a bus left a point X for point Y.two hrs later a car left point X for Y and arrived at Y at the same time as bus.if the car and the bus left simultaneously from the opposite ends X and Y toward each other. they would meet 1.33 hrs after the start. how much time did it take the bus to travel from X to Y?

• let B and C be speed of bus and car respectively
  and time taken by car to travel from X to Y be T
  then
  equating distance
  
  B*(T+2)= C*T= (B+C)*4/3
  solving which T= 2hrs
  and time taken by bus is T+2 i.e. 4 hrs
• 10 years agoHelpfull: Yes(37) No(2)
• bus takes 4 hrs
• 10 years agoHelpfull: Yes(10) No(6)
• speed of bus=XY/t ;t=time taken by bus to cover XY
  speed of car= XY/(t+2)
  
  distance travelled by bus in 1.33hrs + distance travelled by car in 1.33hrs=XY
  
  XY/t * 1.33 + XY/(t+2) *1.33 = XY
  
  then solve the equation
• 10 years agoHelpfull: Yes(4) No(7)
• BUS TAKES 4hrs
• 10 years agoHelpfull: Yes(1) No(2)
• how to slove
  B*(T+2)= C*T= (B+C)*4/3
  and get the value of T
• 9 years agoHelpfull: Yes(1) No(0)
• To complete 1 round of XY together they take 4/3 hrs.
  To complete 2 round of XY together they must take 8/3 hrs.
  therefore t+t-2=8/3
  t=7/3=2hrs 20mins taken by bus.
• 10 years agoHelpfull: Yes(0) No(10)
• 2.66 hr
  by relative velocity concept we can cal the velocity of bus
  and it is given that a bus take t hour to travel a distance
• 10 years agoHelpfull: Yes(0) No(6)
• Choose a convenient value for the dist from x to y: 133 miles
  Let b = the bus speed
  let c - the car speed
  then
  133/b = the bus travel time
  133/c = the car travel time
  :
  "Two hours later a car left x for y and arrived at y at the same time as the bus."
  therefore
  bus time - 2 hrs = car time
  133%2Fb - 2 = 133%2Fc
  mult equation by bc, resulting in:
  133c - 2bc = 133b
  c(133-2b) = 133b
  c = %28133b%29%2F%28%28133-2b%29%29
  :
  " If the car and the bus left simultaneously from the opposite ends x and y towards each other, they would meet 1.33 hours after the start. "
  car dist + bus dist = 133 mi
  1.33b + 1.33c = 133
  
  simplify, divide by 1.33
  b + c = 100
  replace c with %28133b%29%2F%28%28133-2b%29%29
  b + %28133b%29%2F%28%28133-2b%29%29 = 100
  multiply by (133-2b) to get rid of the denominator
  b(133-2b) + 133b = 100(133-2b)
  133b - 2b^2 + 133b = 13300 - 200b
  combine like terms to form a quadratic equation
  -2b^2 + 133b + 133b + 200b - 13300 = 0
  -2b^2 + 466b - 13300 = 0
  simplify divide by -2
  b^2 - 233b + 6650 = 0
  Using the quadratic formula I got two solutions
  b = 199.7, not reasonable
  and
  b = 33.3 mph, as the bus speed
  "How much time for the bus to travel from x to y ?"
  133/33.3 = 4 hrs
  :
  :
  Check this by finding the car speed, we know: b + c = 100
  33.3 + c = 100
  c ~ 66.7 mph is the car speed
  Find the car time
  133/66.7 ~ 2 hrs (started 2 hr later and arrive at the same time)
• 7 years agoHelpfull: Yes(0) No(1)
• Let us assume C and B be speed of Car and Bus and X be the time taken by C to cover the distance.
  
  Now
  
  C*x=B(x+2)
  
  Cx/B+C=4/3
  
  B(x+2)x/2Bx+2B = 4/3
  
  Solving it you''ll get x=2
  
  Time taken by Bus=x+2=4
• 7 years agoHelpfull: Yes(0) No(0)