Last Month Maria Hiked a Total of 90 Miles.
- 0.1 What are altitude word problems?
- 0.2 Two-part and round-trip problems
- 0.3 Intersecting distance problems
- 0.4 Overtaking altitude bug
- 1 Last Month Maria Hiked a Total of 90 Miles
Lesson x: Distance Word Problems
What are altitude word problems?
Altitude discussion problems
are a common type of algebra word issues. They involve a scenario in which you demand to effigy out how
far, or how
one or more objects have traveled. These are often called
because 1 of the most famous types of distance problems involves finding out when ii trains heading toward each other cross paths.
In this lesson, you lot’ll learn how to solve train bug and a few other common types of distance issues. Just get-go, let’s look at some bones principles that use to
The nuts of altitude problems
There are three basic aspects to movement and travel:
charge per unit, and
time. To understand the deviation among these, think about the last time you lot collection somewhere.
you traveled. The
you traveled. The
the trip took.
The relationship amidst these things can exist described past this formula:
distance = rate x time
d = rt
In other words, the
you drove is equal to the
at which you drove times the amount of
you drove. For an case of how this would piece of work in real life, just imagine your last trip was like this:
- Yous drove
- You lot drove an average of
- The drive took you 30 minutes, or 0.5
According to the formula, if we multiply the
charge per unit
fourth dimension, the production should be our distance.
And it is! Nosotros collection
50 ⋅ 0.5
25, which is our altitude.
What if nosotros drove sixty mph instead of fifty? How far could we bulldoze in thirty minutes? We could use the same formula to effigy this out.
0.five is 30, so our distance would be
Solving distance problems
When you solve whatsoever distance problem, yous’ll take to practise what we just did—utilize the formula to find
charge per unit, or
time. Let’s try another simple trouble.
On his day off, Lee took a trip to the zoo. He drove an boilerplate speed of 65 mph, and it took him two-and-a-one-half hours to get from his firm to the zoo. How far is the zoo from his house?
First, we should identify the information we know. Remember, we’re looking for whatever information most altitude, rate, or fourth dimension. According to the problem:
is 65 mph.
is two-and-a-half hours, or 2.five hours.
is unknown—it’due south what we’re trying to find.
You could motion-picture show Lee’s trip with a diagram similar this:
This diagram is a start to understanding this problem, but we still have to effigy out what to practise with the numbers for
fourth dimension. To continue rails of the information in the problem, nosotros’ll set up a table. (This might seem excessive now, simply it’s a good habit for fifty-fifty simple problems and can make solving complicated problems much easier.) Here’s what our table looks like:
We can put this information into our formula:
altitude = rate ⋅ time.
We tin can use the
distance = rate ⋅ fourth dimension
formula to find the altitude Lee traveled.
d = rt
d = rt
looks like this when we plug in the numbers from the problem. The unknown
is represented with the variable
d = 65 ⋅ 2.5
d, all we have to do is multiply 65 and 2.five.
65 ⋅ two.v
d = 162.v
We take an answer to our trouble:
= 162.5. In other words, the distance Lee drove from his house to the zoo is 162.5 miles.
Be conscientious to use the same
units of measurement
for charge per unit and fourth dimension. Information technology’s possible to multiply 65 miles per
because they use the same unit: an
hour. However, what if the fourth dimension had been written in a dissimilar unit, like in
minutes? In that instance, you’d take to catechumen the fourth dimension into hours so it would utilise the same unit as the rate.
Solving for rate and time
In the trouble we just solved we calculated for
distance, but y’all can use the
d = rt
formula to solve for
likewise. For case, take a await at this problem:
After work, Janae walked in her neighborhood for a one-half 60 minutes. She walked a mile-and-a-one-half full. What was her average speed in miles per hr?
We can picture Janae’south walk every bit something like this:
And we can gear up the information from the trouble nosotros know similar this:
The table is repeating the facts we already know from the problem. Janae walked
miles in a half hour, or
As always, we showtime with our formula. Adjacent, nosotros’ll make full in the formula with the information from our table.
d = rt
The rate is represented by
because we don’t nonetheless know how fast Janae was walking. Since we’re solving for
r, we’ll have to get information technology lonely on one side of the equation.
1.5 = r ⋅ 0.5
Our equation calls for
by 0.5, so nosotros can get
lone on 1 side of the equation past
both sides by 0.5:
1.5 / 0.5 =
3 = r
3, so 3 is the reply to our problem. Janae walked
miles per hour.
In the bug on this folio, we solved for
of travel, simply yous can likewise use the travel equation to solve for
time. You tin can even use it to solve certain problems where you’re trying to figure out the distance, rate, or time of 2 or more moving objects. We’ll look at issues like this on the next few pages.
Two-part and round-trip problems
Practice you know how to solve this problem?
Neb took a trip to see a friend. His friend lives 225 miles abroad. He drove in town at an boilerplate of xxx mph, then he drove on the interstate at an average of 70 mph. The trip took 3-and-a-half hours total. How far did Bill drive on the interstate?
This problem is a archetype
two-part trip problem
because it’southward request you to find information about one part of a two-role trip. This problem might seem complicated, but don’t exist intimidated!
You lot can solve it using the same tools we used to solve the simpler problems on the outset folio:
d = rt
to keep track of important data
Let’s first with the
table. Take another expect at the trouble. This time, the data relating to
has been underlined.
Bill took a trip to run across a friend. His friend lives
away. He drove in town at an boilerplate of
30 mph, then he drove on the interstate at an average of
70 mph. The trip took
total. How far did Bill bulldoze on the interstate?
If yous tried to make full in the table the way we did on the terminal page, you might have noticed a trouble: There’due south
information. For case, the problem contains
mph. To include all of this information, let’s create a table with an extra row. The elevation row of numbers and variables will exist labeled
in town, and the bottom row will exist labeled
We filled in the rates, but what most the
time? If you wait back at the trouble, you lot’ll see that these are the
figures, meaning they include both the time in town and on the interstate. And so the
225. This means this is true:
Interstate altitude + in-town distance = Total distance
Together, the interstate distance and in-town distance are equal to the
In any case, we’re trying to find out how far Neb drove on the
interstate, so let’s represent this number with
d. If the interstate distance is
d, information technology ways the in-town distance is a number that equals the total,
d. In other words, information technology’s equal to
Nosotros can fill in our chart similar this:
|altitude||charge per unit||time|
|in boondocks||225 – d||30|
We can use the same technique to fill up in the
cavalcade. The full time is
hours. If we say the time on the interstate is
t, then the remaining time in town is equal to
. Nosotros tin fill in the rest of our chart.
|in town||225 – d||30||3.five – t|
Now we tin work on solving the problem. The main difference betwixt the problems on the start page and this problem is that this trouble involves
equations. Hither’s the 1 for
225 – d = xxx ⋅ (three.5 – t)
And here’south the one for
d = 70t
If you lot tried to solve either of these on its own, you lot might have found information technology impossible: since each equation contains two unknown variables, they can’t exist solved on their own. Endeavor for yourself. If y’all work either equation on its own, you won’t be able to discover a numerical value for
d. In order to find the value of
d, we’ll also take to know the value of
Nosotros tin find the value of
in both problems by combining them. Let’s take another look at our travel equation for interstate travel.
While we don’t know the numerical value of
d, this equation does tell u.s.a. that
is equal to
d = 70t
equal, we can supervene upon
in our equation for interstate travel won’t help usa notice the value of
t—all information technology tells us is that
is equal to itself, which we already knew.
Merely what near our other equation, the ane for in-town travel?
225 – d = thirty ⋅ (3.5 – t)
When we replace the
in that equation with
, the equation suddenly gets much easier to solve.
= 30 ⋅ (three.5 – t)
Our new equation might expect more than complicated, but it’s actually something we tin solve. This is considering it only has one variable:
t. Once we find
t, we can employ information technology to calculate the value of
d—and detect the answer to our problem.
To simplify this equation and discover the value of
t, we’ll have to get the
lonely on one side of the equals sign. Nosotros’ll also have to
the right side as much as possible.
225 – 70t = 30 ⋅ (3.five – t)
Permit’s start with the right side:
105 – 30t
225 – 70t =
105 – 30t
Adjacent, allow’southward abolish out the
. To do this, we’ll subtract
from both sides. On the right side, it means subtracting
105 – 225
– 70t =
Our next step is to
like terms—call up, our eventual goal is to take
side of the equals sign and a number on the
right. We’ll cancel out the
on the right side by
to both sides. On the right side, nosotros’ll add it to
Finally, to get
on its own, nosotros’ll divide each side past its coefficient: -40.
-120 / – 40
t = three
is equal to
3. In other words, the
Bill traveled on the interstate is equal to
3 hours. Recollect, we’re ultimately trying to notice the
Neb traveled on the interstate. Let’southward wait at the
row of our chart again and see if we have enough information to find out.
It looks like nosotros do. At present that nosotros’re but missing one variable, we should exist able to detect its value pretty quickly.
To observe the distance, we’ll use the travel formula
distance = rate ⋅ time.
d = rt
We now know that Bill traveled on the interstate for
mph, and then we can fill in this information.
3 ⋅ 70
Finally, we finished simplifying the right side of the equation.
three ⋅ seventy
d = 210
= 210. We have the answer to our problem! The distance is
210. In other words, Bill drove
on the interstate.
Solving a circular-trip problem
It might take seemed like it took a long time to solve the first trouble. The more do y’all get with these issues, the quicker they’ll go. Let’south effort a like trouble. This one is called a
because it describes a circular trip—a trip that includes a return journey. Even though the trip described in this trouble is slightly different from the one in our first trouble, y’all should be able to solve it the same way. Permit’due south have a wait:
Eva collection to work at an average speed of 36 mph. On the way dwelling house, she hit traffic and simply drove an average of 27 mph. Her total time in the car was one hour and 45 minutes, or 1.75 hours. How far does Eva alive from piece of work?
If y’all’re having trouble understanding this trouble, you might want to visualize Eva’s commute like this:
As e’er, let’due south start by filling in a tabular array with the important information. We’ll brand a row with information well-nigh her trip
to draw the trip from work. (Remember, the
travel fourth dimension is
1.75 hours, then the time
work should equal
From our table, nosotros tin can write 2 equations:
- The trip
to piece of work
can be represented equally
- The trip
can be represented equally
= 27 (1.75 –
In both equations,
represents the total distance. From the diagram, you can see that these ii equations are
to each other—afterwards all, Eva drives the
same distance to and from work.
Just like with the terminal problem we solved, we can solve this one by
the two equations.
We’ll first with our equation for the trip
d = 27 (1.75 – t)
Side by side, nosotros’ll substitute in the value of
. Since the value of
, we can replace any occurrence of
= 27 (ane.75 – t)
Now, permit’south simplify the correct side.
27 ⋅(1.75 –
Next, nosotros’ll abolish out
to both sides of the equation.
Finally, nosotros tin can get
on its ain by dividing both sides by its coefficient:
47.25 / 63
t = .75
is equal to
.75. In other words, the
it took Eva to bulldoze to work is
.75 hours. At present that we know the value of
t, we’ll be able to tin detect the
to Eva’south work.
If y’all guessed that we were going to utilise the
again, you were right. We at present know the value of two out of the 3 variables, which ways we know plenty to solve our problem.
d = rt
Kickoff, let’southward make full in the values we know. We’ll work with the numbers for the trip
to work. We already knew the
36. And we just learned the
36 ⋅ .75
Now all nosotros have to exercise is simplify the equation:
36 ⋅ .75
d = 27
is equal to
27. In other words, the
to Eva’south work is
27 miles. Our trouble is solved.
Intersecting distance problems
An intersecting distance problem is ane where ii things are moving
each other. Here’southward a typical problem:
Pawnee and Springfield are 420 miles apart. A train leaves Pawnee heading to Springfield at the same fourth dimension a railroad train leaves Springfield heading to Pawnee. One train is moving at a speed of 45 mph, and the other is moving threescore mph. How long will they travel earlier they encounter?
This problem is asking you to calculate how long it will accept these two trains moving toward each other to cross paths. This might seem confusing at beginning. Even though it’s a real-world state of affairs, information technology can exist difficult to imagine distance and motility abstractly. This diagram might help you lot get a sense of what this state of affairs looks like:
If you’re still confused, don’t worry! You can solve this trouble the same way you solved the ii-part issues on the final folio. Y’all’ll just demand a
Pawnee and Springfield are
autonomously. A train leaves Pawnee heading toward Springfield at the same time a railroad train leaves Springfield heading toward Pawnee. One railroad train is moving at a speed of
45 mph, and the other is moving
60 mph. How long volition they travel before they run into?
Let’s start past filling in our chart. Here’s the problem again, this time with the important information underlined. We can start by filling in the nigh obvious information:
rate. The trouble gives us the speed of each train. We’ll characterization them
fast railroad train
dull train. The fast train goes
mph. The boring train goes only
Nosotros can also put this data into a table:
We don’t know the altitude each railroad train travels to meet the other nonetheless—we just know the
distance. In order to meet, the trains will cover a combined distance
to the total distance. As y’all can see in this diagram, this is true no affair how far each train travels.
This means that—just similar last time—nosotros’ll represent the distance of i with
and the altitude of the other with the total
So the distance for the fast railroad train will be
, and the distance for the slow railroad train volition be
|fast railroad train||d||60|
|tedious railroad train||420 – d||45|
Because we’re looking for the
both trains travel earlier they meet, the time will be the same for both trains. We can represent information technology with
|slow train||420 – d||45||t|
The table gives us
. Just like we did with the
problems, we can combine these two equations.
The equation for the
train isn’t solvable on its own, but it does tell united states that
is equal to
d = 60t
The other equation, which describes the
train, tin can’t be solved lonely either. However, we can replace the
with its value from the first equation.
Because we know that
is equal to
, we can replace the
in this equation with
. Now we accept an equation nosotros can solve.
To solve this equation, nosotros’ll need to go
and its coefficients on one side of the equals sign and whatsoever other numbers on the other. Nosotros tin first by canceling out the
on the left by
to both sides.
420 = 105t
Now nosotros just need to go rid of the coefficient next to
t. We tin can do this by dividing both sides by
420 / 105
4 = t
= 4. In other words, the
it takes the trains to meet is
4 hours. Our problem is solved!
If you want to exist sure of your reply, yous can
information technology by using the altitude equation with
4. For our fast railroad train, the equation would be
= 60 ⋅ iv.
60 ⋅ iv
240, so the distance our
railroad train traveled would be
For our boring railroad train, the equation would be
= 45 ⋅ 4.
45 ⋅ iv
180, then the altitude traveled by the
railroad train is
Recall how we said the distance the slow railroad train and fast railroad train travel should equal the
240 miles + 180 miles
420 miles, which is the total altitude from our problem. Our answer is correct.
Practise problem 1
Here’s another intersecting distance problem. It’south similar to the one nosotros just solved. See if you lot can solve it on your own. When you’re finished, scroll down to meet the answer and an explanation.
Jon and Dani alive 270 miles autonomously. Ane day, they decided to bulldoze toward each other and hang out wherever they met. Jon drove an average of 65 mph, and Dani drove an average of 70 mph. How long did they drive before they met up?
Problem 1 answer
Hither’s practice problem 1:
Jon and Dani live 270 miles apart. One day, they decided to bulldoze toward each other and hang out wherever they met. Jon collection an average of 65 mph, and Dani drove 70 mph. How long did they drive before they met up?
Allow’s solve this problem similar we solved the others. First, attempt making the chart. It should expect like this:
|Dani||270 – d||lxx||t|
Here’s how we filled in the chart:
Together, Dani and Jon will have covered the
between them by the fourth dimension they meet upwardly. That’south
270. Jon’s altitude is represented past
, so Dani’south distance is
The problem tells united states of america Dani and Jon’due south speeds. Dani drives
mph, and Jon drives
Because Jon and Dani drive the same amount of fourth dimension before they meet upwardly, both of their travel times are represented past
Now we accept 2 equations. The equation for Jon’s travel is
. The equation for Dani’due south travel is
. To solve this trouble, we’ll need to
The equation for Jon tells us that
is equal to
. This ways we can combine the two equations past replacing the
in Dani’s equation with
on ane side of the equation and a number on the other. The first step to doing this is to become rid of
on the left side. We’ll cancel information technology out past
to both sides:
All that’south left to exercise is to become rid of the
adjacent to the
t. Nosotros can do this by dividing both sides by
270 / 135
ii = t
is equal to
ii. We have the respond to our problem: Dani and Jon collection
before they met upwardly.
Overtaking altitude bug
The final type of altitude problem we’ll talk over in this lesson is a problem in which one moving object
passes—another. Here’south a typical overtaking problem:
The Hill family and the Platter family unit are going on a road trip. The Hills left 3 hours before the Platters, but the Platters drive an average of xv mph faster. If information technology takes the Platter family unit xiii hours to catch up with the Loma family unit, how fast are the Hills driving?
You tin can picture the moment the Platter family left for the road trip a little similar this:
The problem tells us that the Platter family volition catch upwardly with the Hill family in xiii hours and asks u.s. to use this information to find the Colina family’s
rate. Like some of the other issues we’ve solved in this lesson, it might non seem similar nosotros have enough information to solve this trouble—but we do. Let’south start making our nautical chart. The
for both the Hills and the Platters—when the Platters catch up with the Hills, both families volition take driven the verbal aforementioned distance.
Filling in the
charge per unit
will require a piffling more thought. We don’t know the rate for either family—recall, that’s what we’re trying to find out. Nevertheless, we do know that the Platters drove 15 mph
than the Hills. This means if the Colina family’south rate is
, the Platter family’due south rate would exist
|the Platters||d||r + fifteen|
Now all that’s left is the time. We know information technology took the Platters
to catch upward with the Hills. Even so, recall that the Hills left
earlier than the Platters—which means when the Platters caught up, they’d been driving
iii hours more
than the Platters.
thirteen + 3
16, and so we know the Hills had been driving
by the time the Platters caught up with them.
|the Platters||d||r + 15||13|
Our nautical chart gives united states of america two equations. The Hill family’southward trip can be described by
⋅ 16. The equation for the Platter family’south trip is
+ 15) ⋅ 13. Just like with our other problems, we can
these equations by replacing a variable in i of them.
The Hill family equation already has the value of
Then nosotros’ll supercede the
in the Platter equation with
⋅ 16. This fashion, it will be an equation we can solve.
r ⋅ xvi = (r + fifteen) ⋅ 13
First, permit’south simplify the right side:
⋅ 16 is
= (r + xv) ⋅ 13
Adjacent, nosotros’ll simplify the correct side and multiply
13r + 195
We can get both
and their coefficients on the left side by
Now all that’due south left to exercise is get rid of the 3 next to the
r. To do this, we’ll carve up both sides by three:
195 / 3
r = 65
And then there’s our answer:
= 65. The Hill family unit collection an average of
Y’all can solve any overtaking problem the same fashion we solved this i. Just retrieve to pay special attention when you’re setting up your chart. Just like the Hill family did in this problem, the person or vehicle who started moving
will ever have a
Do problem two
Try solving this trouble. Information technology’southward similar to the problem we just solved. When you’re finished, scroll down to see the reply and an explanation.
A train moving 60 mph leaves the station at noon. An hour later, a railroad train moving 80 mph leaves heading the same direction on a parallel track. What fourth dimension does the second train catch up to the commencement?
Problem 2 answer
Here’due south practice problem 2:
A train moving 60 mph leaves the station at noon. An hr later, a train moving 80 mph leaves heading the aforementioned direction on a parallel track. What time does the second train catch upward to the first?
To solve this problem, start past making a chart. Here’southward how it should look:
|slow train||d||60||t + 1|
Here’s an caption of the chart:
Both trains will accept traveled the same altitude past the time the fast railroad train catches up with the slow i, so the distance for both is
The problem tells us how fast each train was going. The fast railroad train has a rate of
mph, and the slow railroad train has a charge per unit of
to represent the fast train’s travel time earlier information technology catches upwards. Because the slow train started an hour before the fast one, it will have been traveling one hour more past the time the fast train catches upwardly. It’south
Now we accept two equations. The equation for the fast train is
. The equation for the ho-hum railroad train is
= sixty (t
+ ane). To solve this problem, we’ll need to
The equation for the fast railroad train says
is equal to
. This means nosotros can combine the two equations past replacing the
in the wearisome railroad train’s equation with
= 60 (t + one)
Showtime, let’s simplify the correct side of the equation:
60 ⋅ (t
60t + lx
To solve the equation, we’ll have to become
on one side of the equals sign and a number on the other. We can get rid of
on the correct side by
from both sides: 80t
Finally, we can get rid of the
by dividing both sides by
t = 3
is equal to
3. The fast train traveled for
three hours. However, it’s non the answer to our problem. Let’southward look at the original problem once more. Pay attending to the final sentence, which is the question we’re trying to answer.
A train moving 60 mph leaves the station at noon. An hr afterwards, a railroad train moving 80 mph leaves heading the same management on a parallel track. What time does the second train catch up to the first?
Our problem doesn’t enquire how
either of the trains traveled. It asks
the 2d train catches up with the first.
The problem tells us that the slow train left at apex and the fast 1 left an hour afterward. This means the fast train left at
1 p.m. From our equations, nosotros know the fast train traveled
one + 3
4, so the fast train caught up with the ho-hum one at
4 p.k. The answer to the problem is
Last Month Maria Hiked a Total of 90 Miles