Car A leaves city C at 5pm and is driven at a speed of 40kmph. 2 hours later another car B leaves city C and is driven in the same direction as car A. In how much time will car B be 9 kms ahead of car A if the speed of car is 60kmph

• after 2 hours
  Car A has distance from city C=2*40=80
  hence,Relative speed=60-40=20kmph
  time taken to B to goes 9 kms ahead from Car A=(80+9)/20=4.45hours
  hence Actual time=5+4.45(4 hours And 27 minutes)=9:27 P.M
  
• 9 years agoHelpfull: Yes(18) No(3)
• Let the time is T.
  60T= 40(T+2)+9
  T=4hrs 27mins
• 9 years agoHelpfull: Yes(10) No(1)
• Nobody added that 2 hour later stmt...so the ans is 9:27+2h == 11:27pm
• 9 years agoHelpfull: Yes(7) No(2)
• after 6 hrs both cars A and B be at the same place which is 240 km from the city C.for car A-since in 60 min car A travels 40 km,therefore in 1 min car A travels 2/3=0.67km.
  for carB-since in 60 min car B travels 60 km,therefore in 1 min carB travels 1 km.
  now after applying hit and trial-after 27 min the difference in distance of both the car be 9 kms.
  hence,after 6.27hrs car B will be 9 kms ahead of car A.
• 9 years agoHelpfull: Yes(2) No(3)
• B leaves at 7 pm so it will be 9km ahed after 11:27
• 9 years agoHelpfull: Yes(2) No(0)
• at 9:27 pm,at 9:00pm both will be at same pt. after that relative velocity of car B against A is 20kmph.so,time=(9/20)*60 minutes.so ,the ans is 9:27pm.
• 9 years agoHelpfull: Yes(1) No(2)
• in 2 hr A will travell =80km,
  let x be the time take by b to be 9 km ahead of a
  a/q
  80km+9km+40x=60x
  89/20 hr=4 hr 9/20*60=4hr 27 min
  ans will 7pm+4.27min=11.27min
• 9 years agoHelpfull: Yes(1) No(0)
• 4.495 hrs
  
• 9 years agoHelpfull: Yes(0) No(2)
• 4 hrs and 27 minutes
• 9 years agoHelpfull: Yes(0) No(2)
• something 4.30 min approx
• 9 years agoHelpfull: Yes(0) No(2)
• Car A -- 5pm -> 0kms
  Car A --- 11:30 pm -> 260kms
  
  car B -- 7pm -> 0kms
  car B -- 11:30pm -> 270kms
  
  so, approx.. 4.45 hrs later Car B should be 9kms ahead :)
• 9 years agoHelpfull: Yes(0) No(1)
• A= distance D, speed 40 n time T
  B= distance D+9, speed 60 n time T-2
  solving this we get T=6.45
  thus time taken by B=4.45
  
• 9 years agoHelpfull: Yes(0) No(2)
• The distance covered by a in 2 hrs is :
  
  40 × 2 = 80km
  
  Let the distance to be covered by a at the time b catches up with a be x.
  
  They take the same time to reach the meeting point.
  
  b covers 80 + x
  
  Time taken by a is :
  
  x/40
  
  Time taken by b is :
  
  (x + 80) / 60
  
  x /40 = (x + 80) / 60
  
  60x = 40(x + 80)
  
  60x = 40x + 3200
  
  60x - 40x = 3200
  
  20x = 3200
  
  x = 3200/20
  
  x = 160
  
  The time taken for them to catch up is :
  
  160/40 = 4hrs
  
  The relative distance is 9 km
  
  Relative speed is :
  
  60 - 40 = 20km/h
  
  Time = 9/20 hrs
  
  9/20 × 60 = 27 minutes
  
  The total time is 4hrs + 27mins = 4hrs 27 mins
• 5 years agoHelpfull: Yes(0) No(0)
• At 7PM A's -----> 40 * 2 = 80 Km.
  At 7PM B's -----> 0 Km.
  Total distance B has to travel to get 9 km ahead ---> 80+9 Km.
  relative speed of B w r to A --> (60 - 40) km/hr.
  Total time taken by B to get 9 km ahead ---> 89/20 hr; or 4 hr 27 min.
  since B started at 7 PM so at 7:00 + 4:27 = 11:27 PM will be 9 Km ahead of A
• 4 years agoHelpfull: Yes(0) No(0)