Question

If John walks at the speed of 5 km/h, he reaches his office 7 minutes late. However, if he walks at the speed of 6 km/h, he reaches his office 5 minutes early. How far is his office from his home?

• A. 9
• B. 8
• C. 10
• D. 6

Hint:

Use the direct formula for distance:
$\text{Distance } (d) = \frac{s_1 \times s_2}{s_2 - s_1} \times \Delta t$
$d = \frac{5 \times 6}{6 - 5} \times \frac{12}{60} = 30 \times \frac{1}{5} = 6 \text{ km}$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This problem uses the relationship: $\text{Distance} = \text{Speed} \times \text{Time}$.
We set up an equation based on the difference in travel times at two different speeds.
Key Formula or Approach:
Let the distance be $d$ km.
The time difference ($\Delta t$) between being 7 minutes late and 5 minutes early is: \[ \Delta t = 7 - (-5) = 12 \text{ minutes} \] Convert minutes to hours: \[ \Delta t = \frac{12}{60} = \frac{1}{5} \text{ hour} \]

Step 2: Detailed Explanation:
The equation representing the difference in travel times is: \[ \frac{d}{s_{\text{slow}}} - \frac{d}{s_{\text{fast}}} = \Delta t \] Substitute the given speeds: \[ \frac{d}{5} - \frac{d}{6} = \frac{1}{5} \] Find a common denominator (30): \[ \frac{6d - 5d}{30} = \frac{1}{5} \] \[ \frac{d}{30} = \frac{1}{5} \] \[ d = \frac{30}{5} = 6 \text{ km} \] The office is 6 km away from his home.

Step 3: Final Answer:
The distance is 6 km, matching Option (D).