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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 03 Jul 2008, 09:43
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Question Stats:

57% (01:56) correct 43% (00:43) wrong based on 87 sessions
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

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-dealer-originally-bought-100-identical-batteries-at-a-tota-143748.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 18 Dec 2013, 01:06, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: PS: VIC [#permalink] New post 03 Jul 2008, 09:49
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Cost per battery = q/100 dollars.

Selling price per battery = (q/100) * 1.5 --> as it is 50% more than the cost.
= 1.5q/100 = 15q/1000 = 3q/200.
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Re: PS: VIC [#permalink] New post 03 Jul 2008, 09:50
B. 3q/200

"in term of q" Calculate new value based on q.
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Re: PS: VIC [#permalink] New post 03 Jul 2008, 09:55
vksunder 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+50
e. 150/q

Could one of you please explain what 'in terms of q' mean in the question? Also, could you please solve this problem by plugging numbers. Thanks!


'in terms of q' means 'find the price of each battery as a function of q'

i.e. if p = price of each battery, then what is p expressed in q?

Hope that helps.
solving by plugging you can do the following (though this problem is much easier to solve directly than by plugging in numbers):
q=$100
then p=$1.5

then plug in the numbers
lets try A
3q/200 => 3(100)/200 = 1.5 => matches our answer!
but since we're plugging in numbers we can't stop here, we'd have to try all of them to see if any other ones also fit.
In the end, we'll realize that A is the only one that works, and choose A.


The easier way is to solve directly.
100 batteries cost q dollars => cost of one battery = q/100
price of one batter is 50% more than cost => price of one battery = 1.5*cost = 1.5*(q/100)
=> we get that P = (1.5/100)*q = (3/200)*q = answer choice A
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Re: PS: VIC [#permalink] New post 04 Jul 2008, 22:08
vksunder 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+50
e. 150/q

Could one of you please explain what 'in terms of q' mean in the question? Also, could you please solve this problem by plugging numbers. Thanks!


Cost price of each battery = q/100

Selling price of each battery = q/100 + 50%*q/100
= q/100 + q/200 =3q/200



Selling price of each battery = (q/100) * 1.5 --> as it is 50% more than the cost.
= 1.5q/100 = 15q/1000 = 3q/200.
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Re: PS: VIC [#permalink] New post 10 Jul 2008, 11:49
domleon wrote:
friends, so is the OA now A or B?



I think it should be A. If we calculating in term of cents then 3q/2 would be correct.
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Re: A dealer originally bought 100 identical batteries at a [#permalink] New post 21 Nov 2011, 11:11
I like using a number for the variable and trying to get the target value from the answers. (similar to jasonc did above)

So if we chose q = 400
We get price per battery = 400/100 = 4
New price = 4 + 1/2 (4) = 6 {This is the target value}

if plug q in answer choices, we get 6 for only option A.
Ans: A
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Re: A dealer originally bought 100 identical batteries at a [#permalink] New post 21 Nov 2011, 12:00
We have:
100 Bat -> q $
1 Bat -> q/100 $
each battery sold at profit of 50% (1.5 times original cost)
so 1 Bat sold for 1.5* q/100
multiply numerator and den by 2 we have
each battery sold for 3q/200
Hence A
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Re: A dealer originally bought 100 identical batteries at a [#permalink] New post 21 Nov 2011, 13:27
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+50
e. 150/q

here is how to choose smart numbers and plug in those numbers.
total cost=q=100
so each battery costs $1
and was sold at 50% increase, which means, $1.50...now we need to plug in 100(substitute q with 100) in every answer and the correct choice will be the one that yields $1.50

and A is the correct answe
hope this helps
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Re: A dealer originally bought 100 identical batteries at a [#permalink] New post 20 Nov 2012, 08:29
in plug-in strategy, option E is also 1.5 if i denote q to be 100. am i missing something here???
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Re: A dealer originally bought 100 identical batteries at a [#permalink] New post 17 Dec 2013, 19:14
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Re: A dealer originally bought 100 identical batteries at a tota [#permalink] New post 18 Dec 2013, 01:07
Expert's post
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.

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-dealer-originally-bought-100-identical-batteries-at-a-tota-143748.html
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Re: A dealer originally bought 100 identical batteries at a tota   [#permalink] 18 Dec 2013, 01:07
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