A man is 20 minutes late t 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/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 

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To solve this problem, we can start by determining the distance between the man's starting point and destination using the information provided.

### Step 1: Define the variables

- Let \( d \) be the distance between the two points.
- Let \( t \) be the time he should take to reach the destination on time.

### Step 2: Set up equations based on the given information

1. **When driving at 40 kmph:**
   - The time taken = \( \frac{d}{40} \)
   - He is 20 minutes late, so \( \frac{d}{40} = t + \frac{20}{60} = t + \frac{1}{3} \) hours.

2. **When driving at 50 kmph:**
   - The time taken = \( \frac{d}{50} \)
   - He is 4 minutes early, so \( \frac{d}{50} = t - \frac{4}{60} = t - \frac{1}{15} \) hours.

### Step 3: Solve the equations

From the equations:
\[
\frac{d}{40} = t + \frac{1}{3}
\]
\[
\frac{d}{50} = t - \frac{1}{15}
\]

To eliminate \( t \), subtract the second equation from the first:
\[
\frac{d}{40} - \frac{d}{50} = \left(t + \frac{1}{3}\right) - \left(t - \frac{1}{15}\right)
\]

### Step 4: Simplify the equation

\[
\frac{d(50 - 40)}{40 \times 50} = \frac{1}{3} + \frac{1}{15}
\]
\[
\frac{d \times 10}{2000} = \frac{5 + 1}{15} 
\]
\[
\frac{d}{200} = \frac{6}{15} 
\]
\[
\frac{d}{200} = \frac{2}{5} 
\]
\[
d = 200 \times \frac{2}{5} = 80 \text{ km}
\]

### Step 5: Calculate the time \( t \)

Using \( d = 80 \) km in the first equation:
\[
\frac{80}{40} = t + \frac{1}{3}
\]
\[
2 = t + \frac{1}{3}
\]
\[
t = 2 - \frac{1}{3} = \frac{6}{3} - \frac{1}{3} = \frac{5}{3} \text{ hours} = 1 \text{ hour and } 40 \text{ minutes}
\]

### Step 6: Calculate the time taken at 48 kmph

Now, let's calculate the time taken if he drives at 48 kmph:
\[
\text{Time taken} = \frac{80}{48} \text{ hours} = \frac{5}{3} \text{ hours} = 1 \text{ hour and } 40 \text{ minutes}
\]

Since this is the same as the required time \( t \), the man will arrive exactly on time.

### Conclusion:
- The answer is **d. None of the above**.

Would you like more details or have any questions?

**Relative Questions:**
1. What would happen if the average speed is changed to 60 kmph?
2. How does distance affect the arrival time if speed is kept constant?
3. How can you generalize this problem for different distances?
4. What if the man drives at a varying speed instead of a constant speed?
5. How can this problem be solved using graphical methods?

**Tip:** When solving problems involving time, speed, and distance, always check if the units are consistent before setting up equations.

A man is 20 minutes late 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/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 real-world problem involving time, speed, and distance. A man drives at different speeds and arrives early or late. Discover how to determine if he will be early or late when driving at a specific speed. Suitable for students in Grades 9-12.

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