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Absolute value
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08 Sep 2016, 05:58
08 Sep 2016, 05:58
00:00
Question Stats:
6% (00:07) correct
93% (01:54) wrong based on 15 sessions
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Suppose that |x|<|y+2|<|z|. Suppose further that y>0 and xz>0. Which of the following could be true?
Indicate all statements that apply.
a) 0<y<x<z
b) 0<x<y<z
c) x<z<0<y
d) 0<y+1.5<x<z
e) z<x<0<y
Indicate all statements that apply.
a) 0<y<x<z
b) 0<x<y<z
c) x<z<0<y
d) 0<y+1.5<x<z
e) z<x<0<y
Show: :: OA
A,D,E
Re: Absolute value
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14 Sep 2016, 12:07
14 Sep 2016, 12:07
1
Expert Reply
|x|<|y+2|<|z|. And y>0 and xz>0.
a) 0<y<x<z
b) 0<x<y<z
c) x<z<0<y
d) 0<y+1.5<x<z
e) z<x<0<y
Also |x|<|z| so C is not possible.
a) 0<y<x<z
b) 0<x<y<z
c) x<z<0<y
d) 0<y+1.5<x<z
e) z<x<0<y
for option D take y +1.5 < x; taking modulus both sides |y+1.5|< |x| < |y+2| (And combining with the first inequality).
y=1 x=2.7 and z=5(say). So there are points where D exist.
SO A, D, E is the solution.
a) 0<y<x<z
b) 0<x<y<z
c) x<z<0<y
d) 0<y+1.5<x<z
e) z<x<0<y
Also |x|<|z| so C is not possible.
a) 0<y<x<z
d) 0<y+1.5<x<z
e) z<x<0<y
for option D take y +1.5 < x; taking modulus both sides |y+1.5|< |x| < |y+2| (And combining with the first inequality).
y=1 x=2.7 and z=5(say). So there are points where D exist.
SO A, D, E is the solution.
PS: If you are confused about A and E take modulus both become |y|<|x|<|z| and |x|<|y+2|.
So |y|< |x| < |y+2|.
Say y is 5 as long as |x| is between 5 and 7 you are good.
So a solution can be y=5 x= -6 and z= -10
PPS: Is this really manhattan GRE Question?
So |y|< |x| < |y+2|.
Say y is 5 as long as |x| is between 5 and 7 you are good.
So a solution can be y=5 x= -6 and z= -10
PPS: Is this really manhattan GRE Question?
Re: Absolute value
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15 Sep 2016, 05:09
15 Sep 2016, 05:09
Thanks for the reply Sandy!
And yes it's a Manhattan GRE question.
I don't understand how answer B is not possible.
If I take x=2, y=4, z=7, then I satisfy all three inequalities given.
Also, this question was given in the "hard set" of questions in the book. So, I wanted to know if such level of questions come often in the GRE?
PS: Answer is A, B, D and E
And yes it's a Manhattan GRE question.
I don't understand how answer B is not possible.
If I take x=2, y=4, z=7, then I satisfy all three inequalities given.
Also, this question was given in the "hard set" of questions in the book. So, I wanted to know if such level of questions come often in the GRE?
PS: Answer is A, B, D and E
