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nuthan

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  • 2 weeks later...

From (1) we either have three pos numbers or 2 neg and one pos number

From (2) we have the same idea.

 

Together we see that x and y are either both pos or both neg. if z were neg then x or y would also be neg making the expression neg overall. Therefore, C is correct.

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stmt1) Insuff.

 

stmt2)Insuff.

 

Combining 1) and 2)

 

 

Method1)

 

 

stmt 2 can be written as

 

(X*y*z)* z >0

 

which means both (x*y*z) and z are either positive or negative .

 

stmt 1) says that (x*y*z) is positive , therefore Z is also positive.

 

 

Method 2)

 

On other note , when I tried to solve equation 1 and 2 by adding them , I am not getting correct answer . can any one tell me what’s wrong with this method?

 

 

Adding stmt 1 and stmt 2.

 

xyz+xyz^2>0

 

=> xyz(1+z)>0

 

=> both xyz and (1+z) are either +ive or -ive.

 

since stmt1 tells us xyz is +ive => 1+z is also +ive

 

=> 1+z>0

 

=> z>-1

 

but in this solution value of Z could be 0 or greater then zero .

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  • 4 weeks later...

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