Question: A student walks from his house at a speed of $2 \frac{1}{2}$ km per hour and reaches his school 6 minutes late. The next day he increases his speed by 1 km per hour and reaches 6 minutes before school time. How far is the school from his house?

Choices:

• $1 \frac{1}{4} \text{ km}$

• $2 \frac{1}{4} \text{ km}$ (Correct Answer)

• $1 \frac{3}{4} \text{ km}$

• $2 \frac{3}{4} \text{ km}$

Reasoning and Solution:

To solve this problem, we need to use the concepts of speed, distance, and time. Let us define the variables and analyze the given information:

• Speed on the first day: $2 \frac{1}{2}$ km/h = $\frac{5}{2}$ km/h

• Speed on the second day: $2 \frac{1}{2} + 1$ km/h = $\frac{5}{2} + 1$ km/h = $\frac{7}{2}$ km/h

• Let the actual time needed to reach the school on time be $t$ hours.

• The distance between the student's house and the school be $d$ km.

The given information tells us that when the student walks with a speed of $\frac{5}{2}$ km/h, he reaches 6 minutes late. Thus:

\[ \text{Time taken on first day} = t + \frac{6}{60} \text{ hours} = t + \frac{1}{10} \text{ hours} \]

Using the formula for time ($t = \frac{d}{\text{speed}}$), we get:

\[ t + \frac{1}{10} = \frac{d}{\frac{5}{2}} = \frac{2d}{5} \]

Next, when the student walks with a speed of $\frac{7}{2}$ km/h, he reaches 6 minutes early. Thus:

\[ \text{Time taken on second day} = t - \frac{6}{60} \text{ hours} = t - \frac{1}{10} \text{ hours} \]

Using the same formula, we get:

\[ t - \frac{1}{10} = \frac{d}{\frac{7}{2}} = \frac{2d}{7} \]

Now, we have two equations:

1. $t + \frac{1}{10} = \frac{2d}{5}$

2. $t - \frac{1}{10} = \frac{2d}{7}$

Solving these two equations simultaneously:

Adding both equations: \[ \left(t + \frac{1}{10}\right) + \left(t - \frac{1}{10}\right) = \frac{2d}{5} + \frac{2d}{7} \] \[ 2t = \frac{2d}{5} + \frac{2d}{7} \] \[ 2t = \frac{14d + 10d}{35} \] \[ 2t = \frac{24d}{35} \] \[ t = \frac{12d}{35} \]

Substituting $t$ in the first equation:

\[ \frac{12d}{35} + \frac{1}{10} = \frac{2d}{5} \] \[ \frac{12d}{35} + \frac{1}{10} = \frac{2d}{5} \] \[ \frac{12d}{35} + \frac{1}{10} = \frac{14d}{35} \] \[ \frac{12d}{35} + \frac{1}{10} = \frac{14d}{35} \]

Multiplying both sides by 35 to clear the fraction, we get:

\[ 12d \times 10 + 35 = 14d \times 10 \] \[ 12d + 3.5 = 14d \] \[ 14d - 12d = 3.5 \] \[ d = 2.5 \] \p>Thus, we can confirm our final answer:

Therefore, the correct answer is $2 \frac{1}{4}$ km.