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

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Eternal Intern
Joined: 07 Jun 2003
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A merchant paid \$300 for a shipment of x identical  [#permalink]

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24 Jul 2003, 21:55
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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?

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Manager
Joined: 03 Jun 2003
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25 Jul 2003, 12:25
If I don┬┤t get it wrong, you have \$120 as net result and you charged \$5 above cost for each calculator, so 120/5 = 24

24+2 = 26 (as there were 2 for marketing)

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Director
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25 Jul 2003, 12:52
Ans: 30?

X * C = 300
(X-2) * (C+5) = 300 + 120

Solving these equations, X = 30

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Eternal Intern
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25 Jul 2003, 14:36
Kpadama-- that is two variables with one equation- that is not legitimate.

VT

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Director
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25 Jul 2003, 15:08
Well ! I also fell into the same trap

Look closely and you will see two.

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Eternal Intern
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25 Jul 2003, 19:12
True, but solving equation takes along time, is there a better way here?

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Eternal Intern
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26 Jul 2003, 15:40
Anyone know how to factor this out quickly to solve?

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Director
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11 Jun 2009, 02:18
Let c be the cost of each calculator when shipment was bought.
Let x be the number of calculators in the shipment

Of x calculators, only (x-2) were sold.

Setting up the equation:

(x-2) * (c+5) = 120 + 300

Since c = 300/x
300 - 2*300/x + 5x -10 = 120 + 300
-2*300/x + 5x = 130

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http://gmatclub.com/forum/countdown-beginshas-ended-85483-40.html#p649902

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Intern
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12 Jun 2009, 11:46
30 is the correct answer as equation solved gives you 2 values 30 and -4.

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Re: PS: Tough Algebra   [#permalink] 12 Jun 2009, 11:46
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