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A, B and C walk 1 km in 5 min, 8 min and 10 min, respectively, C starts walking from a points 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 meet B and C at times ?
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- 2 min, 3 min
- 4/3 min, 3 min
- 2 min, 5/3 min
- 1 min, 2 min
Correct Option: C
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.
