Rates, work and speed

One equation behind every rate problem

A rate is an amount per unit of time: kilometres per hour, pages per minute, tanks per hour. Every rate problem uses the same relationship:

$$\text{amount} = \text{rate} \times \text{time}$$

For travel, the amount is distance and the rate is speed: $d = rt$. For work, the amount is the job (often "1 job") and the rate is the fraction of the job done per unit of time. Rearranged, $\text{rate} = \frac{\text{amount}}{\text{time}}$ and $\text{time} = \frac{\text{amount}}{\text{rate}}$.

A small table keeps multi-part problems tidy: one row per person, vehicle or leg, with columns for rate, time and amount. Fill in what you know, and use $d = rt$ along each row.

Average speed is total distance over total time

$$\text{average speed} = \frac{\text{total distance}}{\text{total time}}$$

It is not the average of the speeds, unless equal times are spent at each speed. Drive 120 km at 60 km/h and 120 km back at 40 km/h: the trip takes $2 + 3 = 5$ hours for 240 km, so the average speed is 48 km/h, not 50. You spend longer at the slower speed, so the average leans towards it.

When the two legs cover equal distances, there is a shortcut: the average speed is $\frac{2ab}{a + b}$ for speeds $a$ and $b$. Here $\frac{2 \cdot 60 \cdot 40}{100} = 48$. But the total-over-total method works for every case, so it is the one to rely on.

Work: add the rates, not the times

If a machine finishes a job in 4 hours, its rate is $\frac{1}{4}$ of the job per hour. Two machines working together add their rates:

$$\frac{1}{4} + \frac{1}{6} = \frac{3}{12} + \frac{2}{12} = \frac{5}{12} \text{ of the job per hour}$$

so together they need $\frac{12}{5} = 2.4$ hours. Times never add or average: the combined time must be less than the faster worker's time alone, because the slower one still helps. That sanity check rules out many wrong choices at a glance.

In general, for times $a$ and $b$ alone, the time together is $\frac{ab}{a + b}$. If one worker undoes the work (a drain emptying a tank that a pipe fills), subtract that rate.

If you know the combined time and one worker's time, subtract rates to find the other: together 6 hours and $A$ alone 10 hours means $B$'s rate is $\frac{1}{6} - \frac{1}{10} = \frac{1}{15}$, so $B$ alone takes 15 hours.

Gaps: relative speed

When two objects move along the same line, think about the gap between them.

• Moving towards each other, or apart in opposite directions, the gap changes at the sum of the speeds.
• Moving in the same direction, the gap changes at the difference of the speeds.

Then $\text{time} = \frac{\text{gap to close or open}}{\text{rate of change of the gap}}$. A cyclist at 18 km/h chasing one at 12 km/h who has a 6 km lead closes the gap at 6 km/h, so she catches up in 1 hour.

Units

Keep every quantity in one unit system before you combine. A speed in km/h and a time in minutes must be reconciled first: 40 minutes is $\frac{40}{60} = \frac{2}{3}$ hour. To convert a rate, multiply by fractions equal to 1:

$$72 \ \frac{\text{km}}{\text{h}} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}} = 20 \ \frac{\text{m}}{\text{s}}$$

Unwanted units cancel, which tells you whether to multiply or divide. Read the question's unit last of all: if it asks for minutes and you have hours, convert before you choose.

Machines and people at the same rate

If identical machines work at the same rate, the work done is proportional to (number of machines) $\times$ (time). If 5 machines make 300 parts in 4 hours, one machine makes $\frac{300}{5 \cdot 4} = 15$ parts per hour, and 8 machines make $8 \cdot 15 \cdot 6 = 720$ parts in 6 hours. Find the rate of a single unit first; every variant of the question then follows.