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2 Answers

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2 votes

For compound interest compounded anually,

$A= P\left (1+ \frac{r}{100} \right )^n$

Given $A= 2P$ when $n=6$

$\implies 2P= P\left (1+ \frac{r}{100} \right )^6$

$\implies 2= \left (1+ \frac{r}{100} \right )^6$

$\implies 2^{\frac{1}{6}}= \left (1+ \frac{r}{100} \right )$

$\implies 1.122= 1+ \frac{r}{100}$

$\implies 0.122= \frac{r}{100} $

$\implies r= 12.2\ \% $


Now $A= 16P$  then $n= ? $

$A= P\left (1+ \frac{r}{100} \right )^n$

$\implies 16P= P\left (1+ \frac{12.2}{100} \right )^n$

$\implies 16= \left (1+ \frac{12.2}{100} \right )^n$

$\implies 16= \left (\frac{112.2}{100} \right )^n$

$\implies 16= \left (1.122 \right )^n$

$\implies n= log_{1.122}16 = 24.08 \approx 24$

$\therefore$ Option $B.$ is correct.

Answer:

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