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Difficulty: Sub 505 Level,    Inequalities,                                           
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Re: If x, y, and z are positive numbers, is x > y > z ? [#permalink]
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From S1:if xz>yz
and x,y,z are +ve,we can cancel out z to conclude x>y.But nothing is given about z.So insufficient.

From S2:x>z.Nothing is given about y.Insufficient.

Combining the two we get,x>y and x>z.
But whether y>/<z is not known.So insufficient.

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

We know that all are positive integers so we can divide by the variable without changing the sign.

From1:
xz>yz --> x>y Insufficient

From2:
yx>yz --> x>z Insufficient

Combining both:
x>z and x>y but we don't know anything about how y compares to z . Insufficient.
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Re: If x, y, and z are positive numbers, is x > y > z ? [#permalink]
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Although the correct answer is E, can we apply the following approach? (if not, then why can't we use it?)
At the beginning, we conclude that (1) and (2) (if used separately) are not sufficient. Then, we combine these 2 inequalities, in other words, we form a system of 2 inequalities. Since x,y,z are variables, and we don’t know their signs, we cannot multiply. But we can subtract (2) from (1). As a result, this operation yields: xz-yx>yz-yz => xz-yx>0 => xz>yx => z>y => x>z>y => C
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Re: If x, y, and z are positive numbers, is x > y > z ? [#permalink]
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kronint wrote:
Although the correct answer is E, can we apply the following approach? (if not, then why can't we use it?)
At the beginning, we conclude that (1) and (2) (if used separately) are not sufficient. Then, we combine these 2 inequalities, in other words, we form a system of 2 inequalities. Since x,y,z are variables, and we don’t know their signs, we cannot multiply. But we can subtract (2) from (1). As a result, this operation yields: xz-yx>yz-yz => xz-yx>0 => xz>yx => z>y => x>z>y => C


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

Hope it helps.
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Re: If x, y, and z are positive numbers, is x > y > z ? [#permalink]
When I plug in numbers for

X > Y > Z such as 3 > 2 > 1 or 20 > 10 > 2

the equations from (1) and (2) are always right. Any idea why the answer is still E?

Thanks a lot in advance!
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Re: If x, y, and z are positive numbers, is x > y > z ? [#permalink]
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TomSplashy wrote:
When I plug in numbers for

X > Y > Z such as 3 > 2 > 1 or 20 > 10 > 2

the equations from (1) and (2) are always right. Any idea why the answer is still E?

Thanks a lot in advance!


You plug the numbers that answer YES to the question and then check the equations given in (1) and (2). That#s not correct. When trying to solve an YES/NO DS question with plug-in method you should try to find two sets of numbers (which satisfy given information in the stem and the statements), that give different answer to the question (one YES and another NO). For example, you can check x=10, y=5 and z=1 for an YES answer and x=10, y=1 and z=5 for a NO answer.

Hope it's clear.
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Re: If x, y, and z are positive numbers, is x > y > z ? [#permalink]
if x, y, and z are positive numbers, is x > y > z ?

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

Can you tell me if this method will work ?

divide eq 2 by eq 1

yx/xz>yz/yz

yx/xz>1
yx>xz
and since all are positive
y>z ?

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

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

Can you tell me if this method will work ?

divide eq 2 by eq 1

yx/xz>yz/yz

yx/xz>1
yx>xz
and since all are positive
y>z ?

Bunuel


You cannot divide equations like that. For example:

2 > 1
1 > 1/10

If you divide you'll get 2 > 10, which is not correct.
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Re: If x, y, and z are positive numbers, is x > y > z ? [#permalink]
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Bunuel wrote:
If x, y, and z are positive numbers, is x > y > z ?

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

Solution:

We need to determine whether x > y > z, given that x, y, and z are all positive.

Statement One Alone:

Dividing both sides of the inequality by z, we have:

x > y

However, since we still don’t know whether y > z, statement one alone is not sufficient.

Statement Two Alone:

Dividing both sides of the inequality by y, we have:

x > z

However, since we still don’t know whether y > z, statement two alone is not sufficient.

Statements One and Two Together:

Using both statements, we see that x > y and x > z. However, since we still don’t know whether y > z, both statements together are not sufficient.

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