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Question
| To raise money for an orphanage, students of three schools A, B and C organised an exhibition in their residential colony, where they sold paper bags, scrap books and pastel sheets made by using recycled paper. Student of school A sold 30 paper bags, 20 scrap books and 10 pastel sheets and raised ₹ 410. Student of school B sold 20 paper bags, 10 scrap books and 20 pastel sheets and raised ₹ 290. Student of school C sold 20 paper bags, 20 scrap books and 20 pastel sheets and raised ₹ 440. |
Answer the following question:
- Translate the problem into a system of equations.
- Solve the system of equation by using matrix method.
- Hence, find the cost of one paper bag, one scrap book and one pastel sheet.
Solution
i. Let the cost of one paper bag, one scrap book and one pastel sheet be x, y and z respectively.
30x + 20y + 10z = 410
`\implies` 3x + 2y + z = 41
20x + 10y + 20z = 290
`\implies` 2x + y + 2z = 29
20x + 20y + 20z = 440
`\implies` x + y + z = 22
ii. Given system of equations is equivalent to AX = B
Where `A = [(3, 2, 1),(2, 1, 2),(1, 1, 1)], X = [(x),(y),(z)], B = [(41),(29),(22)]`
|A| = –2 ≠ 0 `\implies` A–1 exists.
adj A = `[(-1, -1, 3),(0, 2, -4),(1, -1, -1)]`
Thus A–1 = `1/|A| adj A`
= `-1/2[(-1, -1, 3),(0, 2, -4),(1, -1, -1)]`
AX = B
`\implies` X = A–1B
= `-1/2[(-1, -1, 3),(0, 2, -4),(1, -1, -1)][(41),(29),(22)]`
= `[(2),(15),(5)]`
∴ x = 2, y = 15, z = 5
iii. The cost of one paper bag, one scrap book and one pastel sheet be Rs. 2, Rs. 15 and Rs. 5 respectively.
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