Question

A trust caring for handicapped children gets Rs 30000 every month from its donors. The trust spends half of the funds received for medical & educational care of the children & for that it charges 2% of the spent amount from them, & deposited the balance amount in a private bank to get the money multiplied so that the trust goes on functioning regularly. What percent of interest should the trust get from the bank so as to get a total of Rs. 1800 every month? Use the matrix method to find the rate of interest.

Answer 1

Hint: Let us assume that the earning of trust from all the charges is x and earning of trust from the interest obtained from the bank is y. It is given that the total earning of trust is Rs. 1800. So we get equation as: $x+y=1800$
Also, it is said that the trust receives Rs 30000 every month out of which it spends half on medical & educational care of the children, i.e. Rs 15000 and deposits rest to the bank. It also charges 2% of this amount to the bank. So, the earning of trust from these charges is: 2% of Rs. 15000, i.e. Rs. 300. So, we have another equation: $x=300$
So, for a system of linear equations in two variables:
$\begin{array}{rl}& ax+by=p\\ & cx+dy=q\end{array}$
We can solve by using matrix method as:
$\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}p\\ q\end{array}\right]$
Solve the matrix for the given equations to get the value of y. Now, assume that the interest rate is r%. So, by using the formula: $\text{interest}=rate\times amount$ , where y is the interest and Rs. 15000 is the amount, get the value of interest rate.

Complete step by step answer:
As we have a system of linear equations in two variables:
$\begin{array}{rl}& x+y=1800......(1)\\ & x=300......(2)\end{array}$
So, by using matrix method, we can write:
$\left[\begin{array}{cc}1& 1\\ 1& 0\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}1800\\ 300\end{array}\right]......(3)$
Now, by separating the variables on one side, we can write:
$$\Rightarrow \left[\begin{array}{c}x\\ y\end{array}\right]={\left[\begin{array}{cc}1& 1\\ 1& 0\end{array}\right]}^{-1}\left[\begin{array}{c}1800\\ 300\end{array}\right]......(4)$$
For a matrix: $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ , we have $$A^{-1}={\displaystyle \frac{adjA}{\left|A\right|}}$$, where $adjA={\left[{C_i}_j\right]}^T$ and ${C_i}_j={(-1)}^{i+j}{M_i}_j$
So, for matrix: $A=\left[\begin{array}{cc}1& 1\\ 1& 0\end{array}\right]$
$\begin{array}{rl}& \left|A\right|=-1\\ & {M_i}_j=\left[\begin{array}{cc}0& 1\\ 1& 1\end{array}\right]\\ & {C_i}_j={(-1)}^{i+j}\left[\begin{array}{cc}0& 1\\ 1& 1\end{array}\right]\\ & =\left[\begin{array}{cc}0& -1\\ -1& 1\end{array}\right]\\ & adjA={\left[\begin{array}{cc}0& -1\\ -1& 1\end{array}\right]}^T\\ & =\left[\begin{array}{cc}0& -1\\ -1& 1\end{array}\right]\end{array}$
Therefore,
$$\begin{array}{rl}& A^{-1}={\displaystyle \frac{1}{-1}}\left[\begin{array}{cc}0& -1\\ -1& 1\end{array}\right]\\ & =\left[\begin{array}{cc}0& 1\\ 1& -1\end{array}\right]\end{array}$$
So, we can write equation (4) as:
$$\Rightarrow \left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{cc}0& 1\\ 1& -1\end{array}\right]\left[\begin{array}{c}1800\\ 300\end{array}\right]......(5)$$

Now, solve the matrix, to get the value of y:
$$\begin{array}{rl}& \Rightarrow \left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}(0\times 1800)+(1\times 300)\\ (1\times 1800)+(-1\times 300)\end{array}\right]\\ & \Rightarrow \left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}300\\ 1500\end{array}\right]\end{array}$$
Therefore, y = Rs. 1500
Now, by using the formula: $\text{interest}=rate\times amount$, we get the interest rate r as:
$\begin{array}{rl}& \Rightarrow 1500={\displaystyle \frac{r}{100}}\times 15000\\ & \Rightarrow r=10\mathrm{\%}\end{array}$

Hence, the interest rate is 10%

Note: There is an alternate method for solving the inverse of a matrix, i.e.
For a matrix: $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ , we have $$A^{-1}={\displaystyle \frac{1}{ad-bc}}\left[\begin{array}{cc}d& -c\\ -b& a\end{array}\right]$$
So, we can write equation (4) as:
$$\begin{array}{rl}& \Rightarrow \left[\begin{array}{c}x\\ y\end{array}\right]={\displaystyle \frac{1}{(1\times 0)-(1\times 1)}}\left[\begin{array}{cc}0& -1\\ -1& 1\end{array}\right]\left[\begin{array}{c}1800\\ 300\end{array}\right]\\ & \Rightarrow \left[\begin{array}{c}x\\ y\end{array}\right]={\displaystyle \frac{1}{-1}}\left[\begin{array}{cc}0& -1\\ -1& 1\end{array}\right]\left[\begin{array}{c}1800\\ 300\end{array}\right]\\ & \Rightarrow \left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{cc}0& 1\\ 1& -1\end{array}\right]\left[\begin{array}{c}1800\\ 300\end{array}\right]\end{array}$$