A sum of money at compound interest amounts to three times of itself i

1.	A sum of money at compound interest amounts to three times of itself in three years. In how many years will it be nine times of itself ? (a) 6 years (b) 5 years (c) 9 years (d) 7 years
Answer» (a) 6 years Given, 3P = P\(\big(1+\frac{r}{100}\big)^3\) \(\Rightarrow\) \(\big(1+\frac{r}{100}\big)^3\) = 3 Let t be the time in years in which the sum will be nine times of itself. Then, 9P = P\(\big(1+\frac{r}{100}\big)^t\) \(\Rightarrow\) \(\big(1+\frac{r}{100}\big)^t\) = 9 = 3² \(\Rightarrow\) \(\big(1+\frac{r}{100}\big)^t\) = \(\Big[\big(1+\frac{r}{100}\big)^3\Big]^2\) (From (i)) \(\Rightarrow\) \(\big(1+\frac{r}{100}\big)^t\) = \(\big(1+\frac{r}{100}\big)^6\) \(\Rightarrow\) t = 6 years.

A sum of money at compound interest amounts to three times of itself in three years. In how many years will it be nine times of itself ? (a) 6 years (b) 5 years (c) 9 years (d) 7 years

Answer»

(a) 6 years

Given, 3P = P\(\big(1+\frac{r}{100}\big)^3\)

\(\Rightarrow\) \(\big(1+\frac{r}{100}\big)^3\) = 3

Let t be the time in years in which the sum will be nine times of itself. Then,

9P = P\(\big(1+\frac{r}{100}\big)^t\) \(\Rightarrow\) \(\big(1+\frac{r}{100}\big)^t\) = 9 = 3²

\(\Rightarrow\) \(\big(1+\frac{r}{100}\big)^t\) = \(\Big[\big(1+\frac{r}{100}\big)^3\Big]^2\) (From (i))

\(\Rightarrow\) \(\big(1+\frac{r}{100}\big)^t\) = \(\big(1+\frac{r}{100}\big)^6\)

\(\Rightarrow\) t = 6 years.