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# Last Sunday a certain store sold copies of Newspaper A for

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VP
Joined: 21 Mar 2006
Posts: 1135
Location: Bangalore
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Last Sunday a certain store sold copies of Newspaper A for [#permalink]  14 Jan 2007, 09:16
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Last Sunday a certain store sold copies of Newspaper A for $1.00 each and copies of Newspaper B for$1.25 each, and the store sold no other newspapers that day. If r percent of the storeâ€™s revenues from newspaper sales was from Newspaper A and if p percent of the newspapers that the store sold were copies of newspaper A, which of the following expresses r in terms of p?
A. 100p / (125 â€“ p)
B. 150p / (250 â€“ p)
C. 300p / (375 â€“ p)
D. 400p / (500 â€“ p)
E. 500p / (625 â€“ p)

Got a little lost in the equations...
Director
Affiliations: FRM Charter holder
Joined: 02 Dec 2006
Posts: 736
Schools: Stanford, Chicago Booth, Babson College
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Kudos [?]: 38 [0], given: 4

Re: PS - Equations [#permalink]  14 Jan 2007, 09:36
kripalkavi wrote:
Last Sunday a certain store sold copies of Newspaper A for $1.00 each and copies of Newspaper B for$1.25 each, and the store sold no other newspapers that day. If r percent of the storeâ€™s revenues from newspaper sales was from Newspaper A and if p percent of the newspapers that the store sold were copies of newspaper A, which of the following expresses r in terms of p?
A. 100p / (125 â€“ p)
B. 150p / (250 â€“ p)
C. 300p / (375 â€“ p)
D. 400p / (500 â€“ p)
E. 500p / (625 â€“ p)

Got a little lost in the equations...

I solved it this way:

A is priced at $1, and sold 50 papers of A. Total revenue from selling A is$50.

B is priced at $1.25, and sold 40 papers of B. Total revenue from selling B is$ 50.

Lets assume that the store sold only A and B. Total revenue by selling A and B is $100. That way r is 50%, and p is 500/9. Then substituted p's value in all the options. D gave back 50% and is the answer. Senior Manager Joined: 24 Nov 2006 Posts: 351 Followers: 1 Kudos [?]: 16 [0], given: 0 Re: PS - Equations [#permalink] 14 Jan 2007, 16:58 aurobindo wrote: I solved it this way: A is priced at$1, and sold 50 papers of A. Total revenue from selling A is $50. B is priced at$1.25, and sold 40 papers of B. Total revenue from selling B is $50. Lets assume that the store sold only A and B. Total revenue by selling A and B is$ 100.

That way r is 50%, and p is 500/9.

Then substituted p's value in all the options. D gave back 50% and is the answer.

I used algebra... quite lengthy for a Gmat question... same result as yours...

For A = number of A newspapers sold; similarly for B:

r/100 = A / (A + 1,25B)
p/100 = A / (A + B)

First, one finds A/B for each equation:

A/B = 1,25r / (100 - r)
A/B = p / (100 - p)

Equating both expressions to the right:

r = 400p / (500 - p) => D.

kripalkavi, could you please tell us the origin (Kaplan, OG, etc) of the question?
VP
Joined: 21 Mar 2006
Posts: 1135
Location: Bangalore
Followers: 2

Kudos [?]: 29 [0], given: 0

I tried to do it the algebrical way and got lost The plugging in way seems to work. I got the question from a set of 'tough' (allegedly) retired GMAT questions that a friend gave me.
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