barroblock wrote:Is x * y^2 * z^3 > 0?

1) xy>0
2) xz>0

What do you think is the answer, B or C?

S2 isn't sufficient on a bit of a technicality. Since, using S2 alone, y could be zero, it's possible that x * y^2 * z^3 = 0, and possible that x * y^2 * z^3 > 0. So we can't be sure of the answer to the question.

When we use S1, we know y can't be zero, and so the two statements together are sufficient: xz, y^2 and z^2 must all then be positive, so their product, x * y^2 * z^3, must be positive.

Ian Stewart wrote:

barroblock wrote:Is x * y^2 * z^3 > 0?

1) xy>0
2) xz>0

What do you think is the answer, B or C?

S2 isn't sufficient on a bit of a technicality. Since, using S2 alone, y could be zero, it's possible that x * y^2 * z^3 = 0, and possible that x * y^2 * z^3 > 0. So we can't be sure of the answer to the question.

When we use S1, we know y can't be zero, and so the two statements together are sufficient: xz, y^2 and z^2 must all then be positive, so their product, x * y^2 * z^3, must be positive.

this is a good solution, but it's a bit up in the clouds: there is essentially 0 chance that your average test-taker is going to think to factor (x)(y^2)(z^3) into xz times y^2 times z^2.
just not gonna happen.

and, more importantly, we don't want to convince students that this problem (which really isn't that hard) depends on such an obscure, try-hard factorization.

so, here's a more ground level solution:

--

statement (1)
means that x and y have the same sign.
we don't know anything about z.

since z is raised to an odd power, reversing the sign of z will reverse the sign of the overall product.

so, insufficient.

--

statement (2)
insufficient for exactly the reasons raised by ian.

--

together:
we know that x and y have the same sign, and are nonzero (from statement 1)
we know that x and z have the same sign, and are nonzero (from statement 2)

therefore, all three of x, y, z have the same sign.

this leaves only 2 possibilities, so just enumerate those possibilities:

* all negative: (neg)(neg^2)(neg^3) = (neg)(pos)(neg) = pos.
* all positive: (pos)(pos^2)(pos^3) = pos.

so, always positive.
sufficient.

ans (c)

lunarpower wrote: this is a good solution, but it's a bit up in the clouds: there is essentially 0 chance that your average test-taker is going to think to factor (x)(y^2)(z^3) into xz times y^2 times z^2.
just not gonna happen.

Let's have just a bit more faith in the 'average test-taker'. There's nothing 'in the clouds' about what I've done above; I'm only using the fact that, if y is not zero, then y^2 is positive, a fact that's tested throughout the GMAT. So if y isn't zero, you can divide by y^2 on both sides of an inequality without worrying about reversing the inequality; you're dividing by a positive number. That's quite a normal thing to do, and in the question above, I didn't spend much time explaining that step because I didn't think that part was what might trap most test takers - I actually did the step in reverse, but the upshot is the same.

Knowing that x^2, x^4, x^6, etc are always positive except when x = 0 lets you rephrase any inequality with products and large powers. For example, if x, y and z are nonzero, and you're asked:

Is x^3 > x^2 ?

we can rephrase the question by dividing by x^2 on both sides:

Is x > 1?

That makes life a lot easier. We can divide by any even power, not only x^2, so if we're asked

Is x^9 > x^8?

We can divide by x^8 to again rephrase the question:

Is x > 1?

And if we have a product, like the following:

Is (x^9)(y^8)(z^7) > 0 ?

We can divide on both sides by x^8, y^8 and z^6, to get the simpler question

Is xz > 0?

And to take one more example, if a question asks if (x^99)*(y^98) > (x^98)(y^99), you can divide by x^98 and by y^98 to rephrase the question as 'is x > y ?'

In each case, we translate what appears to be a complicated question into a straightforward one. The above assumes none of our letters can be zero, which was the only 'trap' in the question in the original post. That's not a trap you often see in these types of questions on the GMAT. I don't think any of this is beyond the capabilities of the 'average test taker'.

Ian Stewart wrote:So if y isn't zero, you can divide by y^2 on both sides of an inequality without worrying about reversing the inequality; you're dividing by a positive number. That's quite a normal thing to do, and in the question above, I didn't spend much time explaining that step because I didn't think that part was what might trap most test takers

ah, i see. i'd thought you had just pulled that out of nowhere, like a "magic factorization".

makes sense.