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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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57% (02:48) wrong based on 92 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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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, 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}. Answer: D.
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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, 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)
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Re: Last Sunday a certain store sold copies of newspaper A for $ [#permalink]
21 May 2012, 00:14
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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
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Last Sunday a certain store sold copies of Newspaper A for [#permalink]
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]
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.
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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?
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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.
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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
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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
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Re: Last Sunday a certain store sold copies of Newspaper A for
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16 May 2013, 11:23
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