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11 Jun 2018, 02:55
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D
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Difficulty:
65%
(hard)
Question Stats:
51% (02:02) correct
49%
(01:24)
wrong
based on 61
sessions
History
Date
Time
Result
Not Attempted Yet
If \(x > y > 0\), is \(y < 2?\)
1) \(1/x = 1/2\)
2) \((1/x)+(1/y) =1\)
1) \(1/x = 1/2\)
2) \((1/x)+(1/y) =1\)
11 Jun 2018, 02:55
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Official Solution:
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.
Since we have 2 variables (\(x\) and \(y\)) and 1 equation \(( x > y\)), D is most likely to be the answer.
Condition 1)
\(1/x = 1/2\) implies that \(x = 2\). Since \(x > y\), we must have \(y < 2\).
Condition 1) is sufficient.
Condition 2)
The original condition \(x > y > 0\) implies that \(1/x < 1/y\).
Using \(1/x + 1/y = 1\) and \(1/x < 1/y\) together, we can see that \(1/y > 1/2\).
Thus, \(0 < y < 2\).
Condition 2) is sufficient.
Therefore, the answer is D.
Answer: D
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.
Since we have 2 variables (\(x\) and \(y\)) and 1 equation \(( x > y\)), D is most likely to be the answer.
Condition 1)
\(1/x = 1/2\) implies that \(x = 2\). Since \(x > y\), we must have \(y < 2\).
Condition 1) is sufficient.
Condition 2)
The original condition \(x > y > 0\) implies that \(1/x < 1/y\).
Using \(1/x + 1/y = 1\) and \(1/x < 1/y\) together, we can see that \(1/y > 1/2\).
Thus, \(0 < y < 2\).
Condition 2) is sufficient.
Therefore, the answer is D.
Answer: D
18 Jan 2019, 02:30
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I think this is a high-quality question and I agree with explanation. good explanation
