Question

A student walks from his house at \( \frac{5}{2} \) km/h and reaches his school late by 6 minutes. Next day, he increases his speed by 1 km/h and reaches 6 minutes before school time. How far is the school from his house?

• A. \( \frac{5}{4} \, \text{km} \)
• B. \( \frac{7}{4} \, \text{km} \)
• C. \( \frac{9}{4} \, \text{km} \)
• D. \( \frac{11}{4} \, \text{km} \)

Hint:

When dealing with time and speed problems, start by writing the formula for time \( \text{time} = \frac{\text{distance}}{\text{speed}} \), then set up equations based on the differences in time to find the unknown distance.

Answer 1

The Correct Option is B

Let the distance to the school be \( D \) km. % First Day Equation On the first day, the time taken by the student is \( \frac{D}{\frac{5}{2}} = \frac{2D}{5} \). % Second Day Equation On the second day, his speed increases by 1 km/h, so the time taken is \( \frac{D}{\frac{7}{2}} = \frac{2D}{7} \). According to the problem, the difference in time is 12 minutes (6 minutes late and 6 minutes early): \[ \frac{2D}{5} - \frac{2D}{7} = \frac{12}{60} = \frac{1}{5} \] Solving for \( D \): \[ \frac{2D}{5} - \frac{2D}{7} = \frac{1}{5} \] Finding a common denominator: \[ \frac{14D - 10D}{35} = \frac{1}{5} \] \[ \frac{4D}{35} = \frac{1}{5} \] Now, solving for \( D \): \[ 4D = 7 \quad \Rightarrow \quad D = \frac{7}{4} = 1.75 \, \text{km} \] Final Answer: The correct answer is (b) \( \frac{7}{4} \, \text{km} \).