Luke53 wrote:Can't solve this one, so please help.
You can see how to do this kind of problem by looking here
Luke53 wrote:A train is travelling from A to B. After it travelled one hour, it stops and then continues it's voyage with a speed that is equal of 3/5 of the initial speed. It arrives at B with a delay of 3 hours . If the train would have stopped 50 km further than the point where it first stopped, it would have arrived 1h 40 min early. Find the initial speed and the distance between A and B.
So pick a variable for the original (fast) speed, like 's'. The distance is some value 'd'.
Make an expression
with the variable so the expression means 'three-fifths of fast speed'.
If there had been no stop, then the time (from 'd=rt') would have been t=d/s.
Now take the two parts of the actual trip. The first part is 1*s = s. The rest of the distance is d-s. The rate will be (3/5)s. Then the time will be the first hour, plus the time (from 'd=rt') for the rest, which is (d-s)/((3/5)s). Since this is three hours more than usual, the time is d/s + 3. Use this to make an equation.
Do the same sort of thing for the 'if the train had stopped sooner' part. This will give you two equations with two variables, so you can solve.
If you get stuck, please reply with what you've done so far. Thanks!