A dealer originally bought 100 identical batteries at a tota

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A dealer originally bought 100 identical batteries at a tota [#permalink]

07 Dec 2012, 04:24

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Re: A dealer originally bought 100 identical batteries at a tota [#permalink]

07 Dec 2012, 04:27

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A dealer originally bought 100 identical batteries at a total cost of q dollars. If each battery was sold at 50 percent above the original cost per battery, then, in terms of q, for how many dollars was each battery sold?

(A) 3q/200
(B) 3q/2
(C) 150q
(D) q/100
(E) 150/q

ALGEBRAIC APPROACH:
The cost of 100 batteries is q dollars, thus the cost of 1 battery is q/100 dollars. Since the selling price is 50% greater than the cost price than the selling price is q/100*1.5=q/100*3/2=3q/200.

Answer: A.

NUMBER PLUGGING APPROACH:
Say q=$200, then the cost of 1 battery is q/100=$2.
The selling price is 2*1.5=$3.

Now, plug q=200 in the answers to see which yields $3. Only answer choice A works.

Answer: A.
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Re: A dealer originally bought 100 identical batteries at a tota [#permalink] 07 Dec 2012, 04:27

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A dealer originally bought 100 identical batteries at a tota

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A dealer originally bought 100 identical batteries at a tota

A dealer originally bought 100 identical batteries at a tota

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