SSC CGL 2024 Tier-I · 10 September 2024, Shift 1 · Quantitative Aptitude · Question 17 of 98

Text as printed on a candidate's response sheet for this shift; answer per SSC's tentative key as marked on that sheet. Solution last reviewed 10 September 2026.

Where this paper and its answers come from

SSC does not publish CGL question papers or a public answer key. The questions and options here are reproduced as printed on a candidate's own response sheet for this shift — the Commission's text, unchanged. Each answer is SSC's tentative key as marked on that sheet; SSC allowed challenges to it and never publishes a final key. The solutions are written by UnlockIAS.

A bus travels of the distance initially at a speed of 40 km/h, the next of the distance 50 km/h, and the final of the distance at 60 km/h. Find the average speed of the bus for the entire journey. (Rounded off to two decimal places)

(a)48.65 km/h
(b)54.13 km/h
(c)51.43 km/h
(d)46.85 km/h

Answer

Why

Correct — A. The three fractions in the stem are printed as images, and each of them is 1⁄3, so the journey splits into three equal thirds covered at 40, 50 and 60 km/h.

Equal distances mean the average speed is the harmonic mean, never the average of the three speeds.

Let the whole distance be 3d, so each leg is d.
Time taken = d⁄40 + d⁄50 + d⁄60

Over the LCM 600: (15d + 12d + 10d)⁄600 = 37d⁄600

Average speed = total distance ÷ total time = 3d ÷ (37d⁄600) = 1800⁄37
1800⁄37 = 48.6486… ≈ 48.65 km/h → option (a).

Why the others are wrong

(b)
54.13 km/h — 54.13 km/h sits above 50, the plain average of 40, 50 and 60. Over equal distances the average speed always comes out below that plain average, so anything over 50 is ruled out at sight.
(c)
51.43 km/h — 51.43 km/h would require a total time of 3d ÷ 51.43, which is 35d⁄600. The three legs actually take 37d⁄600, and more time means a lower average.
(d)
46.85 km/h — 46.85 is 48.65 with the middle two digits swapped. It implies a total time near 38.4d⁄600 against the 37d⁄600 the journey really takes.

Concept

Average speed is total distance ÷ total time, always, and never the average of the speeds. Which mean you end up with depends on what the journey holds equal.

Equal distances on each leg give the harmonic mean: for n legs the average is n ÷ (1⁄v₁ + 1⁄v₂ + … + 1⁄vₙ). Equal times on each leg would give the ordinary arithmetic mean instead.

Here the thirds are equal in distance, so the slow 40 km/h leg occupies the most time and pulls the average under 50.

Each of the three fractions appears as a small image in the question rather than as typed text, and each one is 1⁄3. Those equal thirds are exactly what makes the harmonic mean the right tool.

Key facts

Average speed = total distance ÷ total time, in every case.
For three equal distances at speeds v₁, v₂ and v₃ the average is 3 ÷ (1⁄v₁ + 1⁄v₂ + 1⁄v₃).
1⁄40 + 1⁄50 + 1⁄60 = 37⁄600, so the average speed is 1800⁄37 = 48.6486… km/h.
The harmonic mean of unequal speeds is always below their arithmetic mean, which here is 50.

Study next

Average speed over two equal distances, the 2v₁v₂ ÷ (v₁ + v₂) shortcutJourneys split by equal time rather than by equal distanceRounding a recurring decimal such as 1800⁄37 to two places

Common traps

Averaging 40, 50 and 60 to 50, which ignores that the slow leg lasts longest.
Stretching the two-speed shortcut 2v₁v₂ ÷ (v₁ + v₂) to cover three legs.
Truncating 48.6486… to 48.64 instead of rounding it to 48.65.

SSC asks average speed either as a direct harmonic-mean computation like this one, or buried in a table of distances.

For the table form see 11 Sep 2024, 16:00, Quant Q.17, where an athlete's average speed over the first five hours has to be read off the data.

Related PYQs

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