A person left a city P to go to city Q at 6.30 a.m. He travelled at a speed of 90 km/hr for 2
hours and 30 minutes. After that, the speed was reduced to 50 km/hr. If the distance between
the cities is 345 km, at what time did he reach the city Q?
Answers
Answered by
1
With the speed of 90kmph distance covered is 90* 2.5= 225km
number of km left =345-225=120
with speed of 50kmph time taken =120/50= 2.4 i.e 2hrs and 24mins
so total time of travel= 4hours and 54mins
Answered by
0
Answer:
total distance= 345km
speed for first 2hr 30 min = 90km/h
therefore distance remaining = 345- (90×2.5)=345-225
=120km
speed after 2.5 hr= 50km/h
therefore time taken to travel 120km with speed of 50km/h =120÷50= 2.4hr
total time = 2.5hr + 2.4hr
=2hr 30min + 2hr 24min
=4hr 54min
Thank you
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