JIPMAT 2025 (QA) - The difference between compound and simple interests on a certain sum of money at the interest rate of 10\% per annum for 11/2 years is ₹183, when the interest is compounded semi-annually, then the sum of money is : | PYQs + Solutions | AfterBoards
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JIPMAT 2025

Arithmetic
>
Simple & Compound Interest

Hard

The difference between compound and simple interests on a certain sum of money at the interest rate of 10%10\% per annum for 1121\frac{1}{2} years is ₹183₹183, when the interest is compounded semi-annually, then the sum of money is :

✅ Correct Option: 2
Let the sum of money be x x.
The rate is 10%10 \% per annum, compounded semi-annually, so the rate per half-year is 5% 5\%,
The number of periods in 1.51.5 years is 33.
Using the compound interest formula:
CI=CI = x(1+5100)3−xx\left(1+\dfrac{5}{100}\right)^{3} -x
CI=x(2120)3−xCI =x\left(\dfrac{21}{20}\right)^{3}-x
CI=x(441×218000)−xCI=x\left(\dfrac{441 \times 21}{8000}\right)-x
CI=x(92618000)−xCI=x\left(\dfrac{9261}{8000}\right)-x
CI=x(12618000)CI=x\left(\dfrac{1261}{8000}\right)
Now, simple interest:
SI=x⋅3⋅5100=15x100=3x20SI = \dfrac{x \cdot 3 \cdot 5}{100}=\dfrac{15 x}{100}=\dfrac{3 x}{20}.
The difference between Cl and SI is given as ₹183, so:
⇒1261x8000−3x20=183\Rightarrow \dfrac{1261 x}{8000}-\dfrac{3 x}{20}=183
⇒1261x−1200x8000=183\Rightarrow \dfrac{1261 x-1200 x}{8000}=183
⇒61x8000=183\Rightarrow \dfrac{61 x}{8000}=183
⇒x=3×8000=24,000\Rightarrow x=3 \times 8000=24,000
Hence, the correct option is Option 2.

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