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?
Accepted Solution
A:
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