Math Problem Statement

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

Career

above

Solution

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 dd be the distance between the two points.
  • Let tt 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 = d40\frac{d}{40}
    • He is 20 minutes late, so d40=t+2060=t+13\frac{d}{40} = t + \frac{20}{60} = t + \frac{1}{3} hours.
  2. When driving at 50 kmph:

    • The time taken = d50\frac{d}{50}
    • He is 4 minutes early, so d50=t460=t115\frac{d}{50} = t - \frac{4}{60} = t - \frac{1}{15} hours.

Step 3: Solve the equations

From the equations: d40=t+13\frac{d}{40} = t + \frac{1}{3} d50=t115\frac{d}{50} = t - \frac{1}{15}

To eliminate tt, subtract the second equation from the first: d40d50=(t+13)(t115)\frac{d}{40} - \frac{d}{50} = \left(t + \frac{1}{3}\right) - \left(t - \frac{1}{15}\right)

Step 4: Simplify the equation

d(5040)40×50=13+115\frac{d(50 - 40)}{40 \times 50} = \frac{1}{3} + \frac{1}{15} d×102000=5+115\frac{d \times 10}{2000} = \frac{5 + 1}{15} d200=615\frac{d}{200} = \frac{6}{15} d200=25\frac{d}{200} = \frac{2}{5} d=200×25=80 kmd = 200 \times \frac{2}{5} = 80 \text{ km}

Step 5: Calculate the time tt

Using d=80d = 80 km in the first equation: 8040=t+13\frac{80}{40} = t + \frac{1}{3} 2=t+132 = t + \frac{1}{3} t=213=6313=53 hours=1 hour and 40 minutest = 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: Time taken=8048 hours=53 hours=1 hour and 40 minutes\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 tt, 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.

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Math Problem Analysis

Mathematical Concepts

Time, Speed, and Distance
Algebra

Formulas

Time = Distance / Speed

Theorems

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Suitable Grade Level

Grades 9-12