Question

A total amount of Rs. 7000 is deposited in three different saving bank accounts with interest rates 5%, 8% and $8\dfrac{1}{2}\%$ , respectively. The total annual interest from the three bank accounts is Rs. 550. Equal amounts have been deposited in the 5% and 8% saving accounts. Find the amount deposited in each of the three accounts, with the help of matrices.

Answer 1

Hint: Let the amount invested at 5% interest be Rs. x and at $8\dfrac{1}{2}\%$ interest be Rs. z. As it is given that the amount invested in 5% and 8% are equal, so the amount invested in 8% is Rs.x. The first equation you get is using the point that the total amount invested is Rs. 7000 and the other is the sum of interests is Rs. 550. Solve the equations to get the answer.

Complete step-by-step answer:
Before starting with the question, let us know about interest.
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. In the above question we will deal with simple interests.
To start with the solution to the above question we let the amount invested at 5% interest be Rs. x and at $8\dfrac{1}{2}\%$ interest be Rs. z. As it is given that the amount invested in 5% and 8% are equal, so the amount invested in 8% is Rs.x.
Now it is given that the sum of interest acquired from each of the cases is equal to Rs. 550 and we know that $\text{interest=principal}\times \dfrac{r}{100}$ .
$\dfrac{4}{100}\times x+\dfrac{8}{100}\times x+\dfrac{8\dfrac{1}{2}}{100}\times z=550$
We know that $8\dfrac{1}{2}=\dfrac{17}{2}$ .
$\dfrac{12}{100}\times x+\dfrac{17}{2\times 100}\times z=550$
$\Rightarrow 24x+17z=110000.........(i)$
Also, it is given that the total amount that was invested is equal to Rs. 7000. This can be mathematically represented as:
$x+x+z=2700$
$\Rightarrow 2x+z=7000...........(ii)$
If we multiply both sides of the equation by 12, we get
$12\left( 2x+z \right)=12\times 7000$
\[\Rightarrow 24x+12z=84000..........(iii)\]
Now, we will subtract equation (iii) from equation (i). On doing so, we get
$24x+17z-24x-12z=110000-84000$
$\Rightarrow 5z=26000$
$\Rightarrow z=5200$
If we substitute the value of z in equation (ii), we get
$2x+5200=7000$
$\Rightarrow 2x=1800$
$\Rightarrow x=900$
Therefore, the amount invested is the bank account with 5%, 8% and $8\dfrac{1}{2}\%$ interest are Rs. 900, Rs. 900 and Rs. 5200, respectively.

Note: Be very careful in assuming the variables and using it in forming the equations, as it is a general observation that students assume the variables and then use it correctly in one equation and interchange their values in the next equation by getting confused in what they have assumed, which makes the solution completely wrong. Also, in such questions it is a good practice to put the values of variables you get back in the parent equations and verify your results.