Last Sunday a certain store sold copies of Newspaper A for : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 22 Jan 2017, 22:53

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Manager
Joined: 06 Apr 2010
Posts: 144
Followers: 3

Kudos [?]: 659 [7] , given: 15

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

### Show Tags

26 Sep 2010, 10:41
7
KUDOS
138
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

51% (04:37) correct 49% (04:46) wrong based on 1700 sessions

### HideShow timer Statistics

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
Manager
Joined: 04 Jun 2010
Posts: 113
Concentration: General Management, Technology
Schools: Chicago (Booth) - Class of 2013
GMAT 1: 670 Q47 V35
GMAT 2: 730 Q49 V41
Followers: 14

Kudos [?]: 240 [1] , given: 43

Re: r in terms of P? [#permalink]

### Show Tags

26 Sep 2010, 11:14
1
KUDOS
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?
_________________

Consider Kudos if my post helped you. Thanks!
--------------------------------------------------------------------
My TOEFL Debrief: http://gmatclub.com/forum/my-toefl-experience-99884.html
My GMAT Debrief: http://gmatclub.com/forum/670-730-10-luck-20-skill-15-concentrated-power-of-will-104473.html

Math Expert
Joined: 02 Sep 2009
Posts: 36601
Followers: 7098

Kudos [?]: 93487 [18] , given: 10563

Re: r in terms of P? [#permalink]

### Show Tags

26 Sep 2010, 13:13
18
KUDOS
Expert's post
29
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 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 --> $$r=\frac{100p}{125-0.25p}=\frac{400p}{500-p}$$.

_________________
Retired Moderator
Joined: 02 Sep 2010
Posts: 805
Location: London
Followers: 105

Kudos [?]: 958 [9] , given: 25

Re: r in terms of P? [#permalink]

### Show Tags

26 Sep 2010, 14:41
9
KUDOS
4
This post was
BOOKMARKED
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)
Intern
Joined: 04 Apr 2012
Posts: 3
Followers: 0

Kudos [?]: 5 [2] , given: 0

Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink]

### Show Tags

22 May 2012, 09:53
2
KUDOS
1
This post was
BOOKMARKED
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 Manager Joined: 02 Sep 2010 Posts: 50 Location: India Followers: 0 Kudos [?]: 108 [49] , given: 17 Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 15 Jun 2012, 19:02 49 This post received KUDOS 36 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! Manager Joined: 05 Nov 2012 Posts: 71 Concentration: International Business, Operations Schools: Foster '15 (S) GPA: 3.65 Followers: 1 Kudos [?]: 120 [35] , given: 8 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 02 Mar 2013, 12:09 35 This post received KUDOS 20 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, 10:21, edited 1 time in total. Manager Joined: 04 Dec 2011 Posts: 81 Schools: Smith '16 (I) Followers: 0 Kudos [?]: 20 [2] , given: 13 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 03 May 2013, 09:49 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 Senior Manager Joined: 21 Jan 2010 Posts: 344 Followers: 3 Kudos [?]: 178 [0], given: 12 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 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: 71 Concentration: International Business, Operations Schools: Foster '15 (S) GPA: 3.65 Followers: 1 Kudos [?]: 120 [0], given: 8 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 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 Director Joined: 10 Mar 2013 Posts: 608 Location: Germany Concentration: Finance, Entrepreneurship GMAT 1: 580 Q46 V24 GPA: 3.88 WE: Information Technology (Consulting) Followers: 15 Kudos [?]: 267 [15] , given: 200 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 21 Jul 2013, 08:12 15 This post received KUDOS 3 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 Intern Joined: 08 Jun 2013 Posts: 4 Followers: 0 Kudos [?]: 1 [0], given: 1 Re: r in terms of P? [#permalink] ### Show Tags 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: 36601 Followers: 7098 Kudos [?]: 93487 [0], given: 10563 Re: r in terms of P? [#permalink] ### Show Tags 05 Nov 2013, 05:53 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: 19 Followers: 0 Kudos [?]: 15 [0], given: 16 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 04 Mar 2014, 11:52 2 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. SVP Status: The Best Or Nothing Joined: 27 Dec 2012 Posts: 1858 Location: India Concentration: General Management, Technology WE: Information Technology (Computer Software) Followers: 47 Kudos [?]: 1937 [1] , given: 193 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 04 Mar 2014, 20:10 1 This post received KUDOS 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: $$\frac{r}{1} = (\frac{400p}{(500-p)})$$ D is the correct answer. I got the two equations, but required lot of time to resolve the same in terms of p & r _________________ Kindly press "+1 Kudos" to appreciate Manager Joined: 31 May 2012 Posts: 166 Followers: 5 Kudos [?]: 158 [1] , given: 69 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 09 Apr 2014, 10:18 1 This post received KUDOS 4 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. 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] ### Show Tags 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: 36601 Followers: 7098 Kudos [?]: 93487 [0], given: 10563 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 09 May 2014, 06:05 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. _________________ Moderator Joined: 20 Dec 2013 Posts: 189 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) Followers: 7 Kudos [?]: 66 [0], given: 71 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 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 _________________ Manager Joined: 28 Dec 2013 Posts: 80 Followers: 0 Kudos [?]: 1 [0], given: 3 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 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 Go to page 1 2 3 Next [ 50 posts ] Similar topics Replies Last post Similar Topics: 5 Last Sunday a certain store sold copies of Newspaper 4 09 Aug 2015, 14:25 2 Last Sunday a certain store sold copies of newspaper a for$1.00 each 2 26 Oct 2014, 08:11
2 Last Sunday 4 16 May 2010, 05:08
1 Last Sunday a certain store sold copies of Newspaper A for \$ 4 05 Jan 2010, 06:53
10 There were 36,000 hardback copies of a certain novel sold 9 04 Mar 2008, 22:22
Display posts from previous: Sort by