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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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A sum of money placed at compound interest doubles itself in 3 years. ...
If it doubles in 3 years, it would become 4 times in 6 year and 8 times in 9 years.
Most Upvoted Answer
A sum of money placed at compound interest doubles itself in 3 years. ...
Given:
The sum of money placed at compound interest doubles itself in 3 years.
To Find:
In how many years will it amount to 8 times itself?
Solution:
Let the principal amount be P.
The amount after 3 years will be 2P (as it doubles itself in 3 years).
Let the time period required to amount to 8 times itself be n years.
Then, the amount after n years will be 8P.
Using the compound interest formula: A = P(1 + r/n)^(n*t)
where A is the amount, P is the principal, r is the rate of interest, n is the number of times the interest is compounded per year, and t is the time period.
For the first case, after 3 years:
2P = P(1 + r/n)^(n*3)
For the second case, after n years:
8P = P(1 + r/n)^(n*n)
Dividing the second equation by the first equation, we get:
4 = (1 + r/n)^(n*(n-3))
As we can see, we have two variables (r and n) and one equation. Therefore, we cannot solve for n directly.
Using trial and error method, we can check the options given:
For option A (9 years):
n = 9, then (n-3) = 6
Taking r/n = 100%, we get:
4 = (1 + 1)^(9*6)
4 = 2^54
4 = 4
Hence, option A (9 years) is the correct answer.
Alternatively, we can also use logarithms to solve for n:
Taking logarithm on both sides, we get:
log(4) = n*log(1 + r/n)^(n-3)
log(4) = (n-3)*log(1 + r/n)
log(4)/(n-3) = log(1 + r/n)
e^(log(4)/(n-3)) = 1 + r/n
e^(log(4)/(n-3)) - 1 = r/n
n/r = (e^(log(4)/(n-3)) - 1)^(-1)
Using trial and error method or any numerical method, we can solve for n using the above equation.
The sum of money placed at compound interest doubles itself in 3 years.
To Find:
In how many years will it amount to 8 times itself?
Solution:
Let the principal amount be P.
The amount after 3 years will be 2P (as it doubles itself in 3 years).
Let the time period required to amount to 8 times itself be n years.
Then, the amount after n years will be 8P.
Using the compound interest formula: A = P(1 + r/n)^(n*t)
where A is the amount, P is the principal, r is the rate of interest, n is the number of times the interest is compounded per year, and t is the time period.
For the first case, after 3 years:
2P = P(1 + r/n)^(n*3)
For the second case, after n years:
8P = P(1 + r/n)^(n*n)
Dividing the second equation by the first equation, we get:
4 = (1 + r/n)^(n*(n-3))
As we can see, we have two variables (r and n) and one equation. Therefore, we cannot solve for n directly.
Using trial and error method, we can check the options given:
For option A (9 years):
n = 9, then (n-3) = 6
Taking r/n = 100%, we get:
4 = (1 + 1)^(9*6)
4 = 2^54
4 = 4
Hence, option A (9 years) is the correct answer.
Alternatively, we can also use logarithms to solve for n:
Taking logarithm on both sides, we get:
log(4) = n*log(1 + r/n)^(n-3)
log(4) = (n-3)*log(1 + r/n)
log(4)/(n-3) = log(1 + r/n)
e^(log(4)/(n-3)) = 1 + r/n
e^(log(4)/(n-3)) - 1 = r/n
n/r = (e^(log(4)/(n-3)) - 1)^(-1)
Using trial and error method or any numerical method, we can solve for n using the above equation.
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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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