A bus started from bus stand at 8.00am, and after 30 minutes staying at destination, it returned back to the bus stand. The destination is 27 miles from the bus stand. The speed of the bus is 18mph. In return journey bus travels with 50% fast speed. At what time it returns to the bus stand?
As the Bus is travelling 18mph it takes 1 & 1/2 hr to reach destination. Bus stays there for 1/2 hr. As in the return journy bus travelled with the 50% fast spped meance with the speed of 27mph. meance it took 1 hr to return For total journy it took 1&1/2+1/2+1hr=3 hr. therefore 8+3 =11 Bus returns as 11 a.m.
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RE: A bus started from bus stand at 8.00am, and after 30 minutes staying at destination, it returne...
As the Bus is travelling 18mph it takes 1 & 1/2 hr to reach destination. Bus stays there for 1/2 hr. As in the return journy bus travelled with the 50 fast spped meance with the speed of 27mph. meance it took 1 hr to return For total journy it took 1&1/2+1/2+1hr 3 hr. therefore 8+3 11 Bus returns as 11 a.m.
so time- distance/speed i.e 27/18 hrs ->3/2 ---> 1.5 hrs --> bus will reach the destination at 9.30 am .
Since it waits 30 min in destination .. time (9.30+.30)-->10.00 am
so the bus starts from destination at 10.00 am
while travelling back speed is 50 higher than initiall so 18+50 0f 18 -->18+9-->27 mph so timetaken -->27/27-- 1 hr.. so it will reach back in 10.00 am + 1 hr --> 11.00 am(Ans)
RE: A bus started from bus stand at 8.00am, and after 30 minutes staying at destination, it returned back to the bus stand. The destination is 27 miles from the bus stand. The speed of the bus is 18mph. In return journey bus travels
Total time taken by it is following :
1.5 (to reach at destination) + 0.5 ( talking halt at station) + 1 (return with 50 speed)
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