Here I've made a small correction to my original response:

For any multiple of 8, the value that comprises the last three digits is also a multiple of 8. From the digits given, (0, 2, 4, 6, 8) there these possibilities:
200
208
224
240
248
264
280
288
which totals 8 out of 25 (from 5 x 5), producing an answer of 8/25 (D).

However, the question does not specify whether or not x and y can be the same. For example, if x and y were to be chosen from a set of 5 cards, in using x = 6, then card 6 would not be available for y too. In this latter case, then the choice of digits is limited to:
208
240
248
264
280
where there are no repeats for the last two digits. This would result in an answer of 5/25 = 1/5 (A).

Nonetheless, as there is no mention of any restriction on x being equal to y, then it should be safe to assume that x and y could share the same value. Hence, I conclude that the correct answer is D, but I feel that GMAT should either make the question clearer, or eliminate the alternative 1/5 answer from the multiple choice. After all if you are asked to choose 2 numbers from a selection of digits, how are you to know that you may choose 2 the same?

What say you, GMAT?

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An integer with three or more digits is divisible by 8 if and only if its last three digits from a number that is divisible by 8.

The values of 2xy that are divisible by 8 are 200, 208, 224, 240, 248, 264, 280, and 288.
The total number of 6-digit integers of the form 543,2xy is equal to the number of ways of choosing two digits from 0, 2, 4, 6 and 8, which is 5*5 = 25.
Thus, the probability that 543,2xy is divisible by 8 is 8/25

Therefore, the answer is D.
Answer: D

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

In the 6-digit integer 543, 2xy, x and y are chosen from the digits 0, 2, 4, 6 and 8. What is the probability that 543, 2xy is divisible by 8?

A. $$\frac{1}{5}$$
B. $$\frac{6}{25}$$
C. $$\frac{7}{25}$$
D. $$\frac{8}{25}$$
E. $$\frac{9}{25}$$

Recall that a number is divisible by 8 if its last three digits are divisible by 8. Thus we need to care only about 2xy and ignore 543. Let's assign a digit for x and see how many digits we can assign for y such that 2xy is divisible by 8.

1) If x = 0, then y could be 0 or 8 since 200 and 208 are divisible by 8.
2) If x = 2, then y could only be 4 since 224 is the only such number divisible by 8.
3) If x = 4, then y could be 0 or 8 since 240 and 248 are divisible by 8.
4) If x = 6, then y could only be 4 since 264 is the only such number divisible by 8.
5) If x = 8, then y could be 0 or 8 since 280 and 288 are divisible by 8.

We see that there are 8 numbers that are divisible by 8. Since there are 5 choices for x and 5 choices for y, we could have a total of 5 x 5 = 25 numbers for 2xy. Since out of these 25 numbers, 8 of them are divisible by 8, the probability that 2xy is divisible by 8 is 8/25.

Answer: D