Rate and Work Problems: Practice

Average speed is total distance over total time. The distance is $90 + 90 = 180$ miles. The times are $\frac{90}{45} = 2$ hours out and $\frac{90}{30} = 3$ hours back, a total of $5$ hours.

$\text{average speed} = \frac{180}{5} = 36 \text{ mph}$

The mean of $45$ and $30$ is $37.5$, but more time was spent at the slower $30$ mph, so the true average is lower.

Add the two fill rates and subtract the drain rate, over the common denominator $12$.

$\frac{1}{4} + \frac{1}{6} - \frac{1}{12} = \frac{3}{12} + \frac{2}{12} - \frac{1}{12} = \frac{4}{12} = \frac{1}{3} \text{ per hour}$

The net rate is $\frac{1}{3}$ tank per hour, so the tank fills in $3$ hours.

Subtract the leak rate from the fill rate, over the common denominator $30$.

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

The net rate is $\frac{1}{15}$ pond per hour, so the pond fills in $15$ hours.

In $6$ hours the worker paints $6 \times \frac{1}{9} = \frac{6}{9} = \frac{2}{3}$ of the fence. The unpainted part is what remains of the whole.

$1 - \frac{2}{3} = \frac{1}{3}$

So one third of the fence is still unpainted.

Their combined rate is $\frac{1}{3}$ of the job per hour, so in $1$ hour they finish $\frac{1}{3}$ of it.

$1 - \frac{1}{3} = \frac{2}{3}$

Two thirds of the job is left.

Average speed is total distance over total time. The distance is $24 + 24 = 48$ miles. The times are $\frac{24}{6} = 4$ hours and $\frac{24}{12} = 2$ hours, a total of $6$ hours.

$\text{average speed} = \frac{48}{6} = 8 \text{ mph}$

The mean of $6$ and $12$ is $9$, but more time was spent going the slow $6$ mph, so the average is lower.

The combined rate is $\frac{1}{2}$ per hour. Subtract the two known rates.

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

The third member also works at $\frac{1}{6}$ of the job per hour.

Subtract the drain rate from the fill rate, over the common denominator $12$.

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

The net rate is only $\frac{1}{12}$ tub per hour, so the tub takes $12$ hours to fill.

The gap between them closes at the sum of the two speeds, because both rates work to shrink it.

$40 + 60 = 100 \text{ mph closing speed}$

The time to close $300$ miles is $\frac{300}{100} = 3$ hours.

Add the two fill rates and subtract the drain rate, over the common denominator $24$.

$\frac{1}{6} + \frac{1}{12} - \frac{1}{8} = \frac{4}{24} + \frac{2}{24} - \frac{3}{24} = \frac{3}{24} = \frac{1}{8} \text{ per hour}$

The net rate is $\frac{1}{8}$ tank per hour, so the tank fills in $8$ hours.

The net fill rate equals the pipe rate minus the drain rate, so the drain rate is the pipe rate minus the net rate.

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

The drain empties $\frac{1}{6}$ of the tank per hour, so it would empty a full tank in $6$ hours.

Let each half be $d$ miles. The times are $\frac{d}{60}$ and $\frac{d}{30}$, so the total time is $\frac{d}{60} + \frac{2d}{60} = \frac{3d}{60} = \frac{d}{20}$ for a total distance of $2d$.

$\text{average speed} = \frac{2d}{d/20} = 40 \text{ mph}$

The distance $d$ cancels, and the average lands below the mean of $45$ because more time is spent at $30$ mph.