Arithmetic

Time, Speed & Distance

Easy

A man leaves his home and walks at a speed of $12$ km per hour, reaching the railway station $10$ minutes after the train had departed. If instead he had walked at a speed of $15$ km per hour, he would have reached the station $10$ minutes before the train's departure. The distance (in km) from his home to the railway station is

Entered answer:

✅ Correct Answer: 20

A man walks to a railway station at two different speeds:

Case 1: At 12 km/hr → reaches 10 minutes after train departs

Case 2: At 15 km/hr → reaches 10 minutes before train departs

We need to find the distance from home to the station.

Pretty easy to solve by setting up the equations. D = ST

Both have the same distance, hence:

$12(t+10) = 15(t-10)$

$12t+120 = 15t-150$

$3t = 270$

$t = 90$

Time when speed is 12km/hr = 90 + 10

Distance = Speed × Time

Distance = $12 \times \frac{90+10}{60} = 20$ km

This is a nice question to learn the following relationship:

When distance is constant, speed and time are inversely proportional. If speed increases, time decreases. If speed becomes k times, time becomes $\frac{1}{k}$ times.

Speed comparison: $\frac{15}{12} = \frac{5}{4}$

Since speed becomes $\frac{5}{4}$ times, time becomes $\frac{4}{5}$ times.

If original time = T, then new time = $\frac{4}{5}T$

Time saved = $T - \frac{4}{5}T = \frac{1}{5}T$

Walking at 15 km/hr saves $\frac{1}{5}$ of the original time.

From the problem:

Case 1: 10 minutes late

Case 2: 10 minutes early

Total difference in outcomes = 10 + 10 = 20 minutes

This 20-minute difference equals the time saved, which is $\frac{1}{5}$ of original time.

$\frac{1}{5} \times \text{Original time} = 20$ minutes

Original time = $20 \times 5 = 100$ minutes

In Case 1:

Speed = 12 km/hr

Time = 100 minutes = $\frac{100}{60}$ hours = $\frac{5}{3}$ hours

Distance = Speed × Time

Distance = $12 \times \frac{5}{3} = 20$ km

Therefore, the distance from home to railway station is 20 km.

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