A man travels from his home to his office at a speed of 4 km/h and reaches his office 30 minutes late. If his speed had 6 km/h, he would have reached office 5 minutes early. Find the distance of his office from his home.

The correct answer is 7 km.

Let the actual distance from home to office be d km, and let the actual time to reach exactly on time be T hours.

Setting up the equations:

When the man travels at 4 km/h, he arrives 30 minutes late. This means he takes 30/60 = 1/2 hour more than the scheduled time:

$$\frac{d}{4}=T+\frac{1}{2}\text{... (1)}$$

When the man travels at 6 km/h, he arrives 5 minutes early. This means he takes 5/60 = 1/12 hour less than the scheduled time:

$$\frac{d}{6}=T-\frac{1}{12}\text{... (2)}$$

Eliminating T by subtracting equation (2) from equation (1):

$$\frac{d}{4}-\frac{d}{6}=\left(T+\frac{1}{2}\right)-\left(T-\frac{1}{12}\right)$$

The T cancels out on the right side:

$$\frac{d}{4}-\frac{d}{6}=\frac{1}{2}+\frac{1}{12}$$

Solving the left side — finding a common denominator of 12:

$$\frac{3d}{12}-\frac{2d}{12}=\frac{d}{12}$$

Solving the right side — finding a common denominator of 12:

$$\frac{6}{12}+\frac{1}{12}=\frac{7}{12}$$

Now the equation becomes:

$$\frac{d}{12}=\frac{7}{12}$$

Multiplying both sides by 12:

$$d=7\text{km}$$

Verification:
- At 4 km/h: Time taken = 7/4 = 1.75 hours = 1 hour 45 minutes
- At 6 km/h: Time taken = 7/6 ≈ 1 hour 10 minutes
- Difference = 1h 45min - 1h 10min = 35 minutes = 30 min late + 5 min early ✓

The total time difference between the two scenarios is exactly 35 minutes, which perfectly matches the problem conditions (30 minutes late + 5 minutes early). Therefore, the distance from home to office is confirmed to be 7 km.