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

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26 Sep 2010, 11:41

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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)

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.

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 --> \(r=\frac{100p}{125-0.25p}=\frac{400p}{500-p}\).

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

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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

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

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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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02 Mar 2013, 13:09

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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, 11:21, edited 1 time in total.

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

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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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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.

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

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16 May 2013, 11:23

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]

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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.

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

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04 Mar 2014, 21:10

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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
_________________

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.
_________________

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

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22 Jun 2014, 09:52

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?

gmatclubot

Re: Last Sunday a certain store sold copies of Newspaper A for
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22 Jun 2014, 09:52

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