BTGModeratorVI wrote: ↑

Tue Jul 07, 2020 6:20 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

Given the average (arithmetic mean) cost of the x calculators = $(300/x);

Thus, total sales = (300/x + 5)(x – 2)

=> (300/x + 5)(x – 2) – 300 = 120

(300/x + 5)(x – 2) = 420

Since this is a quadratic equation, it is not advisable to solve by the traditional approach.

Let's apply the plug-in value approach.

Since the options are arranged in the ascending order, we must try with B option first.

B: 25 => x = 25

(300/25 + 5)(25 – 2) = (12 + 5)(23) = 17*23 = 391 < 420

Since at x = 25, the value of (300/x + 5)(x – 2) is less than 420, the correct value of x must be greater than 25.

Let's try with D: 28.

x = 28 cannot be the answer since 300/28 is not an integer; we know that x is an integer; thus, the correct answer must be 30.

Let's check that.

(300/30 + 5)(30 – 2) = (10 + 5)(23) = 15*28 = 420

Correct answer: E

Hope this helps!

-Jay
_________________
Manhattan Review GMAT Prep

Locations: Manhattan IELTS | GMAT Practice Questions | ACT Practice Test | TOEFL Info | and many more...

Schedule your free consultation with an experienced GMAT Prep Advisor! Click here.

BTGModeratorVI wrote: ↑

Tue Jul 07, 2020 6:20 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

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

BTGModeratorVI wrote: ↑

Tue Jul 07, 2020 6:20 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

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