Systems of Equations Application: Rate Word Problems
This section will show you one common applications of systems of equations: problems dealing with rates. Such problems might deal with distance traveled, work done, volumes filled, and so on. For these kinds of problems to be in a “system” of equations, we usually have two things happeneing together (i.e. two people are working at different rates to finish a job together, a pool or tub is fill and leaking/emptying at the same time, two cars or trains are traveling towards each other, etc.). This is a basic application to physics and will be expanded on in calculus, since most rate problems are, in reality, not constant (i.e. a pool will not empty at a constant rate, since the water pressure changes as the pool emplties, causing the pool to drain at a changing rate). Thus, do not consider these problems true to the real world, but merely approximations. Later, we will try to model these problems more accurately with the tools of calculus.
1. Rates and Time: Fundamental Premises
A rate is a mathematical way of relating two quantities that are usually measured in different units. You are likely familiar with some rates in your everyday life: speed of a car is measured in miles per hour or kilometeres per hour ($\text{mi}/\text{hr}$ or $\text{km}/\text{hr}$) or a taxi might charge you some dollar amount per mile or kilometer driven ($dollars/mile$). You are using rates all the time (no pun intended – you’ll get it in a bit)!
Since rates intrinsically deal with time, $t$, we can create a simple equation with time as a our independent variable. In essence, we are “doing a thing” (driving, filling, working, etc.) for some time; the rate of our progress in this “doing” is our “doing” divided by our time; for example, I drive 60 miles in one hour means that I must have been driving (on average) 60 miles per hours (i.e. $\frac{60\ \text{miles}}{1\ \text{hour}}$). If we think of our rate, $r$, as the “thing done” over the “time it took to do the thing”, we can write a basic equation for rate:
$$r=\frac{\text{thing done}}{\text{time taken to do the thing}}$$
or more simply
$$r=\frac{d}{t}$$
In this, $d$ often represents “distance,” since many rate problems deal with distance traveled. However, we can consider $d$ more abstractly and “something done” where $d$ stands for “done”. We can reorganize the above definition and also say that
$$rt=d$$
Both of these equations might be useful under differing circumstances.
A classic rate problem in algebra courses (one often featured as a mind-bending problem in television shows or movies) sends two hypothetical trains rushing towards each other at different speeds, and asks you to determine when/where they will meet. Less exciting, but also common, rate situations involve calculating work rates or the time it takes for a pool to fill or empty. While these problems are often intimidating due to their classical associations with difficult algebra problems, we can solve all rate problems with the same method: creating a system of equations to represent the situation.
2. Speed as a Rate
Given our present understanding of rates as given above, let’s just dive into the most classic of rate problems in Algbera: trains! Consider the following problem:
Two trains leave stations 60 miles apart at the same time heading toward one another on parallel tracks. Train A is traveling 30 miles per hour, while Train B is traveling 50 miles per hour. When do they pass each other?
The mothod to solving rate problems, as previously stated, is to create a system of equations, in these cases a system of rate equations. We have an equation that describes rate, so we should use it! We have two trains traveling at two different speeds (a rate of distance over time), but over the same amount of time. The only thing we do not have is the distance each train travels, two distances which we will need to compare at the end of our problem to understand where/when they will meet.
For any word problem, the most fundamental strategy is to write everything you can say. Given that all these rate problems are, in essence, the same, we can create a standard template to work from to solve these problems: we will set up a table with two rate equations and fill in what we know from the problem given. This will help us organize our data and create a system that we can solve. Let’s make a table with columns for “distance,” “rate,” and “time,” and a row for each train and fill in what we know.
| Object (i.e. train) | Thing Done (i.e. distance traveled) | $=$ | Rate of thing done (i.e. speed) | $\cdot$ | Time it takes to do the thing (i.e. travel a distance) |
| Train A | $=$ | 30 | $\cdot$ | ||
| Train B | $=$ | 50 | $\cdot$ |
Notice that we have left the “Distance” and “Time” columns empty because the specific distances and times that the trains traveled were not provided in the problem. But we can use some mathematical thinking to find these values.
The problem tells us that the stations are 60 miles apart. Let’s use this total distance value and make some algebraic statements about the unknown distance of their meeting point. The trains will meet at some point between the stations, so we can call the distance Train A travels $d$, and the distance Train B travels $60-d$. This can be simply diagramed like in the below image:
One way of thinking about these two distances is “Train A will travel some distance to the meeting point, and Train B will have to travel a different distance, and the total distance will be 60 miles.” Let’s put this information into the chart:
| Object (i.e. train) | Thing Done (i.e. distance traveled) | $=$ | Rate of thing done (i.e. speed) | $\cdot$ | Time it takes to do the thing (i.e. travel a distance) |
| Train A | $d$ | $=$ | 30 | $\cdot$ | |
| Train B | $60-d$ | $=$ | 50 | $\cdot$ | |
Finally, there is the matter of time. The problem asks us to find when these trains will pass each other, so we know that this quantity is unknown. However, if the trains leave at the same time from different stations, then they must have traveled the same amount of time when they pass each other, no matter how fast or far each one goes. So because these times are equivalent, we can use the same variable, $t$ , for both of them.
Based on this new information, our table now looks like this:
| Object (i.e. train) | Thing Done (i.e. distance traveled) | $=$ | Rate of thing done (i.e. speed) | $\cdot$ | Time it takes to do the thing (i.e. travel a distance) |
| Train A | $d$ | $=$ | 30 | $\cdot$ | t |
| Train B | $60-d$ | $=$ | 50 | $\cdot$ | t |
Now we have our system of equations! Train A is represented by the equation $d=30t$, and Train B is represented by the equation $60-d=50t$. We can use the substitution method to substitute $30t$ for $d$ in the second equation, and then solve it for $t$.
$$60-d=50t$$
$$60-30t=50t$$
$$60-30t+30t=50t+30t$$
$$60=80t$$
$$\frac{60}{80}=\frac{80t}{80}$$
$$\frac{3}{4}=t$$
So, since $t=\frac{3}{4}$, the trains will meet in $\frac{3}{4}$ of an hour or 45 minutes after they have begun their journey. Given this time, $t$, we can now calculate how far each train has traveled by substituting the value for $t$ back into each equation:
Train A: $d=30t\rightarrow{d=30(\frac{3}{4})}\rightarrow{d=22.5)$
Train B: $60-d$ is the distance traveled, so $60-22.5=37.5$
Thus, Train A traveled 22.5 miles and Train B traveled 37.5 miles.
Notice how we have made the tables in the above problem with very general labels. This problem can be applied in many different circumstances where a thing is done over time at a certain rate.
Wages can be thought of a rate of pay per hours worked:
$$\text{work}=\text{wage}\cdot\text{time worked}$$
Filling (or emptying) a container (a pool, a tub, etc.) can be thought of as a rate (or negative rate in the case of emptying) of volume filled per time filling:
$$\text{volume of water filled}=\text{rate of water filling}\cdot\text{time spent filling}$$
Notice that in all these cases, the general circumstance is that:
$$\text{a thing done}=\text{rate of the thing being done}\cdot\text{time spent doing the thing}$$
We can expand our general form of rates above to create a general form for a “collective rate” or the rate of multiple things working together or against one another (or travel with or towards one another, etc.)
Let us say that a thing $d$ is done in a time $t_A$ by an object A, which has a rate we will call $A$. Then we can say that
$$A=\frac{d}{t_A}$$
Let us also say that the same thing $d$ is done in a time $t_B$ by an object B, which has a rate we will call $B$. Then we can say that
$$B=\frac{d}{t_B}$$
Then the rate at which both object do something togeter (+) or against one another (-) is
$$A\pm{B}=\text{combined rate of doing by A and B}$$
If we consider the object A and B working together (or against each other) on the same thing $d$, then their combined rates of doing will complete the thing $d$ (or uncomplete the thing $d$) in some joint time $t_{B\bigcup{A}}$:
$$(A\pm{B})t_{B\bigcup{A}}=d$$
Given our equations for $A$ and $B$, we can make a substituteion and say that
$$\left(\frac{d}{t_A}\pm\frac{d}{t_B})=\frac{d\right}{t_{B\bigcup{A}}}$$
With this combined rate equation, we can fill in as much information as we know and easily solve for missing quantities. Often in this equation the value of $d$ is 1, since often 1 job is done or one pool is filled or one room is painted, and so one.
Let’s consider two examples, one where two people are working together and one where two things are working against each other:
Example: Adam can clean a room in 3 hours. If his sister Maria helps, they can clean it in 2.4 hours. How long will it take Maria to do the job alone?
We use the work-rate equation to model the problem, but before doing this, we can display the information on a table if it helps:
| time | job per hour (rate) | |
| Adam | 3 | $\frac{1}{3}$ |
| Maria | t | $\frac{1}{t}$ |
| Together | 2.4 | $\frac{1}{2.4}$ |
Now, let’s set up the equation and solve.
$$\left(\frac{d}{t_A}\pm\frac{d}{t_B}\right)=\frac{d}{t_{B\bigcup{A}}}$$
$$(\frac{1}{3}+\frac{1}{t_\text{Maria}}=\frac{1}{2.4}$$
$$\frac{1}{t}=\frac{1}{\frac{12}{5}}-\frac{1}{3}$$
$$\frac{1}{t}=\frac{5}{12}-\frac{4}{12}$$
$$\frac{1}{t}=\frac{1}{12}$$
And so, $t_\text{Maria}=12$. It would then take Maria 12 hours to clean the room by herself.
Example: A sink can be filled by a pipe in 5 minutes, but it takes 7 minutes to drain a full sink. If both the pipe and the drain are open, how long will it take to fill the sink?
We use the work-rate equation to model the problem, but before doing this, we can display the information on a table if you like:
| time | full per minute (rate) | |
| Fill the sink | 5 | $\frac{1}{5}$ |
| Drain the sink | 7 | $\frac{1}{7}$ |
| Together | t | $\frac{1}{t}$ |
Now, let’s set up the equation and solve. Notice, were are filling the sink and draining it. Since we are draining the sink, we are losing water as the sink fills. Hence, we will subtract the rate in which the sink drains. We first clear denominators, then solve the linear equation as usual.
$$\left(\frac{d}{t_A}\pm\frac{d}{t_B}\right)=\frac{d}{t_{B\bigcup{A}}}$$
$$\frac{1}{5}-\frac{1}{7}=\frac{1}{t}$$
$$35t\bigg(\frac{1}{5}-\frac{1}{7}\bigg)=\bigg(\frac{1}{t}\bigg)35t$$
$$7t-5t=35$$
$$2t=35$$
$$t=17.5$$
Thus, it would take 17.5 minutes to fill the sink.
2. Example Problems with Rates and System of Equations
Click the problem to expand for the solution!
1. Rates with Work Cooperation
Question 1a (unknown combined time): If it takes Felicia 4 hours to paint a room and her daughter Katy 12 hours to paint the same room, then, working together, how long will it take them to paint the room?
Recall that $\text{rate}\cdot\text{time}=\text{work done}$
So we will establish the two seperate rates of Felicia and Katy as follows:
$$\text{Felicia’s rate:}\quad{F_\text{rate}\cdot4\ \text{hours}=1\ \text{room}}$$
$$\text{Katy’s rate:}\quad{K_\text{rate}\cdot12\ \text{hours}=1\ \text{room}}$$
If we isolate for their respective rates, we find that:
$$F_\text{rate}=\frac{1\h\text{room}}{4\h\text{hours}}
$$K_\text{rate}=\frac{1\h\text{room}}{12\h\text{hours}}
To make this into a solvable equation, find the total time $t$ needed for Felicia and Katy to paint the room given their combined rates $r_\text{combined}=F_\text{rate}+K_\text{rate}$
$$r_\text{combined}\cdot{t_\text{combined}}=d_\text{combined work project}$$
$$\big(F_\text{rate}+K_\text{rate}\big)t_\text{combined}=1\ \text{room}$$
$$\big(\frac{1\ \text{room}}{4\ \text{hours}}+\frac{1\ \text{room}}{12\ \text{hours}}\big)t_\text{combined}=1\ \text{room}$$
Now combine and simplify with a common denominator:
$$\big(\frac{3\ \text{rooms}}{12\ \text{hours}}+\frac{1\ \text{room}}{12\ \text{hours}}\big)t_\text{combined}=1\ \text{room}$$
$$\big(\frac{4\ \text{rooms}}{12\ \text{hours}}\big)t_\text{combined}=1\ \text{room}$$
$$\big(\frac{1\ \text{rooms}}{3\ \text{hours}}\big)t_\text{combined}=1\ \text{room}$$
Solve for $t$:
$$\big(\frac{3\ \text{hours}}{1\ \text{rooms}}\big)\big(\frac{1\ \text{rooms}}{3\ \text{hours}}\big)t_\text{combined}=1\ \text{room}\big(\frac{3\ \text{hours}}{1\ \text{rooms}}\big)$$
$$t_\text{combined}=3\ \text{hours}$$
Question 1b (known combined time): Karl can clean a room in 3 hours. If his little sister Kyra helps, they can clean it in 2.4 hours. How long would it take Kyra to do the job alone?
Let $K$ be Karl’s rate and $k$ be his little sister Kyra’s rate:
$$Kt=d\rightarrow{K(3)=1}\rightarrow{K=\frac{1}{3}}$$
$$kt=d\rightarrow{kt=1}$$
Now if we consider the combined efforts, we can say something further since we know something about their combined effors:
$$(K+k)t=d\rightarrow{(\frac{1}{3}+k)(2.4)=1}$$
Now let’s solve for $k$:
$$(\frac{1}{3}+k)(2.4)=1$$
$$(\frac{1}{3}+\frac{3k}{3})(\frac{12}{5})=1$$
$$\frac{1+3k}{3}=1(\frac{5}{12})$$
$$(3)(\frac{1+3k}{3})=(\frac{5}{12})(3)$$
$$1+3k=\frac{15}{12}$$
$$3k=\frac{15}{12}-1$$
$$3k=\frac{15}{12}-\frac{12}{12}$$
$$3k=\frac{3}{12}$$
$$3k(\frac{1}{3})=(\frac{3}{12})(\frac{1}{3})$$
$$k=\frac{1}{12}$$
Now knowing Kyra’s rate, we can use our initial rate formula for Kyra to find her time:
$$kt=d\rightarrow{kt=1}$$
$$(\frac{1}{12})t=1\rightarrow{k=12}$$
Thus, it would take Kyra 12 hours if she worked alone.
Question 1c (related rates): Doug takes twice as long as Becky to complete a project. Together they can complete the project in 10 hours. How long will it take each of them to complete the project alone?
Let $D$ be Doug’s rate and $B$ be Becky’s rate. These rates have a specified relationship, so our rate equations should have this embedded in them:
Recall that $rt=d\rightarrow{r=\frac{d}{t}}$. So,
$$D(2t)=1\rightarrow\text{and}\rightarrow{B(t)=1}$$
Since they both equal 1, we can set them equal to each other:
$$D(2t)=Bt\rightarrow{2D=B}$$
Since together they can complete a project in 10 hours, we can say that:
$$(D+B)(t)=d\rightarrow{(D+2D)(10)=1}$$
$$3D(10)=1$$
$$30D=1$$
$$D=\frac{1}{30)$$
Now use this value to solve for the times it takes each to work alone:
$$Dt=d$$
$$\frac{1}{30}(t)=1\ \text{room}$$
$$t=(1)(\frac{30}{1})=30$$
Thus, it takes Doug 30 hours, and, since we are told it takes Doug twice as long as Becky, we can assume that it will take Becky 15 hours.
Question 1d (quadratic solution): Joey can build a large shed in 10 days less than Cosmo can. If they built it together, it would take them 12 days. How long would it take each of them working alone?
Let $J$ be Joey’s rate and let $C$ be Cosmos’ rate. Since Joey can build a shed in 10 days less than the time it takes Cosmo to build a shed, we can say about Joey that $J(t_\text{Cosmo}-10)=1$ shed. Given this, we can simply say that about Cosmo that $C(t_\text{Cosmo})=1$
Furthermore, since they can build a shed together in 12 days, we can say that their combined rates over their combined time will total one shed as given below:
$$(J+C)t_\text{combined}=1\ \text{shed}\rightarrow{(J+C)(12)=1}$$
Let’s use our rate equations for Joey and Cosmo from above to get better substitutions for $J$ and $C$:
$$J(t_\text{Cosmo}-10)=1\rightarrow{J=\frac{1}{t_\text{Cosmo}-10}}$$
$$C(t_\text{Cosmo})=1\rightarrow{C=\frac{1}{t_\text{Cosmo}}}$$
Now let’s substitute for $J$ and $C$:
$$(J+C)(12)=1$$
$$\left(\frac{1}{t_\text{Cosmo}-10}+\frac{1}{t_\text{Cosmo}}\right)(12)=1$$
$$\frac{1}{t_\text{Cosmo}-10}+\frac{1}{t_\text{Cosmo}}=\frac{1}{12}$$
If we create a common denominator among all these fraction, we would have $(t_\text{Cosmo}-10)(t_\text{Cosmo})(12)$. Multiplying Both sides by this common denominator would give:
$$\bigg((t_\text{Cosmo}-10)(t_\text{Cosmo})(12)\bigg)\bigg(\frac{1}{t_\text{Cosmo}-10}+\frac{1}{t_\text{Cosmo}}\bigg)=\frac{1}{12}\bigg((t_\text{Cosmo}-10)(t_\text{Cosmo})(12)\bigg)$$
$$12t_\text{Cosmo}+12(t_\text{Cosmo}-10)=t_\text{Cosmo}(t_\text{Cosmo}-10)$$
$$12t_\text{Cosmo}+12t_\text{Cosmo}-120={t^2}_\text{Cosmo}-10t_\text{Cosmo}$$
We not need to solve a quadratic equation by setting it equal to zero and factoring:
$${t^2}_\text{Cosmo}-34t_\text{Cosmo}+120=0$$
$$(t_\text{Cosmo}-30)(t_\text{Cosmo}-4)=0$$
Thus Cosmo can build a shed in either 40 days (in which case Joey builds his in 20 days) or in 4 days (in which case Joey will build a shed in -6 days, which is impossible). Thus, the solution is that Cosmo takes 30 days to build a shed and Joey takes 20 days.
Question 1e (abstract situation): “$A$” can do $z$ things in $x$ hours more/less than “$B$”. Together, $A$ and $B$ do the same $z$ things in $y$ hours. How long would it take $A$ and $B$ working alone?
Let $A$ be the rate of A and $B$ be the rate of B.
$$A(t\pm{x})=z\rightarrow{A=\frac{z}{t\pm{x}}}$$
$$Bt=z\rightarrow{B=\frac{z}{t}}$$
$$(A+B)(y)=z$$
$$\bigg(\frac{z}{t\pm{x}}+\frac{z}{t}\bigg)(y)=z$$
$$\bigg(\frac{z}{t\pm{x}}+\frac{z}{t}\bigg)=\frac{z}{y}$$
$$\bigg((t\pm{x})(t)(y)\bigg)\bigg(\frac{z}{t\pm{x}}+\frac{z}{t}\bigg)=\bigg(\frac{z}{y}\bigg)\bigg((t\pm{x})(t)(y)\bigg)$$
$$\bigg(\frac{z(t\pm{x})(t)(y)}{t\pm{x}}+\frac{z(t\pm{x})(t)(y)}{t}\bigg)=\bigg(\frac{z(t\pm{x})(t)(y)}{y}\bigg)$$
$$z(t)(y)+z(t\pm{x})(y)=z(t\pm{x})(t)$$
$$\frac{z(t)(y)}{z}+\frac{z(t\pm{x})(y)}{z}=\frac{z(t\pm{x})(t)}{z}$$
$$(t)(y)+(t\pm{x})(y)=(t\pm{x})(t)$$
$$ty+ty\pm{x}y=t^2\pm{x}t$$
$$2ty\pm{x}y=t^2\pm{x}t$$
$$0=t^2\pm{x}t-2ty\mp{x}y$$
$$0=t^2\pm{t({x}-2y)}\mp{x}y$$
Solve using the quadratic equation where $a=1;\ {b=\pm(x-2y)};\ c=\mp{xy}$
$$t=\frac{-b\pm{\sqrt{b^2-4ac}}}{2a}$$
$$t=\frac{-\bigg(\pm(x-2y)\bigg)\pm{\sqrt{\bigg(\pm(x-2y)\bigg)^2-4\bigg(\mp{xy}\bigg)}}}{2}$$
$$t=\mp\frac{1}{2}(x-2y)\pm\frac{1}{2}{\sqrt{(x-2y)^2\pm4{xy}}}$$
$$t=\mp\frac{1}{2}(x-2y)\pm\frac{1}{2}{\sqrt{(x^2-4xy+y^2)\pm4{xy}}}$$
$$t=\mp\frac{1}{2}(x-2y)\pm\frac{1}{2}{\sqrt{x^2-4xy\pm4{xy}+y^2}}$$
$$t=\mp\frac{1}{2}(x-2y)\pm\frac{1}{2}{\sqrt{x^2-4xy(1\mp1)+y^2}}$$
Thus, B will do $z$ things in $t$ hours as found above, and A will do that amount of time plus or minus $x$ depending on the situation.
2. Rates with Pool/Tub Filling/Emptying
Question 2a: A sink can be filled from the faucet in 5 minutes. It takes only 3 minutes to empty the sink when the drain is open. If the sink is full and both the faucet and the drain are open, how long will it take to empty the sink?
Recall that $rt=d$. For this problem let $d$ be the value of $1\ \text{sink}$ and $t$ will either be the time is takes to fill or empty the sink based on the two rates: $r_\text{fill}$ and $r_\text{drain}$.
Question 2b: It takes 10 hours to fill a pool with the inlet pipe. It can be emptied in 15 hours with the outlet pipe. If the pool is half full to begin with, how long will it take to fill it from there if both pipes are open?
Question 2c: A sink is ¼ full when both the faucet and the drain are opened. The faucet alone can fill the sink in 6 minutes, while it takes 8 minutes to empty it with the drain. How long will it take to fill the remaining ¾ of the sink?
Question 2d: A sink has two faucets: one for hot water and one for cold water. The sink can be filled by a cold-water faucet in 3.5 minutes. If both faucets are open, the sink is filled in 2.1 minutes. How long does it take to fill the sink with just the hot-water faucet open?
Question 2e: A water tank is being filled by two inlet pipes. Pipe A can fill the tank in 4.5 hours, while both pipes together can fill the tank in 2 hours. How long does it take to fill the tank using only pipe B?
3. Rates with Wages
Question 3a: Two roommates saved their wages from weekend jobs to buy a television. One earns 10 dollars per hour, and the other earns 8 dollars per hour. Together they worked for 43 hours and earned exactly 400 dollars. How long did each of them work in order to save the money?
Recall that $rt=d$ where, in this case, $r$ is the wages earned per hour, $t$ is the time worked, and $d$ is the earnings made through working some number of hours.
Let’s make a chart for this problem as done in the initial examples:
| Earnings | $=$ | Wages | $\cdot$ | Time | |
| Roommate 1 | $=$ | 10 | $\cdot$ | ||
| Roommate 2 | $=$ | 8 | $\cdot$ |
The only roommate-specific information that the problem gives us is their hourly wage, which we’ve included in the table. We also know that they earned 400 dollars together and also worked for 43 hours total. How do we include that information in the table?
Let’s start with the 400 dollars total earnings. We know that one roommate earned a portion of that amount, and the other roommate earned the rest. So we can call Roommate 1’s earnings $E$ and Roommate 2’s earnings $400-E$. (Note that this is an arbitrary assignment, and we could have called Roommate 2’s earnings $E$ and used $400-E$ to represent Roommate 1’s earnings.)
We can use the same logic for the hours worked. Together they worked for 43 hours—let’s call Roommate 1’s hours $t$, and Roommate 2’s hours $43-t$. (Assigning the variable $t$ to Roommate 1 and $43-t$ to Roommate 2 has the effect of making substitution a breeze, as you will see in a moment.)
| Earnings | $=$ | Wages | $\cdot$ | Time | |
| Roommate 1 | $E$ | $=$ | 10 | $\cdot$ | $t$ |
| Roommate 2 | $400-E$ | $=$ | 8 | $\cdot$ | $43-t$ |
We have had to introduce two variables into this problem, $E$ and $t$. But that’s okay because we have used them to create a system of two equations: $E=10t$ and $400-E=8(43-t)$. Now that we have our system, we can solve for the variable $t$, by substituting the value of $E$ from the first equation into the second equation:
$$400-E=8(43-t)$$
$$400-10t=8(43-t)$$
$$400-10t=344-8t$$
$$56-10t=-8t$$
$$56=2t$$
$$t=28$$
So, if Roommate 1 worked for 28 hours, then the time that Roommate 2 worked, $43-t$, must be 15 hours.