Question

If a person walks at 4/5th of his usual speed he reaches 40min late. if he walks at his usual speed for how much time does he travel

Answer 1

We know that ,
speed = distance / time
distance = speed x time
d = s x t (1)

If the person walks at 4/5th of his usual speed and he reaches 40min late , then
s1 = ( s x 4 ) / 5
t1 = t + 40
Now by putting this in formula (1)
d1 = s1 x t1
d1 = [( s x 4 ) / 5] x [t + 40]
d1 = ( 4st + 160s ) / 5 (2)
Since distance stays constant, then
By (1) and (2)
s x t = ( 4st + 160s ) / 5
5st = 4st + 160s
st = 160s
Dividing both sides by s
t = 160 min

Answer 2

Answer:

Step-by-step explanation:

Simple short cut for this model:

Numerator×change in time/|numerator-denominator|