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Re: Inequality Help! [#permalink]
21 Jun 2011, 11:32

1

This post received KUDOS

I cannot think of scenario where statement A is not alone sufficient. Statement A should alone give the value of exp as either +ve or -ve. So should be sufficient.

As per state statement 2, the value of the expression can be either 0 or -ve or +ve, so it is not sufficient.

i remember this Q i read somewhere and it was required to determine - x(y+z) >=0 , in that case the both statements will be required and OA should be C.

Last edited by agdimple333 on 21 Jun 2011, 11:45, edited 1 time in total.

Re: Inequality Help! [#permalink]
21 Jun 2011, 11:08

Hmm I don't get the OA..

If xyz is not equal to zero it means neither value, x, y or z can be zero.

1) If we pick values for y and z, -2 and 0,5, then we get 1,5 = 2,5 which obviously doesn't fit. If we take -2 and -0,5 we get 2,5 = 2,5 which fits. In order to fit this y and z must therefore have the same sign. Therefore it can either be negative or positive. Still, whether x(y+z)=0, now depends on the value of x. We can see from the question that if xyz is not equal to zero it means x is also not zero. So, if x can't equal zero, x(y+z), no matter the value of y+z, can not equal zero. Why is it C then, anyone?

Re: Inequality Help! [#permalink]
27 Jun 2011, 10:10

agdimple333 wrote:

I cannot think of scenario where statement A is not alone sufficient. Statement A should alone give the value of exp as either +ve or -ve. So should be sufficient.

As per state statement 2, the value of the expression can be either 0 or -ve or +ve, so it is not sufficient.

i remember this Q i read somewhere and it was required to determine - x(y+z) >=0 , in that case the both statements will be required and OA should be C.

I agree with this. Anyone thinks otherwise? Please give me an example where A alone fails.

If \hspace{2} xyz \ne 0, \hspace{1} is \hspace{3} x(y+z)=0?

1) \hspace{4} |y+z| = |y|+|z|

2) \hspace{4} |x+y| = |x|+|y|

Given : xyz not equal to 0 therefore x not equal to 0 , y not equal to 0 and z not equal to 0.

from condition 1 : |y+z| = |y|+|z| therefore both y and z are either positive or negative but not zero(this is given), therefore (y+z) will never be zero. This condtion is not sufficient beacuse if x=0 then x(y+z) becomes '0' else x(y+z) is non zero.

from condition 2 : |x+y| = |x| + |y| therefore x and y are either positive or negative but not zero(this is given) , therefore neither x =0 nor y = 0 . This condtion is not sufficient because what if z= -y then x(y+z) = 0.

From condition 1 and 2 neither x=0 nor y=-z (because they both are either positive or negative). Hence we can say that x(y+z) will never be zero.

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