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Old 2009 August 18th, 03:05 PM   #1 (permalink)
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Tricky DS, Inequality

Is X > Y?

1. X + Y > 0
2. X^2 > Y^2
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Old 2009 August 18th, 03:22 PM   #2 (permalink)
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IMO A....questions asks whether X-Y > 0

1 is sufficient
2. is insufficient since x > y and x > -y when squared give the same result
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Old 2009 August 18th, 04:16 PM   #3 (permalink)
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Quote:
Originally Posted by tytyros View Post
IMO A....questions asks whether X-Y > 0

1 is sufficient
2. is insufficient since x > y and x > -y when squared give the same result
what about x=-3 y=7
-3+7=4>0 but x is not larger than y
or x=7 y=3 7+3=10 x>y
my vote for C
(2) (x-y)(x+y)>0
x-y>0 and
x>-y
st 1 directly says that x>-y so x>y
C
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Old 2009 August 18th, 04:27 PM   #4 (permalink)
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Quote:
Originally Posted by clock60 View Post
what about x=-3 y=7
-3+7=4>0 but x is not larger than y
or x=7 y=3 7+3=10 x>y
my vote for C
(2) (x-y)(x+y)>0
x-y>0 and
x>-y
st 1 directly says that x>-y so x>y
C
If x>-y, how can you conclude that also x>y, clock60? In my opinion both points are inconclusive an the answer is E.
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Old 2009 August 18th, 04:28 PM   #5 (permalink)
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1. X + Y > 0
X > -Y, INSUFFICIENT
2. X2 > Y2
X2 - Y2 > 0
(X+Y)(X-Y) > 0
Either X > -Y and X > Y
or X < -Y and X < Y
INSUFFICIENT
Combining the two,
(X+Y)(X-Y) > 0 from statement (1)
(X-Y) > 0 (since (X+Y) > 0 from statement (1) )
X > Y SUFFICIENT
Hence C.
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Old 2009 August 18th, 04:40 PM   #6 (permalink)
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Originally Posted by llutsar View Post
If x>-y, how can you conclude that also x>y, clock60? In my opinion both points are inconclusive an the answer is E.
for (x-y)(x+y)>0 two conditions must be maintain
first is x-y>0
and second (x+y)>0
the product of two positive numbers is always positive from the first we are given that x>-y or x+y>0 hence x-y>0 so x>y
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Old 2009 August 18th, 04:54 PM   #7 (permalink)
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1. is not sufficient. as if x + y > 0 is does not give any hint that X > Y!!!!!
2. is needed and it says that x-y> 0 based on above point.
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Old 2009 August 18th, 05:00 PM   #8 (permalink)
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answer C, as both need to make the decision!
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Old 2009 August 18th, 06:08 PM   #9 (permalink)
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Quote:
Originally Posted by 12rk34 View Post
1. X + Y > 0
X > -Y, INSUFFICIENT
2. X2 > Y2
X2 - Y2 > 0
(X+Y)(X-Y) > 0
Either X > -Y and X > Y
or X < -Y and X < Y
INSUFFICIENT
Combining the two,
(X+Y)(X-Y) > 0 from statement (1)
(X-Y) > 0 (since (X+Y) > 0 from statement (1) )
X > Y SUFFICIENT
Hence C.
great explanation thanks
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Old 2009 August 24th, 02:20 AM   #10 (permalink)
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answer is C.
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