Dhiraj goes to his school from his house at a speed of 3.8 km/hr and returns at 2.6 km/ hr. If he takes 5 hours going and coming, the distance between his house and school is:

Correct — D. Let the one-way distance be d km. Time = distance ÷ speed for each leg.

Going: d⁄3.8 h
Returning: d⁄2.6 h
Total time: d⁄3.8 + d⁄2.6 = 5

Common denominator 3.8 × 2.6 = 9.88: d × 6.4⁄9.88 = 5
Multiply out: 6.4d = 5 × 9.88 = 49.4
Divide: d = 49.4 ÷ 6.4 = 7.71875 ≈ 7.72 km → option (d)

Why the others are wrong

• (a)
  6.79 km — 6.79 km takes 6.79⁄3.8 + 6.79⁄2.6 ≈ 1.79 + 2.61 = 4.40 h, well short of the 5 hours in the question.
• (b)
  6.5 km — 6.5 km is 2.6 × 2.5, the return speed times half the total time. The slower leg takes more than half, so 6.5 km fills only 4.21 h.
• (c)
  7 km — 7 km takes 7⁄3.8 + 7⁄2.6 ≈ 1.84 + 2.69 = 4.53 h. The distance must be longer to fill 5 hours.

On a round trip both legs cover the same distance at different speeds, so they take different times.

Add the two times and set the sum to the total: d⁄u + d⁄v = T, which gives d = T × u × v ⁄ (u + v).

Here 5 × 3.8 × 2.6 ⁄ 6.4 = 49.4 ⁄ 6.4 = 7.72 km. The average speed for the whole trip is 2uv⁄(u + v) ≈ 3.09 km/h, not the simple mean of 3.2 km/h.

Key facts

• Round trip over a distance d: total time = d⁄u + d⁄v.
• Average speed over two equal distances = 2uv⁄(u + v), the harmonic mean of the two speeds.
• When the two speeds differ, the harmonic mean is below the simple mean: 3.09 km/h against 3.2 km/h here.

Common traps

• Averaging the speeds to 3.2 km/h. Then 3.2 × 5 = 16 km for the trip and 8 km one way, which is not an option.
• Forgetting to halve. The whole round trip is 15.44 km, and the question asks for house to school.

No directly related past PYQ was found.