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Profit and loss

Expert replies
by vaibhav101 » Mon Jun 05, 2017 5:04 am
The manufacturer of a certain item can sell all he can produce at the selling price of $60 each. It costs him $40 in materials and labour to produce each item and he has overhead expenses of $3000 per week in order to operate the plant. The minimum number of units that he should produce and sell in order to make profit of at least $1000 per week, is
A 200
B 250
C 300
D 400
E 500
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Source: — Problem Solving |

by Brent@GMATPrepNow » Mon Jun 05, 2017 5:12 am
vaibhav101 wrote:The manufacturer of a certain item can sell all he can produce at the selling price of $60 each. It costs him $40 in materials and labour to produce each item and he has overhead expenses of $3000 per week in order to operate the plant. The minimum number of units that he should produce and sell in order to make profit of at least $1000 per week, is
A 200
B 250
C 300
D 400
E 500
Let x = the number of units produced in 1 week.

It costs him $40 in materials and labour to produce each item and he has overhead expenses of $3000 per week in order to operate the plant.
TOTAL EXPENSES for 1 week = 40x + 3000

The manufacturer of a certain item can sell all he can produce at the selling price of $60 each.
So, the manufacturer will sell x units at $60 apiece
TOTAL REVENUE for 1 week = 60x

PROFIT = 60x - ( 40x + 3000)
= 20x - 3000

The minimum number of units that he should produce and sell in order to make profit of at least $1000 per week, is
We can write the equation: 20x - 3000 = 1000
Add 3000 to both sides: 20x = 4000
Solve: x = 200

Answer: A
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by [email protected] » Mon Jun 05, 2017 10:22 am
Hi vaibhav101,

This question can be solved by TESTing THE ANSWERS.

We're told that each unit costs $40 to produce and will be sold for $60. In addition, the production plant costs $3,000 per week to operate. We're asked for the minimum number of units that must be produced in one week to net at least a $1,000 profit.

Let's TEST Answer B: 250 units

250 units would net $20/profit each...
(250)($20) = $5,000
After subtracting the $3,000 needed to run the plant, we have...
$5,000 - $3,000 = $2,000
This is higher than what we were asked for, so 250 units is clearly too many to produce the minimum profit we were after. There's only one answer left that 'fits'

Final Answer: A

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Rich
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by Jay@ManhattanReview » Mon Jun 05, 2017 9:15 pm
vaibhav101 wrote:The manufacturer of a certain item can sell all he can produce at the selling price of $60 each. It costs him $40 in materials and labour to produce each item and he has overhead expenses of $3000 per week in order to operate the plant. The minimum number of units that he should produce and sell in order to make profit of at least $1000 per week, is
A 200
B 250
C 300
D 400
E 500
We see that at $60 selling price and $40 cost price, there is a margin of $20 per unit. If there are x units to be produced, they must bring in the total margin (20x) that can at least cover overhead expense ($3000) and profit ($1000).

Thus, 20x ≥ (3000 + 1000) => 20x ≥ 4000 => x ≥ 200 units.

The correct answer: A

Hope this helps!

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by Matt@VeritasPrep » Mon Jun 05, 2017 11:39 pm
Costs: $40 per item + $3000
Profits: $60 per item

If our guy sells x items and makes $1000, we know that

Profits - Costs = $1000

60x - (40x + 3000) = 1000

20x - 3000 = 1000

20x = 4000

x = 200
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by Jeff@TargetTestPrep » Sat Jun 10, 2017 7:41 am
vaibhav101 wrote:The manufacturer of a certain item can sell all he can produce at the selling price of $60 each. It costs him $40 in materials and labour to produce each item and he has overhead expenses of $3000 per week in order to operate the plant. The minimum number of units that he should produce and sell in order to make profit of at least $1000 per week, is
A 200
B 250
C 300
D 400
E 500
We can let x = the number of items sold.

Thus, the revenue from selling x items is 60x.

The cost of producing x items is 40x and there is a fixed cost of $3,000. Thus, the total cost is 40x + 3000.

Since revenue - cost = profit and we want the profit to be at least $1,000, we can now determine the minimum value of x using the following inequality:

60x - (40x + 3,000) ≥ 1000

60x - 40x - 3,000 ≥ 1000

20x - 3,000 ≥ 1000

20x ≥ 4000

x ≥ 200

Answer: A
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