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If p does not equal q, and pq does not equal 0, then when p is replace

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If p does not equal q, and pq does not equal 0, then when p is replace [#permalink]

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\(\frac{p+q}{p-q}\)

If \(p\neq{q}\) and \(pq\neq{0}\), then when p is replaced by \(\frac{-1}{p}\) and q is replaced by \(\frac{-1}{q}\) everywhere in the expression above, the resulting expression is equivalent to:

A) \(\frac{p+q}{q-p}\)

B) \(\frac{p-q}{p+q}\)

C) \(\frac{p+q}{p-q}\)

D) \(\frac{p^2-q^2}{pq}\)

E) \(\frac{q-p}{p+q}\)
[Reveal] Spoiler: OA

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Last edited by PareshGmat on 17 Nov 2014, 18:30, edited 1 time in total.
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Re: If p does not equal q, and pq does not equal 0, then when p is replace [#permalink]

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New post 17 Nov 2014, 10:41
We can plug in for this problem to make it simple. Let's say that P=5 and Q=10.

p needs to be replaced by -1/5 and q needs to be replaced with -1/10

(-1/5) + (-1/10) in the numerator, (-2/10) + (-1/10) = -3/10
(-1/5) - (-1/10) in the denominator, (-2/10) - (-1/10) = -1/10

(-3/10) / (-1/10) = 3 so we need an answer that equals 3 when we plug in p=5 and q=10.

A. plug in to get 15/5 = 3 bingo!
B. plug in to get -5/15 = -1/3
C. plug in to get 15/-5 = -1/3
D. plug in to get -75/50 = -3/2
E. plug in to get 1/3


Answer A!
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Re: If p does not equal q, and pq does not equal 0, then when p is replace [#permalink]

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New post 17 Nov 2014, 10:56
PareshGmat wrote:
\(\frac{p+q}{p-q}\)

If \(p\neq{q}\) and \(pq\neq{0}\), then when p is replaced by \(\frac{-1}{p}\) and q is replaced by \(\frac{-1}{q}\) everywhere in the expression above, the resulting expression is equivalent to:

A) \(\frac{p+q}{q-p}\)

B) \(\frac{p-q}{p+q}\)

C) \(\frac{p+q}{p-q}\)

D) \(\frac{p^2-q^2}{pq}\)

E) \(\frac{q-p}{p+q}\)


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Re: If p does not equal q, and pq does not equal 0, then when p is replace [#permalink]

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\(\frac{p+q}{p-q}\)

Replace p with \(\frac{-1}{p}\) & q with \(\frac{-1}{q}\)

\(\frac{\frac{-1}{p} - \frac{1}{q}}{\frac{1}{q} - \frac{1}{p}}\)

\(\frac{\frac{1}{p} + \frac{1}{q}}{\frac{1}{p} - \frac{1}{q}}\)

\(\frac{\frac{p+q}{pq}}{\frac{q-p}{pq}}\)

\(\frac{p+q}{q-p}\)

Answer = A
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Re: If p does not equal q, and pq does not equal 0, then when p is replace [#permalink]

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New post 28 May 2017, 23:11
Easy and direct question. Just substitute and calculate. I hope I encounter such questions in final test :)
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Re: If p does not equal q, and pq does not equal 0, then when p is replace [#permalink]

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New post 27 Sep 2017, 03:02
PareshGmat wrote:
\(\frac{p+q}{p-q}\)

Replace p with \(\frac{-1}{p}\) & q with \(\frac{-1}{q}\)

\(\frac{\frac{-1}{p} - \frac{1}{q}}{\frac{1}{q} - \frac{1}{p}}\)

\(\frac{\frac{1}{p} + \frac{1}{q}}{\frac{1}{p} - \frac{1}{q}}\)

\(\frac{\frac{p+q}{pq}}{\frac{q-p}{pq}}\)

\(\frac{p+q}{q-p}\)

Answer = A


Would you be able to explain how you went from: \(\frac{\frac{-1}{p} - \frac{1}{q}}{\frac{1}{q} - \frac{1}{p}}\)
to:\(\frac{\frac{1}{p} + \frac{1}{q}}{\frac{1}{p} - \frac{1}{q}}\)

I was just having difficulty following. Thank you for your help!
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Re: If p does not equal q, and pq does not equal 0, then when p is replace [#permalink]

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New post 27 Sep 2017, 03:21
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imranmah wrote:
PareshGmat wrote:
\(\frac{p+q}{p-q}\)

Replace p with \(\frac{-1}{p}\) & q with \(\frac{-1}{q}\)

\(\frac{\frac{-1}{p} - \frac{1}{q}}{\frac{1}{q} - \frac{1}{p}}\)

\(\frac{\frac{1}{p} + \frac{1}{q}}{\frac{1}{p} - \frac{1}{q}}\)

\(\frac{\frac{p+q}{pq}}{\frac{q-p}{pq}}\)

\(\frac{p+q}{q-p}\)

Answer = A


Would you be able to explain how you went from: \(\frac{\frac{-1}{p} - \frac{1}{q}}{\frac{1}{q} - \frac{1}{p}}\)
to:\(\frac{\frac{1}{p} + \frac{1}{q}}{\frac{1}{p} - \frac{1}{q}}\)

I was just having difficulty following. Thank you for your help!


Take -1 common from both numerator and denominator.

\(\frac{\frac{-1}{p} - \frac{1}{q}}{\frac{1}{q} - \frac{1}{p}}\)

\(\frac{(-1)*(\frac{1}{p} + \frac{1}{q})}{(-1)*(-\frac{1}{q} + \frac{1}{p})}\)

Cancelling off the (-1) from the numerator with that from the denominator, you get

\(\frac{\frac{1}{p} + \frac{1}{q}}{-\frac{1}{q} + \frac{1}{p}}\)

\(\frac{\frac{1}{p} + \frac{1}{q}}{\frac{1}{p}-\frac{1}{q}}\)
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Re: If p does not equal q, and pq does not equal 0, then when p is replace   [#permalink] 27 Sep 2017, 03:21
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