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:

1. Translate the problem into a system of equations.
2. Solve the system of equation by using matrix method.
3. Hence, find the cost of one paper bag, one scrap book and one pastel sheet.

Answer 1

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

⇒ 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

⇒ X = A^–1 B

= `-1/2[(-1, -1, 3),(0, 2, -4),(1, -1, -1)][(41),(29),(22)]`

= `-1/2 [((-1)(41) + (-1)(29) + 3(22)),((0)(41) + (2)(29) + (-4)(22)),((1)(41) + (-1)(29) + (-1)(22))]`

= `-1/2 [(-41-29+66),(0+58-88),(41-29-22)]`

= `-1/2 [(-4),(-30),(-10)]`

= `[((-1/2)(-4)),((-1/2)(-30)),((-1/2)(-10))]`

= `[(2),(15),(5)]`

∴ x = 2, y = 15, z = 5

iii. The cost of one paper bag, one scrap book and one pastel sheet is ₹ 2, ₹ 15 and ₹ 5, respectively.