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

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Manager
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Re: Is (x^7)(y^2)(z^3)>0  [#permalink]

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29 Mar 2014, 07:35
answer should be B.

Group the equation as follows:
(x^4)(x^3)(y^2)(z^3)

y^2 will always be +ive
x^4 will be positive
let x and z be positive (satisfying the second statement): so (x^3)(y^3) will be positive and hence the whole thing will be +ive

Now let x and z be -ve (again satisfying the second statement): so (x^3) will be negative and (z^3) will also be negative. The product will be positive.
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M25-34 GMAT Club Test  [#permalink]

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09 Apr 2014, 14:37
Is X[7]Y[2]Z[3] > 0?
1. yz < 0
2. xz > 0

I am sure I am missing something. Can some one help?
I chose B with the below logic:
xz > 0
so XZ[3] > 0
So the question really becomes (XZ)[3] X[4] Y[2]
From 2, we know that XZ > 0 And X[4] and Y[2] cannot be negative. There by "B" is sufficient?
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Re: M25-34 GMAT Club Test  [#permalink]

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10 Apr 2014, 00:18
Miraarun12345 wrote:
Is X[7]Y[2]Z[3] > 0?
1. yz < 0
2. xz > 0

I am sure I am missing something. Can some one help?
I chose B with the below logic:
xz > 0
so XZ[3] > 0
So the question really becomes (XZ)[3] X[4] Y[2]
From 2, we know that XZ > 0 And X[4] and Y[2] cannot be negative. There by "B" is sufficient?

Statement 2 is sufficient except in case y is 0. If y is 0, x^7*y^2*z^3 will be 0, not positive. If y is not 0, it will be positive. Hence statement 2 alone is not sufficient. Statement 1 tells you that y is not 0 (because yz is less than 0). Hence you need both statements.

(C)
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Re: M25-34 GMAT Club Test  [#permalink]

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10 Apr 2014, 01:12
If there would have been a >= sign, xz>0 should have been sufficient but as it is not, we need the first statement to confirm that Y is not equal to zero.
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Re: Is (x^7)(y^2)(z^3) > 0 ? (1) yz < 0 (2) xz > 0  [#permalink]

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11 Apr 2014, 04:06
tricky question but loving it. thanks for this and for the explanations. I chose 'B' but Burnel nailed it by giving me insight into the reasoning...
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Re: Is (x^7)(y^2)(z^3) > 0 ? (1) yz < 0 (2) xz > 0  [#permalink]

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21 Oct 2014, 13:12
enigma123 wrote:
Is $$(x^7)(y^2)(z^3)>0$$

(1) yz<0

(2) xz>0

Ha Ha Ha Ha..

Another question that caught me off guard.

1. This is insufficient. We don't know anything about the sign of x
2. This one seemed perfect, till I came to know of a possibility of y = 0.

Combining 1 & 2 We would come to know that the prod. will be less than 0 and that none of the values will be 0.
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Re: Is (x^7)(y^2)(z^3) > 0 ? (1) yz < 0 (2) xz > 0  [#permalink]

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17 May 2015, 12:22
Guys, is there any difference between (y^2) and (y)^2. We can be sure that (y)^2 is positive but can we also be sure that (y^2) is positive? (forget about y=0 for a moment)

thank you
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Re: Is (x^7)(y^2)(z^3) > 0 ? (1) yz < 0 (2) xz > 0  [#permalink]

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17 May 2015, 20:35
1
octenisept wrote:
Guys, is there any difference between (y^2) and (y)^2. We can be sure that (y)^2 is positive but can we also be sure that (y^2) is positive? (forget about y=0 for a moment)

thank you

(y^2) and (y)^2 are the same. In both cases, y is squared. y could be anything - the whole of it is squared in both cases.

If y = 2a + b, both representations give (2a + b)^2 = 4a^2 + b^2 + 4ab
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Re: Is (x^7)(y^2)(z^3) > 0 ? (1) yz < 0 (2) xz > 0  [#permalink]

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12 Apr 2019, 11:57
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Re: Is (x^7)(y^2)(z^3) > 0 ? (1) yz < 0 (2) xz > 0   [#permalink] 12 Apr 2019, 11:57

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

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