Question

A bus covered the 400-km distance between points a and b at a certain speed. on the way back the bus traveled at the same speed for 2 hours and then increased the speed by 10 km/hour until it reached a, thus spending 20 fewer minutes on the return trip. how long did the return trip take?

Answer 1

The speed going to B is x, the time t and the distance, 400 km.

xt=400, x kph and t in hours

on the way back, the 2x+(x+10)(t-7/3), where it was total time minus 2 hours already driven minus 1/3 of an hour, the 20 minutes.

2x+xt-(7/3)x+10t-70/3=400

2x+400-(7/3)x+10t-70/3=400, 400 cancels, t=400/x

-(1/3)x+4000/x=70/3

multiply by x

-(1/3)x^2-(70/3)x+4000=0

x^2+70x-12000=0

x=(1/2)(-70+/- sqrt 52900); sqrt term=230

x=80 kph

took 5 hours to go there

on way back 2 hours at 80 kph=160 km

2h40m at 90kph=90*8/3=240 km

They add to 400 km.

The return trip took 4h40m

Answer 2

Let the speed going to B be x, the time taken be t and distance is 400 km
xt=400,
On the way back, the 2x+(x+10)(t-7/3), where it was total time minus 2 hours already driven minus 1/3 of hour, the 20 minutes.
2x+xt-(7/3)x+10t-70/3=400
2x+400-(7/3)x+10t-70/3=400
thus
-(1/3)x+4000/x=70/3
multiplying through by x we get:
-(1/3)x^2-(70/3)x+4000=0
x^2+70x-12000=0
x=(1/2)(-70+/-sqrt52900)
x=80 kmp
thus:
it took 5 hours to make the first trip
on way back 2 hours at 80 kph which means he traveled for 160 km at this speed
2 h 40 min at 90 k/h=90*8/3=240 km