kaplan ps

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!

I am getting these questions from the kaplan CAT's and from the Kaplan online questions that came with the course. thanks for helping out I truly appreciate it.

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

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

perfect!!!

was just wondering ,
what it takes to solve it within 2 mins!!!!

yeah i was wondering that myself.

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!