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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] New post 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] New post 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.

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.

Answer: A.
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Re: A dealer originally bought 100 identical batteries at a tota [#permalink] New post 28 May 2015, 02:42
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noTh1ng wrote:
I don't understand how q can also be used as the price.

I set up an equation like 100 * x = q; with x being the price per battery. Therefore price per battery would be 100x/q...

I just don't get how a equation would be set up without another variable for the price per battery... If anybody could help?


Dear noTh1ng

The correct equation to write from the highlighted part above is: x = q/100.

That is, price per battery = q/100

The mistake that you did was in the red part.

Please let me know if something is still not clear.

Best Regards

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Re: A dealer originally bought 100 identical batteries at a tota [#permalink] New post 06 Jul 2014, 01:47
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: A dealer originally bought 100 identical batteries at a tota [#permalink] New post 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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Re: A dealer originally bought 100 identical batteries at a tota [#permalink] New post 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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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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A dealer originally bought 100 identical batteries at a tota [#permalink] New post 28 May 2015, 01:16
I don't understand how q can also be used as the price.

I set up an equation like 100 * x = q; with x being the price per battery. Therefore price per battery would be 100x/q...

I just don't get how a equation would be set up without another variable for the price per battery... If anybody could help?
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Re: A dealer originally bought 100 identical batteries at a tota [#permalink] New post 28 May 2015, 04:04
EgmatQuantExpert

my god, what a stupid mistake. Thank you very much for clarification, sometimes I just don't see the wood for the trees...
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Re: A dealer originally bought 100 identical batteries at a tota [#permalink] New post 28 May 2015, 10:27
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?

(A) 3q/200
(B) 3q/2
(C) 150q
(D) q/100
(E) 150/q

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Re: A dealer originally bought 100 identical batteries at a tota [#permalink] New post 28 May 2015, 16:38
Expert's post
Hi All,

This question can also be solved by TESTing VALUES.

We're told that 100 identical batteries cost a total of Q dollars.

Let's TEST....
Q = 200
So we bought 100 batteries for $200; that is $2/battery.

Next, we're told that each battery was sold at 50 percent above the original cost per battery...

Buy price = $2/battery
Sell price = (1.5)(2) = $3/battery

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

Final Answer:
[Reveal] Spoiler:
A


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Re: A dealer originally bought 100 identical batteries at a tota [#permalink] New post 28 May 2015, 22:12
I think its easier to calculate like 1.5*q/100 = 3q/200
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Re: A dealer originally bought 100 identical batteries at a tota [#permalink] New post 29 May 2015, 10:01
Expert's post
Hi apoorv601,

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.

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Re: A dealer originally bought 100 identical batteries at a tota [#permalink] New post 30 May 2015, 03:56
Total Cost = Q
Cost per Unit = Q/100
Sale Price = 3/2 * Q/100 = 3Q/200 (A)
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Re: A dealer originally bought 100 identical batteries at a tota   [#permalink] 30 May 2015, 03:56
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