=\text { Rs. } 12,000$
For the payment of $Rs. 17,640$ at the end of second year : $A=Rs. 17,640, n=2$ years and $r=5 \%$
Now, $\mathrm{A}=\mathrm{P}\left(1+\frac{r}{100}\right)^n$
$\Rightarrow 17,640=\mathrm{P}\left(1+\frac{5}{100}\right)^2$
$ \Rightarrow 17,640=\mathrm{P}\left(\frac{21}{20}\right)^2$
$ \Rightarrow \mathrm{P}=\frac{20}{21} \times \frac{20}{21} \times 17,640=\text { Rs. } 16,000$
$\therefore$ Sum borrowed $= Rs. (12,000+16,000)= Rs. 28,000$.
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