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Re: If x, y, and z are positive numbers, is x > y > z ? [#permalink]

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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.

Re: If x, y, and z are positive numbers, is x > y > z ? [#permalink]

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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

Re: If x, y, and z are positive numbers, is x > y > z ? [#permalink]

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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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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 \(5>1\) --> \(3-5<4-1\).

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