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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
Math Expert V
Joined: 02 Sep 2009
Posts: 64939

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Let's talk strategy here. Many explanations of Quantitative questions focus blindly on the math, but remember: the GMAT is a critical-thinking test. For those of you studying for the GMAT, you will want to internalize strategies that actually minimize the amount of math that needs to be done, making it easier to manage your time. The tactics I will show you here will be useful for numerous questions, not just this one. My solution is going to walk through not just what the answer is, but how to strategically think about it. As a result, I might write out some steps that I would normally just do in my head on the GMAT, but I want to make sure everyone sees the complete approach. Ready? Here is the full "GMAT Jujitsu" for this question:

The first thing we need to do is setup our equations. The equation for a percent is simple: $$N=\frac{\%}{100}T$$. Alternately, $$100N=\%T$$.
The problem states that “r percent of the store’s revenue from newspaper sales was from Newspaper A.” Revenue is equal to the number of items sold, multiplied by the sales price of each item. So, let’s plug revenue into our percent formula:

$$100(Revenue_A)=\%(Revenue_{total})$$
$$100(1*A)=r(1*A+\frac{5}{4}B)$$

Notice that I translated $$1.25$$ into its fractional form. I call this strategy, “Fractions Are Your Friends” in my classes. Fractions are almost always easier to mathematically manipulate than decimals are.

The problem also states that “p percent of the newspapers that the store sold were copies of Newspaper A.” Thus,

$$100(Copies_A )=\%(Copies_{total})$$
$$100(A)=p(A+B)$$

We now have 2 equations with 4 unknowns, but the problem isn’t asking us to solve for a number. Our target is “r in terms of p”, which means we need to get rid of variables $$A$$ and $$B$$, while simultaneously solving for $$r$$. At this point, we have two options: (1) just doing the dang algebra, or (2) a strategy I call in my classes “Easy Numbers.” Let's take a look at both strategies, starting with “Easy Numbers”:

Tactic 1: Easy Numbers

Since the problem gives us ratios without ever giving us totals, we can actually chose the “total” values in a way that makes the math simple. The reasons why we can pick values for totals (if the totals aren’t given) are two-fold: first, doing this turns abstract formulas into very concrete, easy-to-understand equations. Second, since the total values clearly drop out of the final solutions – after all, $$A$$ and $$B$$ aren’t in the answer choices – assigning them values that will disappear by the end is totally fine.

With percentage questions without totals, a common “Easy Number” we could pick is $$100$$. The equation $$100A = p(A+B)$$ demonstrates this. If we say that $$A+B=100$$, then $$100A = p(100)$$ and $$A=p$$, allowing us to eliminate $$A$$. But we also know that if $$A=p$$ and $$A+B = 100$$, then $$B = 100–p$$. Plugging in these values into the revenue equation above gets us a single equation in terms of $$r$$ and $$p$$:

$$100(p)=r(1p+\frac{5}{4}(100-p)$$

From this point forward, it’s just math, though our goal shouldn’t be just to randomly move things around – instead we use the answer choices to inform what “shape” the GMAT wants the math to be in. Since the question tells us that we are looking for $$r$$ in terms of $$p$$, we isolate $$r$$ in the equation by dividing both sides by the same value:

$$r=\frac{100(p)}{p+\frac{5}{4}(100)-\frac{5}{4}p}=\frac{100(p)}{\frac{5}{4}(100)-\frac{1}{4}p}$$

Multiplying the fraction by $$\frac{4}{4}$$ (a strategy I call in my classes “Multiply by 1”), keeps the fraction equivalent, but simplifies the denominator so it looks like the answer choices:

$$r=\frac{400p}{500-p}$$

We have our answer. It’s “D”.

Tactic 2: Do the Algebra

Since the GMAT is a critical-thinking test that rewards mental flexibility, there is often more than one way to solve a question. (I guess this means that since the GMAT is a Computer-Adaptive Test, there is more than one way to skin a CAT!) Here is what it would look like algebraically…

First, we know we need to eliminate both $$a$$ and $$b$$ out of the equation, leaving only $$r$$ and $$p$$. One quick way to eliminate variables is to substitute. Solving the first equation for $$r$$, we get:

$$100(1*A)=r(A+\frac{5}{4}B)$$

$$r=\frac{100A}{A+\frac{5}{4}B}$$

Solving the second equation for $$b$$, we get:

$$100(A)=p(A+B)$$

$$100A-pA=pB$$

$$B=\frac{100A-pA}{p}=A(\frac{100-p}{p})$$

Now, substituting the second equation into the first allows us to eliminate $$b$$, while simultaneously getting rid of $$a$$, through a tactic I like to call in my classes “Divide and Conquer.” Watch this:

$$r=\frac{100A}{A+\frac{5}{4}B}=\frac{100A}{A+\frac{5}{4}(A(\frac{100-p}{p}))}$$

Now, our job is to turn what we have into one of the answer choices, so use the answer choices as a guide for how you should think about the math. I call this “Stay on Target.” We know we still need to get rid of $$A$$. Because we are dealing with a fraction, let’s see if we can factor out an $$A$$ out of the top and bottom, thereby cancelling it:

$$r=\frac{100A}{A(1+\frac{5}{4}(\frac{100}{p}-1))}=\frac{100}{1+\frac{5}{4}(\frac{100}{p}-1)} =\frac{100}{1+\frac{500}{4p}-\frac{5}{4}}=\frac{100}{\frac{500}{4p}-\frac{1}{4}}$$

We can multiply both the top and bottom of the fraction by $$4p$$, eliminating the messy fractions in the bottom of the denominator. This simplifies it down to:
$$r=\frac{400p}{500-p}$$

Now, let’s look back at this problem from the perspective of strategy. For those of you studying for the GMAT, it is far more useful to identify patterns in questions than to memorize the solutions of individual problems. This problem can teach us a solid pattern seen throughout the GMAT. If you are given percentages or ratios without totals, one way to cut through this abstraction is to plug in “Easy Numbers” for the totals. Otherwise, you could also solve such problems algebraically, but whatever you do, don't think haphazardly. Your job is to turn what you have been given into what you want to have. Use the answer choices as a guide for what you should do. And that is how you think like the GMAT.
_________________
Aaron Pond
Veritas Prep Teacher of the Year

Visit me at https://www.veritasprep.com/gmat/aaron-pond/ if you would like to learn even more "GMAT Jujitsu"!
Intern  Joined: 02 Sep 2010
Posts: 30
Location: India

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Note that total revenue will be k*(p + (100-p)*1.25) where k is a constant depending on actual number of papers sold

The contribution of type A is kp

So r=100 * kp/k(p + 125 -1.25p)
= 100p/(125-.25p)
= 400p/(500-p)
Intern  Joined: 04 Apr 2012
Posts: 2
Re: Last Sunday a certain store sold copies of Newspaper A for $1.00 each [#permalink] ### Show Tags 8 6 The algebra way is not time taking even..if we proceed as below: (News A) A=$1
(News B) B = $1.25 or$5/4
Total newspaper sold= x
No. of A Newspaper sold = p/100 *x is r% of total revnue
Total revenue: p/100*x*$1 + (100-p)/100*x*$5/4
Equation: px/100=r/100(px/100+(500/4-5p/4)x/100)
px/100=r/100(4px+500x-5px/400)
removing common terms as 100 and x out and keeping only r on RHS
Manager  Joined: 05 Nov 2012
Posts: 63
Schools: Foster '15 (S)
GPA: 3.65

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6
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 = 20/120 = 1/6 = 16.7% and p = 20

A) 100*20/(125-20) -> Incorrect
B) 150*20/(250-20) -> Incorrect
C) 300*20/(375-20) -> Incorrect
D) 400*20/(500-20) = 8/48 = 1/6*100 = 16.7% - > Correct
E) 500*20/(625-20) -> Incorrect

So Ans D

I did tried picking smart nos...mmm..ok may be not smart as yours but basically here is my pick

p=5 (5 papers of A sold) so revenue from A = 5
20 papers of B sold so 20*1.25 so revenue from paper B = 25
Total revenue R = 25+5 = 30
no of A paper sold = P = 5
so revenue = 5/30 or around 16.6% percent-------------->>> till this part I got it right
now try plugin the answer choice D
$$\frac{400*5}{500-5}$$

= $$\frac{2000}{495}$$ is not equal 16.6%.. what's wrong here?
Manager  Joined: 05 Nov 2012
Posts: 63
Schools: Foster '15 (S)
GPA: 3.65
Re: Last Sunday a certain store sold copies of Newspaper A for $1.00 each [#permalink] ### Show Tags 3 nikhil007 wrote: 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 = 20/120 = 1/6 = 16.7% and p = 20 A) 100*20/(125-20) -> Incorrect B) 150*20/(250-20) -> Incorrect C) 300*20/(375-20) -> Incorrect D) 400*20/(500-20) = 8/48 = 1/6*100 = 16.7% - > Correct E) 500*20/(625-20) -> Incorrect So Ans D I did tried picking smart nos...mmm..ok may be not smart as yours but basically here is my pick p=5 (5 papers of A sold) so revenue from A = 5 20 papers of B sold so 20*1.25 so revenue from paper B = 25 Total revenue R = 25+5 = 30 no of A paper sold = P = 5 so revenue = 5/30 or around 16.6% percent-------------->>> till this part I got it right now try plugin the answer choice D $$\frac{400*5}{500-5}$$ = $$\frac{2000}{495}$$ is not equal 16.6%.. what's wrong here? nikhil, the error you are making is in terms of p, since p is the % of A newspapers sold P = 5/30*100 not 5 as you are using. hope that helps Current Student B Joined: 10 Mar 2013 Posts: 448 Location: Germany Concentration: Finance, Entrepreneurship Schools: WHU MBA"20 (A$)
GMAT 1: 580 Q46 V24
GPA: 3.88
WE: Information Technology (Consulting)

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Quote:
$$r=\frac{a}{a + 1.25*\frac{a(100-p)}{p}}*100$$ --> reduce by $$a$$ and simplify --> $$r=\frac{100p}{p+125-1.25p}=\frac{100p}{125-0.25p}$$

Hi Bunel, I don't get the reduction. How do you get rid of the $$a+$$ in the denominator?

I only get this:
$$r=\frac{a}{a + 1.25*\frac{a(100-p)}{p}}*100$$ --> reduces to --> $$r=\frac{100 p}{a+1,25*(100-p)}$$
Math Expert V
Joined: 02 Sep 2009
Posts: 64939

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$$\frac{r}{100} = \frac{a}{(a+1.25b)}$$
$$\frac{p}{100} = \frac{a}{(a+b)}$$
So our problem is how to go about solving with a+b and a+1.25b in the denominator. An easy way out is take the reciprocal.

So,
$$\frac{100}{r} = \frac{(a+1.25b)}{a}$$
and
$$\frac{100}{p} = \frac{(a+b)}{a}$$
So,
$$\frac{100}{r} = 1+ \frac{1.25b}{a}$$ ........(1)
and
$$\frac{100}{p} = 1+ \frac{b}{a}$$
or $$\frac{100}{p} -1 = \frac{b}{a}$$...........(2)
So we have isolated b/a to a corner. Let's substitute for b/a in (1) so that we can have an equation only in r & p which we could solve for r
$$\frac{100}{r} = 1+ \frac{5}{4} * (\frac{100}{p} -1)$$
$$\frac{100}{r} = 1+ \frac{5}{4} * (\frac{100-p}{p})$$
$$\frac{100}{r} = 1+ (\frac{500-5p}{4p})$$
$$\frac{100}{r} = (\frac{500-p}{4p})$$
$$\frac{1}{r} = (\frac{500-p}{400p})$$
Now take reciprocal again to get r:
$$\frac{r}{1} = (\frac{400p}{(500-p)})$$
Manager  Joined: 31 May 2012
Posts: 106
Re: Last Sunday a certain store sold copies of Newspaper A for $1.00 each [#permalink] ### Show Tags 4 6 Here is simplest & quickest way to reach answer: Price of A=$1.
Price of B= $1.25 i.e.$ 5/4

We want revenue(r) in terms of percent of A(p)
To calculate revenue, Assume, p= 20
So, Revenue = 20(1)+80(5/4)=$120 Now, r=$20/$120= 1/6 Now, put p=20 in each option and try to see if you can get 100/6 anywhere. Just by looking at options, I see only Option(D) can serve my purpose. 400p / (500 – p) = 100 X (4X20)/(480) = 100/6. Intern  Joined: 13 Feb 2014 Posts: 6 Re: Last Sunday a certain store sold copies of Newspaper A for$1.00 each  [#permalink]

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Bunuel is there a reason why you chose to isolate B and not A? I tried doing it by isolating A but cant solve it.
Math Expert V
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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)

When you see percent, it is good idea to pick 100.

Let's say store sold 100 newspapers and p is 60% --> 60 (also 60%) A and 40 (also 40%) B.
Then, Rev(A) = 60 * $1 =$60, Rev(B) = 40 * $1.25 =$50. Total Rev --> $60 +$50 = $110. Given that r percent of stores revenue is from newspaper A --> r =$60 / $110 (D) --> (400 * 60%) / (500 - 60) => 240 / 440, which is same as 60 / 110. _________________ If you learn to do things that you need to do - then someday - you can do things that you want to do. Veritas Prep GMAT Instructor V Joined: 16 Oct 2010 Posts: 10622 Location: Pune, India Re: Last Sunday a certain store sold copies of Newspaper A for$1.00 each  [#permalink]

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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)

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

_________________
Karishma
Veritas Prep GMAT Instructor

Math Expert V
Joined: 02 Sep 2009
Posts: 64939

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VeritasPrepKarishma 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)

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

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? Re: Last Sunday a certain store sold copies of Newspaper A for $1.00 each [#permalink] 24 Aug 2014, 09:01 Go to page 1 2 3 Next [ 48 posts ] # Last Sunday a certain store sold copies of Newspaper A for$1.00 each   