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Eternal Intern
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A merchant paid $300 for a shipment of x identical [#permalink] ### Show Tags 24 Jul 2003, 20:55 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?
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25 Jul 2003, 11:25
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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25 Jul 2003, 11:52
Ans: 30?

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

Solving these equations, X = 30
Eternal Intern
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25 Jul 2003, 13:36
Kpadama-- that is two variables with one equation- that is not legitimate.

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25 Jul 2003, 14:08
Well ! I also fell into the same trap

Look closely and you will see two.
Eternal Intern
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25 Jul 2003, 18:12
True, but solving equation takes along time, is there a better way here?
Eternal Intern
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26 Jul 2003, 14:40
Anyone know how to factor this out quickly to solve?
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11 Jun 2009, 01:18
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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12 Jun 2009, 10:46
30 is the correct answer as equation solved gives you 2 values 30 and -4.
Re: PS: Tough Algebra   [#permalink] 12 Jun 2009, 10:46
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