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batteries (OG)

This topic has 4 expert replies and 2 member replies
razorback Junior | Next Rank: 30 Posts Default Avatar
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batteries (OG)

Post 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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Post 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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Post Wed Jun 24, 2015 6:20 am
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Post 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

_________________

Scott Woodbury Stewart Founder & CEO
GMAT Quant Self-Study Course - 500+ lessons 3000+ practice problems 800+ HD solutions
5-Day Free Trial 5-DAY FREE, FULL-ACCESS TRIAL TTP QUANT

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Post Wed Jun 24, 2015 6:20 am
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leumas Senior | Next Rank: 100 Posts
Joined
21 Aug 2011
Posted:
44 messages
Upvotes:
3
Test Date:
12/28/2011
Target GMAT Score:
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GMAT Score:
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Post 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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razorback Junior | Next Rank: 30 Posts Default Avatar
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Post Thu Sep 22, 2011 7:01 pm
Ok that makes a lot of sense, thanks.

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