Writing into math equations would solve the problem


#1

t1- start time of car from home (same in any case)
t2- reaching time of car at home (in regular case)
x- distance walked by joe
w- speed of hoe walking
c- speed of car

in normal case:
2d/c = (t2-t1) —(1)
6+d/c = t2 -----(2)

when he reaches 1hr early
2(d-x)/c=t2-t1-20min
substituting eq-1
x/c=10min ----(3)
5+x/w+(d-x)/c = t2-20 ----(4)
from eq-2 and 4
5+x/w+d/c-x/c = 6+d/c-20min
substituting value of x/c from eq-3
5+x/w-10min = 6-20min
x/c=1-20min=50min
x/c=time he walked
there fore, answer is 50min