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

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
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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

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.


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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.

Answer:
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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:
last-sunday-a-certain-store-sold-copies-of-newspaper-a-for-101739.html
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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)/(500-20) = 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.

Answer: Option D

Check solution attached

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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


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.

Check out our video solution to this problem here: https://www.veritasprep.com/gmat-soluti ... olving_210
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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!

\(p=\frac{a}{a+b}*100\)

Cross-multiply: \(pa+pb=100a\);

Re-arrange: \(pb=100a-pa\);

Divide by p: \(b=\frac{100a-pa}{p}=\frac{a(100-p)}{p}\)

The next step is substituting the value of b here: \(r=\frac{a}{a + 1.25b}*100\) to get \(r=\frac{a}{a + 1.25*\frac{a(100-p)}{p}}*100\).

Number plugging:

How to Do Math on the GMAT Without Actually Doing Math
The Power of Estimation for GMAT Quant
How to Plug in Numbers on GMAT Math Questions
Number Sense for the GMAT
Can You Use a Calculator on the GMAT?
Why Approximate?
GMAT Math Strategies — Estimation, Rounding and other Shortcuts
The 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 GMAT
How to Avoid Tedious Calculations on the Quantitative Section of the GMAT
GMAT Tip of the Week: No Calculator? No Problem.
The Importance of Sorting Answer Choices on the GMAT

Hope it helps.
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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
not-serious self-deprecation. 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

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
\(B = \frac{A(100-P)}{P}\)

3) Isolate A's revenue, R.
You have

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.,
\(B = \frac{A(100-P)}{P}\)

You will get to:

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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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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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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