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

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

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25 Jun 2014, 20:52
Bunnel

Need your help for word translation age question. I am finding them very hard after going through Gmat official guide 12th edition twice.

Do you have some easy notes to understand them or easy strategy to solve age questions.

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

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25 Jun 2014, 23:46
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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)

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

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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 17837 [2], given: 235 Math Expert Joined: 02 Sep 2009 Posts: 42286 Kudos [?]: 132978 [0], given: 12391 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 26 Jun 2014, 01:41 Expert's post 2 This post was BOOKMARKED Kudos [?]: 132978 [0], given: 12391 Intern Joined: 09 Feb 2013 Posts: 23 Kudos [?]: [0], given: 63 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 01 Jul 2014, 00:37 Here is my approach : Let total no. Of newspaper sold bd 100 Then, No. Of A newspaper sold =p No. Of B newspaper sold = 100-p Rev generated by A = p dollars Rev generated by B = (100-p)1.25 Total rev = 125-0.25p Now, R% of (125-0.25p) =p Solving the above will guve soln :D I hope this is quick as well. Let me know your thoughts. Posted from my mobile device Kudos [?]: [0], given: 63 Senior Manager Joined: 15 Aug 2013 Posts: 301 Kudos [?]: 83 [0], given: 23 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 24 Aug 2014, 10:01 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 Answer (D) 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? Kudos [?]: 83 [0], given: 23 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7743 Kudos [?]: 17837 [1], given: 235 Location: Pune, India Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 25 Aug 2014, 00:44 1 This post received 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 Answer (D) 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

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

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28 Sep 2014, 09:47
Assuming total number of papers are 100
Number of paper A sold:P
Number of paper B sold:100-P

Total Revenue earned = (P*1) + 1.25*(100-P) = P+125-1.25P = 125-0.25P

Revenue share of paper A is 'R' which is =P/(125-0.25P)*100
= 100p/(125-0.25P)
Multiplying 4 in both numerator & Denominator
=400p/(500-P)

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

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10 Nov 2014, 03:29
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

You inserted r(A) as a decimal(0,167) and p(A) as a percent(20, not 0,2). Is that right?

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

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10 Jan 2015, 10:39
1
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 Hence, Answer: D Kudos [?]: 40 [1], given: 224 Intern Joined: 20 Dec 2014 Posts: 22 Kudos [?]: 4 [0], given: 31 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 13 Jan 2015, 13:12 hermit84 wrote: 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) Could you break this down further? I dont quite get how you arrived at the answer from this point: percent of revenue from A = r = p/p+[(100-p)5/4)]= 400p / (500 – p) Kudos [?]: 4 [0], given: 31 Intern Joined: 11 Nov 2012 Posts: 12 Kudos [?]: 8 [0], given: 38 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 24 Feb 2015, 14:13 hermit84 wrote: 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) Thanks for the answers and different approaches. But in the red part highlighted above I didn't quite get how to get from p/[p+(100-p)5/4] to the final answer. Could someone please explain? Thanks in advance. Kudos [?]: 8 [0], given: 38 Manager Joined: 22 Jan 2014 Posts: 141 Kudos [?]: 77 [0], given: 145 WE: Project Management (Computer Hardware) Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 21 Mar 2015, 07:53 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) i did it by using the answer choices. let total copies sold of A and B be 50 each => r = (50/(50+62.5)) % = 44.44% now just substitute p (50) in answer choices and whichever gives 44.44 is the correct answer. D does. _________________ Illegitimi non carborundum. Kudos [?]: 77 [0], given: 145 Intern Joined: 13 Nov 2014 Posts: 43 Kudos [?]: 12 [0], given: 178 Location: United States Concentration: Marketing, Finance GMAT 1: 640 Q47 V30 GPA: 3.5 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 10 May 2015, 23:29 hermit84 wrote: 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) Absolute perfect solution. The question asks to express r in terms of P. R is revenue percent for A. Hence R can be expressed as (revenue for A/ total revenue )* 100. I have struggled with this problem for long time. until came up with the above solution. Thanks zaarathelab Kudos [?]: 12 [0], given: 178 SVP Joined: 12 Sep 2015 Posts: 1850 Kudos [?]: 2619 [0], given: 362 Location: Canada Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 29 Mar 2016, 15:35 Quote: 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) If you're not sure how to proceed with this question, or if you're behind on time and you want to catch up, you can give yourself a 50-50 chance in about 20 seconds. To do so, we'll see what happens when we use an EXTREME value for p. Say p = 100 In other words, 100% of the newspapers sold were Newspaper A. This means that 100% of the revenue is from Newspaper A. In other words, when p = 100, then r = 100 At this point, we'll plug in 100 for p and see which one yields a value of 100. Only answer choices B and D work. B) 150(100)/(250-100) = 100 PERFECT D) 400(100)/(500-100) = 100 PERFECT Now take a guess (B or D) and move on. Related Resources The posters have demonstrated two methods (Algebraic and Input-Output) for solving this question type, which I call Variables in the Answer Choices. If you'd like more information on these approaches, we have some free videos: - Variables in the Answer Choices - http://www.gmatprepnow.com/module/gmat- ... /video/933 - Tips for the Algebraic Approach - http://www.gmatprepnow.com/module/gmat- ... /video/934 - Tips for the Input-Output Approach - http://www.gmatprepnow.com/module/gmat- ... /video/935 Cheers, Brent _________________ Brent Hanneson – Founder of gmatprepnow.com Last edited by GMATPrepNow on 21 Apr 2016, 14:33, edited 1 time in total. Kudos [?]: 2619 [0], given: 362 Manager Joined: 23 May 2013 Posts: 189 Kudos [?]: 114 [0], given: 42 Location: United States Concentration: Technology, Healthcare Schools: Stanford '19 (M) GMAT 1: 760 Q49 V45 GPA: 3.5 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 21 Apr 2016, 14:29 Notice you can simplify this question very easily just by looking at the answer choices. This is a GMAT sleuth question more than a "answerable" question. The only two numbers you were given in the question were 1 and 1.25, so the answer should contain that ratio. You can eliminate B immediately. All the others have the same ratio, but it weird amounts - the original fraction should be preserved, so that means A and D are your only realistic choices. You've already narrowed this question down to two choices without doing any algebra. Now, if you think: the ratio of r to p should be similar to the ratio of price A to price B, r/p ~ 1/1.25, so r ~ 4/5p. Realizing this equation should have that inverse relationship will lead you to the correct answer, D. Last edited by eaze on 12 Jan 2017, 14:36, edited 1 time in total. Kudos [?]: 114 [0], given: 42 VP Joined: 09 Jun 2010 Posts: 1393 Kudos [?]: 168 [0], given: 916 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 29 Apr 2016, 03:37 for percentage problem, we pick number there are 100 copies, so, there will be p copies of a there will be (100-p ) copis of b r= p*1/ (P*1+ (100-p)1.25 D. very quick way _________________ visit my facebook to help me. on facebook, my name is: thang thang thang Kudos [?]: 168 [0], given: 916 Math Expert Joined: 02 Aug 2009 Posts: 5222 Kudos [?]: 5868 [0], given: 118 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 29 Apr 2016, 06:32 stonecold wrote: Hey chetan2u it took me good % minutes to get to the answer .. Any Good Approach ? HI stonecold, Sorry missed out on this post.. we have A as$1 and B as $1.25.. To get B in integer, I take B selling 40 copies and A as 60.. Total revenue = 1*60 +1.25*40 = 60+50 =110.. revenue from A = r% of 110 = $$\frac{r}{100}*110 = 60 *1$$.. so$$r = 60*\frac{100}{110} = \frac{600}{11}$$.. lets substitute p as 60% and find if r comes out as$$\frac{600}{11}$$anywhere so we look for 11 in denominator.. we can just check just denominator for 11 and can see ONLY D has 500-60 =440, a multiple of 11.. A.$$\frac{100p}{(125 – p)} = \frac{100*60}{(125-60)}=\frac{6000}{65}.$$. NO B. $$\frac{150p}{(250 – p)}= \frac{150*60}{(250-60)}= \frac{9000}{190}$$.. no C. $$\frac{300p}{(375 – p)}= \frac{300*60}{(375-60)}=\frac{18000}{315}.$$. NO D. $$\frac{400p}{(500 – p)}=\frac{400*60}{(500-60)} = \frac{24000}{440} = \frac{600}{11}$$.. this is what we are looking for .. CORRECT E.$$\frac{500p}{(625 – p} = \frac{500*60}{625-60}=\frac{30,000}{565}$$. No _________________ Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372 Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html Kudos [?]: 5868 [0], given: 118 Target Test Prep Representative Status: Head GMAT Instructor Affiliations: Target Test Prep Joined: 04 Mar 2011 Posts: 1684 Kudos [?]: 908 [1], given: 5 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 29 Apr 2016, 09:50 1 This post received KUDOS Expert's post 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. _________________ Jeffery Miller Head of GMAT Instruction GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions Kudos [?]: 908 [1], given: 5 VP Joined: 09 Jun 2010 Posts: 1393 Kudos [?]: 168 [0], given: 916 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 06 May 2016, 01:39 if problem involve percentage and fraction. pick a number of 100 suppose there are 100 coppies newspaper A has p coppies very easy computation _________________ visit my facebook to help me. on facebook, my name is: thang thang thang Kudos [?]: 168 [0], given: 916 SVP Joined: 12 Sep 2015 Posts: 1850 Kudos [?]: 2619 [0], given: 362 Location: Canada Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 26 Jul 2016, 06:54 Expert's post Top Contributor 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) Let's use the INPUT-OUTPUT approach. Let's say that Newspaper A accounted for 20% of all newspapers sold. In other words, p = 20 This means that Newspaper B accounted for 80% of all newspapers sold. The question asks us to find the value of r, the percentage of newspaper revenue from Newspaper A. To determine this, let's say that 100 newspapers we sold IN TOTAL. This means that 20 Newspaper A's were sold and 80 Newspaper B's were sold. REVENUE: Newspaper A: 20 newspapers at$1 apiece = $20 Newspaper B: 80 newspapers at$1.25 apiece = $100 So, TOTAL revenue =$120

Since Newspaper A accounted for $20 of revenue, we can say that Newspaper A accounted for 16 2/3% of revenue. In other words, r = 16 2/3 Aside: We know this because$20/\$120 = 1/6 = 16 2/3%

So, when we INPUT p = 20, the OUTPUT is r = 16 2/3.
We'll now plug p = 20 into each answer choice and see which one yields an output of = 16 2/3

A. 100(20)/(125 - 20) = 2000/105.
IMPORTANT: If we want, we can use long division to evaluate this fraction (and others), but we can save a lot of time by applying some number sense. Since 2000/100 = 20, we know that 2000/105 will be SLIGHTLY less than 20. So, we can be certain that 2000/105 does not equal 16 2/3. As such, we can ELIMINATE A.

B. 150(20)/(250 - 20) = 3000/230. We know that 3000/200 = 15, so 3000/230 will be less than 15. So, we can be certain that 3000/230 does not equal 16 2/3. As such, we can ELIMINATE B.

C. 300(20)/(375 - 20) = 6000/355. Hmmm, this one is a little harder to evaluate. So,we may need to resort to some long division (yuck!). Using long division, we get 6000/355 = 16.9.... ELIMINATE C.

D. 400(20)/(500 - 20) = 8000/480 = 800/48 = 100/6 = 50/3 = 16 2/3. perfect! KEEP

E. 500(20)/(625 - 20) = 10000/605 = 100/6.05. Notice that, above, we saw that 100/6 = 16 2/3. So, 100/6.05 will NOT equal 16 2/3. ELIMINATE E.

[Reveal] Spoiler:
D

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Re: Last Sunday a certain store sold copies of Newspaper A for   [#permalink] 26 Jul 2016, 06:54

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