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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 [#permalink]  07 Dec 2012, 03:24
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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
[Reveal] Spoiler: OA
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Re: A dealer originally bought 100 identical batteries at a tota [#permalink]  07 Dec 2012, 03:27
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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.

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.

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Re: A dealer originally bought 100 identical batteries at a tota [#permalink]  06 Jul 2014, 01:47
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Re: A dealer originally bought 100 identical batteries at a tota [#permalink]  06 Mar 2015, 02:02
Quote:
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.
\

Hi Bunuel,

According to your approach the answer comes out to be 2q/300 and not 3q/200. Could you kindly explain?

Thanks,
AJ
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Joined: 02 Sep 2009
Posts: 27235
Followers: 4233

Kudos [?]: 41115 [0], given: 5670

Re: A dealer originally bought 100 identical batteries at a tota [#permalink]  06 Mar 2015, 03:53
Expert's post
aj0809 wrote:
Quote:
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.
\

Hi Bunuel,

According to your approach the answer comes out to be 2q/300 and not 3q/200. Could you kindly explain?

Thanks,
AJ

Nothing wrong there: q/100*1.5 = q/100*3/2 = 3q/200.
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Re: A dealer originally bought 100 identical batteries at a tota   [#permalink] 06 Mar 2015, 03:53
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