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y^2 is always positivebhumika.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??
1. yz<0 so eitheris (x^7)(y^2)(z^3)>0?
Statement I .yz<0
Statement II . xz>0
The statement that y^2 is always positive isn't entirely correct.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??
consider situation 1b and 2a then it is negative!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 -
Thanks for pointing out the mistakeBrent 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
VikingWarrior wrote:Excellent point Brent, thanks for pointing it out.
But now shouldn't the answer be E?
consider situation 1b and 2a then it is negative!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 -
ajith wrote:y^2 is always positivebhumika.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??
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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