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The OA is C but I think it's E Statement (1) : clearly insufficient since we don't have the sign of Z (since it has an odd exponent) Statement (2) : clearly insufficient since Y can be 0 Both (1) and (2) : well yes Y can't be 0 but we still can't tell the sign of Z ! Consider this example : X=4 , Y= -1, Z= -1 ; we will have both statements right but the original expression will be negative

The OA is C but I think it's E Statement (1) : clearly insufficient since we don't have the sign of Z (since it has an odd exponent) Statement (2) : clearly insufficient since Y can be 0 Both (1) and (2) : well yes Y can't be 0 but we still can't tell the sign of Z ! Consider this example : X=4 , Y= -1, Z= -1 ; we will have both statements right but the original expression will be negative

Am I wrong ?

Inequality \(x^7*y^2*z^3>0\) to be true: I. \(x\) and \(z\) must be either both positive or both negative, AND II. \(y\) must not be zero.

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

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

(1)+(2) Both conditions are satisfied. Sufficient.

Answer: C.

As for your doubt: we are not interested in the sign of \(z\), we need \(x\) and \(z\) to be be either both positive or both negative. Next, your example is not valid: x=4, y=-1, z=-1 --> yz=4>0 and xz=-4<0 and we are given that \(yz<0\) and \(xz>0\).

Am sorry, in the question is it x exponent z or the product between x and z ??

It is the product.

Is \((x^7)(y^2)(z^3) \gt 0\) ? reduced to is Is \((x)(y^2)(z) \gt 0\) ? -> I have removed the squared values as they do not play any role in changing the sign, but I kept \(y^2\) to consider the y=0 condition. _________________

Inequality \(x^7*y^2*z^3>0\) to be true: I. \(x\) and \(z\) must be either both positive or both negative, AND II. \(y\) must not be zero.

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

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

(1)+(2) Both conditions are satisfied. Sufficient.

Answer: C.

As for your doubt: we are not interested in the sign of \(z\), we need \(x\) and \(z\) to be be either both positive or both negative. Next, your example is not valid: x=4, y=-1, z=-1 --> yz=4>0 and xz=-4<0 and we are given that \(yz<0\) and \(xz>0\).

Hope it helps.

I have a doubt; Why can't B only be true because we care for x and z signs and from II either they both are +ive or -ive and this will provide us the result. _________________

GGG (Gym / GMAT / Girl) -- Be Serious

Its your duty to post OA afterwards; some one must be waiting for that...

Inequality \(x^7*y^2*z^3>0\) to be true: I. \(x\) and \(z\) must be either both positive or both negative, AND II. \(y\) must not be zero.

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

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

(1)+(2) Both conditions are satisfied. Sufficient.

Answer: C.

As for your doubt: we are not interested in the sign of \(z\), we need \(x\) and \(z\) to be be either both positive or both negative. Next, your example is not valid: x=4, y=-1, z=-1 --> yz=4>0 and xz=-4<0 and we are given that \(yz<0\) and \(xz>0\).

Hope it helps.

I have a doubt; Why can't B only be true because we care for x and z signs and from II either they both are +ive or -ive and this will provide us the result.

Please read the solution carefully.

Inequality \(x^7*y^2*z^3>0\) to be true: I. \(x\) and \(z\) must be either both positive or both negative, AND II. \(y\) must not be zero.

So for (2): we know that \(x\) and \(z\) are either both positive or both negative, BUT we don't know whether \(y\) equals to zero, because if it is then \(x^7*y^2*z^3=0\) and not more than zero.

Inequality \(x^7*y^2*z^3>0\) to be true: I. \(x\) and \(z\) must be either both positive or both negative, AND II. \(y\) must not be zero.

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

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

(1)+(2) Both conditions are satisfied. Sufficient.

Answer: C.

As for your doubt: we are not interested in the sign of \(z\), we need \(x\) and \(z\) to be be either both positive or both negative. Next, your example is not valid: x=4, y=-1, z=-1 --> yz=4>0 and xz=-4<0 and we are given that \(yz<0\) and \(xz>0\).

Hope it helps.

I have a doubt; Why can't B only be true because we care for x and z signs and from II either they both are +ive or -ive and this will provide us the result.

Please read the solution carefully.

Inequality \(x^7*y^2*z^3>0\) to be true: I. \(x\) and \(z\) must be either both positive or both negative, AND II. \(y\) must not be zero.

So for (2): we know that \(x\) and \(z\) are either both positive or both negative, BUT we don't know whether \(y\) equals to zero, because if it is then \(x^7*y^2*z^3=0\) and not more than zero.

Hope it's clear.

Sorry to say but still I didn't get this...

Why are we checking Y must not be zero? Why can;t its be X or Z not to be zero _________________

GGG (Gym / GMAT / Girl) -- Be Serious

Its your duty to post OA afterwards; some one must be waiting for that...

Re: Inequality sign [#permalink]
01 Oct 2010, 03:40

from statement 1: it depends on the value of x , x can be either positive or negative so statement 1 not sufficient

from statement 2: xz>0 case1: x +ve & z +ve case 2: x -ve & z-ve we know that y2 is +ve so from case 1:x7y2z3= (+)(+)(+)= + so from case 2:x7y2z3= (-)(+)(-)= + so B alone sufficient

Re: Inequality sign [#permalink]
01 Oct 2010, 04:29

Expert's post

anilnandyala wrote:

from statement 1: it depends on the value of x , x can be either positive or negative so statement 1 not sufficient

from statement 2: xz>0 case1: x +ve & z +ve case 2: x -ve & z-ve we know that y2 is +ve so from case 1:x7y2z3= (+)(+)(+)= + so from case 2:x7y2z3= (-)(+)(-)= + so B alone sufficient

And again: please read the solution above.

The red part is the reason of many mistakes on GMAT.

Square of a number is not positive, it's non-negative --> \(y^2\geq{0}\). So for (2): we know that \(x\) and \(z\) are either both positive or both negative, BUT we don't know whether \(y\) equals to zero, because if it is, then \(x^7*y^2*z^3=0\) and not more than zero.

1) bc<0 2 options : +,- and -,+ but no info about a. Not sufficient

2) ac>0 2 options : +,+ and -,- but no info about b (That could equal 0 here's the trick in my opinion). Not sufficient

1+2) with 1 we know that \(b\neq{0}\) with 2 in both cases a^8*c*3 is > 0 \(+^7*+*3>0\) \(-^7*-^3>0\) as well We are not able to say so by just looking at statement 2 because \(b\) could be \(0\), using both statement we can discard that possibility. Sufficient C _________________

It is beyond a doubt that all our knowledge that begins with experience.

1) bc<0 2 options : +,- and -,+ but no info about a. Not sufficient

2) ac>0 2 options : +,+ and -,- but no info about b (That could equal 0 here's the trick in my opinion)

1+2) with 1 we know that \(b\neq{0}\) with 2 in both cases a^8*c*3 is > 0 \(+^7*+*3>0\) \(-^7*-^3>0\) as well We are not able to say so by just looking at statement 2 because \(b\) could be \(0\), using both statement we can discard that possibility C

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