Last Month Maria Hiked a Total of 90 Miles

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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
fast, how
far, or how
long
one or more objects have traveled. These are often called
train issues
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
any
distance problem.

The nuts of altitude problems

There are three basic aspects to movement and travel:
distance,
charge per unit, and
time. To understand the deviation among these, think about the last time you lot collection somewhere.

The
distance
is how
far
you traveled. The
rate
is how
fast
you traveled. The
time
is how
long
the trip took.

The relationship amidst these things can exist described past this formula:

distance = rate x time

d = rt

In other words, the
distance
you drove is equal to the
rate
at which you drove times the amount of
time
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
  25
  miles—that’s the
  distance.
• You lot drove an average of
  l
  mph—that’s the
  rate.
• The drive took you 30 minutes, or 0.5
  hours—that’s the
  time.

According to the formula, if we multiply the
charge per unit
and
fourth dimension, the production should be our distance.

And it is! Nosotros collection
50
mph for
0.five
hours—and
50 ⋅ 0.5
equals
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.

60
⋅
0.five is 30, so our distance would be
xxx
miles.

Solving distance problems

When you solve whatsoever distance problem, yous’ll take to practise what we just did—utilize the formula to find
distance,
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:

• The
  rate
  is 65 mph.
• The
  time
  is two-and-a-half hours, or 2.five hours.
• The
  distance
  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
altitude,
rate, and
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:

distance	rate	time
d	65	2.5

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

The formula
d = rt
looks like this when we plug in the numbers from the problem. The unknown
distance
is represented with the variable
d.

d = 65 ⋅ 2.5

To find
d, all we have to do is multiply 65 and 2.five.
65 ⋅ two.v
equals
162.5.

d = 162.v

We take an answer to our trouble:
d
= 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
hour
past 2.5
hours
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
rate
and
time
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:

altitude	rate	time
one.5	r	0.5

The table is repeating the facts we already know from the problem. Janae walked
one-and-a-half miles
or
1.five
miles in a half hour, or
0.five
hours.

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
r
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
r
to exist
multiplied
by 0.5, so nosotros can get
r
lone on 1 side of the equation past
dividing
both sides by 0.5:

1.5 / 0.5 =
iii.

3 = r

r
=
3, so 3 is the reply to our problem. Janae walked
3
miles per hour.

In the bug on this folio, we solved for
distance
and
rate
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:

• The
  travel equation
  d = rt
• A
  table
  to keep track of important data

Let’s first with the
table. Take another expect at the trouble. This time, the data relating to
distance,
rate, and
fourth dimension
has been underlined.

Bill took a trip to run across a friend. His friend lives
225 miles
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
three-and-a-one-half hours
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
too much
information. For case, the problem contains
2
rates—30
mph
and
70
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
interstate.

	distance	rate	time
in town		30
interstate		70

We filled in the rates, but what most the
distance
and
time? If you wait back at the trouble, you lot’ll see that these are the
total
figures, meaning they include both the time in town and on the interstate. And so the
total altitude

is
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
total
altitude. See?

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,
225, when
added
to
d. In other words, information technology’s equal to
225 –
d
.

Nosotros can fill in our chart similar this:

	altitude	charge per unit	time
in boondocks	225 – d	30
interstate	d	70

We can use the same technique to fill up in the
time
cavalcade. The full time is
3.five
hours. If we say the time on the interstate is
t, then the remaining time in town is equal to
3.five –
t
. Nosotros tin fill in the rest of our chart.

	altitude	rate	fourth dimension
in town	225 – d	30	3.five – t
interstate	d	70	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
ii
equations. Hither’s the 1 for
in-town travel:

225 – d = xxx ⋅ (three.5 – t)

And here’south the one for
interstate travel:

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
t.

Nosotros tin find the value of
t
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
d
is equal to
70t
.

d = 70t

Since
seventyt

and

d

are
equal, we can supervene upon

d

with
70t
. Substituting
lxxt

for

d

in our equation for interstate travel won’t help usa notice the value of
t—all information technology tells us is that
70t

is equal to itself, which we already knew.

70t
= 70t

Merely what near our other equation, the ane for in-town travel?

225 – d = thirty ⋅ (3.5 – t)

When we replace the
d
in that equation with
70t
, the equation suddenly gets much easier to solve.

225 –
70t
= 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
t
lonely on one side of the equals sign. Nosotros’ll also have to
simplify
the right side as much as possible.

225 – 70t = 30 ⋅ (3.five – t)

Permit’s start with the right side:
30
times
(three.5 –
t)
is
105 – 30t
.

225 – 70t =
105 – 30t

Adjacent, allow’southward abolish out the
225
next to
seventyt
. To do this, we’ll subtract
225
from both sides. On the right side, it means subtracting
225
from
105.
105 – 225
is
-120.

– 70t =
-120
– 30t

Our next step is to
group
like terms—call up, our eventual goal is to take
t
on the
left
side of the equals sign and a number on the
right. We’ll cancel out the
-30t

on the right side by
adding
xxxt

to both sides. On the right side, nosotros’ll add it to
-70t
.
-lxxt
+ 30t

is
-40t
.

– 40t
= -120

Finally, to get
t
on its own, nosotros’ll divide each side past its coefficient: -40.
-120 / – 40
is
iii.

t = three

So
t
is equal to
3. In other words, the
time
Bill traveled on the interstate is equal to
3 hours. Recollect, we’re ultimately trying to notice the
distanc
eastward
Neb traveled on the interstate. Let’southward wait at the
interstate
row of our chart again and see if we have enough information to find out.

	distance	rate	time
interstate	d	seventy	three

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
3
hours at
70
mph, and then we can fill in this information.

d =
3 ⋅ 70

Finally, we finished simplifying the right side of the equation.
three ⋅ seventy
is
210.

d = 210

So


d

= 210. We have the answer to our problem! The distance is
210. In other words, Bill drove
210 miles
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
round-trip problem
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 work
and
from work.

one.75 –


t

to draw the trip from work. (Remember, the
total
travel fourth dimension is
1.75 hours, then the time
to
work and
from
work should equal
1.75.)

From our table, nosotros tin can write 2 equations:

• The trip
  to piece of work
  can be represented equally
  
  d
  = 36t
  .
• The trip
  from work
  can be represented equally
  
  d
  = 27 (1.75 –
  t).

In both equations,
d
represents the total distance. From the diagram, you can see that these ii equations are
equal
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
combining
the two equations.

We’ll first with our equation for the trip
from work.

d = 27 (1.75 – t)

Side by side, nosotros’ll substitute in the value of
d
from our
to work
equation,

d
= 36t
. Since the value of
d
is
36t
, we can replace any occurrence of
d
with
36t
.

36t
= 27 (ane.75 – t)

Now, permit’south simplify the correct side.
27 ⋅(1.75 –
t)
is
47.25.

36t =
47.25
– 27t

Next, nosotros’ll abolish out
-27t

past
adding
27t
to both sides of the equation.
36t
+ 27t

is
63t
.

63t
= 47.25

Finally, nosotros tin can get
t
on its ain by dividing both sides by its coefficient:
63.
47.25 / 63
is
.75.

t = .75

t
is equal to
.75. In other words, the
fourth dimension
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
distance
to Eva’south work.

If y’all guessed that we were going to utilise the
travel equation
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
rate:
36. And we just learned the
fourth dimension:
.75.

d =
36 ⋅ .75

Now all nosotros have to exercise is simplify the equation:
36 ⋅ .75
=
27.

d = 27

d

is equal to
27. In other words, the
distance
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
toward
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
chart
and the
travel formula.

Pawnee and Springfield are
420 miles
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
and
dull train. The fast train goes
60
mph. The boring train goes only
45
mph.

Nosotros can also put this data into a table:

	distance	rate	fourth dimension
fast train		threescore
deadening train		45

We don’t know the altitude each railroad train travels to meet the other nonetheless—we just know the
total
distance. In order to meet, the trains will cover a combined distance
equal
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
d
and the altitude of the other with the total
minus
d.
So the distance for the fast railroad train will be

d
, and the distance for the slow railroad train volition be
420 –
d
.

	distance	rate	fourth dimension
fast railroad train	d	60
tedious railroad train	420 – d	45

Because we’re looking for the
time
both trains travel earlier they meet, the time will be the same for both trains. We can represent information technology with
t.

	distance	rate	time
fast train	d	60	t
slow train	420 – d	45	t

The table gives us
two
equations:

d
= sixtyt

and
420 –
d
= 45t
. Just like we did with the
two-function
problems, we can combine these two equations.

The equation for the
fast
train isn’t solvable on its own, but it does tell united states that
d
is equal to
lxt
.

d = 60t

The other equation, which describes the
slow
train, tin can’t be solved lonely either. However, we can replace the
d
with its value from the first equation.

420 –
d
= 45t

Because we know that

d

is equal to
60t
, we can replace the
d
in this equation with
60t
. Now we accept an equation nosotros can solve.

420 –
60t
= 45t

To solve this equation, nosotros’ll need to go
t
and its coefficients on one side of the equals sign and whatsoever other numbers on the other. Nosotros tin first by canceling out the
-60t

on the left by
adding
60t

to both sides.
45t
+ 60t

is
105t
.

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
105.
420 / 105
is
iv.

4 = t

t
= 4. In other words, the
time
it takes the trains to meet is
4 hours. Our problem is solved!

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?

Answer:
2 hours.

Allow’s solve this problem similar we solved the others. First, attempt making the chart. It should expect like this:

	distance	rate	fourth dimension
Jon	d	65	t
Dani	270 – d	lxx	t

Here’s how we filled in the chart:

• Altitude:
  Together, Dani and Jon will have covered the
  total distance
  between them by the fourth dimension they meet upwardly. That’south
  270. Jon’s altitude is represented past
  
  d
  , so Dani’south distance is
  270 –
  d
  .
• Rate:
  The problem tells united states of america Dani and Jon’due south speeds. Dani drives
  65
  mph, and Jon drives
  70
  mph.
• Time:
  Because Jon and Dani drive the same amount of fourth dimension before they meet upwardly, both of their travel times are represented past
  t.

Now we accept 2 equations. The equation for Jon’s travel is

d
= 65t
. The equation for Dani’due south travel is
270 –
d
= seventyt
. To solve this trouble, we’ll need to
combine
them.

The equation for Jon tells us that
d
is equal to
65t
. This ways we can combine the two equations past replacing the

d

in Dani’s equation with
65t
.

270 –
65t
= 70t

Let’south go
t
on ane side of the equation and a number on the other. The first step to doing this is to become rid of
-65t

on the left side. We’ll cancel information technology out past
adding
65t
to both sides:
lxxt
+ 65t

is
135t.

270 =
135t

All that’south left to exercise is to become rid of the
135
adjacent to the
t. Nosotros can do this by dividing both sides by
135:
270 / 135
is
2.

ii = t

That’s it.
t
is equal to
ii. We have the respond to our problem: Dani and Jon collection
2 hours
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
overtakes—or
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
distance
can exist
d
for both the Hills and the Platters—when the Platters catch up with the Hills, both families volition take driven the verbal aforementioned distance.

	distance	rate	time
the Hills	d
the Platters	d

Filling in the
charge per unit
and
time
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
faster
than the Hills. This means if the Colina family’south rate is

r
, the Platter family’due south rate would exist

r
+ 15.

	distance	rate	fourth dimension
the Hills	d	r
the Platters	d	r + fifteen

Now all that’s left is the time. We know information technology took the Platters
13 hours
to catch upward with the Hills. Even so, recall that the Hills left
3 hours
earlier than the Platters—which means when the Platters caught up, they’d been driving
iii hours more
than the Platters.
thirteen + 3
is
16, and so we know the Hills had been driving
16 hours
by the time the Platters caught up with them.

	altitude	rate	time
the Hills	d	r	16
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

d
=
r
⋅ 16. The equation for the Platter family’south trip is

d
= (r
+ 15) ⋅ 13. Just like with our other problems, we can
combine
these equations by replacing a variable in i of them.

The Hill family equation already has the value of
d
equal to

r
⋅ 16.
Then nosotros’ll supercede the
d
in the Platter equation with

r
⋅ 16. This fashion, it will be an equation we can solve.

r ⋅ xvi = (r + fifteen) ⋅ 13

First, permit’south simplify the right side:
r
⋅ 16 is
16r
.

16r
= (r + xv) ⋅ 13

Adjacent, nosotros’ll simplify the correct side and multiply
(r
+ 15)
by
13.

16r =
13r + 195

We can get both
r
and their coefficients on the left side by
subtracting
xiiir

from
sixteenr
:
sixteenr

–
13r

is
iiir
.

3r
= 195

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
is
65.

r = 65

And then there’s our answer:
r
= 65. The Hill family unit collection an average of
65 mph.

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
outset
will ever have a
greater
travel time.

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?

Answer:
four p.grand.

To solve this problem, start past making a chart. Here’southward how it should look:

	altitude	rate	time
fast train	d	80	t
slow train	d	60	t + 1

Here’s an caption of the chart:

• Distance:
  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
  
  d
  .
• Rate:
  The problem tells us how fast each train was going. The fast railroad train has a rate of
  80
  mph, and the slow railroad train has a charge per unit of
  threescore
  mph.
• Time:
  We’ll use
  t
  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
  
  t
  + 1.

Now we accept two equations. The equation for the fast train is

d
= lxxxt
. The equation for the ho-hum railroad train is

d
= sixty (t
+ ane). To solve this problem, we’ll need to
combine
the equations.

The equation for the fast railroad train says
d
is equal to
80t
. This means nosotros can combine the two equations past replacing the

d

in the wearisome railroad train’s equation with
80t
.

80t
= 60 (t + one)

Showtime, let’s simplify the correct side of the equation:
60 ⋅ (t
+ 1)
is
60t
+ 60.

80t =
60t + lx

To solve the equation, we’ll have to become
t
on one side of the equals sign and a number on the other. We can get rid of
60t

on the correct side by
subtracting
sixtyt
from both sides: 80t
– 60t
is
20t
.

20t
= 60

Finally, we can get rid of the
20
next to
t
by dividing both sides by
20.
60
divided by
twenty
is
three.

t = 3

So
t
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
long
either of the trains traveled. It asks
what time
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
three
hours.
one + 3
is
4, so the fast train caught up with the ho-hum one at
4 p.k. The answer to the problem is
4 p.m.