Jack and Jill went up and down a hill. They started from the bottom and Jack
met jill again 20 miles from the top while returning. Jack completed the race
1 min a head of jill. if the hill is 440 mts high and their speed while down
journey is 1.5 times the up journey. how long it took for the Jack to
complete the race ?

• (20 miles is greater than 440 meter. Assuming that height of the hill is 440 miles )
  
  Let speed of Jack when going up = x miles/minute
  and speed of Jill when going up =y miles/minute
  
  Then speed of Jack when going down = 1.5x miles/minute
  and speed of Jill when going up =1.5y miles/minute
  
  Case 1 : Time taken for Jack to travel 440 miles up and 20 miles down
  = Time taken for Jill to travel 440-20=420 miles up
  
  440/x + 20/1.5x = 420/y
  68/1.5x = 420/y
  68y = 63x
  y = 63x/68 ---(1)
  
  Case 2 : Time taken for Jack to travel 440 miles up and 440 miles down
  = Time taken for Jill to travel 440 miles up and 440 miles down - 1
  
  440/x + 440/1.5x = 440/y + 440/1.5y - 1
  1100/1.5x = 1100/1.5y - 1
  1100/1.5x = 1100/(1.5 * 63x/68) - 1
  1100/1.5x = 68*1100/(1.5 * 63x) - 1
  1100/1.5x - 68*1100/(1.5 * 63x) = -1
  1100/1.5x (1 - 68/63) = -1
  1100/1.5x (-5/63) = -1
  1100/1.5x = 63/5
  x = 1100*5/(1.5*63) = 11000/189
  
  Time taken by Jack to complete the race = 440/x + 440/1.5x
  = 1100/1.5x = 1100/(1.5*11000/189) = 189/15 = 12.6 minutes
• 9 years agoHelpfull: Yes(49) No(0)
• Let speed and time of Jack be s1 & t1 & Jill having s2 & t2.
  
  Time taken for Jack to travel 440 miles up and 20 miles down
  = Time taken for Jill to travel 440-20=420 miles up
  
  440/s1 + 20/1.5s1 = 420/s2
  68/1.5s1= 420/s2
  68s2= 63s1
  s2 = 63s1/68 ---(1)
  
  Now from the second given condition- Jack reaches 1 min before Jill. So
  t2-t1=1
  880/s2 - 880/s1 = 1
  1/s2 - 1/s2 = 1/880
  from equation (1) put value of s2
  Solving and putting value of s1 in t1=d/s1 (d=880)
  t1 = 12.6 min
  
• 9 years agoHelpfull: Yes(5) No(1)