Question

At what rate will the of 64000 be compounded annually to 68921 in 3 years

Answer 1

Answer:

Given :-

• A sum of Rs 64000 will be compounded annually to Rs 68921 in 3 years.

To Find :-

• What is the rate of interest.

Formula Used :-

$\clubsuit$ Amount Formula :

$\bigstar \: \: \sf\boxed{\bold{\pink{A =\: P\bigg(1 + \dfrac{r}{100}\bigg)^n}}}\: \: \: \bigstar$

where,

• A = Amount
• P = Principal
• r = Rate of Interest
• t = Time Period

Solution :-

Given :

• Principal = Rs 64000
• Amount = Rs 68921
• Time Period = 3 years

According to the question by using the formula we get,

$\implies \bf A =\: P\bigg(1 + \dfrac{r}{100}\bigg)^n$

$\implies \sf 68921 =\: 64000\bigg(1 + \dfrac{r}{100}\bigg)^3$

$\implies \sf \dfrac{68921}{64000} =\: \bigg(1 + \dfrac{r}{100}\bigg)^3$

$\implies \sf \bigg(\dfrac{41}{40}\bigg)^3 =\: \bigg(1 + \dfrac{r}{100}\bigg)^3$

$\implies \sf \dfrac{41}{40} =\: 1 + \dfrac{r}{100}$

$\implies \sf \dfrac{41}{40} - 1 =\: \dfrac{r}{100}$

$\implies \sf \dfrac{41 - 40}{40} =\: \dfrac{r}{100}$

$\implies \sf \dfrac{1}{40} =\: \dfrac{r}{100}$

By doing cross multiplication we get,

$\implies \sf 40 \times r =\: 100 \times 1$

$\implies \sf 40r =\: 100$

$\implies \sf r =\: \dfrac{100}{40}$

$\implies \sf\bold{\red{r =\: 2.5\%}}$

$\sf\bold{\purple{\underline{\therefore\: The\: rate\: of\: interest\: is\: 2.5\%\: .}}}$