Prep 13
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Source: Beat The GMAT — Data Sufficiency |
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jackcrystal
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Tryingmybest
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X-Y > -2 => x>-2 +Y
X-2y<-6 => x<-6+2Y => -x> 6-2Y
Add them 0 >4-Y => y>4
x>-2 +Y = > X must be positive since Y >4
so C
X-2y<-6 => x<-6+2Y => -x> 6-2Y
Add them 0 >4-Y => y>4
x>-2 +Y = > X must be positive since Y >4
so C
Last edited by Tryingmybest on Thu Nov 20, 2008 8:33 am, edited 1 time in total.
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logitech
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X> 4 , how did you find this ?Tryingmybest wrote:X-Y > -2 => x>-2 +Y
X-2y<-6 => x<-6+2Y => -x> 6-2Y
Add them 0 >4-Y => y>4
This means X> 4 so C
LGTCH
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"DON'T LET ANYONE STEAL YOUR DREAM!"
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"DON'T LET ANYONE STEAL YOUR DREAM!"
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Tryingmybest
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C
explanation:
for xy>0 you need both x,y with same sign (i.e. both positive or both negative).
(1) x-y > -2 ==> x+2>y
this doesn't lead to x,y with same sign, so A and D are out.
(2) x-2y<-6 ==> x+6<2y
again, doesn't lead to x,y with same sign, so B is out.
need to eliminate C, so let's check both statements together.
start with (2):
x-2y<-6
x+6<2y
x+2+4<2y
use y<x+2 (from 1), and you have:
x+2+4<2y
y+4<x+2+4<2y
y+4<2y
4<2y
2<y
so you have:
y>2
and
x+2>y
x+2>y>2
x+2>2
x>0
so both x, y are positive, thus xy>0 .
explanation:
for xy>0 you need both x,y with same sign (i.e. both positive or both negative).
(1) x-y > -2 ==> x+2>y
this doesn't lead to x,y with same sign, so A and D are out.
(2) x-2y<-6 ==> x+6<2y
again, doesn't lead to x,y with same sign, so B is out.
need to eliminate C, so let's check both statements together.
start with (2):
x-2y<-6
x+6<2y
x+2+4<2y
use y<x+2 (from 1), and you have:
x+2+4<2y
y+4<x+2+4<2y
y+4<2y
4<2y
2<y
so you have:
y>2
and
x+2>y
x+2>y>2
x+2>2
x>0
so both x, y are positive, thus xy>0 .
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logitech
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Dilbert, check your calculations. Be careful with inserting one ineq into another. According your solutiondilbert wrote:C
so you have:
y>2
x>0
so both x, y are positive, thus xy>0 .
y>2 and x>0 BUT y=3 and x=1 WILL NOT satisfy both of equations.
LGTCH
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"DON'T LET ANYONE STEAL YOUR DREAM!"
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"
logitech -
thanks for pointing this out!
indeed, I had a mistake in my calculations. here is the corrected one:
use y<x+2 (from 1), and you have:
x+2+4<2y
y+4<x+2+4<2y
y+4<2y
4<y <== this is corrected line, earlier I had mistake here
so you have:
y>4
and
x+2>y
x+2>y>4
x+2>4
x>2
so the correct solution for the two ineq is
x>2
y>4
and answer is C .
we see from the graph logitech attached that the intersection is (2,4), so x>2;y>4 is the area that solves both ineq.
again - thank you logitech for paying attention to my fault.
this time I was lucky to get C with a mistake, but I won't be that lucky in the real test...
thanks for pointing this out!
indeed, I had a mistake in my calculations. here is the corrected one:
use y<x+2 (from 1), and you have:
x+2+4<2y
y+4<x+2+4<2y
y+4<2y
4<y <== this is corrected line, earlier I had mistake here
so you have:
y>4
and
x+2>y
x+2>y>4
x+2>4
x>2
so the correct solution for the two ineq is
x>2
y>4
and answer is C .
we see from the graph logitech attached that the intersection is (2,4), so x>2;y>4 is the area that solves both ineq.
again - thank you logitech for paying attention to my fault.
this time I was lucky to get C with a mistake, but I won't be that lucky in the real test...
