A student walks from his home to school at a speed of 4 km/h and reaches 15 minutes late. If he increases his speed to 6 km/h, he reaches 5 minutes early. What is the distance (in km) from his home to school?
- (a)4 km
- (b)6 km
- (c)8 km
- (d)10 km
Answer
Why
Correct — A. One walk is late and the other early, so their times differ by both margins added.
Time gap: 15 + 5 = 20 min = 1⁄3 h
Let the distance be d km:
d⁄4 − d⁄6 = 1⁄3
Common denominator 12: 3d⁄12 − 2d⁄12 = d⁄12
d⁄12 = 1⁄3
Multiply by 12: d = 4 km → option (a)
Why the others are wrong
- (b)6 km — 6 km takes 90 min at 4 km/h and 60 min at 6 km/h, a 30-minute gap. Late by 15 and early by 5 allows only 20.
- (c)8 km — 8 km takes 120 min at 4 km/h and 80 min at 6 km/h, a 40-minute gap, double the 20 minutes the stem allows.
- (d)10 km — 10 km takes 150 min at 4 km/h and 100 min at 6 km/h, a 50-minute gap against the required 20.
Concept
When one trip arrives late by a and the other early by b, the two travel times differ by a + b. Had both been late, they would differ by the difference of the margins.
Writing the gap as d⁄v₁ − d⁄v₂ gives d = gap × v₁v₂ ⁄ (v₂ − v₁). Here 1⁄3 × 24 ⁄ 2 = 4 km.
Check: 4 km takes 60 min at 4 km/h and 40 min at 6 km/h. The on-time journey is then 45 min, so the first walk is 15 min late and the second 5 min early, as the stem says.
Key facts
- Late by a at one speed and early by b at another: the travel times differ by a + b.
- Distance = time gap × v₁v₂ ⁄ (v₂ − v₁), with the gap in hours.
- 20 minutes = 1⁄3 hour.
Study next
Common traps
- Subtracting the margins, 15 − 5 = 10 min, which gives d = 2 km.
- Leaving the gap as 20 without converting to hours: d⁄12 = 20 gives 240 km.
This stem gives the speeds in km/h and both margins in minutes, so converting 20 minutes to 1⁄3 hour is part of what it tests.
With whole-kilometre options, timing each option at both speeds, as the notes on the wrong options do, is as quick as solving for d.
Related PYQs
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