A, B, C walk 1km in 5 min ,8 min and 10 min respectively .C starts walking from a point at a certain time,B starts from the same point 1 min later and A starts from the same point 2 min later than C .Then A meets B and C at times

• A covers 1km in 5 min
  Speed of A is 1km/5min=1000m/300sec=10/3 in m/sec
  
  Similarly
  Speed of B is 1km/8min=1000m/480sec=25/12 in m/sec
  
  Speed of C is 1km/10min=1000m/600sec=10/6 in m/sec
  
  As A starts after 1 min of B and 2 min of C:
  Distance covered by B in 1 min=25*60/12=125m
  Distance Covered by C in 2 min=10*120/6=200m
  
  Time taken by A to meet B assuming to be x can be calculated by the below mentioned equation:
  x*10/3=125+(x*25/12)
  solving for x we get [x=100sec]Ans
  
  Time taken by A to meet C assuming to be x can be calculated by the below mentioned equation:
  x*10/3=200+(x*10/6)
  solving for x we get [x=120sec]Ans
• 7 years agoHelpfull: Yes(12) No(2)
• Speed of A is 1km/5min
  Speed of B is 1km/8min
  Speed of C is 1km/10min
  
  A nd C cover same distance when they meet, but A takes 2 min less => t/10 = (t - 2)/5
  A meets C at time => t = 4 min
  
  A nd B cover same distance when they meet, but A takes 1 min less => (t - 1)/8 = (t - 2)/5
  A meets B at time => t = 3.66 min
• 7 years agoHelpfull: Yes(3) No(1)
• 1km=1000m
  s=d/t
  A:1000/5=200m/min
  B:1000/8=125m/min
  c:1000/10=100m/min
  
  When A started B cover 1min*125s=125m And C cover 2min*100s=200m
  Difference between A and C speed:
  200-125=75
  A have to cover 125m
  In 125m A covers 75 m in 1min
  remaining 50 m covers in 40 sec=(50/75*60)convert min to sec
  =100sec
  
  Difference between A and C speed:
  200-10=100
  A have to cover 200m
  In 1min A cover 100m
  Means 200 m covers in 2mi
  =120 sec
• 6 years agoHelpfull: Yes(2) No(0)
• A meets B at 2 min 40 sec and C at 4 min.
  
  Speed of A=1/5 , B=1/8 , C=1/10 km/min.
  Here, A starts 1 min later than B and 2 min later than C.
  Let A meets B after t1 min, then Bt1 = A(t1 - 1) => (1/8)t1 = (1/5)(t1 -1)
  => t1 = 8/3 min= 2 min 40 sec.
  
  Similarly let A meets C after t2 min, then Ct2 = A(t2 - 2) => (1/10)t2 = (1/5)(t2 - 2)
  => t2=4
• 7 years agoHelpfull: Yes(1) No(2)
• A walks 1 km in 5 min.
  Then, distance covered by A in 1 min = 1000/5 = 200 m
  
  B walks 1 km in 8 min.
  Then, distance covered by B in 1 min = 1000/8 = 125 m
  
  C walks 1 km in 10 min.
  Then, distance covered by C in 1 min = 1000/101 = 100 m
  
  Let A meets B and C in x and y min, respectively.
  
  Then, according to the question,
  Distance covered by C in(x + 2) min = Distance covered by A in x min
  ⇒ 100(x + 2) = 200 x
  ⇒100 x + 200 = 200 x
  ⇒ 200 = 100 x
  ∴ x = 200/100 = 2 min
  
  Now, for A and B
  Distance covered by B in (y + 1) min = Distance covered by A in y min
  ⇒ 125(y + 1) = 200y
  ⇒ 125y + 125 = 200y
  ⇒ 125 = 200y - 125y
  ⇒ 125 = 75y
  ∴ y = 125/75 = 5/3 min
• 7 years agoHelpfull: Yes(1) No(1)
• A meet B after 112.5 sec and C after 2 min.
  
  As given in problem A=1km/5min
  that is speed of A=200m/min
  
  that of B=1km/8min
  speed of B=125m/min
  speed of C=1km/10min=100m/min
  C B A
  at 2 min 100*2=200m 125*1=125m not started yet (0 m)
  distance cover after 1 min B
  by each of them started
  
  after 3 min 300m 250m 200m
  after 4 min 400m 375m 400m
  
  
  so clearly it saw that A meet C at 2 min after when A get started.
  and A meet B at before 2 min.....
  so ....A has a speed of 200m/120sec
  so time need to cover 375m is
  (120*375)/400==112.5sec.
• 7 years agoHelpfull: Yes(0) No(1)
• speed of a,b,c respectively 1/5,1/8 and 1/10 in km/min.
  a is late by 2 min tan c-->so,(1/5)(t-2)=(1/10)*t-->t=4min--> in 4 min c travels 400m.
  similarly for b
• 7 years agoHelpfull: Yes(0) No(4)
• no way 1*8
• 3 years agoHelpfull: Yes(0) No(0)