A man traveling with 4545 of his usual speed reaches his distance 70 minutes late. Find the usual time he takes to reach his destination.4 hours423423hours445445hours345345hours

The question seems to be a bit jumbled, but I'll try to answer it based on a common problem in time, speed, and distance problems.

Let's assume the man's usual speed is 's' and the usual time he takes to reach his destination is 't'.

According to the problem, when the man travels at 4/5 of his usual speed, he reaches his destination 70 minutes late.

We know that speed = distance/time. Therefore, distance = speed * time.

In both scenarios (travelling at usual speed and at 4/5 of usual speed), the man travels the same distance. So, we can set up the following equation:

s * t = (4/5)s * (t + 70/60)

We can cancel out 's' from both sides, which leaves us with:

t = (4/5) * (t + 70/60)

Solving this equation will give us the usual time 't' the man takes to reach his destination.

Let's solve it:

5t = 4t + 470/60 5t - 4t = 470/60 t = 4*70/60 t = 280/60 t = 4.67 hours

So, the man usually takes approximately 4.67 hours to reach his destination.