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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]  26 Sep 2010, 11:41
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Difficulty:

80% (hard)

Question Stats:

45% (04:27) correct 54% (02:53) wrong based on 25 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)
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Last Sunday a certain store sold copies of Newspaper A for [#permalink]  15 Jun 2012, 20:02
4
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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: r in terms of P? [#permalink]  26 Sep 2010, 14:13
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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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Re: r in terms of P? [#permalink]  26 Sep 2010, 15:41
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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: r in terms of P? [#permalink]  26 Sep 2010, 12:14
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]  26 Sep 2010, 19:51
ans is 400P/(500-P)
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Last Sunday a certain store sold copies of newspaper A for $[#permalink] 20 May 2012, 23:44 Last Sunday a certain store sold copies of newspaper A for$1.00 each and copies of newspaper B for $1.25 each, and the store sold no other newspapers that day. If r% of the stores revenue from newspaper sales was from Newspaper A and if p% of the newspapers that were sold were copies of newspaper A, which of the following expresses r in terms of p? A (100p)/(125-p) B (150p)/(250-p) C (300p)/(375-p) D (400p)/(500-p) E (500p)/(625-p) GMAT Club team member Joined: 02 Sep 2009 Posts: 11591 Followers: 1798 Kudos [?]: 9582 [0], given: 826 Re: Last Sunday a certain store sold copies of newspaper A for$ [#permalink]  21 May 2012, 00:14
macjas 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% of the stores revenue from newspaper sales was from Newspaper A and if p% of the newspapers that were 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)

Merging similar topics. Please refer to the solutions above.
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Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink]  22 May 2012, 10:53
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 Manager Joined: 05 Nov 2012 Posts: 56 Concentration: International Business, Operations Schools: Haas EWMBA '16 GPA: 3.65 Followers: 1 Kudos [?]: 13 [0], given: 8 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] 02 Mar 2013, 13:09 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 Last edited by pikachu on 16 May 2013, 11:21, edited 1 time in total. Manager Joined: 04 Dec 2011 Posts: 68 Followers: 0 Kudos [?]: 4 [0], given: 8 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] 03 May 2013, 10:49 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: 21 Jan 2010 Posts: 185 Followers: 0 Kudos [?]: 21 [0], given: 9 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] 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. Manager Joined: 04 Dec 2011 Posts: 68 Followers: 0 Kudos [?]: 4 [0], given: 8 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] 03 May 2013, 11:25 appreciate the algebra approach, but im trying to understand where did i go wrong with my approach _________________ 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: 05 Nov 2012 Posts: 56 Concentration: International Business, Operations Schools: Haas EWMBA '16 GPA: 3.65 Followers: 1 Kudos [?]: 13 [0], given: 8 Re: Last Sunday a certain store sold copies of Newspaper A for [#permalink] 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
Re: Last Sunday a certain store sold copies of Newspaper A for   [#permalink] 16 May 2013, 11:23
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