← Quantitative Aptitude · MbPA AEE & EE Civil

Chapter 11 of 17

Time, Speed & Distance

In the MbPA AEE & EE Civil syllabus under Quantitative Aptitude · 2 parts

Time, Speed & Distance · Boats & Streams

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📑 Contents (12 sections)

Part 1 of 2

Time, Speed & Distance

Last reviewed 22 Sept 2026 · 5 min read

The basic relation

Speed = \frac{Distance}{Time}, Distance = Speed \times Time, Time = \frac{Distance}{Speed}

FormulaUnits

1 km/h = \frac{5}{18} m/s, 1 m/s = \frac{18}{5} km/h

Convert before writing the equation: a train's length is in metres and its speed usually in km/h, and mixing the two is the single most common slip in this chapter.

Speed and time are inversely proportional

For the same distance, if the speeds are in the ratio $a : b$ then the times are in the ratio $b : a$ .

FormulaLate and early

If the usual speed is changed to $\frac{a}{b}$ of itself, the time becomes $\frac{b}{a}$ of the usual time, so the extra time is $(\frac{b}{a} - 1) T$ . Setting that equal to the stated delay gives the usual time $T$ in one step.

Average speed

Average speed = \frac{Total distance}{Total time}

For two equal distances at $x$ and $y$ : $\frac{2 x y}{x + y}$ . For three equal parts at $x$ , $y$ , $z$ : $\frac{3}{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}}$ . Halts count as time.

Relative speed

FormulaTwo moving bodies

Opposite directions: relative speed $= a + b$ (they close the gap fast)
Same direction: relative speed $= ∣ a - b ∣$ (overtaking)
Time to meet $= \frac{initial gap}{relative speed}$

Trains

Situation	Distance covered
Crossing a pole, a man standing, a signal	Length of the train
Crossing a platform, a bridge, a tunnel	Length of train + length of platform
Crossing a man walking	Length of train, at relative speed
Two trains crossing each other	Sum of the two lengths, at relative speed

Races

In a race of $d$ metres, "A beats B by $x$ metres" means A has finished while B is $x$ metres short; "A beats B by $t$ seconds" means B needs $t$ more seconds. A "start of $x$ metres" means B runs only $d - x$ .

Worked examples

Worked ExampleExample 1 — units

Convert 72 km/h into m/s, and 25 m/s into km/h.

Solution. $72 \times \frac{5}{18} =$ 20 m/s; $25 \times \frac{18}{5} =$ 90 km/h.

Worked ExampleExample 2 — pole and platform

A train 180 m long runs at 54 km/h. How long does it take to cross a pole, and a platform 270 m long?

Solution. Speed $= 54 \times \frac{5}{18} = 15$ m/s. Pole: $\frac{180}{15} =$ 12 s. Platform: $\frac{180 + 270}{15} =$ 30 s.

Worked ExampleExample 3 — two trains, opposite directions

Trains 120 m and 180 m long run at 54 km/h and 36 km/h towards each other. How long to cross?

Solution. Relative speed $= 90$ km/h $= 25$ m/s; distance $= 300$ m → $\frac{300}{25} =$ 12 s.

Worked ExampleExample 4 — the same two trains, same direction

Solution. Relative speed $= 54 - 36 = 18$ km/h $= 5$ m/s → $\frac{300}{5} =$ 60 s. Same trains, five times the time — direction is everything.

Worked ExampleExample 5 — a train and a walker

A train crosses a man walking at 6 km/h in the same direction in 12 seconds. The train runs at 54 km/h. Find its length.

Solution. Relative speed $= 48$ km/h $= \frac{40}{3}$ m/s; length $= \frac{40}{3} \times 12 =$ 160 m.

Worked ExampleExample 6 — late at a slower speed

Walking at $\frac{5}{6}$ of his usual speed, a man reaches the office 10 minutes late. Find his usual time.

Solution. Time becomes $\frac{6}{5}$ of usual, so the extra $\frac{1}{5} T = 10 \Rightarrow T =$ 50 minutes.

Worked ExampleExample 7 — speed from a time saved

A car covers 240 km. Had its speed been 20 km/h more, it would have taken 2 hours less. Find the speed.

Solution. $\frac{240}{s} - \frac{240}{s + 20} = 2 \Rightarrow 4800 = 2 s (s + 20) \Rightarrow s^{2} + 20 s - 2400 = 0$ . $s = \frac{- 20 + 100}{2} =$ 40 km/h.

Worked ExampleExample 8 — average over three parts

Equal thirds of a journey are covered at 20, 30 and 60 km/h. Find the average speed.

Solution. $\frac{3}{\frac{1}{20} + \frac{1}{30} + \frac{1}{60}} = \frac{3}{\frac{3 + 2 + 1}{60}} = \frac{3 \times 60}{6} =$ 30 km/h.

Worked ExampleExample 9 — meeting

Two people 100 km apart start towards each other at 40 and 60 km/h. When and where do they meet?

Solution. Relative speed $= 100$ km/h → 1 hour, and the first has covered 40 km.

Worked ExampleExample 10 — catching up

A thief running at 8 km/h is spotted by a policeman 1 km behind, who chases at 10 km/h. How long to catch him?

Solution. Relative speed $= 2$ km/h → $\frac{1}{2}$ hour $=$ 30 minutes, by when the thief has run 4 km.

Worked ExampleExample 11 — a race

In a 500 m race A beats B by 50 m. If both keep their speeds, by how much will A beat B in a 1000 m race?

Solution. When A runs 500, B runs 450, so speeds are $10 : 9$ . Over 1000 m, B covers 900 → A wins by 100 m.

Frequently tested points

km/h to m/s: multiply by 5/18; the reverse, by 18/5.
For a fixed distance, speeds in the ratio $a : b$ give times in the ratio $b : a$ .
Average speed is total distance over total time; equal distances at $x$ and $y$ give $2 x y / (x + y)$ .
Relative speed: add for opposite directions, subtract for the same direction.
Crossing a pole covers the train's length; a platform, the sum of the two lengths.
At $\frac{a}{b}$ of the usual speed the time becomes $\frac{b}{a}$ of usual — the whole "late by" family.
In races, convert "beats by $x$ metres" into a ratio of speeds.

Common MistakeCommon mistakes

Mixing metres with km/h in one equation.
Using the plain average of two speeds when the distances are equal.
Subtracting speeds for trains moving towards each other.
Forgetting the train's own length when it crosses a platform.
Treating "10 minutes late" as a speed rather than a time difference.

Revision SummaryChapter summary

Speed, distance and time interconvert directly, and the units must agree before anything else.
For a fixed distance, speed and time are inversely proportional, which solves the late-and-early family in one line.
Average speed is always total distance over total time, never the mean of the speeds.
Relative speed adds for opposite directions and subtracts for the same one.
Train problems are relative-speed problems in which the distance includes the lengths involved.

Part 2 of 2

Boats & Streams

Last reviewed 22 Sept 2026 · 5 min read

Downstream and upstream

Let $b$ be the speed of the boat in still water and $s$ the speed of the stream, with $b > s$ .

FormulaThe four relations

Downstream d = b + s, Upstream u = b - s

b = \frac{d + u}{2}, s = \frac{d - u}{2}

The last two are the whole chapter: any question that gives the two journey speeds gives the boat and the stream at once.

A log, raft or floating object has no speed of its own, so it travels at the speed of the stream, $s$ .

Useful consequences

FormulaShortcuts worth knowing

If upstream takes $n$ times as long as downstream over the same distance, then $u : d = 1 : n$ and

\frac{b}{s} = \frac{n + 1}{n - 1}

Round trip over a distance $D$ each way:

Total time = \frac{D}{b + s} + \frac{D}{b - s} = \frac{2 D b}{b ^{2} - s ^{2}}, Average speed = \frac{b ^{2} - s ^{2}}{b}

A boat that takes the same time to go $x$ km downstream and $y$ km upstream has $\frac{x}{b + s} = \frac{y}{b - s}$ .

The same formulas describe an aeroplane flying with or against a wind, and a swimmer in a current — only the names change.

Worked examples

Worked ExampleExample 1 — both ways

A boat does 12 km/h in still water and the stream flows at 3 km/h. Find the time for 45 km downstream and 45 km upstream.

Solution. Downstream $= 15$ km/h → $\frac{45}{15} =$ 3 hours. Upstream $= 9$ km/h → $\frac{45}{9} =$ 5 hours.

Worked ExampleExample 2 — finding both speeds

A boat covers 30 km downstream in 2 hours and the same 30 km upstream in 3 hours. Find the speed of the boat and of the stream.

Solution. $d = 15$ , $u = 10$ km/h. $b = \frac{15 + 10}{2} =$ 12.5 km/h, $s = \frac{15 - 10}{2} =$ 2.5 km/h.

Worked ExampleExample 3 — n times as long

A boat takes twice as long to go upstream as downstream over the same distance. Find the ratio of the boat's speed to the stream's.

Solution. $\frac{b}{s} = \frac{n + 1}{n - 1} = \frac{3}{1}$ → 3 : 1.

Worked ExampleExample 4 — round trip

A boat does 10 km/h in still water and the stream 2 km/h. It goes 48 km downstream and returns. Find the total time and the average speed.

Solution. $\frac{48}{12} + \frac{48}{8} = 4 + 6 =$ 10 hours for 96 km → average $=$ 9.6 km/h, which is also $\frac{b ^{2} - s ^{2}}{b} = \frac{100 - 4}{10}$ .

Worked ExampleExample 5 — a return journey

A man rows 18 km downstream in 2 hours and returns in 3 hours. Find both speeds.

Solution. $d = 9$ , $u = 6$ → $b =$ 7.5 km/h, $s =$ 1.5 km/h.

Worked ExampleExample 6 — equal times

A boat goes 20 km downstream in the same time as 12 km upstream. The stream runs at 2 km/h. Find the boat's speed.

Solution. $\frac{20}{b + 2} = \frac{12}{b - 2} \Rightarrow 20 b - 40 = 12 b + 24 \Rightarrow 8 b = 64$ → 8 km/h.

Worked ExampleExample 7 — a floating log

A boat does 9 km/h in still water; the stream 3 km/h. How long will a log take to drift 12 km downstream?

Solution. A log moves at the stream's speed: $\frac{12}{3} =$ 4 hours. The boat's speed is irrelevant.

Worked ExampleExample 8 — total time given

A boat whose still-water speed is 8 km/h takes 4 hours 30 minutes to row to a place 16 km away and back. Find the speed of the stream.

Solution. $\frac{16}{8 + s} + \frac{16}{8 - s} = 4.5 \Rightarrow \frac{16 \times 16}{64 - s ^{2}} = 4.5 \Rightarrow 256 = 288 - 4.5 s^{2}$ $\Rightarrow s^{2} = \frac{32}{4.5} = \frac{64}{9}$ → $s =$ 8/3 km/h (about 2.67 km/h).

Worked ExampleExample 9 — a plane and the wind

An aircraft flies 900 km with a 50 km/h tail wind in 3 hours. How long will the return flight take against the same wind?

Solution. Ground speed out $= 300$ km/h, so the still-air speed is $250$ km/h. Return ground speed $= 200$ km/h → $\frac{900}{200} =$ 4.5 hours.

Worked ExampleExample 10 — distance from a difference

A boat's still-water speed is 15 km/h and the stream's 5 km/h. If the upstream trip takes 2 hours longer than the downstream trip over the same distance, find the distance.

Solution. $\frac{D}{10} - \frac{D}{20} = 2 \Rightarrow \frac{D}{20} = 2$ → 40 km.

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