# 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. - Mathematics and Statistics

Sum

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.

#### 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 = 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.

Concept: Elementry Transformations
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