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)} OG 2019 PS03144

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Last Sunday a certain store sold copies of Newspaper A for $1.00 each

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HarshavardhanR

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30 Nov 2024, 23:09

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An algebraic way to approach this question (above).

Harsha

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BelisariusTirto

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30 Dec 2024, 10:09

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Apologies for bumping such an old thread, I was wondering where the 1 in 1+1.25 "goes?" as in the original solution it becomes:

Reduce by $a$ and simplify:

$r=\frac{100p}{p+125-1.25p}=\frac{100p}{125-0.25p}$;

I assumed P goes to the numerator by multiplying it out of the denominator (100-p)/p then it becomes 100p after being multiplied by 100. Forming 100p/1+125-1.25p?

Bunuel

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$r=\frac{a}{a + 1.25*\frac{a(100-p)}{p}}*100$ --> reduce by $a$ and simplify --> $r=\frac{100p}{p+125-1.25p}=\frac{100p}{125-0.25p}$

Hi Bunel, I don't get the reduction. How do you get rid of the $a+$ in the denominator?

I only get this:
$r=\frac{a}{a + 1.25*\frac{a(100-p)}{p}}*100$ --> reduces to --> $r=\frac{100 p}{a+1,25*(100-p)}$

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

Factor out a from the denominator: $r=\frac{a}{a(1 + 1.25*\frac{(100-p)}{p})}*100$.

Reduce it: $r=\frac{1}{1 + 1.25*\frac{(100-p)}{p}}*100$.

Hope it's clear.

Bunuel

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30 Dec 2024, 10:34

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BelisariusTirto

Apologies for bumping such an old thread, I was wondering where the 1 in 1+1.25 "goes?" as in the original solution it becomes:

Reduce by $a$ and simplify:

$r=\frac{100p}{p+125-1.25p}=\frac{100p}{125-0.25p}$;

I assumed P goes to the numerator by multiplying it out of the denominator (100-p)/p then it becomes 100p after being multiplied by 100. Forming 100p/1+125-1.25p?

$p + 125 - 1.25p =$

$= 125 + (p -1.25p)=$

$= 125 -0.25p$

BelisariusTirto

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30 Dec 2024, 11:07

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Ah, maybe I should have phrased it better. I was wondering how the 1 becomes p.

Bunuel

BelisariusTirto

Apologies for bumping such an old thread, I was wondering where the 1 in 1+1.25 "goes?" as in the original solution it becomes:

Reduce by $a$ and simplify:

$r=\frac{100p}{p+125-1.25p}=\frac{100p}{125-0.25p}$;

I assumed P goes to the numerator by multiplying it out of the denominator (100-p)/p then it becomes 100p after being multiplied by 100. Forming 100p/1+125-1.25p?

$p + 125 - 1.25p =$

$= 125 + (p -1.25p)=$

$= 125 -0.25p$

Bunuel

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30 Dec 2024, 11:10

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BelisariusTirto

Ah, maybe I should have phrased it better. I was wondering how the 1 becomes p.

No idea what you mean...

DanTheGMATMan

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21 Oct 2025, 07:42

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nihilveritatis

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25 Jan 2026, 06:19

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Solved using algebra without substituting -

eq 1 - (A+B)*p/100 = A .....(given)
=> (A+B)*p = 100A
=> PA +PB = 100A
PB = A(100 - P)
B/A = (100-P)/ P ....(1)
eq 2 - (A+ 5b/4)*r/100 = A
Manipulate to get
=> B/A = 4(100 - r)/5r ...(2)
Equate the two equations to isolate r in terms of p