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batteries (OG)

by razorback » Mon Sep 19, 2011 5:20 pm
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 (read: 3q all over 200)
(B) 3q/2 (read: 3q all over 2)
(C) 150q
(D) q/100 + 50 (read q over 100, then add 50)
(E) 150/q

Official answer soon.

This question is #84/230 from the Official Guide, which I interpret as "slightly below average" difficulty. I have stared down the explanation in the book and still can't grasp how this problem should be worked out.
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by Anurag@Gurome » Mon Sep 19, 2011 7:02 pm
razorback 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 (read: 3q all over 200)
(B) 3q/2 (read: 3q all over 2)
(C) 150q
(D) q/100 + 50 (read q over 100, then add 50)
(E) 150/q

Official answer soon.

This question is #84/230 from the Official Guide, which I interpret as "slightly below average" difficulty. I have stared down the explanation in the book and still can't grasp how this problem should be worked out.
Cost of 100 batteries = $q
Cost of 1 battery = $ (q/100)
It is given that the selling price of 1 battery is 50 percent above the original cost per battery, so the selling price of 1 battery = (q/100) + (q/100)*(50/100) = (q/100) + (q/200) = 3q/200

The correct answer is A.
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by GMATGuruNY » Mon Sep 19, 2011 7:21 pm
razorback 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 (read: 3q all over 200)
(B) 3q/2 (read: 3q all over 2)
(C) 150q
(D) q/100 + 50 (read q over 100, then add 50)
(E) 150/q

Official answer soon.

This question is #84/230 from the Official Guide, which I interpret as "slightly below average" difficulty. I have stared down the explanation in the book and still can't grasp how this problem should be worked out.
Let q = 200.
Thus, 100 batteries were bought for a total cost of 200.
Cost per battery = 200/100 = 2.
Selling price = 2 + .5(2) = 3. This is our target.

Now we plug q=200 into the answers to see which yields our target of 3.

Only answer choice A works:
3q/200 = (3*200)/200 = 3.

The correct answer is A.
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by razorback » Thu Sep 22, 2011 7:01 pm
Ok that makes a lot of sense, thanks.
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by leumas » Fri Sep 23, 2011 8:14 am
razorback 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 (read: 3q all over 200)
(B) 3q/2 (read: 3q all over 2)
(C) 150q
(D) q/100 + 50 (read q over 100, then add 50)
(E) 150/q

Official answer soon.

This question is #84/230 from the Official Guide, which I interpret as "slightly below average" difficulty. I have stared down the explanation in the book and still can't grasp how this problem should be worked out.
Cost = q/100
Profit =50%
Selling price=1.5q/100
Answer doesn't have this option: Multiply by 2.

3q/200= Answer
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by Scott@TargetTestPrep » Wed Jun 24, 2015 3:16 am
razorback 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 (read: 3q all over 200)
(B) 3q/2 (read: 3q all over 2)
(C) 150q
(D) q/100 + 50 (read q over 100, then add 50)
(E) 150/q

Official answer soon.

This question is #84/230 from the Official Guide, which I interpret as "slightly below average" difficulty. I have stared down the explanation in the book and still can't grasp how this problem should be worked out.
Solution:

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

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by Brent@GMATPrepNow » Wed Jun 24, 2015 6:20 am
This is a question type that I call "Variables in Answer Choices" (VIACs)
The above posters have demonstrated 2 approaches: Algebraic and Input-Output
Here are a few more VIAC questions to practice with:

https://www.beatthegmat.com/ps-rate-times-t276107.html
https://www.beatthegmat.com/help-me-solv ... 81418.html
https://www.beatthegmat.com/to-find-the- ... 73338.html
https://www.beatthegmat.com/what-mistake ... 76293.html
https://www.beatthegmat.com/car-dealer-s ... 74136.html
https://www.beatthegmat.com/a-better-exp ... 79396.html

Also, I recently wrote two articles about solving VIAC questions:
https://www.gmatprepnow.com/articles/var ... approaches
https://www.gmatprepnow.com/articles/var ... 93-part-ii

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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