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Last Sunday a certain store sold copies of Newspaper A for [#permalink]
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26 Sep 2010, 11:41
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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)
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Re: r in terms of P? [#permalink]
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26 Sep 2010, 12:14
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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 23 minutes. What's the source?
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Re: r in terms of P? [#permalink]
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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 \(p=\frac{a}{a+b}*100\) > \(b=\frac{a}{p}*100a=\frac{a(100p)}{p}\) > \(r=\frac{a}{a + 1.25*\frac{a(100p)}{p}}*100\) > reduce by \(a\) and simplify > \(r=\frac{100p}{p+1251.25p}=\frac{100p}{1250.25p}\) > multiply by 4/4 > \(r=\frac{100p}{1250.25p}=\frac{400p}{500p}\). Answer: D.
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Re: r in terms of P? [#permalink]
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26 Sep 2010, 15:41
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Note that total revenue will be k*(p + (100p)*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/(500p)



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Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink]
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22 May 2012, 10:53
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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 + (100p)/100*x*$5/4 Equation: px/100=r/100(px/100+(500/45p/4)x/100) px/100=r/100(4px+500x5px/400) removing common terms as 100 and x out and keeping only r on RHS p=r(500p)/400 or r=400p/(500p)..Answer..D



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Last Sunday a certain store sold copies of Newspaper A for [#permalink]
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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 $1.25 each, and the store sold no other newspapers that day. If r percent of the store’s revenues from newspaper sales was from Newspaper A and if p percent of the newspapers that the store sold were copies of newspaper A, which of the following expresses r in terms of p? A. 100p / (125 – p) B. 150p / (250 – p) C. 300p / (375 – p) D. 400p / (500 – p) E. 500p / (625 – p)
What is the simplest way to solve this?? Let the total copies of newspaper(A+B) sold be 100 so the number of copies of A sold is p number of copies of B sold is 100p thus revenue from A = p*1$ = p$ revenue from B = (100p)5/4; because 1.25 = 5/4 percent of revenue from A = r = p/p+[(100p)5/4)]= 400p / (500 – p)
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Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink]
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02 Mar 2013, 13:09
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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/(12520) > Incorrect B) 150*20/(25020) > Incorrect C) 300*20/(37520) > Incorrect D) 400*20/(50020) = 8/48 = 1/6*100 = 16.7%  > Correct E) 500*20/(62520) > Incorrect So Ans D
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Last edited by pikachu on 16 May 2013, 11:21, edited 1 time in total.



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Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink]
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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/(12520) > Incorrect B) 150*20/(25020) > Incorrect C) 300*20/(37520) > Incorrect D) 400*20/(50020) = 8/48 = 1/6*100 = 16.7%  > Correct E) 500*20/(62520) > 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 \(\frac{400*5}{5005}\) = \(\frac{2000}{495}\) 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]
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03 May 2013, 11:21
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. 100p is % sales of paper B. Therefore : 1p = 100b/(a+b) ........3
Now to make life simpler divide 3 by 2 : (100p)/p = 100b/(a+b) x (a+b)/100a  > b/a = (100p)/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.



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Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink]
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16 May 2013, 11:23
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/(12520) > Incorrect B) 150*20/(25020) > Incorrect C) 300*20/(37520) > Incorrect D) 400*20/(50020) = 8/48 = 1/6*100 = 16.7%  > Correct E) 500*20/(62520) > 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 \(\frac{400*5}{5005}\) = \(\frac{2000}{495}\) 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
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Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink]
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............A ..... B Price: .... 1 .... 1,25 Amount: . P .... 100P  Revenue: P+1,25*(100p) > P + 125  1,25P= 125  0,25P > Revenue of A / Total Revenue: P / 125  0,25P = P/ ((500p)/4)) = 4P/500P > r/100 = 4P/500P > r = 400P / 500P ...................Correct Answer is (D)Hope that helps
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Re: r in terms of P? [#permalink]
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05 Nov 2013, 03:55
Quote: \(r=\frac{a}{a + 1.25*\frac{a(100p)}{p}}*100\) > reduce by \(a\) and simplify > \(r=\frac{100p}{p+1251.25p}=\frac{100p}{1250.25p}\)
Hi Bunel, I don't get the reduction. How do you get rid of the \(a+\) in the denominator? I only get this: \(r=\frac{a}{a + 1.25*\frac{a(100p)}{p}}*100\) > reduces to > \(r=\frac{100 p}{a+1,25*(100p)}\)



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Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink]
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\(\frac{r}{100} = \frac{a}{(a+1.25b)}\) \(\frac{p}{100} = \frac{a}{(a+b)}\) 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 \(\frac{100}{r} = 1+ \frac{5}{4} * (\frac{100}{p} 1)\) \(\frac{100}{r} = 1+ \frac{5}{4} * (\frac{100p}{p})\) \(\frac{100}{r} = 1+ (\frac{5005p}{4p})\) \(\frac{100}{r} = (\frac{500p}{4p})\) \(\frac{1}{r} = (\frac{500p}{400p})\) Now take reciprocal again to get r: \(\frac{r}{1} = (\frac{400p}{(500p)})\) D is the correct answer.



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Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink]
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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 \(\frac{100}{r} = 1+ \frac{5}{4} * (\frac{100}{p} 1)\) \(\frac{100}{r} = 1+ \frac{5}{4} * (\frac{100p}{p})\) \(\frac{100}{r} = 1+ (\frac{5005p}{4p})\) \(\frac{100}{r} = (\frac{500p}{4p})\) \(\frac{1}{r} = (\frac{500p}{400p})\) Now take reciprocal again to get r: \(\frac{r}{1} = (\frac{400p}{(500p)})\) D is the correct answer. 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]
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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.



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Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink]
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09 May 2014, 06:06
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.



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Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink]
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09 May 2014, 15:47
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
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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 (A) = 20/120 = 1/6 = 16.7% and p (A) = 20/100 *100 = 20 A) 100*20/(12520) > Incorrect B) 150*20/(25020) > Incorrect C) 300*20/(37520) > Incorrect D) 400*20/(50020) = 8/48 = 1/6*100 = 16.7%  > Correct E) 500*20/(62520) > Incorrect So Ans D Question : How did you get b = 80 exactly for # sold?




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