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A sum of money doubles itself at a certain rate of compound interest in 12 years when the interest is compounded annually. In how many more years will it become eight times of itself?
24
48
12
36
24
In compound interest, if money becomes x times in t years, it becomes xn times in n×t years. Here, 8=23, so total time =3×12=36 years. Since 12 years have passed, additional time =36−12=24 years.
A sum of money doubles in 12 years with annual compound interest. We need to find the number of additional years for it to become 8 times.
A=P(1+r)n
In compound interest, if money becomes x times in t years, it becomes xn times in n×t years. Here, 8=23, so total time =3×12=36 years. Since 12 years have passed, additional time =36−12=24 years.
Students often misinterpret 'how many more years' and select the total time (36 years) instead of the additional time (24 years).
Establish the base relationship
Given the sum doubles in 12 years, we can write 2P=P(1+r)12, which simplifies to (1+r)12=2.
(1+r)12=2
Relate to the target amount
We need the sum to become 8 times itself, so we solve 8P=P(1+r)T for the total time T.
8=(1+r)T
Solve for total time
Substitute 8=23 into the equation. Since 2=(1+r)12, we have 8=((1+r)12)3=(1+r)36. Thus, T=36 years.
T=36
Calculate additional years
The total time required is 36 years. The question asks for 'how many more years' after the initial 12 years, so we subtract the initial time.
36−12=24
A is correct because the total time to reach 8 times is 36 years, and subtracting the initial 12 years leaves 24 more years.
This concept of power relations in growth is identical to half-life calculations in Physics or radioactive decay problems.
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