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Re: Last Sunday a certain store sold copies of Newspaper A for
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12 Oct 2016, 20:51
pikachu wrote: udaymathapati wrote: Last Sunday a certain store sold copies of Newspaper A for $1.00 each and copies of Newspaper B for $1.25 each, and the store sold no other newspapers that day. If r percent of the store’s revenues from newspaper sales was from Newspaper A and if p percent of the newspapers that the store sold were copies of newspaper A, which of the following expresses r in terms of p?
A. 100p / (125 – p) B. 150p / (250 – p) C. 300p / (375 – p) D. 400p / (500 – p) E. 500p / (625 – p) This problem can be easily solved by picking numbers. The explanation given in the OG can be very laborious. Lets say the number of newspaper A sold = 20, so revenue from A = 20 and the number of newspaper sold from B = 80, so revenue from B = 100. Now total revenue =120 out of which 20 came from A. So r (A) = 20/120 = 1/6 = 16.7% and p (A) = 20/100 *100 = 20 A) 100*20/(12520) > Incorrect B) 150*20/(25020) > Incorrect C) 300*20/(37520) > Incorrect D) 400*20/(50020) = 8/48 = 1/6*100 = 16.7%  > Correct E) 500*20/(62520) > Incorrect So Ans D Thanks for providing the solution. Great, number picking is always the best approach in these types of hard problem. Just popped up a query: may be a lame one In terms of picking numbers, where it is not mentioned A or B which one is greater/lesser. In this case, picking the value of A as lesser/greater than B can be an issue to reach the right answer?



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Re: Last Sunday a certain store sold copies of Newspaper A for
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12 Jan 2017, 14:32
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



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Re: Last Sunday a certain store sold copies of Newspaper A for
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16 Jan 2017, 00:21
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: lastsundayacertainstoresoldcopiesofnewspaperafor101739.html
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Re: Last Sunday a certain store sold copies of Newspaper A for
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16 May 2017, 13:26
pikachu wrote: udaymathapati wrote: Last Sunday a certain store sold copies of Newspaper A for $1.00 each and copies of Newspaper B for $1.25 each, and the store sold no other newspapers that day. If r percent of the store’s revenues from newspaper sales was from Newspaper A and if p percent of the newspapers that the store sold were copies of newspaper A, which of the following expresses r in terms of p?
A. 100p / (125 – p) B. 150p / (250 – p) C. 300p / (375 – p) D. 400p / (500 – p) E. 500p / (625 – p) This problem can be easily solved by picking numbers. The explanation given in the OG can be very laborious. Lets say the number of newspaper A sold = 20, so revenue from A = 20 and the number of newspaper sold from B = 80, so revenue from B = 100. Now total revenue =120 out of which 20 came from A. So r (A) = 20/120 = 1/6 = 16.7% and p (A) = 20/100 *100 = 20 A) 100*20/(12520) > Incorrect B) 150*20/(25020) > Incorrect C) 300*20/(37520) > Incorrect D) 400*20/(50020) = 8/48 = 1/6*100 = 16.7%  > Correct E) 500*20/(62520) > Incorrect So Ans D This problem was definitely something to work on. I was able to understand it with Pikachu’s help, although, the calculations were not very clear (substitution of P in % and in Number). So, I am going to try and put it in my own vision of explanation, using number pickings as suggested, since algebraic calculations is really long and seems very confusing to me. Please correct my mistakes. So what we have here is: Newspapers A sold – $1/piece Newspaper B sold – $1.25/piece r  % of revenue generated from A newspapers sales p  % of A newspapers sold. Let’s use a and b as numbers of A & B np sold. a + b = a total number of newspapers sold. % of something = (part/whole)*100%. => p = (a/(a+b))*100% now r % = (Revenue of A/Revenue A + Revenue B)*100% => R(total) = RevenueA +RevenueB. (revenue = price * number of pieces sold) Having price np A sold and 1$/piece and np B sold 1.25$/piece the formula is as follows: R(total) =1$*a + 1.25$b or = a + 1.25b r = (a/(a+1.25b))*100% Now let’s use numbers: 20 np A sold 80 np B sold. Total np sold = 100. p (% of np A sold) = 20% Total revenue R = (1$*20 + 1.25$*80) = 20 + 100 = 120$ r (% of revenue A sold) = (20$/120$)*100% = 16.7% Now lest try and plug in the numbers. We are looking for r = 16.7%, the final result is to be multiplied to 100%. Also in numerator we will be p in % as 0.2, and denominator – we are using p – as a number – 20 (number of A newspapers sold in pieces and not in %) (a) (100(pieces)*0.2)/(125(pieces) – 20(pieces)) = (20/105)*100 = 19%  incorrect (b) (150*0.2)/(250 – 20) = 30/230 = 3/23 = 13% => incorrect (c) (300*0.2)/(375 – 20) = 12/71 = 16,9 % (we are looking for 16,7%) => incorrect (d) (400*0.2)/(50020) = 1/6 = 16.7%  correct. (e) (500*0.2)/(635 – 20) = 100/615 = 50/123 = 40%. => incorrect Now let’s check it and pick other numbers: 40 np A sold 60 np B sold Total np sold are 100. R total = 115$ p = 40%, 40 in $ r = 34.7% or ~ 35%. (a) 100*0.4/(125 – 40) = 8/17= 47% => incorrect (b) 150*0.4/(250 – 40) = 6/21 = 29% => incorrect (c) 300*0.4/(375 – 40) = 24/17 ~ 36% (again close but incorrect) (d) 400*0.4/(500 – 40) = 8/23 = 34.7% correct (e) 500*0.4/(625 – 40) = 40/117 = 34% Seems like D is a correct answer for both. Please correct me if I am making any errors in understanding the concept of how to approach this problem, since, the way I see it – it is a key to solving it in 2 minutes.



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Re: Last Sunday a certain store sold copies of Newspaper A for
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31 Jul 2017, 09:30
udaymathapati wrote: Last Sunday a certain store sold copies of Newspaper A for $1.00 each and copies of Newspaper B for $1.25 each, and the store sold no other newspapers that day. If r percent of the store’s revenues from newspaper sales was from Newspaper A and if p percent of the newspapers that the store sold were copies of newspaper A, which of the following expresses r in terms of p?
A. 100p/(125 – p) B. 150p/(250 – p) C. 300p/(375 – p) D. 400p/(500 – p) E. 500p/(625 – p) This Q no doubt will take around 3 mins IF YOU KNOW WHAT YOU ARE DOING! Anyway here's my approach  Say there are total of 100 copies of A and B Lets pick a nice number for p [Remember to check the options  all are multiples of 25] p = 50%. No. of copies of A = 50; B = 50 Rev from A = 1$ * 50 = 50$ Rev from B = 1.25$ * 50 = 62.5$ [you could also pick a total as 1000 and avoid decimals  whichever suits best] Total rev = 50 + 62.5 = 112.5 So r = 50/112.5 = 4/9 [both numerator and denom are multiples of 25  easy to simplify.] I just plugged it in D coz the denom is 500 p that will give me 450 and thought maybe it will yield me a 9 in denom.
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Re: Last Sunday a certain store sold copies of Newspaper A for
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31 Jul 2017, 13:03
if we try to express Rev ratio in terms of price*#, we get  r / (100r) = 1*p / 1.25*(100p) solving this for r gets the answer r=400p/(500p)



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Re: Last Sunday a certain store sold copies of Newspaper A for
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09 Sep 2017, 18:50
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) Answer: Option D Check solution attached
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Re: Last Sunday a certain store sold copies of Newspaper A for
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21 Sep 2017, 21:26
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) Hi, Thought process: There are two ways to calculate the revenue of the newspaper A: 1. Direct revenue calculation, and 2. Using total revenue. Let the number of the newspaper sold = 100 Attachment:
OG_Newspaper.jpg [ 16.7 KiB  Viewed 1041 times ]
Revenue from A = p  (1) If r percent of the store’s revenues from newspaper sales was from Newspaper A => \(\frac{(125  0.25p)*r}{100}\)  (2) From (1) and (2) we have following: \(\frac{(125  0.25p)*r}{100} = p \Rightarrow r = \frac{100p}{125  0.25p} = \frac{400p}{500  p}\) Thanks.



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Re: Last Sunday a certain store sold copies of Newspaper A for
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22 Sep 2017, 07:57
for percent problem, try to pick a specific number, which is 100
suppose there is 100 coppies, so, there is p copies from A there is 100p copies from b revenue from A is 1. p revernue from B is 1.25 x (100p)
r/100= p/ p+ 1.25(100p)
very fast



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Last Sunday a certain store sold copies of Newspaper A for
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29 Sep 2017, 10:13
You can use smart numbers here.
Because P is a percentage it is best to pick 100 for the total number of hypothetical newspaper sales. This allows us to transform the percentages into real numbers.
Let's make p =20 for ease. if p =20 then 1p =80
Now let's apply the prices for newspapers A and B: $1*20 + $1.25*80 = $20 + $100 = $120 this is the total revenue
p = Revenue A/ Total Revenue*100 = $20/$120*100 = 100/6
Looking at the answer choices we want a resulting denominator that is divisible by 6. Answer choices (A), (C) and (E) can be eliminated straightaway because they all have units digits of 5 which can never be divisible by 6. So we are down to (B) and (D). Looking at the denominator of (B) 25020 =230 this is not divisible by 6! So the correct answer must be (D).



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Last Sunday a certain store sold copies of Newspaper A for
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07 Nov 2017, 22:47
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. \(\frac{100p}{(125 – p)}\)
B. \(\frac{150p}{(250 – p)}\)
C. \(\frac{300p}{(375 – p)}\)
D. \(\frac{400p}{(500 – p)}\)
E. \(\frac{500p}{(625 – p)}\) This problem is definitely confusing and can screw one up, especially at the beginning of the exam. The best way to solve is to plug in numbers and if one is rushing, the best answer would be D, i.e it contains multiples of \(4, 5 & 100.\) Now, if we assume that\(r:p = 1:1\) then \(r=p=50\) Then, B =\(\frac{125}{(125+100)}*100\) = \(\frac{500}{9}\) and \(A = \frac{400}{9}\) Plug in this in the options and only D would suffice. As I said earlier I would try D first as it contains similar numbers as derived above. Plugging in B in D would give 50, hence D is the answer. i.e \(\frac{400p}{(500 – p)}\) = \(400*\frac{500}{9}\) * \(\frac{9}{4000}\) \(=50\) Likewise plugging in 50 in D would give us \(A\), the contribution of \(r\), i.e \(\frac{400*50}{50050}\)\(\) \(=\frac{400}{9}\)



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Re: Last Sunday a certain store sold copies of Newspaper A for
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15 Dec 2017, 00:41
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. \(\frac{100p}{(125 – p)}\)
B. \(\frac{150p}{(250 – p)}\)
C. \(\frac{300p}{(375 – p)}\)
D. \(\frac{400p}{(500 – p)}\)
E. \(\frac{500p}{(625 – p)}\) Check out our video solution to this problem here: https://www.veritasprep.com/gmatsoluti ... olving_210
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Re: Last Sunday a certain store sold copies of Newspaper A for
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23 Dec 2017, 09:34
If you choose p as 100, you know that r has to be 100 as well. D is the only answer that gives us this outcome, so it is the right answer



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Last Sunday a certain store sold copies of Newspaper A for
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25 Dec 2017, 15:38
Here is the solution using equations: T= total revenue, NA = Number of papers sold from stand A, NB = Number of papers sold at Stand B Total Number N = NA + NB; Total revenue T = [1.0 x NA] + [1.25 X NB]; = [1.0 x pN/100] + [1.25 x N x (1p)/100] => [1.25N/100  0.25pN/100];  equation 1 Revenues from A = 1.00 x NA = r/100 x T ; and NA = p/100 x N so equating NA we get rT = pN => r = pN/T in this substiting equation 1 r = pN/[1.25N/100  0.25pN/100] => 100p/[1.25  0.25p] => 400p/[500p]



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Re: Last Sunday a certain store sold copies of Newspaper A for
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06 Jan 2018, 03:34
Bunuel 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 question can be solved by number plugging: just try some numbers for # of newspaper A sold and the # of newspaper B sold. Below is algebraic approach: Let the # of newspaper A sold be \(a\) and the # of newspaper B sold be \(b\). Then: \(r=\frac{a}{a + 1.25b}*100\) and \(p=\frac{a}{a+b}*100\) > \(b=\frac{a}{p}*100a=\frac{a(100p)}{p}\) > \(r=\frac{a}{a + 1.25*\frac{a(100p)}{p}}*100\) > reduce by \(a\) and simplify > \(r=\frac{100p}{p+1251.25p}=\frac{100p}{1250.25p}\) > multiply by 4/4 > \(r=\frac{100p}{1250.25p}=\frac{400p}{500p}\). Answer: D. Hello Bunuel, this problem seems so confusing, even when person knows algebra very well, i think during actual exam it will be time consuming to use algebraic approach As for number plugging any tips for using this strategy? not all numbers will give correct answer, /quick solution when applying number plugging in such kind of questions, no ? By the way i am trying to understand your algebraic approach how did you derive from this \(p=\frac{a}{a+b}*100\) this equation and how do you call this process/step > \(b=\frac{a}{p}*100a=\frac{a(100p)}{p}\) and from that how how got this one \(r=\frac{a}{a + 1.25*\frac{a(100p)}{p}}*100\) and how call this process / step ? thanks!
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Re: Last Sunday a certain store sold copies of Newspaper A for
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06 Jan 2018, 03:53
dave13 wrote: Bunuel 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 question can be solved by number plugging: just try some numbers for # of newspaper A sold and the # of newspaper B sold. Below is algebraic approach: Let the # of newspaper A sold be \(a\) and the # of newspaper B sold be \(b\). Then: \(r=\frac{a}{a + 1.25b}*100\) and \(p=\frac{a}{a+b}*100\) > \(b=\frac{a}{p}*100a=\frac{a(100p)}{p}\) > \(r=\frac{a}{a + 1.25*\frac{a(100p)}{p}}*100\) > reduce by \(a\) and simplify > \(r=\frac{100p}{p+1251.25p}=\frac{100p}{1250.25p}\) > multiply by 4/4 > \(r=\frac{100p}{1250.25p}=\frac{400p}{500p}\). Answer: D. Hello Bunuel, this problem seems so confusing, even when person knows algebra very well, i think during actual exam it will be time consuming to use algebraic approach As for number plugging any tips for using this strategy? not all numbers will give correct answer, /quick solution when applying number plugging in such kind of questions, no ? By the way i am trying to understand your algebraic approach how did you derive from this \(p=\frac{a}{a+b}*100\) this equation and how do you call this process/step > \(b=\frac{a}{p}*100a=\frac{a(100p)}{p}\) and from that how how got this one \(r=\frac{a}{a + 1.25*\frac{a(100p)}{p}}*100\) and how call this process / step ? thanks! \(p=\frac{a}{a+b}*100\) Crossmultiply: \(pa+pb=100a\); Rearrange: \(pb=100apa\); Divide by p: \(b=\frac{100apa}{p}=\frac{a(100p)}{p}\) The next step is substituting the value of b here: \(r=\frac{a}{a + 1.25b}*100\) to get \(r=\frac{a}{a + 1.25*\frac{a(100p)}{p}}*100\). Number plugging: How to Do Math on the GMAT Without Actually Doing MathThe Power of Estimation for GMAT QuantHow to Plug in Numbers on GMAT Math QuestionsNumber Sense for the GMATCan You Use a Calculator on the GMAT?Why Approximate?GMAT Math Strategies — Estimation, Rounding and other ShortcutsThe 4 Math Strategies Everyone Must Master, Part 1 (1. Test Cases and 2. Choose Smart Numbers.) The 4 Math Strategies Everyone Must Master, part 2 (3. Work Backwards and 4. Estimate) Intelligent Guessing on GMATHow to Avoid Tedious Calculations on the Quantitative Section of the GMATGMAT Tip of the Week: No Calculator? No Problem.The Importance of Sorting Answer Choices on the GMATHope it helps.
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Last Sunday a certain store sold copies of Newspaper A for
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25 Jan 2018, 00:00
THE BEST SOLUTION for this Question:
Let total newspapers sold sale was 100 Newspaper A sold = p; Newspaper B sold = 100 – p
Revenue from Newspaper A sold: = p*1 = P Revenue from Newspaper B sold: = (100p) *1.25 Total revenue from Newspaper A & Newspaper B sold: = P + (100p) *1.25 = 125 0.25p It is given that r percent of the store’s revenues from newspaper sales was from Newspaper A; P = r % of [125 0.25p] P = (r/100) of [125 0.25p] 100P = r [125 0.25p] r = 100P/ [125 0.25p] r = 400P/ [500 – p] The Correct Answer D:



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Last Sunday a certain store sold copies of Newspaper A for
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14 Apr 2018, 02:40
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. \(\frac{100p}{(125 – p)}\)
B. \(\frac{150p}{(250 – p)}\)
C. \(\frac{300p}{(375 – p)}\)
D. \(\frac{400p}{(500 – p)}\)
E. \(\frac{500p}{(625 – p)}\) Hello generis , pushpitkc I find this problem really hard i tried to tackle it but got into pitstop here is what i did: Let the price of newspapers \(A\) be \($ 1,00\) Let the price of newspapers \(B\) be \($ 1,25\) Let total number of newspaper \(A\) sold be\(A\) Let total number of newspaper \(B\) sold be \(B\) Let total number of newspapers sold be \(T\) \(T = A+B\) if Total Revenue is \(($1 *A ) + ($1.25 *B)\) > then \(R\) percent of the store’s revenues from newspaper sales was from Newspaper \(A\) > \(\frac{R}{100}\) * \(($1 *A ) + ($1.25 *B)\) Also, if total number of newspapers sold is \(A+B\), > then \(P\) percent of the newspapers that the store sold were copies of newspaper \(A\) > \(\frac{P}{100}\) *\((A+B)\) ok, after the above mentioned steps I got stuck, what to do now I would appreciate explanation for dummies have an awesome weekend
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14 Apr 2018, 11:40
dave13 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. \(\frac{100p}{(125 – p)}\)
B. \(\frac{150p}{(250 – p)}\)
C. \(\frac{300p}{(375 – p)}\)
D. \(\frac{400p}{(500 – p)}\)
E. \(\frac{500p}{(625 – p)}\) Hello generis , pushpitkc I find this problem really hard i tried to tackle it but got into pitstop here is what i did: Let the price of newspapers \(A\) be \($ 1,00\) Let the price of newspapers \(B\) be \($ 1,25\) Let total number of newspaper \(A\) sold be\(A\) Let total number of newspaper \(B\) sold be \(B\) Let total number of newspapers sold be \(T\) \(T = A+B\) if Total Revenue is \(($1 *A ) + ($1.25 *B)\) > then \(R\) percent of the store’s revenues from newspaper sales was from Newspaper \(A\) > \(\frac{R}{100}\) * \(($1 *A ) + ($1.25 *B)\) Also, if total number of newspapers sold is \(A+B\), > then \(P\) percent of the newspapers that the store sold were copies of newspaper \(A\) > \(\frac{P}{100}\) *\((A+B)\) ok, after the above mentioned steps I got stuck, what to do now I would appreciate explanation for dummies have an awesome weekend dave13 , an explanation for dummies? That quip had better be notserious selfdeprecation. If not: Boo. You are on the right track of what I think is lengthy algebra. I like "pit stop" better. Made me laugh. Levity is good. The algebra may or may not be straightforward. Some posters say the algebra is easy or is the best way. Other posters say that picking numbers is easier. I do not see a monopoly on truth here. Do you? If you continue your algebraic route . . . 1) isolate P; 2) isolate B; 3) isolate R; 4) substitute B's value and 5) solve In the end, we must have two variables only: P and R 1) Isolate P Add A to LHS of this part Quote: then \(P\) percent of the newspapers sold were copies of \(A\) > \(\frac{P}{100}\) *\((A+B)\) Make that arrow an equal sign. Include what you are defining. We want P. NOT A. Rearrange until P is on LHS Alone A as a percent of total NUMBER of copies sold \(A = \frac{P}{100} * (A+B)\)
\(\frac{A}{(A+B)} =\frac{P}{100}\)
\(P =\frac{100A}{(A+B)}\)2) Isolate B (B on LHS alone) \(\frac{P}{100}=\frac{A}{A+B}\)  Cross multiply \(PA + PB = 100A\)  Subtract PA \(PB = 100A  PA\)  Factor out A \(PB = A(100  P)\)  Divide by P \(B = \frac{A(100P)}{P}\)3) Isolate A's revenue, R. You have Quote: if Total Revenue is \(($1 *A ) + ($1.25 *B)\) > then \(R\) percent of revenues [from A]> \(\frac{R}{100}\) * \(($1 *A ) + ($1.25 *B)\) Same as above. R on LHS. Use an equals sign. R = (# of copies * price per copy) \(R_A = (A * 1) = A\) \(R_B = (B * 1.25) = 1.25B\)R as A's PERCENT of revenue \(= \frac{R_A}{R_A + R_B} * 100\) \(R = \frac{A}{A + 1.25B}\)4) Then follow Bunuel and rewrite A's percent of revenue, R with value for B that has been substituted: \(\frac{A}{A + 1.25B}\) \(1.25 B\) cannot stay \(R_B = 1.25B\) => substitute B's value from #2, i.e., \(B = \frac{A(100P)}{P}\) You will get to: Quote: The next step is . . . \(r=\frac{a}{a + 1.25*\frac{a(100p)}{p}}*100\) 5) Solve Would you please consider substituting values? I did. Please? Have you tried it? Tenacity is excellent. So is flexibility.
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Last Sunday a certain store sold copies of Newspaper A for
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27 Apr 2018, 05:16
Bunuel 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 question can be solved by number plugging: just try some numbers for # of newspaper A sold and the # of newspaper B sold. Below is algebraic approach: Let the # of newspaper A sold be \(a\) and the # of newspaper B sold be \(b\). Then: \(r=\frac{a}{a + 1.25b}*100\) and \(p=\frac{a}{a+b}*100\) > \(b=\frac{a}{p}*100a=\frac{a(100p)}{p}\) > \(r=\frac{a}{a + 1.25*\frac{a(100p)}{p}}*100\) > reduce by \(a\) and simplify > \(r=\frac{100p}{p+1251.25p}=\frac{100p}{1250.25p}\) > multiply by 4/4 > \(r=\frac{100p}{1250.25p}=\frac{400p}{500p}\). Answer: D. Bunuel Please can you show the Plug in approach? This algebraic method bounced over my head.




Last Sunday a certain store sold copies of Newspaper A for &nbs
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