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 plugin 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
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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