Question

A sum of INR 27800 is invested for 2 years in 2 different schemes, both on simple interest. The rates of interest are 14% and 11% in the 1st and 2nd schemes respectively. The total interest after 2 years is INR 7016. What sum was invested in the 2nd scheme?

• A. 12800
• B. 16200
• C. 14600
• D. None of these

Hint:

When solving problems involving simple interest, use the formula \( I = \frac{P \times R \times T}{100} \) and set up equations based on the total sum and interest to solve for the unknowns.

Answer 1

The Correct Option is A

Let the sum invested in the 1st scheme be \( P_1 \) and in the 2nd scheme be \( P_2 \). The total sum invested is:
\[ P_1 + P_2 = 27800 \] The total interest is given by the formula for simple interest:
\[ I = \frac{P \times R \times T}{100} \] where \( P \) is the principal, \( R \) is the rate of interest, and \( T \) is the time in years.
For the 1st scheme:
\[ I_1 = \frac{P_1 \times 14 \times 2}{100} = \frac{28P_1}{100} \] For the 2nd scheme:
\[ I_2 = \frac{P_2 \times 11 \times 2}{100} = \frac{22P_2}{100} \] The total interest is INR 7016:
\[ I_1 + I_2 = 7016 \] \[ \frac{28P_1}{100} + \frac{22P_2}{100} = 7016 \] Multiplying through by 100:
\[ 28P_1 + 22P_2 = 701600 \] From \( P_1 + P_2 = 27800 \), we can express \( P_1 \) as:
\[ P_1 = 27800 - P_2 \] Substitute into the equation:
\[ 28(27800 - P_2) + 22P_2 = 701600 \] \[ 778400 - 28P_2 + 22P_2 = 701600 \] \[ 778400 - 6P_2 = 701600 \] \[ 6P_2 = 76800 \] \[ P_2 = \frac{76800}{6} = 12800 \] Thus, the sum invested in the 2nd scheme is INR 12800.