Question:
The difference between Compound Interest (compounded annually) and Simple Interest on a certain sum of money at \(10\%\) per annum for \(3\) years is ₹155. Find the principal sum.
The difference between Compound Interest (compounded annually) and Simple Interest on a certain sum of money at \(10\%\) per annum for \(3\) years is ₹155. Find the principal sum.
Show Hint
For 3 years,
\[
\text{CI}-\text{SI}
=
P\left(\frac{r}{100}\right)^2
\left(3+\frac{r}{100}\right).
\]
This shortcut saves a lot of calculation.
Updated On: Jun 8, 2026
- ₹4000
- ₹5000
- ₹6000
- ₹5500
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Concept:
For three years,
\[
\text{CI}-\text{SI}
=
P\left(\frac{r}{100}\right)^2
\left(3+\frac{r}{100}\right).
\]
This formula directly gives the difference between Compound Interest and Simple Interest.
Step 1: Substitute the given rate.
Given, \[ r=10\% \] and \[ \text{CI}-\text{SI}=155. \] Therefore, \[ 155 = P\left(\frac{10}{100}\right)^2 \left(3+\frac{10}{100}\right). \]
Step 2: Simplify the expression.
\[ 155 = P\left(\frac1{10}\right)^2 \left(\frac{31}{10}\right). \] \[ 155 = P\left(\frac{31}{1000}\right). \]
Step 3: Find the principal.
\[ P = 155\times\frac{1000}{31}. \] \[ P=5000. \] Hence, \[ \boxed{₹5000} \]
Step 1: Substitute the given rate.
Given, \[ r=10\% \] and \[ \text{CI}-\text{SI}=155. \] Therefore, \[ 155 = P\left(\frac{10}{100}\right)^2 \left(3+\frac{10}{100}\right). \]
Step 2: Simplify the expression.
\[ 155 = P\left(\frac1{10}\right)^2 \left(\frac{31}{10}\right). \] \[ 155 = P\left(\frac{31}{1000}\right). \]
Step 3: Find the principal.
\[ P = 155\times\frac{1000}{31}. \] \[ P=5000. \] Hence, \[ \boxed{₹5000} \]
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