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# Last Sunday a certain store sold copies of Newspaper A for

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26 Sep 2010, 10:41
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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. 100p/(125 – p)
B. 150p/(250 – p)
C. 300p/(375 – p)
D. 400p/(500 – p)
E. 500p/(625 – p)
[Reveal] Spoiler: OA
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15 Jun 2012, 19:02
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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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02 Mar 2013, 12:09
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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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Last edited by pikachu on 16 May 2013, 10:21, edited 1 time in total.
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Re: r in terms of P? [#permalink]

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26 Sep 2010, 13:13
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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 --> $$r=\frac{100p}{125-0.25p}=\frac{400p}{500-p}$$.

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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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26 Sep 2010, 14:41
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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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22 May 2012, 09:53
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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 Manager Joined: 04 Dec 2011 Posts: 81 Schools: Smith '16 (I) Followers: 0 Kudos [?]: 25 [2] , given: 13 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 03 May 2013, 09:49 2 This post received KUDOS 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? _________________ Life is very similar to a boxing ring. Defeat is not final when you fall down… It is final when you refuse to get up and fight back! 1 Kudos = 1 thanks Nikhil Manager Joined: 04 Jun 2010 Posts: 113 Concentration: General Management, Technology Schools: Chicago (Booth) - Class of 2013 GMAT 1: 670 Q47 V35 GMAT 2: 730 Q49 V41 Followers: 15 Kudos [?]: 249 [1] , given: 43 Re: r in terms of P? [#permalink] ### Show Tags 26 Sep 2010, 11:14 1 This post received KUDOS 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? _________________ Consider Kudos if my post helped you. Thanks! -------------------------------------------------------------------- My TOEFL Debrief: http://gmatclub.com/forum/my-toefl-experience-99884.html My GMAT Debrief: http://gmatclub.com/forum/670-730-10-luck-20-skill-15-concentrated-power-of-will-104473.html SVP Status: The Best Or Nothing Joined: 27 Dec 2012 Posts: 1858 Location: India Concentration: General Management, Technology WE: Information Technology (Computer Software) Followers: 50 Kudos [?]: 1994 [1] , given: 193 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 04 Mar 2014, 20:10 1 This post received KUDOS prsnt11 wrote: $$\frac{r}{100} = \frac{a}{(a+1.25b)}$$ $$\frac{p}{100} = \frac{a}{(a+b)}$$ 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. I got the two equations, but required lot of time to resolve the same in terms of p & r _________________ Kindly press "+1 Kudos" to appreciate Manager Joined: 31 May 2012 Posts: 166 Followers: 5 Kudos [?]: 161 [1] , given: 69 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 09 Apr 2014, 10:18 1 This post received KUDOS 4 This post was BOOKMARKED 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. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7194 Location: Pune, India Followers: 2175 Kudos [?]: 14064 [1] , given: 222 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 25 Jun 2014, 22:46 1 This post received KUDOS Expert's post 1 This post was BOOKMARKED 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) _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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

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24 Aug 2014, 23:44
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russ9 wrote:
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?

When the question says "r in terms of p", it implies that r should be on the left hand side of the equation and everything on the right should only in terms of p. There should be no other variable on the right hand side. Then either you can solve algebraically or plug in values.

The question says you need to find r. What is r? It is "r percent of the store’s revenues from newspaper sales was from Newspaper A"
It means: Revenue from Newspaper A = (r/100)* Total revenue
So r = (Revenue from Newspaper A/Total revenue) * 100
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Joined: 03 Jan 2015 Posts: 68 Concentration: Strategy, Marketing WE: Research (Pharmaceuticals and Biotech) Followers: 2 Kudos [?]: 32 [1] , given: 224 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 10 Jan 2015, 09:39 1 This post received KUDOS 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

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

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29 Apr 2016, 08:50
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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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Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink]

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15 Jan 2017, 23:21
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stampap wrote:
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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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Senior Manager Joined: 21 Jan 2010 Posts: 344 Followers: 3 Kudos [?]: 180 [0], given: 12 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 03 May 2013, 10:21 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. Manager Joined: 05 Nov 2012 Posts: 71 Concentration: International Business, Operations Schools: Foster '15 (S) GPA: 3.65 Followers: 1 Kudos [?]: 127 [0], given: 8 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] ### Show Tags 16 May 2013, 10:23 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 _________________ ___________________________________________ Consider +1 Kudos if my post helped Intern Joined: 08 Jun 2013 Posts: 4 Followers: 0 Kudos [?]: 1 [0], given: 1 Re: r in terms of P? [#permalink] ### Show Tags 05 Nov 2013, 02:55 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 --> $$r=\frac{100 p}{a+1,25*(100-p)}$$ Math Expert Joined: 02 Sep 2009 Posts: 37154 Followers: 7277 Kudos [?]: 96860 [0], given: 10808 Re: r in terms of P? [#permalink] ### Show Tags 05 Nov 2013, 05:53 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. _________________ Re: r in terms of P? [#permalink] 05 Nov 2013, 05:53 Go to page 1 2 3 Next [ 50 posts ] Similar topics Replies Last post Similar Topics: 5 Last Sunday a certain store sold copies of Newspaper 4 09 Aug 2015, 14:25 2 Last Sunday a certain store sold copies of newspaper a for$1.00 each 2 26 Oct 2014, 08:11
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