Algebra PoW Packet
Time Trial Triumphs
December 13, 2010
Welcome!
•
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This packet contains a copy of the problem, the “answer check,” our solutions, teaching suggestions,
and some samples of the student work we received in October 2006, when Time Trial Triumphs was
first presented. It is Library Problem #3952. The text of the problem is included below. A printfriendly version is available using the “Print this Problem” link on the current AlgPoW problem page.
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Standards
In Time Trial Triumphs students are asked to compare the distance traveled of two bikers over time
based on the given information about start times and speed. The key concept is to create a
representation of the situation that allows the student to see when the bikers will be in the same place.
This representation may include equations with variables or a table and students will need to
understand percent difference and the use of the distance formula.
If your state has adopted the Common Core State Standards, this alignment may be helpful:
Algebra: Creating Equations
1. Create equations and inequalities in one variable and use
them to solve problems.
Algebra: Reasoning with Equations and Inequalities
1. Explain each step in solving a simple equation as following from the equality of numbers
asserted at the previous step, starting from the assumption that the original equation has a
solution. Construct a viable argument to justify a solution method.
Algebra: Mathematical Practices
1. Make sense of problems and persevere in solving them.
4. Model with mathematics.
5. Use appropriate tools strategically.
6. Attend to precision.
Additional alignment information can be found through the Write Math with the Math Forum service,
where teachers can browse by NCTM and individual state standards, as well as popular textbook
chapters, to find related problems.
The Problem
Time Trial Triumphs
Johnny West and his little brother, Ernesto, are participating in a 30 mile
charity time trial on their bicycles. After Lance Armstrong won yet another Tour
de France bicycle race, the local community organization decided to run a time
trial instead of a regular bike-a-thon in order to raise money for the American
Cancer Society.
A “time trial” is a racing format where you are racing against the clock, instead of directly against the
other competitors. Instead of all starting at the same time, riders start at regular intervals. In this time
trial, a new rider will leave the starting line every 30 seconds.
At 9:05 a.m. Ernesto starts riding at a speed of 440 yards per minute. At 9:20 a.m. Johnny starts riding,
but he’s going 20% faster than his little brother.
At what time will Johnny catch Ernesto? How many miles will they each have ridden at that time?
Extra: When Johnny reaches the finish line after 30 miles, how many miles behind is Ernesto? How
much longer will it take Ernesto to finish?
After students submit their solution, they can choose to “check” their work by looking at the answer
that we provide. Along with the answer itself (which never explains how to actually get the answer) we
provide hints and tips for those whose answer doesn’t agree with ours, as well as for those whose
answer does. You might use these as prompts in the classroom to help students who are stuck and
also to encourage those who are correct to improve their explanation:
Answer Check
Johnny will catch Ernesto at 10:35 a.m. They will each have ridden 22.5 miles at that time.
If your answers don’t match ours:
Did you use a rate of 528 yards/minute for Johnny?
Did you realize that Johnny started 15 minutes later than Ernesto?
Did you use the relationship distance = rate x time for each of the bikers?
Have you incorporated into your equation the fact that they rode the same distance in different
amounts of time?
• Did you check your arithmetic?
•
•
•
•
If any of those ideas help you, you might revise your answer, and then leave a comment that tells
us what you did. If you’re still stuck, leave a comment that tells us where you think you need help
or where you’re having trouble.
If your answers do match ours,
• Is your explanation of your work clear and complete? Will someone reading it fully understand
what you did and why you did it?
• Did you make any mistakes along the way? If so, how did you find and fix them?
• Are there any hints that you would give another student?
• Did you try the Extra question?
Revise your work if you have any ideas to add. Otherwise leave us a comment that tells us how
you think you did - you might answer one or more of the questions above.
This problem is intended to have students use a variable and the relationship between distance, rate
and time to write and solve an equation. Some students will assign the variable to be the number of
minutes Ernesto cycles before Johnny catches up while others may assign the variable to be the
number of minutes Johnny cycles. Some students may solve this problem by creating a chart.
Our Solutions
Method 1a (Using Distance = Rate x Time, using yards/min)
Here’s how I tackle problems like this. I want to make a distance, rate, time chart that represents all of
the information that I have. It ends up looking like this:
Let t = the number of minutes Ernesto has cycled.
Since Johnny started 15 minutes later, he has cycled t - 15 minutes.
Since Johnny is going 20% faster than Ernesto, I can calculate. That means Jimmy’s rate is 120% of
Ernesto’s rates and we can get Jimmy's rate from Ernesto's rate by multiplying it by 1.2.
440 yards
528 yards
•1.2 =
min
min
!
© 2010 Drexel University
2
I made this chart to help me think about what was going on:
Ernesto
Rate
Time
Distance
440 yds/min
t
440t
t-15
528(t - 15)
440 yds/min + 20%
Johnny
528 yds/min
Since we are asked to figure out when Jimmy catches Ernesto, we know that at that moment, they will
have cycled the same distance. So we can set the distances equal to each other and solve for t.
440t = 528(t " 15)
440t = 528t " 7920
7920 = 88t
90 = t
We know that Johnny catches Ernesto after Ernesto has been cycling 90 minutes. The problem
actually
! asks us what time he catches him. Ernesto started cycling at 9:05, so 90 minutes later it’s
10:35.
To figure out many miles they have cycled, we can substitute 90 in for t (since t was time in minutes) in
the formula for Ernesto’s distance, and then convert the answer from yards to miles.
440 yards
1 mile
• 90 min = 39600 yards
= 22.5 miles
min
1760 yards
Johnny and Ernesto have traveled 22.5 miles at the point when Jimmy catches Ernesto.
!
Method 1b (Using Distance = Rate x Time, using miles/min)
Since one of the questions I am answering asks about how many miles the two brothers will have
biked, I decided to start by changing their rates into miles/min. Since 1 mile = 1760 yards, I can
convert Ernesto’s speed to miles/min like this:
Ernesto :
440 yards
1 mile
0.25 miles
•
=
min
1760 yards
min
Since Johnny is going 20% faster than Ernesto, I can calculate. That means Jimmy’s rate is 120% of
Ernesto’s
rates and we can get Jimmy’s rate from Ernesto's rate by multiplying it by 1.2.
!
Johnny :
0.25 miles
0.3 yards
•1.2 =
min
min
I know distance = rate x time. I know the rates for both cyclists, and I know that Johnny starts 15
minutes
! after Ernesto so I am going to let:
let t = Ernesto’s time in minutes
let (t – 15) = Johnny’s time in minutes
For Ernesto, I know that
d=
0.25 miles
•t
min
For Johnny, I know that
!
© 2010 Drexel University
!
d=
0.3 miles
•(t " 15)
min
3
Since I want to know what time Johnny catches up to Ernesto, I want to know what the value of t is
when their distances are equal, so:
0.25 miles
0.3 miles
•(t) =
•(t " 15)
min
min
multiply both sides by 100 to get rid of the decimals
25t = 30(t " 15)
25t = 30t " 450
450 = 5t
!
90 = t
So after 90 minutes, Johnny will catch up with Ernesto. 90 minutes after 9:05 is 1 hour and 30 minutes
after 9:05 which is 10:35.
!
To see how far Ernesto has cycled, I can use distance = rate x time:
0.25 miles
• 90
min
d = 22.5 miles
d=
So Ernesto and Johnny will have cycled 22.5 miles when Johnny catches up.
! To check my answer, I can make sure it works with Johnny’s time. That means Johnny will have been
cycling for 75 minutes:
0.3 miles
• 75
min
d = 22.5 miles
d=
I got 22.5 miles for both of them, so my answer checks.
!
Method 2 (Start with a table)
I knew that Ernesto was cycling at a rate of 440 yards/min. I also knew that Johnny was cycling 20%
faster than Ernesto, so I began by calculating Johnny’s speed.
Johnny’s speed = Ernesto’s speed * 1.2
Johnny’s speed = 440 yards/min * 1.2 = 528 yards/min
Then I decided I wanted to make a chart to see their distance cycled as time passed. Even though I
was using Excel, I didn’t want to go minute by minute unless I needed to, so I looked at where they
were every five minutes. I converted their rates into how far each of them could go in 5 minutes.
Ernersto’s speed ! 2200 yards/5 min
Johnny’s speed ! 2640 yards/5 min
9:05
9:10
9:15
9:20
9:25
9:30
9:35
9:40
9:45
9:50
9:55
10:00
10:05
10:10
© 2010 Drexel University
Ernesto
Johnny
0
2200
4400
6600
8800
11000
13200
15400
17600
19800
22000
24200
26400
28600
0
0
0
0
2640
5280
7920
10560
13200
15840
18480
21120
23760
26400
4
10:15
10:20
10:25
10:30
10:35
30800
33000
35200
37400
39600
29040
31680
34320
36960
39600
10:40
41800
42240
The row in red, at 10:35 am, is where they have cycled the same distance, so Johnny will catch
Ernesto at 10:35am. This is 90 minutes after Ernesto started cycling.
As I look at this table, I wonder if there is a way I could do this more efficiently. I noticed that every five
minutes Ernesto is going 2200 yards, and every 5 minutes Johnny is going 2640 yards. Also, I
remember that Johnny started 15 minutes after Ernesto.
Since I want to know how long it is going to take until Johnny reaches Ernesto, I am going to
let t = the time in minutes Johnny is biking before he catches Ernesto
Looking at my table, I can see that Ernesto bikes for 15 minutes before Johnny starts.
Let Ernesto’s time = be t + 15
I know their distances are going to be the same and I know distance = rate x time.
So, Ernesto’s distance is:
2200 yards
• t + 15
5 min
(
)
And, Johnny’s distance is:
2640 yards
•t
5 min
!
Since their distances are equal, I can use this to solve for the time:
2200 yards
2640 yards
• t + 15 =
•t
5 min
5 min
2200
2640
• t + 15 =
•t
5
5
(
(
!
)
)
To get rid of the 5 minutes, I multiply both sides by 5:
(
)
2200 t + 15 = 2640 •t
!
2200t + 33000 = 2640t
33000 = 440t
75 = t
So Johnny cycles for 75 minutes, or 1 hour and 15 minutes before catches up to Ernesto. At the time
Johnny catches up with Ernesto, Ernesto has cycled 15 minutes longer than Johnny or 90 minutes.
!
To figure out many miles they have cycled, we can multiply 90 minutes by Ernesto’s rate, and then
convert the answer from yards to miles (or we can multiply 75 minutes by Johnny’s rate).
440 yards
1 mile
• 90 min = 39600 yards •
= 22.5 miles
min
1760 yards
Johnny and Ernesto have traveled 22.5 miles at the point when Jimmy catches Ernesto.
!
Method 3 (Use the rate at which Johnny makes up distance once he starts moving)
I started by figuring out how far ahead Ernesto was when Johnny starts. If Ernesto is going 440
yards/min and starts at 9:05 and cycles for 15 minutes before Johnny starts then to calculate Ernesto’s
distance traveled we multiply his rate, 440 yards/min by the times, 15 min.
© 2010 Drexel University
5
440 yards
•15 min = 6600 yards
min
Since I know that Johnny is traveling 20% faster than Ernesto, I calculated how fast he was going.
So Johnny’s rate is:
!
440 yards
528 yards
•1.2 =
min
min
So Johnny is travelling 88 yards/min faster than Ernesto, which means once he starts cycling, he will
make up 88 yards each minute. Since he needs to make up 6600 yards, I set up this equation and
! it:
solved
Let x = the number of minutes it takes Johnny to catch up
88 x = 6600
x = 75
Since Johnny left at 9:20, 75 minutes later is 10:35. So Johnny will catch up with Ernesto at 10:35.
!
To figure out many miles they have cycled, we can multiply 75 minutes by Johnny’s rate, and then
convert the answer from yards to miles.
528 yards
1 mile
• 75 min = 39600 yards
= 22.5 miles
min
1760 yards
Johnny and Ernesto have traveled 22.5 miles at the point when Jimmy catches Ernesto.
!
Extra
To tackle the extra I decided to change their rates to miles/hour.
Ernesto :
440 yards
1 mile
60 min 15 miles
•
•
=
min
1760 yards
hour
hour
Johnny :
528 yards
1 mile
60 min 18 miles
•
•
=
min
1760 yards
hour
hour
!
This lets me see that it’s going to take Ernesto 2 hours and Johnny 1 hour and 40 minutes. Since
Johnny starts 15 minutes after Ernesto, and it takes him 20 minutes less time to complete the 30 miles,
he’ll!finish 5 minutes before Ernesto.
Now I need to figure out how far Ernesto travels in 5 minutes.
Ernesto :
440 yards
1 mile
•
• 5 min = 1.25 miles
min
1760 yards
So Ernesto is 1.25 miles behind Johnny when Johnny finished 5 minutes before him.
!
Teaching
Suggestions
As noted earlier in the Our Solutions section, the goal of the problem is to have students continue to
practice and improve their ability to choose a variable, write an equation or a model that represents the
problem situation, and use that to solve the problem.
One nice thing about this problem is that you can use a variable and an algebraic representation to
solve it, but you can also solve it by using a table and charting the cyclists’ distance over time based
on their rates. It might be nice to have a discussion with your students about what each strategy
allows you to learn about the problem – for example, using the chart allows you to see distance cycled
as time passes, while setting up an equations allows you to solve for a specific but unknown distance
efficiently and quickly. The table method lends itself to a discussion about how to move from a table to
an algebraic expression. Students can make a table, talk about what they learned and then use the
table and their observations to think about how to write and equation and how an equation can be
more efficient than a table.
© 2010 Drexel University
6
You will probably notice that some students will use yards and then change convert the distance
traveled to miles at the end of all their calculations while other students will start out converting the
rates to miles in the beginning before setting up their equations. If a student starts out by converting to
miles, the rates are less than one. Some students like working with decimals, while others may not. A
conversation about how to eliminate decimals by multiplying by 10 or 100 might be a nice reminder
about how to work with equations and what you’re “allowed to do” to make them easier to manipulate
and solve.
The Extra question is nice because it helps students think about what’s happening with this problem
and how the cyclists’ rates really dictate who finishes first even though they start at different times. It
also give students more practice attending to units and converting units.
Some students may struggle with how to use the distance = rate x time formula and using a table to
organize the relationships for each biker may be helpful. The table method may also be useful for
weaker students. In the case illustrated we used intervals of 5 minutes, but any intervals could be
used and converting the rates based on the desired intervals is a good exercise.
The problem is a good chance to practice the “Make a Table” strategy and activities as outlined in our
Activity Series area. A link to that page is always available in the left menu when you’re logged in, and
links to details and examples of that strategy can be found in the blue box on the problem page. I’ve
also modeled what that approach might look like in the Our Solutions section above.
One way to get students thinking about the “scenario” of the problem instead of jumping right to
finding “the answer” is to remove the question when you present the problem. To help with that, we
make a version of the problem without the question available - just look for the “Scenario Only” in the
Teacher Support Materials. In this case, students might begin to actually solve the problem, since it’s
largely presented in the scenario. What’s not included is the time Johnny starts; this will allow
students to play with the relationships between the two cyclists’ speeds without knowing how far apart
they started.
The Online Resources Page for this problem contains links to related problems in the Problem Library
and to other web-based resources.
If you would like one page to find all of the resources for the Current Problems as we add them
throughout the 2010-2011 season, consider bookmarking this page:
http://mathforum.org/pow/support/
Student
Solutions
Focus on
Completeness
In the solutions below, we’ve focused on students’ Completeness of the problem, meaning that they
have shown the steps they have taken to solve the problem in way that one can follow their thinking.
Our hope is that these student solutions help provide insight into conversations you might have with
your students as they work to improve their problem solving and communication. The table below is an
excerpt from the rubric for this problem showing the guidelines for scoring in Completeness:
Novice
Has written very little
that tells or shows
how they found their
answer.
Apprentice
Submitted explanation
without work or work
without explanation.
Leaves out enough
details that another
student couldn’t follow or
learn from the
explanation.
Practitioner
Expert
Explains all of the important steps
taken to solve the problem,
including stating which variable
expression is the larger number.
Adds in useful
extensions and further
explanation of some
of the ideas involved
Shows equations, formulas, and
calculations used and explains the
rationale behind them.
The additions are
helpful, not just “I’ll
say more to get more
credit.”
Defines variable(s).
Does not define
variable(s).
For each solution, I’ve included a comment about why I would score it as shown, as well as what I’d
ask the student to work on when they revised their solution to help them move forward with solving the
problem or improving their write-up of their work. I hope they are both informative and fun.
© 2010 Drexel University
7
Donna
age 15
Johnny will catch Ernesto at 10:35 a.m. and they will have each traveled
2.5 miles.
Set up the equation and solved for the time.
Completeness
Novice
Ashok
Paul
age 10
15
johnny would catch up after 132 minutes after he started.
Ernesto started at 9:05 and was 6600 miles further than Johnny when he
started. 6600/528 is 132.
Completeness
Interpretation
Apprentice
Novice
Andrew
Paul
age 14
15
Completeness
Interpretation
Apprentice
at 59 minutes is when Johnny will catch Ernesto. Ernesto will have ridden
7.4 miles, and Johnny will have ridden 4.4 miles.
First, I figured out how many yards johnny can travel per minute. then i
figured out how many yards ernesto traveled by the time Johnny starting
riding. then, i subtracted 88 yards for each minute until I came to 59
minnutes since there is a 20% difference in their speeds.
© 2010 Drexel University
Although Donna has come
up with the right answer,
she has offered us no insight
into the steps she took to
get that answer. I would ask
her to share with me the
equation she set up and the
variables she used. I would
encourage her to think
about how she would write
about what she did in way
that another student could
learn from it if she were not
there to explain it.
Ashok has left out some
steps that would allow us to
follow his thinking. I would
ask him to offer more detail
about where he came up
with 6600 and 528 and to
include units with the
numbers shares to help
others understand his
thinking.
Andrew has a great start
and a good idea for a
strategy. It’s true that for
each minute they travel at
the same time, Johnny is
gaining 88 yards on Ernesto.
However, in terms of
completeness, it’s hard to
know what his calculations
look like and where we
might be able to pinpoint a
misstep. I would ask Andrew
to tell me how he came up
with 59 minutes. Was he
using a table, an equation,
or what did his thinking look
like?
8
Katie
Paul
age 14
15
Completeness
Interpretation
Apprentice
Dallin
Paul
age 13
15
Completeness
Interpretation
Practitioner
Apprentice
Johnny will catch Ernesto at 10:35. They will have ridden 22.5 miles.
Here is my work:
440 ypm
528 ypm
528x=440(x+15)
528x=440x+6600
88x=6600
X=75 min
9:20+75 min=10:35
440*90=39600
528*75=39600
5280 ft per mile
5280/3=1760 yds per mile
39600/1760=22.5 miles
Johnny will catch Ernesto one hour, twenty-two minutes, and thirty
seconds (75 minutes) after Johnny leaves, or in other words at 10:35:00
a.m. They will have ridden twenty-four and three eighths of a mile (22.5
miles).
My original equation was:
if m=minutes, y=yards, and x=time traveled at meeting point
then:
[15m(440y/m)]+440y/m(x)=528y/m(x)
I then isolated x:
[15m(440y/m)]+440y/m(x)=528y/m(x)
6600y+440y/m(x)=528y/m(x)
[6600y+440y/m(x)]/528y/m=[528y/m(x)]/(520y/m)
165/13m+(11x)/13=x
[165/13m+(11x)/13](13)=13x
165m+11x=13x
165m+11x-11x=13x-11x
165m=2x
165m/2=2x/2
75m=x
x=75m
Katie’s approach is great
and I would give her a
practitioner in strategy,
interpretation and accuracy.
However, in terms of
completeness, I would ask
her to define her variables,
since she has not told me
what x is. I would also ask
her to write a bit more
explaining her thinking. How
did she know
528x = 440(x+15)? I would
encourage her to write a
little more so that if one of
her classmates wanted to
learn from her approach
they could understand not
just what she did, but why.
Dallin has clearly and
completely laid out his work
and his thinking. He defines
his variables and units nicely
and shows his work and his
thinking behind it. The one
question I would have for
Dallin is if he could explain
more about this calculation
in the middle of his work:
D=(520yards/minute)(82.5mi
nutes). It is unclear to me
where those numbers came
from.
With x, I had the length of time Johnny had traveled and was thus
able to add 75 minutes to Johnny’s time of departure.
9:20:00+1:15:00=10:35:00
I then used the value of x (75 minutes) and Johnny’s speed (520
yards per minute) to find the distance traveled.
If d=distance, r=rate, and t=time
Then:
D=r(t)
D=(520yards/minute)(82.5minutes)
© 2010 Drexel University
9
I proceeded by solving for D.
D=(528yards/minute)(75minutes)
D=(528yards/minute)(75minutes)
D=528yards(75)
D=39600yards
I finished by dividing the value of D by the number of yards in a
mile (1760).
39600yards/1760yards=22.5miles.
Torpaskolan
Paul
age 15
Completeness
Interpretation
Apprentice
Practioner
Johnny will catch up with Ernesto at 10:35 am. They will have ridden 22.5
miles then. When Johnny reaches the finish line, Ernesto will be 1.25 mile
behind and will not reach the finish line until 5 minutes later, so it will take
him 20 minutes lon
Johnny will catch Ernesto at 10:35. They will have ridden 22.5 miles.
Johnny bikes 20% faster:
440/5 = 88
440 + 88 = 528
He travels with a speed of 528 yards/minute.
He will be getting close every minute:
528 - 440 = 88 <- Every minute he will get 88 yards closer.
Ernesto has already ridden for 15 minutes:
15 * 440 = 6600 yards <- He's ridden 6600 yards at that time.
Torpaskolan has done an
excellent job of explaining
his thinking with a narrative
and calculations. His
thinking is clear and another
student could follow his
logic and work well.
Although he’s a practitioner
in completeness, I would
rate him as an apprentice in
strategy because I would
encourage him to push his
thinking towards an
algebraic representation of
the problem.
Since Johnny is riding 88 yards more every minute, check how long
time it will take for him to reach Ernesto:
6600 / 88 = 75 <- It will take him 75 minutes.
He started 9:20:
9:20 + 1:15 = 10:35 am <- The time when he's reachen Ernesto.
Now calculate how many miles he's been travelling:
1 miles = 1 760 yards.
75 * 528 = 39 600 <- Yards that he's ridden, minutes * speed.
39 600 / 1 760 = 22.5 <- Convert to miles
They had been riding for 22.5 miles!
Scoring Rubric
A problem-specific rubric, to help in assessing student solutions, is available in the Teacher Support
Materials on the Problem page when you are logged in as a teacher. As shown above, we consider
each category separately when evaluating the students’ work, thereby providing more focused
information regarding the strengths and weaknesses in the work. A generic student-friendly rubric
can be downloaded from the Teaching with PoWs link in the left menu (when you are logged in). We
encourage you to share it with your students to help them understand our criteria for good solutions.
We hope these packets are useful in helping you make the most of the Algebra Problems of the Week.
Please let me know if you have ideas for making them more useful.
~ Valerie
© 2010 Drexel University
([email protected])
10