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by hk » Thu May 21, 2009 6:57 pm
Is (x^7)(y^2)(z^3)>0?

1. yz<0
2. xz>0

OA after some discussions.
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by aj5105 » Thu May 21, 2009 7:52 pm
Together, [spoiler](C)[/spoiler]

[We get YES]
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by PAB2706 » Thu May 21, 2009 8:26 pm
IMO B


y^2 will always be positive irrespective of the y being positive or negetive

State 2 gives us xz>0 ie either both x and Z negetive or both positive

so x^7*y^3 will always be positive and y^2 is positive

thus stat 2 sufficient
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by aj5105 » Thu May 21, 2009 8:49 pm
I think (B) makes sense.

y^2 is anyway positive.

aj5105 wrote:Together, [spoiler](C)[/spoiler]

[We get YES]
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by vpr » Fri May 22, 2009 6:27 am
x^7*y^2*z^3 = x^6*x*y^2*z^2*z = (x^6*y^2*z^2)*x*z.
Since x^6*y^2*z^2 is always positive, xz>0. You can not say anything about yz so answer is B.
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by hk » Fri May 22, 2009 8:03 am
aj5105 you nailed it.

There is a small trap here and most of us fell for it (ya i selected B too.)

B would be suffient as y^2 is always going to be positive. However, B will hold true only if y is not equal to zero. Coz if y is zero then the eqn will not be >0.

From St.1 we know that y is not equal to zero. So the answer is C
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by sanjay_dce » Fri May 22, 2009 11:22 am
C it is , 2nd stmt misses the case when the product= 0
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by raleigh » Fri May 22, 2009 1:59 pm
(1) is insufficient.

Consider x = 1, y = 1, z = -1. You get that the product is < 0.
Now consider x = 1, y = -1, z = 1. You get that the product is > 0.

(2) is insufficient.

Consider x = 1, y = 0, z = 1. You get that the product is zero.
Now consider x = 1, y = 1, z = 1. The product is > 0.

(1) and (2) together are sufficient. Together, x, y, and z must be non-zero.
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