IPMAT Indore 2023 · Quantitative Ability · Question 28

Step 1 — write both interests for 3 years. With principal P and rate r=10\%:

SI=P\times 0.1\times 3=0.3P

CI=P{\left( 1.1 \right)}^{3}-P=P\left( 1.331-1 \right)=0.331P

Step 2 — take the difference.

CI-SI=0.331P-0.3P=0.031P

Step 3 — solve.

0.031P=527 \Rightarrow P=\frac{527}{0.031}=\frac{527000}{31}=17000

Step 4 — verify. SI=0.3\times 17000=5100; CI=0.331\times 17000=5627; difference =527 ✔

Answer: Option 2 — ₹17,000

Shortcut. Memorise the 3-year difference formula:

CI-SI{|}_{3 \text{yr}}=\frac{P{r}^{2}\left( r+300 \right)}{{10}^{6}}

At r=10 this is \frac{P\left( 100 \right)\left( 310 \right)}{{10}^{6}}=0.031P, so P=\frac{527}{0.031}=17000.

Fast mental check: 527=17\times 31, and the multiplier is 31 thousandths — so P=17\times 1000. The setter chose 527 precisely so that the 31 cancels.

For reference, the 2-year version is CI-SI=P{\left( r/100 \right)}^{2}, which is tested more often; knowing both saves the full expansion every time.