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11 Nov 2019, 04:08
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
69% (01:28) correct
31%
(01:45)
wrong
based on 227
sessions
History
Date
Time
Result
Not Attempted Yet
Is \(p > q\) ?
(1) \(p + r > q + s\) and \(r > s\)
(2) \(p + s < q + r\) and \(s > r\)
(1) \(p + r > q + s\) and \(r > s\)
(2) \(p + s < q + r\) and \(s > r\)
This Question is Locked Due to Poor Quality
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The question you've reached has been archived due to not meeting our community quality standards. No more replies are possible here.
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11 Nov 2019, 08:01
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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
(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
12 Nov 2019, 08:59
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Bunuel
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.
