The difference between the compound interest and the simple interest accrued on an amount of ` 18,000 in 2 years was ` 405. What was the rate of interest p.c.p.a.?
Answer
Answer (as printed):
Explanation
Let the rate be R% p.a. Then, $\left[18000\left(1+\frac{R}{100}\right)^{2}-18000\right]-\left(\frac{18000 \times R \times 2}{100}\right)=405$ $\Leftrightarrow 18000\left[\frac{(100+R)^{2}}{10000}-1-\frac{2R}{100}\right]=405$ $\Leftrightarrow 18000\left[\frac{(100+R)^{2}-10000-200R}{10000}\right]=405$ $\Leftrightarrow \frac{9}{5}R^{2}=405 \Leftrightarrow R^{2}=\left(\frac{405 \times 5}{9}\right)=225 \Leftrightarrow R=15$. ∴ Rate $=15\%$.
Explanation as extracted from the printed page; notation may be imperfect.