Answer: Option A
Let the annual installment be Rs. x.
Then,
$$ \Rightarrow \left[ {x + \left( {\frac{{x \times 3 \times 4}}{{100}}} \right)} \right] + $$     $$\left[ {x + \left( {\frac{{x \times 2 \times 4}}{{100}}} \right)} \right] + $$     $$\left[ {x + \left( {\frac{{x \times 1 \times 4}}{{100}}} \right)} \right] + $$     $$x = 848$$
$$\eqalign{
& \Leftrightarrow \frac{{28x}}{{25}} + \frac{{27x}}{{25}} + \frac{{26x}}{{25}} + x = 848 \cr
& \Leftrightarrow 106x = 848 \times 25 \cr
& \Leftrightarrow 106x = 21200 \cr
& \Leftrightarrow x = 200 \cr} $$
Short Cut Method : The annual payment that will discharge a debt of Rs. A due in t years at the rate of interest r % p.a. is.
$$\eqalign{
& \frac{{100{\text{A}}}}{{100t + \frac{{rt\left( {t - 1} \right)}}{2}}} \cr
& \therefore {\text{Annual installment}} \cr
& = {\text{Rs}}{\text{.}}\left[ {\frac{{100 \times 848}}{{100 \times 4 + \frac{{4 \times 4 \times 3}}{2}}}} \right] \cr
& = {\text{Rs}}{\text{.}}\left( {\frac{{100 \times 848}}{{424}}} \right) \cr
& = {\text{Rs}}{\text{. }}200 \cr} $$