In a kilometre race, if A gives B a 40 m start, A wins by 19 s. But if A gives B a 30 s start, B wins by 40 m. Find the time taken by B to run 5,000 m?
a. 150 s
b. 450 s
c. 750 s
d. 825 s

• 750 secs
  
  When B is given a 40 m start then B runs 960m in T secs and A runs 1000m in T-19 secs.
  When B is given a 30 second start then B runs 1000m in t secs and A runs 960m in t - 30 secs.
  Assume that the speed of A (Va) and B (Vb) is constant in both situations.
  Then Vb = 960 ÷ T = 1000 ÷ t, and Va = 1000 ÷ (T - 19) = 960 ÷ (t - 30)
  So 960t = 1000T, and 1000t - 30000 = 960T - 18240. Then 40t = -40T + 11760, t = -T + 294.
  Substituting for t in 960t = 1000T
  gives : -960T + 282240 = 1000T
  1960T = 282240
  T = 144
  B 's speed is 960 / T = 960 / 144 = 6.67 metres per second.
  Assuming the same constant speed to run 5000 metres then the time taken is
  5000 / 6.67 = 750 seconds
• 13 years agoHelpfull: Yes(67) No(18)
• 1)if A gives B a 40 m start, A wins by 19 :-
  distance=while A runs 1000m,B runs 960m
  time=A time is unknown;B takes 19s more than A to cover 960m
  speed=Let 'a' be speed of A ; 'b' be the speed of B
  
  Time=Distance/Speed
  Time taken by A=1000/a;Time taken by B=960/b
  from 1)=> 1000/a + 19 = 960/b
  
  2)if A gives B a 30 s start, B wins by 40 m
  while B runs 1000m , A runs 960m
  B takes 30s more than A to cover the 1000m
  
  Distance of A=1000m ; Distance of B=960m
  Time taken by A and B is unknown.B takes 30s more than A
  speed of A='a' ;B='b'
  
  Time taken by A=960/a ; Time taken by B=1000/b
  From 2)=>960/a + 30 = 1000/b
  
  1)=> 1000/a = 960/b -19 ; 2) 960/a = 1000/b - 30
  Divide 1/2 => 1000/960 = 960 - 19b / 1000 - 30b
  25/24 = 960 - 19b / 1000 - 30b
  25(1000 - 30b) = 24(960 - 19b)
  
  solving,we get b=20/3
  sub b in 1)
  1000/a = 960/(20/3) - 19
  a = 8
  Time taken by b to cover 5000m = 5000/b
  = 5000/(20/3)
  = 750 seconds
  
  
• 10 years agoHelpfull: Yes(18) No(3)
• b. 750 s
  EA gives B a 40metre start then B runs 960m in T secs and A runs 1000m in T-19 secs.
  
  2.A gives B a 30 second start then B runs 1000m in t secs and A runs 960m in (t-30) secs.
  Assuming speed of A=X and B=Y
  Then
  X = 960 ÷ T = 1000 ÷ t,
  960t = 1000T
  Y = 1000 ÷ (T - 19) = 960 ÷ (t - 30)
  1000t - 30000 = 960T - 18240
  Then 40t = -40T + 11760,
  t = -T + 294.
  Substituting for t in 960t = 1000T gives
  -960T + 282240 = 1000T
  1960T = 282240
  T = 144
  B 's speed is 960 ÷ T = 960 ÷ 144 = 6.67 metres per second.
  to run 5000 metres the time taken is 5000 ÷ 6.67 = 750 seconds = 12 minutes 30 seconds
• 13 years agoHelpfull: Yes(10) No(15)
• 750 s by the mathematical order.
• 13 years agoHelpfull: Yes(5) No(18)
• 750 s by the mathematical order.
• 10 years agoHelpfull: Yes(2) No(4)
• When B is given a 40metre start then B runs 960m in T secs and A runs 1000m in T-19 secs.
  When B is given a 30 second start then B runs 1000m in t secs and A runs 960m in t - 30 secs.
  Assume that the speed of A (Va) and B (Vb) is constant in both situations.
  Then Vb = 960 ÷ T = 1000 ÷ t, and Va = 1000 ÷ (T - 19) = 960 ÷ (t - 30)
  So 960t = 1000T, and 1000t - 30000 = 960T - 18240. Then 40t = -40T + 11760, t = -T + 294.
  Substituting for t in 960t = 1000T gives : -960T + 282240 = 1000T : 1960T = 282240 : T = 144
  B 's speed is 960 ÷ T = 960 ÷ 144 = 6.67 metres per second.
  Assuming the same constant speed to run 5000 metres then the time taken is 5000 ÷ 6.67 = 750 seconds = 12 minutes 30 seconds.
• 11 years agoHelpfull: Yes(1) No(5)
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• 1 year agoHelpfull: Yes(0) No(0)
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• 1 year agoHelpfull: Yes(0) No(0)