Rate and Work Problems: Practice

Add the rates over the common denominator $30$.

$\frac{1}{10} + \frac{1}{15} = \frac{3}{30} + \frac{2}{30} = \frac{5}{30} = \frac{1}{6} \text{ per hour}$

The combined rate is $\frac{1}{6}$ per hour, so together they take $6$ hours.

Time is distance divided by rate.

$t = \frac{270}{45} = 6 \text{ hours}$

Add the rates over the common denominator $60$.

$\frac{1}{20} + \frac{1}{30} = \frac{3}{60} + \frac{2}{60} = \frac{5}{60} = \frac{1}{12} \text{ per hour}$

The combined rate is $\frac{1}{12}$ per hour, so together they finish in $12$ hours.

The drain is a negative rate, so subtract it from the fill rate.

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

The net rate is $\frac{1}{3}$ tank per hour, so the tank fills in $3$ hours. Adding the rates instead would wrongly give $1.5$ hours.

Distance is rate times time.

$d = 16 \times 2.5 = 40 \text{ miles}$

Add the rates over the common denominator $20$.

$\frac{1}{5} + \frac{1}{20} = \frac{4}{20} + \frac{1}{20} = \frac{5}{20} = \frac{1}{4} \text{ per hour}$

The combined rate is $\frac{1}{4}$ per hour, so the time is $4$ hours.

Distance is rate times time.

$d = 52 \times 3 = 156 \text{ miles}$

Each paints $\frac{1}{8}$ of the room per hour, so add the rates.

$\frac{1}{8} + \frac{1}{8} = \frac{2}{8} = \frac{1}{4} \text{ per hour}$

The combined rate is $\frac{1}{4}$ per hour, so together they finish in $4$ hours.

Subtract the known rate from the combined rate.

$\frac{1}{T_2} = \frac{1}{4} - \frac{1}{12} = \frac{3}{12} - \frac{1}{12} = \frac{2}{12} = \frac{1}{6}$

The other tap fills $\frac{1}{6}$ per hour, so it takes $6$ hours alone.

Subtract the drain rate from the fill rate.

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

The net rate is $\frac{1}{4}$ tank per hour, so the cistern fills in $4$ hours.

Average speed is total distance over total time. The distances are $30 \times 2 = 60$ and $50 \times 3 = 150$ miles, so the total is $210$ miles in $5$ hours.

$\text{average speed} = \frac{210}{5} = 42 \text{ mph}$

The plain mean of $30$ and $50$ is $40$, but more time was spent at $50$ mph, so the average is higher.

Distance is rate times time.

$d = 55 \times 4 = 220 \text{ miles}$