## A merchant paid $300 for a shipment of x identical calculators. ##### This topic has expert replies Legendary Member Posts: 1223 Joined: 15 Feb 2020 Followed by:1 members ### A merchant paid$300 for a shipment of x identical calculators.

by BTGModeratorVI » Tue Jul 07, 2020 6:20 am

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A

B

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D

E

## Global Stats

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 ### GMAT/MBA Expert GMAT Instructor Posts: 3008 Joined: 22 Aug 2016 Location: Grand Central / New York Thanked: 470 times Followed by:32 members ### Re: A merchant paid$300 for a shipment of x identical calculators.

by [email protected] » Tue Jul 07, 2020 10:24 pm
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

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

Hope this helps!

-Jay
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### Re: A merchant paid $300 for a shipment of x identical calculators. by [email protected] » Wed Jul 08, 2020 5:35 am 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

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 Brent Hanneson - Creator of GMATPrepNow.com ### GMAT/MBA Expert GMAT Instructor Posts: 6365 Joined: 25 Apr 2015 Location: Los Angeles, CA Thanked: 43 times Followed by:26 members ### Re: A merchant paid$300 for a shipment of x identical calculators.

by [email protected] » Sat Jul 11, 2020 1:09 pm
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.

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### Re: A merchant paid $300 for a shipment of x identical calculators. by Harvey L. Walker » Wed Sep 22, 2021 11:22 pm 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.