Question

Sagarika divides her savings of 10000 rupees to invest across two schemes A and B. Scheme A offers an interest rate of 10% per annum, compounded half-yearly, while scheme B offers a simple interest rate of 12% per annum. If at the end of the first year, the value of her investment in scheme B exceeds the value of her investment in scheme A by 2310 rupees, then the total interest, in rupees, earned by Sagarika during the first year of investment is:

• A. 1100
• B. 1130
• C. 1111
• D. 1000

Hint:

For compound interest and simple interest problems, be sure to apply the correct formulas for each type and handle the algebra carefully. Use transformations to simplify the problem if needed.

Answer 1

The Correct Option is B

Step 1: Let the investment in scheme A be \( x \) rupees. Then, the investment in scheme B will be \( 10000 - x \). 
Step 2: Interest earned in scheme A. The interest rate for scheme A is 10% per annum, compounded half-yearly. For compound interest, the formula is: \[ A = P \left( 1 + \frac{r}{2} \right)^{2t}, \] where: - \( P \) is the principal (investment), - \( r \) is the annual interest rate (10% = 0.1), - \( t \) is the time in years (1 year). Thus, the interest earned in scheme A in the first year is: \[ A_A = x \left( 1 + \frac{0.1}{2} \right)^{2 \times 1} = x \left( 1 + 0.05 \right)^2 = x \times (1.05)^2 = x \times 1.1025. \] The interest earned in scheme A is: \[ \text{Interest in A} = x \times 1.1025 - x = x(1.1025 - 1) = x \times 0.1025. \] Step 3: Interest earned in scheme B. For simple interest, the formula is: \[ I = \frac{P \times r \times t}{100}. \] Thus, the interest earned in scheme B is: \[ I_B = \frac{(10000 - x) \times 12 \times 1}{100} = (10000 - x) \times 0.12. \] The interest earned in scheme B is: \[ \text{Interest in B} = (10000 - x) \times 0.12 = 1200 - 0.12x. \] Step 4: Set up the equation using the condition. At the end of the first year, the value of her investment in scheme B exceeds that in scheme A by 2310 rupees. This means: \[ \text{Value in B} - \text{Value in A} = 2310. \] The values are: \[ (10000 - x) \times 1.12 - x \times 1.1025 = 2310. \] Expanding and solving for \( x \): \[ 11200 - 1.12x - 1.1025x = 2310 \quad \Rightarrow \quad 11200 - 2.2225x = 2310 \quad \Rightarrow \quad 2.2225x = 8890 \quad \Rightarrow \quad x = 4000. \] Step 5: Calculate the total interest earned. Now that we know \( x = 4000 \), the interest earned in scheme A is: \[ \text{Interest in A} = 4000 \times 0.1025 = 410. \] The interest earned in scheme B is: \[ \text{Interest in B} = (10000 - 4000) \times 0.12 = 6000 \times 0.12 = 720. \] Thus, the total interest earned is: \[ \text{Total interest} = 410 + 720 = 1130. \] Thus, the correct answer is (B) 1130.