rishab starts for a weeding venue at 6pm and drives at a speed of 60km/hr.Ramesh starts for the same venue at 6.30 pm,and drives at a speed of 75km/hr.When will both reach the venue,provided they reach at the same time.
(a)8.00pm(b)9.30pm(c)9pm(d)8.30pm

• d)
  let time to reach destination be t hours @ 6pm
  
  since distance covered is same
  therefore
  60*t=75(t-0.5)
  => t=2.5 i.e, 2 and a half hour
  hence time =6+2.5=8:30
• 9 years agoHelpfull: Yes(16) No(1)
• s1:s2=60:75 = 4:5
  => t1:t2 = 5:4
  t1-t2= 1 unit = (6:30-6:00)= 1/2 hr
  => 4 parts =4*1/2=2 hrs from 6:30 i.e, 8:30
• 9 years agoHelpfull: Yes(5) No(0)
• I have put the options in the equatin speed*time = distance.
  for rishabh.. it will be 60* (8:30-6:30) = 60*2.5 = 150km
  for ramesh, the distance remaining same and he covers it in 2hrs.. so, 75*2=150km
  
  so (d)8:30
• 9 years agoHelpfull: Yes(4) No(1)
• 8.30
  time rishab ramesh
  7 60 37.5
  8 120 112.5
  8.30 150 150
• 9 years agoHelpfull: Yes(0) No(0)
• 8.30
  time rishab ramesh
  7 60 37.5
  8 120 112.5
  8.30 150 150
• 9 years agoHelpfull: Yes(0) No(0)
• Taking distance as d which is same in both case
  =>60xd=75(x-30/60) (Ramesh is 30 mins late)
• 9 years agoHelpfull: Yes(0) No(0)