A sum of money at compound interest amounts to twice itself in four years. In how many years will it be four times itself?

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Solution

The correct option is C
8

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

$2 P = P (1 + \frac{r}{100})^{4}$

$N o w, 4 P = P (1 + \frac{r}{100})^{n}$

$T h u s, n = 8$

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A sum of money at compound interest amounts to twice itself in four years. In how many years will it be four times itself?

The correct option is C 8 A=P(1+r100)n 2P=P(1+r100)4 Now, 4P=P(1+r100)n Thus, n=8

The correct option is C
8

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

$2 P = P (1 + \frac{r}{100})^{4}$

$N o w, 4 P = P (1 + \frac{r}{100})^{n}$

$T h u s, n = 8$