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kevincan
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If |z| > |y| > |x|, is x^3*y^3*z > x^2*y^4*z ? (1) Updated on: 16 Aug 2006, 02:55
Originally posted by kevincan on 16 Aug 2006, 01:55.
Last edited by kevincan on 16 Aug 2006, 02:55, edited 1 time in total.
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If |z| > |y| > |x|, is x^3*y^3*z > x^2*y^4*z ?

(1) z > y > x
(2) y+z > 0



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AK
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If |z| > |y| > |x|, is x^3*y^3*z > x^2*y^4*z ? (1) 16 Aug 2006, 02:22
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should be E ( tried by puting different combinations of x,y,z)
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Professor
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If |z| > |y| > |x|, is x^3*y^3*z > x^2*y^4*z ? (1) 16 Aug 2006, 06:30
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kevincan
If |z| > |y| > |x|, is x^3*y^3*z > x^2*y^4*z ?

(1) z > y > x
(2) y+z > 0
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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.
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If |z| > |y| > |x|, is x^3*y^3*z > x^2*y^4*z ? (1) 16 Aug 2006, 10:03
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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
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If |z| > |y| > |x|, is x^3*y^3*z > x^2*y^4*z ? (1) 16 Aug 2006, 10:59
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kevincan
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kevincan
If |z| > |y| > |x|, is x^3*y^3*z > x^2*y^4*z ?

(1) z > y > x
(2) y+z > 0

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 inequality 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.

How?
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hmmmmmm........ i did not calculate. thats not correct. when simplified, the inequality becomes x<y. A is it... :twisted: :twisted:
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Natalya Khimich
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If |z| > |y| > |x|, is x^3*y^3*z > x^2*y^4*z ? (1) 17 Aug 2006, 19:26
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I agree with 2time solution regarding condition 1, but from from the solution of condition 2 I came to different conclusion :

Quote:


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
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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 x<y, which contradicts x>y

if y<0 , to hold x^3*y^3*z > x^2*y^4*z true, should be x<y
from lyl>lxl x>y, which contradicts x<y

Thus it sufficient to tell that x^3*y^3*z > x^2*y^4*z is not true.

The answer is D.
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If |z| > |y| > |x|, is x^3*y^3*z > x^2*y^4*z ? (1) 17 Aug 2006, 20:36
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2times

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

Show more


2times, Could you please explain how you got x<y in the second case ? Are you considering x to be < 0 as well ? I did not understand this part. Kevincan / Proffesor - your inputs will be highly appreciated as well.

But I agree with the 2 cases above, i.e.
if y > 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 ? :-)
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If |z| > |y| > |x|, is x^3*y^3*z > x^2*y^4*z ? (1) 17 Aug 2006, 21:06
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[quote="prashrash"][quote="2times"]
[u]now if y>0[/u]
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 [u]either case x has to be < y[/u]

[u]now if y<0[/u]
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 [u]either case x has to be > y [/u]

[/quote]

2times, Could you please explain how you got x<y in the second case ? Are you considering x to be < 0 as well ? I did not understand this part. Kevincan / Proffesor - your inputs will be highly appreciated as well.

But I agree with the 2 cases above, i.e.
if y > 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 ? :-)[/quote][quote]
[b]
I have a same point with this.

y > 0 is ture.
y < 0 is NOT true.

Therefore, no.2 condition is NOT sufficient.

So, A is the answer, I think. :-D[/quote][/b]
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If |z| > |y| > |x|, is x^3*y^3*z > x^2*y^4*z ? (1) 19 Aug 2006, 13:48
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Natalya Khimich
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 x<y, which contradicts x>y

if y<0 , to hold x^3*y^3*z > x^2*y^4*z true, should be x<y
from lyl>lxl x>y, which contradicts x<y
this part of statement 2 gives us z>0 and y<0; so you can re write the given statement xyz>y^2z as xz<yz as z>0; x<y - so it does *not* contradict x<y

Thus it sufficient to tell that x^3*y^3*z > x^2*y^4*z is not true.

The answer is D.
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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.
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If |z| > |y| > |x|, is x^3*y^3*z > x^2*y^4*z ? (1) 19 Aug 2006, 15:03
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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 [b]x<y, which is false

Thus (2) is sufficient

OA D Thanks to everybody who tried. I appreciate your posting answers, as it really helps me to make new questions. I hope you found this question helpful/enjoyable



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