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Re M2130
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16 Sep 2014, 00:15
Official Solution:Inequality \(x^2*y^3*z \gt 0\) to be true: 1. \(y\) and \(z\) must be either both positive or both negative, so they must have the same sign (in this case \(y^3*z\) will be positive); AND 2. \(x\) must not be zero (in this case \(x^2\) will be positive). (1) \(yz \gt 0\). From this statement it follows that \(y\) and \(z\) are either both positive or both negative, so the first condition is satisfied. But we don't know about \(x\) (the second condition). Not sufficient. (2) \(xz \lt 0\). From this statement it follows that \(x \ne 0\), so the second condition is satisfied. Don't know about the signs of \(y\) and \(z\) (the first condition). Not sufficient. (1)+(2) Both conditions are satisfied. Sufficient. Answer: C
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Re: M2130
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12 Feb 2016, 00:34
great question
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Re: M2130
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15 Mar 2016, 21:26
Bunuel wrote: Is \(x^2*y^3*z \gt 0\)?
(1) \(yz \gt 0\)
(2) \(xz \lt 0\) Hi Can you pls help with testing both statements combined. I tried bringing them both down to the same inequality sign that mostly works in linear equations bur getting stuck in nonlinear set up here. yz>0 or yz<0 1) xz<0 2) Tried dividing 1) and 2) but end up losing z and any clue on its sign. Pls advice on how this process can be made to work. Thanks.



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Re: M2130
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21 Mar 2016, 05:45
even if we don't know about x, in case of both positive or negative x will not affect since it is in square. So, how A is wrong? Please explain. Thanks, Bunuel wrote: Official Solution:
Inequality \(x^2*y^3*z \gt 0\) to be true:
1. \(y\) and \(z\) must be either both positive or both negative, so they must have the same sign (in this case \(y^3*z\) will be positive);
AND
2. \(x\) must not be zero (in this case \(x^2\) will be positive).
(1) \(yz \gt 0\). From this statement it follows that \(y\) and \(z\) are either both positive or both negative, so the first condition is satisfied. But we don't know about \(x\) (the second condition). Not sufficient. (2) \(xz \lt 0\). From this statement it follows that \(x \ne 0\), so the second condition is satisfied. Don't know about the signs of \(y\) and \(z\) (the first condition). Not sufficient. (1)+(2) Both conditions are satisfied. Sufficient.
Answer: C



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Re: M2130
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21 Mar 2016, 05:53
sun01 wrote: even if we don't know about x, in case of both positive or negative x will not affect since it is in square. So, how A is wrong? Please explain. Thanks, Bunuel wrote: Official Solution:
Inequality \(x^2*y^3*z \gt 0\) to be true:
1. \(y\) and \(z\) must be either both positive or both negative, so they must have the same sign (in this case \(y^3*z\) will be positive);
AND
2. \(x\) must not be zero (in this case \(x^2\) will be positive).
(1) \(yz \gt 0\). From this statement it follows that \(y\) and \(z\) are either both positive or both negative, so the first condition is satisfied. But we don't know about \(x\) (the second condition). Not sufficient. (2) \(xz \lt 0\). From this statement it follows that \(x \ne 0\), so the second condition is satisfied. Don't know about the signs of \(y\) and \(z\) (the first condition). Not sufficient. (1)+(2) Both conditions are satisfied. Sufficient.
Answer: C If x is 0, then \(x^2y^3z\) will be 0, not more than 0.
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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: M2130
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07 Sep 2016, 02:08
Is x^2∗y^3∗z>0? (1) yz>0 (2) xz<0
Question breakup/understanding: x^2∗y^3∗z>0 it can be possible only if we know below two facts: I) x NOT EQUAL to 0 II) y and z are of same sign.
Statement 1: yz> 0 it means both y and z can be either positive or negative. But no info on X Thus not sufficient.
Statement 2: xz<0 It means either x and z are of opposite sign It also tells the x NOT EQUAL to 0. But no info on y. Thus not sufficient.
Combine both Statement we get X NOT EQUAL to 0. Y and Z of same sign. Thus sufficient.
other way yz>0 (Eq 1) xz<0 (Eq 2) yzxz>0 z(yx) >0 Thus z >0 and yx>0 this ensures z,y same sign. Already we know x not equal to 0 received from Eq 2 Thus Sufficient.
Please correct me if i am wrong anywhere. Ans: C



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13 Dec 2017, 07:49
+1 for C. To find the answer to this question, we need to know a) if any of x,y,or z is equal to zero or not. b) yz > 0 or not. (xy)^2*(yz). Statement 1 gives yz>0 and y,z are not equal to 0. > Not sufficient Statement 2 gives x and z are not equal to 0. > Not sufficient Both 1 and 2 > Sufficient. We come to know that x,y,and z are non zero and that yz>0. Hence option C.
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24 Apr 2018, 19:34
I think this is a highquality question and I agree with explanation.



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Re: M2130
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25 Aug 2018, 00:55
HI Bunuel,
Please help me with the below approach:
If X=1/4 Y=1 Z=1 then answer will be a no.
But if X=2 Y=2 Z=2 then answer will be a Yes.
No where it has been said that all three are integers.
Please guide:)



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25 Aug 2018, 02:16










