Shortcut Trick
In Compound Interest, if a sum becomes 'n' times in 'T' years, it becomes 'nk' times in 'k × T' years.
Here, the sum doubles (2 times) in 10 years (n = 2, T = 10).
To become four times (4 = 22), the power k = 2.
Time required = k × T = 2 × 10 = 20 years.
∴ The correct answer is 20 years.
Alternate Method
Given:
Initial Sum = P
Amount after 10 years = 2P
Target Amount = 4P
Formula Used:
A = P(1 + r/100)t
Calculations:
⇒ 2P = P(1 + r/100)10
⇒ 2 = (1 + r/100)10 ---- (i)
⇒ Let the required time be 't' years for the sum to become 4P.
⇒ 4P = P(1 + r/100)t
⇒ 4 = (1 + r/100)t
⇒ (2)2 = (1 + r/100)t
⇒ Substituting value of 2 from (i):
⇒ [(1 + r/100)10]2 = (1 + r/100)t
⇒ (1 + r/100)20 = (1 + r/100)t
⇒ By comparing powers, t = 20 years.
∴ The correct answer is 20 years.
Additional Information
Effective Rate of Interest
When interest is compounded more than once a year, the effective annual rate is (1 + r/n)n − 1, where n is the frequency of compounding.
Rule of 72
A quick way to estimate how long it takes to double money at CI: Time ≈ 72 ÷ Interest Rate.
SI vs CI for Doubling
In Simple Interest, if a sum doubles in T years, it becomes n times in (n−1) × T years. For 4 times, it would take (4−1) × 10 = 30 years.