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Algebra

by eitijan » Thu Apr 28, 2016 8:40 am

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Last Sundaya certain store sold copies of Newspaper A for $1.00 each and copies of NewspaperBfor $1.25
each, and the store sold no other newspapers that day. If r percent of the store's revenue from newspaper sales was from Newspaper Aand 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

How to solve such questions in less time. Algebra is taking too long. For plugging, cant find intelligent numbers to make the calculations easy.
Experts, please help.
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by GMATGuruNY » Thu Apr 28, 2016 8:46 am
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 revenue 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
Plug in values that make the math easy.

Let p=50, implying that 50% of the newspapers sold are A, while the other half are B.
In other words, the store sells an EQUAL NUMBER of each type of newspaper.
Since the price of each newspaper B = 1.25 = 5/4, let the store sell 4 OF EACH TYPE of newspaper.

Revenue from 4 copies of newspaper B = 4(1.25) = 5.
Revenue from 4 copies of newspaper A = 4*1 = 4.
Since (revenue from A)/(total revenue) = 4/9, r = (4/9)* 100 = 400/9. This is our target.

Now plug p=50 into the answers to see which yields our target of 400/9.

Answer choice D looks like a good bet, since it includes 400:
400p/(500 - p) = (400*50)/(500-50) = (400*50)/450 = 400/9.

The correct answer is D.
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by DavidG@VeritasPrep » Thu Apr 28, 2016 8:50 am
eitijan wrote:Last Sundaya certain store sold copies of Newspaper A for $1.00 each and copies of NewspaperBfor $1.25
each, and the store sold no other newspapers that day. If r percent of the store's revenue from newspaper sales was from Newspaper Aand 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

How to solve such questions in less time. Algebra is taking too long. For plugging, cant find intelligent numbers to make the calculations easy.
Experts, please help.
Say you sold 100 papers of each type. Well, p = 50%.
Revenue for A: $1*100= $100; Revenue for B = $1.25* 100= 125. Total Revenue = 225.
r = 100/225 = 4/9 = about 44.44%

Now plug 50 in place of p until you hit 44.44.
D) 400*50/(500-50) = 20,000/450 = 400/9 = 44.44, so that's our answer
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by [email protected] » Thu Apr 28, 2016 9:06 am
Hi eitijan,

There was a full discussion of this question here:

https://www.beatthegmat.com/algebra-t282913.html

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by Brent@GMATPrepNow » Thu Apr 28, 2016 9:22 am
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 10 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.

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by Matt@VeritasPrep » Thu Apr 28, 2016 2:38 pm
Plugging in certainly helps, but let's learn a little algebra too (you never know when it'll come in handy!)

Suppose we sold x of Newspaper A and y of Newspaper B.

We made $1x (or x) from A and $1.25y (or 1.25y) from B.

Our first equation says that

x = (r/100) * (x + 1.25y)

Our second equation says that

x = (p/100) * (x + y)

Since we want r in terms of p, we want to get rid of x and y somehow. Let's start by noticing that each equation = x, so they must equal each other.

(r/100)*(x + 1.25y) = (p/100)*(x + y)

Multiply both sides by 100 to clear the denominators:

r*(x + 1.25y) = p*(x + y)

rx + 1.25ry = px + py

Now we need x in terms of y. Let's use our second equation:

x = (p/100)*(x + y)

x = px/100 + py/100

100x = px + py

100x - px = py

(100 - p)*x = py

x*(100 - p)/p = y

Now we'll sub x*(100 - p)/p in for y in the blue equation.

rx + 1.25ry = px + py =>

rx + 1.25rx*(100-p)/p = px + px*(100-p)/p =>

r + 1.25r*(100 - p)/p = p + 100 - p =>

rp + 1.25r*(100 - p) = 100p =>

rp + 125r - 1.25rp = 100p =>

125r - .25rp = 100p =>

r * (125 - .25p) = 100p =>

r * (500 - p) = 400p =>

r = 400p / (500 - p)
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by GMATGuruNY » Thu Apr 28, 2016 2:45 pm
Another option is combine plugging in with algebra:

Let the number of newspapers sold = 100.
Since p% are A, the number of copies of A sold = p/100 * 100 = p.
Thus, the number of copies of B sold = 100-p.

Revenue from A = p*1 = p.
Revenue from B = (100-p)(1.25) = 125 - 1.25p.
Total revenue = p + (125 - 1.25p) = 125 - 0.25p.

Since r% of the total revenue comes from A:
r = revenue from A/total revenue * 100 = p/(125 - 0.25p) * 100 = 100p / (125 - 0.25p).

Since the answer choices do not include any decimals, we must clear 0.25 by multiplying by 4/4:
(100p)(4) / (125 - .25p)(4) = 400p/(500-p).

The correct answer is D.
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