It takes the high-speed train x hours to travel the z miles from Town A to Town B at a constant rate, while it takes the regular train y hours to travel the same distance at a constant rate. If the high-speed train leaves Town A for Town B at the same time that the regular train leaves Town B for Town A, how many more miles will the high-speed train have traveled than the regular train when the two trains pass each other?
z(y – x) / (x + y)
z(x – y) / (x + y)
z(x + y) / (y – x)
xy(x – y) / (x + y)
xy(y – x) / (x + y)
Haha wow... this one is nuts to try and solve with variables. I sat at my desk for about 15min... not proud of that.
I'm not the biggest fan of using numbers for rate problems, but this is def one of those problems.
Took me 1min 15 sec when I did it this way.
Thus rate of the fast train is 50 and that of the slow train is 20.
t= time for both
100-d = 500/7
Answer is 300/7.
Just plug away. I started with A:
100(5-2)/(5+2) --> 300/7 Heyyyyyyyyy look at that theres our answer.
Stupid problem >:-(