Q. A man walking at 45 km/hr, late to to the school by 5 min and walking at 54 km/hr, early to the school by 2 mins, with what speed he should walk to reach the school on time?

• Ans is 51.08 Km/hr.
  Lets assume the man takes time T if he walks with speed S.
  if he walks with 45km/hr then time=T+5/60 (in hours) and
  if he walks with 54km/hr then time=T-2/60 (in hours)
  
  Now 45(T+5/60)=54(T-2/60)====>by solving this----T=37/60 hrs.
  Now Distance=45(T+5/60)=====>D=63/2 KM
  
  Now speed he should walk to reach school on time S=D/T===>(63/2)/(37/60)
  By solving this---S=51.08 km/hr.
• 11 years agoHelpfull: Yes(3) No(0)
• (54-45)=9, from late by 2 mins to 5 mins early swing of 7 because (-2 to 5). So now dividing 9/7 we get 1.28 approx. Now we can follow either of the two steps 1. late by 5 mins so 5*1.28=6.4 add this to 45 we get 51.4 or early by 2 mins so substract 2*1.28=2.56 from 54 54-2.56=51.4(approx)
• 11 years agoHelpfull: Yes(0) No(0)