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Is xy>0?

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Is xy>0? [#permalink]

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New post 09 Nov 2017, 19:57
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Is \(xy>0\)?

(1) \(|x|+y=|x+y|\)
(2) \(x+|y|=|x+y|\)



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[Reveal] Spoiler: OA

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Re: Is xy>0? [#permalink]

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New post 09 Nov 2017, 22:13
Is xy > 0?

St1: |x| + y = |x + y|
|1| + 0 = |1 + 0| --> xy = 0
|1| + 1 = |1 + 1| --> xy = 1
Not Sufficient

St2: x + |y| = |x + y|
1 + |0| = |1 + 0| --> xy = 0
1 + |1| = |1 + 1| --> xy = 1
Not Sufficient

Since both St1 and St2 result in xy = 0 (x = 1, y = 0) and xy = 1 (x = 1, y = 1), combining both the statements is not sufficient.

Answer: E

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Re: Is xy>0? [#permalink]

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New post 11 Nov 2017, 13:21
chetan2u wrote:
Is \(xy>0\)?

(1) \(|x|+y=|x+y|\)
(2) \(x+|y|=|x+y|\)



self made



Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Since we have 2 variables and 0 equation, C is most likely to be the answer.
Then we need to check both conditions together first by VA(Variable Approach) method.

Conditions 1) & 2)
x = y = 1: Yes
x = y = 0: No

Thus the answer is E.

Normally for cases where we need 2 more equations, such as original conditions with 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, we have 1 equation each in both 1) and 2). Therefore C has a high chance of being the answer, which is why we attempt to solve the question using 1) and 2) together. Here, there is 70% chance that C is the answer, while E has 25% chance. These two are the key questions. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer according to DS definition, we solve the question assuming C would be our answer hence using 1) and 2) together. (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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Is xy>0? [#permalink]

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New post 11 Nov 2017, 21:23
chetan2u wrote:
Is \(xy>0\)?

(1) \(|x|+y=|x+y|\)
(2) \(x+|y|=|x+y|\)



self made


Well number plugging method has already been explained so I'll go the algebraic way ;)

For \(xy>0\), both \(x\) & \(y\) have to be of the same sign and \(x\) & \(y\) should not both be equal to \(0\)

Statement 1: LHS of the equation is a mod function hence it will always be \(≥0\)

so we have \(|x|+y≥0 => |x|≥-y\). For this to be true \(y≤0\) and \(x≤0\) or \(x≥0\). Hence multiple scenarios possible. Insufficient

Statement 2: again the same logic applies \(x+|y|≥0\), so \(|y|≥-x\). For this to be true \(x≤0\) and \(y≤0\) or \(y≥0\). Hence multiple scenarios possible. Insufficient

Combining statement 1 & 2 we know that either \(x≤0\) or \(x≥0\) and \(y≤0\) or \(y≥0\). So both \(x\) & \(y\) can be positive / negative or both can be \(x=y=0\). Multiple values possible. Hence Insufficient

Option \(E\)

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Is xy>0?   [#permalink] 11 Nov 2017, 21:23
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