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Is p > q ? (1) p + r > q + s and r > s (2) p + s < q + r and s > r

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Is p > q ? (1) p + r > q + s and r > s (2) p + s < q + r and s > r  [#permalink]

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New post 11 Nov 2019, 04:08
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71% (01:22) correct 29% (00:53) wrong based on 21 sessions

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Re: Is p > q ? (1) p + r > q + s and r > s (2) p + s < q + r and s > r  [#permalink]

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New post 11 Nov 2019, 08:01
Is \(p > q\) ?

(1) \(p + r > q + s\) and \(r > s\)
say r=-2 and s=-3.......\(p -2> q -3......p>q-1\) ..here p>q when p=4 and q=2 OR p<q when p=4 and q=4.5

(2) \(p + s < q + r\) and \(s > r\)
Add \(p + s < q + r\) and \(s > r\)....\(p + s+r < q + r+s......p<q\)
Suff as our answer is NO

B
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Re: Is p > q ? (1) p + r > q + s and r > s (2) p + s < q + r and s > r  [#permalink]

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New post 12 Nov 2019, 08:59
Bunuel wrote:
Is \(p > q\) ?

(1) \(p + r > q + s\) and \(r > s\)
(2) \(p + s < q + r\) and \(s > r\)


Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.

Condition 2) is equivalent to \(p > q\) for the following reason.
Since \(r < s\), we have \(p + r < p + s\).
Then we have \(p + r < p + s < q + r\) or \(p < r\), since \(p + s < q + r\).
Thus condition 2) is sufficient.

Condition 1)
If \(p = 2, q = 1, r = 2\) and \(s = 1\), then \(p > q\) and the answer is 'yes'.
If \(p = 1, q = 2, r = 100\) and \(s = 1\), then \(p < q\) and the answer is 'no'.
Since condition 1) does not yield a unique solution, it is not sufficient

Therefore, B is the answer.
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Re: Is p > q ? (1) p + r > q + s and r > s (2) p + s < q + r and s > r   [#permalink] 12 Nov 2019, 08:59
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