BrookG: Math help please? Train A and train B. Thank you!!!!!!?
Trains A and B are traveling in the same direction on parallel tracks. Train A is traveling at 80mph and train B is traveling at 88mph. Train A passes the station at 5:10am. If train B passes the same station at 5:40am, what time will train B catch up to train A?
Answers and Views:
Answer by Brian
This can be solved using a bit of common sense.
Since Train B is a half hour behind Train A, Train B is 1/2 * 80 = 40 miles behind Train A. Since Train B is gaining on Train A at 8mph, we see that the required time is 40/8 = 5 hours.
I hope this helps!
Answer by MthTchrIf all you need is the answer, mental math will get you there the fastest.
Between 5:10am and 5:40am, Train A travels D= RT = (80)(1/2) = 40 miles.
B gains 8 miles on A each hour after that, so it will take it 5 hours to gain 40 miles and catch up.
5:40am + 5 hours = 10:40am
BUT, if you are supposed to use algebra to solve this (one of those, “if you don’t use algebra to solve this, you’ll be lost when we get to the ones that you CAN’T solve using common sense later” kinds of things)… then you will have to do it this way.
Both trains are traveling the same distance between the station and the place where they are even.
Let D₁ = Distance of Train A
Let D₂ = Distance of Train B
Then
(a) D₁ = D₂
Substitute RT (rate x time) for each distance.
(b) R₁T₁ = R₂T₂
What do we know about the rates and times?
(c) R₁ = 80
(d) R₂ = 88
(e) T₁ = T₂ + 1/2
Fit those into the equation (b).
(80)(T₂ + 1/2) = (88)(T₂)
80T₂ + 40 = 88T₂
40 = 8T₂
T₂ = 5
Now add 5 to Train B’s beginning time:
5:40am + 5 hours = 10:40am
Leave a Reply