Question

11. Joseph starts from his house to his office on motorcycle daily at a fixed time. If he covers
this distance at 40 km/h, he is late by 11 minutes. However, if he covers this distance a
50 km/h, he is late by 5 minutes only. Find the distance between his house and
the office.

Answer 1

Answer:

case 1 let the fixed time be t.

so s= d/t

t = d/s

case 2

s is 40

t= d/40

late by 11 minutes (11/60 hour)

so t = d/40 - 11/60....(1)

case 3

s is 50

t= d/50

late by 5 minutes ( 5/60hour)

so t is d/50 - 5/60....(2)

from 1 & 2

d/40 - 11/60= d/50 - 1/12

we get d=20 km which is the distance between his house and the office.

kindly like and rate .If any query ask it in comments....Thanks