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If the amount is 3(3)/8 times the sum af...

If the amount is `3(3)/8` times the sum after 3 years at compound interest compounded annually, then the rate of interest per annum is

A

`25%`

B

`50%`

C

`16(2)/3%`

D

`33(1)/3%`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem step by step, we will use the formula for compound interest and the information given in the question. ### Step 1: Understand the given information We are given that the amount after 3 years is \( \frac{27}{8} \) times the principal. We need to find the rate of interest per annum. ### Step 2: Assume the principal Let us assume the principal amount (P) is \( 8x \). ### Step 3: Calculate the amount using the given ratio According to the problem, the amount (A) after 3 years is: \[ A = \frac{27}{8} \times P = \frac{27}{8} \times 8x = 27x \] ### Step 4: Use the compound interest formula The formula for the amount in compound interest is: \[ A = P \left(1 + \frac{r}{100}\right)^n \] where: - \( A \) is the amount after time \( n \), - \( P \) is the principal, - \( r \) is the rate of interest, - \( n \) is the number of years. Substituting the values we have: \[ 27x = 8x \left(1 + \frac{r}{100}\right)^3 \] ### Step 5: Simplify the equation Dividing both sides by \( 8x \): \[ \frac{27x}{8x} = \left(1 + \frac{r}{100}\right)^3 \] This simplifies to: \[ \frac{27}{8} = \left(1 + \frac{r}{100}\right)^3 \] ### Step 6: Take the cube root of both sides To solve for \( 1 + \frac{r}{100} \), we take the cube root: \[ 1 + \frac{r}{100} = \sqrt[3]{\frac{27}{8}} \] ### Step 7: Calculate the cube root Calculating the cube root: \[ \sqrt[3]{\frac{27}{8}} = \frac{\sqrt[3]{27}}{\sqrt[3]{8}} = \frac{3}{2} \] ### Step 8: Solve for \( r \) Now we have: \[ 1 + \frac{r}{100} = \frac{3}{2} \] Subtracting 1 from both sides: \[ \frac{r}{100} = \frac{3}{2} - 1 = \frac{3}{2} - \frac{2}{2} = \frac{1}{2} \] Multiplying both sides by 100: \[ r = 100 \times \frac{1}{2} = 50 \] ### Final Answer Thus, the rate of interest per annum is \( 50\% \). ---
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