Answer:

Tracey lives 18 miles from work.
***EXTRA: Tracey must go 60 miles per hour to obtain her
average.

Explanation:
I decided to solve this problem using a table with distance, rate, and time. Under rate, I put 40 miles per hour and 45 miles per hour, the two rates given in the problem. Under time, I used x for the time it takes Tracey at 40 miles per hour, and x-2 for her time at 45 miles per hour. This is because she arrives one minute late at 40 mph, and one minute early at 45 mph, a difference of 2 minutes. For distance, I used 40x and 45(x-2). This is because I multiplyed across the table, and got these two products. To solve the problem, I set up an equation of 40x = 45(x-2). This distributes to be 40x = 45x-90. I solved it, and got an answer of 18 miles to work. I set up the equation in the way that I did because the distance to her house, and to work is the same, assuming that she takes the same route. ******EXTRA: I started off with multiplying 40 by 2 to get 80. This is because when you find the average of a set of numbers, there is always a specific value that you get that corresponds to the average that you get depending on the number of numbers you are averaging. For example, a set of three numbers with an average of 10 will always have a sum of 30. Then, I subtracted 20 from 80, because I know that 20 is part of the sum. That left me with 60. Therefore, Tracey must go 60 mph to acheive the average of 40 mph.

[The student looked at answer 1]

Answer:

Tracey lives 18 miles from work.
***EXTRA: Tracey must go 60 miles per hour to obtain her
average.

Explanation:
I decided to solve this problem using a table with distance, rate, and time. Under rate, I put 40 miles per hour and 45 miles per hour, the two rates given in the problem. Under time, I used x for the time it takes Tracey at 40 miles per hour, and x-2 for her time at 45 miles per hour. This is because she arrives one minute late at 40 mph, and one minute early at 45 mph, a difference of 2 minutes. For distance, I used 40x and 45(x-2). This is because I multiplyed across the table, and got these two products. To solve the problem, I set up an equation of 40x = 45(x-2). This distributes to be 40x = 45x-90. I solved it, and got an answer of 18 miles to work. I set up the equation in the way that I did because the distance to her house, and to work is the same, assuming that she takes the same route. ******EXTRA: I started off with multiplying 40 by 2 to get 80. This is because when you find the average of a set of numbers, there is always a specific value that you get that corresponds to the average that you get depending on the number of numbers you are averaging. For example, a set of three numbers with an average of 10 will always have a sum of 30. Then, I subtracted 20 from 80, because I know that 20 is part of the sum. That left me with 60. Therefore, Tracey must go 60 mph to acheive the average of 40 mph.

[The student looked at answer 1]

Actual written comment:
I completed the problem, but I got the wrong answer, and I cannot figure out
what is wrong with my reasoning. I wondering if it is my problem solving
techniques. I always either get an answer of 18 or 16. What should I do?

Hi Jenna,

Welcome to our Problem of the Week project.  We hope you will enjoy solving
our problems this year.

You've done some good thinking here, for sure, & I think I see your
confusion.  Look: what did you call "x" at the start?  The time at the slow
speed, right?  But after all your work, you forgot that, and when your
equation gave you 18, you called it "miles".    :>)

That should help a good bit.  But, you still have more work to do.

As to the 'extra', it too is quite wrong, & tricky.  Let's get the main
question fixed up first though, before we take that on.

Thank you so much.  We hope to see your revision soon.

P.S.  You checked the html button for your "Explanation" when you sent your
submission, but this should only be checked if you include html tags for
formatting your solution - otherwise it just runs all of your text together
(and makes it very hard to read).  Please leave text checked in the future.
Thanks!

Summary:

         Problem Solving  Interpretation: Practitioner
                          Strategy:       Apprentice
                          Accuracy:       Apprentice

         Communication    Completeness:   Apprentice
                          Clarity:        Apprentice
                          Reflection:     Apprentice

(for an explanation of scores see: http://mathforum.org/pow/scoring.html)

-- Terrel Trotter, for the Algebra Problem of the Week

Answer:

Tracey lives 12 miles from work.
***EXTRA: Tracey must go 60 miles per hour to obtain her
average. ( i know this is wrong)

Explanation:
I decided to solve this problem using a table with distance, rate, and time columns. In the rate column, I put 40 miles per hour in one row, and 45 miles per hour in another, the two rates given in the problem. In the time column, I used x for the time it takes Tracey at 40 miles per hour,(I put it is the column next to 40 mph) and x-2 for her time at 45 miles per hour (put in the column next to 45 mph). This is because she arrives one minute late at 40 mph, and one minute early at 45 mph, a difference of 2 minutes. For distance, I used 40x and 45(x-2). This is because I multiplyed across the table, and got these two products. To solve the problem, I set up an equation of 40x = 45(x-2). This distributes to be 40x = 45x-90. I solved it by subtracting 45x from both sides, resulting in -5x =-90. I divided both sides by negative five and got an answer of 18 minutes for her to get to work. I divided by negative five in order to get x by itself. However, this is not the answer. I then found how many miles she can obtain in one minute: 40 miles in 60 minutes, or 2/3 of a mile per minute. I multiplyied this by her time of 18 minutes, and I got an answer of 12 miles, assuming that she takes the same route both ways. ******EXTRA: I started off with multiplying 40 by 2 to get 80. This is because when you find the average of a set of numbers, there is always a specific value that you get that corresponds to the average that you get depending on the number of numbers you are averaging. For example, a set of three numbers with an average of 10 will always have a sum of 30. Then, I subtracted 20 from 80, because I know that 20 is part of the sum. That left me with 60. Therefore, Tracey must go 60 mph to acheive the average of 40 mph.

Answer:

Tracey lives 12 miles from work.
***EXTRA: Tracey must go 60 miles per hour to obtain her
average. ( i know this is wrong)

Explanation:

I decided to solve this problem using a table with distance, rate,
and time columns. In the rate column, I put 40 miles per hour in
one row, and 45 miles per hour in another, the two rates given in
the problem. In the time column, I used x for the time it takes
Tracey at 40 miles per hour,(I put it is the column next to 40 mph)
and x-2 for her time at 45 miles per hour (put in the column next
to 45 mph). This is because she arrives one minute late at 40
mph, and one minute early at 45 mph, a difference of 2 minutes.
For distance, I used 40x and 45(x-2). This is because I
multiplyed across the table, and got these two products. To solve
the problem, I set up an equation of 40x = 45(x-2). This
distributes to be 40x = 45x-90. I solved it by subtracting 45x from
both sides, resulting in -5x =-90. I divided both sides by negative
five and got an answer of 18 minutes for her to get to work. I
divided by negative five in order to get x by itself. However, this is
not the answer. I then found how many miles she can obtain in
one minute: 40 miles in 60 minutes, or 2/3 of a mile per minute. I
multiplyied this by her time of 18 minutes, and I got an answer of
12 miles, assuming that she takes the same route both ways.

******EXTRA: I started off with multiplying 40 by 2 to get 80. This
is because when you find the average of a set of numbers, there
is always a specific value that you get that corresponds to the
average that you get depending on the number of numbers you
are averaging. For example, a set of three numbers with an
average of 10 will always have a sum of 30. Then, I subtracted
20 from 80, because I know that 20 is part of the sum. That left
me with 60. Therefore, Tracey must go 60 mph to acheive the
average of 40 mph.

Thanks for the correction, Jenna.  Now we can work on your overall
presentation, & of course, the extra.

Your explanation is sorta hard to follow.  You presented all your thoughts
in one long gigantic paragraph, which makes it rather difficult to follow &
understand.  The equations are easier to read & comprehend if they are not
buried among a lot of wordy sentences, you see.  (Check out some past
problems & how kids wrote their work.  For more about this aspect, please
go to our Guide for Writing Answers at
http://mathforum.com/algpow/writing_answers.html

Could you also show your table that you spoke of?  That would help a great
deal.

As to the extra, one thing that might help in such "round trip" questions
is to recall that "average speed" for a whole trip is found by
dividing "total distance" by "total time".  Many folks think it is just the
average of the two one-way speeds, which is never true.

I await your revision.  See ya'.

Summary:

         Problem Solving  Interpretation: Practitioner
                          Strategy:       Practitioner
                          Accuracy:       Practitioner

         Communication    Completeness:   Practitioner
                          Clarity:        Apprentice
                          Reflection:     Apprentice

(for an explanation of scores see: http://mathforum.org/pow/scoring.html)

-- Terrel Trotter, for the Algebra Problem of the Week

Answer:

Tracey lives 12 miles from work.
***EXTRA: There is no possible speed for Tracey to go to get her to
an average speed of 40 mph

Explanation:

1.I decided to solve this problem using a table with distance, rate,
and time columns. In the rate column, I put 40 miles per hour in
one row, and 45 miles per hour in another, the two rates given in
the problem. In the time column, I used x for the time it takes
Tracey at 40 miles per hour,(I put it is the column next to 40 mph)
and x - 2 for her time at 45 miles per hour (put in the column next
to 45 mph). This is because she arrives one minute late at 40
mph, and one minute early at 45 mph, a difference of 2 minutes.
For distance, I used 40 x and 45 (x - 2). This is because I
multiplyed across the table, and got these two products. This is the
table :

     Rate    time  distance
    ______________________
     40  I   x  I  40 x
    ______________________
     45  Ix + 2 I 45 (x + 2)

2.To solve
the problem, I set up an equation of 40 x = 45 (x - 2). This
distributes to be 40 x = 45 x - 90. I solved it by subtracting 45 x
from
both sides, resulting in -5 x = -90. I divided both sides by negative
five and got an answer of 18 minutes for her to get to work. I
divided by negative five in order to get x by itself.
    40 x = 45 x - 90
    -5 x = -90
       x = 18

3. However, this is not the answer. I then found how many miles she
can obtain in one minute: 40 miles in 60 minutes, or 2/3 of a mile
per minute. I multiplyied this by her time of 18 minutes, and I got
an answer of 12 miles, assuming that she takes the same route both
ways.
       40/60 = 2/3
    2/3 * 18 = 12

******EXTRA:
1.Now that I know that Tracey lives 12 miles from her work, I know
that she will travel a total of 24 miles round trip. I have already
figured out that each way, at 40 mph, it should take her 18 minutes.
This is useful because she wants an average speed of 40 mph.
Therefore, it should take her 36 minutes round trip.

2. I decided that an equation would be the bast way to find out how
long it takes her to travel 12 miles at a rate of 20 mph. The
equation d = r t would be suitable. The distance is 12 miles. The
rate is 20 mph. We are trying to figure out the times, so it is x.
        d = r t
       12 = 20 x
      0.6 = x
This gave me an answer of 0.6 hours, or 36 minutes. It takes her
twice as long at half the speed. This makes sense.

3. She has already exceeded her 36 minutes round trip. Therefore,
there is no attainable speed for her to go to achieve an average of
40 mph.

4. This answer seems very unlikely, but according to my work, it is a
right answer. I proved it by doing this work:
   d = r t
  24 = 40 x
 0.6 = x
This proves that for traveling 24 miles, and trying for an average
speed of 40 mph, you may only travel for 36 minutes, which she has
already traveled.

[The student looked at answer 1]

Thx for the revision, Jenna.  It looks much better now.  May I suggest one
last thing?  You have separated the parts of many terms, the numerical
coefficient & the letter.  Like "40 x", etc.  Traditionally, we don't do
that.  Rather we write them like this --> 40x.  Hence, your other
expressions & equations should do the same, like

45(x - 2)

40x = 45x - 90

    40x = 45x - 90
     -x = -90
      x = 18

and so on.  Ok?

Your work for the extra is now good.  Interesting, wasn't it?

See you later now.

Summary:

         Problem Solving  Interpretation: Practitioner
                          Strategy:       Practitioner
                          Accuracy:       Practitioner

         Communication    Completeness:   Practitioner
                          Clarity:        Practitioner
                          Reflection:     Apprentice

(for an explanation of scores see: http://mathforum.org/pow/scoring.html)

-- Terrel Trotter, for the Algebra Problem of the Week

Answer:

Tracey lives 12 miles from work.
***EXTRA: There is no possible speed for Tracey to go to get her to
an average speed of 40 mph

Explanation:

1.I decided to solve this problem using a table with distance, rate,
and time columns. In the rate column, I put 40 miles per hour in
one row, and 45 miles per hour in another, the two rates given in
the problem. In the time column, I used x for the time it takes
Tracey at 40 miles per hour,(I put it is the column next to 40 mph)
and x - 2 for her time at 45 miles per hour (put in the column next
to 45 mph). This is because she arrives one minute late at 40
mph, and one minute early at 45 mph, a difference of 2 minutes.
For distance, I used 40x and 45 (x - 2). This is because I
multiplied across the table, and got these two products. This is the
table :

     Rate  time  distance
    ______________________
     40  I   x  I  40x
    ______________________
     45  Ix + 2 I 45(x + 2)

2.To solve the problem, I set up an equation of 40x = 45(x - 2). This
distributes to be 40x = 45x - 90. I solved it by subtracting 45x
from both sides, resulting in -5x = -90. I divided both sides by
negative five and got an answer of 18 minutes for her to get to work.
I divided by negative five in order to get x by itself.
    40x = 45x - 90
    -5x = -90
       x = 18

3. However, this is not the answer. I then found how many miles she
can obtain in one minute: 40 miles in 60 minutes, or 2/3 of a mile
per minute. I multiplied this by her time of 18 minutes, and I got
an answer of 12 miles, assuming that she takes the same route both
ways.
       40/60 = 2/3
    2/3 * 18 = 12

4. This answer seems very reasonable as the number of miles that
Tracey lives from work. I checked my work by substituting the answers
back into my equations
    40(18) = 45(18)- 90
       720 = 810 - 90
       720 = 720
It checked to be correct, so I can assume that my work is correct.
This problem took a long time to do, and I had to correct it many
times, but overall, it was fun and challenging, and I enjoyed being
challenged. My math class recently learned how to do these problems,
and without that knowledge of the process, I would not have been able
to do this problem, or it would have been MUCH more challenging for
me.

******EXTRA:
1.Now that I know that Tracey lives 12 miles from her work, I know
that she will travel a total of 24 miles round trip. I have already
figured out that each way, at 40 mph, it should take her 18 minutes.
This is useful because she wants an average speed of 40 mph.
Therefore, it should take her 36 minutes round trip.

2. I decided that an equation would be the bast way to find out how
long it takes her to travel 12 miles at a rate of 20 mph. The
equation d = r t would be suitable. The distance is 12 miles. The
rate is 20 mph. We are trying to figure out the times, so it is x.
        d = rt
       12 = 20x
      0.6 = x
This gave me an answer of 0.6 hours, or 36 minutes. It takes her
twice as long at half the speed. This makes sense.

3. She has already exceeded her 36 minutes round trip. Therefore,
there is no attainable speed for her to go to achieve an average of
40 mph.

4. This answer seems very unlikely, but according to my work, it is a
right answer. I proved it by doing this work:
   d = rt
  24 = 40x
 0.6 = x
This proves that for traveling 24 miles, and trying for an average
speed of 40 mph, you may only travel for 36 minutes, which she has
already traveled. This was a very challenging problem, and I needed
to put much more work and thought into it than I had originally
thought. The answer was a very strange one, that even doing my work
and checking it many times, I did not believe was a correct one.

Thx, Jenna.  You're improving a lot now.  See you again soon, I hope.



Summary:

         Problem Solving  Interpretation: Practitioner
                          Strategy:       Practitioner
                          Accuracy:       Practitioner

         Communication    Completeness:   Practitioner
                          Clarity:        Practitioner
                          Reflection:     Practitioner

(for an explanation of scores see: http://mathforum.org/pow/scoring.html)

--
I'm Terry Trotter, a retired math teacher at the middle school and
high school levels.  I write problems and mentor students for the
Algebra Problem of the Week.  I'm originally from Kansas and
Illinois, but have lived and worked in El Salvador for the past 21
years.  Website: www.trottermath.com
--

Actual written comment:
Hi! You said that I was improving a lot, but I don't see how I can
improve my answer more. Can you assist me on improving my answer.
Thank you for helping me so much on this problem. I feel that it has
helped me a lot.

Jenna,

You made nice progress.  I was pleased that you took my advice; some kids
don't, you know.  :>(

I'll show you an outline of how I solved it, so you can compare, ok?
-------------
d = rt

40 mph = 40/60 miles/minute   &  45 mph = 45/60 miles/minute.

Let t = time in minutes, to arrive 'on time'
Then t + 1 = late time
And t Ð 1 = early time

d = 40/60 * (t + 1) = 45/60 * (t Ð 1)

2/3 * (t +1) = 3/4 * (t -1)

multiply both sides by 12.

8(t + 1) = 9(t -1)

8t + 8 = 9t Ð 9

t = 17

d = 2/3 * (17 +1) = 2/3 *18 = 12

d = 3/4 * (17 Ð 1) = 3/4 * 16 = 12.

So the distance is 12 miles.

-------------
That's another option.  Later on, come back to this Pow & read other kids'
solutions when they are published, ok?

Get ready for the next AlgPow...

bye,

Terry

p.s. in the meantime, why not check out my personal math website.  see
below...


Summary:

         Problem Solving  Interpretation: Practitioner
                          Strategy:       Practitioner
                          Accuracy:       Practitioner

         Communication    Completeness:   Practitioner
                          Clarity:        Practitioner
                          Reflection:     Practitioner

(for an explanation of scores see: http://mathforum.org/pow/scoring.html)

--
I'm Terry Trotter, a retired math teacher at the middle school and
high school levels.  I write problems and mentor students for the
Algebra Problem of the Week.  I'm originally from Kansas and
Illinois, but have lived and worked in El Salvador for the past 21
years.  Website: www.trottermath.com
--