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Last Sunday a certain store sold copies of Newspaper A for [#permalink]
26 Sep 2010, 10:41

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A

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Difficulty:

65% (hard)

Question Stats:

53% (04:09) correct
47% (03:17) wrong based on 490 sessions

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)

Re: r in terms of P? [#permalink]
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.

Re: r in terms of P? [#permalink]
26 Sep 2010, 13:13

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Expert's post

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}.

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

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

The world ain't all sunshine and rainbows. It's a very mean and nasty place and I don't care how tough you are it will beat you to your knees and keep you there permanently if you let it. You, me, or nobody is gonna hit as hard as life. But it ain't about how hard ya hit. It's about how hard you can get it and keep moving forward. How much you can take and keep moving forward. That's how winning is done!

Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink]
02 Mar 2013, 12:09

17

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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 _________________

___________________________________________ Consider +1 Kudos if my post helped

Last edited by pikachu on 16 May 2013, 10:21, edited 1 time in total.

Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink]
03 May 2013, 09:49

1

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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? _________________

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!

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

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

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

Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink]
04 Mar 2014, 20:10

1

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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 _________________

Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink]
09 May 2014, 06:05

Expert's post

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. _________________

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

gmatclubot

Re: Last Sunday a certain store sold copies of Newspaper A for
[#permalink]
22 Jun 2014, 08:52