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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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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

Answer (D)
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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)

Hope that helps
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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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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)


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:

Originally posted by BrentGMATPrepNow on 26 Jul 2016, 06:54.
Last edited by BrentGMATPrepNow on 19 Jan 2020, 13:51, edited 1 time in total.
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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)


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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Quote:
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)


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 - https://www.gmatprepnow.com/module/gmat- ... /video/933
- Tips for the Algebraic Approach - https://www.gmatprepnow.com/module/gmat- ... /video/934
- Tips for the Input-Output Approach - https://www.gmatprepnow.com/module/gmat- ... /video/935


Cheers,
Brent

Originally posted by BrentGMATPrepNow on 29 Mar 2016, 15:35.
Last edited by BrentGMATPrepNow on 21 Apr 2016, 14:33, edited 1 time in total.
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Lets say the total number of copies sold is 100
Number of newspaper A copies sold is P and then the newspaper B copies sold is 100 -p
We can form this equation.
\(\frac{R}{100} [ P*1+ 1.25(100-P)] = P *1\)


\(\frac{R}{100}(P + 1.25(100-P) = P\)
R(P+ 125 -1.25P) = 100P
\(125 R – .25RP= 100P\)
\(125R - \frac{1}{4}RP = 100P\)
\(500 R – RP = 400P\)
\(R(500-P) = 400P\)

\(R= \frac{400P}{(500-P)}\\
\)
Option D is the answer

Or as an alternate approach, we can assume values.
Let say, total no of newspapers be 100.
Assume P =60%
Let the No of Newspaper A sold = 60 % of 100= 60
Then no of Newspaper B sold is 40
We assumed P as 60 % because you get 40 as the no of Newspaper B sold and when 40 multiplied by 1.25, you will get an integer and your calculations will be easy (Or you can start by assuming P=20 %)
Revenue from Newspaper A= 60*1 = 60 $
Revenue from Newspaper B = 40*1.25= 50 $

Then R= % revenue from sales of Newspaper A = \(\frac{60}{110}* 100 = 54.54 \)%
(* Use Fraction-% conversion table, 1/11= 9.09 % then 6/11 = 6*9.09= 54.54 %)

In each option, you can substitute P = 60 and check whether you are getting R = 54 %
You don’t need to calculate exact value in each option, instead you can check whether it’s close to
50 % or not
For example, in Option A.
R = 100*60/65 = (60/65) *100
We can straightaway eliminate as its close to 100 %

In Option b,
\(R= 150*60/190 = \frac{90}{190}*100\)
Since 90 is less than 50 % of 190 . we can eliminate option B

in option C ,
\(R= 300*60/315= \frac{180}{315}*100\)
we can eliminate as its more than 55 % of 315
Try to split the percentage as 50% + 5 % of 315 to get approximate value faster.

in Option D,
\(R =400*60/440 = \frac{24}{44 }* 100 = 6/11* 100 = 54.54 \)%

Option D is the answer

Thanks,
Clifin J Francis
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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)


Answer: Option D

Check solution attached
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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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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

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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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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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 $1.00 each [#permalink]
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\(r=\frac{a}{a + 1.25*\frac{a(100-p)}{p}}*100\) --> 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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Re: Last Sunday a certain store sold copies of Newspaper A for $1.00 each [#permalink]
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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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\(\frac{r}{100} = \frac{a}{(a+1.25b)}\)
\(\frac{p}{100} = \frac{a}{(a+b)}\)
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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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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