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

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Manager
Joined: 07 Jul 2003
Posts: 56

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

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

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10 Feb 2004, 13:29
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

Could someone tell me how to solve this with an equation..I tried backsolving and got the answer.

Thanks.

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SVP
Joined: 30 Oct 2003
Posts: 1788

Kudos [?]: 112 [0], given: 0

Location: NewJersey USA

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10 Feb 2004, 13:48
Cost per calculator = 300/x
New cost = 5 + 300/x
No of calculators = (x-2)
Total cost = 300+120 = 420
(5+300/x)(x-2) = 420
x^2-26x-120 = 0
so x = 30
So there were 30 calculators in the original shipment

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Manager
Joined: 28 Jan 2004
Posts: 202

Kudos [?]: 28 [0], given: 4

Location: India

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11 Feb 2004, 01:56
Same solution as anandnk. 30 is the answer.

Kudos [?]: 28 [0], given: 4

11 Feb 2004, 01:56
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