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

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03 Jul 2008, 10:43

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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 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+50 e. 150/q

Could one of you please explain what 'in terms of q' mean in the question? Also, could you please solve this problem by plugging numbers. Thanks!

'in terms of q' means 'find the price of each battery as a function of q'

i.e. if p = price of each battery, then what is p expressed in q?

Hope that helps. solving by plugging you can do the following (though this problem is much easier to solve directly than by plugging in numbers): q=$100 then p=$1.5

then plug in the numbers lets try A 3q/200 => 3(100)/200 = 1.5 => matches our answer! but since we're plugging in numbers we can't stop here, we'd have to try all of them to see if any other ones also fit. In the end, we'll realize that A is the only one that works, and choose A.

The easier way is to solve directly. 100 batteries cost q dollars => cost of one battery = q/100 price of one batter is 50% more than cost => price of one battery = 1.5*cost = 1.5*(q/100) => we get that P = (1.5/100)*q = (3/200)*q = answer choice A

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+50 e. 150/q

Could one of you please explain what 'in terms of q' mean in the question? Also, could you please solve this problem by plugging numbers. Thanks!

Cost price of each battery = q/100

Selling price of each battery = q/100 + 50%*q/100 = q/100 + q/200 =3q/200

Selling price of each battery = (q/100) * 1.5 --> as it is 50% more than the cost. = 1.5q/100 = 15q/1000 = 3q/200.

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

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21 Nov 2011, 13:00

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We have: 100 Bat -> q $ 1 Bat -> q/100 $ each battery sold at profit of 50% (1.5 times original cost) so 1 Bat sold for 1.5* q/100 multiply numerator and den by 2 we have each battery sold for 3q/200 Hence A
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Re: A dealer originally bought 100 identical batteries at a [#permalink]

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21 Nov 2011, 14:27

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+50 e. 150/q

here is how to choose smart numbers and plug in those numbers. total cost=q=100 so each battery costs $1 and was sold at 50% increase, which means, $1.50...now we need to plug in 100(substitute q with 100) in every answer and the correct choice will be the one that yields $1.50

and A is the correct answe hope this helps
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Re: A dealer originally bought 100 identical batteries at a [#permalink]

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17 Dec 2013, 20:14

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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.

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