Math Core

Trains leaving stations, runners catching up, pipes filling pools: these problems all rest on one formula, $d = rt$ , plus two ideas about how rates combine. Once you see that, "work" problems turn out to be distance problems in disguise.

Distance equals rate times time

$d = r \cdot t \qquad r = \frac{d}{t} \qquad t = \frac{d}{r}$

Keep units consistent. If the speed is in miles per hour, the time must be in hours: $20$ minutes is $\dfrac{1}{3}$ hour.

Relative speed

When two objects move along the same line, you can often pretend one of them stands still.

Toward each other (or apart in opposite directions): the gap between them changes at the sum of their speeds.
Same direction: the gap changes at the difference of their speeds. This is the catch-up rule.

Worked example: Toward each other

Two trains are $300$ miles apart and head toward each other at $70$ mph and $80$ mph. How long until they meet?

The gap shrinks at $70 + 80 = 150$ mph, so they meet after $300 \div 150 = 2$ hours.

Worked example: Catching up

Ana leaves on her bike at $12$ mph. One hour later, Ben follows the same road at $18$ mph. How far from the start does Ben catch her?

When Ben starts, Ana is $12$ miles ahead. Ben gains $18 - 12 = 6$ miles each hour, so he catches her $12 \div 6 = 2$ hours after he leaves. He has gone $18 \cdot 2 = 36$ miles.

On a distance-time graph, each rider is a line whose slope is the speed. Ben's steeper line starts an hour late and meets Ana's line at $(3, 36)$ .

Distance (miles) against hours after Ana leaves. The steeper line is Ben.Open in grapher →

Average speed

Common mistake

Average speed is total distance divided by total time. It is not the average of the speeds. Driving somewhere at $30$ mph and back at $60$ mph does not average $45$ mph, because you spend twice as long at the slow speed.

Worked example: There and back

Kim drives $60$ miles to her aunt's house at $30$ mph and drives back at $60$ mph. What is her average speed for the round trip?

Going takes $60 \div 30 = 2$ hours; returning takes $1$ hour. The total is $120$ miles in $3$ hours: $40$ mph.

Work problems

Treat a whole job (painting a fence, filling a tank) as "distance" $1$ . Someone who finishes a job in $6$ hours works at a rate of $\dfrac{1}{6}$ job per hour.

Rates add

When people or machines work together, their rates add (not their times). If one pipe fills a tank in $a$ hours and another in $b$ hours, together they fill

$\frac{1}{a} + \frac{1}{b} \text{ tanks per hour,}$

and the time to fill one tank together is $1$ divided by that sum.

Worked example: Two pipes

Pipe A fills a tank in $6$ hours and pipe B fills it in $3$ hours. How long do they take together?

Together they fill $\dfrac{1}{6} + \dfrac{1}{3} = \dfrac{1}{2}$ of the tank per hour, so the job takes $2$ hours.

For jobs done by several identical workers, count worker-days (or worker-hours): $4$ workers for $6$ days is $24$ worker-days of work, no matter how it's split up.

Tip

Sanity check a work answer: working together must be faster than the fastest worker alone, but no faster than if everyone were as fast as the fastest worker.