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Last Sunday a certain store sold copies of Newspaper A for

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Manager
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Last Sunday a certain store sold copies of Newspaper A for [#permalink]

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05 Aug 2006, 08:23
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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)
Senior Manager
Joined: 22 May 2006
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Location: Rancho Palos Verdes
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05 Aug 2006, 12:08
ong 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)

Let n = total number of sold
Total revenue R = 1n(p/100) + 1.25n(1-p/100)
R*r/100 = 1n(p/100)
R*r = n*p
[1n(p/100) + 1.25n(1-p/100)]*r = np
[np+1.25n(100-p)]r = 100np
r = 100p/[125-0.25p]
r = 400p/(500-p)

Hence D.
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The only thing that matters is what you believe.

Manager
Joined: 08 Oct 2005
Posts: 95
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Kudos [?]: 12 [0], given: 0

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05 Aug 2006, 20:40
freetheking wrote:
ong 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)

Let n = total number of sold
Total revenue R = 1n(p/100) + 1.25n(1-p/100)
R*r/100 = 1n(p/100)
R*r = n*p
[1n(p/100) + 1.25n(1-p/100)]*r = np
[np+1.25n(100-p)]r = 100np
r = 100p/[125-0.25p]
r = 400p/(500-p)

Hence D.

Sorry I am confusing with this part. Could explain more. The rest is fine
R*r/100 = 1n(p/100)
R*r = n*p
Senior Manager
Joined: 22 May 2006
Posts: 369
Location: Rancho Palos Verdes
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05 Aug 2006, 21:51
ong wrote:
freetheking wrote:
ong 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)

Let n = total number of sold
Total revenue R = 1n(p/100) + 1.25n(1-p/100)
R*r/100 = 1n(p/100)
R*r = n*p
[1n(p/100) + 1.25n(1-p/100)]*r = np
[np+1.25n(100-p)]r = 100np
r = 100p/[125-0.25p]
r = 400p/(500-p)

Hence D.

Sorry I am confusing with this part. Could explain more. The rest is fine
R*r/100 = 1n(p/100)
R*r = n*p

R*r/100 = 1n(p/100)
multiply by 100 both side
R*r = n*p
_________________

The only thing that matters is what you believe.

Manager
Joined: 08 Oct 2005
Posts: 95
Followers: 1

Kudos [?]: 12 [0], given: 0

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06 Aug 2006, 17:37
freetheking wrote:
ong wrote:
freetheking wrote:
ong 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)

Let n = total number of sold
Total revenue R = 1n(p/100) + 1.25n(1-p/100)
R*r/100 = 1n(p/100)
R*r = n*p
[1n(p/100) + 1.25n(1-p/100)]*r = np
[np+1.25n(100-p)]r = 100np
r = 100p/[125-0.25p]
r = 400p/(500-p)

Hence D.

Sorry I am confusing with this part. Could explain more. The rest is fine
R*r/100 = 1n(p/100)
R*r = n*p

R*r/100 = 1n(p/100)
multiply by 100 both side
R*r = n*p

R*r/100 = 1n(p/100) -----again,where is it from? thanks you
Manager
Joined: 07 Aug 2005
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06 Aug 2006, 21:03
ong 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)

r = (Revenue from A/ Total revenue from paper)*100
= p *1.00 / (p*1.00 + (100 -p)*1.25) *100
=100p/(p + 125 - 1.25p
=100p/125 -.25p
=400p/500-p
Re: Newspaper   [#permalink] 06 Aug 2006, 21:03
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