Math · College algebra · Worked example

Solve a work-rate problem

One pipe fills a tank in $4 hours$ , and a second pipe fills it in $6 hours$ . How long do they take together?

\frac{1}{4} + \frac{1}{6} = \frac{1}{t}

Turn each time into a rate

A pipe that takes $4 hours$ fills $1 / 4$ of the tank each hour. The second fills $1 / 6$ of the tank each hour. Rates can be added. Times cannot, because two pipes clearly finish sooner than either one alone.

\frac{1}{4} tank per hour, \frac{1}{6} tank per hour

\begin{array}{c} \frac{1}{4} tank per hour \\ \frac{1}{6} tank per hour \end{array}

Write the equation

If together they take $t$ hours, their combined rate is $1 / t$ of the tank per hour. The two rates add up to it. The only excluded value is $t = 0$ .

\frac{1}{4} + \frac{1}{6} = \frac{1}{t}, t \neq 0

Multiply every term by the LCD

The LCD of $4$ , $6$ and $t$ is $12 t$ . Then $12 t / 4 = 3 t$ , $12 t / 6 = 2 t$ and $12 t / t = 12$ .

3 t + 2 t = 12

Collect the terms

$3 t + 2 t = 5 t$ .

5 t = 12

Solve and check the excluded value

Divide by $5$ to get $t = 12 / 5$ . It is not the excluded value $0$ , so it stands.

t = \frac{12}{5}

Interpret and check

$12 / 5$ hours is $2.4 hours$ , or $2 hours$ $24 minutes$ . That is less than $4 hours$ , as two pipes should be faster than the quicker one alone. In $2.4 hours$ the first pipe fills $2.4 / 4 = 0.6$ of the tank and the second fills $2.4 / 6 = 0.4$ , which make one full tank.

\frac{2.4}{4} + \frac{2.4}{6} = 0.6 + 0.4 = 1

Result

Together the pipes take $12 / 5 = 2.4 hours$ , which is $2 hours$ $24 minutes$ .

t = \frac{12}{5}

Your turn

1. One like the example

One painter takes $3 hours$ to paint a room and another takes $6 hours$ . How long do they take together?

Show the answer and explanation

$2 hours$ .

Their rates are $1 / 3$ and $1 / 6$ of the room per hour, so $1 / 3 + 1 / 6 = 1 / t$ . Multiplying every term by $6 t$ gives $2 t + t = 6$ , so $t = 2$ . Check: in $2 hours$ they paint $2 / 3 + 1 / 3 = 1$ room.

\begin{array}{l} \frac{1}{3} + \frac{1}{6} = \frac{1}{t} \\ 2 t + t = 6 \\ t = 2 \end{array}

2. A variation

A pipe fills a tank in $3 hours$ , but a drain empties it in $5 hours$ . With both open, how long does the tank take to fill?

Show the answer and explanation

$7.5 hours$ .

The drain works against the pipe, so its rate is subtracted: $1 / 3 - 1 / 5 = 1 / t$ . Multiplying every term by $15 t$ gives $5 t - 3 t = 15$ , so $2 t = 15$ and $t = 15 / 2 = 7.5$ . It is longer than $3 hours$ , as it should be with the drain open.

\begin{array}{l} \frac{1}{3} - \frac{1}{5} = \frac{1}{t} \\ 5 t - 3 t = 15 \\ 2 t = 15 \\ t = \frac{15}{2} \end{array}

3. Somewhere new

A boat moves at $5 km / h$ in still water. It travels $12 km$ upstream and $12 km$ back in $5 hours$ . How fast is the current?

Show the answer and explanation

$1 km / h$ .

With current $c$ , the speeds are $5 - c$ upstream and $5 + c$ downstream, and time is distance over speed. The excluded values are $c = 5$ and $c = - 5$ . Multiplying every term by $(5 - c) (5 + c)$ gives $120 = 125 - 5 c^{2}$ , so $c^{2} = 1$ . The current is a speed between $0$ and $5$ , so $c = 1 km / h$ . Check: $12 / 4 + 12 / 6$ $= 3 + 2$ $= 5 hours$ .

\begin{array}{l} \frac{12}{5 - c} + \frac{12}{5 + c} = 5 \\ 12 (5 + c) + 12 (5 - c) = 5 (25 - c^{2}) \\ 120 = 125 - 5 c^{2} \\ c^{2} = 1 \\ c = 1 or c = - 1 \end{array}

Keep exploring

Open the steps in Math and change $1 / 6$ to $1 / 12$ . The LCD becomes $12 t$ , and the time together becomes $3 hours$ .

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Your existing work stays on this device. Examples open as editable copies.

Turn each time into a rate

Write the equation

Multiply every term by the LCD

Collect the terms

Solve and check the excluded value

Interpret and check

Result

Your turn

1. One like the example

2 hours2 hours.

2. A variation

7.5 hours7.5 hours.

3. Somewhere new

1 km⁢/⁢h1 km/h.

Keep exploring

Study it, then try it

$2 hours$ .

$7.5 hours$ .

$1 km / h$ .