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Last Sunday a certain store sold copies of Newspaper A for

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

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New post 10 Aug 2018, 14:56
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!
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Re: Last Sunday a certain store sold copies of Newspaper A for  [#permalink]

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New post 17 Dec 2018, 21:35
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)/500-p

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

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New post 17 Dec 2018, 22:19
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)/500-p

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

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New post 18 Dec 2018, 10:11
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)/500-p

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 nice-looking 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  [#permalink]

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New post 17 Feb 2019, 05:08
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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 \(p=\frac{a}{a+b}*100\) --> \(b=\frac{a}{p}*100-a=\frac{a(100-p)}{p}\) --> \(r=\frac{a}{a + 1.25*\frac{a(100-p)}{p}}*100\) --> reduce by \(a\) and simplify --> \(r=\frac{100p}{p+125-1.25p}=\frac{100p}{125-0.25p}\) --> multiply by 4/4 --> \(r=\frac{100p}{125-0.25p}=\frac{400p}{500-p}\).

Answer: D.


Hi Bunuel,

Many thanks for the solution.

Do you happen to have recommendations of similar problems?

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

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New post 09 Jun 2019, 05:25
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 \(p=\frac{a}{a+b}*100\) --> \(b=\frac{a}{p}*100-a=\frac{a(100-p)}{p}\) --> \(r=\frac{a}{a + 1.25*\frac{a(100-p)}{p}}*100\) --> reduce by \(a\) and simplify --> \(r=\frac{100p}{p+125-1.25p}=\frac{100p}{125-0.25p}\) --> multiply by 4/4 --> \(r=\frac{100p}{125-0.25p}=\frac{400p}{500-p}\).

Answer: D.



Shouldn't the denominator be same for both r & p i.e., a + 1.25b.

For r, you are using a + 1.25b, but for P you are using a+b in the denominator.

Can anyone explain why a+b in denominator for P ?
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Re: Last Sunday a certain store sold copies of Newspaper A for   [#permalink] 09 Jun 2019, 05:25

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