x,y,z all positive,Is Z not between x and y ?
i.e x<z<y ?
(1) x <2z <y
here we can say x<2z & 2z < y
so z <y but z may be less than x INSUFF
(2) 4x <z <4y
4x < z & z <4y> x but z may be > y INSUFF
Combine 2z < y & 4x < z
so x < z < y SUFF
C
number properties
This topic has expert replies
Source: Beat The GMAT — Data Sufficiency |
-
samirpandeyit62
- Master | Next Rank: 500 Posts
- Posts: 460
- Joined: Sun Mar 25, 2007 7:42 am
- Thanked: 27 times
-
silivest60
- Junior | Next Rank: 30 Posts
- Posts: 29
- Joined: Sat Oct 27, 2007 9:49 pm
- Location: Vancouver, BC, Canada
- Thanked: 1 times
I think a little different....
(1) If you divide x<2z<y by 2, you get x/2<z<y/2. If you sketch this latest inequality on a number line, you can see that you can have z between x and y, but you can also have z outside x and y. Therefore, not sufficient
(2) Using the number line again, you can see that 4x and 4y will always be to the right of x and y , because x and y are pozitive numbers. So, z will always be outside (on the right) of x and y. Thus the answer to the question will always be YES: Z is not between x and y.
Therefore B is the right choice.
(1) If you divide x<2z<y by 2, you get x/2<z<y/2. If you sketch this latest inequality on a number line, you can see that you can have z between x and y, but you can also have z outside x and y. Therefore, not sufficient
(2) Using the number line again, you can see that 4x and 4y will always be to the right of x and y , because x and y are pozitive numbers. So, z will always be outside (on the right) of x and y. Thus the answer to the question will always be YES: Z is not between x and y.
Therefore B is the right choice.
