Question

Anish borrowed Rs. 15000 at the rate of 12% and another amount at the rate of 15% for two years. The total interest paid by him was Rs. 9000. How much did he borrow?

Answer 1

Hint: We first assume the amount for the second installment. We describe the formula which gives us the interest for a specific principal. We use that to find the interests for both amounts of principal. We add them and equate it with 9000. We solve the equation to find the amount of the second installment.

Complete step by step answer:
We have two different amounts of money borrowed for Anish at a different rates for the same amount of time span.
Anish borrowed Rs. 15000 at the rate of 12% and another amount at the rate of 15% for two years. The total interest paid by him was Rs. 9000. Let the other principal amount be Rs. x.
If the total interest for the principal amount P with rate r% for the time n years be A then we can say $ A=\dfrac{n\Pr }{100} $ .
So, for the first amount of Rs. 15000 at the rate of 12% for 2 years will be $ {{A}_{1}} $ . We place the values $ n=2,P=15000,r=12 $ .
The interest will be $ {{A}_{1}}=\dfrac{2\times 15000\times 12}{100}=3600 $ .
So, for the second amount of Rs. x at the rate of 15% for 2 years will be $ {{A}_{2}} $ . We place the values $ n=2,P=x,r=15 $ .
The interest will be $ {{A}_{2}}=\dfrac{2\times x\times 15}{100}=\dfrac{3x}{10} $ .
Total interest be $ {{A}_{1}}+{{A}_{2}}=3600+\dfrac{3x}{10} $ . This is equal to 9000.
Forming the equation, we get $ {{A}_{1}}+{{A}_{2}}=3600+\dfrac{3x}{10}=9000 $ .
 $ \begin{align}
  & \Rightarrow 3600+\dfrac{3x}{10}=9000 \\
 & \Rightarrow \dfrac{3x}{10}=9000-3600=5400 \\
 & \Rightarrow x=\dfrac{5400\times 10}{3}=18000 \\
\end{align} $
Therefore, he borrowed Rs. 18000 in the second installment.

Note:
The amount of 9000 was only for the interest. The principal was not added to that. That’s why we didn’t add the principal in the formula. If not mentioned otherwise, we will always use the concept of simple interest, not compound interest.