Seeking Advice Picking Smart Numbers

[jasonthomasyee]

I hope it's alright that I post the entirety of this Manhattan CAT question:

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?

Answer: (z(y-x))/(x+y)

The explanation suggests picking x=2 (ie 15 mph), y=3 (ie 10 mph), and z=30.

After the first hour both trains have traveled a combined 25 miles, leaving a distance of 5 miles between them. Since out of these initial 25 miles X traveled 3/5 of the distance (15 miles) and X traveled 2/5 of the distance (10 miles) we expect that ratio to hold true for the remaining 5 miles. Therefore out of the 30 miles between them to start, X traveled 18 miles and y traveled 12 miles.

However, I chose numbers X=3 (20 mph), Y=4 (15 mph), Z=60. After one hour, X travels 20 miles and Y travels 15 miles for a combined 35 miles. In this case, X travels 4/7 of the 35 miles and Y travels 3/7 of the 35 miles. However, I hit a wall because I don't know what to do with the 25 remaining miles since 25 is not divisible by 7 so I can't set up a clean ratio.

My question is... How do I mitigate running into this wall? I made sure to choose small numbers for X and Y as Stacey recommends... is there another strategy I should keep in mind as well?

Thanks,

Jason

[RonPurewal]

If you can't set up a clean ratio, then go ahead and set up a dirty ratio.
Sure, the fractions might make scary faces at you, but they're just fractions. You could do them in seventh grade"”so you can do them now, too.
(:

You've already made the biggest realization here: With your numbers, the fast train covers 4/7 of the distance, while the slow train covers the remaining 3/7 of the distance. Yes. That.

I'm not quite sure, though, why you do what you do then: You split off the first 35 miles (= the first hour).
There's no reason to do that; your fractions 4/7 and 3/7 apply to any distance the trains might travel together. (If there were additional complications"”e.g., one of the trains starts before the other one"”then you may want to try an hour-by-hour breakdown, but here there's no reason to do so.)

So, with your numbers, the fast train travels 4/7 of 60 miles, and the slow one travels 3/7 of 60 miles. The difference between these is 1/7 of 60 miles, or 60/7 miles.
Plug your numbers into the answer choice and you'll get 60(1)/7. Perfect.

Who's afraid of the big bad fractions?

[RonPurewal]

By the way"”Please read the forum rules next time; you should post all answer choices that accompany a problem, even if you're only asking about one of them.