neeti2711 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 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
