VJesus12 wrote: ↑Fri Feb 19, 2021 6:17 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: Official Guide
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
