Word problem

[griz]

Can you help me with this word problem?

On the first part of a trip to Carmel driving on the freeway Sue averaged 70 mph. And the rest of the trip, which was 25 miles longer than the first part, she averaged 60 mph. Find the total distance to Carmel if the second part of the trip took 30 minutes more than the first part.

 

[MarkFL]

Hello, and welcome to FMH!

Let's let $d_1$ be the distance of the first part of the trip, in miles. And let's also let $t_1$ be the time, in hours, that it took to cover the first part. So, we may state:

[MATH]d_1=70t_1[/MATH]
And for the 2nd part, we may state:

[MATH]d_1+25=60\left(t_1+\frac{1}{2}\right)[/MATH]
You now have 2 equations and 2 unknowns. I would suggest solving the first for $t_1$ and then substitute into the second to get an equation in $d_1$ only, and from that, you can answer the question.

 

[Otis]

I substituted 70t₁ for d₁ in Mark's second equation:

70t₁ + 25 = 60t₁ + 30

?

 

[mathdad]

Hello, and welcome to FMH!

Let's let $d_1$ be the distance of the first part of the trip, in miles. And let's also let $t_1$ be the time, in hours, that it took to cover the first part. So, we may state:

[MATH]d_1=70t_1[/MATH]
And for the 2nd part, we may state:

[MATH]d_1+25=60\left(t_1+\frac{1}{2}\right)[/MATH]
You now have 2 equations and 2 unknowns. I would suggest solving the first for $t_1$ and then substitute into the second to get an equation in $d_1$ only, and from that, you can answer the question.

What you did here has been my dream, my desire. I read this question 3 times and did not see two equations in two unknowns hiding within the words. Why did you use d_1 and t_1? A lot of students, including myself, prefer no sub text. Why not use d and t or x and y or a and b?

 

[mathdad]

Hello, and welcome to FMH!

Let's let $d_1$ be the distance of the first part of the trip, in miles. And let's also let $t_1$ be the time, in hours, that it took to cover the first part. So, we may state:

[MATH]d_1=70t_1[/MATH]
And for the 2nd part, we may state:

[MATH]d_1+25=60\left(t_1+\frac{1}{2}\right)[/MATH]
You now have 2 equations and 2 unknowns. I would suggest solving the first for $t_1$ and then substitute into the second to get an equation in $d_1$ only, and from that, you can answer the question.

1. Can you make a table using D, r and t for me to understand this question just a bit more clearly?

2. Can you also translate this question to algebra one sentence at a time showing me how you came up with the two equations in two unknowns?

3. Is there another, more simple way to solve this problem?

4. I like distance applications.

 

[JeffM]

What you did here has been my dream, my desire. I read this question 3 times and did not see two equations in two unknowns hiding within the words. Why did you use d_1 and t_1? A lot of students, including myself, prefer no sub text. Why not use d and t or x and y or a and b?

If you use mnemonics to help keep in mind what your variables mean, it is helpful to use d for distance, t for time, and r for rate. Then, if a problem involves more than one distance, [MATH]d_1[/MATH] and [MATH]d_2[/MATH] represent a very easy to understand notation.

That is an argument for subscripts in presentation. I personally do not like to use subscripts when solving a problem because it is too easy to forget the subscripts while actually thinking.

 

[mathdad]

If you use mnemonics to help keep in mind what your variables mean, it is helpful to use d for distance, t for time, and r for rate. Then, if a problem involves more than one distance, [MATH]d_1[/MATH] and [MATH]d_2[/MATH] represent a very easy to understand notation.

That is an argument for subscripts in presentation. I personally do not like to use subscripts when solving a problem because it is too easy to forget the subscripts while actually thinking.

What do you use in place of subscripts?

 

[mathdad]

I decided to create a similar question using different numbers. Let me know if my set up is correct and if it makes sense.

Question:

On the first part of a trip to Coney Island driving on the freeway, Rita averaged 65 mph. And the rest of the trip, which was 25 miles longer than the first part, she averaged 60 mph. Find the total distance to Coney Island if the second part of the trip took 20 minutes more than the first part.

Let x = first part of trip

Let y = time of trip

x = 65y....Equation 1

x + 25 = 60(1/3 + y)...Equation 2

Yes or no?

Thanks.

 

[Denis]

On the first part of a trip to Carmel driving on the freeway Sue averaged 70 mph.
And the rest of the trip, which was 25 miles longer than the first part, she averaged 60 mph.
Find the total distance to Carmel if the second part of the trip took 30 minutes more than the first part.

I find this kind of "breakdown" useful.

Let u = hours for 1st part; then u+1/2 = hours for 2nd part
Representative "diagram":
@70mph.............(70u)............>(u hours)@60mph.................(70u+25).................>(u+1/2 hours)

Now apply ye olde speed formula: speed = distance / time

 

[mathdad]

I find this kind of "breakdown" useful.

Let u = hours for 1st part; then u+1/2 = hours for 2nd part
Representative "diagram":
@70mph.............(70u)............>(u hours)@60mph.................(70u+25).................>(u+1/2 hours)

Now apply ye olde speed formula: speed = distance / time

Ok. Nice work. Is my set incorrect?

 

[Dr.Peterson]

I decided to create a similar question using different numbers. Let me know if my set up is correct and if it makes sense.

Question:

On the first part of a trip to Coney Island driving on the freeway, Rita averaged 65 mph. And the rest of the trip, which was 25 miles longer than the first part, she averaged 60 mph. Find the total distance to Coney Island if the second part of the trip took 20 minutes more than the first part.

Let x = first part of trip

Let y = time of trip

x = 65y....Equation 1

x + 25 = 60(1/3 + y)...Equation 2

Yes or no?

Thanks.

The first problem is that you didn't clearly define your variables. Is x the time or distance of the first part of the trip? Is y the time for the whole trip, as I would assume?

Next, assuming x is the distance for the first part, equation 1 implies that y must be the time for the first part only.

Now, I can restate the definitions as you apparently intended them:

x = distance for first part of the trip, in miles
y = time for first part of the trip, in hours

(That illustrates how I like to state definitions -- as fully as possible, including units.)

With those definitions, your equations are valid. Unfortunately, I find that y = -1, so the problem itself is not well-designed. (When I make up a problem, I generally start with the solution, to make sure the problem makes sense.)

So those are two lessons: make valid problems, and give detailed definitions of variables. Otherwise, your work is good (especially converting minutes to hours).

There are, of course, many different ways to solve the problem, but the best way is generally the one you think of, not the one you think a teacher would consider right! So I don't tend to push for doing things my way. Having said that, I probably would have used only one variable, as Denis did; on the other hand, there are many cases where using more variables than needed, as MarkFL did, can be a good idea. This is especially true when a problem is of an unfamiliar type and you have no idea how to start; you can just translate each statement to an equation using as many variables as there are quantities in the problem. The downside of this is that it can't easily be taught to beginners, who would be overwhelmed by four or five variables!

 

[JeffM]

I decided to create a similar question using different numbers. Let me know if my set up is correct and if it makes sense.

Question:

On the first part of a trip to Coney Island driving on the freeway, Rita averaged 65 mph. And the rest of the trip, which was 25 miles longer than the first part, she averaged 60 mph. Find the total distance to Coney Island if the second part of the trip took 20 minutes more than the first part.

Let x = first part of trip

Let y = time of trip

x = 65y....Equation 1

x + 25 = 60(1/3 + y)...Equation 2

Yes or no?

Thanks.

Yes, it works, but the problem makes no sense.

[MATH]x = 65y \text { and } x + 25 = 60(1/3 + y) \implies 65y + 25 = 20 + 60y \implies[/MATH]
[MATH]5y = -\ 5 \implies y = -\ 1.[/MATH]
What does a negative number of hours mean?

In any case, this is NOT the way that I recommend beginning students approach such problems. I suggest this way: start by identifying and naming EACH potentially relevant but unknown number

[MATH]\text d = \text{first distance.}[/MATH]
[MATH]\text e = \text{second distance.}[/MATH]
[MATH]\text t = \text{first time.}[/MATH]
[MATH]\text u = \text{second time.}[/MATH]
Four unknowns so I need four equations. Write those down in mathematical language.

[MATH]d = 65t.[/MATH]
[MATH]e = 60u.[/MATH]
[MATH]d + 25 = e[/MATH]
[MATH]t + \dfrac{20}{60} = u.[/MATH]
In an earlier post, you asked how to find two equations in the verbiage. My approach is not to worry about equations until after you have explicitly identified all unknowns. Then you know how many equations you must find and can start looking for them. Now combine and simplify.

[MATH]e = d + 25 \text { and } e = 60u \implies 60u = d + 25.[/MATH]
[MATH]\text {BUT } u = t + \dfrac{20}{60} \ implies 60 * \left ( t + \dfrac{20}{60} \right ) = d + 25 \implies[/MATH]
[MATH]60t + 20 = d + 25 \implies 60t - d = 5.[/MATH]
[MATH]d = 65t \implies 60t - 65t = 5 \implies -\ 5t = 5 \implies t = -\ 1.[/MATH]
It is a bit slower, but it is absolutely sure. Very organized and logical.

 

[mathdad]

Thank you everyone for your help.

 

[mathdad]

Can you help me with this word problem?

On the first part of a trip to Carmel driving on the freeway Sue averaged 70 mph. And the rest of the trip, which was 25 miles longer than the first part, she averaged 60 mph. Find the total distance to Carmel if the second part of the trip took 30 minutes more than the first part.

I gotta say something here. Several members have volunteered their time to help you with YOUR application. The least you can do is to thank each one for placing YOU on the right track to find the answer. Let me be the first one to make known the fact that this site expects members to show effort or math work to posted questions. Take a look at my questions. I do my best to show some effort.