Is this even possible? Woman runs N @ 6-min-mile pace, then S @ 7-min-mile pace; find total miles.

[allegansveritatem]

Here is the problem:

62. Jogging rates: A woman begins jogging at 3:00pm, running due north at a 6-minute-mile pace. Later, she reverses direction and runs due south at a 7-minute-mile pace. If she returns to her starting point at 3:45pm, find the total number of miles run.


Here is what I did with it:


This can't be right but it is all I could come up with. The 10 and 8.57 are the hourly rates. the .75 is the time in hours. I think the true solution here would yield something between 6.5 and 7 miles.

 

[pka]

Here is the problem:

62. Jogging rates: A woman begins jogging at 3:00pm, running due north at a 6-minute-mile pace. Later, she reverses direction and runs due south at a 7-minute-mile pace. If she returns to her starting point at 3:45pm, find the total number of miles run.

Let's look at the wording. Note that there are two different rates. The out rate is one mile every six minutes, while the return rate is one mile every seven minutes. So the out-distance equals the return-distance. If the out-time is \(\displaystyle T \text{ minutes}\) then the return-time is \(\displaystyle 45-T \text{ minutes}\).

 

[Dr.Peterson]

Here is the problem:

62. Jogging rates: A woman begins jogging at 3:00pm, running due north at a 6-minute-mile pace. Later, she reverses direction and runs due south at a 7-minute-mile pace. If she returns to her starting point at 3:45pm, find the total number of miles run.

Here is what I did with it:
View attachment 11638

This can't be right but it is all I could come up with. The 10 and 8.57 are the hourly rates. the .75 is the time in hours. I think the true solution here would yield something between 6.5 and 7 miles.

You're closer than you realize. You didn't write the definition of x (WHICH YOU SHOULD ALWAYS DO!), but from the equation, it is clear that x is the one-way distance, not the total number of miles run, which was requested. So in 45 minutes, you found that she ran 2*3.461 = 6.922 miles, which is right in the range you expected.

I probably would have avoided decimals by doing everything in fractional miles per minute, the rates being 1/6 and 1/7; then your equation becomes x/(1/6) + x/(1/7) = 45, or 6x + 7x = 45. That gives the same solution (in fraction form, 45/13 = 3 6/13 miles. The answer, then, is 6 12/13 miles total.

It was great that you checked for feasibility against an estimated range. You could also have checked by finding the actual time each way (the two terms of your equation) and seeing that they do add up to 45 minutes; in doing so, you might have realized what your x means, and answered your own question.

 

[allegansveritatem]

Let's look at the wording. Note that there are two different rates. The out rate is one mile every six minutes, while the return rate is one mile every seven minutes. So the out-distance equals the return-distance. If the out-time is \(\displaystyle T \text{ minutes}\) then the return-time is \(\displaystyle 45-T \text{ minutes}\).

I did not read the problem closely enough. Yes, m there is an equivalence here.

 

[allegansveritatem]

You're closer than you realize. You didn't write the definition of x (WHICH YOU SHOULD ALWAYS DO!), but from the equation, it is clear that x is the one-way distance, not the total number of miles run, which was requested. So in 45 minutes, you found that she ran 2*3.461 = 6.922 miles, which is right in the range you expected.

I probably would have avoided decimals by doing everything in fractional miles per minute, the rates being 1/6 and 1/7; then your equation becomes x/(1/6) + x/(1/7) = 45, or 6x + 7x = 45. That gives the same solution (in fraction form, 45/13 = 3 6/13 miles. The answer, then, is 6 12/13 miles total.

It was great that you checked for feasibility against an estimated range. You could also have checked by finding the actual time each way (the two terms of your equation) and seeing that they do add up to 45 minutes; in doing so, you might have realized what your x means, and answered your own question.

I worked this out again today and got the right thing. I have not yet downloaded my photo but when I do I will post the work Thanks for comment. I changed the rate from mph to mpm.

 

[allegansveritatem]

Here is how I finally got this right:

You're closer than you realize. You didn't write the definition of x (WHICH YOU SHOULD ALWAYS DO!), but from the equation, it is clear that x is the one-way distance, not the total number of miles run, which was requested. So in 45 minutes, you found that she ran 2*3.461 = 6.922 miles, which is right in the range you expected.

I probably would have avoided decimals by doing everything in fractional miles per minute, the rates being 1/6 and 1/7; then your equation becomes x/(1/6) + x/(1/7) = 45, or 6x + 7x = 45. That gives the same solution (in fraction form, 45/13 = 3 6/13 miles. The answer, then, is 6 12/13 miles total.

It was great that you checked for feasibility against an estimated range. You could also have checked by finding the actual time each way (the two terms of your equation) and seeing that they do add up to 45 minutes; in doing so, you might have realized what your x means, and answered your own question.

I hit upon this same formulation (mpm) before I read your post. What I don't quite get is, why didn't it work when I was trying to work it with mph?

 

[Dr.Peterson]

Here is how I finally got this right:
View attachment 11651

I hit upon this same formulation (mpm) before I read your post. What I don't quite get is, why didn't it work when I was trying to work it with mph?

But your original version, as I said, was fine! You merely forgot that x was only half the total distance. It did work!

The method pka suggested is an alternative, and in some senses not as good as what you or I did (you with decimals, I with fractions). In your new work, based on pka, x is the time for the first direction, and you have to do extra work at the end to find the distance.

Your work here has some rounding errors; the two distances, which have to be the same, are actually both 45/13 = 3 6/13 = 3.4615... . The total distance is twice this, 6 12/13 = 6.923... .

I do think my version, in post #3, is the easiest and cleanest.

 

[pka]

The method pka suggested is an alternative, and in some senses not as good as what you or I did (you with decimals, I with fractions). In your new work, based on pka, x is the time for the first direction, and you have to do extra work at the end to find the distance.
I do think my version, in post #3, is the easiest and cleanest.

Well, isn't that strictly in the view of the beholder?

 

[allegansveritatem]

But your original version, as I said, was fine! You merely forgot that x was only half the total distance. It did work!

The method pka suggested is an alternative, and in some senses not as good as what you or I did (you with decimals, I with fractions). In your new work, based on pka, x is the time for the first direction, and you have to do extra work at the end to find the distance.

Your work here has some rounding errors; the two distances, which have to be the same, are actually both 45/13 = 3 6/13 = 3.4615... . The total distance is twice this, 6 12/13 = 6.923... .

I do think my version, in post #3, is the easiest and cleanest.

Yes, I see that now. X represents half the distance.Part of my problem is I did not read the problem closely and so somehow didn't catch the fact that the disetances were equal. When I read it again I saw that. So...that is why I set it up the first time as I did rather than go straight for an equivalence format.
Pretty dumb to miss that. I think I will have to study these problems more methodically and get clearly in mind what the variables are going to stand for.

 

[Dr.Peterson]

Yes, I see that now. X represents half the distance.Part of my problem is I did not read the problem closely and so somehow didn't catch the fact that the disetances were equal. When I read it again I saw that. So...that is why I set it up the first time as I did rather than go straight for an equivalence format.
Pretty dumb to miss that. I think I will have to study these problems more methodically and get clearly in mind what the variables are going to stand for.

Interestingly, you did use the fact that the distances are equal, in calling them both x. But you're right that it is important to consciously be aware of what you are doing!

Also, I've been presuming that you used the method you did because it was used in examples you have seen. Some books expect that method for this kind of problem because it (almost) directly gives the value requested, but also because it illustrates a different way of thinking. The other way, starting with the time as a variable, is more routine (that is, it uses d = rt rather than t=d/r, which students are often used to from other problems); I often point out to students who use either method that the other method is available, to encourage flexible thinking. Neither is "dumb" (and which is "better" is, indeed, in the eye of the beholder, even when one does turn out to be less work).