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is (x^7)(y^2)(z^3)>0?

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by ajith » Sun Jan 24, 2010 9:55 am
bhumika.k.shah wrote:is (x^7)(y^2)(z^3)>0?

Statement I .yz<0
Statement II . xz>0

OA C

should i not consider the value of y^2 as it will always be positive??
y^2 is always positive

so is z^2 and x^6

So the question becomes

xz>0?

Which can be solved using Statement 2 alone

(and my answer differs with OA)

[Assumption: x,y,z are real]
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by VikingWarrior » Sun Jan 24, 2010 12:43 pm
is (x^7)(y^2)(z^3)>0?

Statement I .yz<0
Statement II . xz>0
1. yz<0 so either
a. y - , z+ thus true but if
b. y+, z- it is not true unless x is -

2. xz >0 so either
a. x +, z + thus true since y^2 must be +
b. x - , z - still true since x^7 is - and z^3 is -

For me answer is B
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by VikingWarrior » Sun Jan 24, 2010 12:44 pm
Ajith, you are a champ at Math man! :)
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by ajith » Sun Jan 24, 2010 12:46 pm
VikingWarrior wrote:Ajith, you are a champ at Math man! :)
Thanks a ton :)
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by Brent@GMATPrepNow » Sun Jan 24, 2010 1:00 pm
bhumika.k.shah wrote:is (x^7)(y^2)(z^3)>0?
Statement I .yz<0
Statement II . xz>0
OA C
should i not consider the value of y^2 as it will always be positive??
The statement that y^2 is always positive isn't entirely correct.
We might say that y^2 is usually positive. However, y^2 is not positive when y=0.

So, statement (2) doesn't help rule out the possibility that y might equal 0.
Statement (1) tells us that y does not equal zero.

So, we need both statements to answer the question.
Answer = C
Brent Hanneson - Creator of GMATPrepNow.com
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by VikingWarrior » Sun Jan 24, 2010 1:10 pm
Excellent point Brent, thanks for pointing it out.
But now shouldn't the answer be E?
1. yz<0 so either
a. y - , z+ thus true but if
b. y+, z- it is not true unless x is -

2. xz >0 so either
a. x +, z + thus true since y^2 must be +
b. x - , z - still true since x^7 is - and z^3 is -
consider situation 1b and 2a then it is negative!
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by ajith » Sun Jan 24, 2010 1:12 pm
Brent Hanneson wrote: The statement that y^2 is always positive isn't entirely correct.
We might say that y^2 is usually positive. However, y^2 is not positive when y=0.

So, statement (2) doesn't help rule out the possibility that y might equal 0.
Statement (1) tells us that y does not equal zero.

So, we need both statements to answer the question.
Answer = C
Thanks for pointing out the mistake :)
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by VikingWarrior » Sun Jan 24, 2010 1:15 pm
sorry, of course I am wrong 1b and 2 a are mutually exclusive! Answer IS C
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by bhumika.k.shah » Sun Jan 24, 2010 10:19 pm
No,
u will require both statements which will finally conclude that y is not equal to zero.

Regards,
Bhumika
VikingWarrior wrote:Excellent point Brent, thanks for pointing it out.
But now shouldn't the answer be E?
1. yz<0 so either
a. y - , z+ thus true but if
b. y+, z- it is not true unless x is -

2. xz >0 so either
a. x +, z + thus true since y^2 must be +
b. x - , z - still true since x^7 is - and z^3 is -
consider situation 1b and 2a then it is negative!
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by bhumika.k.shah » Sun Jan 24, 2010 10:21 pm
So now that we know the answer Option is C...Ajith is this question of urs really valid ?
I wanna know what the main question is been asked here !

Is the question that u derived correct ???
ajith wrote:
bhumika.k.shah wrote:is (x^7)(y^2)(z^3)>0?

Statement I .yz<0
Statement II . xz>0

OA C

should i not consider the value of y^2 as it will always be positive??
y^2 is always positive

so is z^2 and x^6

So the question becomes

xz>0?

Which can be solved using Statement 2 alone

(and my answer differs with OA)

[Assumption: x,y,z are real]
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