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

• Qn is not clear.
  
  What is slow speed and what is fast speed ?
  Wht is duration/distance for fast and slow speed /
• 13 years agoHelpfull: Yes(12) No(2)
• The ques is incomplete...the speed of 18mph is also given
  bus started after 30min ie .5hrs
  Now,
  Time=distance/speed ie Time=27/18 during going ie 1.5 hrs
  now during the return journey speed increasd by 50% so new speed 27mph
  Time=27/27 ie 1(RETURN journey)
  so total time=1+.5+1.5=3hrs
  so the bus will reach at 11am
• 13 years agoHelpfull: Yes(10) No(4)
• It is not mentioned that speed is 50% more in return journey.
  
  Qn is incomplete.
  
  One can make assumption depending on options given and solve according to his understanding.
  which is not correct.
• 13 years agoHelpfull: Yes(4) No(0)
• here a/c to question the time in going is half of time in coming because distace is same so a/c to the given options only option c is satisfying the relation for 1hr going and half an hr coming time so the total time consume is 2 hr so c is the right ans. otherwise none of the rest options can match or question is incomplete
• 13 years agoHelpfull: Yes(2) No(0)
• Speed is not given question is incomplete.....
  
• 8 years agoHelpfull: Yes(0) No(0)
• 28a+30b+31c=365
  28*1+30*4+31*7=365
  therefore a=1,b=4,c=7(ac to days in months)
  so a+b+c
  1+4+7=12.
  
• 8 years agoHelpfull: Yes(0) No(0)