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07 Oct 2023, 05:15
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C
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:
56% (01:49) correct
44%
(01:48)
wrong
based on 41
sessions
History
Date
Time
Result
Not Attempted Yet
07 Oct 2023, 09:45
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Question: Is \(x^2 > y^2\) ?
Statement 1: \(x < y\)
Case 1: \(x\) and \(y\) are positive.
If \(x\) and \(y\) are positive and \(x < y\), then \(x^2\) will always be lesser than \(y^2\)
Therefore, the answer is No.
Case 2: \(x = 0\) and \(y\) is positive.
If \(x = 0\) and \(y\) is positive and \(x < y\), then \(x^2\) will always be lesser than \(y^2\)
Therefore, the answer is No.
Case 3: \(x\) is negative and \(y\) is positive.
If \(x\) is negative and \(y\) is positive and \(x < y\), then \(x^2\) may or may not be less than \(y^2\)
e.g: \(x = -1\) and \(y = 10\), \(x^2 = 1\) and \(y^2 = 100\). Therefore, the answer is No.
However, if \(x = -1\) and \(y = 0.5\), \(x^2 = 1\) and \(y^2 = 0.25\). Therefore, the answer is Yes.
As you are getting inconsistent answers with statement 1, you can eliminate statement 1.
Statement 2: \(-y > x\)
This can be re written as \(x < -y\).
Case 1: \(x\) is positive.
If \(x\) is postive, then \(y\) will be less than \(-x\).
e.g: If \(x = 10\), then \(y\) will be will be less than -10. Hence, \(x^2 = 100\) and \(y^2\) will be greater than 100.
Therefore, the answer is No.
Case 2: \(x = 0\).
If \(x = 0\), \(y\) will be less than 0. Hence, \(x^2\) will always be lesser than \(y^2\).
Therefore, the answer is No.
Case 3: x is negative.
If \(x\) is negative, then \(y\) will be less than \(-x\). Hence \(x^2\) may or may not be greater than \(y^2\).
e.g: If \(x = -3\) and \(y = -1\), then \(x^2\) will be greater than \(y^2\). Therefore, the answer is Yes.
However, if \(x = -3\) and \(y = -8\), then \(x^2\) will not be greater than \(y^2\). Therefore, the answer is No.
As you are getting inconsistent answers with statement 2, you can eliminate statement 2.
Combining Statement 1 and Statement 2:
Statement 1: \(x < y\) and Statement 2: \(-y > x\)
Therefore, \(x\) should be less than \(-y\). Hence, \(x^2\) will be greater than \(y^2\).
Therefore, the correct option is Option C.
Statement 1: \(x < y\)
Case 1: \(x\) and \(y\) are positive.
If \(x\) and \(y\) are positive and \(x < y\), then \(x^2\) will always be lesser than \(y^2\)
Therefore, the answer is No.
Case 2: \(x = 0\) and \(y\) is positive.
If \(x = 0\) and \(y\) is positive and \(x < y\), then \(x^2\) will always be lesser than \(y^2\)
Therefore, the answer is No.
Case 3: \(x\) is negative and \(y\) is positive.
If \(x\) is negative and \(y\) is positive and \(x < y\), then \(x^2\) may or may not be less than \(y^2\)
e.g: \(x = -1\) and \(y = 10\), \(x^2 = 1\) and \(y^2 = 100\). Therefore, the answer is No.
However, if \(x = -1\) and \(y = 0.5\), \(x^2 = 1\) and \(y^2 = 0.25\). Therefore, the answer is Yes.
As you are getting inconsistent answers with statement 1, you can eliminate statement 1.
Statement 2: \(-y > x\)
This can be re written as \(x < -y\).
Case 1: \(x\) is positive.
If \(x\) is postive, then \(y\) will be less than \(-x\).
e.g: If \(x = 10\), then \(y\) will be will be less than -10. Hence, \(x^2 = 100\) and \(y^2\) will be greater than 100.
Therefore, the answer is No.
Case 2: \(x = 0\).
If \(x = 0\), \(y\) will be less than 0. Hence, \(x^2\) will always be lesser than \(y^2\).
Therefore, the answer is No.
Case 3: x is negative.
If \(x\) is negative, then \(y\) will be less than \(-x\). Hence \(x^2\) may or may not be greater than \(y^2\).
e.g: If \(x = -3\) and \(y = -1\), then \(x^2\) will be greater than \(y^2\). Therefore, the answer is Yes.
However, if \(x = -3\) and \(y = -8\), then \(x^2\) will not be greater than \(y^2\). Therefore, the answer is No.
As you are getting inconsistent answers with statement 2, you can eliminate statement 2.
Combining Statement 1 and Statement 2:
Statement 1: \(x < y\) and Statement 2: \(-y > x\)
Therefore, \(x\) should be less than \(-y\). Hence, \(x^2\) will be greater than \(y^2\).
Therefore, the correct option is Option C.
08 Oct 2023, 06:42
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Bunuel
\(x^2> y^2\)
\(x^2-y^2>0 \)
\((x-y)(x+y)>0\)
So either (x-y) and (x+y) are both positive or both negative
Statement 1: x<y
\(x-y<0 \)
but we know nothing about\( x+y\)
statement 1 is not sufficent
Statement 2: -y>x
\(x+y<0\)
but we know nothing about \(x-y\)
statement 2 is not sufficent
statement 1+ 2 :
\(x-y<0 and x+y<0\)
Sufficient
answer choice C
