At what rate per cent compound interest, does a sum of money become 2.25 times of itself in 2 years?

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Solution

$A = P (1 + \frac{R}{100})^{T}$
According to the question
P=x, A=2.25x, T=2 years, R=?
$2.25 x = x (1 + \frac{R}{100})^{2}$
$(1 + \frac{R}{100})^{2} = \frac{2.25 x}{x}$
$(1 + \frac{R}{100})^{2} = 2.25$
$1 + \frac{R}{100} = 1.5$
$\frac{R}{100} = 1.5 - 1$
$\frac{R}{100} = .5$
$R = 50$ %

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At what rate per cent compound interest, does a sum of money become 2.25 times of itself in 2 years?

A=P(1+R100)TAccording to the questionP=x, A=2.25x, T=2 years, R=?2.25x=x(1+R100)2(1+R100)2=2.25xx(1+R100)2=2.251+R100=1.5R100=1.5−1R100=.5R=50%

$A = P (1 + \frac{R}{100})^{T}$
According to the question
P=x, A=2.25x, T=2 years, R=?
$2.25 x = x (1 + \frac{R}{100})^{2}$
$(1 + \frac{R}{100})^{2} = \frac{2.25 x}{x}$
$(1 + \frac{R}{100})^{2} = 2.25$
$1 + \frac{R}{100} = 1.5$
$\frac{R}{100} = 1.5 - 1$
$\frac{R}{100} = .5$
$R = 50$ %