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If 0 < q ≤ p, then for all possible values of p and q, p must be great

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Joined: 29 Dec 2018
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If 0 < q ≤ p, then for all possible values of p and q, p must be great  [#permalink]

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New post 04 Mar 2019, 05:28
2
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

27% (00:56) correct 73% (02:04) wrong based on 15 sessions

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If 0 < q ≤ p, then for all possible values of p and q, p must be greater than which of the following?

A. \(\frac{1}{p+q}\)

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

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

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

E. \(\frac{pq+q^2}{p}\)
Intern
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Joined: 21 Feb 2019
Posts: 36
Re: If 0 < q ≤ p, then for all possible values of p and q, p must be great  [#permalink]

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New post 04 Mar 2019, 06:43
To solve this quickly you should be able to recognize as soon as possible that:

\(\frac{p^2 - q^2}{p + q} = \frac{(p + q)(p - q)}{p + q} = p - q\)

\(p > p - q\) everytime. To be accurate,

\(- q < 0\)

\(q > 0\)

Verified.

Plug-in strategy is good to discard first options IMO.
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Re: If 0 < q ≤ p, then for all possible values of p and q, p must be great   [#permalink] 04 Mar 2019, 06:43
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