Question 243046: 2 trains, A and B travelling at a constant speeds left 2 different stations 640km apart. Train A started at noon and train B at 1pm. Travelling on parallel tracks towards each other, they passed each other at 4pm. If train A travelled at a speed of x km/h and train B was 20km/h slower than train A,
(i) write an expression, in terms of x, for the speed of Train B
(ii)write an expression, in terms of x, for the distance covered by train A just before it passed train B
Answer by oberobic(2304)   (Show Source):
You can put this solution on YOUR website! Fundamentally, this is a distance equation, so w start with D = RT as our basic distance equation.
Train A setup:
D = 640 - distance travelled by Train B
R = x
T = 4 hrs when it passes Train B coming from the opposite direction.
Train B setup:
D = 640 - distance travelled by Train A
R = x - 20
T = 3 hrs when it passes Train A coming from the opposite direction
(i) Speed of Train B = x - 20
(ii) Distance travelled by Train A = 4x
Since the moment they pass each other they have covered the 640 km distance. We can take our solution farther by substituting what we know.
4x + 3(x-20) = 640
4x + 3x - 60 = 640
7x = 700
x = 100
So Train A is going 100 km/hr.
We are told Train B is going 20 km/hr slower, so it is going 80.
We can check the solution by substituting these values. And we can calculate answers to questions that have not (yet) been asked.
Train A ran 100 km/hr for 4 hrs, so it had travelled 400 km at the moment it passed Train B.
Train B ran 80 km/hr for 3 hrs, so it travelled 240 km at the moment it passed Train A.
When they pass each other, their combined distance travelled logically has to be 640, which it is.
So, we're done.
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