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For a manufacturer, the cost of producing x units is given the expression y +kx, where y and k are constants. If the

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by VJesus12 » Thu Jun 04, 2020 7:04 am
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For a manufacturer, the cost of producing x units is given the expression y +kx, where y and k are constants. If the manufacturer sells 100 units at $50 each, his per-unit profit for these 100 units is equal to 64 percent of the selling price per unit and if he sells 250 units at $40 each, his per-unit profit for these 250 units is equal to 70 percent of the selling price per unit. How much profit in dollars would the manufacturer make if he sold 50 units at $70 each?

A) 1400
B) 1500
C) 2000
D) 2100
E) 2500

[spoiler]OA=D[/spoiler]

Source: e-GMAT
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Source: — Problem Solving |

VJesus12 wrote: ↑
Thu Jun 04, 2020 7:04 am
For a manufacturer, the cost of producing x units is given the expression y +kx, where y and k are constants. If the manufacturer sells 100 units at $50 each, his per-unit profit for these 100 units is equal to 64 percent of the selling price per unit and if he sells 250 units at $40 each, his per-unit profit for these 250 units is equal to 70 percent of the selling price per unit. How much profit in dollars would the manufacturer make if he sold 50 units at $70 each?

A) 1400
B) 1500
C) 2000
D) 2100
E) 2500

[spoiler]OA=D[/spoiler]

Solution:

We can break this problem into 4 parts: (1) Selling 100 units @ $50; (2) Selling 250 units @ $40; (3) Determining the values of k and y for the cost function y + kx; and (4) Determining the profit when 50 units are sold @ $70.

We note that the cost function is given as Cost = y + kx, where y and k are constants and x is the number of units sold.

We will use the formula Cost = Revenue - Profit


Part 1: Selling 100 units @ $50

Cost = Revenue - Profit

Cost = (100 x 50) - (0.64 x 50 x 100)

Cost = 5,000 - 3,200

Cost = 1,800

Part 2: Selling 250 units @ $40

Cost = Revenue - Profit

Cost = (250 x 40) - (0.70 x 40 x 250)

Cost = 10,000 - 7,000

Cost = 3,000

Part 3: Determining the values of k and y for the cost function y + kx

When we sell x = 100 units, the cost is 1,800. Thus, the cost function Cost = y + kx is:

1,800 = y + 100k (Eq. 1)

When we sell x = 250 units, the cost is 3,000. Thus the cost function Cost = y + kx is:

3,000 = y + 250k (Eq. 2)

Subtracting Eq. 1 from Eq. 2, we obtain:

1,200 = 0y + 150k

8 = k

We know that k = 8, so we substitute this value into Eq. 1 to solve for y:

1,800 = y + 100 x 8

1,000 = y

Thus, the cost equation Cost = y + kx is Cost = 1,000 + 8x

Part 4: Determining the profit when 50 units are sold @ $70.

The cost of 50 units is Cost = 1,000 + 8 x 50 = 1,400

The revenue from selling 50 units @ $70 is Revenue = 50 x 70 = 3,500.

Thus the profit from this sale is:

Profit = Revenue - Cost

Profit = 3,500 - 1,400

Profit = $2,100

Answer: D

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VJesus12 wrote: ↑
Thu Jun 04, 2020 7:04 am
For a manufacturer, the cost of producing x units is given the expression y +kx, where y and k are constants. If the manufacturer sells 100 units at $50 each, his per-unit profit for these 100 units is equal to 64 percent of the selling price per unit and if he sells 250 units at $40 each, his per-unit profit for these 250 units is equal to 70 percent of the selling price per unit. How much profit in dollars would the manufacturer make if he sold 50 units at $70 each?

A) 1400
B) 1500
C) 2000
D) 2100
E) 2500

[spoiler]OA=D[/spoiler]

Source: e-GMAT
Sale price for \(100\) units, \(\$5,000,\) so the cost of production will be \(5,000\cdot 0.36 = \$1800 = y + 100k\)

Sale price for \(250\) units, \(\$10,000,\) the cost of production is \(\$3,000 = y + 250k\)

Solving the two equations, we get \(k =8\) and \(y= 1000.\)

So the cost of producing \(50\) units is \(\$1,400\) and the sale price is \(\$3,500.\)

Therefore, profit is \(\$2,100.\)
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