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Manager  Joined: 02 Dec 2012
Posts: 173
A dealer originally bought 100 identical batteries at a tota  [#permalink]

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27 00:00

Difficulty:   45% (medium)

Question Stats: 65% (01:31) correct 35% (01:34) wrong based on 1193 sessions

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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
Math Expert V
Joined: 02 Sep 2009
Posts: 59632
Re: A dealer originally bought 100 identical batteries at a tota  [#permalink]

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3
5
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.

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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##### General Discussion
Intern  Joined: 17 May 2012
Posts: 33
Re: A dealer originally bought 100 identical batteries at a tota  [#permalink]

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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
Math Expert V
Joined: 02 Sep 2009
Posts: 59632
Re: A dealer originally bought 100 identical batteries at a tota  [#permalink]

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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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Manager  Joined: 07 Apr 2015
Posts: 152
A dealer originally bought 100 identical batteries at a tota  [#permalink]

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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?
e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 3158
Re: A dealer originally bought 100 identical batteries at a tota  [#permalink]

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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

Japinder
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Manager  Joined: 07 Apr 2015
Posts: 152
Re: A dealer originally bought 100 identical batteries at a tota  [#permalink]

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EgmatQuantExpert

my god, what a stupid mistake. Thank you very much for clarification, sometimes I just don't see the wood for the trees...
Manager  Joined: 27 Dec 2013
Posts: 193
Re: A dealer originally bought 100 identical batteries at a tota  [#permalink]

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

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
EMPOWERgmat Instructor V
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 15685
Location: United States (CA)
GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: A dealer originally bought 100 identical batteries at a tota  [#permalink]

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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

GMAT assassins aren't born, they're made,
Rich
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Manager  Joined: 21 May 2015
Posts: 213
Concentration: Operations, Strategy
GMAT 1: 750 Q50 V41 Re: A dealer originally bought 100 identical batteries at a tota  [#permalink]

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1
I think its easier to calculate like 1.5*q/100 = 3q/200
_________________
Apoorv

I realize that i cannot change the world....But i can play a part EMPOWERgmat Instructor V
Status: GMAT Assassin/Co-Founder
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Joined: 19 Dec 2014
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Location: United States (CA)
GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: A dealer originally bought 100 identical batteries at a tota  [#permalink]

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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.

GMAT assassins aren't born, they're made,
Rich
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Senior Manager  B
Joined: 10 Mar 2013
Posts: 461
Location: Germany
Concentration: Finance, Entrepreneurship
Schools: WHU MBA"20 (A$) GMAT 1: 580 Q46 V24 GPA: 3.88 WE: Information Technology (Consulting) Re: A dealer originally bought 100 identical batteries at a tota [#permalink] ### Show Tags Total Cost = Q Cost per Unit = Q/100 Sale Price = 3/2 * Q/100 = 3Q/200 (A) Target Test Prep Representative V Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 8656 Location: United States (CA) Re: A dealer originally bought 100 identical batteries at a tota [#permalink] ### Show Tags 2 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. _________________ # Scott Woodbury-Stewart Founder and CEO Scott@TargetTestPrep.com See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Current Student B Joined: 31 Jan 2016 Posts: 19 Schools: Rotman '19 (A) A dealer originally bought 100 identical batteries at a tota [#permalink] ### Show Tags Could someone chime in and explain if this method works as well? q/100 + (q/100 * 50/100) and simplify from here. I get the following after simplification: q/100 + 50q/10000 100q/10000 + 50q/10000 150q/10000 15q/1000 (divided by 5) 3q/200 A Manager  S Joined: 24 Oct 2013 Posts: 127 Location: India Concentration: General Management, Strategy WE: Information Technology (Computer Software) Re: A dealer originally bought 100 identical batteries at a tota [#permalink] ### Show Tags from problem stem 100 (x) = q so assume x = 2 and q = 200 50% increase => x = 3 and q = 300 substitute new q and x values in each option, Option A satisfies Manager  G Joined: 30 Dec 2015 Posts: 81 GPA: 3.92 WE: Engineering (Aerospace and Defense) Re: A dealer originally bought 100 identical batteries at a tota [#permalink] ### Show Tags 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 cost price per battery = $$\frac{q}{100}$$ selling price per batter = $$1.5 * (\frac{q}{100}) = \frac{3}{2} * (\frac{q}{100}) = \frac{3q}{200}$$ _________________ If you analyze enough data, you can predict the future.....its calculating probability, nothing more! Intern  B Joined: 02 Mar 2018 Posts: 5 Re: A dealer originally bought 100 identical batteries at a tota [#permalink] ### Show Tags BrainLab wrote: Total Cost = Q Cost per Unit = Q/100 Sale Price = 3/2 * Q/100 = 3Q/200 (A) Why we have taken sales price =3/2 ? Sent from my MI 5 using GMAT Club Forum mobile app Math Expert V Joined: 02 Sep 2009 Posts: 59632 Re: A dealer originally bought 100 identical batteries at a tota [#permalink] ### Show Tags Akshusaya wrote: BrainLab wrote: Total Cost = Q Cost per Unit = Q/100 Sale Price = 3/2 * Q/100 = 3Q/200 (A) Why we have taken sales price =3/2 ? Sent from my MI 5 using GMAT Club Forum mobile app We are told that each battery was sold at 50 percent above the original cost per battery: 50 percent above = times 1.5, so times 3/2. _________________ Intern  B Joined: 02 Mar 2018 Posts: 5 Re: A dealer originally bought 100 identical batteries at a tota [#permalink] ### Show Tags Bunuel wrote: Akshusaya wrote: BrainLab wrote: Total Cost = Q Cost per Unit = Q/100 Sale Price = 3/2 * Q/100 = 3Q/200 (A) Why we have taken sales price =3/2 ? Sent from my MI 5 using GMAT Club Forum mobile app We are told that each battery was sold at 50 percent above the original cost per battery: 50 percent above = times 1.5, so times 3/2. Yes.. Thank you so much Bunuel Sent from my MI 5 using GMAT Club Forum mobile app VP  D Joined: 09 Mar 2016 Posts: 1229 Re: A dealer originally bought 100 identical batteries at a tota [#permalink] ### Show Tags Bunuel 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 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.

hello niks18,

q/100*1.5 this part means that selling price of each item is increaed by 50% - shouldnt 1.5q/100 be the answer? ;?

and this one q/100*3/2 what does it mean ? where from do we get 2/3 and what is it  Re: A dealer originally bought 100 identical batteries at a tota   [#permalink] 21 Mar 2018, 14:20

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