MarkusKarl wrote:

Donnie84 wrote:

I think there is something wrong with this question. The stem says x, y, and z are positive integers. I got y/z = -3 which is not possible.

The stem may have been edited. However, it now says xyz is positive. Either all three are positive or two are negative.

Posted from my mobile device Thanks for the clarification.

So we have xyz > 0. This implies either 2 are negative integers or all 3 are positive integers.

Let's look at the statements.

1. x + y = 2z

This means that z is the average of x and y.

x, z, and y can take any values. Examples -> -10, -2, 6 or 1, 1, 1 or 5, 6, 7 and so on.

St1 is insufficient.

2. 2x + 3y = z

There can be many solutions to this equation. Example ->x = 1, y = 1, z = 5 or x = 2, y = 1, z = 7 and so on.

St2 is insufficient.

St1 and St2:

x + y = 2z

2x + 3y = z

Multiply 1st equation by 2 -> 2x + 2y = 4z

Now, subtract the 2nd equation from the 1st -> -y = 3z or y = -3z or y/z = -3.

Take y = -3z and plug it in the 1st equation -> x - 3z = 2z or x = 5z or x/z = 5.

y/z = -3

x/z = 5

x:y:z = 5:3:1

Answer (C.)

This next bit is not needed to answer this question but may help in other questions.

y/z = -3

x/z = 5

Either y is negative or z is negative. As per stem, 2 integers can be negative.

Either both x and z are negative or both x and z are positive. As per stem, 2 integers can be negative or all 3 can be positive.

If y is negative, then z must be positive, and then x must be positive. This would violate the stem conditions. Rejected.

So y must be positive which would make both z and x negative. This would not violate the stem conditions. Accepted.

Experts,

I think I have the right answer but I don't think this is a good question. Though ratios are not dependent on +/-, I haven't come across an official question where ratios involve a combination positive and negative numbers. In that context, what purpose does xyz > 0 serve?