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Last Sunday a certain store sold copies of Newspaper A for

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

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New post 25 Jun 2014, 23:46
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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)


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

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New post 01 Jul 2014, 00:37
Here is my approach :

Let total no. Of newspaper sold bd 100

Then,
No. Of A newspaper sold =p
No. Of B newspaper sold = 100-p

Rev generated by A = p dollars
Rev generated by B = (100-p)1.25
Total rev = 125-0.25p
Now,
R% of (125-0.25p) =p
Solving the above will guve soln :D

I hope this is quick as well. Let me know your thoughts.

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

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New post 24 Aug 2014, 10:01
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?
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Re: Last Sunday a certain store sold copies of Newspaper A for  [#permalink]

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New post 25 Aug 2014, 00:44
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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
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Re: Last Sunday a certain store sold copies of Newspaper A for  [#permalink]

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New post 28 Sep 2014, 09:47
Assuming total number of papers are 100
Number of paper A sold:P
Number of paper B sold:100-P

Total Revenue earned = (P*1) + 1.25*(100-P) = P+125-1.25P = 125-0.25P

Revenue share of paper A is 'R' which is =P/(125-0.25P)*100
= 100p/(125-0.25P)
Multiplying 4 in both numerator & Denominator
=400p/(500-P)
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Re: Last Sunday a certain store sold copies of Newspaper A for  [#permalink]

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New post 10 Nov 2014, 03:29
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


You inserted r(A) as a decimal(0,167) and p(A) as a percent(20, not 0,2). Is that right?
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Re: Last Sunday a certain store sold copies of Newspaper A for  [#permalink]

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New post 10 Jan 2015, 10:39
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Price of A = $1.00
Price of B = $1.25

Assume:
Number of A sold = 100
Number of B sold = 0

Therefore:
r = Revenue = $100 (all from A)
p = Percent of A sold = 100%

Now, plug in values to see which option returns r = 100. Only D satisfies this.
r = [(400*100)/(500-100] = 100

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

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New post 13 Jan 2015, 13:12
hermit84 wrote:
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)

Could you break this down further? I dont quite get how you arrived at the answer from this point:

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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New post 24 Feb 2015, 14:13
hermit84 wrote:
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)


Thanks for the answers and different approaches.
But in the red part highlighted above I didn't quite get how to get from p/[p+(100-p)5/4] to the final answer.
Could someone please explain? Thanks in advance.
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Re: Last Sunday a certain store sold copies of Newspaper A for  [#permalink]

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New post 21 Mar 2015, 07:53
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)


i did it by using the answer choices.

let total copies sold of A and B be 50 each

=> r = (50/(50+62.5)) % = 44.44%

now just substitute p (50) in answer choices and whichever gives 44.44 is the correct answer.
D does.
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Re: Last Sunday a certain store sold copies of Newspaper A for  [#permalink]

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New post 10 May 2015, 23:29
hermit84 wrote:
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)



Absolute perfect solution.

The question asks to express r in terms of P. R is revenue percent for A. Hence R can be expressed as (revenue for A/ total revenue )* 100.

I have struggled with this problem for long time. until came up with the above solution. Thanks zaarathelab :)
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Re: Last Sunday a certain store sold copies of Newspaper A for  [#permalink]

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New post Updated on: 21 Apr 2016, 14:33
Quote:
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)


If you're not sure how to proceed with this question, or if you're behind on time and you want to catch up, you can give yourself a 50-50 chance in about 20 seconds.

To do so, we'll see what happens when we use an EXTREME value for p.
Say p = 100
In other words, 100% of the newspapers sold were Newspaper A.
This means that 100% of the revenue is from Newspaper A.
In other words, when p = 100, then r = 100

At this point, we'll plug in 100 for p and see which one yields a value of 100.
Only answer choices B and D work.
B) 150(100)/(250-100) = 100 PERFECT
D) 400(100)/(500-100) = 100 PERFECT

Now take a guess (B or D) and move on.

Related Resources
The posters have demonstrated two methods (Algebraic and Input-Output) for solving this question type, which I call Variables in the Answer Choices.
If you'd like more information on these approaches, we have some free videos:
- Variables in the Answer Choices - http://www.gmatprepnow.com/module/gmat- ... /video/933
- Tips for the Algebraic Approach - http://www.gmatprepnow.com/module/gmat- ... /video/934
- Tips for the Input-Output Approach - http://www.gmatprepnow.com/module/gmat- ... /video/935


Cheers,
Brent
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Originally posted by GMATPrepNow on 29 Mar 2016, 15:35.
Last edited by GMATPrepNow on 21 Apr 2016, 14:33, edited 1 time in total.
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Re: Last Sunday a certain store sold copies of Newspaper A for  [#permalink]

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New post Updated on: 12 Jan 2017, 14:36
Notice you can simplify this question very easily just by looking at the answer choices. This is a GMAT sleuth question more than a "answerable" question.

The only two numbers you were given in the question were 1 and 1.25, so the answer should contain that ratio. You can eliminate B immediately. All the others have the same ratio, but it weird amounts - the original fraction should be preserved, so that means A and D are your only realistic choices. You've already narrowed this question down to two choices without doing any algebra.

Now, if you think: the ratio of r to p should be similar to the ratio of price A to price B, r/p ~ 1/1.25, so r ~ 4/5p. Realizing this equation should have that inverse relationship will lead you to the correct answer, D.

Originally posted by eaze on 21 Apr 2016, 14:29.
Last edited by eaze on 12 Jan 2017, 14:36, edited 1 time in total.
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Re: Last Sunday a certain store sold copies of Newspaper A for  [#permalink]

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New post 29 Apr 2016, 03:37
for percentage problem, we pick number
there are 100 copies,
so, there will be p copies of a
there will be (100-p ) copis of b

r= p*1/ (P*1+ (100-p)1.25

D.

very quick way
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New post 29 Apr 2016, 06:32
stonecold wrote:
Hey chetan2u it took me good % minutes to get to the answer ..
Any Good Approach ?


HI stonecold,
Sorry missed out on this post..
we have A as $1 and B as $1.25..
To get B in integer, I take B selling 40 copies and A as 60..

Total revenue = 1*60 +1.25*40 = 60+50 =110..
revenue from A = r% of 110 = \(\frac{r}{100}*110 = 60 *1\)..
so\(r = 60*\frac{100}{110} = \frac{600}{11}\)..

lets substitute p as 60% and find if r comes out as\(\frac{600}{11}\)anywhere
so we look for 11 in denominator..
we can just check just denominator for 11 and can see ONLY D has 500-60 =440, a multiple of 11..

A.\(\frac{100p}{(125 – p)} = \frac{100*60}{(125-60)}=\frac{6000}{65}.\). NO

B. \(\frac{150p}{(250 – p)}= \frac{150*60}{(250-60)}= \frac{9000}{190}\).. no

C. \(\frac{300p}{(375 – p)}= \frac{300*60}{(375-60)}=\frac{18000}{315}.\). NO

D. \(\frac{400p}{(500 – p)}=\frac{400*60}{(500-60)} = \frac{24000}{440} = \frac{600}{11}\).. this is what we are looking for .. CORRECT

E.\(\frac{500p}{(625 – p} = \frac{500*60}{625-60}=\frac{30,000}{565}\). No
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Re: Last Sunday a certain store sold copies of Newspaper A for  [#permalink]

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New post 29 Apr 2016, 09:50
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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)


We are given that newspaper A sold for $1 and that newspaper B sold for $1.25.

Next we are are given that that p percent of the newspapers that sold were copies of newspaper A. However, we are not given the total number of copies of both newspapers sold. We can let T = the total copies of both newspapers sold. This means:

(p/100)T = copies of newspaper A sold

This also means that:

(1 – p/100)T = copies of newspaper B sold

We are finally given that r percent of the revenue came from newspaper A. We can translate this into an expression and simplify from there.

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

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New post 26 Jul 2016, 06:54
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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)


Let's use the INPUT-OUTPUT approach.

Let's say that Newspaper A accounted for 20% of all newspapers sold. In other words, p = 20
This means that Newspaper B accounted for 80% of all newspapers sold.

The question asks us to find the value of r, the percentage of newspaper revenue from Newspaper A.
To determine this, let's say that 100 newspapers we sold IN TOTAL.
This means that 20 Newspaper A's were sold and 80 Newspaper B's were sold.

REVENUE:
Newspaper A: 20 newspapers at $1 apiece = $20
Newspaper B: 80 newspapers at $1.25 apiece = $100
So, TOTAL revenue = $120

Since Newspaper A accounted for $20 of revenue, we can say that Newspaper A accounted for 16 2/3% of revenue. In other words, r = 16 2/3
Aside: We know this because $20/$120 = 1/6 = 16 2/3%

So, when we INPUT p = 20, the OUTPUT is r = 16 2/3.
We'll now plug p = 20 into each answer choice and see which one yields an output of = 16 2/3

A. 100(20)/(125 - 20) = 2000/105.
IMPORTANT: If we want, we can use long division to evaluate this fraction (and others), but we can save a lot of time by applying some number sense. Since 2000/100 = 20, we know that 2000/105 will be SLIGHTLY less than 20. So, we can be certain that 2000/105 does not equal 16 2/3. As such, we can ELIMINATE A.

B. 150(20)/(250 - 20) = 3000/230. We know that 3000/200 = 15, so 3000/230 will be less than 15. So, we can be certain that 3000/230 does not equal 16 2/3. As such, we can ELIMINATE B.

C. 300(20)/(375 - 20) = 6000/355. Hmmm, this one is a little harder to evaluate. So,we may need to resort to some long division (yuck!). Using long division, we get 6000/355 = 16.9.... ELIMINATE C.

D. 400(20)/(500 - 20) = 8000/480 = 800/48 = 100/6 = 50/3 = 16 2/3. perfect! KEEP

E. 500(20)/(625 - 20) = 10000/605 = 100/6.05. Notice that, above, we saw that 100/6 = 16 2/3. So, 100/6.05 will NOT equal 16 2/3. ELIMINATE E.

Answer:

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

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New post 26 Jul 2016, 06:55
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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)


If you're not sure how to proceed with this question, or if you're behind on time and you want to catch up, you can give yourself a 50-50 chance in about 15 seconds.

To do so, we'll see what happens when we use an EXTREME value for p.
Say p = 100
In other words, 100% of the newspapers sold were Newspaper A.
This means that 100% of the revenue is from Newspaper A.
In other words, when p = 100, then r = 100

At this point, we'll plug in 100 for p and see which one yields a value of 100.
Only answer choices B and D work.
B) 150(100)/(250-100) = 100 PERFECT
D) 400(100)/(500-100) = 100 PERFECT

Now take a guess (B or D) and move on.

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

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New post 26 Jul 2016, 07:18
for percent problem, pick a specific number which is 100

suppose there are 100 newspaper. p percent mean p/100 so the number of newspaper A is

100*p/100= p

the number of newspaper B will be 100-p

doing this way will make our calculation simpler much.
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Re: Last Sunday a certain store sold copies of Newspaper A for &nbs [#permalink] 26 Jul 2016, 07:18

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