A sum of money at compound interest doubles itself in 4 years. In how many years does the sum become 8 times of itself at the same rate of interest?

The correct answer is 12.

Step-by-step explanation:

Let the principal sum of money be $P$ and the annual compound interest rate be $r$%.

The formula for the total amount $A$ under compound interest after $t$ years is:

$A=P{(1+\frac{r}{100})}^t$

Step 1: Express the condition for doubling the sum

The sum doubles itself ($A=2P$) in $t=4$ years:

$2P=P{(1+\frac{r}{100})}^4$

Dividing both sides by $P$:

$2={(1+\frac{r}{100})}^4$    --- (Equation 1)

Step 2: Express the condition for the sum to become 8 times

Let $t$ be the time in years for the sum to become 8 times of itself ($A=8P$):

$8P=P{(1+\frac{r}{100})}^t$

Dividing both sides by $P$:

$8={(1+\frac{r}{100})}^t$    --- (Equation 2)

Step 3: Relate 8 and 2

We know that:

$8=2^3$

Substitute the expression for $2$ from Equation 1:

$8={\left[{(1+\frac{r}{100})}^4\right]}^3$

$8={(1+\frac{r}{100})}^{4\times 3}$

$8={(1+\frac{r}{100})}^{12}$

Step 4: Solve for $t$

Comparing this with Equation 2:

${(1+\frac{r}{100})}^t={(1+\frac{r}{100})}^{12}$

Since the bases are equal, equating the exponents gives:

$t=12$ years

Alternative Shortcut Method:

Under compound interest, growth occurs exponentially over equal time intervals:

• In 4 years, the sum becomes 2 times the original amount.

• In another 4 years (total 8 years), it doubles again: $2\times 2=4$ times.

• In another 4 years (total 12 years), it doubles again: $4\times 2=8$ times.

Thus, the sum becomes 8 times of itself in 12 years.