Home  >>  CBSE XII  >>  Math  >>  Matrices

# A trust fund has Rs 30,000 that must be invested in two different types of bonds. The first bond pays 5 $\%$ interest per year, and the second bond pays 7 $\%$ interest per year. Using matrix multiplication, determine how to divide Rs 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of: b) Rs. 2000

This question has 2 parts and each part has been answered separately here.

Toolbox:
• With these type of word problems, we need to set up the correct matrix multiplication and solve for the unknown variables.
• Multiplication of two matrices is defined only if the number of columns of the left matrix is the same as the number of rows of the right matrix.
• If A is an m-by-n matrix and B is an n-by-p matrix, then their matrix product AB is the m-by-p matrix whose entries are given by dot product of the corresponding row of A and the corresponding column of B: $\begin{bmatrix}AB\end{bmatrix}_{i,j} = A_{i,1}B_{1,j} + A_{i,2}B_{2,j} + A_{i,3}B_{3,j} ... A_{i,n}B_{n,j}$
Let $Rs.30,000$ be divided into two parts. Assume that the amount invested in the first bond as $x$, then the amount invested in the second one is $Rs(30,000-x)$.
We are given that the first bond pays $5\%$, i,e., $0.05$ in interest and the second bond $7\%$, i.e., $0.07$.
We can represent the problem using matrix multiplication as the following $1\times 2$ matrix: $\begin{bmatrix}x & 30000-x\end{bmatrix}\begin{bmatrix}0.05\\0.07\end{bmatrix}$ which is equal to the interest earned $Rs. 2000$.
Given $\begin{bmatrix}x & 30000-x\end{bmatrix}\begin{bmatrix}0.05\\0.07\end{bmatrix} = 2000$:
$\begin{bmatrix}x\times (0.05)+(30000-x)\times 0.07\end{bmatrix} = 2000$
$0.05x + 2100 - 0.07x = 2000$
$2100 - 0.02x = 2000$
$2100-2000 = 0.02x$
$0.02x = 100 \rightarrow x = \frac{1}{0.02}100$
Solving for $x$, $x = 5,000$ and $30,000-x = 25,000$.
Hence the amount has to be divided into two sums of $Rs.5,000$ and $Rs.25,000$.