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Home  >>  CBSE XII  >>  Math  >>  Matrices
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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: $$ \text{(a) Rs 1800} \qquad \qquad \text{(b) Rs 2000} $$

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  • 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 Rs30,000 be divided into two parts ,Assume one bond as x,then second one is Rs(30,000-x).Let us represent this by $1\times 2$ matrix.[x(30000-x)]
Given Rate of interest as 0.05&0.07 per rupee.Let us denote with matrix R of order $2\times 1$
$R=\begin{bmatrix}0.05\\0.07\end{bmatrix}$
$(i)\begin{bmatrix}x & 30000-x\end{bmatrix}\begin{bmatrix}0.05\\0.07\end{bmatrix}$
$\Rightarrow \begin{bmatrix}x\times (0.05)+(30000-x)\times 0.07\end{bmatrix}=1800$
$\frac{5x}{100}+\frac{7(30000-x)}{100}=1800$
Multiplying by 100
5x+210000-7x=180000
2x=210000-180000=30000
x=15000
$\Rightarrow 30-x=30000-15000=15000.$
Thus the two part are 15000 each.
(ii)Given total interest 2000
$\begin{bmatrix}x & 30000-x\end{bmatrix}\begin{bmatrix}0.05\\0.07\end{bmatrix}=2000$
$\frac{5x}{100}+\frac{7(30000-x)}{100}=2000$
Multiplying by 100 on both sides
5x+210000-7x=200000
2x=210000-200000
x=5000
30000-x=30000-5000=25000
Thus the two parts are 25000&5000.
answered Mar 4, 2013 by sharmaaparna1
 

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