Select the correct answerWalking 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?Options 318 214 314

To solve this problem, we can use the concept of relative speed and time.

Step 1: Let's assume the usual speed of the man is S (km/h) and the usual time he takes to reach his destination is T (hours).

Step 2: According to the problem, when the man walks at 60% of his usual speed, he takes 1 hour and 40 minutes (which is 1.67 hours) more than his usual time to reach his destination.

Step 3: We know that time is equal to distance divided by speed. So, we can set up the following equation according to the problem:

T + 1.67 = Distance / (0.6S)

Step 4: We also know that under normal circumstances (at his usual speed), the time the man takes to reach his destination is T. So, we have another equation:

T = Distance / S

Step 5: Now we have a system of two equations, and we can solve it by substitution or elimination.

From Step 4, we get Distance = TS. Substituting this into the equation from Step 3, we get:

T + 1.67 = TS / (0.6S)

Step 6: Simplifying this equation, we get:

T = 2.78 hours

So, the man usually takes 2.78 hours to reach his destination.

Looking at the options, the closest one is 2.78, which is not listed. There seems to be a mistake in the options provided.