Rate and Work Problems: Additional Practice (Set 2)

Add all three rates over the common denominator $6$.

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

The combined rate is $\frac{2}{3}$ per hour, so the time is the reciprocal, $\frac{3}{2} = 1.5$ hours.

The drain rate is the fill rate minus the net rate, over the common denominator $12$.

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

The drain empties $\frac{1}{12}$ of the basin per hour, so it takes $12$ hours alone.

Subtract A's rate from the combined rate, over the common denominator $60$.

$\frac{1}{12} - \frac{1}{20} = \frac{5}{60} - \frac{3}{60} = \frac{2}{60} = \frac{1}{30}$

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

Average speed is total distance over total time. The total distance is $150 + 90 = 240$ miles. The times are $\frac{150}{50} = 3$ hours and $\frac{90}{45} = 2$ hours, a total of $5$ hours.

$\text{average speed} = \frac{240}{5} = 48 \text{ mph}$

Average speed is total distance over total time. The total distance is $100 + 120 = 220$ miles. The times are $\frac{100}{50} = 2$ hours and $\frac{120}{40} = 3$ hours, so the total is $5$ hours.

$\text{average speed} = \frac{220}{5} = 44 \text{ mph}$

Each tiles $\frac{1}{10}$ of the floor per hour, so add the rates.

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

The combined rate is $\frac{1}{5}$ per hour, so together they finish in $5$ hours.

Average speed is total distance over total time. The distance is $60 + 60 = 120$ miles. The times are $\frac{60}{30} = 2$ hours and $\frac{60}{20} = 3$ hours, a total of $5$ hours.

$\text{average speed} = \frac{120}{5} = 24 \text{ mph}$

In its $2$-hour head start the first car covers $40 \times 2 = 80$ miles. The second car closes that gap at $60 - 40 = 20$ mph.

$t = \frac{80}{20} = 4 \text{ hours}$

So the second car catches up $4$ hours after it starts.

Average speed is total distance over total time. The distance is $120 + 120 = 240$ miles. The times are $\frac{120}{60} = 2$ hours and $\frac{120}{40} = 3$ hours, a total of $5$ hours.

$\text{average speed} = \frac{240}{5} = 48 \text{ mph}$

The answer falls below the mean of $50$ because more time was spent at $40$ mph.

Average speed is total distance over total time. The distance is $40 + 40 = 80$ miles. The times are $\frac{40}{40} = 1$ hour out and $\frac{40}{10} = 4$ hours back, a total of $5$ hours.

$\text{average speed} = \frac{80}{5} = 16 \text{ mph}$

The answer is far below the mean of $25$ because the return leg at $10$ mph ate up most of the time.

The three equal rates add to the combined rate $\frac{1}{4}$ per hour, so each pump is one third of that.

$\text{one pump} = \frac{1}{3} \times \frac{1}{4} = \frac{1}{12} \text{ per hour}$

A single pump empties $\frac{1}{12}$ of the tank per hour, so it takes $12$ hours alone.

Subtract the known rate from the combined rate, over the common denominator $24$.

$\frac{1}{T_2} = \frac{1}{8} - \frac{1}{12} = \frac{3}{24} - \frac{2}{24} = \frac{1}{24}$

The other worker does $\frac{1}{24}$ of the job per hour, so it takes $24$ hours alone.