speed of the bus including stoppages = 45 kmph
Loss in speed when including stoppages = 54 – 45 = 9kmph
=> In 1 hour, bus covers 9 km less due to stoppages
Hence, time that the bus stop per hour = time taken to cover 9 km
Speed and time are inversely proportional.
Hence new time = 7/6 of usual time
Hence, 7/6 of usual time – usual time = 12 minutes
=> 1/6 of usual time = 12 minutes
=> usual time = 12 x 6 = 72 minutes
= 1 hour 12 minutes
the distance travelled by it when it meets the slower train be y miles. Time taken by the faster train to cover y miles
= Time taken by the slower train to cover (287-y) miles
= 3.5 hours (y/x) = (287-y)/(x-6) = 3.5 Solving, x = 44 miles/hr
If ‘d’ is the distance between X and Y, then ‘d’ is the distance between Y and Z.
Now the total time for the batsman to row from X to Z is 4 hours. Therefore, time to row from X to Y is 2 hours.
Also the time for the boats man to row from X to Y and back is 10 hours. Hence, time required to row from Y to X is 8 hours.
If, a: speed of boats man in still water
b: speed of the river
d/(a + b) = 2; d/(a – b) = 8
2*(a + b) = 8*(a – b)
a + b = 4a – 4b
3a = 5b
a:b = 5:3
usual speed be x kmph
usual travel time be t hours
distance to office be d km d=xt d=(0.8x)*[t+(10/60)] xt=0.8xt+(2x/15) Solving,
t=40 minutes