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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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Updated on: 17 Nov 2014, 19:30
2
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Difficulty:

45% (medium)

Question Stats:

63% (01:26) correct 37% (01:21) wrong based on 135 sessions

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

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Originally posted by PareshGmat on 30 Oct 2014, 22:10.
Last edited by PareshGmat on 17 Nov 2014, 19:30, 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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17 Nov 2014, 11:41
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

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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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17 Nov 2014, 11:56
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) $$\frac{q-p}{p+q}$$

Similar question to practice from MGMAT: if-x-does-not-equal-y-and-xy-does-not-equal-0-then-when-x-104108.html
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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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17 Nov 2014, 20:01
2
$$\frac{p+q}{p-q}$$

Replace p with $$\frac{-1}{p}$$ & q with $$\frac{-1}{q}$$

$$\frac{\frac{-1}{p} - \frac{1}{q}}{\frac{1}{q} - \frac{1}{p}}$$

$$\frac{\frac{1}{p} + \frac{1}{q}}{\frac{1}{p} - \frac{1}{q}}$$

$$\frac{\frac{p+q}{pq}}{\frac{q-p}{pq}}$$

$$\frac{p+q}{q-p}$$

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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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29 May 2017, 00:11
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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27 Sep 2017, 04:02
PareshGmat wrote:
$$\frac{p+q}{p-q}$$

Replace p with $$\frac{-1}{p}$$ & q with $$\frac{-1}{q}$$

$$\frac{\frac{-1}{p} - \frac{1}{q}}{\frac{1}{q} - \frac{1}{p}}$$

$$\frac{\frac{1}{p} + \frac{1}{q}}{\frac{1}{p} - \frac{1}{q}}$$

$$\frac{\frac{p+q}{pq}}{\frac{q-p}{pq}}$$

$$\frac{p+q}{q-p}$$

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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27 Sep 2017, 04:21
2
imranmah wrote:
PareshGmat wrote:
$$\frac{p+q}{p-q}$$

Replace p with $$\frac{-1}{p}$$ & q with $$\frac{-1}{q}$$

$$\frac{\frac{-1}{p} - \frac{1}{q}}{\frac{1}{q} - \frac{1}{p}}$$

$$\frac{\frac{1}{p} + \frac{1}{q}}{\frac{1}{p} - \frac{1}{q}}$$

$$\frac{\frac{p+q}{pq}}{\frac{q-p}{pq}}$$

$$\frac{p+q}{q-p}$$

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 &nbs [#permalink] 27 Sep 2017, 04:21
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