ABC26GN4083 · Simple Interest

Let the annual instalment be ₹ $x$. Then, $$\left[x+\left(\frac{x \times 3 \times 4}{100}\right)\right]+\left[x+\left(\frac{x \times 2 \times 4}{100}\right)\right]$$ $+\left[x+\left(\frac{x \times 1 \times 4}{100}\right)\right]+x=848$ $$\begin{aligned} & \Leftrightarrow \quad \frac{28 x}{25}+\frac{27 x}{25}+\frac{26 x}{25}+x=848 \\ & \Leftrightarrow \quad 106 x=848 \times 25=21200 \Leftrightarrow x=200 . \end{aligned}$$ Short Cut Method: The annual payment that will discharge a debt of ₹ $A$ due in $t$ years at the rate of interest $r \%$ p.a. is $\frac{100 A}{100 t+\frac{r t(t-1)}{2}}$. ∴ Annual instalment $$=₹\left[\frac{100 \times 848}{100 \times 4+\frac{4 \times 4 \times 3}{2}}\right]=₹\left(\frac{100 \times 848}{424}\right)=₹ 200 .$$