Time, Speed, Distance, Trains, Boats & Calendar

There is only one definition, and it is not negotiable: Average speed = Total distance ÷ Total time. Speeds themselves are never averaged; only distances and times are added.

Worked example 6. A car covers 60 km at 30 km/h and the next 60 km at 60 km/h. Total distance = 60 + 60 = 120 km. Time for the first leg = 60 ÷ 30 = 2 h; for the second = 60 ÷ 60 = 1 h; total = 3 h. Average speed = 120 ÷ 3 = 40 km/h — not 45. The car spent twice as long crawling as it did speeding, so the slow speed gets twice the weight.

Because that pattern — equal distances at two speeds — appears in nearly every paper, it deserves a formula. Let each leg be d at speeds x and y. Total distance = 2d; total time = d/x + d/y = d(x + y)/xy. Dividing, the d cancels and you get the harmonic mean:

Average speed = 2xy / (x + y)

Re-checking example 6: 2 × 30 × 60 ÷ (30 + 60) = 3600 ÷ 90 = 40 km/h ✓. Note the d cancelled — so this answer does not depend on how long the legs were, only that they were equal.

Three equal stretches at three speeds extend the same idea: Average speed = 3xyz / (xy + yz + zx).

Worked example 7. A train covers three equal stretches at 20, 30 and 60 km/h. Numerator = 3 × 20 × 30 × 60. Take it in steps: 20 × 30 = 600; 600 × 60 = 36,000; × 3 = 1,08,000. Denominator = (20 × 30) + (30 × 60) + (60 × 20) = 600 + 1800 + 1200 = 3600. Average = 1,08,000 ÷ 3600 = 30 km/h.

Long check, taking each stretch as 60 km: times are 60/20 = 3 h, 60/30 = 2 h, 60/60 = 1 h → 6 h for 180 km → 180 ÷ 6 = 30 km/h ✓.

Now the mirror case, where the plain average IS correct: equal times at two speeds.

Worked example 8. A man drives for 2 hours at 40 km/h and then for 2 hours at 60 km/h. Distances = 40 × 2 = 80 km and 60 × 2 = 120 km, total 200 km in 4 hours → 200 ÷ 4 = 50 km/h, which is exactly the arithmetic mean (40 + 60)/2. It works because equal times give each speed equal weight. So the deciding question is never the formula — it is "was the distance equal, or was the time equal?"

Worked example 9 — same direction (subtract). Two cars leave the same point together travelling the same way at 40 km/h and 55 km/h. How far apart are they after 4 hours? Relative speed = 55 − 40 = 15 km/h, so gap = 15 × 4 = 60 km. Check the long way: they have covered 40 × 4 = 160 km and 55 × 4 = 220 km, and 220 − 160 = 60 km ✓.

Worked example 10 — opposite directions (add). Two men start from two towns 90 km apart and walk towards each other at 4 km/h and 5 km/h. When do they meet? Relative speed = 4 + 5 = 9 km/h, so time = 90 ÷ 9 = 10 hours. Check: in 10 hours they cover 40 km and 50 km, and 40 + 50 = 90 km ✓ — they meet 40 km from the first town.

Trains add exactly one new idea: a train is not a point, it has length. "Crossing" something means the engine reaches it and the last coach leaves it. So the distance is measured from the front of the train to the back of the train.

Worked example 11 — crossing a pole or a standing man. A train 180 m long runs at 54 km/h. How long to cross a telegraph pole? A pole has no length, so the train must cover only its own 180 m. Speed = 54 × 5/18 = 15 m/s. Time = 180 ÷ 15 = 12 seconds. The same 12 seconds applies to a standing man, a signal post or a tree — all are points.

Worked example 12 — crossing a platform, bridge or tunnel. The same 180 m train at 54 km/h (= 15 m/s) crosses a platform 270 m long. Now the front must travel the whole platform and pull the whole train clear of it, so distance = 180 + 270 = 450 m. Time = 450 ÷ 15 = 30 seconds.

A bridge and a tunnel behave identically — the word changes, the arithmetic does not. And read the question backwards too: if that train crosses a platform in 30 s, the platform = (15 × 30) − 180 = 450 − 180 = 270 m ✓.

Worked example 13 — two trains crossing each other. Trains of lengths 140 m and 160 m run at 60 km/h and 48 km/h. Total length to clear = 140 + 160 = 300 m in both cases; only the speed changes.

Opposite directions: relative speed = 60 + 48 = 108 km/h = 108 × 5/18 = 30 m/s. Time = 300 ÷ 30 = 10 seconds.

Same direction: relative speed = 60 − 48 = 12 km/h = 12 × 5/18 = 10/3 m/s. Time = 300 ÷ (10/3) = 300 × 3/10 = 90 seconds.

Ten seconds against ninety — the same two trains, a nine-fold difference. Always read the direction before touching the calculator.

Worked example 14 — a train overtaking or meeting a person. A train 150 m long travels at 45 km/h. A cyclist moves at 9 km/h. The person is a point, so the distance is always just the train's 150 m.

Cyclist going the same way (train overtakes him): relative speed = 45 − 9 = 36 km/h = 10 m/s. Time = 150 ÷ 10 = 15 seconds.

Cyclist coming towards the train: relative speed = 45 + 9 = 54 km/h = 15 m/s. Time = 150 ÷ 15 = 10 seconds.

Moving against the train, he is passed in 10 s; moving with it, he stays alongside for 15 s. That extra 5 seconds is exactly what "with or against" is worth.

A boat is a train with the wind in its favour. Let u be the boat's speed in still water and v the speed of the stream. Going downstream the current helps: speed = u + v. Going upstream it resists: speed = u − v.

Worked example 15. A boat rows at 10 km/h in still water in a stream flowing at 2 km/h. Downstream speed = 10 + 2 = 12 km/h; upstream speed = 10 − 2 = 8 km/h. To go 36 km downstream takes 36 ÷ 12 = 3 hours; the same 36 km upstream takes 36 ÷ 8 = 4.5 hours. The river gave him an hour and a half.

Exams usually give you the two journey speeds and ask for the two hidden speeds. Call them D (downstream) and U (upstream). Since D = u + v and U = u − v, adding the two equations kills v and subtracting them kills u:

u = (D + U) / 2 and v = (D − U) / 2

Worked example 16. A boat goes downstream at 15 km/h and upstream at 9 km/h. Then u = (15 + 9)/2 = 24/2 = 12 km/h in still water, and v = (15 − 9)/2 = 6/2 = 3 km/h for the stream. Check both ways: 12 + 3 = 15 ✓ and 12 − 3 = 9 ✓. In words — the boat is the average of the two, the stream is half their difference.

Worked example 17 — round-trip average speed. The boat of example 15 (D = 12, U = 8) travels 24 km downstream and returns. Time down = 24 ÷ 12 = 2 h; time up = 24 ÷ 8 = 3 h; total 5 h for 48 km. Average = 48 ÷ 5 = 9.6 km/h.

This is the harmonic mean again: 2 × 12 × 8 ÷ (12 + 8) = 192 ÷ 20 = 9.6 ✓. There is also a pretty shortcut using u and v directly: (u² − v²)/u = (10² − 2²)/10 = (100 − 4)/10 = 96/10 = 9.6 ✓. And note it is below the still-water speed of 10 — a round trip on a river always loses time, no matter which way you start.

Worked example 18 — the "n times" type. A boat takes three times as long to go upstream as to come downstream over the same distance. If the boat does 12 km/h in still water, find the speed of the stream.

Same distance, so time is inversely proportional to speed. Time up = 3 × time down means speed down = 3 × speed up:

u + v = 3(u − v) → u + v = 3u − 3v → 4v = 2u → v = u/2 = 12/2 = 6 km/h.

Check: downstream = 12 + 6 = 18 km/h, upstream = 12 − 6 = 6 km/h, and 18 = 3 × 6 ✓ — so a 36 km leg takes 2 h down and 6 h up, exactly three times. The general result is u : v = (n + 1) : (n − 1); here n = 3 gives 4 : 2 = 2 : 1 ✓.

Worked example 19 — "to a place and back". A man rows 6 km/h in still water; the river flows at 2 km/h. He rows to a place and returns in 3 hours. How far is the place?

Downstream = 6 + 2 = 8 km/h, upstream = 6 − 2 = 4 km/h. Let the distance be d:

d/8 + d/4 = 3 → d/8 + 2d/8 = 3 → 3d/8 = 3 → d = 8 km.

Check: 8 ÷ 8 = 1 h down, 8 ÷ 4 = 2 h up, total 3 h ✓.

One sentence unlocks every race question: "A beats B by 20 m in a 100 m race" means that at the instant A finishes 100 m, B has run only 80 m. Both ran for the same time, so their speeds are in the ratio of the distances, 100 : 80 = 5 : 4.

Worked example 20. In a 100 m race A beats B by 20 m, and in a 100 m race B beats C by 25 m. By how much does A beat C in a 100 m race?

When A runs 100, B runs 80. When B runs 100, C runs 75. So when B runs only 80, C runs 75 × (80/100) = 60 m. Hence when A finishes 100 m, C is at 60 m: A beats C by 40 m. (Ratio check: A : B = 100 : 80 and B : C = 100 : 75, so A : C = 100 : (80 × 75/100) = 100 : 60 ✓.)

Worked example 21 — "beats by t seconds", and converting it to metres. A runs 100 m in 12 s and B runs 100 m in 15 s. A beats B by 15 − 12 = 3 seconds. To express the same lead in metres, ask how far B (the loser) travels in those 3 seconds: B's speed = 100 ÷ 15 = 20/3 m/s, so in 3 s he covers (20/3) × 3 = 20 m. So A beats B by 3 seconds or 20 metres — the same fact in two units.

The common slip is using A's speed here. A's speed is 100/12 = 8.33 m/s, which would give 25 m — wrong, because it is the loser who is still on the track during those 3 seconds.

Worked example 22 — a head start ("A gives B a start of x metres"). A start of 10 m means B begins 10 m ahead, so in a 100 m race B has only 90 m to run.

Suppose in a 100 m race A gives B a start of 10 m and still beats him by 5 m. When A completes his 100 m, B is 5 m short of the finish, so B has covered 90 − 5 = 85 m. Same time, so speeds are in the ratio A : B = 100 : 85 = 20 : 17.

Check: if A runs at 20 m/s and B at 17 m/s, A takes 100/20 = 5 s. In 5 s B covers 17 × 5 = 85 m, and 85 + 10 (his start) = 95 m — 5 m short of 100 ✓.

Worked example 23 — dead heat. A race is a dead heat when both runners reach the finishing line at the same instant. A covers 100 m in 20 s, B in 25 s. What start must A give B for a dead heat?

A finishes in 20 s, so B must also finish within 20 s. B's speed = 100 ÷ 25 = 4 m/s, so in 20 s B can run only 4 × 20 = 80 m. He therefore needs a start of 100 − 80 = 20 m.

Check: B starts 20 m ahead and runs 80 m at 4 m/s = 20 s; A runs 100 m in 20 s; both cross the line together ✓.

Worked example 24 — first meeting on a circular track. A and B run on a 400 m circular track at 5 m/s and 3 m/s, starting together from the same point. On a closed loop, one lap of relative distance separates one meeting from the next, so the distance to be closed is always the track length, 400 m.

Opposite directions: relative speed = 5 + 3 = 8 m/s → first meeting after 400 ÷ 8 = 50 seconds. (In 50 s A runs 250 m and B runs 150 m; 250 + 150 = 400 ✓.)

Same direction: relative speed = 5 − 3 = 2 m/s → first meeting after 400 ÷ 2 = 200 seconds. (In 200 s A runs 1000 m and B runs 600 m; A is exactly one full lap of 400 m ahead ✓.)

Worked example 25 — meeting again at the starting point. This is a different question, and it needs an LCM, not a relative speed, because both runners must be at the same fixed spot at once.

With the same pair on the 400 m track: A's lap time = 400 ÷ 5 = 80 s; B's lap time = 400 ÷ 3 = 133⅓ s. They are both at the start whenever the time is a whole-number multiple of both. Multiples of 80: 80, 160, 240, 320, 400. Multiples of 133⅓: 133⅓, 266⅔, 400. So they first meet at the starting point after 400 seconds — A having done 5 laps and B 3 laps.

Compare: they meet somewhere on the track after 200 s (same direction), but they only meet at the start line after 400 s. Read which one the question wants.

A week has 7 days, so weekdays repeat every 7 days. Therefore only the remainder after dividing by 7 matters — and that remainder is called the number of odd days. Complete weeks are thrown away because they change nothing.

An ordinary year has 365 days = (52 × 7) + 1 = 52 weeks + 1 day → 1 odd day. So a date shifts forward by one weekday each year: if 1 January is a Monday this year, next year it is a Tuesday.

A leap year has 366 days = 52 weeks + 2 days → 2 odd days, because February gains a 29th day.

Worked example 26 — what day of the week was 15 August 1947? Independence Day. We count the odd days of everything before that date.

Step 1 — completed years: 1946 = 1600 + 300 + 46.
• 1600 years → 0 odd days (multiple of 400).
• 300 years → 1 odd day.
• 46 years (1901–1946): leap years are 1904, 1908, … 1944, that is (1944 − 1904)/4 + 1 = 11 leap years, leaving 46 − 11 = 35 ordinary years. Odd days = (35 × 1) + (11 × 2) = 35 + 22 = 57; and 57 ÷ 7 = 8 weeks remainder 1.
• Subtotal = 0 + 1 + 1 = 2 odd days.

Step 2 — days elapsed in 1947 up to 15 August. 1947 is not divisible by 4, so February has 28 days. 31 (Jan) + 28 (Feb) + 31 (Mar) + 30 (Apr) + 31 (May) + 30 (Jun) + 31 (Jul) + 15 (Aug) = 227 days. Now 227 ÷ 7: 7 × 32 = 224, remainder 3 odd days.

Step 3 — add and read off. Total odd days = 2 + 3 = 5. From the day-code table, 5 = Friday. And indeed India became independent on a Friday ✓.

Worked example 27 — when does a calendar repeat? Two years share the same calendar when (a) the total odd days between them is 0, and (b) they are of the same type — a leap year cannot match an ordinary year, because February differs.

Which year repeats the calendar of 2023? Add odd days year by year until the running total is a multiple of 7. 2023 (ordinary) 1 → total 1; 2024 (leap) 2 → 3; 2025 → 4; 2026 → 5; 2027 → 6; 2028 (leap) 2 → 8; 2029 → 9; 2030 → 10; 2031 → 11; 2032 (leap) 2 → 13; 2033 → 14, which is 2 × 7. So the calendar of 2023 returns in 2034 — and 2034 is an ordinary year, so condition (b) also holds ✓.

For a leap year the answer is almost always the neat one: the calendar repeats after 28 years (2024 → 2052). Reason: those 28 years contain 7 leap years and 21 ordinary ones → (21 × 1) + (7 × 2) = 35 odd days, and 35 ÷ 7 = 0 exactly. The 28-year rule breaks only if a non-leap century year such as 1900 or 2100 lies in between.

Almost nobody loses marks in this chapter because the method was unknown. Marks go because a minute was treated as a decimal or a length was left out. So make it mechanical: before writing a single number, force the whole question into one system of units — either km and hours, or metres and seconds. Never both.

Two conversions do all the work: 10 minutes = 10/60 = 1/6 hour (not 0.10), and 45 minutes = 45/60 = 3/4 hour (not 0.45). And if any length in the question is in metres, convert every speed to m/s at the very start and be done with it.

• Averaging speeds by adding and halving when the distances are equal. 40 and 60 give 48, never 50.
• Forgetting the train's own length for a platform, bridge or tunnel. The distance is train + platform.
• Adding the train's length when the object is a point — a pole, tree, signal or standing man has no length.
• Adding speeds for the same direction and subtracting for opposite — it is exactly the other way round.
• Writing 10 minutes as 0.10 hour. It is 1/6 hour. Similarly 20 min = 1/3 h, 30 min = 1/2 h, 40 min = 2/3 h.
• Subtracting the late and early minutes instead of adding them. 10 min late and 5 min early is a gap of 15 minutes.
• Confusing still-water speed with downstream speed in boat problems — decide which one the question handed you.
• Reading "A beats B by 20 m" as B running 120 m. B runs 80 m — the loser always runs less.
• Calling 1900 a leap year because it divides by 4. Century years need division by 400.
• Counting the given year itself in the calendar method. For 15 August 1947 you use the 1946 completed years.