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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]

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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) 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]

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07 Dec 2012, 04:27
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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.

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

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28 May 2015, 03:42
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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

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

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06 Jul 2014, 02:47
Hello from the GMAT Club BumpBot!

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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.
\

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]

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06 Mar 2015, 04:53
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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A dealer originally bought 100 identical batteries at a tota [#permalink]

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28 May 2015, 02:16
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?
Manager
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Re: A dealer originally bought 100 identical batteries at a tota [#permalink]

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28 May 2015, 05:04
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]

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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.

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]

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28 May 2015, 17:38
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

[Reveal] Spoiler:
A

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

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30 May 2015, 04:56
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]

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07 Jun 2016, 10:58
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

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

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10 Aug 2016, 17:04
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
A dealer originally bought 100 identical batteries at a tota   [#permalink] 10 Aug 2016, 17:04
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# A dealer originally bought 100 identical batteries at a tota

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