Calculators

[neeti2711]

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 x calculators. If the total revenue form 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) 28
(D) 30
(E) Cannot be determined

[GMATGuruNY]

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 x calculators. If the total revenue form 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) 28
(D) 30
(E) Cannot be determined

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 D (30) are factors of 300.
Since D implies an average cost of $10 per calculator -- a nice, round number -- we should start with D.

Answer choice D: 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 D.

[Brent@GMATPrepNow]

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 x calculators. If the total revenue form 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) 28
(D) 30
(E) Cannot be determined

Mitch's plug-in-the-answers solution is great.

Here's the algebraic solution.

If it costs $300 to purchase x calculators, then the average cost per calculator is 300/x

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 + 5x^2 - 10x = 420x
Simplify: 5x^2 - 130x - 600 = 0
Divide both sides by 5: x^2 - 26x - 120 = 0
Factor: (x - 30)(x + 4) = 0
So, x = 30 or x = -4

Since x can't be negative, x = [spoiler]30 = D[/spoiler]

Cheers,
Brent