12 Days of Christmas GMAT Competition - Day 10: If x and y are

Interesting question.

Rewrite target as y=-5/3x+500

I) if x+y=500, it will only be equal to the target at the point x=0. Therefore, since x and y are non-zero integers, this will never occur given the constraints.
Sufficient.

II) y=20+x
20+x=-5/3x+500
8/3x=480
x=180

Therefore, there is a possible intersection of x and y, but we cannot determine this with the given info. yes and no answer, therefore B is insuff.

Let's deal with this one from the task itself. What does it mean when the given equation equals to 10?
\(\frac{x}{30} + \frac{y}{50} = 10\)

\(\frac{5x}{150} + \frac{3y}{150} = 10\) |*150

\(5x+3y=1500\)

Okay then, so now let's look at the conditions.

(1) \(x+y=500\)
Well, if we multiply this equation by 3, we will get: \(3x+3y=1500\)
And if we subtract it from the original equation, we get:

\(5x+3y=1500\)
-
\(3x+3y=1500\)
_________________
\(2x = 0\)

From the task we know that X must be non-zero. Therefore, this scenario is simply impossible!
So, Condition 1 is sufficient by itself, because we can give a definite answer 'no' to the original question.

(2) \(y–x=20\)
Unfortunately, this one is neither here nor there. We can, of course, just select such X and Y that the original question is answered as 'yes':
\(5x+3y=1500\) and \(y=20+x\)

So, \(5x + 3(20+x) = 8x + 60 = 1500\)

Then \(x = \frac{1440}{8} = 180\), and \(y=200\)
However, as far as Condition 2 goes, X can also equal to 1, and Y can equal to 21, with which we'll get nowhere near the answer above.
Therefore, Condition 2 is insufficient by itself.

The answer is A.

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