Official Solution:


If \(xyz < 0\), is \(x < 0\)?

\(xyz < 0\) implies that either all three unknowns are negative or one is negative and the remaining two are positive.

(1) \(x - y < 0\). This means that \(x < y\). Could \(x\) be negative? Yes, if \(x\), \(y\), and \(z\) are negative. Could \(x\) be positive? Yes, if \(x\) and \(y\) are positive and \(z\) is negative. Not sufficient.

(2) \(x - z < 0\). Basically the same here. We have that \(x < z\). Could \(x\) be negative? Yes, if \(x\), \(y\), and \(z\) are negative. Could \(x\) be positive? Yes, if \(x\) and \(z\) are positive and \(y\) is negative. Not sufficient.

(1)+(2) We have that \(x < y\) and \(x < z\). Could \(x\) be positive? No, because if \(x\) is positive then from \(x < y\) and \(x < z\), both \(y\) and \(z\) must be positive but in this case \(xyz\) will be positive not negative as given in the stem. Therefore \(x\) must be negative. Sufficient.


Answer: C
_________________

Lovely question ! That's the reason I do these gmatclub tests ! Questions cover all the fundamentals.

Hi, sorry, i keep getting confused with the concept of shifting variables from one side of the equation to the other - when to do it and when to not without affecting the inequality - please could you help me out here

Hi,
you can always add the variables on either side of the equation and rather the correct method is to get them on one side and then move ahead..
\(xy - y > 2y +7......\)would become \(xy-y-2y>7.........y(x-3)>7......\)

Do not cross multiply or divide teh variables on two sides unless you knoe they are positive..
\(x^2>x.............. x^2-x>0 ......... x(x-1)>0..........\)now either BOTH x and x-1 are +ive or BOTH are -ive...
Youi cannot convert \(x^2>x ..............as ... x>1\)
_________________

Thank you - very helpful!

Why can't we consider x as the lone negative in this being less than y and z (where y and z are positive) thus xyz < 0

Hi, why can´t all three be negative?

xyz<0 implies that odd number of variables are negative, so either 1 or all three.

(1) \(x - y < 0\)
Just tells us that x<y.
Possible that z<0<x<y or z<x<y<0.
Insufficient

(2) \(x - z < 0\)
Just tells us that x<z.
Possible that y<0<x<z or y<x<z<0.
Insufficient

Combined
We know x is less than y and less than z, so x is the least.
Now xyz<0 means only one or all three are negative.
Be it any of the two possibilities, the LOWEST will surely be negative. Thus, x<0.
Sufficient


C
_________________