The difference between compound interest (CI) and simple interest (SI) on a sum for $4$ years is ₹ $1282$. Find the sum.
I. Amount of simple interest accrued after $4$ years is ₹ $4000$.
II. Rate of interest is $10\%$ per annum.
When $r$ is given, CI-SI over multiple years becomes a simple multiplier of $P$. Use it to solve for $P$ directly from the given difference.
I + II together necessary.
For $n = 4$ years at rate $r$, the difference (CI – SI) equals
\[\Delta = P \left[ \left(1 + \tfrac{r}{100}\right)^4 - \left(1 + \tfrac{4r}{100}\right) \right].\]
With $r = 10\%$ (II),
\[\Delta = P(1.1^4 - 1.4) = P(1.4641 - 1.4) = 0.0641P.\]
Given $\Delta = 1282 \Rightarrow P = \tfrac{1282}{0.0641} = \text{₹}\,20000.$
So II alone is sufficient.
I alone gives SI (simple interest) = ₹4000 = $P \cdot \tfrac{4r}{100}$, but $r$ is unknown $\Rightarrow$ not sufficient.
Rather than substituting directly into the general \( n \)-year CI-SI difference formula in one shot, build up the year-by-year gap between compound and simple interest, assuming the rate found in Statement II, and see which statement(s) let this be pinned down.
With rate \( r=10\% \), track the cumulative difference \( D_n=P\big[(1.1)^n-1-0.1n\big] \) year by year: \[ D_1=0,\quad D_2=0.01P,\quad D_3=0.031P,\quad D_4=0.0641P. \]
The incremental year-by-year buildup of the CI-SI gap, using only the rate from Statement II, pins down the sum on its own.
Hence, the correct answer is If II alone is sufficient but I alone is not sufficient.
