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25. A bus started from bus stand at 8.00a m and after 30 min staying at destination, it returned back to the bus stand. The destination is 27 miles from the bus stand. The speed of the bus 50 percent fast speed. At what time it returns to the bus stand.
(a) 11a.m (b) 12a.m (c) 10a.m (d) 10.30p.m
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- da answer is 11 a.m. ..........let da speed of bus is V.....so it takes time in going T1=(27/V) and speed increases 50 % ,while returning time takenT2 =(27/1.5V=18/V) and it stops for 30 mins i.e., (.5 hr).therefore total time taken T=(27/V+18/V+.5).
den average speed ,V=total dist.travld upon total time
i.e.,V=2*27/T on solving we get V = 18 kmph
hence finally time taken while going =1.5hr +staying time (.5hr.)+time in returning with increased speed of 27kmph(1hr.)
hence total time taken=3 hrs. hence 11a.m.
dat's all I can do fr ur help ............. - 13 years agoHelpfull: Yes(21) No(6)
- Data is incomplete.
- 13 years agoHelpfull: Yes(8) No(10)
- some plz solve and explain tcs#M40001159
- 13 years agoHelpfull: Yes(1) No(6)
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