. A man drives at a speed of 40 miles/hr. His wife left 30 mins. late with 50 miles/hr speed. when will they meet ?
Ans. 2 hours

• Ans. 2 hrs.
  In 30 minutes the man would have droven 20 miles, when his wife starts. Now both are driving in the same direction. So speed of wife related to man is 10(50-40) miles/hr. So wife will cover 20 miles in 2 hours to meet the man.
• 12 years agoHelpfull: Yes(19) No(2)
• min M W
  30 20 0
  60 40 25
  90 60 50
  120 80 75
  150 100 100
  
  ans =150-30=120min = 2hrs
  
• 9 years agoHelpfull: Yes(8) No(0)
• Ans. 2 1/2 hr.
  
  Suppose man meets woman after x hrs. driving.
  In x hrs. he would have drived a distance of 40x miles.
  His wife left 1/2 hrs late. So she would have driven a distance of 50(x-1/2) miles in (x-1/2) hours to meet the man.
  As both of them will meet so these both distances will be same.
  So 40x = 50(x-1/2) => 10x = 25 => x = 5/2 hrs. = 2 1/2 hrs.
• 10 years agoHelpfull: Yes(7) No(9)
• Man would have drove 20 miles in 30 mins ie: when his wife starts.
  
  W--------------M--------------------meeting point
  
  Equating their time
  
  (X+20)/50 = X/40
  solving we get X=80
  
  therefore D/S= 80/40 = 2 hours
  
  
• 7 years agoHelpfull: Yes(2) No(1)
• in 30 min(1/2 hour) A covers=40*1/2=20 miles.
  wife will meet at= 20 /relative speed=20/10=2 hours.
• 7 years agoHelpfull: Yes(2) No(0)
• 60-40
  1-40/60
  30-40*30/60
  in 1/2 hr man cover 20 miles
  relative speed=50-40=10miles/hr.
  s=d/t=t=d/s=20/10=2hr
• 10 years agoHelpfull: Yes(1) No(3)