Walking at 60% of his usual speed, a man reaches his destination 1 hour 40 minutes late. His usual time (In hours) to reach the destination is?

To solve this problem, we need to understand that speed and time are inversely proportional to each other. This means that if the speed decreases, the time taken to cover the same distance increases and vice versa.

Let's denote:

• The usual speed of the man as 's' (in km/h or any other unit)
• The usual time taken to reach the destination as 't' (in hours)
• The slower speed of the man as 0.6s (since he is walking at 60% of his usual speed)
• The time taken to reach the destination at the slower speed as 't + 1.67' (since he is 1 hour 40 minutes late, which is approximately 1.67 hours)

Since the distance to the destination remains the same, we can set up the following equation using the formula distance = speed * time:

s * t = 0.6s * (t + 1.67)

We can simplify this equation by dividing both sides by 's' (since 's' is not equal to zero):

t = 0.6 * (t + 1.67)

Next, we can solve this equation for 't':

t = 0.6t + 1

0.4t = 1

t = 1 / 0.4

t = 2.5 hours

So, the man's usual time to reach the destination is 2.5 hours.