a man walks at 40kmph and reaches the distance in 10 min less than that of usual . by walking at 60 kmph he reaches 10 minutes earlier. the distance covered by him is.?

• let x be distance in km ,let t be usual time
  given 40km/hr=40/60km/min=2/3km/min 10 min less t+10
  given 60km/hr=60/60km/min=1km/min 10 min early t-10
  since speed=distance/time
  2/3=x/t+10.......eq1
  1=x/t-10.....eq2
  divide eq 1&2
  2/3 = t-10/t+10 => t=50min
  x=50-10=40km answer
• 12 years agoHelpfull: Yes(2) No(0)
• Total distance=x
  x/40-x/60=20/60
  (3*x-2*x)/120=1/3
  x/120=1/3
  x=40 km
• 8 years agoHelpfull: Yes(1) No(0)