two men start together from their respective positions and start walking towards each other. one walks at one-third the speed of the other. they meet at point Y at 4 pm. if the slower one leaves 5 minutes earlier and the faster one 10 minutes later. they would meet at a point that is at a distance of 1 km from Y. At what time would they meet now?
Answers
Answer:
Step-by-step explanation:
Let say Speed of Slower = S km/minute
then Speed of Faster = 3S km/minute
Let say they started T minutes before 4PM
Distance Covered by Slower = ST
Distance Covered by Faster = 3ST
Total Distance = 4ST
Let say now they meet after M minutes of original start time
Slower one leaves 5 min before =>
Distance Covered = S(M + 5) = ST + 1
Faster one Leaves after 10 mins
=> Distance Covered by Faster = 3S(M - 10) = 3ST - 1
S(M + 5) + 3S(M - 10) = ST + 1 + 3ST - 1
=>S(4M - 25) = 4ST
=> 4M = 4T + 25
=> M = T + 6.25
=> M - T = 6.25 minutes
Time they would meet now = 4PM - T + M
= 4PM + M - T
= 4PM + 6.25
= 4 :06:15
They will meet at 4 :06:15
Additional info
3S(M - 10) - 3S(M + 5) = 3ST - 1 - 3ST -3
=> -30S -15S = -4
=> -45S = -4
=> S = 4/45 km/min = 16/3 km/hr
3S = 4/15 km/min = 16 km/hr
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