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

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07 Dec 2012, 04:24

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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 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.

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

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06 Jul 2014, 02:47

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

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06 Mar 2015, 03:02

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.

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.

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

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28 May 2015, 11:27

nice question. But I doubt this can be 700 level question especially when considering some of Bunuel and Egmatquant questions, which are absoulte killers.

Walkabout wrote:

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?

We're asked for how many dollars each battery was sold.....so we're looking for an answer that equals 3 when Q = 200...

Answer A: 3Q/200 = 600/200 = 3 This IS a match Answer B: 3Q/2 = 600/2 = 300 NOT a match Answer C: 150Q = 30,000 NOT a match Answer D: Q/100 = 200/100 = 2 NOT a match Answer E: 150/Q = 150/200 = 3/4 NOT a match

One of the great things about most GMAT questions is that they can be approached in a variety of different ways. Ultimately, when dealing with any individual question, you want to know MORE than one way to answer the question AND you want to use whatever method is fastest and easiest for you. If you think that an Algebra-based approach is easiest for this prompt, then that's fine, BUT an Algebra-based approach will NOT be the easiest approach for every prompt. To deal with that contingency on Test Day, you have to be practicing other approaches now, so that you'll be flexible enough to not get 'stuck' during the actual Exam.

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

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07 Jun 2016, 10:58

Walkabout wrote:

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

We are given that 100 batteries cost a TOTAL of q dollars. We are also given that EACH battery was sold at 50% above the original cost. The first thing we must do is create an equation for q. Remember that q is the TOTAL COST. So if we make b = the original cost per battery we can say:

100 x b = q

b = q/100

We now have the original cost per battery in terms of q. Next, we determine the selling price when we increase the cost by 50%. To calculate this increase we simply multiply q/100 by 1.5. We have:

(q/100) x 1.5

(q/100) x 3/2 = 3q/200

Answer: A

If you don't like working with variables, you could instead substitute a convenient number for q. Normally I would not suggest the plugging-in method in a problem such as this; however, since we have only one variable, the method will be sufficient.

Let's say we make the total cost of the 100 batteries q = 200. This is a convenient number that will work well with the numbers presented in the problem. We can now set up a similar equation to what we did above, where b = the original price per battery.

100 x b = 200

b = 200/100

b = 2

The cost per battery is $2. Now we need to show the selling price per battery by increasing $2 by 50%.

2 x 1.5 = 3

The answer 3 is the selling price of the battery.

The last step is to now plug our value of q = 200 into each answer choice to see which one provides us with a value of 3. This will yield the correct answer.

(A) 3q/200

(3 x 200)/200 = 3

This IS equal to 3.

(B) 3q/2

(3 x 200)/2 = 600/2 = 300

This IS NOT equal to 3.

(C) 150q

150 x 200 = 30,000

This IS NOT equal to 3.

(D) q/100

200/100 = 2

This IS NOT equal to 3.

(E) 150/q

150/200 = 15/20 = ¾

This IS NOT equal to 3.

Answer choice A is the only one that is equal to 3. _________________

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