NCHMCT JEE 2020 Question Paper with solutions for online practice

Let the annual instalment be ? $\mathit{x}$.
$\therefore \left(\mathit{x}+\frac{\mathit{x}\times 3\times 5}{100}\right)+\left(\mathit{x}+\frac{\mathit{x}\times 2\times 5}{100}\right)+\left(\mathit{x}+\frac{\mathit{x}\times 1\times 5}{100}\right)+\mathit{x}$
$\Rightarrow =6450$
$\Rightarrow \frac{115\mathit{x}}{100}+\frac{110\mathit{x}}{100}+\frac{105\mathit{x}}{100}+\mathit{x}=6450$
$\Rightarrow 115\mathit{x}+110\mathit{x}+105\mathit{x}+100\mathit{x}=6450\times 100$
$\Rightarrow 430\mathit{x}=6450\times 100$
$\therefore \mathit{x}=\frac{6450\times 100}{430}=\mp 1500$
Alternate Method
The annual payment that will discharge a debt of ₹ A due in $\mathit{t}\mathit{y}\mathit{r}$ at the rate of interest $\mathit{r}\%$ per annum, is
$\frac{100\mathit{A}}{100\mathit{t}+\frac{\mathit{r}\mathit{t}\left(\mathit{t}-1\right)}{2}}$
Here, $\mathit{A}=6450$
$\mathit{t}=4\mathit{y}\mathit{r}$
$\mathit{r}=5\%$
Annual instalment
$=\frac{100\times 6450}{100\times 4+\frac{5\times 4\left(4-1\right)}{2}}$
$=\frac{100\times 6450}{400+30}=\frac{100\times 6450}{430}$