Rate and Work Problems: Practice

A whole job finished in $T$ units of time is done at the rate $\frac{1}{T}$ per unit. Here $T = 8$ hours.

$\text{rate} = \frac{1}{8} \text{ of the batch per hour}$

The number $8$ is the time, not the rate; the rate is its reciprocal.

Each pipe fills $\frac{1}{4}$ of the tank per hour, and rates add when they run together.

$\frac{1}{4} + \frac{1}{4} = \frac{1}{2} \text{ of the tank per hour}$

Together they fill half the tank each hour.

Add the two rates over the common denominator $6$.

$\frac{1}{6} + \frac{1}{3} = \frac{1}{6} + \frac{2}{6} = \frac{1}{2} \text{ per hour}$

Rates add directly; you do not add the fractions across the top and bottom.

Speed is distance divided by time, $r = \dfrac{d}{t}$.

$r = \frac{24}{3} = 8 \text{ miles per hour}$

Amount of work is rate times time. In $2$ hours she does twice her hourly rate.

$2 \times \frac{1}{5} = \frac{2}{5} \text{ of the fence}$

Time is distance divided by rate, $t = \dfrac{d}{r}$.

$t = \frac{180}{60} = 3 \text{ hours}$

Add the rates over the common denominator $12$.

$\frac{1}{4} + \frac{1}{12} = \frac{3}{12} + \frac{1}{12} = \frac{4}{12} = \frac{1}{3} \text{ per hour}$

Distance is rate times time, $d = rt$.

$d = 15 \times 4 = 60 \text{ miles}$

Speed is distance divided by time.

$r = \frac{40}{5} = 8 \text{ miles per hour}$

A road paved in $12$ days is paved at the rate $\frac{1}{12}$ per day.

$\text{rate} = \frac{1}{12} \text{ of the road per day}$

Distance is rate times time, $d = rt$.

$d = 400 \times 6 = 2{,}400 \text{ miles}$

Add the two equal rates.

$\frac{1}{10} + \frac{1}{10} = \frac{2}{10} = \frac{1}{5} \text{ per hour}$

Rates add, so two equal helpers double the rate rather than halving or squaring it.