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

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.

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?

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

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 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.

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 ---> \(b=\frac{a}{p}*100-a=\frac{a(100-p)}{p}\) and from that how how got this one \(r=\frac{a}{a + 1.25*\frac{a(100-p)}{p}}*100\) and how call this process / step ?

thanks!

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.

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.

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.