2.5 Average Speed

Figure: For equal distances, average speed is a harmonic mean, not a simple average.

This is the topic most prone to a fatal misconception: you cannot simply average two speeds unless the times are equal. The correct method depends on what is equal — distance or time.

(a) Equal distances covered at two different speeds x and y: Average speed = (2xy)/(x+y) (This is the Harmonic Mean of x and y.)

For three equal distances at speeds x, y, z: Average speed = (3xyz)/(xy+yz+zx)

(b) Equal times spent travelling at two (or more) different speeds: Average speed = (x+y)/(2) (simple average) For n equal time intervals at speeds x_1, x_2, …, x_n: Average speed = (x_1+x_2+…+x_n)/(n).

Why this happens: Average speed is always Total Distance ÷ Total Time, never an average of the speed values directly. When distances are equal, more time is spent at the slower speed, which pulls the true average down below the simple mean — hence the harmonic mean (which is always ≤ the arithmetic mean) applies. When times are equal, both speeds contribute equally to total distance, so the simple average works correctly.

Solved Example 13: A man travels a certain distance at 40 km/h and returns the same distance at 60 km/h. Find his average speed for the entire journey.

Since distance is equal: Average speed = 2xy/(x+y) = 2×40×60/(100) = 4800/100 = 48 km/h. (Note: NOT (40+60)/2 = 50 km/h — this is the single most common mistake in this chapter.)

Solved Example 14: A car travels for 2 hours at 50 km/h and for another 2 hours at 70 km/h. Find its average speed.

Since time is equal (2 hrs each): Average speed = (50+70)/2 = 60 km/h. (Verify: Total distance = 100+140 = 240 km; Total time = 4 hrs; Average speed = 240/4 = 60 km/h. ✓)

Solved Example 15: A cyclist covers equal distances at speeds of 10 km/h, 20 km/h, and 30 km/h. Find the average speed for the whole journey.

Average speed = 3xyz/(xy+yz+zx) = 3×10×20×30/(200+600+300) = 18000/1100 = 16.36 km/h (approx).

Solved Example 15(a): A man goes to office at 20 km/h and returns home along the same route at 30 km/h. Find his average speed for the entire to-and-fro journey.

Distances are equal (same route both ways). Average speed = 2xy/(x+y) = 2×20×30/(20+30) = 1200/50 = 24 km/h.

Solved Example 15(b): A train covers half its journey at 60 km/h and the remaining half at a certain speed, resulting in an average speed of 48 km/h for the whole journey. Find the speed for the second half.

Distances (each half) are equal, so 2xy/(x+y) = 48, with x = 60. So: 2×60×y / (60+y) = 48 120y = 48(60+y) 120y = 2880 + 48y 72y = 2880 y = 40 km/h.

This “reverse” application — given the average speed, find one of the two component speeds — is a favourite SSC CGL Tier-II and RRB NTPC CBT-2 twist. Always set up the 2xy/(x+y) equation directly and solve algebraically; do not try to guess-and-check under time pressure.

Solved Example 15(c): A car travels 40 km at 20 km/h, then 75 km at 25 km/h, and finally 90 km at 30 km/h. Find its average speed for the entire journey.

Here the three legs cover unequal distances at different speeds, so neither the 2xy/(x+y) nor the simple-average shortcut applies directly — this is exactly the situation the “Total Distance ÷ Total Time” rule (Section 2.5, opening note) is built for. Time for leg 1 = 40/20 = 2 h. Time for leg 2 = 75/25 = 3 h. Time for leg 3 = 90/30 = 3 h. Total distance = 40+75+90 = 205 km. Total time = 2+3+3 = 8 h. Average speed = 205/8 = 25.625 km/h.

Solved Example 15(d): A train travels for 4 hours at 60 km/h, then halts at a station for 1 hour, and then travels for 2 hours at 90 km/h. Find its average speed for the entire duration, including the halt.

This tests the edge case of a stoppage (speed = 0 for part of the total time) — a classic SSC trick where students forget to include the halted time in the denominator. Distance covered = (4×60) + (0, during the halt) + (2×90) = 240 + 0 + 180 = 420 km. Total time elapsed = 4 + 1 + 2 = 7 h. Average speed = 420/7 = 60 km/h. (If the halt were wrongly excluded from the time, one would get 420/6 = 70 km/h — a common but incorrect shortcut; the “average speed” of a journey with a stoppage must always account for the full elapsed time, not just the moving time, unless the question explicitly asks for “average speed while moving.”)