A train covered the distance of 400 km between A and B at a certain speed. On the way back it covered 2/5 of the distance at that same speed and then it decreased its speed by 20 km/hour. Find the speed of the train at the end of its journey from B back to A, if the entire trip took 11 hours.

Answer 1

I did this for you here

http://www.jiskha.com/display.cgi?id=1458920959

hoping you would be able to find the equation yourself

then Steve finished it for you here:
http://www.jiskha.com/display.cgi?id=1458939085

Answer 2

I cant solve it though

Answer 3

400/s + 160/s + 240/(s-20) = 11

ok, hint:
multiply each term by s(s-20) to get a quadratic.
Hint: it comes out to be a nice whole number

Answer 4

400(s-20)+160(s-20)+240(s)=11(s)(s-20)
(400+160)(s-20)+240s=11(s^2-20s)

560(s-20)+240s=11s^2-220s
560s-11200+240s=11s^2-220
800s-11200=11s^2-220
−11s2+1020s−11200=0
(−11s+140)(s−80)=0
s=80? or 140/11?

It still does not work