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# Ps-calculators

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Manager
Joined: 09 Feb 2003
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28 Nov 2003, 11:59
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A merchant paid $300 for a shipment of x identical calculators. The merchant used 2 of the calculators as demonstrators and sold each of the other for 55 more than the average (arithmetic mean) cost of x calculators. If the total revenue from the sale of the calculators was$ 120 more than the cost of the shipment, how many calculators were in the shipment

24
25
26
28
30
The answer is 30. Can anybody explain why?
Director
Joined: 13 Nov 2003
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28 Nov 2003, 13:20
the eq. will be:
(x-2)(300/x + 55) = 120

solve for x - 30
Manager
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28 Nov 2003, 13:55
Are you sure about the equation? If you substitute 30 in the equation you get
(x-2)(300/x + 55) = 120

(30-2)((300/30)+55)

28*65= 1820 doesnтАЩt equal to 120
Also what happens to the part of the question that says тАЬthe total revenue from the sale of the calculators was $120 more than the cost of the shipmentтАЭ I agree with LHS of the equation, but not sure about the RHS. Director Joined: 13 Nov 2003 Posts: 964 Location: Florida Followers: 1 Kudos [?]: 133 [0], given: 0 ### Show Tags 28 Nov 2003, 14:18 oops...my bad (x-2)(300/x + 5) = 420 28*15 = 420 btw, merchant must have sold it for$5 more than average price.. then only we can relate these eqs.
Intern
Joined: 24 Nov 2003
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28 Nov 2003, 18:03

Here is the explaination.

x calculators for \$300
2 used for Demo so x-2 calcs left for sale.

Average cost of each calc = 300/x
Sale price of each calc = 300/x + 55

Total revenue = Total Cost + 120 = 300 + 120 = 420

So,

(x-2)*(300/x + 55) = 420

We get the equation: 55x^2 -230x -600 = 0

Solving we get a negative number and a positive number 6, since objects cannot be -ve, we except 6.
28 Nov 2003, 18:03
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