A man jogs at 6 mph over a certain journey and walks over the same route at 4 mph. What is his average speed for the journey?
(a) 2.4 mph (b) 4 mph (c) 4.8 mph (d) 5 mph

• we know x=6mph y=4mph
  so average journey=2xy/(x+y)=48/10=4.8
• 13 years agoHelpfull: Yes(55) No(0)
• 4.8 mph
  
  suppose he travels 24 miles in one side.
  he takes 4 hrs for jogging and 6 hrs for walking total distance of 48 miles.
  Then av speed =48/10=4.8 mph
• 13 years agoHelpfull: Yes(25) No(5)
• (2*6*4)/(6+4)=4.8
• 13 years agoHelpfull: Yes(11) No(1)
• avg speed= 2xy/ (x+y);
  2*6*4/ (6+4)= 4.8.
  if two different people have their different speed than we calculate avg speed by = (x+y)/2;
  6+4 /2=5.
  
• 13 years agoHelpfull: Yes(7) No(3)
• t1=x/6 & t2=x/4
  t1+t2=t=x/6 + x/4
  hence,avg speed=(x/t)=2.4 mph
• 13 years agoHelpfull: Yes(3) No(24)
• (2*x*y)/(x+y)=28
• 13 years agoHelpfull: Yes(1) No(9)
• given a man walking from one point to another point at a certain speed s1km/hr and then walking back the same route with a speed s2 km/hr then average speed for the journey is
  
  average speed = (2s1s2)/(s1+s2)
  i.e = (2*6*4)/(6+4)
  = 4.8 mph
  therefore c is the rt answer
• 9 years agoHelpfull: Yes(1) No(0)
• avg speed= 2xy/ (x+y);
  2*6*4/ (6+4)= 4.8.
  if two different people have their different speed than we calculate avg speed by = (x+y)/2;
  6+4 /2=5.
  
• 13 years agoHelpfull: Yes(0) No(3)
• v1v2/v1+v2=
  2.4
• 9 years agoHelpfull: Yes(0) No(2)
• when distance is constant,
  avg speed=2*s1*s2/s1*s2;
  avg speed=2*6*4/6+4;
  =4.8
• 8 years agoHelpfull: Yes(0) No(0)
• Average speed=(2*x*y)/(x+y)
  4.8
• 5 years agoHelpfull: Yes(0) No(0)