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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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Question Stats:

51% (04:24) correct
49% (05:31) wrong based on 1030 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)

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

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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: r in terms of P? [#permalink]
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 --> \(r=\frac{100p}{125-0.25p}=\frac{400p}{500-p}\).

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

2

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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: r in terms of P? [#permalink]
26 Sep 2010, 11:14

1

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

Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink]
25 Jun 2014, 22:46

1

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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)

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

Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink]
24 Aug 2014, 23:44

1

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Expert's post

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

Answer (D)

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 _________________

Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink]
22 May 2012, 09:53

1

This post was BOOKMARKED

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

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

2

This post was BOOKMARKED

\(\frac{r}{100} = \frac{a}{(a+1.25b)}\) \(\frac{p}{100} = \frac{a}{(a+b)}\) 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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