A sum becomes three times itself at compound interest in eight years. In how many years will the sum become twenty seven times itself?

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Solution

The correct option is A 24
When interest is compounded, Amount $A = P (1 + \frac{R}{100})^{n}$
Given, $3 P = P (1 + \frac{R}{100})^{8}$
$=> 3 = (1 + \frac{R}{100})^{8}$ --- (1)
And, $27 P = P (1 + \frac{R}{100})^{n}$
$=> 27 = (1 + \frac{R}{100})^{n}$
$=> 3^{3} = (1 + \frac{R}{100})^{n}$
From eqn 1
$=> ((1 + \frac{R}{100})^{8})^{3} = (1 + \frac{R}{100})^{n}$
$=> (1 + \frac{R}{100})^{24} = (1 + \frac{R}{100})^{n}$
$=> n = 24$

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A sum becomes three times itself at compound interest in eight years. In how many years will the sum become twenty seven times itself?

The correct option is A 24When interest is compounded, Amount A=P(1+R100)nGiven, 3P=P(1+R100)8=>3=(1+R100)8 --- (1)And, 27P=P(1+R100)n=>27=(1+R100)n =>33=(1+R100)n From eqn 1=>((1+R100)8)3=(1+R100)n =>(1+R100)24=(1+R100)n =>n=24