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

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Last Sunday a certain store sold copies of Newspaper A for [#permalink]  26 Sep 2010, 10:41
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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 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)
[Reveal] Spoiler: OA
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Last Sunday a certain store sold copies of Newspaper A for [#permalink]  15 Jun 2012, 19:02
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zaarathelab 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)

What is the simplest way to solve this??

Let the total copies of newspaper(A+B) sold be 100
so the number of copies of A sold is p
number of copies of B sold is 100-p
thus revenue from A = p*1$= p$
revenue from B = (100-p)5/4; because 1.25 = 5/4
percent of revenue from A = r = p/p+[(100-p)5/4)]= 400p / (500 – p)
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Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink]  02 Mar 2013, 12:09
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udaymathapati 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)

This problem can be easily solved by picking numbers. The explanation given in the OG can be very laborious.

Lets say the number of newspaper A sold = 20, so revenue from A = 20 and the number of newspaper sold from B = 80, so revenue from B = 100. Now total revenue =120 out of which 20 came from A. So

r (A) = 20/120 = 1/6 = 16.7% and p (A) = 20/100 *100 = 20

A) 100*20/(125-20) -> Incorrect
B) 150*20/(250-20) -> Incorrect
C) 300*20/(375-20) -> Incorrect
D) 400*20/(500-20) = 8/48 = 1/6*100 = 16.7% - > Correct
E) 500*20/(625-20) -> Incorrect

So Ans D
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Last edited by pikachu on 16 May 2013, 10:21, edited 1 time in total.
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Re: r in terms of P? [#permalink]  26 Sep 2010, 13:13
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udaymathapati 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)

This question can be solved by number plugging: just try some numbers for # of newspaper A sold and the # of newspaper B sold.

Below is algebraic approach:

Let the # of newspaper A sold be a and the # of newspaper B sold be b.

Then:
r=\frac{a}{a + 1.25b}*100 and p=\frac{a}{a+b}*100 --> b=\frac{a}{p}*100-a=\frac{a(100-p)}{p} --> r=\frac{a}{a + 1.25*\frac{a(100-p)}{p}}*100 --> reduce by a and simplify --> r=\frac{100p}{p+125-1.25p}=\frac{100p}{125-0.25p} --> multiply by 4/4 --> r=\frac{100p}{125-0.25p}=\frac{400p}{500-p}.

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Re: r in terms of P? [#permalink]  26 Sep 2010, 14:41
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Note that total revenue will be k*(p + (100-p)*1.25) where k is a constant depending on actual number of papers sold

The contribution of type A is kp

So r=100 * kp/k(p + 125 -1.25p)
= 100p/(125-.25p)
= 400p/(500-p)
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Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink]  21 Jul 2013, 08:12
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............A..... B
Price:.... 1.... 1,25
Amount:. P.... 100-P
------------------------------
Revenue: P+1,25*(100-p) ---> P + 125 - 1,25P= 125 - 0,25P

---> Revenue of A / Total Revenue: P / 125 - 0,25P = P/ ((500-p)/4)) = 4P/500-P
---> r/100 = 4P/500-P --> r = 400P / 500-P ...................Correct Answer is (D)

Hope that helps
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Re: r in terms of P? [#permalink]  26 Sep 2010, 11:14
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Wow this is a hard question no doubt.
How do you solve this one quickly? I tried pluging numbers instead on r and p but the result was really not comfortable no matter what numbers I used.
Also solving it with pure algebra is far from being simple or done in under 2-3 minutes.

What's the source?
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Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink]  03 May 2013, 09:49
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pikachu wrote:
udaymathapati 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)

This problem can be easily solved by picking numbers. The explanation given in the OG can be very laborious.

Lets say the number of newspaper A sold = 20, so revenue from A = 20 and the number of newspaper sold from B = 80, so revenue from B = 100. Now total revenue =120 out of which 20 came from A. So

r = 20/120 = 1/6 = 16.7% and p = 20

A) 100*20/(125-20) -> Incorrect
B) 150*20/(250-20) -> Incorrect
C) 300*20/(375-20) -> Incorrect
D) 400*20/(500-20) = 8/48 = 1/6*100 = 16.7% - > Correct
E) 500*20/(625-20) -> Incorrect

So Ans D

I did tried picking smart nos...mmm..ok may be not smart as yours but basically here is my pick

p=5 (5 papers of A sold) so revenue from A = 5
20 papers of B sold so 20*1.25 so revenue from paper B = 25
Total revenue R = 25+5 = 30
no of A paper sold = P = 5
so revenue = 5/30 or around 16.6% percent-------------->>> till this part I got it right
now try plugin the answer choice D
\frac{400*5}{500-5}

= \frac{2000}{495} is not equal 16.6%.. what's wrong here?
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Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink]  04 Mar 2014, 20:10
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prsnt11 wrote:
\frac{r}{100} = \frac{a}{(a+1.25b)}
\frac{p}{100} = \frac{a}{(a+b)}
So our problem is how to go about solving with a+b and a+1.25b in the denominator. An easy way out is take the reciprocal.

So,
\frac{100}{r} = \frac{(a+1.25b)}{a}
and
\frac{100}{p} = \frac{(a+b)}{a}
So,
\frac{100}{r} = 1+ \frac{1.25b}{a} ........(1)
and
\frac{100}{p} = 1+ \frac{b}{a}
or \frac{100}{p} -1 = \frac{b}{a}...........(2)
So we have isolated b/a to a corner. Let's substitute for b/a in (1) so that we can have an equation only in r & p which we could solve for r
\frac{100}{r} = 1+ \frac{5}{4} * (\frac{100}{p} -1)
\frac{100}{r} = 1+ \frac{5}{4} * (\frac{100-p}{p})
\frac{100}{r} = 1+ (\frac{500-5p}{4p})
\frac{100}{r} = (\frac{500-p}{4p})
\frac{1}{r} = (\frac{500-p}{400p})
Now take reciprocal again to get r:
\frac{r}{1} = (\frac{400p}{(500-p)})

I got the two equations, but required lot of time to resolve the same in terms of p & r
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Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink]  22 May 2012, 09:53
The algebra way is not time taking even..if we proceed as below:
(News A) A= $1 (News B) B =$1.25 or $5/4 Total newspaper sold= x No. of A Newspaper sold = p/100 *x is r% of total revnue Total revenue: p/100*x*$1 + (100-p)/100*x*$5/4 Equation: px/100=r/100(px/100+(500/4-5p/4)x/100) px/100=r/100(4px+500x-5px/400) removing common terms as 100 and x out and keeping only r on RHS p=r(500-p)/400 or r=400p/(500-p)..Answer..D Senior Manager Joined: 21 Jan 2010 Posts: 348 Followers: 0 Kudos [?]: 67 [0], given: 12 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] 03 May 2013, 10:21 My train of thoughts : Let paper A sold = a. Let paper B sold = b. Now r=a/(a+1.25b) x 100 .....1 p=100a/(a+b) ....2 Now we see 2 equations and 3 variables. We must find another equation to get to the answer. if p = % sales of paper A. 100-p is % sales of paper B. Therefore : 1-p = 100b/(a+b) ........3 Now to make life simpler divide 3 by 2 : (100-p)/p = 100b/(a+b) x (a+b)/100a - > b/a = (100-p)/p......4 Divide 1 by a at numerator and denominator. r = 100/(1+1.25(b/a)...........5 If you substitute the value of b/a from 4 into 5, you get D. Manager Joined: 05 Nov 2012 Posts: 72 Concentration: International Business, Operations Schools: Foster '15 (S) GPA: 3.65 Followers: 1 Kudos [?]: 42 [0], given: 8 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] 16 May 2013, 10:23 nikhil007 wrote: pikachu wrote: udaymathapati 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) This problem can be easily solved by picking numbers. The explanation given in the OG can be very laborious. Lets say the number of newspaper A sold = 20, so revenue from A = 20 and the number of newspaper sold from B = 80, so revenue from B = 100. Now total revenue =120 out of which 20 came from A. So r = 20/120 = 1/6 = 16.7% and p = 20 A) 100*20/(125-20) -> Incorrect B) 150*20/(250-20) -> Incorrect C) 300*20/(375-20) -> Incorrect D) 400*20/(500-20) = 8/48 = 1/6*100 = 16.7% - > Correct E) 500*20/(625-20) -> Incorrect So Ans D I did tried picking smart nos...mmm..ok may be not smart as yours but basically here is my pick p=5 (5 papers of A sold) so revenue from A = 5 20 papers of B sold so 20*1.25 so revenue from paper B = 25 Total revenue R = 25+5 = 30 no of A paper sold = P = 5 so revenue = 5/30 or around 16.6% percent-------------->>> till this part I got it right now try plugin the answer choice D \frac{400*5}{500-5} = \frac{2000}{495} is not equal 16.6%.. what's wrong here? nikhil, the error you are making is in terms of p, since p is the % of A newspapers sold P = 5/30*100 not 5 as you are using. hope that helps _________________ ___________________________________________ Consider +1 Kudos if my post helped Intern Joined: 08 Jun 2013 Posts: 4 Followers: 0 Kudos [?]: 0 [0], given: 1 Re: r in terms of P? [#permalink] 05 Nov 2013, 02:55 Quote: r=\frac{a}{a + 1.25*\frac{a(100-p)}{p}}*100 --> reduce by a and simplify --> r=\frac{100p}{p+125-1.25p}=\frac{100p}{125-0.25p} Hi Bunel, I don't get the reduction. How do you get rid of the a+ in the denominator? I only get this: r=\frac{a}{a + 1.25*\frac{a(100-p)}{p}}*100 --> reduces to --> r=\frac{100 p}{a+1,25*(100-p)} Math Expert Joined: 02 Sep 2009 Posts: 19020 Followers: 3358 Kudos [?]: 24331 [0], given: 2676 Re: r in terms of P? [#permalink] 05 Nov 2013, 05:53 Expert's post Marcoson wrote: Quote: r=\frac{a}{a + 1.25*\frac{a(100-p)}{p}}*100 --> reduce by a and simplify --> r=\frac{100p}{p+125-1.25p}=\frac{100p}{125-0.25p} Hi Bunel, I don't get the reduction. How do you get rid of the a+ in the denominator? I only get this: r=\frac{a}{a + 1.25*\frac{a(100-p)}{p}}*100 --> reduces to --> r=\frac{100 p}{a+1,25*(100-p)} r=\frac{a}{a + 1.25*\frac{a(100-p)}{p}}*100; Factor out a from the denominator: r=\frac{a}{a(1 + 1.25*\frac{(100-p)}{p})}*100. Reduce it: r=\frac{1}{1 + 1.25*\frac{(100-p)}{p}}*100. Hope it's clear. _________________ Intern Joined: 02 Mar 2010 Posts: 15 Followers: 0 Kudos [?]: 8 [0], given: 16 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] 04 Mar 2014, 11:52 \frac{r}{100} = \frac{a}{(a+1.25b)} \frac{p}{100} = \frac{a}{(a+b)} So our problem is how to go about solving with a+b and a+1.25b in the denominator. An easy way out is take the reciprocal. So, \frac{100}{r} = \frac{(a+1.25b)}{a} and \frac{100}{p} = \frac{(a+b)}{a} So, \frac{100}{r} = 1+ \frac{1.25b}{a} ........(1) and \frac{100}{p} = 1+ \frac{b}{a} or \frac{100}{p} -1 = \frac{b}{a}...........(2) So we have isolated b/a to a corner. Let's substitute for b/a in (1) so that we can have an equation only in r & p which we could solve for r \frac{100}{r} = 1+ \frac{5}{4} * (\frac{100}{p} -1) \frac{100}{r} = 1+ \frac{5}{4} * (\frac{100-p}{p}) \frac{100}{r} = 1+ (\frac{500-5p}{4p}) \frac{100}{r} = (\frac{500-p}{4p}) \frac{1}{r} = (\frac{500-p}{400p}) Now take reciprocal again to get r: \frac{r}{1} = (\frac{400p}{(500-p)}) D is the correct answer. Manager Joined: 31 May 2012 Posts: 166 Followers: 4 Kudos [?]: 62 [0], given: 69 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] 09 Apr 2014, 10:18 Here is simplest & quickest way to reach answer: Price of A=$1.
Price of B= $1.25 i.e.$ 5/4

We want revenue(r) in terms of percent of A(p)
To calculate revenue, Assume, p= 20
So, Revenue = 20(1)+80(5/4)=$120 Now, r=$20/$120= 1/6 Now, put p=20 in each option and try to see if you can get 100/6 anywhere. Just by looking at options, I see only Option(D) can serve my purpose. 400p / (500 – p) = 100 X (4X20)/(480) = 100/6. Intern Joined: 13 Feb 2014 Posts: 7 Followers: 0 Kudos [?]: 0 [0], given: 9 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] 09 May 2014, 05:06 Bunuel is there a reason why you chose to isolate B and not A? I tried doing it by isolating A but cant solve it. Math Expert Joined: 02 Sep 2009 Posts: 19020 Followers: 3358 Kudos [?]: 24331 [0], given: 2676 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] 09 May 2014, 06:05 Expert's post gciftci wrote: Bunuel is there a reason why you chose to isolate B and not A? I tried doing it by isolating A but cant solve it. In r=\frac{a}{a + 1.25b}*100 we have b only in one place while a there is represented twice. So, it's better to substitute b there. Though you should get the same answer no matter whether you substitute a or b. _________________ Manager Joined: 20 Dec 2013 Posts: 126 Followers: 0 Kudos [?]: 14 [0], given: 46 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] 09 May 2014, 14:47 great question/practice, thanks for posting it r/100 = Qa/(Qa+1.25Qb) p/100 * Q = Qa or (1- p)/100 * Q = Qb using these equations, the answer is D Intern Joined: 28 Dec 2013 Posts: 35 Followers: 0 Kudos [?]: 0 [0], given: 3 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] 22 Jun 2014, 08:52 pikachu wrote: udaymathapati 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)

This problem can be easily solved by picking numbers. The explanation given in the OG can be very laborious.

Lets say the number of newspaper A sold = 20, so revenue from A = 20 and the number of newspaper sold from B = 80, so revenue from B = 100. Now total revenue =120 out of which 20 came from A. So

r (A) = 20/120 = 1/6 = 16.7% and p (A) = 20/100 *100 = 20

A) 100*20/(125-20) -> Incorrect
B) 150*20/(250-20) -> Incorrect
C) 300*20/(375-20) -> Incorrect
D) 400*20/(500-20) = 8/48 = 1/6*100 = 16.7% - > Correct
E) 500*20/(625-20) -> Incorrect

So Ans D

Question : How did you get b = 80 exactly for # sold?
Re: Last Sunday a certain store sold copies of Newspaper A for   [#permalink] 22 Jun 2014, 08:52
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