If a man covers a distance of 20 km. at the speed of 5 km./hr. then he reaches 40 min late. If he will walk with a speed of 8 km./hr., how much time will he reach before than the scheduled time ?

Options

A. 30 min
B. 40 min
C. 50 min
D. 60 min

Answer

To solve the problem, we need to determine the scheduled time for the journey and then find out how much earlier the man will arrive if he walks at a speed of 8 km/hr instead of 5 km/hr.

### Step 1: Calculate the time taken at 5 km/hr
The distance covered is 20 km and the speed is 5 km/hr.

Using the formula:
\[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} \]
we can calculate the time taken:
\[ \text{Time} = \frac{20 \text{ km}}{5 \text{ km/hr}} = 4 \text{ hours} \]

### Step 2: Determine the scheduled time
Since the man arrives 40 minutes late, we need to convert 40 minutes into hours:
\[ 40 \text{ minutes} = \frac{40}{60} \text{ hours} = \frac{2}{3} \text{ hours} \approx 0.67 \text{ hours} \]
Thus, the scheduled time (the time he was supposed to take) is:
\[ \text{Scheduled Time} = \text{Actual Time} - \text{Late Time} \]
\[ \text{Scheduled Time} = 4 \text{ hours} - \frac{2}{3} \text{ hours} = \frac{12}{3} \text{ hours} - \frac{2}{3} \text{ hours} = \frac{10}{3} \text{ hours} \approx 3.33 \text{ hours} \]

### Step 3: Calculate the time taken at 8 km/hr
Now, we calculate the time taken if he walks at a speed of 8 km/hr:
\[ \text{Time} = \frac{20 \text{ km}}{8 \text{ km/hr}} = 2.5 \text{ hours} \]

### Step 4: Determine how much earlier he arrives
To find out how much earlier he arrives compared to the scheduled time, we subtract the time taken at 8 km/hr from the scheduled time:
\[ \text{Time Difference} = \text{Scheduled Time} - \text{Time at 8 km/hr} \]
\[ \text{Time Difference} = \frac{10}{3} \text{ hours} - 2.5 \text{ hours} \]
Converting 2.5 hours into a fraction:
\[ 2.5 \text{ hours} = \frac{5}{2} \text{ hours} = \frac{15}{6} \text{ hours} \]
Now, converting \(\frac{10}{3}\) to a common denominator of 6:
\[ \frac{10}{3} = \frac{20}{6} \]
Now we can subtract:
\[ \text{Time Difference} = \frac{20}{6} - \frac{15}{6} = \frac{5}{6} \text{ hours} \]
Converting \(\frac{5}{6}\) hours back to minutes:
\[ \frac{5}{6} \text{ hours} = \frac{5 \times 60}{6} \text{ minutes} = 50 \text{ minutes} \]

### Final Answer
Thus, if he walks at a speed of 8 km/hr, he will arrive **50 minutes earlier** than the scheduled time.

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