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If x, y, and z are positive numbers, is x > y > z ?
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30 Jan 2014, 22:46
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The Official Guide For GMAT® Quantitative Review, 2ND EditionIf x, y, and z are positive numbers, is x > y > z ? (1) xz > yz (2) yx > yz Data Sufficiency Question: 69 Category: Algebra Inequalities Page: 158 Difficulty: 600 GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition  Quantitative Questions ProjectEach 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. Thank you!
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Re: If x, y, and z are positive numbers, is x > y > z ?
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Re: If x, y, and z are positive numbers, is x > y > z ?
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30 Jan 2014, 23:56
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 EBunuel wrote: The Official Guide For GMAT® Quantitative Review, 2ND EditionIf x, y, and z are positive numbers, is x > y > z ? (1) xz > yz (2) yx > yz Data Sufficiency Question: 69 Category: Algebra Inequalities Page: 158 Difficulty: 600 GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition  Quantitative Questions ProjectEach 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. Thank you!
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Re: If x, y, and z are positive numbers, is x > y > z ?
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31 Jan 2014, 01:05
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 ?
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31 Jan 2014, 12:53
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 ?
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Re: If x, y, and z are positive numbers, is x > y > z ?
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05 Feb 2014, 22:54
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: xzyx>yzyz => xzyx>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 ?
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05 Feb 2014, 23:11
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: xzyx>yzyz => xzyx>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 \(<\)) > \(ac>bd\) (take the sign of the inequality you subtract from). Example: \(3<4\) and \(5>1\) > \(35<41\). Adding/subtracting/multiplying/dividing inequalities: helpwithaddsubtractmultdividmultipleinequalities155290.htmlHope it helps.
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Re: If x, y, and z are positive numbers, is x > y > z ?
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05 Feb 2014, 23:26
Thank you. You are right!



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Re: If x, y, and z are positive numbers, is x > y > z ?
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08 Jan 2017, 10:44
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 ?
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08 Jan 2017, 10:54
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 plugin 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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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: If x, y, and z are positive numbers, is x > y > z ?
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08 Jan 2017, 14:51
@Bunel: Thanks for the explanation!



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Re: If x, y, and z are positive numbers, is x > y > z ?
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24 Jan 2017, 14:56
1, xz>yz ; x > y Insufficient 2, yx > yz ; x > z Insufficient 3, No data about y and z Insufficient E
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Re: If x, y, and z are positive numbers, is x > y > z ?
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22 Apr 2018, 21:08
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 ?
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22 Apr 2018, 21:16



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Re: If x, y, and z are positive numbers, is x > y > z ?
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22 Apr 2018, 21:20
Bunuel wrote: 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 ? BunuelYou cannot divide equations like that. For example: 2 > 1 1 > 1/10 If you divide you'll get 2 > 10, which is not correct. can you tell me under an example of when it will be okay to divide inequalities ?




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