Rate and Work Problems: Additional Practice (Set 1)

Average speed is total distance over total time. The distance is $150 + 150 = 300$ miles. The times are $\frac{150}{50} = 3$ hours out and $\frac{150}{30} = 5$ hours back, a total of $8$ hours.

$\text{average speed} = \frac{300}{8} = 37.5 \text{ mph}$

The mean of $50$ and $30$ is $40$, but more time was spent at the slower $30$ mph.

Add the three rates over the common denominator $12$.

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

The combined rate is $\frac{1}{3}$ per hour, so the time is the reciprocal, $3$ hours.

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

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

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

Subtract A's rate from the combined rate, over the common denominator $30$.

$\frac{1}{10} - \frac{1}{15} = \frac{3}{30} - \frac{2}{30} = \frac{1}{30}$

B works at $\frac{1}{30}$ of the job per hour, so B takes $30$ hours alone.

Subtract the drain rate from the fill rate, over the common denominator $6$.

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

The net rate is $\frac{1}{6}$ tank per hour, so the tank fills in $6$ hours.

The combined rate is $\frac{1}{2}$ per hour. Subtract the two known rates, over the common denominator $10$.

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

The third worker does $\frac{1}{10}$ of the job per hour.

Time is distance divided by rate.

$t = \frac{150}{60} = 2.5 \text{ hours}$

Add the rates over the common denominator $60$.

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

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

Distance is rate times time.

$d = 18 \times 2.5 = 45 \text{ miles}$

Subtract the drain rate from the fill rate, over the common denominator $24$.

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

The net rate is $\frac{1}{12}$ tank per hour, so the tank fills in $12$ hours.

Average speed is total distance over total time. The distances are $30 \times 1 = 30$ and $55 \times 4 = 220$ miles, a total of $250$ miles in $5$ hours.

$\text{average speed} = \frac{250}{5} = 50 \text{ mph}$

The average sits well above the mean of $42.5$ because far more time was spent at $55$ mph.

Let the one-way distance be $d$ miles. The times are $\frac{d}{6}$ and $\frac{d}{3}$, so the total time is $\frac{d}{6} + \frac{2d}{6} = \frac{3d}{6} = \frac{d}{2}$ for a total distance of $2d$.

$\text{average speed} = \frac{2d}{d/2} = 4 \text{ mph}$

The distance cancels, and the average falls below the mean of $4.5$ because more time is spent walking at $3$ mph.