Question
The difference between the compound interest and simple interest on a certain sum at $15\%$ per annum for $3$ years is $Rs. 283.50$. Find the sum.
✓
Answer
Given: $CI − SI = Rs. 283.50\ R = 15\%\ n = 3$ years Let the sum be $Rs. x$.
We know that: $\text{A}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{x}\Big(1+\frac{15}{100}\Big)^{3}$
$=\text{x}(1.15)^{3}\ ...(1)$ Also, $\text{SI}=\frac{\text{PRT}}{100}=\frac{\text{x}(15)(3)}{100}=0.45\text{x}$
$\text{A}=\text{SI}+\text{P}=1.45\text{x}\ ...(2)$
Thus, we have: $\text{x}(1.15)^{3}-1.45\text{x}=283.50$ [From $(1)$ and $(2)$]
$1.523\text{x}-1.45\text{x}=283.50$
$0.070875\text{ x}=283.50$
$\text{x}=\frac{283.50}{0.070875}$
$=4,000$
Thus, the sum is $Rs. 4,000.$
We know that: $\text{A}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{x}\Big(1+\frac{15}{100}\Big)^{3}$
$=\text{x}(1.15)^{3}\ ...(1)$ Also, $\text{SI}=\frac{\text{PRT}}{100}=\frac{\text{x}(15)(3)}{100}=0.45\text{x}$
$\text{A}=\text{SI}+\text{P}=1.45\text{x}\ ...(2)$
Thus, we have: $\text{x}(1.15)^{3}-1.45\text{x}=283.50$ [From $(1)$ and $(2)$]
$1.523\text{x}-1.45\text{x}=283.50$
$0.070875\text{ x}=283.50$
$\text{x}=\frac{283.50}{0.070875}$
$=4,000$
Thus, the sum is $Rs. 4,000.$
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