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24 Mar 2022, 01:30
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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:
55%
(hard)
Question Stats:
63% (02:15) correct
37%
(02:34)
wrong
based on 215
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If \(x + y > 0\), is \(xy < 0\) ?
(1) \(x^{2y} < 1\)
(2) \(x + 2y < 0\)
Are You Up For the Challenge: 700 Level Questions
(1) \(x^{2y} < 1\)
(2) \(x + 2y < 0\)
Are You Up For the Challenge: 700 Level Questions
25 Mar 2022, 00:47
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If \(x + y > 0\), is \(xy < 0\) ?
We are looking for an answer to - Do x and y have opposite sign?
(1) \(x^{2y} < 1\)
We can have numerous combinations.
x=2 and y=-1.....Answer is YES.
x=1/2 and y=3/2......Answer is NO.
(2) \(x + 2y < 0\)
Since x+y>0, we can write x+y>0>x+2y.
We can also add similar sides to get x+y+0>x+2y+0 or x+y>x+2y
\(x+y>x+2y\) or \(y<0.\)
But x+y>0, so surely x>0.
Thus x*y < 0.
Sufficient
B
We are looking for an answer to - Do x and y have opposite sign?
(1) \(x^{2y} < 1\)
We can have numerous combinations.
x=2 and y=-1.....Answer is YES.
x=1/2 and y=3/2......Answer is NO.
(2) \(x + 2y < 0\)
Since x+y>0, we can write x+y>0>x+2y.
We can also add similar sides to get x+y+0>x+2y+0 or x+y>x+2y
\(x+y>x+2y\) or \(y<0.\)
But x+y>0, so surely x>0.
Thus x*y < 0.
Sufficient
B
vverma1995
30 Mar 2022, 07:15
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Question: is xy <0?
Key answer: Do x and y have opposite signs?
1. Statement 1: x^2y < 1
Since x is raised to an even power, the result will always be positive. To evaluate if x and y have different signs, we can consider two cases: y>0 and y<0
If y <0, the statement would be true for all values of x such that x > 1 or x < -1
If y >0, the statement would be true for all values of x between - 1 and 1
Hence, we cannot determine if the signs are opposite - INSUFFICIENT
Statement 2: x + 2y <0
This can be rearranged to read:
x < -2y
Now we have three possibilities
If y >0, x <0
If y <0, x >0
If y = 0, x <0 but from the question, we know that x + y >0, therefore this case cannot be considered true (This is important, I missed it in my first shot at the question!)
Hence, x and y will have opposite signs
Hence, xy<0
SUFFICIENT
Answer is B
Key answer: Do x and y have opposite signs?
1. Statement 1: x^2y < 1
Since x is raised to an even power, the result will always be positive. To evaluate if x and y have different signs, we can consider two cases: y>0 and y<0
If y <0, the statement would be true for all values of x such that x > 1 or x < -1
If y >0, the statement would be true for all values of x between - 1 and 1
Hence, we cannot determine if the signs are opposite - INSUFFICIENT
Statement 2: x + 2y <0
This can be rearranged to read:
x < -2y
Now we have three possibilities
If y >0, x <0
If y <0, x >0
If y = 0, x <0 but from the question, we know that x + y >0, therefore this case cannot be considered true (This is important, I missed it in my first shot at the question!)
Hence, x and y will have opposite signs
Hence, xy<0
SUFFICIENT
Answer is B
