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M21-30

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M21-30 [#permalink]

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New post 16 Sep 2014, 01:15
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29% (00:55) correct 71% (00:35) wrong based on 143 sessions

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Re M21-30 [#permalink]

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New post 16 Sep 2014, 01:15
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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: M21-30 [#permalink]

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New post 12 Feb 2016, 01:34
great question
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Re: M21-30 [#permalink]

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New post 15 Mar 2016, 22:26
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 non-linear 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: M21-30 [#permalink]

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New post 21 Mar 2016, 06:45
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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: M21-30 [#permalink]

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New post 21 Mar 2016, 06:53
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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New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

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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: M21-30 [#permalink]

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New post 07 Sep 2016, 03:08
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)
yz-xz>0
z(y-x) >0
Thus
z >0 and y-x>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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Re: M21-30 [#permalink]

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New post 13 Dec 2017, 08:49
+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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Re: M21-30   [#permalink] 13 Dec 2017, 08:49
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