Good one! Which Kaplan book are u getting these questions from?
The question stem provides all the info
1) xyz not equal t 0 implies x not 0 y not 0 z not 0
2) xy > 0
x and y are both positive or both negative
3) x^2+xy+xz<0
Take common factor x out
x (x+y+z)<0
Either x<0 or x+y+z<0
I. x(y+z)<0
From 3) we have either x is negative or x+y+z is negative
Lets say x and y are both positive from 2)
For x(x+y+z) to be < 0 from 3) z has to be negative and its absolute value greater than x+y for x ( x+y+z) < 0
Since z is greater than x+y z is definitely greater tha y making y+z negative and therefore x(y+z) negative
i.e positive * negative = negative
i.e. x (y+z) < 0
Lets say x and y are both negative from 2)
For x ( x+y+z) < 0 in 3) z has to be positive and its absolute value greater than x+y for x ( x+y+z) < 0 in 3)
So z is definitely greater than y which would make y+z positive and therefore x( y+z) negative
i.e negative ( positive) = negative
i.e x( y+z) < 0
Hence x (y+z) < 0 MUST BE TRUE
II) x+y+z<0
We have showed in 3) that either x is < 0 or x+y+z<0 . So x+y+z could be less than 0 but not necessary(I.E IF X < 0 THEN X+Y+Z > 0 for x ( x+y+z) to be less than 0). S
So eliminate this answer choice
III) If x <0 the z>0
We have showed in I's explanation that if x is negative then z has to be positive for 3) to be true
If x <0 the z>0MUST BE TRUE
Hence C)
I have tried to explain as detailed as possible but if this got a bit confusing, let me know. (i.e. if u still have questions)
Good luck!
kaplan ps
This topic has expert replies
Source: Beat The GMAT — Problem Solving |
xy>0
x +ve y +ve
x -ve y -ve
x^2+xy+xz < 0
x^2 > 0 always
xy > 0 given
so xz < 0
hence
x +ve y +ve z -ve
x -ve y -ve z +ve
x^2+xy+xz < 0
x(x+y+z) < 0
x +ve y +ve z -ve (x+y+z) -ve
x -ve y -ve z +ve (x+y+z) +ve
Now lets look at options
I. x(y+z)<0
Given
x^2+xy+xz < 0
x^2 always > 0
so
xy+xz < 0
x(y+z) < 0
hence TRUE
II. x+y+z<0
x +ve (x+y+z) -ve
x -ve (x+y+z) +ve
hence not true
If x <0 the z>0
x -ve y -ve z +ve (x+y+z) +ve
TRUE
hence ans is C
x +ve y +ve
x -ve y -ve
x^2+xy+xz < 0
x^2 > 0 always
xy > 0 given
so xz < 0
hence
x +ve y +ve z -ve
x -ve y -ve z +ve
x^2+xy+xz < 0
x(x+y+z) < 0
x +ve y +ve z -ve (x+y+z) -ve
x -ve y -ve z +ve (x+y+z) +ve
Now lets look at options
I. x(y+z)<0
Given
x^2+xy+xz < 0
x^2 always > 0
so
xy+xz < 0
x(y+z) < 0
hence TRUE
II. x+y+z<0
x +ve (x+y+z) -ve
x -ve (x+y+z) +ve
hence not true
If x <0 the z>0
x -ve y -ve z +ve (x+y+z) +ve
TRUE
hence ans is C
perfect!!!xy>0
x +ve y +ve
x -ve y -ve
x^2+xy+xz < 0
x^2 > 0 always
xy > 0 given
so xz < 0
hence
x +ve y +ve z -ve
x -ve y -ve z +ve
x^2+xy+xz < 0
x(x+y+z) < 0
x +ve y +ve z -ve (x+y+z) -ve
x -ve y -ve z +ve (x+y+z) +ve
Now lets look at options
I. x(y+z)<0
Given
x^2+xy+xz < 0
x^2 always > 0
so
xy+xz < 0
x(y+z) < 0
hence TRUE
II. x+y+z<0
x +ve (x+y+z) -ve
x -ve (x+y+z) +ve
hence not true
If x <0 the z>0
x -ve y -ve z +ve (x+y+z) +ve
TRUE
hence ans is C
was just wondering ,
what it takes to solve it within 2 mins!!!!
-
cramya
- Legendary Member
- Posts: 2467
- Joined: Thu Aug 28, 2008 6:14 pm
- Thanked: 331 times
- Followed by:11 members
I know sometimes the explanations look daunting. It took me about 2 -3 minutes to solve this problem and definitely I woudnt be writing all this explanation on the scratch sheet nor would stop@800 (speaking for him hope he doesnt mind)
The key take aways are in the question stem and then figuring out the statements themselves are just a matter of applying info provided in the stem.
Good luck!
The key take aways are in the question stem and then figuring out the statements themselves are just a matter of applying info provided in the stem.
Good luck!
