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Each week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution.

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

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31 Jan 2014, 00:56

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We are given x , y, z > 0, all the variables are positive, we also know that multiplying/dividing each side of an inequality with a positive integer does not change the direction of the inequality. So let's analyze the statements.

Statement (1)

xz > yz z can cancel each other without affecting the direction of inequality we can deduce x > y but don't know anything about z in relation to if it is smaller or greater than the other variables. Not Sufficient

Statement (2)

Similarly we can deduce x > z but we can't place y so Statement 2 alone is not sufficient.. (Be careful NOT to carry over statement 1 on to this one yet!!!)

Now look at:

Statement (1) & (2)

x > y and x > z so now we know is is the greatest but still we don't know whether y > z or y < z

So either of the statements together NOT sufficient. Answer E

Each week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution.

We'll be glad if you participate in development of this project: 1. Please provide your solutions to the questions; 2. Please vote for the best solutions by pressing Kudos button; 3. Please vote for the questions themselves by pressing Kudos button; 4. Please share your views on difficulty level of the questions, so that we have most precise evaluation.

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

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05 Feb 2014, 23:54

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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

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

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

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07 Oct 2015, 03:52

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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.