Question

In how many years a sum placed on compound interest becomes 8 times of itself, if it becomes twice in 6 years?

• A. 20
• B. 18
• C. 15
• D. 12

Answer 1

The Correct Option is B

To solve the problem, we need to find the number of years required for a sum to become 8 times its original value at compound interest, given that it becomes double in 6 years. Let's use the formula for compound interest:

\( A = P(1 + r)^n \)

where:

\( A \) = final amount

\( P \) = principal amount

\( r \) = rate of interest

\( n \) = number of years

We know that the sum becomes twice in 6 years:

\( 2P = P(1 + r)^6 \)

Which simplifies to:

\( 2 = (1 + r)^6 \)

Taking the logarithm on both sides:

\( \log(2) = 6 \log(1 + r) \)

Thus:

\( \log(1 + r) = \frac{\log(2)}{6} \)

Now we need the sum to be 8 times:

\( 8P = P(1 + r)^n \)

It simplifies to:

\( 8 = (1 + r)^n \)

Taking the logarithm on both sides:

\( \log(8) = n \log(1 + r) \)

We know:

\( \log(8) = 3 \log(2) \)

Using the expression for \(\log(1 + r)\) from above, we have:

\( 3 \log(2) = n \cdot \frac{\log(2)}{6} \)

Solving for \( n \):

\( n = 3 \times 6 = 18 \)

Thus, the sum becomes 8 times of itself in 18 years.