At what rate per cent compound interest, does a sum of money become 1.44 times of itself in 2 years ?

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Solution

$A = P (1 + \frac{r}{100})^{n} 1.44 p = p (\frac{r}{100})^{2} \sqrt{1.44} = (\frac{r}{100}) 1.2 = 1 + \frac{r}{100} 1.2 - 1 = \frac{r}{100} 0.2 \times 100 = r r = 20 %$

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At what rate per cent compound interest, does a sum of money become 1.44 times of itself in 2 years ?

A=P(1+r100)n1.44p=p(r100)2√1.44=(r100)1.2=1+r1001.2−1=r1000.2×100=rr=20%

$A = P (1 + \frac{r}{100})^{n} 1.44 p = p (\frac{r}{100})^{2} \sqrt{1.44} = (\frac{r}{100}) 1.2 = 1 + \frac{r}{100} 1.2 - 1 = \frac{r}{100} 0.2 \times 100 = r r = 20 %$