It is currently 22 Oct 2017, 05:09

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# A merchant paid \$300 for a shipment of x identical

Author Message
Manager
Joined: 05 Sep 2007
Posts: 144

Kudos [?]: 50 [0], given: 0

Location: New York
A merchant paid \$300 for a shipment of x identical  [#permalink]

### Show Tags

22 Mar 2008, 15:52
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

Kudos [?]: 50 [0], given: 0

Senior Manager
Joined: 01 Feb 2005
Posts: 270

Kudos [?]: 109 [0], given: 1

### Show Tags

22 Mar 2008, 17:57

I chose numbers

300/x = average price of each calculator.

Chose x = 30
Therefore each calculator costs 10 dollars. If the person kept 2 calculators and sold the remaining for 5 more dollars than he brought it (15 dollars)

Then 28 calculators * 15 = 420.

420 - 300 = 120 dollars profit which is the additional revenue he made (from the statement of the problem).

Kudos [?]: 109 [0], given: 1

Manager
Joined: 05 Sep 2007
Posts: 144

Kudos [?]: 50 [0], given: 0

Location: New York

### Show Tags

22 Mar 2008, 19:03
is there another way to solve it?

Kudos [?]: 50 [0], given: 0

Manager
Joined: 02 Mar 2008
Posts: 205

Kudos [?]: 52 [0], given: 1

Concentration: Finance, Strategy

### Show Tags

22 Mar 2008, 19:23
300 =x.cost
420 = (x-2)(5+300/x)
-> reduce to x-120/x=26
Then choose x>26 and x is divisible by 120 ->30 (or u solve the quadratic eqn)

Kudos [?]: 52 [0], given: 1

Re: PS: Calculators   [#permalink] 22 Mar 2008, 19:23
Display posts from previous: Sort by