A merchant paid \(\$300\) for a shipment of \(x\) identical calculators. The merchant used \(2\) of the calculators as

M7MBA wrote: ↑

Fri Aug 06, 2021 11:09 am

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

Here's one approach:

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 + 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