A man is 20 minutes late to his office, if he drives at an average speed of 40 kmph and is four minutes eary, if he drives at an average speed of 50 kmph. By how much time will he be early late, if he drives at an average speed of 48 kmph? a minute early b. 1 minute late c. 2 minutes late

 d. Noneo uhể above

Let the distance from the man's house to his office be \( d \) kilometers and the usual time taken to reach his office be \( t \) hours.

### Information given:
1. If the man drives at 40 km/h, he is 20 minutes late.
2. If he drives at 50 km/h, he is 4 minutes early.

Using this information, we can set up two equations based on the relationship between time, speed, and distance:

- Time taken at 40 km/h: \( \frac{d}{40} \) hours
- Time taken at 50 km/h: \( \frac{d}{50} \) hours

#### Equation 1:
If he is 20 minutes late when driving at 40 km/h, the time taken is 20 minutes more than his usual time, or \( t + \frac{20}{60} = t + \frac{1}{3} \). Thus:
\[
\frac{d}{40} = t + \frac{1}{3}
\]

#### Equation 2:
If he is 4 minutes early when driving at 50 km/h, the time taken is 4 minutes less than his usual time, or \( t - \frac{4}{60} = t - \frac{1}{15} \). Thus:
\[
\frac{d}{50} = t - \frac{1}{15}
\]

Now, we have two equations:
1. \( \frac{d}{40} = t + \frac{1}{3} \)
2. \( \frac{d}{50} = t - \frac{1}{15} \)

### Solving for \( d \) and \( t \):

1. Solve both equations for \( t \):
   - From the first equation:
   \[
   t = \frac{d}{40} - \frac{1}{3}
   \]
   - From the second equation:
   \[
   t = \frac{d}{50} + \frac{1}{15}
   \]

2. Set these expressions for \( t \) equal to each other:
\[
\frac{d}{40} - \frac{1}{3} = \frac{d}{50} + \frac{1}{15}
\]

3. Clear the fractions by multiplying through by the least common denominator, which is 600:
\[
600 \left( \frac{d}{40} - \frac{1}{3} \right) = 600 \left( \frac{d}{50} + \frac{1}{15} \right)
\]
\[
15d - 200 = 12d + 40
\]

4. Simplify and solve for \( d \):
\[
15d - 12d = 240
\]
\[
3d = 240
\]
\[
d = 80 \text{ km}
\]

### Find \( t \):
Using \( d = 80 \) in either equation for \( t \), we get:
\[
t = \frac{80}{50} + \frac{1}{15} = 1.6 + 0.0667 = 1.6667 \text{ hours} = 1 \text{ hour } 40 \text{ minutes}
\]

### Case for 48 km/h:
Now, we calculate the time taken if the man drives at 48 km/h:
\[
\text{Time taken} = \frac{80}{48} = 1.6667 \text{ hours} = 1 \text{ hour } 40 \text{ minutes}
\]
This is exactly the same as the usual time \( t \), so the man will arrive **on time**.

### Answer: 
The correct option is **d. None of the above**.

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Do you want more details or have any questions? 

Here are 5 related questions:
1. How would the time change if he drives at 60 km/h?
2. Can we calculate the distance if only the time differences were given?
3. What happens if the road distance increases by 20 km?
4. How does changing the speed impact the time taken for the same distance?
5. What if the man alternates between 40 km/h and 50 km/h during the trip?

**Tip:** To solve problems involving speed, distance, and time, always start by identifying key relationships using the formula \( \text{Distance} = \text{Speed} \times \text{Time} \).

A man is 20 minutes late to his office, if he drives at an average speed of 40 kmph and is four minutes early, if he drives at an average speed of 50 kmph. By how much time will he be early or late if he drives at an average speed of 48 kmph? a. 1 minute early b. 1 minute late c. 2 minutes late d. None of the above

Solve a time and speed problem where a man arrives 20 minutes late driving at 40 km/h and 4 minutes early at 50 km/h. Learn how to calculate if he will be early or late at 48 km/h using the relationship between speed, distance, and time. Suitable for Grades 8-10.

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