Question

A sum of money placed at compound interest doubles itself in 3 years. in how many years will it amount to 8 times itself?

Answer 1

Answer:

In 9 years the amount becomes 8 times itself.

Step-by-step explanation:

i )Let the principal = P

Time (T) = 3 years,

Number of times interest paid (n) = 3,

Amount = 2P

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

$\implies 2P = P\left(1+\frac{R}{100}\right)^{3}$

$\implies 2= \left(1+\frac{R}{100}\right)^{3}\:---(1)$

ii )Let the principal = P

Time (T) = ?

Number of times interest paid (n) =?

Amount = 8P

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

$\implies 8P = P\left(1+\frac{R}{100}\right)^{n}$

$\implies 8= \left(1+\frac{R}{100}\right)^{n}$

$\implies 2^{3}= \left(1+\frac{R}{100}\right)^{n}$

$\implies \left(\big(1+\frac{R}{100}\big)^{3}\right)^{3}=\left(1+\frac{R}{100}\right)^{n}$

$\implies \left(1+\frac{R}{100}\right)^{9}=\left(1+\frac{R}{100}\right)^{n}$

$\implies n = 9$

Therefore,

In 9 years the amount becomes 8 times itself.

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Answer 2

Answer:

9 years

Step-by-step explanation:

Let, 100 be the amount.

100..............200...............3 years

100..............400...............+3 years

100..............800...............+3 years

So, Total Time = 9 years.