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Is x^7*y^2*z^3 > 0 ? (1) yz < 0 (2) xz > 0

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Is x^7*y^2*z^3 > 0 ? (1) yz < 0 (2) xz > 0 [#permalink] New post 16 Feb 2012, 14:42
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Is x^7*y^2*z^3 > 0 ?

(1) yz < 0

(2) xz > 0

From the Gmatclub Math 25 practice test:
http://gmatclub.com/tests/m25#q34

Response:
x^7 . y^2 . z^3 is same as:
(xz) . (x^6) . (y^2) . (z^2)
xz is positive based on assumption #2.
x^6 is (x^2)^3, since x^2 is positive for all real numbers, x^6 is also positive.
z^2 is positive for all real numbers.
y^2 is positive for all real numbers.

So product of 4 positive real numbers is also positive. SUFFICIENT.

Hence, B - Statement 2 Alone is Sufficient.

Per the gmatclub math25 test, the response is C - both statements together are sufficient. What incorrect assumptions am I making?

Thanks!
[Reveal] Spoiler: OA

Last edited by Bunuel on 16 Feb 2012, 15:02, edited 1 time in total.
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Re: GmatClub Math25 qn 34 - is it positive? [#permalink] New post 16 Feb 2012, 15:01
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fxsunny wrote:
From the Gmatclub Math 25 practice test:
http://gmatclub.com/tests/m25#q34

Is x^7*y^2*z^3> 0?

(1) yz < 0
(2) xz > 0

Response:
x^7*y^2*z^3 is same as:
(xz)*(x^6)*(y^2)*(z^2)
xz is positive based on assumption #2.
x^6 is (x^2)^3, since x^2 is positive for all real numbers, x^6 is also positive.
z^2 is positive for all real numbers.
y^2 is positive for all real numbers.

So product of 4 positive real numbers is also positive. SUFFICIENT.

Hence, B - Statement 2 Alone is Sufficient.

Per the gmatclub math25 test, the response is C - both statements together are sufficient. What incorrect assumptions am I making?

Thanks!


The red parts are not correct. Square of a number is nonnegative and not positive as you've written. So for (2) if y=0 then x^7*y^2*z^3=0.

Complete solution:
Is x^7*y^2*z^3 > 0 ?

Inequality x^7*y^2*z^3>0 to be true x and z must be either both positive or both negative (in order x^7*z^3 to be positive) AND y must not be zero (in order x^7*y^2*z^3 not to equal to zero).

(1) yz<0 --> y\neq{0}. Don't know about x and z. Not sufficient.

(2) xz>0 --> x and z are either both positive or both negative. Don't know about y. Not sufficient.

(1)+(2) Sufficient.

Answer: C.

Similar question to practice: m21-q30-96613.html?hilit=conditions#p744188

Hope it helps.
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new [#permalink] New post 22 Mar 2012, 20:41
In this case we need to be certain that the product of x and z are positive (they are the same sign), and that none of the numbers are zero.
The combination of the two show that this will hold true, but not individually.
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Re: Is x^7*y^2*z^3 > 0 ? (1) yz < 0 (2) xz > 0 [#permalink] New post 23 Mar 2012, 05:59
x^7*y^2*z^3>0
y is positive, we need to know regarding x and z
1) no information about x
insufficient

2) xz>0

either of them is negative hence

the answer to our question is no

sufficient

hence B
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Re: Is x^7*y^2*z^3 > 0 ? (1) yz < 0 (2) xz > 0 [#permalink] New post 23 Mar 2012, 06:03
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kraizada84 wrote:
x^7*y^2*z^3>0
y is positive, we need to know regarding x and z
1) no information about x
insufficient

2) xz>0

either of them is negative hence

the answer to our question is no

sufficient

hence B


OA for this question is C, not B.

(2) xz>0 means thatx and z are either both positive or both negative. So, x^7*z^3 > 0 but y can still be zero and in this case x^7*y^2*z^3=0, hence this statement is not sufficient.

Refer to the complete solution above.

Hope it helps.
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Re: Is x^7*y^2*z^3 > 0 ? (1) yz < 0 (2) xz > 0 [#permalink] New post 30 Jul 2012, 22:25
tricky problem. Totally forgot that Y could be 0 when I'm testing B.
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Re: doubt on DS Question [#permalink] New post 04 Oct 2012, 13:35
solerenegade wrote:
Hi,

I was taking a practice quiz and came across this question where my answer did not match the given answer. Further I wasn't convinced with the explanation offered.
What do you guys think is the correct solution?

Is (x^7)(y^2)(z^3)>0 ?
(1) yz<0

(2) xz>0


Merging similar questions. Please refer to the posts above and ask if anything remains unclear.

Also, please read and follow: rules-for-posting-please-read-this-before-posting-133935.html (pay attention to the rule #3).
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Re: Is x^7*y^2*z^3 > 0 ? (1) yz < 0 (2) xz > 0 [#permalink] New post 04 Oct 2012, 13:39
My Bad! On missing the fact that the product of the 3 could be zero and for forgetting Rule# 3 :oops:
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Is (x^7)(y^2)(z^3)>0 ? [#permalink] New post 02 Nov 2012, 07:31
Is (x^7)(y^2)(z^3)>0 ?
(1) yz<0

(2) xz>0

have a doubt here not statement 1 says y*x < 0 this means that either y has to be positive and x has to be negative or x has to be positive and y has to be negative but we have no info about Z so insufficient

statement 2 says X * Z > 0 so both X,Z positive or both X,Z negative

now we plug this info back in the question -X^7 = -ve number similary -Z^3 = -ve number now the question is about y since Y is raised to the power 2 it will always be positive so this won't matter to us so -ve * -ve positive number sufficient

same case if we assume both X and Z positive. So why is the answer C? Is my reasoning incorrect?
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Re: Is (x^7)(y^2)(z^3)>0 ? [#permalink] New post 02 Nov 2012, 08:42
nelz007 wrote:
Is (x^7)(y^2)(z^3)>0 ?
(1) yz<0

(2) xz>0

have a doubt here not statement 1 says y*x < 0 this means that either y has to be positive and x has to be negative or x has to be positive and y has to be negative but we have no info about Z so insufficient

statement 2 says X * Z > 0 so both X,Z positive or both X,Z negative

now we plug this info back in the question -X^7 = -ve number similary -Z^3 = -ve number now the question is about y since Y is raised to the power 2 it will always be positive so this won't matter to us so -ve * -ve positive number sufficient

same case if we assume both X and Z positive. So why is the answer C? Is my reasoning incorrect?


Merging similar topics. Please refer to the solutions above.

As, for your doubt, see here: is-x-7-y-2-z-3-0-1-yz-0-2-xz-127692.html#p1063962

Hope it helps.
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Re: Is x^7*y^2*z^3 > 0 ? (1) yz < 0 (2) xz > 0 [#permalink] New post 02 Nov 2012, 08:55
Thanks bunuel very silly mistake to not consider 0.
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Re: Is x^7*y^2*z^3 > 0 ? (1) yz < 0 (2) xz > 0   [#permalink] 02 Nov 2012, 08:55
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