x<y. Is x<0?
1) x^2>y^2 2) y>1
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123 is x<0
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Rahul@gurome
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(1) x^2>y^2. Let us take examples for this.
If x = -4 and y = 3, it implies x < y but x^2 = 16 and y^2 = 9, so x^2 > y^2 in this case.
If x = 4 and y = 3, it implies x > y but x^2 = 16 and y^2 = 9, so x^2 > y^2 in this case.
So, x may or may not be less than 0.
Hence, (1) is NOT SUFFICIENT.
(2) y > 1 implies x < 1 as x < y, but still we are not sure whether x < 0 as can take any other value between 0 and 1.
Hence, (2) is NOT SUFFICIENT.
Combining (1) and (2), we get that x < 0 always as x^2 > y^2.
The correct answer is (C).
If x = -4 and y = 3, it implies x < y but x^2 = 16 and y^2 = 9, so x^2 > y^2 in this case.
If x = 4 and y = 3, it implies x > y but x^2 = 16 and y^2 = 9, so x^2 > y^2 in this case.
So, x may or may not be less than 0.
Hence, (1) is NOT SUFFICIENT.
(2) y > 1 implies x < 1 as x < y, but still we are not sure whether x < 0 as can take any other value between 0 and 1.
Hence, (2) is NOT SUFFICIENT.
Combining (1) and (2), we get that x < 0 always as x^2 > y^2.
The correct answer is (C).
Rahul Lakhani
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Rahul@gurome wrote:(1) x^2>y^2. Let us take examples for this.
If x = -4 and y = 3, it implies x < y but x^2 = 16 and y^2 = 9, so x^2 > y^2 in this case.
If x = 4 and y = 3, it implies x > y but x^2 = 16 and y^2 = 9, so x^2 > y^2 in this case.
So, x may or may not be less than 0.
Hence, (1) is NOT SUFFICIENT.
(2) y > 1 implies x < 1 as x < y, but still we are not sure whether x < 0 as can take any other value between 0 and 1.
Hence, (2) is NOT SUFFICIENT.
Combining (1) and (2), we get that x < 0 always as x^2 > y^2.
The correct answer is (C).
: if x<y or x-y<0 (or is negative), is x<0 .... hmmm
1. x^2>y^2 means x^2-y^2 > 0 or (x-y)(x+y) > 0;
given that x-y<0, it can only mean for (x-y)(x+y) to be positive, (x+y) is negative or x+y<0!!
INSUFF
2. y>1 (i.e. positive)
INSUFF
combining 1&2: given that x-y<0, we ONLY need to worry about x+y<0 (from statement 1).
if y>1, we'll have to add to it a very big negative x to get a positive (x-y)(x+y) or (x-y)(x+y) > 0!
ANS (C)
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analyst218
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well in the question it gives you X<YRahul@gurome wrote:(1) x^2>y^2. Let us take examples for this.
If x = -4 and y = 3, it implies x < y but x^2 = 16 and y^2 = 9, so x^2 > y^2 in this case.
If x = 4 and y = 3, it implies x > y but x^2 = 16 and y^2 = 9, so x^2 > y^2 in this case.
So, x may or may not be less than 0.
Hence, (1) is NOT SUFFICIENT.
(2) y > 1 implies x < 1 as x < y, but still we are not sure whether x < 0 as can take any other value between 0 and 1.
Hence, (2) is NOT SUFFICIENT.
Combining (1) and (2), we get that x < 0 always as x^2 > y^2.
The correct answer is (C).
i say A..
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Rahul@gurome
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Agree. Since x < y so the example x = 4 and y = 3 is not possible. Hence, x < 0 always.
The correct answer should be (A).
The correct answer should be (A).
Rahul Lakhani
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)
