Jake and Paul each walk 10 km. The speed of jack is 1.5 faster than paul speed . Jack reaches the destination 1.5 hrs before paul .Then Jake's speed is equal to

• Let pouls speed be xkmph
  Then jacks speed is (x+1.5)kmph
  (10/x)-(10/1.5+x)=1.5
  X=2.5
  Jacks speed=2.5+1.5=4kmph
  
  
  
  
  
• 10 years agoHelpfull: Yes(17) No(5)
• Let speed of jake = x kmph.
  So, speed of paul = ( x - 1.5 ) kmph.
  Time difference =1.5hours
  Total Distance = 10km
  10/(x-1.5) -10/x = 1.5
  By solving the above eq.
  x = 4 kmph
• 7 years agoHelpfull: Yes(4) No(0)
• (10/x)-(10/1.5x)=1.5
  =4.44
• 9 years agoHelpfull: Yes(2) No(4)
• Let speed of Paul = x km/h
  then , speed of Jack = (x + 1.5 ) km/h
  
  a/c to question, Jack reaches the destination 1.5 hrs before Paul.
  e.g., time taken by Jack + 1.5hrs = time taken by Paul
  
  time taken by Jack = distance covered by Jack/speed of Jack
  = 10km/(x + 1.5) hrs
  
  time taken by Paul = distance covered by Paul/speed of Paul
  = 10km/x hrs
  
  now, 10/(x + 1.5) + 1.5 = 10/x
  
  => 1.5 = 10/x - 10/(x + 1.5)
  
  => 1.5 = 10{(x + 1.5 - x)}/x(x + 1.5)
  
  => 1.5 = 10 × 1.5/(x² + 1.5x)
  
  => x² + 1.5x = 10
  
  => x² + (3/2)x - 10 = 0
  
  => 2x² + 3x - 20 = 0
  
  => 2x² + 8x - 5x - 20 = 0
  
  => 2x(x + 4) -5(x + 4) = 0
  
  => (x + 4)(2x - 5) = 0
  
  x = -4 , 5/2
  
  hence, speed of Paul = 2.5 km/h
  so, speed of Jack = x + 1.5 = 2.5 + 1.5 = 4 kmph
• 5 years agoHelpfull: Yes(2) No(1)
• (10/x)-(10/1.5+x)=1.5 derive this line
• 7 years agoHelpfull: Yes(0) No(2)