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SCPREP [#permalink]

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New post 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)
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New post 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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Re: SCPREP [#permalink]

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New post 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 :-D
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New post 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.

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Re: SCPREP [#permalink]

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