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Re: Last Sunday a certain store sold copies of Newspaper A for
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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?

Re: Last Sunday a certain store sold copies of Newspaper A for
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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
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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

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29 Apr 2016, 06:32

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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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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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26 Jul 2016, 06:54

Top Contributor

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.

Re: Last Sunday a certain store sold copies of Newspaper A for
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16 Jan 2017, 00:21

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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).

Re: Last Sunday a certain store sold copies of Newspaper A for
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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 %)

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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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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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 [ 16.7 KiB | Viewed 2096 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:

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

Re: Last Sunday a certain store sold copies of Newspaper A for
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15 Dec 2017, 00:41

1

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?

Re: Last Sunday a certain store sold copies of Newspaper A for
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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 ?

Re: Last Sunday a certain store sold copies of Newspaper A for
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06 Jan 2018, 03:53

3

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 \(r=\frac{a}{a + 1.25*\frac{a(100-p)}{p}}*100\).

Re: Last Sunday a certain store sold copies of Newspaper A for
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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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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?

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14 Apr 2018, 11:40

1

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?

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

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

So is flexibility.
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