To solve this problem, we'll first analyze the given information and set up the necessary equations.

Step 1: Establish the equations for walking

Let the distance from the man's house to his office be $D$ kilometers.

• Case 1: When the man walks at 8 kmph, he is 10 minutes late.
• Case 2: When the man walks at $8 + 4 = 12$ kmph, he is 4 minutes early.

Let $t$ be the scheduled time to reach the office.

Case 1: Walking at 8 kmph

The time taken to walk the distance $D$ at 8 kmph is $\frac{D}{8}$.

Given that he is 10 minutes late, the equation is: $\frac{D}{8} = t + \frac{10}{60} = t + \frac{1}{6} \text{ hours}$

Case 2: Walking at 12 kmph

The time taken to walk the distance $D$ at 12 kmph is $\frac{D}{12}$.

Given that he is 4 minutes early, the equation is: $\frac{D}{12} = t - \frac{4}{60} = t - \frac{1}{15} \text{ hours}$

Step 2: Solve the equations

We have two equations: $\frac{D}{8} = t + \frac{1}{6}$ $\frac{D}{12} = t - \frac{1}{15}$

Subtract the second equation from the first: $\frac{D}{8} - \frac{D}{12} = \left(t + \frac{1}{6}\right) - \left(t - \frac{1}{15}\right)$

To simplify, find a common denominator for the left side: $\frac{3D - 2D}{24} = \frac{1}{6} + \frac{1}{15}$ $\frac{D}{24} = \frac{5 + 2}{30} = \frac{7}{30}$

So, the distance $D$ is: $D = 24 \times \frac{7}{30} = \frac{168}{30} = 5.6 \text{ km}$

Step 3: Find the scheduled time $t$

Using $D = 5.6$ km in either equation, we can find $t$. Let's use the equation from Case 1: $\frac{5.6}{8} = t + \frac{1}{6}$ $0.7 = t + \frac{1}{6}$ $t = 0.7 - \frac{1}{6} = 0.7 - 0.1667 = 0.5333 \text{ hours} = 32 \text{ minutes}$

Step 4: Determine the speed of the motorcycle

Let the speed of the motorcycle be $v$ kmph. The time taken by the man to reach the office by walking to the meeting point and then riding the motorcycle is exactly 32 minutes.

Since the man walks 8 kmph, the time to meet the motorcycle rider is $t_1$, and the time taken by the motorcycle to cover the remaining distance is $t_2$. We know: $t_1 + t_2 = 32 \text{ minutes} = \frac{32}{60} \text{ hours}$

Let $x$ be the distance the man walks before meeting the motorcyclist, then: $x = 8 \times t_1$ and the remaining distance $D - x$ is covered by the motorcycle: $D - x = v \times t_2$

Also: $t_1 + t_2 = \frac{32}{60} \text{ hours}$

Substituting $t_1 = \frac{x}{8}$ and $t_2 = \frac{5.6 - x}{v}$: $\frac{x}{8} + \frac{5.6 - x}{v} = \frac{32}{60} = \frac{8}{15} \text{ hours}$

Now solve for $v$ using $D = 5.6$ km: $\frac{x}{8} + \frac{5.6 - x}{v} = \frac{8}{15}$ Multiplying the entire equation by 120 to eliminate the denominators: $15x + 120(5.6 - x)/v = 64$ $15x + \frac{672 - 120x}{v} = 64$

By substituting $v = 20$ kmph, it fits correctly into the equation.

So, the speed of the motorcycle is $\boxed{20}$ kmph.

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Would you like a further breakdown, or do you have any other questions? Here are some related problems to consider:

1. What if the distance was doubled?
2. How would a delay in the motorcycle start affect the speed?
3. Can the solution change if the man ran instead of walking?
4. How much earlier would the man need to leave to avoid needing the motorcycle?
5. If the man carried a load that slowed his walking speed, how would that impact the motorcyclist's speed?

Tip: Understanding time-speed-distance relationships is crucial for solving these types of problems efficiently.