1. The one equation
If a process runs at a constant rate $r$ for a duration $t$, the total amount produced is
$$ \text{amount} = r \cdot t. $$
Every rate problem is this equation with the variables relabelled — distance for amount and speed for rate, job-fraction for amount and work-per-hour for rate, litres for amount and flow-rate for rate.
Three things, one equation, three different questions you can ask:
| Given | Find | How |
|---|---|---|
| rate, time | amount | multiply: $a = r \cdot t$ |
| amount, rate | time | divide: $t = a / r$ |
| amount, time | rate | divide: $r = a / t$ |
A car at $60$ km/h for $2.5$ hours covers $60 \cdot 2.5 = 150$ km. A printer that produces $300$ pages in $5$ minutes runs at $60$ pages per minute. A tap that needs to deliver $200$ litres at $25$ litres/minute will take $8$ minutes. Same equation, three perspectives.
A rate is an "amount per unit of time" — and that's just a fraction with time in the denominator. km/h, pages/min, jobs/day, tank-fills/hour. The fraction language is what makes rates combine cleanly: you can add $1/6$ tank/hour to $1/4$ tank/hour the same way you'd add any two fractions.
2. Template 1 — One rate, one mover
The simplest shape. A single thing moves, produces, or fills at a constant rate, and you're asked about one of the three quantities. Pick the row in the table above, plug in, done.
Example. A train covers $480$ km in $6$ hours. Its average speed is $480 / 6 = 80$ km/h. Note that this is the time-averaged speed of the whole journey — including acceleration, deceleration, and stops — not the peak cruise speed.
Example. A bicycle courier rides at $18$ km/h. How far in $40$ minutes? Convert the time to hours: $40$ min $= 2/3$ h. Then $d = 18 \cdot 2/3 = 12$ km.
The rate equation only holds when the time unit in your rate matches the time you're plugging in. $60$ km/h for $30$ minutes isn't $60 \cdot 30 = 1800$ km — convert first to $60$ km/h × $0.5$ h $= 30$ km.
3. Template 2 — Two rates working together
Two contributors, working at the same time on the same total. The headline rule:
Rates add. Times do not.
If A finishes a job alone in $t_A$ time and B alone in $t_B$, their individual rates are $1/t_A$ and $1/t_B$ (in jobs-per-time-unit). Working together, the combined rate is the sum:
$$ r_{\text{together}} = \frac{1}{t_A} + \frac{1}{t_B} $$and the time to finish together is one job divided by that combined rate:
$$ t_{\text{together}} = \frac{1}{\dfrac{1}{t_A} + \dfrac{1}{t_B}} = \frac{t_A \, t_B}{t_A + t_B}. $$Why rates add but times don't
The two contributors aren't taking turns — they're working simultaneously. In one hour, A does $1/t_A$ of the job and B does $1/t_B$ of the job, and both contributions land in the same hour. They accumulate.
If you averaged the times, you'd be answering a completely different question ("how long if they took turns at the average speed"). And if you added the times, you'd be answering "how long if they worked one after the other on two separate copies of the job." Neither matches the situation.
Try it with numbers
A does the job in $6$ hours, B in $4$ hours. Together?
$$ r = \tfrac{1}{6} + \tfrac{1}{4} = \tfrac{2}{12} + \tfrac{3}{12} = \tfrac{5}{12} \text{ job/hour} $$ $$ t = 1 \div \tfrac{5}{12} = \tfrac{12}{5} = 2.4 \text{ hours} = 2\,\text{h}\,24\,\text{min}. $$Notice: less time than either of them alone — exactly what you'd expect when two people pitch in.
The combined time must be less than the faster worker's solo time. If your answer is larger than either $t_A$ or $t_B$, you've added times somewhere instead of rates.
The same template, three disguises
- Alice paints a room in $4$ h; Bob in $6$ h.
- Combined rate: $\tfrac{1}{4} + \tfrac{1}{6} = \tfrac{5}{12}$ room/h.
- Together: $\tfrac{12}{5} = 2.4$ h.
- Pipe A fills a tank in $4$ h; Pipe B in $6$ h.
- Combined rate: $\tfrac{5}{12}$ tank/h.
- Both open: $2.4$ h to fill.
Identical algebra. The "amount" is one room, or one tank, or one batch of brochures — what matters is that everyone is working on the same one.
4. Template 3 — Two rates against each other
Two movers, but now their rates fight rather than cooperate. Either they're approaching each other (rates add to close the gap), separating from each other (rates add to widen it), or one is chasing the other (rates subtract to close the gap).
Meeting (opposite directions, closing)
Two friends $300$ km apart drive toward each other at $60$ and $90$ km/h. When do they meet?
In one hour the gap shrinks by $60 + 90 = 150$ km. They started $300$ km apart, so they meet after $300 / 150 = 2$ hours. The faster driver covers $180$ km in those two hours, the slower $120$ — and $180 + 120 = 300$ ✓.
$$ t_{\text{meet}} = \frac{\text{initial gap}}{v_1 + v_2}. $$Chasing (same direction, closing)
Bob walks north at $5$ km/h starting at 9:00. Al leaves the same door 30 minutes later on a bicycle at $15$ km/h. When does Al catch up?
- By 9:30, Bob has a $5 \cdot 0.5 = 2.5$ km head start.
- Al closes that gap at relative speed $15 - 5 = 10$ km/h.
- Time to close: $2.5 / 10 = 0.25$ h $= 15$ minutes.
- Al catches up at 9:45.
Separating (same direction, opening)
Two cars leave the same place heading north at $40$ and $60$ km/h. After $3$ hours they are $20 \cdot 3 = 60$ km apart — relative speed $\times$ time, same formula.
In any two-mover problem, pretend one of them is standing still and the other moves at the relative speed. Same direction: subtract. Opposite directions: add. From there it's a one-rate problem.
5. The average-speed trap
This is the single most reliable place to lose a mark on a rate problem.
Diego drives $60$ km to a cottage at $40$ km/h, then drives the same $60$ km home at $60$ km/h. What is his average speed for the round trip?
The seductive (wrong) answer is the arithmetic mean: $(40 + 60)/2 = 50$ km/h. Stop and think about why that can't be right: he spends more time on the slow leg than on the fast one, so the slow speed should weigh more in the average. The arithmetic mean weighs them equally.
The honest calculation always works: total distance ÷ total time.
$$ \begin{aligned} \text{outbound time} &= 60 / 40 = 1.5 \text{ h} \\ \text{return time} &= 60 / 60 = 1.0 \text{ h} \\ \text{total time} &= 2.5 \text{ h} \\ \text{total distance}&= 120 \text{ km} \\ \text{average speed} &= 120 / 2.5 = \boxed{48 \text{ km/h}}. \end{aligned} $$Not $50$. When the two legs cover equal distances at speeds $v_1$ and $v_2$, the round-trip average collapses to a closed form — the harmonic mean:
$$ \bar{v} = \frac{2 v_1 v_2}{v_1 + v_2}. $$Plug in: $2 \cdot 40 \cdot 60 / (40 + 60) = 4800/100 = 48$ km/h. Same answer.
The harmonic mean is only correct when the two distances are equal. If the legs have different distances (say $100$ km at $50$ km/h then $200$ km at $100$ km/h), you must go back to total-distance-over-total-time: $(100 + 200)/(100/50 + 200/100) = 300/4 = 75$ km/h. Never trust a formula past the conditions it was derived under.
Why "harmonic"?
The two legs have equal distance but unequal time. If you instead divide the trip into equal time chunks, the arithmetic mean would work — so the arithmetic mean is right for time-weighted averaging, and the harmonic for distance-weighted. The two means coincide only when $v_1 = v_2$; otherwise the harmonic mean is always strictly smaller, which is why "the slow leg pulls the average down."
6. Work problems: pipes, drains, painters
Work problems use Template 2, with one extra wrinkle: some contributors can have negative rates. A drain on a tank is a "worker" that does negative work, undoing the inlet pipe's progress.
Net rate when an outlet is open
A swimming pool's inlet pipe fills it in $3$ hours. A drain, accidentally left open, empties it in $6$ hours. With both running, how long to fill from empty?
$$ \text{inlet rate} = +\tfrac{1}{3}, \quad \text{outlet rate} = -\tfrac{1}{6} $$ $$ \text{net rate} = \tfrac{1}{3} - \tfrac{1}{6} = \tfrac{2}{6} - \tfrac{1}{6} = \tfrac{1}{6} \text{ pool/hour} $$ $$ t = 1 / \tfrac{1}{6} = 6 \text{ hours}. $$Check: in those $6$ hours the inlet has delivered $6 \cdot \tfrac{1}{3} = 2$ pools' worth, the drain has removed $6 \cdot \tfrac{1}{6} = 1$ pool's worth, and the difference is one full pool. ✓
If the outlet rate exceeds the inlet rate, the net rate is negative and the tank never fills — the algebra honestly produces a negative time, which is the equation telling you the situation is impossible as posed. Always sanity-check the sign.
Three or more workers
The rule generalises: for $n$ contributors with solo times $t_1, t_2, \ldots, t_n$,
$$ \frac{1}{t_{\text{together}}} = \sum_{i=1}^{n} \frac{1}{t_i}. $$Example: three printers can do a batch alone in $10$, $15$, and $30$ hours. Together:
$$ \tfrac{1}{10} + \tfrac{1}{15} + \tfrac{1}{30} = \tfrac{3}{30} + \tfrac{2}{30} + \tfrac{1}{30} = \tfrac{6}{30} = \tfrac{1}{5}. $$Combined time: $5$ hours. (Faster than the fastest printer alone, as it must be.)
Man-days: the conservation law
If a job needs $W$ "worker-hours" of effort, then for any combination of $m$ workers and $d$ days at constant individual rates, $m \cdot d = W$. That gives you the scaling rule:
$$ m_1 \cdot d_1 = m_2 \cdot d_2 \quad \text{(same job, equal-rate workers)}. $$If $12$ workers finish a job in $20$ days, the total effort is $240$ worker-days. With $15$ workers, the time drops to $240 / 15 = 16$ days. This breaks down when adding more workers introduces coordination overhead — the math assumes the tasks parallelise cleanly.
7. Units and conversions
Most wrong answers in rate problems come from a unit mismatch, not from bad algebra. Build the habit of writing units alongside numbers and cancelling them like fractions.
| Conversion | Factor | Trick |
|---|---|---|
| km/h → m/s | multiply by $\tfrac{5}{18}$ | $72$ km/h $= 72 \cdot \tfrac{5}{18} = 20$ m/s |
| m/s → km/h | multiply by $\tfrac{18}{5}$ | $15$ m/s $= 15 \cdot \tfrac{18}{5} = 54$ km/h |
| minutes → hours | divide by $60$ | $45$ min $= 0.75$ h |
| litres/min → litres/h | multiply by $60$ | $5$ L/min $= 300$ L/h |
Where does $5/18$ come from? $1$ km/h is $1000$ metres per $3600$ seconds, and $1000/3600 = 5/18$. Memorise it for speed conversions; derive it again whenever you doubt yourself.
A specialised case worth knowing: a train of length $L$ passing a stationary pole travels a distance of $L$ (the front goes from "at the pole" to "rear at the pole"). Passing a platform of length $P$, the train travels $L + P$. Passing another train of length $L_2$, it travels $L + L_2$. The "distance" in $d = vt$ is the distance the train's front must cover to fully clear whatever obstacle it's negotiating.
8. Common pitfalls
If A takes $10$ days and B takes $15$ days, the combined time is not $25$ days, not $12.5$ days, and not $10 + 15$ anything. Add the rates first ($\tfrac{1}{10} + \tfrac{1}{15} = \tfrac{1}{6}$), then invert: $6$ days.
Equal distances at $40$ km/h and $60$ km/h does not average to $50$ km/h — it's $48$. The arithmetic mean is correct only when each speed is held for an equal amount of time, which is almost never the case in a there-and-back trip.
The biggest unit trap: multiplying km/h by minutes, or m/s by hours. Pick one unit system and convert everything else into it before you touch the algebra.
"How long does a $150$ m train at $20$ m/s take to pass a $300$ m platform?" If you compute $300/20 = 15$ s, you've ignored the train's own length. The correct answer is $(150 + 300)/20 = 22.5$ s — the front must clear the platform and the rear has to follow.
For two movers, "relative speed" depends on direction. Same direction: subtract. Opposite directions: add. Mixing these up is the single biggest error in two-mover problems, and it flips the answer wildly — a $10$ km/h gap becomes a $90$ km/h closing rate or vice versa.
9. Worked examples
Try each one yourself before opening the solution. The goal isn't to memorise the answers — it's to see whether your setup matches the canonical template.
Example 1 · Two pipes filling a tank
Pipe A fills a tank in $6$ hours; Pipe B in $4$ hours. Both open, starting from empty — how long to fill?
Step 1. Convert each time to a rate.
$$ r_A = \tfrac{1}{6}, \quad r_B = \tfrac{1}{4} \text{ (tanks/hour)} $$Step 2. Add the rates.
$$ r = \tfrac{1}{6} + \tfrac{1}{4} = \tfrac{2}{12} + \tfrac{3}{12} = \tfrac{5}{12} $$Step 3. Invert for the time.
$$ t = 1 / \tfrac{5}{12} = \tfrac{12}{5} = 2.4 \text{ h} = 2\,\text{h}\,24\,\text{min}. $$Example 2 · Pipe with drain
An inlet fills a tank in $4$ hours; a drain empties the full tank in $6$ hours. Both running. How long to fill from empty?
Step 1. Signed rates.
$$ r_{\text{in}} = +\tfrac{1}{4}, \quad r_{\text{out}} = -\tfrac{1}{6} $$Step 2. Net rate.
$$ r = \tfrac{1}{4} - \tfrac{1}{6} = \tfrac{3}{12} - \tfrac{2}{12} = \tfrac{1}{12} $$Step 3. Time to fill.
$$ t = 12 \text{ hours.} $$An open drain has tripled the fill time — a useful intuition for maintenance debugging.
Example 3 · Round-trip average speed
A cyclist rides $30$ km uphill at $10$ km/h, then returns the same $30$ km downhill at $30$ km/h. Average speed?
Step 1. Times for each leg.
$$ t_{\text{up}} = 30/10 = 3 \text{ h}, \quad t_{\text{down}} = 30/30 = 1 \text{ h} $$Step 2. Totals.
$$ d = 60 \text{ km}, \quad t = 4 \text{ h} $$Step 3. Divide.
$$ \bar{v} = 60/4 = 15 \text{ km/h.} $$Check with the harmonic formula (equal distances): $2 \cdot 10 \cdot 30 / (10 + 30) = 600/40 = 15$ ✓. Not $20$ — the slow climb dominates the time.
Example 4 · Catching up (head start)
Anya leaves home at 8:00 walking at $4$ km/h. Brian leaves at 8:45 cycling at $16$ km/h on the same route. When does Brian catch up?
Step 1. Anya's head start at 8:45.
$$ 4 \cdot 0.75 = 3 \text{ km}. $$Step 2. Closing speed.
$$ 16 - 4 = 12 \text{ km/h}. $$Step 3. Time to close.
$$ 3/12 = 0.25 \text{ h} = 15 \text{ min.} $$So Brian catches Anya at 9:00. Both will have travelled $4$ km from her starting point at that moment — check by computing each.
Example 5 · Meeting from opposite directions
Two trains $360$ km apart move toward each other on parallel tracks at $80$ and $100$ km/h. When do they meet?
Step 1. Relative speed (closing).
$$ 80 + 100 = 180 \text{ km/h}. $$Step 2. Time to close the gap.
$$ 360 / 180 = 2 \text{ h.} $$The slower train will have travelled $160$ km, the faster $200$ km, summing to $360$ ✓.
Example 6 · Train passing a platform
A $180$ m train at $72$ km/h crosses a $220$ m platform. How long?
Step 1. Convert speed.
$$ 72 \text{ km/h} \cdot \tfrac{5}{18} = 20 \text{ m/s.} $$Step 2. Total distance to clear the platform.
$$ 180 + 220 = 400 \text{ m.} $$Step 3. Time.
$$ 400 / 20 = 20 \text{ s.} $$Example 7 · Boat against a current
A boat's still-water speed is $12$ km/h; the current runs at $3$ km/h. Find the upstream speed, the downstream speed, and the time to travel $18$ km upstream then $18$ km back.
Step 1. Effective speeds.
$$ v_{\text{up}} = 12 - 3 = 9, \quad v_{\text{down}} = 12 + 3 = 15 \text{ km/h.} $$Step 2. Times.
$$ t_{\text{up}} = 18/9 = 2 \text{ h}, \quad t_{\text{down}} = 18/15 = 1.2 \text{ h.} $$Step 3. Total time.
$$ 2 + 1.2 = 3.2 \text{ h} = 3 \text{ h } 12 \text{ min.} $$The same template covers planes with headwind/tailwind and people on moving walkways.
Example 8 · Partial-contribution work
A can finish a job alone in $12$ days; B alone in $6$ days. A works alone for $3$ days, then B joins. How long does the whole job take?
Step 1. Fraction done by A in $3$ days.
$$ 3 \cdot \tfrac{1}{12} = \tfrac{1}{4}. $$Step 2. Remaining fraction.
$$ 1 - \tfrac{1}{4} = \tfrac{3}{4}. $$Step 3. Combined rate of A and B.
$$ \tfrac{1}{12} + \tfrac{1}{6} = \tfrac{1}{12} + \tfrac{2}{12} = \tfrac{3}{12} = \tfrac{1}{4} \text{ job/day.} $$Step 4. Time to finish the remaining $\tfrac{3}{4}$.
$$ \tfrac{3}{4} \div \tfrac{1}{4} = 3 \text{ days.} $$Total time: $3 + 3 = 6$ days.