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The total cost of 3 T.V. and 2 V.C.R. is ₹ 35,000. The shopkeeper wants profit of ₹1000 per television and ₹ 500 per V.C.R. He can sell 2 T.V. and 1 V.C.R. and get the total revenue as ₹ 21,500. Find - Mathematics and Statistics

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The total cost of 3 T.V. and 2 V.C.R. is ₹ 35,000. The shopkeeper wants profit of ₹1000 per television and ₹ 500 per V.C.R. He can sell 2 T.V. and 1 V.C.R. and get the total revenue as ₹ 21,500. Find the cost price and the selling price of a T.V. and a V.C.R.

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Solution

Let the cost of a T.V. be ₹ x and the cost of a V.C.R. be ₹ y.

According to the first condition,

3x + 2y = 35000      ......(i)

The required profit per T.V. is ₹ 1000 and per V.C.R. is ₹ 500.

∴ Selling price of a T.V. is ₹ (x + 1000) and selling price of a V.C.R. is ₹ (y + 500).

According to the second condition,

2(x + 1000) + 1(y + 500) = 21500

∴ 2x + 2000 + y + 500 = 21500

∴ 2x + y = 21500 – 2500

∴ 2x + y = 19000                       ......(ii)

Matrix form of equations (i) and (ii) is

`[(3, 2),(2, 1)] [(x),(y)] = [(35000),(19000)]`

Applying  R2 → 2R2 – R1, we get

`[(3, 2),(1, 0)] [(x),(y)] = [(35000),(3000)]`

Applying  R1 ↔ R2 – R1, we get

`[(1, 0),(3, 2)] [(x),(y)] = [(300),(35000)]`

Hence, the original matrix is reduced to a lower triangular matrix.

∴ `[(x + 0),(3x + 2y)] = [(3000),(35000)]`

∴ By equality of matrices, we get

x = 3000                        ......(iii)

3x + 2y = 35000            ......(iv)

Substituting x = 3000 in equation (iv), we get

3(3000) + 2y = 35000

∴ 2y = 35000 – 9000

∴ y = `(35000 - 9000)/(2)`

= `(26000)/(2)`

= 13000.

∴ The cost price of a T.V. is ₹ 3,000 and the cost price of a V.C.R. is ₹ 13,000.

Hence, the selling price of a T.V.

= ₹ (3,000 + 1,000)

= ₹ 4,000

and the selling price of a V.C.R.

= ₹ (13,000 + 500)

= ₹ 13,500.

Concept: Application of Matrices
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