Question

A sum of money invested at compound interest doubles itself in 4 years, interest being payable annually. in how much time will it be eight times?

Answer 1

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Answer 2

$\huge\red{Answer\: is}$

Let the Sum is P
and R = R% p.a
T = 4 years = n = 4 (Compounded annually)
A = 2p
As we know the formula

A = P(1+ R/100)^n
$2p = p(1 + \frac{r}{100}) {}^{4} \\ \frac{2p}{p} = (1 + \frac{r}{100} ) {}^{4} \\ 2 = (1 + \frac{r}{100} ) {}^{4} \\ 2 {}^{ \frac{1}{4} } = (1 + \frac{r}{100} )$
Now
Sum = P
A = 8P
T = n years
As we know that
A = P(1+R/100)^n
$8p = p(1 + \frac{r}{100} ) {}^{n} \\ \frac{8p}{p} = (1 + \frac{r}{100} ) {}^{n} \\ 8 =( 2 {}^{ \frac{1}{4} } ) {}^{n} \\ 2 {}^{3} = 2 {}^{ \frac{n}{4} } \\ as \: on \: the \: same \: base \\ 3 = \frac{n}{4} \\ n = 4 \times 3 \\ n = 12 \: years$

Hence the Sum is eight times in 12 years

$<marquee>Hope it helps you$