Jake and paul start together to walk 10kms jake walks 1.5kmph faster than paul and arrives at his destination 1.5 hours earlier than paul jakes speed walking is a)6kmph b)2kmph c)4 kmph d)8 kmph

• Let speed of jake(Sj)= x kmph. then speed of paul (Sp)=( x-1.5 )kmph.
  time difference =1.5hours
  total distance =10km
  10/Sj -10/Sp=1.5
  10/x-1.5 -10/x=1.5
  solving above equation
  x=4kmph
  ans=C
  
• 9 years agoHelpfull: Yes(13) No(0)
• ans-3.175
  ts=(t-1.5)(s+1.5)&st=10
• 9 years agoHelpfull: Yes(0) No(3)
• [1.5hr * 60 min =90 min earlier. hence 10km/90=0.1 km/ min walk faster by jake]
  for an hour, 6km/hr faster. now 10-6 = 4 kms for 90 min taken by paul .
  4km=90 min
  hence 1km = 22.5 min
  then calc for 1 hr, 60/22.5= 2.66 kmph taken by paul.
  Jake 1.5kmph faster than paul
  Finally 2.66*1.5= 4 kmph taken by Jake
• 9 years agoHelpfull: Yes(0) No(0)