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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 01 Jul 2014, 00:37
1
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
1
1
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 10 Jan 2015, 10:39
2
1
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 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,
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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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Last Sunday a certain store sold copies of Newspaper A for  [#permalink]

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New post 29 Apr 2016, 06:32
1
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 12 Jan 2017, 14:32
the question mentions news papers of A and (copies) of newspapers of A , where do we count B? i got really confused with this question please help
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Re: Last Sunday a certain store sold copies of Newspaper A for  [#permalink]

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New post 16 Jan 2017, 00:21
1
stampap wrote:
the question mentions news papers of A and (copies) of newspapers of A , where do we count B? i got really confused with this question please help


Note what the question says:

"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?"

r% of the revenue is from A
p% of the papers sold are A

There is no conflict here. It is possible that say 50% (p%) of the papers sold were A and that accounted for 80% (r%) of the revenue (because A is more expensive).

Now check out the solutions here:
last-sunday-a-certain-store-sold-copies-of-newspaper-a-for-101739.html
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Re: Last Sunday a certain store sold copies of Newspaper A for  [#permalink]

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New post 16 May 2017, 13:26
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




This problem was definitely something to work on. I was able to understand it with Pikachu’s help, although, the calculations were not very clear (substitution of P in % and in Number).

So, I am going to try and put it in my own vision of explanation, using number pickings as suggested, since algebraic calculations is really long and seems very confusing to me. Please correct my mistakes.

So what we have here is:
Newspapers A sold – $1/piece
Newspaper B sold – $1.25/piece

r - % of revenue generated from A newspapers sales
p - % of A newspapers sold.

Let’s use a and b as numbers of A & B np sold.
a + b = a total number of newspapers sold.

% of something = (part/whole)*100%. => p = (a/(a+b))*100%

now r % = (Revenue of A/Revenue A + Revenue B)*100% =>
R(total) = RevenueA +RevenueB. (revenue = price * number of pieces sold)
Having price np A sold and 1$/piece and np B sold 1.25$/piece the formula is as follows:

R(total) =1$*a + 1.25$b or = a + 1.25b
r = (a/(a+1.25b))*100%

Now let’s use numbers:
20 np A sold
80 np B sold.

Total np sold = 100.
p (% of np A sold) = 20%
Total revenue R = (1$*20 + 1.25$*80) = 20 + 100 = 120$
r (% of revenue A sold) = (20$/120$)*100% = 16.7%

Now lest try and plug in the numbers. We are looking for r = 16.7%, the final result is to be multiplied to 100%. Also in numerator we will be p in % as 0.2, and denominator – we are using p – as a number – 20 (number of A newspapers sold in pieces and not in %)

(a) (100(pieces)*0.2)/(125(pieces) – 20(pieces)) = (20/105)*100 = 19% - incorrect
(b) (150*0.2)/(250 – 20) = 30/230 = 3/23 = 13% => incorrect
(c) (300*0.2)/(375 – 20) = 12/71 = 16,9 % (we are looking for 16,7%) => incorrect
(d) (400*0.2)/(500-20) = 1/6 = 16.7% - correct.
(e) (500*0.2)/(635 – 20) = 100/615 = 50/123 = 40%. => incorrect

Now let’s check it and pick other numbers:
40 np A sold
60 np B sold

Total np sold are 100. R total = 115$
p = 40%, 40 in $
r = 34.7% or ~ 35%.

(a) 100*0.4/(125 – 40) = 8/17= 47% => incorrect
(b) 150*0.4/(250 – 40) = 6/21 = 29% => incorrect
(c) 300*0.4/(375 – 40) = 24/17 ~ 36% (again close but incorrect)
(d) 400*0.4/(500 – 40) = 8/23 = 34.7% correct
(e) 500*0.4/(625 – 40) = 40/117 = 34%

Seems like D is a correct answer for both. Please correct me if I am making any errors in understanding the concept of how to approach this problem, since, the way I see it – it is a key to solving it in 2 minutes.
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Re: Last Sunday a certain store sold copies of Newspaper A for  [#permalink]

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New post 09 Sep 2017, 18:50
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)


Answer: Option D

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

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New post 21 Sep 2017, 21:26
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)


Hi,

Thought process: There are two ways to calculate the revenue of the newspaper A: 1. Direct revenue calculation, and 2. Using total revenue.

Let the number of the newspaper sold = 100
Attachment:
OG_Newspaper.jpg
OG_Newspaper.jpg [ 16.7 KiB | Viewed 1516 times ]


Revenue from A = p --- (1)

If r percent of the store’s revenues from newspaper sales was from Newspaper A => \(\frac{(125 - 0.25p)*r}{100}\) --- (2)

From (1) and (2) we have following:

\(\frac{(125 - 0.25p)*r}{100} = p \Rightarrow r = \frac{100p}{125 - 0.25p} = \frac{400p}{500 - p}\)

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

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New post 15 Dec 2017, 00:41
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. \(\frac{100p}{(125 – p)}\)

B. \(\frac{150p}{(250 – p)}\)

C. \(\frac{300p}{(375 – p)}\)

D. \(\frac{400p}{(500 – p)}\)

E. \(\frac{500p}{(625 – p)}\)


Check out our video solution to this problem here: https://www.veritasprep.com/gmat-soluti ... olving_210
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Re: Last Sunday a certain store sold copies of Newspaper A for  [#permalink]

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New post 06 Jan 2018, 03:34
Bunuel 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 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.


Hello Bunuel, this problem seems so confusing, even when person knows algebra very well, i think during actual exam it will be time consuming to use algebraic approach :? As for number plugging any tips for using this strategy? not all numbers will give correct answer, /quick solution when applying number plugging in such kind of questions, no ? :?

By the way i am trying to understand your algebraic approach how did you derive from this \(p=\frac{a}{a+b}*100\) this equation and how do you call this process/step ---> \(b=\frac{a}{p}*100-a=\frac{a(100-p)}{p}\) and from that how how got this one \(r=\frac{a}{a + 1.25*\frac{a(100-p)}{p}}*100\) and how call this process / step ?

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

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New post 06 Jan 2018, 03:53
dave13 wrote:
Bunuel 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 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.


Hello Bunuel, this problem seems so confusing, even when person knows algebra very well, i think during actual exam it will be time consuming to use algebraic approach :? As for number plugging any tips for using this strategy? not all numbers will give correct answer, /quick solution when applying number plugging in such kind of questions, no ? :?

By the way i am trying to understand your algebraic approach how did you derive from this \(p=\frac{a}{a+b}*100\) this equation and how do you call this process/step ---> \(b=\frac{a}{p}*100-a=\frac{a(100-p)}{p}\) and from that how how got this one \(r=\frac{a}{a + 1.25*\frac{a(100-p)}{p}}*100\) and how call this process / step ?

thanks!


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

Cross-multiply: \(pa+pb=100a\);

Re-arrange: \(pb=100a-pa\);

Divide by p: \(b=\frac{100a-pa}{p}=\frac{a(100-p)}{p}\)

The next step is substituting the value of b here: \(r=\frac{a}{a + 1.25b}*100\) to get \(r=\frac{a}{a + 1.25*\frac{a(100-p)}{p}}*100\).

Number plugging:


How to Do Math on the GMAT Without Actually Doing Math
The Power of Estimation for GMAT Quant
How to Plug in Numbers on GMAT Math Questions
Number Sense for the GMAT
Can You Use a Calculator on the GMAT?
Why Approximate?
GMAT Math Strategies — Estimation, Rounding and other Shortcuts
The 4 Math Strategies Everyone Must Master, Part 1 (1. Test Cases and 2. Choose Smart Numbers.)
The 4 Math Strategies Everyone Must Master, part 2 (3. Work Backwards and 4. Estimate)
Intelligent Guessing on GMAT
How to Avoid Tedious Calculations on the Quantitative Section of the GMAT
GMAT Tip of the Week: No Calculator? No Problem.
The Importance of Sorting Answer Choices on the GMAT

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

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New post 25 Jan 2018, 00:00
1
THE BEST SOLUTION for this Question:

Let total newspapers sold sale was 100
News-paper A sold = p; News-paper B sold = 100 – p

Revenue from News-paper A sold: = p*1 = P
Revenue from News-paper B sold: = (100-p) *1.25
Total revenue from News-paper A & News-paper B sold: = P + (100-p) *1.25 = 125 -0.25p
It is given that r percent of the store’s revenues from newspaper sales was from
Newspaper A;
 P = r % of [125 -0.25p]
 P = (r/100) of [125 -0.25p]
 100P = r [125 -0.25p]
 r = 100P/ [125 -0.25p]
 r = 400P/ [500 – p]
The Correct 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 14 Apr 2018, 02:40
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. \(\frac{100p}{(125 – p)}\)

B. \(\frac{150p}{(250 – p)}\)

C. \(\frac{300p}{(375 – p)}\)

D. \(\frac{400p}{(500 – p)}\)

E. \(\frac{500p}{(625 – p)}\)



Hello generis , pushpitkc

I find this problem really hard :) i tried to tackle it but got into pitstop :)

here is what i did:

Let the price of newspapers \(A\) be \($ 1,00\)

Let the price of newspapers \(B\) be \($ 1,25\)

Let total number of newspaper \(A\) sold be\(A\)

Let total number of newspaper \(B\) sold be \(B\)

Let total number of newspapers sold be \(T\)

\(T = A+B\)

if Total Revenue is \(($1 *A ) + ($1.25 *B)\) --->

then \(R\) percent of the store’s revenues from newspaper sales was from Newspaper \(A\) --> \(\frac{R}{100}\) * \(($1 *A ) + ($1.25 *B)\)

Also, if total number of newspapers sold is \(A+B\), --->

then \(P\) percent of the newspapers that the store sold were copies of newspaper \(A\) ---> \(\frac{P}{100}\) *\((A+B)\)

ok, after the above mentioned steps I got stuck, what to do now :? I would appreciate explanation for dummies :)

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

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New post 14 Apr 2018, 11:40
dave13 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. \(\frac{100p}{(125 – p)}\)

B. \(\frac{150p}{(250 – p)}\)

C. \(\frac{300p}{(375 – p)}\)

D. \(\frac{400p}{(500 – p)}\)

E. \(\frac{500p}{(625 – p)}\)

Hello generis , pushpitkc

I find this problem really hard :) i tried to tackle it but got into pitstop :)

here is what i did:

Let the price of newspapers \(A\) be \($ 1,00\)

Let the price of newspapers \(B\) be \($ 1,25\)

Let total number of newspaper \(A\) sold be\(A\)

Let total number of newspaper \(B\) sold be \(B\)

Let total number of newspapers sold be \(T\)

\(T = A+B\)

if Total Revenue is \(($1 *A ) + ($1.25 *B)\) --->

then \(R\) percent of the store’s revenues from newspaper sales was from Newspaper \(A\) --> \(\frac{R}{100}\) * \(($1 *A ) + ($1.25 *B)\)

Also, if total number of newspapers sold is \(A+B\), --->

then \(P\) percent of the newspapers that the store sold were copies of newspaper \(A\) ---> \(\frac{P}{100}\) *\((A+B)\)

ok, after the above mentioned steps I got stuck, what to do now :? I would appreciate explanation for dummies :)

have an awesome weekend :)

dave13 , an explanation for dummies?
That quip had better be
not-serious self-deprecation. If not: Boo.
You are on the right track
of what I think is lengthy algebra.

I like "pit stop" better. Made me laugh. Levity is good.

The algebra may or may not be straightforward.
Some posters say the algebra is easy or is the best way.

Other posters say that picking numbers is easier.
I do not see a monopoly on truth here. Do you? ;)

If you continue your algebraic route . . .

1) isolate P; 2) isolate B; 3) isolate R; 4) substitute B's value
and 5) solve

In the end, we must have
two variables only: P and R

1) Isolate P

Add A to LHS of this part
Quote:
then \(P\) percent of the newspapers sold were copies of
\(A\) --->
\(\frac{P}{100}\) *\((A+B)\)

Make that arrow an equal sign.

Include what you are defining.

We want P. NOT A. Rearrange until P is on LHS Alone

A as a percent of total NUMBER of copies sold
\(A = \frac{P}{100} * (A+B)\)

\(\frac{A}{(A+B)} =\frac{P}{100}\)

\(P =\frac{100A}{(A+B)}\)


2) Isolate B (B on LHS alone)

\(\frac{P}{100}=\frac{A}{A+B}\) - Cross multiply
\(PA + PB = 100A\) - Subtract PA
\(PB = 100A - PA\) - Factor out A
\(PB = A(100 - P)\) - Divide by P
\(B = \frac{A(100-P)}{P}\)

3) Isolate A's revenue, R.
You have
Quote:
if Total Revenue is \(($1 *A ) + ($1.25 *B)\) --->
then \(R\) percent of revenues
[from A]--> \(\frac{R}{100}\) * \(($1 *A ) + ($1.25 *B)\)

Same as above. R on LHS. Use an equals sign.

R = (# of copies * price per copy)
\(R_A = (A * 1) = A\)
\(R_B = (B * 1.25) = 1.25B\)

R as A's PERCENT of revenue \(= \frac{R_A}{R_A + R_B} * 100\)
\(R = \frac{A}{A + 1.25B}\)


4) Then follow Bunuel and rewrite A's percent of revenue, R
with value for B that has been substituted:

\(\frac{A}{A + 1.25B}\)
\(1.25 B\)
cannot stay
\(R_B = 1.25B\) => substitute B's value from #2, i.e.,
\(B = \frac{A(100-P)}{P}\)

You will get to:
Quote:
The next step is . . . \(r=\frac{a}{a + 1.25*\frac{a(100-p)}{p}}*100\)

5) Solve :dazed :dazed :dazed :dazed :dazed :dazed

Would you please consider substituting values?
I did. Please? :) Have you tried it?
Tenacity is excellent.

So is flexibility. :thumbup:
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Last Sunday a certain store sold copies of Newspaper A for  [#permalink]

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New post 13 Jun 2018, 14:26
Can you please check if my approach is correct:

Assumption: I have assumed P= 50% and the total copies sold as 200.


Therefore Revenue from A= 1*100 =100 and B=1.25*100=125.
Total revenue= 225

Hence r= (100/225)* 100 =400/9.

If we plug in P=50 in all the options only D gives the answer.
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Last Sunday a certain store sold copies of Newspaper A for   [#permalink] 13 Jun 2018, 14:26

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