Question

In how many years will Rs.6250 amount to Rs.7290 at 8%per annum, compounded anually?

Answer 1

Answer:

Time (T) = 2 years

Step-by-step explanation:

Given Principal (P) = Rs 6250,

Amount (A) = Rs7290,

Rate of interest (r) = 8%,

Let Time = T years

Number of times interest paid = n,

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

$\implies 7290 = 6250\left(1+\frac{8}{100}\right)^{n}$

$\implies \frac{7290}{6250}=\left(1+\frac{2}{25}\right)^{n}$

$\implies \frac{729}{625}=\left(\frac{25+2}{25}\right)^{n}$

$\implies \left(\frac{27^{2}}{25^{2}}\right)=\left(\frac{27}{25}\right)^{n}$

$\implies \left(\frac{27}{25}\right)^{2}=\left(\frac{27}{25}\right)^{n}$

$\implies 2=n$

Therefore,

Time (T) = 2 years

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Answer 2

SOLUTION:-

Given:

•Principal,(P)= Rs.6250

•Amount, (A)= Rs.7290

•Rate, (R)= 8%

To find:

The time (years).

Explanation:

We know that, formula of the compound Interest;

C.I. = Amount - Principal.

&

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

So,

$= > 7290 = 6250(1 + \frac{8}{100} ) {}^{n} \\ \\ = > \frac{7290}{6250} = (1 + \frac{2}{25} ) {}^{n} \\ \\ = > \frac{729}{625} = ( \frac{25 + 2}{25} ) {}^{n} \\ \\ = > \sqrt{ \frac{729}{625} } = (\frac{27}{25} ) {}^{n} \\ \\ = >( \frac{27}{25} ) {}^{2} = ( \frac{27}{25} ) {}^{n} \\ \\ = > n = 2 \: years$

Thus,

The time is 2 years.

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