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by simplyjat » Wed Feb 13, 2008 12:44 am
These are confusing question. People tend to think A/B/C/D only if the answer is yes; i.e. only if the two statements prove that x(y+z) = 0. But you have to look at the trap here; even if you are able to deduce that x(y+z) != 0 , you know the answer.

Statement 1 can be interpreted as "y and z" have same sign...
Statement 2 can be interpreted as "x and y" have same sign...

Combining these two, we know that x, y, & z; all have the same sign, and from the stem you know that none of them is zero. So it can be deduced that the result will never be zero...
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by gabriel » Wed Feb 13, 2008 4:57 am
simplyjat wrote:These are confusing question. People tend to think A/B/C/D only if the answer is yes; i.e. only if the two statements prove that x(y+z) = 0. But you have to look at the trap here; even if you are able to deduce that x(y+z) != 0 , you know the answer.

Statement 1 can be interpreted as "y and z" have same sign...
Statement 2 can be interpreted as "x and y" have same sign...

Combining these two, we know that x, y, & z; all have the same sign, and from the stem you know that none of them is zero. So it can be deduced that the result will never be zero...
If the question is "is x(y+z)=0" then A should be enough to answer the question.

We know that x,y,z is not equal to 0, so the only way x(y+z) = 0 is if y+z is equal to 0 for which y should be = -z.

From statement 1 we know that y and z are of the same sign, that means y+z is not 0. hence we can conclude that x(y+z) is not equal to 0.

But I am not sure what the question is, I dont see any "=" in the equation. Magical Cook can u take a look at the question again.

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by magical cook » Wed Feb 13, 2008 8:13 am
sorry, it's supposed to be "x (y + z) >= 0 ?" so it could be 0> or >0 since we just know all of the sign are the same? that's my theory to get E...
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