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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 $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. \(\frac{500p}{(625 – p)}\)



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

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New post 26 Sep 2010, 14:13
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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)


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

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New post 04 Oct 2018, 10:09
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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!) :grin: 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:
\(r=\frac{400p}{500-p}\)
The answer is still “D”.

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

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New post 15 Jun 2012, 20:02
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zaarathelab 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)

What is the simplest way to solve this??

Let the total copies of newspaper(A+B) sold be 100
so the number of copies of A sold is p
number of copies of B sold is 100-p
thus revenue from A = p*1$ = p$
revenue from B = (100-p)5/4; because 1.25 = 5/4
percent of revenue from A = r = p/p+[(100-p)5/4)]= 400p / (500 – p)
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Re: Last Sunday a certain store sold copies of Newspaper A for  [#permalink]

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New post 26 Sep 2010, 12:14
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Wow this is a hard question no doubt.
How do you solve this one quickly? I tried pluging numbers instead on r and p but the result was really not comfortable no matter what numbers I used.
Also solving it with pure algebra is far from being simple or done in under 2-3 minutes.

What's the source?
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Re: Last Sunday a certain store sold copies of Newspaper A for  [#permalink]

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New post 26 Sep 2010, 15:41
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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)
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Re: Last Sunday a certain store sold copies of Newspaper A for  [#permalink]

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New post 22 May 2012, 10:53
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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
p=r(500-p)/400 or r=400p/(500-p)..Answer..D
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Re: Last Sunday a certain store sold copies of Newspaper A for  [#permalink]

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New post Updated on: 16 May 2013, 11:21
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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)


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/(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
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Originally posted by pikachu on 02 Mar 2013, 13:09.
Last edited by pikachu on 16 May 2013, 11:21, edited 1 time in total.
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Re: Last Sunday a certain store sold copies of Newspaper A for  [#permalink]

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New post 03 May 2013, 10:49
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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?
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Re: Last Sunday a certain store sold copies of Newspaper A for  [#permalink]

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New post 03 May 2013, 11:21
My train of thoughts :

Let paper A sold = a.
Let paper B sold = b.

Now r=a/(a+1.25b) x 100 .....1

p=100a/(a+b) ....2

Now we see 2 equations and 3 variables. We must find another equation to get to the answer.
if p = % sales of paper A. 100-p is % sales of paper B. Therefore :
1-p = 100b/(a+b) ........3

Now to make life simpler divide 3 by 2 :
(100-p)/p = 100b/(a+b) x (a+b)/100a - > b/a = (100-p)/p......4

Divide 1 by a at numerator and denominator.
r = 100/(1+1.25(b/a)...........5

If you substitute the value of b/a from 4 into 5, you get D.
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Re: Last Sunday a certain store sold copies of Newspaper A for  [#permalink]

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New post 16 May 2013, 11:23
2
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
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Re: Last Sunday a certain store sold copies of Newspaper A for  [#permalink]

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New post 21 Jul 2013, 09:12
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............A..... B
Price:.... 1.... 1,25
Amount:. P.... 100-P
------------------------------
Revenue: P+1,25*(100-p) ---> P + 125 - 1,25P= 125 - 0,25P

---> Revenue of A / Total Revenue: P / 125 - 0,25P = P/ ((500-p)/4)) = 4P/500-P
---> r/100 = 4P/500-P --> r = 400P / 500-P ...................Correct Answer is (D)

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

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New post 05 Nov 2013, 03:55
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 --> \(r=\frac{100 p}{a+1,25*(100-p)}\)
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New post 05 Nov 2013, 06:53
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Marcoson wrote:
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 --> \(r=\frac{100 p}{a+1,25*(100-p)}\)


\(r=\frac{a}{a + 1.25*\frac{a(100-p)}{p}}*100\);

Factor out a from the denominator: \(r=\frac{a}{a(1 + 1.25*\frac{(100-p)}{p})}*100\).

Reduce it: \(r=\frac{1}{1 + 1.25*\frac{(100-p)}{p}}*100\).

Hope it's clear.
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PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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

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New post 04 Mar 2014, 12:52
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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:
\(\frac{r}{1} = (\frac{400p}{(500-p)})\)
D is the correct answer.
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Re: Last Sunday a certain store sold copies of Newspaper A for  [#permalink]

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New post 09 Apr 2014, 11:18
3
4
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.
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Re: Last Sunday a certain store sold copies of Newspaper A for  [#permalink]

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New post 09 May 2014, 06:06
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.
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Re: Last Sunday a certain store sold copies of Newspaper A for  [#permalink]

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New post 09 May 2014, 07:05
gciftci wrote:
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.


In \(r=\frac{a}{a + 1.25b}*100\) we have b only in one place while a there is represented twice. So, it's better to substitute b there. Though you should get the same answer no matter whether you substitute a or b.
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Re: Last Sunday a certain store sold copies of Newspaper A for  [#permalink]

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New post 09 May 2014, 15:47
great question/practice, thanks for posting it

r/100 = Qa/(Qa+1.25Qb)
p/100 * Q = Qa or (1- p)/100 * Q = Qb

using these equations, the answer is D
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Re: Last Sunday a certain store sold copies of Newspaper A for  [#permalink]

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New post 25 Jun 2014, 20:00
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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.
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Re: Last Sunday a certain store sold copies of Newspaper A for &nbs [#permalink] 25 Jun 2014, 20:00

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