George walks 36 Kms partly at a speed of 4kms per hour and partly at 3 km per hour. If he had walked at a speed of 3km per hour when he had walked at 4 and 4 Km per when he had walked at 3 he would have walked only 34 kms . The time (in hours ) spent by George walking was:

• he question can be solved with the help of two equations...
  4x+3y=36----equ(1)
  3x+4y=34----equ(2)
  by solving these two equations
  x=6 and y=4
  so total time is 10hours.
  
• 8 years agoHelpfull: Yes(16) No(0)
• let he walked "t1" time with the speed of 3kmphr and for "t2" 4kmphr then
  3t1+4t2=36..........(1)
  now according to the question, 4t1+3t2=34......(2)
  (1)-(2)
  t1-t2=2
  put this value in (1)
  get answer t1+t2=10hr
• 8 years agoHelpfull: Yes(2) No(0)
• We have to hit nd trial. We cn make george walking 36 km either by 4*3+3*8=36 or 4*6+3*4=36. Bt d condition given is satisfied by only the 2nd.. Tht means 3*6+4*4=34. So the total time taken =(6+4)=10 hr
• 8 years agoHelpfull: Yes(1) No(0)
• 4 t1 + 3 t2 = 36
  3 t1 + 4 t2 =34
  On solving , t1=6 and t2=4
  So, total time = t1+ t2= 6 +4= 10 hours
• 8 years agoHelpfull: Yes(0) No(0)
• 4x + 3y= 36......(1)
  3y + 4y= 34......(2)
  (1)+(2) gives
  7(x+y)=70
  Therefore
  x+y=10
• 8 years agoHelpfull: Yes(0) No(0)