X runs 3/2 times as fast as Y. If X gives Y a starts of 300 meter, how must X run before he cateches up with Y?

• let speed of y = z
  then speed of x = 3z/2
  if y runs i meter before x catches up
  then x have to run i + 300 meter
  equating the time we get :
  (300+i)/(3z/2)=i/z
  so i==600 meters
• 10 years agoHelpfull: Yes(2) No(0)
• let speed of X be sx, & speed of Y be sy. then, sx/sy = 3/2. let X catches Y after running x m. then,the time at which X run (300+x) m, at the same time, Y will run x m. speed=dis/time => time= dis/speed.
  =>sx/sy = (300+x)/x. =3/2.=> x=500m answer.
• 10 years agoHelpfull: Yes(1) No(3)
• Ans- 900
  Y=300 meter so X=3/2*Y=3*300/2=450
  So Y is already 300 meter so next 300 meter X=450
  So y=300+300+300 and X=450+450
  So X cateches Y 900 meters
• 10 years agoHelpfull: Yes(1) No(1)
• let y speed is k m/s then X have 3/2 k m/s
  let Y run x meter and they met
  so time is equal but X run 300 +xand Y only x meter
  so x/k = (300+x)* 3/2k
  ans is 600 m
• 10 years agoHelpfull: Yes(0) No(0)