Question

A person invested in all Rs. 2600 at 4%, 6% and 8% per annum simple interest. At the end of the year, he got the same interest in all the three cases. The money invested at 4% is:\[\]
(a) Rs. 200\[\]
(b) Rs. 600\[\]
(c) Rs. 800\[\]
(d) Rs. 1200\[\]

Answer 1

Hint: Let the amount invested at 4% interest be Rs. x, at 6% interest be Rs. y and at 8% interest be Rs. z. The first equation you get is using the point that the total of x, y and z is Rs. 2600 and the other is the interest of one year for all three investments were equal.\[\]

Complete step-by-step answer:
We know that interest in the financial term is the amount that a borrower pays to the lender along with the repayment of the actual principal amount. Broadly, there are two kinds of interest, first is the simple interest, and the other is the compound interest. We will deal with simple interest In the above question. \[\]
To start with the solution to the above question we let the amount invested at 4% interest be Rs. x, at 6% interest be Rs. y and at 8% interest be Rs. z. Now it is given that the interest acquired from each of the cases are equal and we know that interest$=$principal$\times \dfrac{r}{100}$ . So we have,
\[\dfrac{4}{100}\times x=\dfrac{6}{100}\times y=\dfrac{8}{100}\times z\]
If we take first two terms at a time and we get:
\[\begin{align}
  & \dfrac{4}{100}\times x=\dfrac{6}{100}\times y \\
 & \Rightarrow \dfrac{2}{3}\times x=y.......(i) \\
\end{align}\]

We take the second and third term and get,
\[\begin{align}
  & \dfrac{4}{100}\times x=\dfrac{8}{100}\times z \\
 & \Rightarrow z=\dfrac{1}{2}x.............(ii) \\
\end{align}\]

Also, it is given that the total amount that was invested is equal to Rs. 2600. This can be mathematically represented as:
\[x+y+z=2600\]
If we substitute y and z from equation (i) and (ii), we get
\[x+\dfrac{2}{3}x+\dfrac{1}{2}x=2600\]
We Take 6 as the LCM of the denominator on the left hand side and the terms to get,

\[\begin{align}
  & \dfrac{\left( 4+3+6 \right)x}{6}=2600 \\
 & \Rightarrow 13x=2600\times 6 \\
 & \Rightarrow x=1200 \\
\end{align}\]
Therefore, the answer to the above question is option (d).\[\]

So, the correct answer is “Option d”.

Note: Don’t get confused and take 2600 to be the principal amount in each case. Also, remember whenever you are dealing with a system of equations in three variables, try to represent two of the variables in terms of the third one and substitute in the untouched equation to get the solution. The variable in whose terms you are expressing will decide the complexity and time for solving the system.