Since we're given some solid restrictions on what x and y could be, picking numbers is a quick and easy strategy.

x>y>1, and x and y are integers. Let's try x=3 and y=2. So z=x/y = 3/2

Looking at statements I and II, we can see that they are mutually exclusive. z cannot be both > and < a number at the same time. So right away, we know that either I or II must be true, but never both. So B is out.

Now, using our numbers for x and y, we can set up a little chart to see what happens when we add or subtract 1 from the numerator and denominator.

x-----y-----z-----(x+1)/(y+1)-----(x-1)/(y-1)3-----2----3/2--------4/3--------------2/1

------------------------<z---------------->z

We can see that when 1 is added to the numerator and denominator, the value decreases, and when it is subtracted, the value increases.

Looking at the statements then, we can see that II and III are true, and I is not.

Answer: C

Now, in general, it is good to know what happens to fractions when you add or subtract a constant from both the numerator and denominator. As long as we're dealing with positive numbers, adding a constant on top and bottom of a fraction will cause the value of the fraction to move closer to 1, and subtracting a constant from the top and bottom a fraction will cause the value of the fraction to get farther from 1. Notice I didn't say increase or decrease, because that will depend on whether the fraction is greater than or less than 1.

Examples:

\(\frac{20-10}{100-10} < \frac{20}{100} < \frac{20+100}{100+100}\)

\(\frac{50+50}{10+50}< \frac{50}{10} < \frac{50-5}{10-5}\)

So in general:

If x>y --> \(\frac{x+a}{y+a} < \frac{x}{y} < \frac{x-a}{y-a}\)

And if x<y --> \(\frac{x-a}{y-a} < \frac{x}{y} < \frac{x+a}{y+a}\)

In our case, x>y, so adding a constant will decrease the fraction (statement III) and subtracting a constant will increase the fraction (statement II).

_________________

Dave de Koos

GMAT aficionado