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
Answers
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]
A meets B & C after 5/3 & 2 Mins respectively
Step-by-step explanation:
A start 2 mins later than C
B start 1 mins Later than C
=> A start 1 min Later than B
Let say A meets B after t mins
Distance covered by A in t mins = Distance Covered by B in t + 1 mins
=> t * /5 = (t + 1)/8
=> 8t = 5t + 5
=> 3t = 5
=> t = 5/3
Let say A meets C after t mins
Distance covered by A in t mins = Distance Covered by C in t + 2 mins
=> t * /5 = (t + 2)/10
=> 2t = t + 2
=> t = 2
A meets B & C after 5/3 & 2 Mins
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