If |z| |y| |x|, is x^3*y^3*z x^2*y^4*z ? (1)

Question

If |z| > |y| > |x|, is x^3*y^3*z > x^2*y^4*z ?

(1) z > y > x
(2) y+z > 0

AK · Answer

should be E ( tried by puting different combinations of x,y,z)

Professor · Answer

st (1): z and y are +ves but x could be both. also they all could be in fractions.

since z is a +ve, the enequality can be reduced to "x^3 y^3 > x^2 y^4".

if x is +ve, x^3 y^3 > x^2 y^4 can be true.
if x is -ve, x^3 y^3 < x^2 y^4. NSF

st (2) tell z is + ve but y and x could be either. so NSF.

E. combining togather is also NSF.

2times · Answer

required to solve this....x^3*y^3*z > x^2*y^4*z

now x^2*y^2 is >0

so our statement = x*y*z > y^2*z ...(A)

Now statement 1)

z>y>x also the question says |z| > |y| > |x|, this is possible only when atleast z>y>0 ; X can be negative example 5,4,1 & 5,4,-1

so as Z> 0 & y> 0 we can rewrite statement (A) as....
x >y

this contradicts z>y>x - so statement 1 is SUFF

Statement 2)

y+z>0
lets put in values of Z ; say Z = 5 Y = -1 , -2... or Y = 1, 2...
lets put in values of Z ; say Z = -5 Y has to be > 5 ... 6,7, 8 this contradicts |z| > |y| > |x|

So, given y+z>0 & |z| > |y| > |x| ; z> 0

which means we can rewrite statement (A) as
xy>y^2 ..(B)

now if y>0
then B reduces to x> y
and we know that |y| > |x| ; so if y is a number like 5; X can be 1 or -1
in either case x has to be < y

now if y<0
then B reduces to x< y
and we know that |y| > |x| ; so if y is a number like -5; X can be 1 or -1
in either case x has to be > y

so statement 2.. gives us two possibilities - INSUFF

therefore ans ...A

would be great if someone could work and see if this is correct and/or provide an alternate solution

Professor · Answer

hmmmmmm........ i did not calculate. thats not correct. when simplified, the inequality becomes x :twisted:

Natalya Khimich · Answer

I agree with 2time solution regarding condition 1, but from from the solution of condition 2 I came to different conclusion : Why INSUFF ? if y>0 , to hold x^3*y^3*z > x^2*y^4*z true, should be x>y from lyl>lxl xy if y<0 , to hold x^3*y^3*z > x^2*y^4*z true, should be xlxl x>y, which contradicts x x^2*y^4*z is not true. The answer is D.

prashrash · Answer

2times, Could you please explain how you got x 0, x will be < y (and thus inequality x>y will be untrue) if y < 0, x will be > y (and thus inequality x>y will be true). The above 2 inferences have been drawn from the inequality |z|>|y|>|x|. Hence stmt (2) is insufficient. IMO answer is (A). What is the OA ? A, D or E ? :-)

AREA88 · Answer

2times · Answer

kevincan, your questions can only be of help to the community here if you post the OA and if no one has been able to get the answer, then maybe also a hint to the solution.

Thanks.

kevincan · Answer

Sorry if the wait has been counterproductive. There are so many talented people that I knew that you would get to the answer sooner or later. Plus, after the suspense, you won't easily forget how to tackle these questions! Notice that the question is complicated. SIMPLIFY AS SOON AS POSSIBLE! Turn a hard DS into an easier one! We know that neither z nor y are zero, so Y^2 is a positive number Thus, the question reduces to " Is x^3*yz>x^2*y^2*z ? " (1) says that z and y are positive but x could be zero or nonzero If x=0, the answer is no/false If x is not =0 , we can simplify further: divide both sides by x^2*y*z (which is +ve ) to get an even simpler question... Is x>y ? We know that this is no/false ! So, (1) is sufficient (2) Now all we know is that z is positive. y could be positive or negative Remember if x=0 , answer to question is no/false If y>0 , see (1) false If yy (since |x| < |y|) and dividing both sides of the inequality by x^2*z*y^3, which is <0, we get x

If |z| > |y| > |x|, is x^3y^3z > x^2y^4z ? (1)

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If |z| > |y| > |x|, is x^3*y^3*z > x^2*y^4*z ? (1)