Ramu wants to catch a bus. if he walks at the rate of 5kmph, he misses the bus by 7 minutes. however, if walks at a speed of 6 kmph he reaches 5 minutes early. what shoud be his approximate speed such that he reaches the station on time
Answers
let the distance traveled to reach the station be "d" and time left in leaving of the bus be "t"
Case 1 :
speed of the person = v = 5 km/h
time taken to reach the station = t' = distance/speed = d/v = d/5
given that : t' = t + (7/60)
(d/5) = t + (7/60)
multiplying both side by "5"
d = 5 t + (7/12) eq-1
Case 2 :
speed of the person = v' = 6 km/h
time taken to reach the station = t'' = distance/speed = d/v' = d/6
given that : t'' = t - (5/60)
(d/6) = t - (1/12)
multiplying both side by "6"
d = 6 t - (1/2) eq-2
using eq-1 and eq-2
5 t + (7/12) = 6 t - (1/2)
t = (7/12 + 1/2)
t = 13/12 hours
using eq-1
d = 5 t + (7/12)
d = 5 (13/12) + (7/12)
d = 6 km
speed is given as
speed = distance/time = d/t
Speed = 6/(13/12) = 72/13 km/h = 5.54 km/h