A merchant paid $300 for a shipment of x identical calculators. The merchant used 2 of the calculators as demonstrators and sold each of the others for $5 more than the average (arithmetic mean) cost of the x calculators. If the total revenue from the sale of the calculators was $120 more than the cost of the shipment, how many calculators were in the shipment?
A. 24
B. 25
C. 26
D. 28
E. 30
Answer: E
Source: GMAT paper tests
A merchant paid $300 for a shipment of x identical calculators.
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Given the average (arithmetic mean) cost of the x calculators = $(300/x);BTGModeratorVI wrote: ↑Tue Jul 07, 2020 6:20 amA merchant paid $300 for a shipment of x identical calculators. The merchant used 2 of the calculators as demonstrators and sold each of the others for $5 more than the average (arithmetic mean) cost of the x calculators. If the total revenue from the sale of the calculators was $120 more than the cost of the shipment, how many calculators were in the shipment?
A. 24
B. 25
C. 26
D. 28
E. 30
Answer: E
Source: GMAT paper tests
Thus, total sales = (300/x + 5)(x – 2)
=> (300/x + 5)(x – 2) – 300 = 120
(300/x + 5)(x – 2) = 420
Since this is a quadratic equation, it is not advisable to solve by the traditional approach.
Let's apply the plug-in value approach.
Since the options are arranged in the ascending order, we must try with B option first.
B: 25 => x = 25
(300/25 + 5)(25 – 2) = (12 + 5)(23) = 17*23 = 391 < 420
Since at x = 25, the value of (300/x + 5)(x – 2) is less than 420, the correct value of x must be greater than 25.
Let's try with D: 28.
x = 28 cannot be the answer since 300/28 is not an integer; we know that x is an integer; thus, the correct answer must be 30.
Let's check that.
(300/30 + 5)(30 – 2) = (10 + 5)(23) = 15*28 = 420
Correct answer: E
Hope this helps!
-Jay
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If it costs $300 to purchase x calculators, then the average cost per calculator is 300/xBTGModeratorVI wrote: ↑Tue Jul 07, 2020 6:20 amA merchant paid $300 for a shipment of x identical calculators. The merchant used 2 of the calculators as demonstrators and sold each of the others for $5 more than the average (arithmetic mean) cost of the x calculators. If the total revenue from the sale of the calculators was $120 more than the cost of the shipment, how many calculators were in the shipment?
A. 24
B. 25
C. 26
D. 28
E. 30
Answer: E
Source: GMAT paper tests
Later, the calculators are sold for $5 more than the average purchase cost of 300/x dollars
So, the resell price is (300/x) + 5
How many were sold? Well, the merchant began with x calculators, but used 2 as demonstrators, so the merchant sold x - 2 calculators.
Finally, the merchant's profit was $120 (after a $300 investment). So, the revenue was $420
We can now write an equation: [(300/x) + 5](x - 2) = 420
IMPORTANT: This is an awful equation to solve. At this point, it may be faster to try plugging in the answer choices.
Or we can solve the equation.
[(300/x) + 5](x - 2) = 420
Expand: 300 - (600/x) + 5x - 10 = 420
Multiply both sides by x: 300x - 600 + 5x² - 10x = 420x
Simplify: 5x² - 130x - 600 = 0
Divide both sides by 5: x² - 26x - 120 = 0
Factor: (x - 30)(x + 4) = 0
So, x = 30 or x = -4
Since x can't be negative, x = 30
Answer: E
Cheers,
Brent
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Solution:BTGModeratorVI wrote: ↑Tue Jul 07, 2020 6:20 amA merchant paid $300 for a shipment of x identical calculators. The merchant used 2 of the calculators as demonstrators and sold each of the others for $5 more than the average (arithmetic mean) cost of the x calculators. If the total revenue from the sale of the calculators was $120 more than the cost of the shipment, how many calculators were in the shipment?
A. 24
B. 25
C. 26
D. 28
E. 30
Answer: E
The average price of the cost of the x calculators is 300/x dollars.
(x - 2) calculators were sold for 300/x + 5 dollars each, for a total revenue of:
(x - 2)(300/x + 5) = 300 + 5x - 600/x - 10 = (300x - 600)/x + 5x - 10
Since the total revenue from the sale of the calculators was $120 more than the cost of the shipment:
(300x - 600)/x + 5x - 10 = 120 + 300
Multiplying by x, we have:
300x - 600 + 5x^2 - 10x = 120x + 300x
5x^2 - 130x - 600 = 0
x^2 - 26x - 120 = 0
(x - 30)(x + 4) = 0
x = 30 or x = -4
Since x can’t be negative, x = 30.
Answer: E
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The average price of the cost of the x calculators is 300/x dollars.
(x - 2) calculators were sold for 300/x + 5 dollars each, for a total revenue of:
(x - 2)(300/x + 5) = 300 + 5x - 600/x - 10 = (300x - 600)/x + 5x - 10
Since the total revenue from the sale of the calculators was $120 more than the cost of the shipment
(300x - 600)/x + 5x - 10 = 120 + 300
Multiplying by x, we have:
300x - 600 + 5x^2 - 10x = 120x + 300x
5x^2 - 130x - 600 = 0
x^2 - 26x - 120 = 0
(x - 30)(x + 4) = 0
x = 30 or x = -4
Since x can’t be negative, x = 30.
Answer: E
Regards
letracking.com
(x - 2) calculators were sold for 300/x + 5 dollars each, for a total revenue of:
(x - 2)(300/x + 5) = 300 + 5x - 600/x - 10 = (300x - 600)/x + 5x - 10
Since the total revenue from the sale of the calculators was $120 more than the cost of the shipment
(300x - 600)/x + 5x - 10 = 120 + 300
Multiplying by x, we have:
300x - 600 + 5x^2 - 10x = 120x + 300x
5x^2 - 130x - 600 = 0
x^2 - 26x - 120 = 0
(x - 30)(x + 4) = 0
x = 30 or x = -4
Since x can’t be negative, x = 30.
Answer: E
Regards
letracking.com