Multiple choice

Starting from his house one day, a student walks at a speed of $\displaystyle 2\dfrac {1}{2}$ km/hour and reaches his school $6$ minutes late. Next day he increased his speed by $1$ km/hr and reaches the school $6$ minutes early. How far is the school from his house?

  1. $1$ km
  2. $1 \displaystyle \dfrac {1}{2}$ km
  3. $1 \displaystyle \dfrac {3}{4}$ km
  4. $2$ km
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let d be the distance. Time taken at 2.5 km/hr is d/2.5, and at 3.5 km/hr is d/3.5. The difference in time is 12 minutes (0.2 hours). Solving d/2.5 - d/3.5 = 0.2 leads to d = 1.75 km.

AI explanation

Let the distance to the school be d kilometers and the correct time be t hours. Using the formula distance equals speed multiplied by time, we write 2.5 = d / (t - 0.1) and 3.5 = d / (t + 0.1). Multiplying both sides of the first equation by 3.5 and the second by 2.5 gives 8.75 = 3.5d / (t - 0.1) and 8.75 = 2.5d / (t + 0.1), which simplifies to 3.5 / (t - 0.1) = 2.5 / (t + 0.1). Solving 3.5t + 0.35 = 2.5t - 0.25 yields t = -0.6 hours, implying the initial assumption of t being late is reversed, so we re-evaluate using the actual time difference of 12 minutes (0.2 hours) to find d = 1.75 km.

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