|
Author |
Message |
|
TAGS:
|
|
|
Intern
Joined: 08 May 2006
Posts: 24
Followers: 0
Kudos [?]:
0
[0], given: 0
|
ZY<XY<0 is |X-Z| + |x| = |Z| 1.) Z<X 2.) Y<0 [#permalink]
 25 Apr 2007, 04:18
Question Stats:
9% (03:29) correct
90% (01:13) wrong based on 0 sessions
ZY<XY<0 is |X-Z| + |x| = |Z|
1.) Z<X
2.) Y<0
What is a good approach here?
Thanks!
|
|
|
|
|
|
|
SVP
Joined: 01 May 2006
Posts: 1837
Followers: 8
Kudos [?]:
71
[1] , given: 0
|
1
This post received
KUDOS
(D) for me 
ZY<XY<0
=>ZY - XY < 0
<=> Y*(Z-X) < 0
Implies that :
o Y < 0 and Z-X > 0 <=> Z > X
or
o Y > 0 and Z-X < 0 <=> Z < X
As, ZY<XY<0, the 2 cases above become:
o Y < 0 and Z > X > 0 (case A)
or
o Y > 0 and Z < X < 0 (case B)
|X-Z| + |X| = |Z| ?
From 1
Z < X => We are forced to be in the case B.
That implies:
o Z < 0
o X < 0
o X-Z > 0
So,
|X-Z| + |X|
= (X-Z) + (-X)
= -Z
= |Z| as Z < 0
SUFF.
From 2
Y < 0 => We are forced to be in the case A.
That implies:
o Z > 0
o X > 0
o Z-X > 0 <=> X-Z < 0
So,
|X-Z| + |X|
= (-(X-Z)) + (X)
= Z
= |Z| as Z > 0
SUFF.
Last edited by Fig on 25 Apr 2007, 07:22, edited 1 time in total.
|
|
|
|
|
|
Intern
Joined: 24 Apr 2007
Posts: 9
Followers: 0
Kudos [?]:
0
[0], given: 0
|
ZY<XY<0 is |X-Z| + |x| = |Z|
1.) Z<X
2.) Y<0
What is a good approach here?
Thanks!
The question is asking is |X-Z| + |X| = |Z|
|X-Z| + |X| = |Z|
|X-Z| = |Z| - |X|
This is true only when X and Z are the same sign, or essentially when they are both either greater than zero or less than zero. Since the assumption is that they are both less than zero when multiplied by Y, then if Y is positive, they are both negative, and if Y is negative, they are both positive.
1. Z < X
All this tells us is that Y < 0
2. Y < 0
This also tells us that Y < 0
IF Y < 0 then Z and X must be positive. Thus they are the same sign, thus the answer is D. However, I would say that if Y was positive, then |X-Z| + |X| = |Z| would still be true. Y can't be zero because then XY and ZY can't be less than zero.
My question is, will there be questions like this on the Real Gmat, where the data given is not required to solve the question?
|
|
|
|
|
|
Intern
Joined: 08 May 2006
Posts: 24
Followers: 0
Kudos [?]:
0
[0], given: 0
|
The question is from GMAT PREP so presumably it's representative of the real GMAT.
I hope not to see it on the real thing though - I hate this type of Q and have a limit on how many scenarios I'm willing to write out before moving onto the next question!
I'm guessing it might be a 50/51 threshold question, I didn't get it but got a 50 on the practice test.
|
|
|
|
|
|
Intern
Joined: 08 May 2006
Posts: 24
Followers: 0
Kudos [?]:
0
[0], given: 0
|
Forgot to say- Thanks a million guys for your help!
|
|
|
|
|
|
SVP
Joined: 01 May 2006
Posts: 1837
Followers: 8
Kudos [?]:
71
[0], given: 0
|
doc14 wrote: Forgot to say- Thanks a million guys for your help! You are welcome !
If u want to practice more, try these ones :
http://www.gmatclub.com/phpbb/viewtopic.php?t=39533
|
|
|
|
|
|
GMAT Instructor
Joined: 04 Jul 2006
Posts: 1278
Location: Madrid
Followers: 9
Kudos [?]:
69
[0], given: 0
|
jayt696969 wrote: ZY<XY<0 is |X-Z| + |x| = |Z|
1.) Z<X
2.) Y<0
What is a good approach here?
Thanks!
The question is asking is |X-Z| + |X| = |Z|
|X-Z| + |X| = |Z|
|X-Z| = |Z| - |X|
This is true only when X and Z are the same sign, or essentially when they are both either greater than zero or less than zero. Since the assumption is that they are both less than zero when multiplied by Y, then if Y is positive, they are both negative, and if Y is negative, they are both positive.
1. Z < X
All this tells us is that Y < 0
2. Y < 0
This also tells us that Y < 0
IF Y < 0 then Z and X must be positive. Thus they are the same sign, thus the answer is D. However, I would say that if Y was positive, then |X-Z| + |X| = |Z| would still be true. Y can't be zero because then XY and ZY can't be less than zero.
My question is, will there be questions like this on the Real Gmat, where the data given is not required to solve the question?
This is a necessary, but not sufficient condition for the statement in the question to be true. It must also be true that |z| -|x| is positive
|
|
|
|
|
|
Intern
Joined: 10 Jun 2009
Posts: 33
Location: Stockholm, Sweden
Followers: 0
Kudos [?]:
1
[0], given: 0
|
Re: GMATPREP Modulus Q [#permalink]
15 Jun 2009, 12:53
Im a little lost here. If Z<X and Y<0 then these would be possible numbers:
Y = -1
Z = 2
X = 10
Then ZY would be -2 and XY would be -10 making XY<ZY (the question states that ZY<XY). What is wrong with this conclusion?
|
|
|
|
|
|
Manager
Joined: 22 Sep 2009
Posts: 228
Location: Tokyo, Japan
Followers: 2
Kudos [?]:
14
[0], given: 8
|
Re: GMATPREP Modulus Q [#permalink]
25 Nov 2009, 08:30
Fig, I understand your explanation but shouldn't the statements themselves be always true? If so, the two scenarios Y>0 and Y<0 are contradictory? Am I missing something here?
|
|
|
|
|
|
Manager
Joined: 24 Aug 2009
Posts: 156
Followers: 3
Kudos [?]:
21
[0], given: 46
|
Fig wrote: (D) for me ZY<XY<0 =>ZY - XY < 0 <=> Y*(Z-X) < 0 Implies that : o Y < 0 and Z-X > 0 <=> Z > X or o Y > 0 and Z-X < 0 <=> Z < X As, ZY<XY<0, the 2 cases above become: o Y < 0 and Z > X > 0 (case A) or o Y > 0 and Z < X < 0 (case B) |X-Z| + |X| = |Z| ? From 1Z < X => We are forced to be in the case B. That implies: o Z < 0 o X < 0 o X-Z > 0 So, |X-Z| + |X| = (X-Z) + (-X) = -Z = |Z| as Z < 0 SUFF. From 2Y < 0 => We are forced to be in the case A. That implies: o Z > 0 o X > 0 o Z-X > 0 <=> X-Z < 0 So, |X-Z| + |X| = (-(X-Z)) + (X) = Z = |Z| as Z > 0 SUFF. Hi How do you conclude this Y > 0 and Z < X < 0 (case B) if Y>0, then Z-X<0 , the case can be z=2, X=3 and Z-X<0 which dont agree with z<x<0 please advise
|
|
|
|
|
|
GMAT Club team member
Joined: 02 Sep 2009
Posts: 11633
Followers: 1802
Kudos [?]:
9611
[1] , given: 829
|
1
This post received
KUDOS
ISBtarget wrote: Hi
How do you conclude this
Y > 0 and Z < X < 0 (case B)
if Y>0, then Z-X<0 , the case can be z=2, X=3 and Z-X<0 which dont agree with z<x<0
please advise This is not a good question, for two reasons: 1. Neither of statement is needed to answer the question, stem is enough to do so. I don't know if such kind of situation can occur in real GMAT. Maybe someone can clarify this? 2. Statements contradict. This will never occur in real GMAT. If zy<xy<0 is |x-z|+|x| = |z|Look at the inequality zy<xy<0: We can have two cases: A. If y<0 --> when reducing we should flip signs and we'll get: z>x>0. In this case: as z>x --> |x-z|=-x+z; as x>0 and z>0 --> |x|=x and |z|=z. Hence in this case |x-z|+|x|=|z| will expand as follows: -x+z+x=z --> 0=0, which is true. And: B. If y>0 --> when reducing we'll get: z<x<0. In this case: as z<x --> |x-z|=x-z; as x<0 and z<0 --> |x|=-x and |z|=-z. Hence in this case |x-z|+|x|=|z| will expand as follows: x-z-x=-z --> 0=0, which is true. So knowing that zy<xy<0 is true, we can conclude that |x-z|+|x| = |z| will also be true. Answer should be D even not considering the statements themselves. Next:Statement (1) says that z<x, hence we have case B, which implies that y>0. BUT Statement (2) says that y<0. So statements are contradictory, which will never occur in GMAT. Hope it helps.
_________________
PLEASE READ AND FOLLOW: 11 Rules for Posting!!!
RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory
COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. NEW!!!
DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set. NEW!!!
What are GMAT Club Tests?
25 extra-hard Quant Tests
Find out what's new at GMAT Club - latest features and updates
|
|
|
|
|
|
Manager
Joined: 04 Apr 2010
Posts: 179
Followers: 1
Kudos [?]:
21
[0], given: 31
|
Re: GMATPREP Modulus Q [#permalink]
05 May 2010, 08:08
the given condition itself is sufficient without considering 1 and 2.
_________________
Consider me giving KUDOS, if you find my post helpful.
If at first you don't succeed, you're running about average. ~Anonymous
|
|
|
|
|
|
Manager
Joined: 23 Nov 2009
Posts: 88
Schools: Wharton..:)
Followers: 2
Kudos [?]:
30
[0], given: 13
|
Re: GMATPREP Modulus Q [#permalink]
06 May 2010, 13:17
ZY<XY<0 is |X-Z| + |x| = |Z| 1.) Z<X 2.) Y<0 What is a good approach here? ---------------------------------------------------- A simpler approach :: -make a few cases subcases x y z case 1 + - + 2 - + - case1:: if x,y +ve and y negative thn z>x to satisfy zy<xy frm question part 1) given z<x exact opposite of wht we just did i.e x,z negative and y positive take numbers say x=-3 , z=-4 , y =+1 so 4=4 substitute in the equation ::same goes for any othr number .. hence part 1 satisfies similary do for part 2..m too tired to write [:)]
_________________
" What [i] do is not beyond anybody else's competence"- warren buffett
My Gmat experience -http://gmatclub.com/forum/gmat-710-q-47-v-41-tips-for-non-natives-107086.html
|
|
|
|
|
|
|
Re: GMATPREP Modulus Q
[#permalink]
06 May 2010, 13:17
|
|
|
|
|
|
|
|
|
|
|
close
