The average speed of a man running a lap is 150 mph. He covers first two fifth of the lap at a speed of 123 mph and the second two fifth distance by 164 mph, so what is the time taken by him to cover the remaining one fifth distance?

• (123+164+x)/3=150;
  123+164+x=450;
  287+x=450;
  x=163
• 8 years agoHelpfull: Yes(44) No(15)
• Let us take the time taken to cover the reamaining one fifth as 'T'
  Since time is inversely proportional to speed,
  1/150= {(2/5)/123} + {(2/5)/164} + T
  On simplifying, we get T=0.002439 hrs = 0.002439*60*60 = 8.78 seconds.
• 8 years agoHelpfull: Yes(27) No(9)
• (123+164+x)/3 = 150
  287+x = 450
  x=173 mph
• 8 years agoHelpfull: Yes(19) No(61)
• Answer: 205 mph.
  
  I have had a lot of queries on this puzzle and I feel that an more detailed answer is required. Many methods have been suggested but only the one below gives the correct answer.
  
  If we take the number of miles in a lap to be x miles and using time = distance / speed.
  The first 2/5 are covered at 123 mph. Therefore, the time taken was [(2/5) * x] / 123 hours.
  The next 2/5 are covered at 164 mph. Therefore, the time taken was [(2/5) * x] / 164 hours.
  The last 1/5 is covered at y mph. Therefore, the time taken was [(1/5) * x] / y hours.
  The entire trip is covered at an average of 150 mph. we can conclude that:
  2x 2x x x
  ----- + ----- + --- = ---
  5*123 5*164 5*y 150
  
  Working this out we get:
  
  1 -2 -2 1
  -- - --- - --- = -
  30 123 164 y
  
  The means that y = 1 / (1/30 - 2/123 - 2/164)
  Therefore y = 205
• 8 years agoHelpfull: Yes(14) No(5)
• let as assume average speed to be 'x' and x=150
  1st 2/5 lap covered at a speed of 123 mph so it can be taken as 2/5*123
  2nd 2/5 lap covered at a speed of 164 mph so it is 2/5*164
  remaining 1/5 lap can be taken as x/5
  so,2/5*123+2/5*164+x/5=150
  on solving this we obtain x=176
• 8 years agoHelpfull: Yes(13) No(12)
• (2*123+2*164+x)/5=150;
  246+328+x=750;
  574+x=750;
  x=176;
• 8 years agoHelpfull: Yes(5) No(2)
• avg speed=150
  fifth distance=x;
  (123+164+x)/3=150
  287+x=450
  x=450-287
  x=163
• 8 years agoHelpfull: Yes(3) No(4)
• 176
  
  (150*5) - ((123*2)+(164*2)) = 176
• 7 years agoHelpfull: Yes(3) No(1)
• time=distance/speed
  speed: (123+164+x)/3=150
  287+x=450
  x=163
  time=((1/5)/163) *60*60 seconds
  4.42 seconds
• 6 years agoHelpfull: Yes(3) No(3)
• time=distance/time
  2x/(5*123) + 2x/(5*164) + x/(5*y) = x/150
  multiply by 5
  2x/(123) + 2x/(164) + x/(y) = x/30
  divide by x
  2/123 + 2/164 + 1/y = 1/30
  y=205mph
• 5 years agoHelpfull: Yes(1) No(0)
• ans is (1000/48)
• 8 years agoHelpfull: Yes(0) No(13)
• (123+164+x)/3=150
  287+x=450
  x=163 mph
• 8 years agoHelpfull: Yes(0) No(5)
• this will be an easy way
  say the total length is 1 unit
  it is div into 3 units (2/5, 2/5, 1/5)
  2/5=123, 2/5=164, needed is 1/5
  so
  1/5+2/5+2/5=1
  so 1/5+164+123=150*3
  ans is 163
  mul by 3 is for avg
• 8 years agoHelpfull: Yes(0) No(4)
• ((123+164+x)/3) = 150
  287+x=450
  x=450-287
  x=163
  x=173mph
• 7 years agoHelpfull: Yes(0) No(1)
• ((123*2)+(164*2)) = 574(first 4 lap)
  150*5 =750 (total)
  750-574=176 mph (last lap)
• 6 years agoHelpfull: Yes(0) No(1)
• we cannot tell the time it is a wrong question...we can tell the speed though
  205mph
• 6 years agoHelpfull: Yes(0) No(2)
• let the time taken for the last 1/5th dist be x hrs
  total= 123+164+x/3=150
  287+x/3=150
  287+x/3=450
  x=173
• 5 years agoHelpfull: Yes(0) No(1)
• Average speed =150
  => (123 + 164 + x)/3 =150
  =>287 + x = 450
  => x = 163
• 5 years agoHelpfull: Yes(0) No(0)
• The Question is for "TIME" not the "SPEED" of the final lap.
  
  since t=d/s
  and avg Speed = 150 mph
  total distance be 5*d.
  => total speed = 5* 150
  (avg speed = ((123*2)+2(164)+final lap speed)/5)
  let T bet time of final lap
  
  =>
  total time taken
  5/(5*150)={(2/5)/123}+{(2/5)/164}+T
  1/150=(2/615)+(2/820)+T
  T= 0.000975 hours
  T= 0.000975 *60*60 sec = 3.512 sec
  
  I think it's better to convert h to sec in first place
• 3 years agoHelpfull: Yes(0) No(0)