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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 [#permalink] New post 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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Re: r in terms of P? [#permalink] New post 26 Sep 2010, 11:14
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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: r in terms of P? [#permalink] New post 26 Sep 2010, 13:13
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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)


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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Re: r in terms of P? [#permalink] New post 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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Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] New post 22 May 2012, 09:53
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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Last Sunday a certain store sold copies of Newspaper A for [#permalink] New post 15 Jun 2012, 19:02
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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)

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] New post 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: Last Sunday a certain store sold copies of Newspaper A for [#permalink] New post 03 May 2013, 09:49
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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] New post 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.
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Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] New post 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
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Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] New post 21 Jul 2013, 08:12
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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)

Hope that helps
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Re: r in terms of P? [#permalink] New post 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)}
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Re: r in terms of P? [#permalink] New post 05 Nov 2013, 05:53
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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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Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] New post 04 Mar 2014, 11:52
\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] New post 04 Mar 2014, 20:10
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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
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Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] New post 09 Apr 2014, 10:18
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] New post 09 May 2014, 05:06
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] New post 09 May 2014, 06:05
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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.


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] New post 09 May 2014, 14:47
Expert's post
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] New post 22 Jun 2014, 08:52
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 (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


Question : How did you get b = 80 exactly for # sold?
Re: Last Sunday a certain store sold copies of Newspaper A for   [#permalink] 22 Jun 2014, 08:52
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