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If x, y, and z are positive numbers, is x > y > z ?

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If x, y, and z are positive numbers, is x > y > z ? [#permalink] New post 06 May 2006, 00:05
00:00
A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

64% (01:45) correct 36% (00:47) wrong based on 91 sessions
If x, y, and z are positive numbers, is x > y > z ?

(1) xz > yz
(2) yx > yz

[color=#ff0000][b]OPEN DISCUSSION OF THIS QUESTION IS HERE: if-x-y-and-z-are-positive-numbers-is-x-y-z-166907.html[/b][/color]
[Reveal] Spoiler: OA

Last edited by Bunuel on 09 Mar 2014, 23:28, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to DS forum.
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 [#permalink] New post 06 May 2006, 00:10
E!

Doenst tell us abt Y!
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 [#permalink] New post 19 Jun 2006, 18:00
No info on Y, could be X>Y>Z or X>Z>Y, so E.
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 [#permalink] New post 19 Jun 2006, 19:34
just to clarify something here:


in the equation: xz > yz you can't simply cancel out the z's right?

because it's a unknown variable, it hides the sign, in which case you don't know if z is positive or negative... is that correct logic?
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 [#permalink] New post 19 Jun 2006, 20:55
consultinghokie wrote:
just to clarify something here:


in the equation: xz > yz you can't simply cancel out the z's right?

because it's a unknown variable, it hides the sign, in which case you don't know if z is positive or negative... is that correct logic?


that's a good point if you dont know whether 'z' is +ve or -ve but in this case the stem already tells us that z is positive, so you are right if you cancel out z on both sides.
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 [#permalink] New post 19 Jun 2006, 22:15
E)

1) not suff x>y z-?
2) not suff x>z y-?
if we combine, we have
x>y and x>z but whether y>z or y<z we do not know, therefore E
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Re: is x>y>z? [#permalink] New post 27 Oct 2011, 00:45
amansingla4 wrote:
If x, y, and z are positive numbers, is x > y > z ?

(1) xz > yz
(2) yx > yz

Is this correct way to solve ?

1) xz - yz > 0

z(x-y) > 0

z> 0 or x > y

No info about z in relation with x and y ---> Insufficient

2) y(x-z)>0

y > 0 or x > z

No info about y in terms of x and x ---> Insufficient

1 + 2 combined,

z > 0 and x > y OR z<0 and x<y

OR

y>0 and x>z OR y< 0 and x < z


Insufficient


Hence E
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Re: is x>y>z? [#permalink] New post 27 Oct 2011, 20:16
We know that x > y and x > z but do not have any information about y relative to z

Therefore answer should be E, as everyone seems to agree
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Re: is x>y>z? [#permalink] New post 27 Oct 2011, 21:26
runitback wrote:
We know that x > y and x > z but do not have any information about y relative to z

Therefore answer should be E, as everyone seems to agree




Is this correct way to solve ?

1) xz - yz > 0

z(x-y) > 0

z> 0 or x > y

No info about z in relation with x and y ---> Insufficient

2) y(x-z)>0

y > 0 or x > z

No info about y in terms of x and x ---> Insufficient

1 + 2 combined,

z > 0 and x > y OR z<0 and x<y

OR

y>0 and x>z OR y< 0 and x < z


Insufficient


Hence E
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Re: is x>y>z? [#permalink] New post 28 Oct 2011, 00:06
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You can cancel out the z's in this case, because the question tells us they are all positive numbers.

1) x > y => INSUFF
2 x > z => INSUFF

Together
x > y
x > z

Don't know anything about r'ship between y and z

Therefore E
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Re: is x>y>z? [#permalink] New post 31 Oct 2011, 07:02
siddhans wrote:
runitback wrote:
We know that x > y and x > z but do not have any information about y relative to z

Therefore answer should be E, as everyone seems to agree




Is this correct way to solve ?

1) xz - yz > 0

z(x-y) > 0

z> 0 or x > y

No info about z in relation with x and y ---> Insufficient

2) y(x-z)>0

y > 0 or x > z

No info about y in terms of x and x ---> Insufficient

1 + 2 combined,

z > 0 and x > y OR z<0 and x<y

OR

y>0 and x>z OR y< 0 and x < z


Insufficient


Hence E



We are told that x,y and z are positive numbers. So you don't have to consider the second case in which the variables are negative. This approach is also fine. But you will realize that for the given conditions, it is easier to cancel out the common variable
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Re: is x>y>z? [#permalink] New post 31 Oct 2011, 08:18
siddhans wrote:
runitback wrote:
We know that x > y and x > z but do not have any information about y relative to z

Therefore answer should be E, as everyone seems to agree




Is this correct way to solve ?

1) xz - yz > 0

z(x-y) > 0

z> 0 or x > y

No info about z in relation with x and y ---> Insufficient

2) y(x-z)>0

y > 0 or x > z

No info about y in terms of x and x ---> Insufficient

1 + 2 combined,

z > 0 and x > y OR z<0 and x<y

OR

y>0 and x>z OR y< 0 and x < z


Insufficient


Hence E


Yes, that approach is fine. However given the time constraints, it is easier to cancel out the common variable and spend time on the tougher questions.
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Re: is x>y>z? [#permalink] New post 31 Oct 2011, 21:35
Here is a video explanation:

http://www.gmatquantum.com/gmat-quantit ... -ds69.html

Dabral

amansingla4 wrote:
If x, y, and z are positive numbers, is x > y > z ?

(1) xz > yz
(2) yx > yz

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Re: is x>y>z? [#permalink] New post 15 Nov 2011, 11:12
I am confused here.

If we simplify the two statements we will end up with the following
from (1), we will have x > y
from (2), we will have x > z

We multiply them and end up with this
xz > xy

Divide by x and we have z > y so this proves that x > y > z is NOT true hence the answer is (C).

Or is it because we don't know the values of y and z, we would not be able to determine if our cross out multiplication is correct?
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Re: is x>y>z? [#permalink] New post 15 Nov 2011, 11:42
Lstadt wrote:
I am confused here.

If we simplify the two statements we will end up with the following
from (1), we will have x > y
from (2), we will have x > z

We multiply them and end up with this
xz > xy

Divide by x and we have z > y so this proves that x > y > z is NOT true hence the answer is (C).

Or is it because we don't know the values of y and z, we would not be able to determine if our cross out multiplication is correct?


I am confused as to how you got to your statement (I highlighted in color).
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Re: is x>y>z? [#permalink] New post 15 Nov 2011, 13:37
bobbynakka wrote:
Lstadt wrote:
I am confused here.

If we simplify the two statements we will end up with the following
from (1), we will have x > y
from (2), we will have x > z

We multiply them and end up with this
xz > xy

Divide by x and we have z > y so this proves that x > y > z is NOT true hence the answer is (C).

Or is it because we don't know the values of y and z, we would not be able to determine if our cross out multiplication is correct?


I am confused as to how you got to your statement (I highlighted in color).


Not sure If I am allowed to do it but I basically cross multiplied (1) by (2).
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Re: is x>y>z? [#permalink] New post 21 Nov 2011, 09:38
Lstadt wrote:
bobbynakka wrote:
Lstadt wrote:
I am confused here.

If we simplify the two statements we will end up with the following
from (1), we will have x > y
from (2), we will have x > z

We multiply them and end up with this
xz > xy

Divide by x and we have z > y so this proves that x > y > z is NOT true hence the answer is (C).

Or is it because we don't know the values of y and z, we would not be able to determine if our cross out multiplication is correct?


I am confused as to how you got to your statement (I highlighted in color).


Not sure If I am allowed to do it but I basically cross multiplied (1) by (2).


Nevermind, I don't think I am allowed to cross multiply inequalities like that :D
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Re: If x, y, and z are positive numbers, is x > y > z ? [#permalink] New post 09 Mar 2014, 20:12
if we combine 1 and 2:

is it possible to have

xz>yz so x>y
yx>yz so x>z

and then subtract both sides of the inequation

xz-yx>yz-yz
xz-yx>0
xz>yx
z>y ?
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Re: If x, y, and z are positive numbers, is x > y > z ? [#permalink] New post 09 Mar 2014, 23:29
Expert's post
biancaneri wrote:
if we combine 1 and 2:

is it possible to have

xz>yz so x>y
yx>yz so x>z

and then subtract both sides of the inequation

xz-yx>yz-yz
xz-yx>0
xz>yx
z>y ?


No, that's not correct. You cannot subtract yx > yz from xz > yz because their signs are in the same direction (> and >).

ADDING/SUBTRACTING INEQUALITIES:


You can only add inequalities when their signs are in the same direction:

If a>b and c>d (signs in same direction: > and >) --> a+c>b+d.
Example: 3<4 and 2<5 --> 3+2<4+5.

You can only apply subtraction when their signs are in the opposite directions:

If a>b and c<d (signs in opposite direction: > and <) --> a-c>b-d (take the sign of the inequality you subtract from).
Example: 3<4 and 5>1 --> 3-5<4-1.

Adding/subtracting/multiplying/dividing inequalities: help-with-add-subtract-mult-divid-multiple-inequalities-155290.html

SOLUTION

If x, y, and z are positive numbers, is x > y > z ?

(1) xz > yz. Since z is a positive number we can safely reduce by it: x > y. Not sufficient.

(2) yx > yz. Since y is a positive number we can safely reduce by it: x > z. Not sufficient.

(1)+(2) We know that x is greater than both y and z but we don't know whether y > z. Not sufficient.

Answer: E.

Hope this helps.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-x-y-and-z-are-positive-numbers-is-x-y-z-166907.html
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Re: If x, y, and z are positive numbers, is x > y > z ?   [#permalink] 09 Mar 2014, 23:29
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