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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]

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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. $$\frac{100p}{(125 – p)}$$

B. $$\frac{150p}{(250 – p)}$$

C. $$\frac{300p}{(375 – p)}$$

D. $$\frac{400p}{(500 – p)}$$

E. $$\frac{500p}{(625 – p)}$$
[Reveal] Spoiler: OA

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Re: Last Sunday a certain store sold copies of Newspaper A for [#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 2-3 minutes.

What's the source?
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Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink]

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26 Sep 2010, 14:13
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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: Last Sunday a certain store sold copies of Newspaper A for [#permalink]

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26 Sep 2010, 15:41
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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]

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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 + (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 Kudos [?]: 6 [2], given: 0 Intern Joined: 02 Sep 2010 Posts: 46 Kudos [?]: 141 [61], given: 17 Location: India Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 15 Jun 2012, 20:02 61 This post received KUDOS 53 This post was BOOKMARKED 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) _________________ The world ain't all sunshine and rainbows. It's a very mean and nasty place and I don't care how tough you are it will beat you to your knees and keep you there permanently if you let it. You, me, or nobody is gonna hit as hard as life. But it ain't about how hard ya hit. It's about how hard you can get it and keep moving forward. How much you can take and keep moving forward. That's how winning is done! Kudos [?]: 141 [61], given: 17 Manager Joined: 05 Nov 2012 Posts: 70 Kudos [?]: 152 [42], given: 8 Concentration: International Business, Operations Schools: Foster '15 (S) GPA: 3.65 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 02 Mar 2013, 13:09 42 This post received KUDOS 33 This post was BOOKMARKED 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 _________________ ___________________________________________ Consider +1 Kudos if my post helped Last edited by pikachu on 16 May 2013, 11:21, edited 1 time in total. Kudos [?]: 152 [42], given: 8 Manager Joined: 04 Dec 2011 Posts: 80 Kudos [?]: 27 [2], given: 13 Schools: Smith '16 (I) Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 03 May 2013, 10:49 2 This post received KUDOS 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? _________________ Life is very similar to a boxing ring. Defeat is not final when you fall down… It is final when you refuse to get up and fight back! 1 Kudos = 1 thanks Nikhil Kudos [?]: 27 [2], given: 13 Senior Manager Joined: 21 Jan 2010 Posts: 329 Kudos [?]: 222 [0], given: 12 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 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. 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. Kudos [?]: 222 [0], given: 12 Manager Joined: 05 Nov 2012 Posts: 70 Kudos [?]: 152 [1], given: 8 Concentration: International Business, Operations Schools: Foster '15 (S) GPA: 3.65 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 16 May 2013, 11:23 1 This post received KUDOS 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 Kudos [?]: 152 [1], given: 8 Director Joined: 10 Mar 2013 Posts: 595 Kudos [?]: 434 [19], given: 200 Location: Germany Concentration: Finance, Entrepreneurship GMAT 1: 580 Q46 V24 GPA: 3.88 WE: Information Technology (Consulting) Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 21 Jul 2013, 09:12 19 This post received KUDOS 5 This post was BOOKMARKED ............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 _________________ When you’re up, your friends know who you are. When you’re down, you know who your friends are. Share some Kudos, if my posts help you. Thank you ! 800Score ONLY QUANT CAT1 51, CAT2 50, CAT3 50 GMAT PREP 670 MGMAT CAT 630 KAPLAN CAT 660 Kudos [?]: 434 [19], given: 200 Intern Joined: 08 Jun 2013 Posts: 4 Kudos [?]: 1 [0], given: 1 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 05 Nov 2013, 03:55 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 --> $$r=\frac{100 p}{a+1,25*(100-p)}$$ Kudos [?]: 1 [0], given: 1 Math Expert Joined: 02 Sep 2009 Posts: 41698 Kudos [?]: 124664 [2], given: 12079 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 05 Nov 2013, 06:53 2 This post received KUDOS 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. _________________ Kudos [?]: 124664 [2], given: 12079 Intern Joined: 02 Mar 2010 Posts: 19 Kudos [?]: 18 [0], given: 16 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 04 Mar 2014, 12:52 3 This post was BOOKMARKED $$\frac{r}{100} = \frac{a}{(a+1.25b)}$$ $$\frac{p}{100} = \frac{a}{(a+b)}$$ 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. Kudos [?]: 18 [0], given: 16 Manager Joined: 31 May 2012 Posts: 157 Kudos [?]: 191 [1], given: 69 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 09 Apr 2014, 11:18 1 This post received KUDOS 5 This post was BOOKMARKED 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. Kudos [?]: 191 [1], given: 69 Intern Joined: 13 Feb 2014 Posts: 7 Kudos [?]: [0], given: 9 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 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. Kudos [?]: [0], given: 9 Math Expert Joined: 02 Sep 2009 Posts: 41698 Kudos [?]: 124664 [0], given: 12079 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 09 May 2014, 07:05 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. _________________ Kudos [?]: 124664 [0], given: 12079 Moderator Joined: 20 Dec 2013 Posts: 185 Kudos [?]: 74 [0], given: 71 Location: United States (NY) GMAT 1: 640 Q44 V34 GMAT 2: 710 Q48 V40 GMAT 3: 720 Q49 V40 GPA: 3.16 WE: Consulting (Venture Capital) Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 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 _________________ Kudos [?]: 74 [0], given: 71 Manager Joined: 28 Dec 2013 Posts: 79 Kudos [?]: 3 [1], given: 3 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 22 Jun 2014, 09:52 1 This post received KUDOS 1 This post was BOOKMARKED 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? Kudos [?]: 3 [1], given: 3 Manager Joined: 21 Nov 2011 Posts: 72 Kudos [?]: 6 [0], given: 32 Location: United States Concentration: Accounting, Finance GMAT Date: 09-10-2014 GPA: 3.98 WE: Accounting (Accounting) Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 25 Jun 2014, 20:00 2 This post was BOOKMARKED 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) When you see percent, it is good idea to pick 100. Let's say store sold 100 newspapers and p is 60% --> 60 (also 60%) A and 40 (also 40%) B. Then, Rev(A) = 60 *$1 = $60, Rev(B) = 40 *$1.25 = $50. Total Rev -->$60 + $50 =$110.
Given that r percent of stores revenue is from newspaper A --> r = $60 /$110

(D) --> (400 * 60%) / (500 - 60) => 240 / 440, which is same as 60 / 110.
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Re: Last Sunday a certain store sold copies of Newspaper A for   [#permalink] 25 Jun 2014, 20:00

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