a man walks @ 5km/hr & reaches bus stop 5minutes late, if he walks @ 6km/hr reaches 5mts early. wht is distnce

Question

if a man walks @ 5km/hr, he reaches bus stop 5minutes late, if he walks @ 6km/hr, he reaches bus stop 5minutes early. what is distance of the bus stop? please provide me the solution with the method of solving it.

Answer 1

5(t+1/12)=X
6(t-1/12)=X
5t +5/12 = 6t -1/2

solve for t and this will be the # hours time,... then plug into either equation to get the distance in Km.

Answer 2

At 5 kmh, it takes 1/5*60 minutes to travel 1 km. (= 12 minutes)
At 6 kmh, it takes 1/6*60 minutes to travel 1 km. (= 10 minutes)

This is a total time difference of 2 minutes (5 kmh is 1 minute slow, 6 kmh is 1 minute fast).

To make a total difference of ten minutes (+5 and -5), the distance must be 5 times as long.

Therefore the distance to be traveled to the bus stop is 5 km. It takes the slower walked 60 minutes to get there, and the faster walker arrives in 50 minutes. Both started with 55 minutes to the bus' arrival time.

Answer 3

distance is 5 Kms.

let the distance be X kms and the time to arrive in time be Y minutes.
then ( use time in minutes so speed also is 1/12 per minute and 1/10 per minute.)
so distance / speed = time.

12X = Y+5
and
10X = Y-5

solve the two equations
hence 2X= 10
or X=5
and the time is 55 minutes ( to arrive without delay or too early)

Answer 4

distance in both the cases is same, so:-
5*(t+5/60)=6(t-5/60)
=>5t+5/12=6t-1/2
=>(5+6)/12=t
=>t=11/12 hrs.
=>distance = 5*(11/12+5/60)=5*(11+1)/60=5*12/60=1 km.

Answer 5

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