PoW-TER Problem Packet
Car Rental Quandary (Author: Ashley Miller)
1. The Problem: Car Rental Quandary [Problem #554]
I need to rent a car for my upcoming trip. Rent-A-Gem charges
$20.25 per day plus 14 cents per mile. Super Saver Rentals
charges $18.25 per day plus 22 cents per mile.
At first glance it looks as if I should go with Super Saver Rentals.
Still, I'm concerned about the per-mile charge because I plan to do a
lot of driving during all three days of my trip.
How many miles would I have to drive to make the cost of renting a car from Rent-A-Gem the
same as the cost of renting a car from Super Saver Rentals?
Please read the problem carefully. Be sure that you give all the details of your solution, including
all the comparisons that you make in order to determine the answer.
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2. About the Problem
This question is a great question to either introduce linear equations or systems of equations. It
could also be used as an alternative assessment to see if students understand how linear equations
are applied.
Linear equations tend to be a very difficult topic for many students. Students may struggle with
linear equations because they do not see the need for them. Car Rental Quandary shows how
having a knowledge of equations could save a consumer money and help a consumer make
smarter decisions.
One could use this question as a way to introduce consumer topics that are often left out of many
math curricula. Another idea would be to hold a class discussion about how people decide
between companies such as car rentals or cell phone companies (A Phone-y Deal is a great
example of this one).
This question could be used early in the year to show the importance of tables and equations.
Then this problem could be reintroduced when students are ready for more sophisticated solving
methods such as solving linear systems. Reintroducing this problem could serve several different
purposes. Students could do this problem a second time to see if they have grown in their
approach to problem solving later in the school year. It might also be helpful to try this problem
in groups the first time and then individually to effectively evaluate student problem solving
abilities.
Car Rental Quandary 1
As an alternative to just writing the solution in a journal would be to have students present this
from the standpoint of an advertiser trying to convince a consumer to use one company over the
other. This is a way for students to get practice in communicating and representing math.
This problem is engaging and a quick way for teachers to assess whether students understand
equations and equality.
3. Math Goals (6-8)
•
•
•
•
•
•
•
Tables
Equations
Linear Equations
Break Even Point
Consumer Vocabulary
Inequality words (less than, greater than)
Basic Operations
4. NCTM Mathematics Standards (6-8)
represent, analyze, and generalize a variety of patterns with tables, graphs, words, and,
when possible, symbolic rules;
• develop an initial conceptual understanding of different uses of variables;
• explore relationships between symbolic expressions and graphs of lines, paying particular
attention to the meaning of intercept and slope;
• use symbolic algebra to represent situations and to solve problems, especially those that
involve linear relationships;
• recognize and generate equivalent forms for simple algebraic expressions and solve linear
equations
• model and solve contextualized problems using various representations, such as graphs,
tables, and equations.
• solve problems involving scale factors, using ratio and proportion;
• solve simple problems involving rates and derived measurements for such attributes as
velocity and density
Problem Solving NCTM Standards (6-8)
• build new mathematical knowledge through problem solving; • solve problems that arise in mathematics and in other contexts; • apply and adapt a variety of appropriate strategies to solve problems; • monitor and reflect on the process of mathematical problem solving. Communication NCTM Standards (6-8)
• organize and consolidate their mathematical thinking through communication; • communicate their mathematical thinking coherently and clearly to peers, teachers, and others; • analyze and evaluate the mathematical thinking and strategies of others; • use the language of mathematics to express mathematical ideas precisely. •
Car Rental Quandary 2
Representation NCTM Standards (6-8)
• create and use representations to organize, record, and communicate mathematical ideas; • select, apply, and translate among mathematical representations to solve problems; • use representations to model and interpret physical, social, and mathematical phenomena. 5. Common Misconceptions
1. How many days?
Several students submitted answers that gave the answer if the cars were rented for only one day.
This is a common mistake because students assume that the price given is the price needed
instead of reading closer to see that this particular customer wanted the cars for 3 days instead of
one. I would ask students directly how many days the car was needed and if the answer would
change if the rental was for multiple days.
The answer to the problem is 25 miles.
I first made a table that tells you the total price per mile. For Rent-A-Car, I added 14 cents per
mile to 20.25. I added 22 cents ever mile to 18.25 for Super Saver Rentals. The table looked
something like this.
Mi. Rent-A-Car Super Saver Rentals
1
20.39
18.47
2
20.53
18.69
3
20.67
18.91
4
20.81
19.13
5
20.95
19.35
6
21.09
19.57
7
21.23
19.79
8
21.37
20.01
9
21.51
20.23
10
21.65
20.45
11
21.79
20.64
12
21.93
20.89
13
22.07
21.11
14
22.21
21.33
15
22.35
21.55
16
22.49
21.77
17
22.63
21.99
18
22.77
22.21
19
22.91
22.43
20
23.05
22.65
21
23.19
22.87
22
23.33
23.09
23
23.47
23.31
24
23.61
23.53
25
23.75
23.75
Car Rental Quandary 3
2. Are we looking for the price or the miles?
Students will often be so focused on finding out when the two companies are equal they will
forget they are being asked for how long it takes to get to that price and not what the price
actually is. At this point it would be a good idea to ask students to notice and wonder to see if
they notice that the question is how many days and not looking for a price. If not, I would ask
directly how many days it would take to get to the price. In this case they also had the first
misconception of how many days the car was rented for in the first place.
it was 23.75 for both of them
i took the 14 cents +25 miles and a 20.25 and i did the samething in revers 14cent times 25+ 3.50
+20.25=and 23.75 if you round it an then 22 cents times 18.25= 23.75 and I rounded it the reason
why i added the cents and miles was because you needed to know how many miles and how
much would it cost and so I took the amount of the car price and the 3 days with the miles and
counted 14 cent to the 25 miles and the same with thee other one.
3. Are both companies changing?
This student submitted nine miles because they came up with a price for the first company and
then added miles to the other until they “matched”. This does not consider equal mileage for
both companies. If a student had this misconception it may help to have them make a table of at
least the first five miles to see that both companies would be increasing in price. Hopefully they
would then see that when both companies had been used for nine miles it would not be an equal
price.
The answer to the problem is 9 miles.
The way i figured out the answer is i wrote the two companies down on a sepreate piece of paper
into to columns. Next I wrote Reant-A-Gen in the first column and then wrote down the price
and how many cents per mile. I did the same for the ither column but instead it is Super Saver
Rentals. after that i figured out the math beacues at first i thought about the three days and then
multiplied the prices by three but then when i added the 22 cents per mile the total of the Super
Saver Rentals didnt equal the Rent-A-Gen's total so i knew what the question really was. After
that i figured out the math by adding 22 cents until i got to Rent-A-gen's total and i knew i had
the answer of nine mile. that is how i figured out the problem.
For Super Saver Rentals you would get 9 miles for $2.24 and Rent a Gem 14 miles for $2.24 so
rent a gem would be better if you were goning to drive alot. I know that is you were going to
drive a lot and you needed to rent a car you shouold go to Rent a Gem so you would pay $.14
cents per mile so you could drive further and pay less. you can go 14 miles with REnt a Gem and
9 miles with Super Saver REntals.
The following student solutions show a progression of how students might approach this
problem. The solutions are organized to give you a sample of simple solution methods and
sophisticated methods. This will allow you, as the teacher, to see many of the possible ways
Car Rental Quandary 4
students will approach this problem. This is not an exhaustive list and there may be other ways
to solve this problem.
Add and add and add some more
75 miles is the answer I got. the service charge is 20.25 for RAG. It is 18.25 for service charge
for SSR. I added 14 tto 20.25 and i added 22 cents to 18.25.. Until I got the same amont....It was
25. 25 was for one day. So for three day I multiplied 25x3 and it gave me 75.
This solution is a good starter for discussion since it is the least efficient and likely the most
common strategy used by students who are not comfortable with problem solving. They do not
explain their method in detail just that they added until they found the same number. The fact
that they found the right answer leads me to believe that they understood what they were doing
but may not be comfortable explaining their process. I would use this solution as a springboard
to finding a more detailed way to present this solution. Teachers could have a discussion about
whether or not this method would be ideal every time. Students would possibly see that solving
all problems by adding and adding and adding some more would be tiresome and not efficient.
Over under with guess and check
If a person is going on a trip for 3 days , and they think that they are going to go
more than 76 miles then it is better to have Rent-AGem, if they think they are
goin to use less then 74 miles than it is cheaper to have Super Saver Rentals, f
For this problem the firsty thing i did was figure out how much each company would cost for 3
days with no miles, i did this by takeing the original price and timed it by 3
(20.25x3=60.75)then i did it with the other company
(18.25x3=54.75).Next I took rent-agem and figured out how much it would cost for 3 days and
50 miles
(3days=60.75+.14x50=67.75), then i did that with Super saver
Rentals(3days=54.75+.22x50=65.75). Then i did the same thing going up by another
50(60.75+.14x100=74.75)then i did it with Super saver
(54.75+.22x100=76.75).Since
1000 was too high because Super Saver came out higher so i knew that I needed a smaller
number.So i went with 75 (60.75+.14x75=71.25) then i did the same thing with Supoer
Saver(54.75+.22x75=71.25). I got the same anwser for 75 miles , which makes it if you are
going to travel more then 75 miles then Super Saver Rentals is cheaper , if you arew going to
travel more then Rent-A-Gem would be cheaper.
This person used guess and check and chose 50 for the first attempt. He recognized that it
wasn’t big enough so then he tried 100. This time the solution was too big so he chose to split
the difference and try 75 which worked. This solution method is a little risky because all
answers are not nice clean pretty whole numbers. If the answer would have been a decimal it
may have taken a lot of guesses to find the answer which is frustrating for the guesser. When
frustration sets in students often will give up. Therefore, I would use this solution to show
Car Rental Quandary 5
students that using a more organized method of solving can be less frustrating and more efficient
for most types of problems.
Organized Guess and Check using a Table
The answer I got is 75 miles.
What is the question?
How many miles would I have to drive to make the cost of renting a car from Rent-A-Gem the
same cost of renting a car from Super Saver Rentals?
What I know?
I know that Rent-A-Gem (RAM) charges $20.25 per day plus 14 cents per mile. I also know that
Super Saver Rentals (SSR) charges $18.25 per day plus 22 ents per mile.The last thing that I
know is that their going on a three day trip.
How did I get the answer?
Since I know that it is a three day trip I firts multiplied the amounts of each company by 3:
RAG-$20.75*3=$60.75 for three days
SSR-$18.25*3=54.75 for three days
Since the totals are not the same and I am not going to drive for
more than three days I have to add miles until I reach the same amount for both companies.To
find out how many miles in total I would drive I made a chart to not loose track.For RAG I had
to add 14 cents per mile, and for SSR I had to add 22 cents per mile.
Miles RAG SSR
0 $60.75 $54.75
1 60.89 54.97
2 60.03 55.19
3 61.17 55.41
4 61.31 55.63
5 61.45 55.85
Since there still far apart by a large amount I started to go by 5's instead.
10 62.15 56.95
15 62.85 58.05
20 63.55 59.15
25 64.25 60.25
Sice I saw that they are still far apart I skipped to 60 miles because they are still far apart.
60 69.15 62.95
Since RAG is still more expensive than SSR Iknow that I have to still continue, but since their
not that far apart I decided to go by 5's again.
65 69.85 69.05
70 70.55 70.15
75 71.25 71.25
FINALLY at 75 miles they are both the same amount of money ($71.25)
So, the answer is 75 miles to be the same amount of money.
This student organized their solution by asking and answering questions. They used a table to
show their attempts at a solution but did notice the distance between the answers and used skip
Car Rental Quandary 6
counting to keep from having to try all of the numbers from 1-75. While this solution is not the
most efficient it is effective and for most students would be easier to understand. This solution
shows how an understanding of Algebra is not a prerequisite to solving this problem and that
Algebra would only help make it more efficient.
The teacher could remind students how easy it was to follow this solution and how important it is
to verbalize every step so that the reader is not lost.
A Difference of______?
It would take the driver to go 75 miles to make Rent-A-Gem and Super Saver Rentals to be the
same cost in same number of miles.
First, I realized that I had to find the difference in cost of the two renting services in three days.
So since the difference between the two rentals services for the first day was two dollars I
multiplied that by three to get a difference of six dollars in three days between the two services.
Next, I knew that the difference between the amounts of cents per mile was eight cents per mile
because Rent-A-Gem was 14 cents per mile and a daily cost of $20.25 and Super Saver Rentals
were 22 cents with the cost per day of $18.25. Then I knew that I had to multiply both 22 and 14
by the same number which would equal the number of miles to make Super Saver Rentals to be
six dollars greater than Rent-A-Gem. So I divided 600 by eight, which represents the difference
between the cents per mile ratio between the two services and got 75 which represents the miles
in which you would have to travel to make the two rental cars have the same mileage and the
same amount of days to cost the same price.
This was an interesting solution and would be helpful for students to see who may think that
there is only one method to solve any problem. This solution didn’t use any Algebra to solve but
instead found out the difference between the two companies up front. This would be a great way
to introduce equations because you could show students the equations and show how that when
you solve equations you find the difference between the two companies in a more organized
method.
This student didn’t consider other methods of solving or at least they didn’t verbalize it here it
would be a great place to have a conversation with students about the multiple ways to solve a
problem. There were parts of this solution that were unclear because the student assumed the
reader would understand. Therefore, I would discuss the importance of being clear and thorough
with solutions with students using this solution.
Car Rental Quandary 7
Equations minus the decimals
Student A:
75 When first looking at this problem i initially thought two make two columns for the two
different car rental agency's.
Rent A Gems Super Savers $20.25 a day and 14 cents per mile $18.25 a day and 22 cents per
mile Since he is trying to find the best deal for a three day time period i multiplied the day by 3
so 20.25 would be 60.75. And 18.25 would be 54.75. From here i took my information and made
a formula to solve.
60.75+.14x=54.75+.22x to make it easier i multiplied the whole equation by 100 to get rid of the
decimals.
6075+14x=5475+22x
-14x -14x
6075=5475+8x
-5475 -5475
600=8x
/8 /8
x=75
So the two company's become equal at 75 miles..and since hes clearly going to be driving more
than 75 miles he has to choose which one will be cheaper. Since Rent A Gems has less sence per
mile he will save money by choosing them.
Student B:
The total miles to drive to make both companies cost the same is 75 miles.
First I read the problem and pulled all of the important information out. Rent-A-Gem charges
$20.25 per day plus 14 cents per mile and Super Saver Rentals charges $18.25 per day plus 22
cents per mile. Another important piece of information is that the trip will happen over a course
of three days. I then made an equation equalling RentA-Gem and Super Saver Rentals where "x" represented the amount of miles needed to travel to
make both companies cost the same. This equation was:
20.25(3)+.14x=18.25(3)+.22x
An explanation for the numbers is as follows: 20.25 and 18.25 are the amount that both
companies charge per day for the rental of the vehicle; the 3 is for the amount of days that the
trip will occur over; and the .14 and .22 are the cost per mile. The x's represent the amount of
miles that the person will be driving.
I then solved the problem by multiplying the whole problem by 100 to remove the decimals. The
problem then looked as follows:
2025(3)+14x=1825(3)+22x
I then multiplied 2025 by 3 and 1825 by 3 to find out how much it would cost for all three days
to rent the car. The problem then looked as follows:
6075+14x=5475+22x
I then subtracted 14x from both sides and 5475 from both sides. The problem then looked like
this: 600=8x
Car Rental Quandary 8
I then divided both sides by 8 and got x=75. X represents the amount of miles to be driven to
make both companies cost the same.
Therefore the total miles to drive to make both companies cost the same is 75 miles.
After I found that the miles needed to drive to make both companies the same is 75 miles, I then
checked my answer. I plugged 75 in for x. The equation then looked like:
20.25(3)+0.14(75)=18.25(3)+0.22(75)
71.25=71.25
This proves that 75 is the right answer because when both sides of the equation are solved, they
equal 71.25.
The total miles to drive to make both companies cost the same is 75 miles.
Student C:
For the cost of renting to be the same for Rent-A-Gem and Super Saver Rentals you would have
to drive 75 miles.
We are told that someone wants to rent a car for a 3 day trip. They want to see how many miles
they would have to drive for the cost of the two rental companies to be the same. There is two
ways to do this.
1) You take the per day charge of Rent-A-Gem which is $20.25, add that to the per mile charge
which is 14 cents and times that buy a number but we don’t know it so we will make it X that
will equal the total miles per day. That will be equal to the per day charge of Super Saver Rentals
which is $18.25, add that to the per mile charge which is 22 cents and times that buy a number
but we don’t know it so we will make it X also. Then you have to solve for X but that’s not the
answer because that’s only for one day and the trip is for 3 days so you must multiple whatever
X is to 3. This way you must drive 25 miles a day for it to work. My assumption would be that if
you drive for 25 miles a day for 3 days the cost will be the same.
20.25 + .14x = 18.25 + .22x
[20.25 + .14x = 18.25 + .22x] * 10
2025 + 14x = 1825 + 22x
2025 = 1825 + 8x
200 = 8x
25 = x
25 * 3 = 75
2) This way is similar the only thing different is that at the beginging you times Rent-A-Gems
per day charge by 3 and Super Saver Rentals per day charge by 3. Also X is equal to the total
miles you have to drive. This does not mean that you have to drive it 25 miles a day like the
other example just that the total miles has to add up to be 75 miles.
3(20.25) + .14x = 3(18.25) + .22x
[60.75 + .14x = 54.75 + .22x] * 10
6075 + 14x = 5475 + 22x
6075 = 5475 + 8x
600 = 8x
75 = x
The three solutions above all show different ways of using equations to solve this problem. They
Car Rental Quandary 9
are all similar but slightly different in their method. It was interesting that all three of them chose
to remove the decimals by multiplying by 100. However, one of them said they multiplied by 10
but show they multiplied by 100. It is good to show students that it is important to show what
you do step by step so the reader can follow your thoughts and solution method.
Using an equation is more sophisticated than just guess and check and should be a natural
progression in an Algebra classroom. By the time students are exposed to this question in 8th or
9th grade they should be expected to either start with equations or be able to understand how
equations are a more efficient way to solve this problem.
Graphing Calculator
You would have to drive 75 miles for 'Rent-A-Gem' and
'Super Saver Rentals' to have the same cost.
As with any math problem, initially you must read the problem, and identity exactly what is
being asked. The questing being asked was: How many miles would I have to drive to make the
cost of renting a car from Rent-A-Gem the same cost of renting a car form the company Super
Saver Rentals?
At first glance, Super Saver Rentals (SSR) looks cheaper due to the lower initial cost, has a
higher mileage rate then Rent-A-Gem (RAG), therefore further analysis of the problem was
necessary.
Respectively, both SSR and RAG had a formula to calculate the mileage needed for the costs to
be equal.
RAG = $60.75* + .14x
SSR = $54.75* + .22x
*Note: Stated above in the problem, the charges per day were $20.25 (RAG) and $18.25(SSR).
Both of these costs were multiplied by a quantity of 3. They were multiplied because this is a 3day trip.
To find where these equations intersected, a TI-83 Plus graphing calculator was used. By
imputing these formulas in the Y= function, it will find the intersection oh where x, the mileage,
is equal.
To input this into the calculator:
[Y=], enter 60.75+.14, [X,T,θ,n], [⇓], enter 54.75+.22, [X,T,θ,n], [GRAPH].
Once graphed, you are able to trace the intersections of the too lines. This intersection at 75. This
means that at 75 miles, SSR and RAG both will be equal.
This solution is important because it shows the effective utilization of technology. Students do
need to be exposed to the technology but in this particular problem, especially if this is the first
time students have been exposed to a problem of this type, they should not jump to the calculator
first. This could be enrichment to the problem after it is solved in another way. This problem is
another example for students to see the multiple ways students can solve one problem. This
student clearly understood the question and solved it correctly while giving details about each
step. This solution was put last because students should be discouraged from jumping to the
calculator without first thinking through the question mathematically.
Car Rental Quandary 10
6. Discussion of Student Solutions
Students would need to explore how to make two equations equal and would explore the
multiple ways of solving a problem. The solutions above show a progression of sophistication of
solving with the first being simply adding until an answer is found. This is effective but not
efficient. The final solution showing not only a development of equations but the use of graphing
calculators to efficiently find the intersection of two equations.
Teachers could use the solutions to see the many ways students could tackle this question and
anticipate students’ struggles based on the misconceptions these students had. Teachers could
also show how learning multiple strategies is helpful to be more successful on all types of
problems.
However, it is important for teachers to stress that while the last method may be more efficient it
may be better to have an understanding of the other methods to be able to recognize when
careless errors have occurred. Students who jump to using calculators without understanding the
algorithms they are using, often just memorize buttons. The disadvantage is students do not think
through their solution to see if their answer is reasonable. This revising and evaluation step is
crucial in the problem solving process.
In addition to teaching students that there are often multiple ways to solve any problem; this
problem is also a makes solving equations with two variables realistic and demonstrates the daily
use of it. Students can relate to having to compare prices and will see the value in finding a
solution. Students will see with the solutions above that solutions near the end are using
equations and in general are easier to follow than the ones that just tried numbers until they
found a number that worked.
This problem would be a great introduction to all of the methods for solving systems of
equations and could be a springboard for the substitution method or the elimination method. In a
Pre-Algebra class this problem could lead to a more in depth look at the parts of the equation
such as the slope and the y-intercept.
7. Closing Comments
While this question seems like a question that should be saved until students fully understand
linear equations or even systems of equations, it really could be used to bridge the arithmeticalgebraic gap that often confuses students. In many cases students see arithmetic and algebra as
two separate entities but if this problem is used early in the year and then reintroduced later in
the year, students will be more likely to see the connection.
Car Rental Quandary 11