Max@Math Revolution wrote:[GMAT math practice question]

$$Is\ \frac{x}{yz}>0？$$

$$\left(1\right)\ yz\ >\ x^2$$
$$\left(2\right)\ x\ <\ y+z$$

I tried to solve it but I don't know if it is well.

The question $$\frac{x}{yz} > 0$$ is equivalent to $$xyz > 0 \ ？$$ $$\left(1\right)\ yz\ >\ x^2$$ implies that yz have the same sign because $$if\ \ yz>x^2\ \ \ then\ \ \ \ \ yz\ge0\ \Rightarrow\ \ \ both\ are\ positive\ or\ both\ are\ negative.$$ But we don't know anything about x. Then it is not Sufficient.

$$\left(2\right)\ x\ <\ y+z$$ it doesn't say anything about the sign of each variable. INSUFFICIENT.

Using both statements together we get that: y+z has the same sign that y (and z). Hence, we have two cases:

If y and z are negative, then y+z is negative and therefore x is negative. Hence xyz is not greater than 0.

If y and z are positive then y+z is positive, but it doesn't tell us anything about the sign of x. Therefore, is not sufficient.

After all these cases, the conclusion is that the answer is the option E.

I don't know if this is the best way to solve it.

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Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 3 variables (x, y and z) and 0 equations, E is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2) together:
If x = 1, y = 2 and z = 3, then the answer is 'yes'.
If x = -1, y = 2 and z = 3, then the answer is 'no'.

Since we don't have a unique answer, both conditions together are not sufficient.

Therefore, E is the answer.

Answer: E

Vincen wrote:

Max@Math Revolution wrote:[GMAT math practice question]

$$Is\ \frac{x}{yz}>0？$$

$$\left(1\right)\ yz\ >\ x^2$$
$$\left(2\right)\ x\ <\ y+z$$

I tried to solve it but I don't know if it is well.

The question $$\frac{x}{yz} > 0$$ is equivalent to $$xyz > 0 \ ？$$ $$\left(1\right)\ yz\ >\ x^2$$ implies that yz have the same sign because $$if\ \ yz>x^2\ \ \ then\ \ \ \ \ yz\ge0\ \Rightarrow\ \ \ both\ are\ positive\ or\ both\ are\ negative.$$ But we don't know anything about x. Then it is not Sufficient.

$$\left(2\right)\ x\ <\ y+z$$ it doesn't say anything about the sign of each variable. INSUFFICIENT.

Using both statements together we get that: y+z has the same sign that y (and z). Hence, we have two cases:

If y and z are negative, then y+z is negative and therefore x is negative. Hence xyz is not greater than 0.

If y and z are positive then y+z is positive, but it doesn't tell us anything about the sign of x. Therefore, is not sufficient.

After all these cases, the conclusion is that the answer is the option E.

I don't know if this is the best way to solve it.

There is a flaw in your logic.

x/yz > 0 is NOT equivalent to xyz > 0.

If yz > 0, then x/yz > 0 is equivalent to xyz > 0.
But, if yz < 0, then x/yz > 0 is equivalent to xyz < 0.