Discussion

At the end of 2 years difference between simple and compound interest is Rs. 2166. If the principal is Rs. 60000, then what is the rate of interest?

Answer: C

\[ A = P\left(1 + \frac{R}{100}\right)^{n}, \qquad \text{CI} = A - P, \qquad \text{SI} = \frac{PRn}{100} \]

For two years the difference is

\[ \text{CI} - \text{SI} = P\left(\frac{R}{100}\right)^{2} \]

\[ 2166 = 60000 \times \frac{R^{2}}{10000} = 6R^{2} \]

\[ R^{2} = 361 \quad\Longrightarrow\quad R = 19\% \]

The rate appears squared, so a perfect square is the signal that the arithmetic is on track - here \( 19^{2} = 361 \).

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