A bus covers the first half of a certain distance with speed v₁ and the second half with a speed v₂. The average speed during the whole journey is :

Question

UnlockIAS · Accepted Answer

Correct — B, 2v₁v₂/(v₁ + v₂). Average speed is defined as total distance divided by total time — never as the average of the speeds, unless each speed happens to be held for the same length of time. Do the algebra. Let the whole journey be D, so each half is D/2. The first half takes t₁ = (D/2)/v₁ and the second takes t₂ = (D/2)/v₂, giving a total time T = (D/2)(1/v₁ + 1/v₂) = (D/2)·(v₁ + v₂)/(v₁v₂). The average speed is then D ÷ T = D · 2v₁v₂ / [D(v₁ + v₂)] = 2v₁v₂/(v₁ + v₂). Notice that D cancels — the answer does not depend on how long the journey was, which is why the question can be answered without any distance being given. This expression is the harmonic mean of the two speeds, and it is the correct average whenever equal distances are covered at different speeds. Two checks confirm it in seconds. Put v₁ = v₂ = v: the formula gives 2v²/2v = v, exactly right, since a bus that never changes speed averages that speed. Put v₁ = 40 and v₂ = 60 km/h: the formula gives 2 × 40 × 60 / 100 = 48 km/h, which is below the intuitive 50 — correct, because the bus spends more time crawling through the slow half than it does racing through the fast one, and time is what the average weights.

Why the other options are wrong — (a) Exactly half the right answer, and it comes from one specific slip: dividing by the total time but forgetting that the total distance is two halves, not one. In other words, computing (D/2) ÷ T instead of D ÷ T. The equal-speed test destroys it instantly — set v₁ = v₂ = v and the formula returns v²/2v = v/2, which claims a bus travelling steadily at 60 km/h averages 30 km/h. Any candidate who tests a formula on the trivial case will never mark this option.  (c) The arithmetic mean — the intuitive answer, and the one the question is built to punish. It is not wrong in general; it is wrong here. The arithmetic mean is the correct average speed when the two speeds are maintained for equal TIMES, not equal distances. Because the bus takes longer over the slower half, that half must carry more weight in the average, and the arithmetic mean gives both halves equal weight. It also always overstates: by the AM ≥ HM inequality, (v₁ + v₂)/2 is greater than 2v₁v₂/(v₁ + v₂) for every pair of unequal speeds. With 40 and 60 km/h it gives 50 where the truth is 48.  (d) The geometric mean. It is a cleverer trap than (c) because it survives the equal-speed test — put v₁ = v₂ = v and it correctly returns v — so the quick sanity check does not eliminate it. What eliminates it is that no step of the derivation produces it, and that by the standard AM ≥ GM ≥ HM chain it too always overstates the true average for unequal speeds: with 40 and 60 km/h it gives √2400 ≈ 48.99 against a true 48. The geometric mean belongs to problems about compounding growth rates and ratios, not to distance-time averaging.

Concept: Average speed is total path length divided by total elapsed time. It is a weighted average of the individual speeds in which the weights are the times spent at each speed — which is why the shape of the answer depends entirely on what is held constant. If the segments are of equal DISTANCE, more time is spent at the lower speed, and the correct average is the harmonic mean: for two segments, 2v₁v₂/(v₁ + v₂), and for n equal segments, n ÷ (1/v₁ + 1/v₂ + … + 1/vₙ). If instead the segments are of equal TIME, the weights are equal and the correct average is the plain arithmetic mean, (v₁ + v₂)/2. Distinguish average speed from average velocity as well: velocity uses displacement rather than distance, so a bus that returns to its starting point has an average velocity of zero while its average speed is perfectly non-zero.

State PCS 2024 Q71 — Science & Technology PYQ with Answer

State PCS 2024 Q71 — Science & Technology PYQ with Answer