the average speed of a train is 20% less on the return journey than on the onward journey.the train halts for half an hour at the return journey.if the total is 23 hours,covering a distance of 1000 km, the speed of the train on the return journey is...............
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speed =x
t=distance/speed
return speed=80/100*x=4/5*x
22(1/2)=1000/x+(1000*5/4x)
45/2=1000((1/x)+(5/4x))
45/2=1000(9/4x)
x=100 km/h
t=distance/speed
return speed=80/100*x=4/5*x
22(1/2)=1000/x+(1000*5/4x)
45/2=1000((1/x)+(5/4x))
45/2=1000(9/4x)
x=100 km/h
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