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# Ps: Quick Calc Needed!

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Eternal Intern
Joined: 07 Jun 2003
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24 Jul 2003, 18:32
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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? Why doesn't this work, baby? x-2 * ( 5 + 300/ x) =420 Manager Joined: 24 Jun 2003 Posts: 146 Location: India Followers: 2 Kudos [?]: 2 [0], given: 0 Re: Ps: Quick Calc Needed! [#permalink] ### Show Tags 24 Jul 2003, 22:58 Curly05 wrote: 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?

Why doesn't this work, baby?

x-2 * ( 5 + 300/ x) =420

Curly,

you don't need to use the longer route. We know that $120 profit is made from selling (X-2) calculators at a profit of$5 on each. Therefore, the equation to be used is

(x-2)*5=120 which gives x=26.
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25 Jul 2003, 08:39
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?

Your equation was almost right, Curly05.
(x-2) * ( 5 + 300/ x) =420
=> 5x - 600/x - 10 + 300 = 420
=> 5x^2 - 130x - 600=0
=> x^2 - 26x - 120 = 0

Solving we get, x = 30 or x = -4
x cannot be negative as it is the number of calculators
So, there were 30 calculators.
_________________

P K Das

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