Okay, don't know if you solved it or not, but here's a general approach to this type of problem. One things for sure; you WILL see the dreaded train question again at some point (they just love trains for some reason)!
The first thing to note is that they are asking you for the speed of both trains, so automatically you know you have two unknowns, and so you need two equations to solve it.
The part that started to confuse me was all the units, so you should quickly note what units you will be dealing with. You have miles/hr, miles, and hours involved.
Now set up what the variables will stand for; in this case, you already know that you're looking for the speeds so it's straightforward: A = speed of train A, B = speed of train B. In some cases you will solve for something they didn't directly ask for, like time, and plug it back in to get your answer.
There are two statements that give you info, so there's a good chance that each one stands for an equation. Looking at the first one:
Train A leaves the station at noon and Train B leaves the station at 2:00 going 50mph faster than Train A
The fact that train B leaves later is significant, but only for the second equation. This was a big confusing, as I was trying to find a way to incorporate this info in the first equation. However, the important thing to pull out is that train B travels 50 mph faster than train A, so the first equation is:
B = A +50. Always make sure units match; in this case, A and B are in units of mph, so adding the 50 mph here works.
Now you take the second statement:
One hour later at 3:00 Train B is 10 miles from Train A
What I saw right away is that the second equation would somehow be in the form of B=A-10. "Train B is..." translates as "B=...." But once again, units must match, so to get the equation in units of miles, you have to multiply the rate by the time (mile/hr * hr = miles). So, after train B travels for 1 hour (2:00 - 3:00), at it's rate, it will be 10 miles behind train A, which has been traveling for 3 hours (12:00 - 3:00).
B*1 hour = A*3hours - 10 miles
Two equations, two unknowns. Now substitute the first equation into the second so you will have everything in terms of A or B (doesn't matter which).
The first thing to note is that they are asking you for the speed of both trains, so automatically you know you have two unknowns, and so you need two equations to solve it.
The part that started to confuse me was all the units, so you should quickly note what units you will be dealing with. You have miles/hr, miles, and hours involved.
Now set up what the variables will stand for; in this case, you already know that you're looking for the speeds so it's straightforward: A = speed of train A, B = speed of train B. In some cases you will solve for something they didn't directly ask for, like time, and plug it back in to get your answer.
There are two statements that give you info, so there's a good chance that each one stands for an equation. Looking at the first one:
Train A leaves the station at noon and Train B leaves the station at 2:00 going 50mph faster than Train A
The fact that train B leaves later is significant, but only for the second equation. This was a big confusing, as I was trying to find a way to incorporate this info in the first equation. However, the important thing to pull out is that train B travels 50 mph faster than train A, so the first equation is:
B = A +50. Always make sure units match; in this case, A and B are in units of mph, so adding the 50 mph here works.
Now you take the second statement:
One hour later at 3:00 Train B is 10 miles from Train A
What I saw right away is that the second equation would somehow be in the form of B=A-10. "Train B is..." translates as "B=...." But once again, units must match, so to get the equation in units of miles, you have to multiply the rate by the time (mile/hr * hr = miles). So, after train B travels for 1 hour (2:00 - 3:00), at it's rate, it will be 10 miles behind train A, which has been traveling for 3 hours (12:00 - 3:00).
B*1 hour = A*3hours - 10 miles
Two equations, two unknowns. Now substitute the first equation into the second so you will have everything in terms of A or B (doesn't matter which).
