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

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

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

Solution:

The average price of the cost of the x calculators is 300/x dollars.

(x - 2) calculators were sold for (300/x) + 5 dollars each, for a total revenue of:

(x - 2)[(300/x) + 5] = 300 + 5x - (600/x) - 10 = (300x - 600)/x + 5x - 10

Since the total revenue from the sale of the calculators was $120 more than the cost of the shipment:

(300x - 600)/x + 5x - 10 = 120 + 300

Multiplying by x, we have:

300x - 600 + 5x^2 - 10x = 120x + 300x

5x^2 - 130x - 600 = 0

x^2 - 26x - 120 = 0

(x - 30)(x + 4) = 0

x = 30 or x = -4

Since x can’t be negative, x = 30.

Answer: E