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 https://gmatclub.com/AppTrack

 It is currently 28 Feb 2017, 07:10

### 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

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

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

Kudos [?]: 677 [8] , given: 15

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

### Show Tags

26 Sep 2010, 10:41
8
KUDOS
140
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

51% (04:36) correct 49% (04:43) wrong based on 1753 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: 02 Sep 2010
Posts: 50
Location: India
Followers: 0

Kudos [?]: 112 [49] , given: 17

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

### Show Tags

15 Jun 2012, 19:02
49
KUDOS
39
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 [?]: 127 [36] , given: 8

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

### Show Tags

02 Mar 2013, 12:09
36
KUDOS
24
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.
Math Expert
Joined: 02 Sep 2009
Posts: 37154
Followers: 7277

Kudos [?]: 96860 [20] , given: 10808

Re: r in terms of P? [#permalink]

### Show Tags

26 Sep 2010, 13:13
20
KUDOS
Expert's post
31
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 --> $$r=\frac{100p}{125-0.25p}=\frac{400p}{500-p}$$.

_________________
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 [?]: 293 [16] , given: 200

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

### Show Tags

21 Jul 2013, 08:12
16
KUDOS
4
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

Retired Moderator
Joined: 02 Sep 2010
Posts: 805
Location: London
Followers: 108

Kudos [?]: 974 [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: 04 Dec 2011 Posts: 81 Schools: Smith '16 (I) Followers: 0 Kudos [?]: 25 [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 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: 15 Kudos [?]: 249 [1] , given: 43 Re: r in terms of P? [#permalink] ### Show Tags 26 Sep 2010, 11:14 1 This post received 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 SVP Status: The Best Or Nothing Joined: 27 Dec 2012 Posts: 1858 Location: India Concentration: General Management, Technology WE: Information Technology (Computer Software) Followers: 50 Kudos [?]: 1994 [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 [?]: 161 [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. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7194 Location: Pune, India Followers: 2175 Kudos [?]: 14064 [1] , given: 222 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 25 Jun 2014, 22:46 1 This post received KUDOS Expert's post 1 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) Yes, you can solve this question by assuming a value for p. Note that the options are such that they will involve heavy calculations for most values of p. Easiest should be putting p = 100. Now you might think that two types of newspapers are sold so p = 100 will not be accurate but it is possible that p is approximately equal to 100. Say the store sold 1 million newspapers such that only 1 newspaper was of type B while all others were of type A. In that case, p would be approximately equal to 100%. Of course if almost all newspapers sold were of type A, all the revenue would also come from type A newspapers. So we are looking for the option which gives 100 when you put p = 100. A. 100p/(125 – p) If you put p = 100, you will get 100*100/25 (much more than 100) B. 150p/(250 – p) If you put p = 100, you will get 150*100/150 = 100 C. 300p/(375 – p) If you put p = 100, you will get (300/275)*100 (more than 100) D. 400p/(500 – p) If you put p = 100, you will get 400*100/400 = 100 E. 500p/(625 – p) If you put p = 100, you get (500/525)*100 (less than 100) So answer should be one of (B) and (D). Put p = 50. If 50% newspapers were A and 50% were B, say 100 type A papers were sold and 100 type B such that fraction of revenue from type A papers = (100/225)* 100 = 400/9 B. 150p/(250 – p) Put p = 50, we get 150*50/200. There is no 9 in the denominator here so answer must be (D). Just to verify, we can calculate for (D) as well. D. 400p/(500 – p) Put p = 50, we get 400*50/450 = 400/9 Answer (D) _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

Veritas Prep Reviews

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7194
Location: Pune, India
Followers: 2175

Kudos [?]: 14064 [1] , given: 222

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

### Show Tags

24 Aug 2014, 23:44
1
KUDOS
Expert's post
1
This post was
BOOKMARKED
russ9 wrote:
VeritasPrepKarishma 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)

Yes, you can solve this question by assuming a value for p. Note that the options are such that they will involve heavy calculations for most values of p. Easiest should be putting p = 100. Now you might think that two types of newspapers are sold so p = 100 will not be accurate but it is possible that p is approximately equal to 100. Say the store sold 1 million newspapers such that only 1 newspaper was of type B while all others were of type A. In that case, p would be approximately equal to 100%. Of course if almost all newspapers sold were of type A, all the revenue would also come from type A newspapers.
So we are looking for the option which gives 100 when you put p = 100.

A. 100p/(125 – p)
If you put p = 100, you will get 100*100/25 (much more than 100)

B. 150p/(250 – p)
If you put p = 100, you will get 150*100/150 = 100

C. 300p/(375 – p)
If you put p = 100, you will get (300/275)*100 (more than 100)

D. 400p/(500 – p)
If you put p = 100, you will get 400*100/400 = 100

E. 500p/(625 – p)
If you put p = 100, you get (500/525)*100 (less than 100)

So answer should be one of (B) and (D). Put p = 50. If 50% newspapers were A and 50% were B, say 100 type A papers were sold and 100 type B such that fraction of revenue from type A papers = (100/225)* 100 = 400/9

B. 150p/(250 – p)
Put p = 50, we get 150*50/200. There is no 9 in the denominator here so answer must be (D). Just to verify, we can calculate for (D) as well.

D. 400p/(500 – p)
Put p = 50, we get 400*50/450 = 400/9

Hi Karishma,

Interesting approach and it makes total sense in hindsight. That being said, the hardest part about this problem was trying to figure out WHAT the problem was asking. It said express r in terms of P, so I just solved for R% = (N-number of newspapers sold by A) * p/100 and didn't know where to go after.

What in this problem is indicative that we are trying to solve for Revenue from A/ Total Revenue? I don't see that despite reading this over and over?

When the question says "r in terms of p", it implies that r should be on the left hand side of the equation and everything on the right should only in terms of p. There should be no other variable on the right hand side. Then either you can solve algebraically or plug in values.

The question says you need to find r. What is r? It is "r percent of the store’s revenues from newspaper sales was from Newspaper A"
It means: Revenue from Newspaper A = (r/100)* Total revenue
So r = (Revenue from Newspaper A/Total revenue) * 100
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Joined: 03 Jan 2015 Posts: 68 Concentration: Strategy, Marketing WE: Research (Pharmaceuticals and Biotech) Followers: 2 Kudos [?]: 32 [1] , given: 224 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 10 Jan 2015, 09:39 1 This post received KUDOS Price of A =$1.00
Price of B = $1.25 Assume: Number of A sold = 100 Number of B sold = 0 Therefore: r = Revenue =$100 (all from A)
p = Percent of A sold = 100%

Now, plug in values to see which option returns r = 100. Only D satisfies this.
r = [(400*100)/(500-100] = 100

Director
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 595
Followers: 24

Kudos [?]: 259 [1] , given: 2

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

### Show Tags

29 Apr 2016, 08:50
1
KUDOS
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)

We are given that newspaper A sold for $1 and that newspaper B sold for$1.25.

Next we are are given that that p percent of the newspapers that sold were copies of newspaper A. However, we are not given the total number of copies of both newspapers sold. We can let T = the total copies of both newspapers sold. This means:

(p/100)T = copies of newspaper A sold

This also means that:

(1 – p/100)T = copies of newspaper B sold

We are finally given that r percent of the revenue came from newspaper A. We can translate this into an expression and simplify from there.

_________________

Jeffrey Miller
Jeffrey Miller
Head of GMAT Instruction

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7194
Location: Pune, India
Followers: 2175

Kudos [?]: 14064 [1] , given: 222

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

### Show Tags

15 Jan 2017, 23:21
1
KUDOS
Expert's post
stampap wrote:
the question mentions news papers of A and (copies) of newspapers of A , where do we count B? i got really confused with this question please help

Note what the question says:

"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?"

r% of the revenue is from A
p% of the papers sold are A

There is no conflict here. It is possible that say 50% (p%) of the papers sold were A and that accounted for 80% (r%) of the revenue (because A is more expensive).

Now check out the solutions here:
last-sunday-a-certain-store-sold-copies-of-newspaper-a-for-101739.html
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Senior Manager Joined: 21 Jan 2010 Posts: 344 Followers: 3 Kudos [?]: 180 [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 [?]: 127 [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 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: 37154 Followers: 7277 Kudos [?]: 96860 [0], given: 10808 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. _________________ Re: r in terms of P? [#permalink] 05 Nov 2013, 05:53 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

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

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.