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Re: Last Sunday a certain store sold copies of Newspaper A for $1.00 each
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18 Dec 2018, 09:11
VeritasKarishma wrote: jadorexox wrote: I'm having a hard time with this problem.
I chose
100 units sold of A 20 units sold B
Revenue = (100)(1) + (1.25)(20) = 125 r% = 100/125 r=80
p% = 100/120 p= 83.33
Plug in numbers to get "r"
(400*83.33)/500p
My question is, how do you know what numbers to pick to get you easy to work with numbers? Additionally, I understand the whole concept of choosing 100 and 100 or saying that you don't sell any B books, but isn't there a possibility that somehow picking the same number with "cancel" itself out?
Appreciate any response! To plug in numbers, go for the easy ones such as 0, 1, 100 or 100%, 50%, 0% (if percentages are required). For example, in this question, my first instinct would be to use 100%. Say p% is 100% i.e. only newspaper A copies were sold. Then r would also be 100 since the entire revenue will come from newspaper A. So I will just put p = 100 and look for the option(s) that give me 100. (B) and (D) give me 100. Now, I assume that both papers are sold in equal numbers i.e. p = 50. Then the revenue collected will be in the ratio of the cost of the papers. So revenue from A will be 1/(1+1.25) = 4/9th of the total revenue. So I check which of (B) and (D) gives me 400/9 when I put p = 50. I see option (D) does that so it must be the answer. The only thing I would add to Karishma's excellent response is to point out that she focused on using an easy value for p specifically. There is a good reason for this: p is the number we're actually plugging into the answer choices. So, we don't want to just choose nicelooking numbers for A and B; we want to choose values for A and B that give us an easy value for p. So, choosing A and B so that p = 100, or 50, or 20, or a similar nice value would work well. Also, to address your concern about the same value for A and B "canceling" out, again notice that we are plugging p into the answer choices, rather than A or B. So, because choosing A = B gives us a nice value for p (50), we know we can plug this in for p and don't need to worry about A and B canceling. Please let me know if you have any more questions!
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Re: Last Sunday a certain store sold copies of Newspaper A for $1.00 each
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21 Jul 2019, 11:00
JeffYin wrote: dassamik89 wrote: Can you please check if my approach is correct:
Assumption: I have assumed P= 50% and the total copies sold as 200.
Therefore Revenue from A= 1*100 =100 and B=1.25*100=125. Total revenue= 225
Hence r= (100/225)* 100 =400/9.
If we plug in P=50 in all the options only D gives the answer. First, I realize this is a pretty late reply, so my apologies for that. The good news is that your approach was correct, dassamik89! I appreciate the lively discussion about algebra vs. number picking on this thread. You all have demonstrated that it can be done either way! I'm writing to weigh in with my own GMAT Timing Tip on this question. My advice is: Unless you have seen a relatively fast way to do the algebra, I recommend number picking on a question like this where the algebra looks so ugly. So, here is my GMAT Timing Tip (the link has a growing list of questions that you can use to practice applying this tip): Pick smart numbers to plug into variables in answer choices: While you are ultimately picking a value for p, notice that you are really picking values for A and B that will lead to a nice value for p. Since B is multiplied by 5/4, 4 is a good, easy choice for B. Choices for A that lead to nice values of p are A=1 (p=20) and A=4 (p=50); p=20 is easier to plug into the answer choices, so let’s go with A=1. This means that r=100/6, so we plug in for p and see which answer choice gives us 100/6 for r. Once you have plugged the values into the answer choices, eliminate the answer choices that don't have a multiple of 3 in the denominator (B, C, E), and see that A is not equal to 100/6, but D is. Please let me know if you have any questions about my timing tip, or if you want me to post a video solution! Thanks, JeffYin. Do you know a quick way to tell whether the algebra will be ugly? I was already 1.5 minutes into the problem when I saw that the algebra was ugly.



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Re: Last Sunday a certain store sold copies of Newspaper A for $1.00 each
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26 Jul 2019, 04:48
ChiefsFan wrote: Thanks, JeffYin. Do you know a quick way to tell whether the algebra will be ugly? I was already 1.5 minutes into the problem when I saw that the algebra was ugly. Good question! I'll answer with some more general thoughts as well. First, as a first step on a word problem, I recommend translating the words into math and then looking at what you have. In this question, we end up with: \(r = \frac{A}{A + 1.25B}\) and \(p = \frac{A}{A + B}\) Next, before you jump into doing any calculations, look at the answer choices. When you see variables in answer choices, you know that you could picking numbers, so you may want to consider doing this unless you see another quick way to do the question. If you are still considering doing the algebra at this point, I would look at the equations that you have, and consider what you need to do: Find r in terms of p. I would ask: Do I see a fast way of doing this? I would not see one, so I would want to fall back on picking numbers. Another hint that the algebra could be ugly is that you don't have linear equations. Instead, you have fractions with variables in both the numerator and denominator of the fractions. Please let me know if you have more questions!
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Re: Last Sunday a certain store sold copies of Newspaper A for $1.00 each
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24 Sep 2019, 07:46
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. \(\frac{500p}{(625 – p)}\) OG 2019 PS03144 Given: 1. 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. 2. 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 Asked: Which of the following expresses r in terms of p? p% of the newspapers that the store sold were copies of newspaper A (100p)% of the newspapers that the store sold were copies of newspaper B revenues from newspaper A / revenues form newspapers B = \(\frac{$p}{$1.25(100p)} = \frac{p}{(125 1.25p)} = \frac{r}{100r}\) r % of revenues come from newspaper A = \(\frac{p}{(125 .25p)} = \frac{4p}{(500p)} = \frac{r}{100}\) \(r = \frac{400p}{(500p)}\) IMO D
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Re: Last Sunday a certain store sold copies of Newspaper A for $1.00 each
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05 Jan 2020, 16:13
SP Newspaper A is $1.00 each Let 'a' be the number of copies sold of Newspaper A SP Newspaper B is $1.25 each ( Recall that 1.25 is \(\frac{5}{4}\) Let 'b' be the number of copies sold of Newspaper B Revenue earned by selling Newspaper A = number of copies sold * price per copy = a*1 Revenue earned by selling Newspaper B = number of copies sold * price per copy = b*\(\frac{5}{4}\) % Revenue earned by selling copies of Newspaper A will be \(\frac{(a*1)}{(a*1+ b*\frac{5}{4})}\) *100 = r r= \(\frac{(4a)}{(4a+ 5b)}\) *100 (I) % copies sold of Newspaper A will be \(\frac{(a)}{(a+ b)}\) *100 = p (II) Look at equation I. Should we want to simplify that we need to express all the variable terms ( a ,b) in such a way that we have all the terms in a or b . So from equation (II) we can get the value of b and substitute it in equation (I) . All the terms in equation (I) ( Numerator and denominator) will cancel out and also we will have a relation between r and p So from Equation (II) we have b =\(\frac{a}{p}(1001)\) (III) Substituting (III) in (I) we have r= \(\frac{(4a)}{(4a+ 5(\frac{a}{p}(1001)))}\) *100 r= \(\frac{(4p)}{(4p+ 5(100p)}\) *100 which is Answer D
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Re: Last Sunday a certain store sold copies of Newspaper A for $1.00 each
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04 Feb 2020, 05:55
Hello, I would appreciate some help on this one , the way I approached this question was by picking some numbers,so: Price of A: 1$ Price of B: 1.25$ Now let's say that 8 B newspapers were sold and 2 A newspapers were sold , so: Total revenue 10$ and total production 10 pieces. r%=2/10= 20 p%=2/10=20 By checking each choice I couldn't find a suitable answer choice could someone show me the flaw of my approach?



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Re: Last Sunday a certain store sold copies of Newspaper A for $1.00 each
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12 May 2020, 05:01
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. AaronPondBunuelCan't we just take r to be 100, in that case p=100 (if 100% revenue comes from A, then no. of copies of A sold should also be 100% with this we can eliminate a,c and e Is there a similar approach to eliminate "b' through such simple calculation(no algebra)



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Last Sunday a certain store sold copies of Newspaper A for $1.00 each
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31 May 2020, 01:21
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 I'd argue plugging numbers in this question is not straight forward because options D and E end up being very close. The numbers you chose happened to allow for simplification of (D) to 1/6 but the numbers I chose (6 of A and 4 of B) didn't allow for such a nice simplification and it was hard to discern (or 'guesstimate') which was correct out of (D) and (E). My view is the algebraic approach suits this question better, even though I normally prefer the number plugging method.




Last Sunday a certain store sold copies of Newspaper A for $1.00 each
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