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A sum-of money placed at compound interest doubles itself in 3 years. In how many years will it amount to 8 times itself?
  • a)
    9 years
  • b)
    8 years
  • c)
    27 years
  • d)
    7 years
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
A sum-of money placed at compound interest doubles itself in 3 years....
Let sum = x
⇒ 8 = [21/3]T
⇒ 23 = 2T/3
⇒ T/3 = 3
∴ T = 9 yrs.
Alternative:
If a certain sum of money becomes ‘m’ times in ‘y’ years. Then it will become (mn) times in ‘n x y’ years. Hence 23 in 3 x 3 = 9 years,
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Most Upvoted Answer
A sum-of money placed at compound interest doubles itself in 3 years....
Solution:

Let the principal be P and rate of interest be R.

Given, P(1+R/100)^3 = 2P

Simplifying the above equation, we get:

(1+R/100)^3 = 2

Taking cube root on both sides, we get:

1+R/100 = 2^(1/3)

R/100 = 2^(1/3) - 1

R = 100(2^(1/3) - 1)

Now, we need to find the number of years it takes for the principal to become 8 times itself.

Let the required number of years be n.

Then, P(1+R/100)^n = 8P

Simplifying the above equation, we get:

(1+R/100)^n = 8

Taking logarithm on both sides, we get:

n*log(1+R/100) = log(8)

n = log(8)/log(1+R/100)

Substituting the value of R, we get:

n = log(8)/log(2^(1/3))

n = 3*log(8)/log(2)

n = 9

Therefore, the required number of years is 9 years.

Answer: (a) 9 years
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A sum-of money placed at compound interest doubles itself in 3 years. In how many years will it amount to 8 times itself?a)9 yearsb)8 yearsc)27 yearsd)7 yearsCorrect answer is option 'A'. Can you explain this answer?
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