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Average speed there and back

A man goes from A to B at 40 km/h and returns at 60 km/h. What is his average speed for the whole journey?

Show the answer and the worked solution

Correct answer: B.

48 km/h

Worked solution

  1. Take the distance as 120 km each way. Times: 120/40 = 3 h and 120/60 = 2 h.
  2. Total = 240 km in 5 h.
  3. Average = 48 km/h, matching 2 × 40 × 60/100.

Answer: 48 km/h.

The idea

Equal distances give the harmonic mean 2ab/(a + b), always below the arithmetic mean. Equal times would give the arithmetic mean of 50 instead.

Shortcut. More time is spent at the slower 40, so the average falls below 50 — striking out 50 and 52 — and 2 × 40 × 60 ÷ 100 = 48 picks it from 45.

Common trap. 50 km/h — the plain mean of 40 and 60, which would need equal times; with equal distances the slower leg lasts longer and drags the average down.

Average speed there and back — Aptitude Question

Average speed there and back: A man goes from A to B at 40 km/h and returns at 60 km/h. What is his average speed… Answer and worked solution on CodeKairo.

Question

A man goes from A to B at 40 km/h and returns at 60 km/h. What is his average speed for the whole journey?

Options

  1. 45 km/h
  2. 48 km/h
  3. 50 km/h
  4. 52 km/h
Show the answer and the worked solution

Correct answer: B. 48 km/h

Worked solution

  1. Take the distance as 120 km each way. Times: 120/40 = 3 h and 120/60 = 2 h.
  2. Total = 240 km in 5 h.
  3. Average = 48 km/h, matching 2 × 40 × 60/100.

Answer: 48 km/h.

The idea

Equal distances give the harmonic mean 2ab/(a + b), always below the arithmetic mean. Equal times would give the arithmetic mean of 50 instead.

Shortcut. More time is spent at the slower 40, so the average falls below 50 — striking out 50 and 52 — and 2 × 40 × 60 ÷ 100 = 48 picks it from 45.

Common trap. 50 km/h — the plain mean of 40 and 60, which would need equal times; with equal distances the slower leg lasts longer and drags the average down.

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