A bus started from bu standat 8.00a m and after 30 min staying at destination, it returned back to the bu stand. the destination is 27 miles from the bu stand. the speed of the bus 50 percent fast speed. at what time it returns to the bu stand?
Submitted by: Administratora bus cover 27miles with 18mph in=27/18 =1hr30min and it wait at stand=30min.
after this speed of return increases by 50%. so 50% of 18mph=9mph
total speed of returning=18+9=27
then in return it take 27/27=1hr.
then total time in journey=1+1.30+0.30=3hr
so it will come at 8+3hr=11am.
so ans=11am
Submitted by: Administrator
after this speed of return increases by 50%. so 50% of 18mph=9mph
total speed of returning=18+9=27
then in return it take 27/27=1hr.
then total time in journey=1+1.30+0.30=3hr
so it will come at 8+3hr=11am.
so ans=11am
Submitted by: Administrator
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