UPPSC 2024 Prelims GS Paper-I · Question 71 of 150

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.

• (a)v₁v₂ / (v₁ + v₂) — 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)(v₁ + v₂) / 2 — 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)√(v₁v₂) — 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.

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.

Every wrong option here is a real mean, offered to see whether the candidate has actually derived the result or merely recognised a familiar shape. The reliable exam method is not to memorise the formula but to run two tests on whatever you have written. First, the equal-speed test: substitute v₁ = v₂ = v and see whether the expression collapses to v — options (a) fails outright. Second, a numeric test with easy numbers such as 40 and 60 km/h, where the true answer of 48 separates the harmonic mean from the arithmetic (50) and geometric (≈ 49) means. Underlying both is one physical intuition worth carrying: over a fixed distance, slowing down costs more time than speeding up saves, so the true average always leans towards the smaller speed. That is also the reason the harmonic mean is the smallest of the three means.

• Average speed = total distance ÷ total time. It equals the simple arithmetic mean of the speeds only when each speed is held for the same amount of time.
• Equal distances at v₁ and v₂ → harmonic mean, 2v₁v₂/(v₁ + v₂). For n equal-distance segments the general form is n ÷ (1/v₁ + 1/v₂ + … + 1/vₙ).
• Equal times at v₁ and v₂ → arithmetic mean, (v₁ + v₂)/2. The two cases give different answers and examiners switch between them deliberately.
• AM ≥ GM ≥ HM for positive numbers, with equality only when the numbers are equal — so (v₁ + v₂)/2 and √(v₁v₂) always overstate the true equal-distance average.
• The distance cancels out of the derivation, which is why the answer is independent of how far the bus travelled and why such questions can be set without any distance being given.
• Worked check: equal halves at 40 and 60 km/h give 2 × 40 × 60 / 100 = 48 km/h, not 50 km/h.
• Average speed uses distance; average velocity uses displacement. A complete round trip therefore has a non-zero average speed but zero average velocity.

• Averaging the two speeds. (v₁ + v₂)/2 is right only for equal times; this question specifies equal halves of the DISTANCE.
• Dropping the factor of 2. Option (a) is exactly half the correct expression and is what you get if you divide only one half of the distance by the total time.
• Assuming a distance is missing. It cancels out — questions of this type are deliberately set without one, and any assumed value gives the same answer.

UPPSC puts this in Paper-I as a symbolic formula question, asking for the expression rather than a number. UPSC has asked the identical concept several times as a worked numerical in the aptitude portion — a round trip at two speeds, or a journey split into thirds — so be ready to produce both the algebraic form and an arithmetic answer from it.

• practice — not a real PYQ

  A car covers the first half of a distance at 30 km/h and the second half at 60 km/h. Its average speed for the whole journey is :

  • (a)45 km/h
  • (b)40 km/h
  • (c)42 km/h
  • (d)50 km/h

  Answer(b) 40 km/h — using the harmonic mean, 2 × 30 × 60 / (30 + 60) = 3600/90 = 40 km/h. The tempting 45 km/h is the arithmetic mean, which would be right only if the two speeds were maintained for equal times rather than equal distances.

• practice — not a real PYQ

  A body travels for the first half of the total TIME with speed v₁ and for the second half of the time with speed v₂. Its average speed for the whole journey is :

  • (a)2v₁v₂ / (v₁ + v₂)
  • (b)√(v₁v₂)
  • (c)(v₁ + v₂) / 2
  • (d)v₁v₂ / (v₁ + v₂)

  Answer(c) (v₁ + v₂)/2 — with equal times t each, distance = v₁t + v₂t and total time = 2t, so the average is (v₁ + v₂)/2, the arithmetic mean. Equal times give the arithmetic mean; equal distances give the harmonic mean. Reading which one the question specifies is the whole task.