Please solve the following question:
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Two tourists start cycling towards each other from two points A and B. The first tourist leaves point A 6 hours later than the second tourist leaves B.They meet after a certain time and the first tourist has covered 12 Km less than the second tourist at the time of their meeting.After their meeting they keep cycling and the first tourist arrives at B 8 hours later and the second tourist arrives at A 9 hours later. Find the speed of the faster tourist.
a) 4 kmph
b) 6 kmph
c) 9 kmph
d) 2 kmph
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Please solve the above question in the shortest way possible such that the solving it takes the least time possible.
Thanks

• assume,
  t1=x+6 and t2=x
  d1=y-12 and d2=y
  given that after meeting A covers y km in 8 hrs and B covers y-12 km in 9 hrs
  therefore,
  SA=y/8 and SB=(y-12)/9
  by using both the equation we find that
  y=8SA and y=9SB+12
  8SA-9SB=12
  which implies A is more faster than B
  now take the inter values and get the speed of A
  it is 2km/hr
• 11 years agoHelpfull: Yes(1) No(1)
• speed of faster tourist is 6 km/hr
• 11 years agoHelpfull: Yes(1) No(0)
• Answer is 6kmph. A0B Here 0 is the meeting point and B has covered x+12km and A has covered x km (12km less)then given that A is covering (x+12)km in 8hrs and B is covering xkm in 9hrs. => (x+12)/S1=8 and x/S2=9 where S1 and S2 are speeds of A and B respectively. From 2nd eq. x=9*S2 plugging this value in 1st eq.(9*S2+12)/S1=8 =>9*S2+12=8S1=> 8S1-9S2=12 => S1>S2 This eq holds for S1=6 and S2=4 so, the speed of the faster one is=6kmph.
• 11 years agoHelpfull: Yes(1) No(0)