Solving the Distance, Speed, and Time Problem

This problem asks us to find the distance between Akhil's home and office using information about his travel times at different average speeds. We need to use the relationship between distance, speed, and time: Distance = Speed × Time.

Understanding the Variables

Let's define the variables we'll use:

• Let $d$ represent the distance from home to the office in kilometers (km).
• Let $t$ represent the usual time taken to travel from home to the office in hours (h).

Setting up Equations Based on Given Information

We are given two scenarios:

1. Scenario 1: Faster Speed

   Akhil drives at an average speed of $50$ km/h and reaches 10 minutes early.

   • The time taken in this scenario is $t - 10$ minutes. We need to convert minutes to hours: $10 \text{ minutes} = \frac{10}{60} \text{ hours} = \frac{1}{6} \text{ hours}$.
   • So, the time taken is $t - \frac{1}{6}$ hours.
   • Using the formula Distance = Speed × Time, we get:

     $d = 50 \times (t - \frac{1}{6})$

     $d = 50t - \frac{50}{6}$

     $d = 50t - \frac{25}{3}$

2. Scenario 2: Slower Speed

   Akhil drives at an average speed of $30$ km/h and reaches 10 minutes late.

   • The time taken in this scenario is $t + 10$ minutes. Converting to hours: $10 \text{ minutes} = \frac{1}{6} \text{ hours}$.
   • So, the time taken is $t + \frac{1}{6}$ hours.
   • Using the formula Distance = Speed × Time, we get:

     $d = 30 \times (t + \frac{1}{6})$

     $d = 30t + \frac{30}{6}$

     $d = 30t + 5$

Solving the Equations

We now have a system of two equations with two variables ($d$ and $t$):

1. $d = 50t - \frac{25}{3}$
2. $d = 30t + 5$

We can solve this system. One way is to express $t$ in terms of $d$ from both equations and then equate them.

From equation (2):

$30t = d - 5$

$t = \frac{d - 5}{30}$

Substitute this expression for $t$ into equation (1):

$d = 50 \times \left(\frac{d - 5}{30}\right) - \frac{25}{3}$

$d = \frac{50(d - 5)}{30} - \frac{25}{3}$

$d = \frac{5(d - 5)}{3} - \frac{25}{3}$

To eliminate the fractions, multiply the entire equation by 3:

$3d = 5(d - 5) - 25$

$3d = 5d - 25 - 25$

$3d = 5d - 50$

Now, rearrange the equation to solve for $d$:

$50 = 5d - 3d$

$50 = 2d$

$d = \frac{50}{2}$

$d = 25$

Alternative Method: Using Time Difference

Another approach is to consider the difference in time taken for the two journeys.

Time taken at 50 km/h = $\frac{d}{50}$ hours.

Time taken at 30 km/h = $\frac{d}{30}$ hours.

The difference in time between these two journeys is the sum of the early arrival time and the late arrival time.

Total time difference = 10 minutes (early) + 10 minutes (late) = 20 minutes.

Convert 20 minutes to hours: $20 \text{ minutes} = \frac{20}{60} \text{ hours} = \frac{1}{3} \text{ hours}$.

The difference in time is:

Time at slower speed - Time at faster speed = Total time difference

$\frac{d}{30} - \frac{d}{50} = \frac{1}{3}$

Find a common denominator for 30 and 50, which is 150:

$\frac{5d}{150} - \frac{3d}{150} = \frac{1}{3}$

$\frac{5d - 3d}{150} = \frac{1}{3}$

$\frac{2d}{150} = \frac{1}{3}$

Simplify the fraction $\frac{2d}{150}$ to $\frac{d}{75}$:

$\frac{d}{75} = \frac{1}{3}$

Solve for $d$:

$d = \frac{75}{3}$

$d = 25$

Conclusion

Both methods show that the distance from Akhil's home to his office is 25 km.