|
Author |
Message |
|
TAGS:
|
|
|
Director
Joined: 14 Jan 2007
Posts: 787
Followers: 1
Kudos [?]:
32
[0], given: 0
|
w, x, y, and z are integers. If w > x > y > z > [#permalink]
 20 May 2007, 06:35
Question Stats:
0% (00:00) correct
 0% (00:00) wrong based on 0 sessions
w, x, y, and z are integers. If w > x > y > z > 0, is y a common divisor of w and x?
(1)w/x = (1/z) + (1/x)
(2) w^2 - wy - 2w = 0
OA later
|
|
|
|
|
|
|
Manager
Joined: 11 Mar 2007
Posts: 70
Followers: 1
Kudos [?]:
2
[0], given: 0
|
Is it C? If so, let me know and I'll give my reasoning. If not, I don't want to look like an idiot...lol
|
|
|
|
|
|
Director
Joined: 14 Jan 2007
Posts: 787
Followers: 1
Kudos [?]:
32
[0], given: 0
|
I don't want to tell OA now. Let others apply their minds.
|
|
|
|
|
|
Senior Manager
Joined: 04 Mar 2007
Posts: 449
Followers: 1
Kudos [?]:
7
[0], given: 0
|
For me it is B.
w-y-2=0
w=y+2
hence
x=y+1
so: y, x and w are consecutives.
And they are not 1,2 and 3 as there is still z which is greater than 0.
y is not a common divisor of w and x
|
|
|
|
|
|
Manager
Joined: 13 Feb 2007
Posts: 63
Followers: 1
Kudos [?]:
0
[0], given: 0
|
i found the same answer with the same thinking as Caas. I think B.
|
|
|
|
|
|
Manager
Joined: 12 Apr 2007
Posts: 171
Followers: 1
Kudos [?]:
1
[0], given: 0
|
I'm gonna go with D on this one.
For both statements, you can prove that w and x are consecutive integers so that there can be no common divisor that is an integer.
If you look at statement 1:
w/x = (1/z) + (1/x)
Since you know that w > x, then w/x must be >1. The last term, (1/x), has to be a fraction because x is >1, therefore in order to make this equation true, z must equal 1.
So now, we have:
w/x = 1 + (1/x)
Multiply both sides by x and we get:
w = x + 1. Therefore, w and x are consecutive integers.
Statement 2:
w^2 - wy - 2w. Factor out a w and you end up with w-y-2=0.
Like Caas explained above, this is the same as w=y+2. thus, y,x, and w are consecutive integers and they are not 1, 2, and 3 because z needs to be a positive integer as well.
Therefore, D?
|
|
|
|
|
|
CEO
Joined: 29 Mar 2007
Posts: 2618
Followers: 13
Kudos [?]:
142
[0], given: 0
|
Im lost 
|
|
|
|
|
|
Manager
Joined: 12 Apr 2007
Posts: 171
Followers: 1
Kudos [?]:
1
[0], given: 0
|
just curious -- how realistic is it to get one of these types of questions on the actual GMAT?
I have no idea how I would be able to figure out a problem like this in under 2 minutes and I'd probably end up guessing.
This problem took me a good 10 minutes to get a good grasp of, so I'm pretty scared now haha
|
|
|
|
|
|
Manager
Joined: 23 Mar 2007
Posts: 177
Followers: 0
Kudos [?]:
1
[0], given: 0
|
plaguerabbit wrote: I'm gonna go with D on this one.
For both statements, you can prove that w and x are consecutive integers so that there can be no common divisor that is an integer.
If you look at statement 1:
w/x = (1/z) + (1/x)
Since you know that w > x, then w/x must be >1. The last term, (1/x), has to be a fraction because x is >1, therefore in order to make this equation true, z must equal 1.
So now, we have:
w/x = 1 + (1/x)
Multiply both sides by x and we get:
w = x + 1. Therefore, w and x are consecutive integers.
Statement 2:
w^2 - wy - 2w. Factor out a w and you end up with w-y-2=0.
Like Caas explained above, this is the same as w=y+2. thus, y,x, and w are consecutive integers and they are not 1, 2, and 3 because z needs to be a positive integer as well.
Therefore, D?
I too thought on the same lines. It has to be D . Can we have the OA please ?
|
|
|
|
|
|
Director
Joined: 14 Jan 2007
Posts: 787
Followers: 1
Kudos [?]:
32
[0], given: 0
|
plaguerabbit wrote: I'm gonna go with D on this one.
For both statements, you can prove that w and x are consecutive integers so that there can be no common divisor that is an integer.
If you look at statement 1:
w/x = (1/z) + (1/x)
Since you know that w > x, then w/x must be >1. The last term, (1/x), has to be a fraction because x is >1, therefore in order to make this equation true, z must equal 1.
So now, we have:
w/x = 1 + (1/x)
Multiply both sides by x and we get:
w = x + 1. Therefore, w and x are consecutive integers.
Statement 2:
w^2 - wy - 2w. Factor out a w and you end up with w-y-2=0.
Like Caas explained above, this is the same as w=y+2. thus, y,x, and w are consecutive integers and they are not 1, 2, and 3 because z needs to be a positive integer as well.
Therefore, D?
PERFECT explanation. OA is 'D'
|
|
|
|
|
|
Director
Joined: 14 Jan 2007
Posts: 787
Followers: 1
Kudos [?]:
32
[0], given: 0
|
plaguerabbit wrote: I'm gonna go with D on this one.
For both statements, you can prove that w and x are consecutive integers so that there can be no common divisor that is an integer.
If you look at statement 1:
w/x = (1/z) + (1/x)
Since you know that w > x, then w/x must be >1. The last term, (1/x), has to be a fraction because x is >1, therefore in order to make this equation true, z must equal 1.
So now, we have:
w/x = 1 + (1/x)
Multiply both sides by x and we get:
w = x + 1. Therefore, w and x are consecutive integers.
Statement 2:
w^2 - wy - 2w. Factor out a w and you end up with w-y-2=0.
Like Caas explained above, this is the same as w=y+2. thus, y,x, and w are consecutive integers and they are not 1, 2, and 3 because z needs to be a positive integer as well.
Therefore, D?
PERFECT explanation. OA is 'D'
|
|
|
|
|
|
Manager
Joined: 11 Mar 2007
Posts: 70
Followers: 1
Kudos [?]:
2
[0], given: 0
|
Makes sense now. My error was that in statement 1, i did not catch the fact they were consecutive int's. Glad to have you guys to fall back on.
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Similar topics |
Author |
Replies |
Last post |
|
Similar
Topics:
|
 |
|
|
If x < y, y <z> w, which of the following
|
andrehaui |
1 |
30 May 2007, 06:11 |
|
|
1
|
|
If W>X>Y>Z>0, is Z>4 1) 1/W +1/X +1/Y +1/Z =
|
dreambeliever |
7 |
20 Mar 2011, 14:14 |
 |
10
|
 |
w, x, y, and z are integers. If z > y > x > w, is |w| > x^2
|
enigma123 |
13 |
01 Dec 2011, 01:52 |
|
|
5
|
|
w, x, y, and z are integers. If w >x>y>z>0, is y a common
|
enigma123 |
6 |
26 Jan 2012, 22:19 |
|
|
1
|
|
If w>x>y>z>0, is z>4?
|
BN1989 |
1 |
01 Apr 2012, 03:47 |
|
|
|
|
|
|
close
