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# SCPREP

Author Message
SVP
Joined: 04 May 2006
Posts: 1890

Kudos [?]: 1344 [0], given: 1

Schools: CBS, Kellogg

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07 Feb 2009, 03:31
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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 store'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?
100p/(125-p)
150p/(250-p)
300p/(375-p)
400p/(500-p)
500p/(625-p)
_________________

Kudos [?]: 1344 [0], given: 1

SVP
Joined: 29 Aug 2007
Posts: 2473

Kudos [?]: 832 [0], given: 19

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07 Feb 2009, 15:08
sondenso 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 store'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-p)
B. 150p/(250-p)
C. 300p/(375-p)
D. 400p/(500-p)
E. 500p/(625-p)

a = no. of newspaper a sold
b = no. of newspaper b sold
total = a+b

1. No. of paper sold:

$$\frac{p%}{100}=\frac{a}{a+b}$$

$$b=\frac{100a - pa}{p}$$

2. Sales revenue:

$$\frac{r%}{100}=\frac{a}{a+1.25b}$$

put the value of b in 2.

$$\frac{r%}{100} = \frac{a}{a + \frac{1.25(100a - pa)}{p}$$

solve for r,

$$r = \frac{400p}{500-p}$$

So it is D.

It took lot of time while i was solving but I now realize that its not that much time consuming.
_________________

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GT

Kudos [?]: 832 [0], given: 19

SVP
Joined: 04 May 2006
Posts: 1890

Kudos [?]: 1344 [0], given: 1

Schools: CBS, Kellogg

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07 Feb 2009, 19:27
GMAT TIGER wrote:
sondenso 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 store'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-p)
B. 150p/(250-p)
C. 300p/(375-p)
D. 400p/(500-p)
E. 500p/(625-p)

a = no. of newspaper a sold
b = no. of newspaper b sold
total = a+b

1. No. of paper sold:

$${(p%)/100} = {a/(a+b)}$$
$${b = (100a - pa)/p"$$

2. Sales revenue:

$${(r%)/100 = a/(a+1.25b)}$$

put the value of b in 2.
$${(r%)/100} = {a/{(a + 1.25(100a - pa)/p}}$$

solve for r,

$${r = 400p/(500-p)}$$

D. It took lot of time while i was solving but I now realize that its not that much time consuming.

Thanks Tiger, what is that realization? Like your, my way consumes a lot of time
_________________

Kudos [?]: 1344 [0], given: 1

Director
Joined: 29 Aug 2005
Posts: 855

Kudos [?]: 472 [0], given: 7

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08 Feb 2009, 05:20
1. pick value for the number of A and B, say, 5 and 4.
2. Calculate r and p
3. Replace the value for p in answer choices.
D.

Kudos [?]: 472 [0], given: 7

SVP
Joined: 07 Nov 2007
Posts: 1795

Kudos [?]: 1033 [0], given: 5

Location: New York

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08 Feb 2009, 21:10
sondenso 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 store'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?
100p/(125-p)
150p/(250-p)
300p/(375-p)
400p/(500-p)
500p/(625-p)

assume p = 100% then r should be 100%

only B and D left.

say p = 50% then r = 50*100/(50+50*1.25) = 100/2.25 = 400/9

This possible only in D case.
_________________

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Kudos [?]: 1033 [0], given: 5

Re: SCPREP   [#permalink] 08 Feb 2009, 21:10
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