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A sum of money becomes eight times in 3 years, if the rate is compounded annually. In how much time will the same amount at the same compound rate become sixteen times ?
- A6 years
- B4 years
- C8 years
- D5 years
Answer not set
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CCommunity
Formula for compound interest:-
$(A = P{\left( {1 + \frac{R}{{100}}} \right)^T})$, CI = A – P
Where CI is compound interest, A is amount, P is principal, R is rate, T is time
Given, Sum of money becomes 8 times in 3 years.
∴ A = 8P
$(\Rightarrow 8{\rm{P}} = P{\left( {1 + \frac{R}{{100}}} \right)^3})$
$(\Rightarrow {\rm{\;}}8 = {\left( {1 + \frac{R}{{100}}} \right)^3})$
$(\Rightarrow {2^3} = {\left( {1 + \frac{R}{{100}}} \right)^3})$
⇒ 2 = 1 + R/100
⇒ R = 100 %
Let the time be T in which sum of money becomes 16 times of itself.
∴ A = 16P
$(\Rightarrow 16{\rm{P\;}} = P{\left( {1 + \frac{R}{{100}}} \right)^T})$
$(\Rightarrow 16{\rm{\;}} = {\left( {1 + \frac{{100}}{{100}}} \right)^T})$
⇒ 24 = 2T
By property of indices ; ab = ac, ⇒ b = c
⇒ T = 4 years
$(A = P{\left( {1 + \frac{R}{{100}}} \right)^T})$, CI = A – P
Where CI is compound interest, A is amount, P is principal, R is rate, T is time
Given, Sum of money becomes 8 times in 3 years.
∴ A = 8P
$(\Rightarrow 8{\rm{P}} = P{\left( {1 + \frac{R}{{100}}} \right)^3})$
$(\Rightarrow {\rm{\;}}8 = {\left( {1 + \frac{R}{{100}}} \right)^3})$
$(\Rightarrow {2^3} = {\left( {1 + \frac{R}{{100}}} \right)^3})$
⇒ 2 = 1 + R/100
⇒ R = 100 %
Let the time be T in which sum of money becomes 16 times of itself.
∴ A = 16P
$(\Rightarrow 16{\rm{P\;}} = P{\left( {1 + \frac{R}{{100}}} \right)^T})$
$(\Rightarrow 16{\rm{\;}} = {\left( {1 + \frac{{100}}{{100}}} \right)^T})$
⇒ 24 = 2T
By property of indices ; ab = ac, ⇒ b = c
⇒ T = 4 years