Question

Q.2 A principal of Rs 24,000 becomes Rs 40,560 in 2 years at a certain rate of compound interest per annum. Find the rate of interest.​

Answer 1

Given:

A principal of Rs. 24,000 becomes Rs. 40,560 in 2 years at a certain rate of compound interest per annum. Find the rate of interest.​

To find:

The rate of interest

Solution:

We know,

$\boxed{\bold{A = P [1 + \frac{R}{100} ]^n}}$

The sum of money, P = Rs. 24000

The amount, A = Rs. 40560

The no. of years, n = 2 years

Let "R" be the rate of interest.

Therefore, on substituting the given values of A, P and n in the formula above, we get

$40560 = 24000 [1 + \frac{R}{100} ]^2$

$\implies \frac{40560}{24000} = [1 + \frac{R}{100} ]^2$

$\implies 1.69 = [1 + \frac{R}{100} ]^2$

$\implies \sqrt{1.69} = \sqrt{ [1 + \frac{R}{100} ]^2}$

$\implies 1.3 = [1 + \frac{R}{100} ]$

$\implies 1.3 - 1 = \frac{R}{100}$

$\implies 0.3= \frac{R}{100}$

$\implies R = 0.3 \times 100$

$\implies \bold{R = 30\%}$

Thus, the rate of interest is → 30%.

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Answer 2

Solution :-

Let us assume that, the rate of interest is R% per annum .

So,

→ Amount = Principal[1 + (R/100)]^(Time)

putting given values now, we get,

→ 40560 = 24000[1 + (R/100)]²

→ (40560/24000) = [1 + (R/100)]²

→ (4056/2400) = [1 + (R/100)]²

→ (338/200) = [1 + (R/100)]²

→ (169/100) = [1 + (R/100)]²

→ (13/10)² = [1 + (R/100)]²

square root both sides,

→ (13/10) = 1 + (R/100)

→ (R/100) = (13/10) - 1

→ (R/100) = (13 - 10)/10

→ (R/100) = (3/10)

→ R = 30% (Ans.)

Hence, the rate of interest is 30% per annum .

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