If you don't know how to solve a question that has variables in the answers, you don't always have to do the algebra. One approach is to plug in, to build the model as you want it to be.
The variables are
x=hours it takes the fast train to travel z miles.
y=hours it takes the slow train to travel z miles.
z=distance between towns.
Let's say that the fast train travels at 10 miles per hour.
Let's stay that the slow train travels at 5 miles per hout.
Let's say that they will meet each other after 2 hours. This means that when they meet each other, the fast train has traveled 20 miles and the slow train 10 miles. Therefore the total distance is 30 miles; z=30
Since the fast train travels at 10mph, the time it takes to travel the total distance is 3hrs. x=3.
Since the slow train travels at 5mph, the time it takes to travel the total distance is 6hrs. y=6.
The question asks how many more miles will the fast train have traveled when they meet. Since the fast train travels 20 miles and the slow train travels 10 miles to the meet, the answer is 20-10= 10 more miles.
Go through the answers looking for 10. Remember that x=3, y=6, z=30.
(A) z(y-x)/x+y = 30(3)/9 = 10.
(B) z(x-y)/x+y = a negative number (x-y is negative)
(C) z(x+y)/y-x = 30(9)/3 = too big.
(D) Doesn't even use z
(E) Doesn't even use z
Pick A.
