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calculators

by grandh01 » Sat Aug 18, 2012 5:41 pm
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
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by truplayer256 » Sat Aug 18, 2012 5:57 pm
Average arithmetic cost of the x calculators = 300/x
Total amount received from selling all the calculator = (300/x + 5)(x - 2)= Total Revenue
From the problem:
(300/x + 5)(x - 2)= 300 + 120 = 420
300 - 600/x + 5x - 10 = 420
-600/x + 5x = 130
5x^2 - 130x - 600 = 0
x^2 - 26x - 120 = 0
x = 30
Choose E.
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by GMATGuruNY » Sat Aug 18, 2012 7:39 pm
grandh01 wrote: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
Total revenue = total cost + 120 = 300+120 = 420.

We can plug in the answers, which represent the number of calculators.
The correct answer almost certainly is a factor of 300.
Only B (25) and E (30) are factors of 300.
Since E implies an average cost of $10 per calculator -- a nice, round number -- we should start with E.

Answer choice E: 30
Average cost per calculator = 300/30 = 10.
Since 2 calculators are used, 28 are sold.
Selling price = average cost + 5 = 10+5 = 15.
Revenue from 28 calculators sold for $15 each = 28*15 = 420.
Success!

The correct answer is E.
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