Q. A man started to walk from Town A to Town B at 1200 hrs, and his friend started at two P.M. on the same day to walk from B to A. They met on the road at five minutes past four o'clock, and each man reached his destination at exactly the same time. Can you tell at what time they both arrived at their destination?

• 7 PM
  
  A...................X.................B
  
  Suppose the person starting from A started with speed A and person starting from B started with speed B.
  
  They met at X at five minutes past four o'clock,.
  Each man reached his destination at exactly the same time say after x mins from 5 mins past 4 PM.
  So 245*A= B*x .................(1)
  and
  A*x= B*125 .................(2)
  
  Dividing (2) with (1), we get
  
  x/245 = 125/x
  
  or
  x^2 = 245 *125
  
  x= 175 mins
  
  so they met after 175 mins( 2 hr 55 mins) from 4:05 means
  
  they met at 7 PM.
  
• 14 years agoHelpfull: Yes(3) No(1)
• 7 PM
• 14 years agoHelpfull: Yes(0) No(2)
• Both of them would reach their destination at 6 PM.
  Reason: At 4.05 PM % of distance covered by first man and his friend is 58.33 (245 min*100/420 min) and 41.67 (125 min*100/300 min) respectively which would amount to 100%. ie, traveled time so far *100 / total traveled time.
• 14 years agoHelpfull: Yes(0) No(2)
• they will reach the destination by 6:10 pm
• 14 years agoHelpfull: Yes(0) No(1)
• 6:10 pm
• 14 years agoHelpfull: Yes(0) No(1)
• speed of A=x,speed of B=y.
  total dis =4.05x+2.05y=49/12x+25/12y
  let at u p.m they reach the destination
  ux=49/12x+25/12y
  and ux=(u-2)y
  by solving we get u=7p.m
• 14 years agoHelpfull: Yes(0) No(0)
• 1810 hrs
• 14 years agoHelpfull: Yes(0) No(1)
• they will reach the destination by 7 pm
  
• 10 years agoHelpfull: Yes(0) No(0)