Rate, distance, time and work problems
Many word problems can be solved with rational equations. On this page we look at two common types: distance, rate and time problems involving a current, and work problems in which two people or machines work together.
Distance, rate and time problems
Distance, rate and time are related by
$d=r\cdot t$
where $d$ is the distance, $r$ is the rate (speed), and $t$ is the time. Solving for the time gives
$t=\dfrac{d}{r}$
When a boat travels in a river, the current changes its speed. Going downstream (with the current), the current adds to the boat's speed, so the rate is the boat's still-water speed plus the current. Going upstream (against the current), the current slows the boat down, so the rate is the boat's still-water speed minus the current.
Example 1: A boat travels $12$ miles upstream (against the current) in the same amount of time it takes to travel $20$ miles downstream (with the current). The current flows at $2$ miles per hour. What is the rate of the boat in still water?
Solution:
Let $x$ be the rate of the boat in still water, in miles per hour. Then
the downstream rate is $x+2$, and the upstream rate is $x-2$.
Since time equals distance divided by rate, $t=\dfrac{d}{r}$, the two times are
downstream time $=\dfrac{20}{x+2}$, and upstream time $=\dfrac{12}{x-2}$.
The problem says these two times are equal, so we set up a proportion:
$\dfrac{20}{x+2}=\dfrac{12}{x-2}$
Cross-multiply:
$20(x-2)=12(x+2)$
Distribute:
$20x-40=12x+24$
Collect the variable terms on one side and the constants on the other:
$\begin{align*}20x-12x&=24+40\\8x&=64\\x&=8\end{align*}$
So the rate of the boat in still water is $8$ miles per hour.
Example 2: A boat travels $12$ miles per hour in still water. It can travel $6$ miles upstream in the same amount of time it takes to travel $10$ miles downstream. Find the rate of the river current.
Solution:
Let $x$ be the rate of the current, in miles per hour. Then
the downstream rate is $12+x$, and the upstream rate is $12-x$.
Using $t=\dfrac{d}{r}$, the two times are
downstream time $=\dfrac{10}{12+x}$, and upstream time $=\dfrac{6}{12-x}$.
These times are equal, so
$\dfrac{10}{12+x}=\dfrac{6}{12-x}$
Cross-multiply:
$10(12-x)=6(12+x)$
Distribute:
$120-10x=72+6x$
Collect the variable terms on one side and the constants on the other:
$\begin{align*}120-72&=6x+10x\\48&=16x\\x&=3\end{align*}$
So the rate of the current is $3$ miles per hour.
Work problems
In a work problem, two people or machines do the same job at different rates. If one alone finishes the job in a time $t_1$ and the other alone finishes it in a time $t_2$, and together they finish it in a time $t$, then the times are related by
$\dfrac{1}{t}=\dfrac{1}{t_1}+\dfrac{1}{t_2}$
This says that the amount of work each does per unit time adds up to the amount of work they do together per unit time.
Example 1: One water faucet takes $18$ minutes to fill a bathtub, and a second faucet takes $9$ minutes to fill the same tub. How long will it take to fill the tub with both faucets working together?
Solution:
Let $x$ be the time, in minutes, for both faucets working together. Here $t_1=18$ and $t_2=9$, so
$\dfrac{1}{x}=\dfrac{1}{18}+\dfrac{1}{9}$
The least common denominator of $18$ and $9$ is $18$. Multiply every term by $18x$ to clear the fractions:
$\dfrac{1}{x}\cdot 18x=\dfrac{1}{18}\cdot 18x+\dfrac{1}{9}\cdot 18x$
Simplifying each term:
$18=x+2x$
Combine like terms:
$\begin{align*}18&=3x\\x&=6\end{align*}$
So both faucets working together fill the tub in $6$ minutes.
Example 2: Antonio and Bob are installing a new roof. Working alone, Antonio can do the job in $15$ hours. Working together, they can complete the job in $10$ hours. How long would it take Bob to install the roof working alone?
Solution:
Let $x$ be the time, in hours, for Bob working alone. Here the together time is $t=10$, Antonio's time is $15$, and Bob's time is $x$, so
$\dfrac{1}{10}=\dfrac{1}{15}+\dfrac{1}{x}$
The least common denominator of $10$, $15$ and $x$ is $30x$. Multiply every term by $30x$ to clear the fractions:
$\dfrac{1}{10}\cdot 30x=\dfrac{1}{15}\cdot 30x+\dfrac{1}{x}\cdot 30x$
Simplifying each term:
$3x=2x+30$
Collect the variable terms on one side:
$\begin{align*}3x-2x&=30\\x&=30\end{align*}$
So it would take Bob $30$ hours to install the roof working alone.