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805+ Level|   Word Problems|                                 
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HarshavardhanR
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Last Sunday a certain store sold copies of Newspaper A for $1.00 each 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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Last Sunday a certain store sold copies of Newspaper A for $1.00 each 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?



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

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\(p + 125 - 1.25p =\)

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

\(= 125 -0.25p\)
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Last Sunday a certain store sold copies of Newspaper A for $1.00 each 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?



\(p + 125 - 1.25p =\)

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

\(= 125 -0.25p\)
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Last Sunday a certain store sold copies of Newspaper A for $1.00 each 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.
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No idea what you mean...
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Last Sunday a certain store sold copies of Newspaper A for $1.00 each 21 Oct 2025, 07:42
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