Answer Posted / diffraction

let us assume



speed of dog D speed of hare H
leap of dog X leap of hare Y

now
a dog takes 4 leaps for every 5 leaps of a hare
let time T
in time T

dog =4X/D=T hare=5Y/H=T


3 leaps of a dog are equal to 4 leaps of the hare

3Y=4X

so

3Y/D=5Y/H=T

so D/H=3/5

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