A and B run 1 k.m race.if A gives B a start of 50meter,A wins by 14 sec and if A gives B a start of 22 sec, B wins by 20 meter.find the time taken by A to run 1k.m...?

• So here the only given quantity is 14 sec.So express equations in terms of time.
  Let a and b be the speed of A and B respectively.
  Given,
  When A gives B a head start of 50m,
  950/b-1000/a=14-----(1)
  The second condition:
  980/a+22=1000/b----(2)
  Let 1/a=x 1/b=y
  then,
  950y-1000x=14
  980x+22=1000y--->1000y-22=980x
  solving we get..
  a=10m/s
  time taken by A to run 1000km=1000/10=100s
• 10 years agoHelpfull: Yes(40) No(3)
• Let times taken by A and B to run a km, be x and y, respectively.
  When B gets a start of 50 m, B runs
  1000 – 50 = 950 m while A runs 1000 m.
  (950/1000)y - x = 14,
  i.e. 0.95y - x = 14
  and, When B gets a start of 22 sec, A runs (y - 22) sec, while B runs for y sec.
  1000-(1000/x)(y-22)=20
  50y - 49x = 1100.
  3.45y = 414
  y = 120 sec.
  x = 0.95 x 120 - 14 = 100 sec.
  so the answer is 100 sec
• 10 years agoHelpfull: Yes(8) No(7)
• could you plz inform me with the latest TCS 2013-2014 criteria?
• 10 years agoHelpfull: Yes(5) No(11)
• in first case,
  suppose time taken by A is 'T'.
  After B is '50m' ahead in beginning ,still B takes '14sec' more than 'T' to reach destination.
  That means,time taken by B to reach from '50m' to '1000m' is 'T+14'.
  That means,time taken by B to travel 950m is 'T+14'.
  So ,total time taken by B from '0m to 1000m' = 'T+14+(time taken to travel 50m );
  which is ='T+14+{(T+14)*50)/950};
  In second case ,
  B is '22sec' ahead.
  Now A reaches in 'T sec'.
  In the same 'T sec' B reaches '20m' ahead.
  So the time taken to travel '1000m' by B =[22+T-{time taken by B to travel '20m'
  
  So the time taken to travel '1000m' by B={22+T-(20*(T+14)/950)};
  
  [time taken by B to travel '20m' is achieved from case 1];;;
  
  equating eqn in case 1 and 2 ,we get
  
  T=102.5
  
  
• 10 years agoHelpfull: Yes(5) No(6)
• the answer is 100 sec
  From the first part we calculate the speeds of A and B
  if in t sec A covers 1000 m then B covers 950 m in (t+14) sec
  thus A(speed)= 1000/t
  B(speed)=950/(t+14)
  It is further given that A gives B a head start of 22 sec. B covers 950 m in t+14 sec thus in 22 sec it will cover (950*22)/(t+14). To complete the race B has to cover 1000-(the distance covered in 22 sec as calculated earlier). and this time will be equal to the time taken by A to cover 980 m.
• 10 years agoHelpfull: Yes(3) No(6)
• Plz explain me properly.bcoz here first is camparision between distance n time then in time and distance..i m totally blank to solve this
• 10 years agoHelpfull: Yes(2) No(0)
• 950B-1000A=14
  980A+22=1000B
  
  SOLVE ABOVE EQUATIONS AND YOU GET A=10;
  SO A=1000/10= 100M
• 7 years agoHelpfull: Yes(2) No(0)