MAT Exam 2020 Paper

Concept:Let the amount borrowed at 12% be $x$.The remaining amount is borrowed at 14%.Use simple interest for one year: $\text{Interest} = \frac{P \times R \times T}{100}$.Explanation:Loan at 12% per annum = $x$ ₹.Loan at 14% per annum = $(1500 - x)$ ₹.Interest at 12% for 1 year = $\frac{x \times 12 \times 1}{100}$.Interest at 14% for 1 year = $\frac{(1500 - x) \times 14 \times 1}{100}$.Total interest given is ₹186.So, $\frac{12x}{100} + \frac{14(1500 - x)}{100} = 186$.Multiplying by 100: $12x + 14(1500 - x) = 18600$.$12x + 21000 - 14x = 18600$$-2x = 18600 - 21000 = -2400$$x = \frac{2400}{2} = 1200$Answer:Ashok borrows ₹1200 at the rate of 12%.Thus, the correct option is A. ₹1200.