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Question
A salesman has the following record of sales during three months for three items A, B and C which have different rates of commission
| Month | Sale of units | Total commission drawn (in Rs) |
||
| A | B | C | ||
| Jan | 90 | 100 | 20 | 800 |
| Feb | 130 | 50 | 40 | 900 |
| March | 60 | 100 | 30 | 850 |
Find out the rates of commission on items A, B and C by using determinant method.
Solution
Let x, y and z be the rates of commission on items A, B and C respectively. Based on the given data, we get
\[90x + 100y + 20z = 800\]
\[130x + 50y + 40z = 900\]
\[60x + 100y + 30z = 850\]
Dividing all the equations by 10 on both sides, we get
\[9x + 10y + 2z = 80\]
\[13x + 5y + 4z = 90\]
\[6x + 10y + 3z = 85\]
\[D = \begin{vmatrix}9 & 10 & 2 \\ 13 & 5 & 4 \\ 6 & 10 & 3\end{vmatrix} \left[\text{ Expressing the equation as a determinant }\right]\]
\[ = 9(15 - 40) - 10(39 - 24) + 2(130 - 30)\]
\[ = 9( - 25) - 10(15) + 2(100)\]
\[ = - 175\]
\[ D_1 = \begin{vmatrix}80 & 10 & 2 \\ 90 & 5 & 4 \\ 85 & 10 & 3\end{vmatrix}\]
\[ = 80(15 - 40) - 10(270 - 340) + 2(900 - 425)\]
\[ = 80( - 25) - 10( - 70) + 2(475)\]
\[ = - 350\]
\[ D_2 = \begin{vmatrix}9 & 80 & 2 \\ 13 & 90 & 4 \\ 6 & 85 & 3\end{vmatrix}\]
\[ = 9(270 - 340) - 80(39 - 24) + 2(1105 - 540)\]
\[ = 9( - 70) - 80(15) + 2(565)\]
\[ = - 700\]
\[ D_3 = \begin{vmatrix}9 & 10 & 80 \\ 13 & 5 & 90 \\ 6 & 10 & 85\end{vmatrix}\]
\[ = 9(425 - 900) - 10(1105 - 540) + 80(130 - 30)\]
\[ = 9( - 475) - 10(565) + 80(100)\]
\[ = - 1925\]
Thus,
\[x = \frac{D_1}{D} = \frac{- 350}{- 175} = 2\]
\[y = \frac{D_2}{D} = \frac{- 700}{- 175} = 4\]
\[z = \frac{D_3}{D} = \frac{- 1925}{- 175} = 11\]
Therefore, the rates of commission on items A, B and C are 2, 4 and 11, respectively.
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