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Question
The total cost of 3 T.V. sets and 2 V.C.R.’s is ₹ 35,000. The shopkeeper wants a profit of ₹ 1000 per T.V. set and ₹ 500 per V.C.R. He sells 2 T.V. sets and 1 V.C.R. and gets the total revenue as ₹ 21,500. Find the cost price and the selling price of a T.V. set and a V.C.R.
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Solution
Let the cost of each T.V. set be ₹ x and each V.C.R. be ₹ y. Then the total cost of 3 T.V. sets and 2 V.C.R.’s is ₹ (3x + 2y) which is given to be ₹35,000.
∴ 3x + 2y = 35000
The shopkeeper wants profit of ₹ 1000 per T.V. set and of ₹ 500 per V.C.R.
∴ the selling price of each T.V. set is ₹ (x + 1000) and each V.C.R. is ₹ (y + 500).
∴ selling price of 2 T.V. set and 1 V.C.R. is
₹ [2(x + 1000) + (y + 500)] which is given to be ₹ 21500
∴ 2(x + 1000) + (y + 500) = 21500
∴ 2x + 2000 + y + 500 = 21500
∴ 2x + y = 21500 - 2500
∴ 2x + y = 19000
Hence, the system of linear equations is
3x + 2y = 35000
2x + y = 19000
These equations can be written in the matrix form as:
`[(3,2),(2,1)] [("x"),("y")] = [(35000),(19000)]`
By R1 ↔ R2, we get,
`[(2,1),(3,2)] [("x"),("y")] = [(19000),(35000)]`
By R2 - 2R1, we get,
`[(2,1),(-1,0)] [("x"),("y")] = [(19000),(- 3000)]`
∴ `[(2"x" + "y"),(- "x" + 0)] = [(19000),(- 3000)]`
By equality of matrices,
2x + y = 19000 ....(1)
- x = - 3000 ....(2)
From (2), x = 3000
Substituting x = 3000 in (1), we get,
2(3000) + y = 19000
∴ y = 13000
∴ the cost price of one T.V. set is ₹ 3000 and of one V.C.R. is ₹ 13000 and the selling price of one T.V. set is ₹ 4000 and of one V.C.R. is ₹ 13500.
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