Persons A and B residing 5 kilometres apart decide to meet on the way by walking towards each other’s residence. A
walks at the speed of 5 km/hr while B does so at 4 km/hr. A starts walking 10 minutes after B has started. For how long will
B walk before meeting A?
Answers
Solution :-
→ Speed of B = 4 km/h
So,
→ Distance covered by B in 10 minutes = S * T = 4 * (10/60) = (2/3) km.
then,
→ Distance left to be covered by both before meeting = 5 - (2/3) = (13/3) km .
→ Usual speed { opposite direction } = 5 + 4 = 9km/h .
→ Time taken to m e e t = D/S = (13/3)/9 = (13/3) * (1/9) = (13/27) hours .
therefore,
→ Distance covered by B before meeting A = S * T = 4 * (13/27) = (52/27) km .
hence,
→ Total distance covered by B = (2/3) + (52/27) = (18 + 52)/27 = (70/27) km (Ans.)
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