Assessing Problem Solving Using
Constructed-Response Problems
Steve Weimar
[email protected]
Annie Fetter
[email protected]
NCSM 2006
St. Louis
How Submissions are "Scored"
We look for good problem solving and strong mathematical communication when reading
submissions to our Problems of the Week. Solutions should include enough information to
help another student understand the steps taken to solve the problem and the decisions
made in the process.
Submissions are scored using the following categories:
Problem Solving
Interpretation: Interpret the problem correctly and attempt to solve all of the parts.
Strategy:
Pick a good strategy and apply it well, achieve success through skill
instead of luck.
Accuracy:
Get the calculations and details correct, including writing correct
statements and equations.
Communication
Completeness: Explain all the steps taken to solve the problem.
Clarity:
Explain the steps in such a way that a fellow student would understand,
and make an effort to check formatting, vocabulary, and spelling.
Reflection:
Check the answer, reflect on its reasonableness, summarize the
process, and connect it to prior knowledge and experience.
Submissions are scored using four levels of performance:
Novice:
Apprentice:
Practitioner:
Expert:
Just starting out
On the right track, but not quite there
Got it
Wow! Above expectations in some way
Scoring Grids
General versions of the grid are available for each of our
services, which can be accessed from the main page of each
service. The Pre-Algebra rubric is included on the next page of
this document.
Math Forum mentors, volunteers, and Teacher Members who
choose to mentor their own students’ solutions use problemspecific scoring guidelines to help them apply the rubric as
consistently as possible.
A grid like the one shown on the right is used by mentors
when viewing an individual submission. The “scores” are then
automatically pasted into the reply to the student (and can
easily be removed if desired) and are available to teachers
when managing student work.
pre-alg rubric goes here – see separate document
Cereal Consumption Comparison (Grade 9)
Mikey and Lily eat different kinds of cereal but their cereals come in
the same size box.
Every day Lily has the same big bowl of cereal and she eats four
times as much as Mikey.
On Monday Lily starts with a new box and Mikey still has half a box to finish.
Later in the week it looks like they have the same amount left in their boxes. How much
of each box has cereal in it?
From: Sasha
Date: 2006-02-06 12:04:23
1/6 box is full
I set the equations 1 - 4x and .5 - x since the amount left in the
boxes is the same after a few days. I solved the equation and the
answer was 1/6, therefore, 1/6 of the boxes are full when they each
have the same amount.
Actual written comment:
if I want to know when the boxes have the same amount of cereal, why
wouldn't I set the two equations representing how the cereal is being used
equal to each other to obtain an answer?
Date: 2006-02-08 21:37:43
Hi Sasha,
I think it does make sense to write equations that you set equal to each
other. You don't really show your work or explain the equations that you set
up. That would help, but I think you are doing very nice work here.
Perhaps most importantly, you probably want to think about what your result
of 1/6 represents. What is "x" in your equation? And how will this x help
you figure out how much cereal is left in each box?
I look forward to reading a revised solution.
-- Tyrone, The Math Forum
From: Sasha
Date: 2006-02-09 10:47:46
1/3 of both the boxes is left.
I set the equations 1 - 4x and .5 - x since the amount left in the
boxes is the same after a few days. I solved the equation and the
answer was x = 1/6. Since X = 1/6, I substituted the value into each
equation and I see that the quanty on each side of the equal symbols
is 1/3. My solution:
1 - 4 * X = .5 - X
1 - 4 * 1/6 = .5 - 1/6
1 - 4/6 = .5 - 1/6
1 - 2/3 = 3/6 - 1/6
and finally, 2/6 = 2/6 or 1/3 = 1/3
Thank you for your hints!
Date: 2006-02-09 12:31:26
Hi Sasha,
Substituting the result back in was a good way to find the answer. It also
works well to see your calculations written out so completely and nicely.
I am guessing that on the PSSA you can get a better score if you also say
what your expressions and equations mean. These are things which are obvious
to you, so you don't usually bother.
For example: where did 1-4x come from and what does it represent? Clearly
you know, but if you had written it out, and if you had written out what "x"
represents, then, not only would you get a better score, but you might have
realized your error without any hints. Right?
This is the one of the main reasons mathematicians write everything out.
When they do, they see things that they missed. And it makes easier for
other people to add ideas as well.
Thank you so much for doing this problem-solving with me. I look forward to
more.
-- Tyrone
Falling Leaves (Grade 6)
Last weekend I was watching leaves fall from a tree. As the wind
picked up and the leaves fell faster, I noticed an interesting
pattern:
The first minute, only one leaf fell. During each passing minute,
the number of leaves that fell was equal to the total on the ground
plus one.
If this pattern continued, how many leaves fell during the 12th minute? during the 24th
minute?
Extra: Write an expression to represent how many leaves would fall during the nth
minute.
From: Chad
Date: 2006-01-27 09:45:16
On the 12th minute it would have been 23 leaves that would of fell because
for 11 minutes it would had been 22 leaves so you would add one more. On the
24th minute it would have been 46 leaves because i justed doulbed from 12
and got 46
Date: 2006-01-28 10:07:25
Hi Chad,
Thanks for starting to work on this problem. I wonder if making
a chart might help?
The problem says, "The first minute, only one leaf fell. During
each passing minute, the number of leaves that fell was equal
to the total on the ground plus one."
A chart might look like:
minutes
--------1
2
3
fell
---1
on the ground
------------1
How many leaves will fall during the second minute? How many
leaves will be on the ground then?
Talk to you again soon.
Sophia
From: Chad
Date: 2006-01-29 14:51:29
On the 12th minute 2048 or 2 to the 11th power fell.
minute 8,388,608 or 2 to the 23rd power fell.
On the 24th
I made a chart to the 12th minute then used the pattern to get the
answer for the 24th minute.
Minutes
1
2
3
4
5
6
7
Fell
1
2
4
8
16
32
64
Ground
1
3
7
15
31
63
127
8
9
10
11
12
128
256
512
1024
2048
255
511
1023
2047
2095
Extra Credit
number of falling leaves per minunte = 2(to the power of number of
minutes-1)
Date: 2006-01-29 16:39:43
Hi Chad,
Thanks for working on the problem more! You've done a great job
to improve your solution.
There are a few minor things to consider this time. One, is that you
might like to use the "power" notation that we use when typing. You
can type 2^11 instead of "2 to the 11th power." In your chart you
might want to reconsider the number of leaves that you've listed for
the "Ground" at the 12th minute.
It would be interesting, too, if you could explain how you saw the
pattern. Is there anything that helped? (I'm just thinking that if
another student reads your solution, anything you can explain to
help them see what you see would be good.)
Talk to you again soon.
Sophia
What Number Am I? (Grade 3)
What number does the above picture stand for?
How do you know?
Challenge: Each letter stands for a number:
X=1
Y = 10
Z = 100
What number does this stand for?
ZZ YYYYY XXX
How do you know?
From: Lela
Date: 2006-03-17 11:59:50
To figure out the answer you had to add 200+50+3 which =253. Therefor
the answer is 253.
Date: 2006-03-17 12:27:48
Hi, Lela.
Great job finding the answer!
You are right on.
Could you explain how you knew it was 200, 50, and 3?
classmate who can't solve the problem?
What would you tell a
Adding that will make your explanation stronger.
-Nadine
From: Lela
Date: 2006-03-22 11:10:28
I would tell him/her that one flat = 100 one long = 10 and
cube =1.
To figure out the answer I had add z+z+y+y+y+y+y+x+x+x. Since one z
=100 one y = 10 one x = 1,and there're two z's which = 200 five y's
which = 50 three x's = 3. Therefor 200 + 50 + 3 = 253.
Date: 2006-03-22 12:34:50
Hi, Lela. That's super! Thanks for working on it some more.
have added would really help another student understand.
-Nadine
What you
Congruent Rectangles (HS Geometry and Middle School)
These seven congruent rectangles form a larger rectangle.
Question: If the area of the larger rectangle is 756 units2, what is its perimeter?
James
The perimeter is 166 units.
I knew that area was length times width. So I knew that 27 times 28
gave me 756. So i added 28 plus 28 plus 27 plus 27 to get my answer
of 166.
Jessula
The perimeter is 27.5 units.
You must take the square root of 756 units squared and that is the
perimeter.
Dre
The perimeter of the rectangle is 111.2 units^2.
First I tried to solve it algebraicly. I used xy=756 but there were
too many answers so I tried finding the area of one of the smaller
squares. I did this by dividing the area of the big square by seven
because there are seven little triangles in the big one. 756/7 = 108
but that got me no where.
Then I saw the answer. It takes 3 of the rectangles layed across the
long way to go from one side to the other the long way but it takes 4
rectangles stacked up the tall way to get from one side to the other
the long way.
Hmmm I thought I now have 2 equations and 2 unknowns. They are
3/4x = y and xy = 756.
All I need to do now is solve.
I plugged in 3/4x into y in the seccond
equation. I now have x(3/4x) = 756. I then multiplied x*3/4x and
got 3/4x^2 = 756. Then I divided 756 by 3/4 and got x^2 = 1008. Next I
took the square root of 1008 and got x = aprrox. 31.8. To solve for
y you must multiply by 3/4 because 3/4x = y. Therefore y = approx 23.8.
There are 2 x's and 2 y's for both sides of the rectangle. When you add
those up you get 31.8*2+23.8*2 = 111.2. The perimeter of the rectangle
is 111.2 units^2.
Matteo
3024 units
If 756 units2 is the area, and area is calculated by multiplying
width times length, then the width and height most both be 756
units. Then, if perimeter is calculated by 2L+2W the perimeter must
be 3024 units.
Lissa
My answer is 230.
I divided 756 by 7, since there are 7 equal triangles. I got 108. I
added 108 and 7 and got 115. I doubled that and got 230, since the
formula for perimeter is (2)l+w.
Chung
The perimeter of the larger rectangle is 57 units.
Let the dimensions of the small rectangles be a, b(length a, width
b), then the area of the large rectangle is:
4b(a+b)=756
From the graph, we can see that 4b=3a, so
b=(3/4)a, then we have
3a(a+(3/4)a)=756
(7/4)a^2=252
a^2=144
a=12 ( -12 rejected)
b=(3/4)a=(3/4)*12=9
The perimeter of the larger rectangle is
4b+a+b
=a+5b
=12+5*9
=12+45
=57
John’s Analysis of the Students’ Solutions
(submitted for an assignment to analyze the above solutions)
1. For James and Jessica's equation to be set up properly, both the large and the smaller
rectangle would have to be squares.
2. Dre put a lot of good thinking into her problem and made real good progress. In fact,
this is how I first figured this problem out. However, I knew that this thinking was
making an improper assumption of the large rectangle sides. Even the picture shows
that four widths of a length are being compared to three length of a same distance. The
only proper way to solve this problem is to first calculate the widths and lengths of the
smaller rectangles.
3. Matteo doesn't understand the basic of substituting a problem into a formula. The 2L +
2W = 756.
4. Lisa was on the right track by properly calculating the area of the smaller triangles.
However, at this point in solving this problem, she needed to stand back and rethink
her problem. Sometimes students aren't used to problems having multi-steps.
5. Chung has a lot of fancy algebra expression but they are incorrectly applied. He needs
to slow down a bit and make sure that his algebraic equations accurately state the
problem. He failed to see that there are several steps in solving this problem.
Erika
The perimeter would be 114 units.
I am trying to find the solution to the perimeter of the figure shown.
Since I knew the rectangle was divided into 7 congruent rectangles, I
took the total of the area, 756 un^2, and divided it by 7 and got that
each smaller rectangle had an area of 108 un^2. After I did that, I
saw that the 4 smaller sides had to equal the 3 bigger sides becasue
they were opposite sides and they had to be equal becasue it was a
rectangle. I also saw that the smaller number and the larger number
multiplied together had to equal 108 un^2(which would be the area).
Then I made the equation 4S=3B and that S*B=108 (S being the smaller
side and B being the bigger side). I then used guess and check and
plugged in various numbers. I started by plugging in 3 for S and 6 in
for B and got 12=18, which was wrong. Then 6 in for S and 9 in for B
and got 24=27, which was also worng. Then I plugged in 9 for S and 12
for B and gor 36=36, which was correct. So I found that 9 was equal to
S and 12 was equal to B. Then, looking back at the rectangle, there
are 5 larger sides(B) on the perimeter and 6 smaller sides(S) on the
perimeter. So (5*12)+(6*9) would equal the perimeter, which is 114
units.
Marissa
The perimeter of the large triangle is 114 units.
The large rectangle is made of seven smaller, congruent rectangles.
Since the area of the large rectangle is 756 units squared, the area
of each small rectangle is 756/7 (108 units squared).
If we assign (x) to be the length of the shorter side of each small
rectangle and (y) to be the length of each longer side, we see that
4x=3y, meaning the ratio of short side to long side is 3:4.
Using the ratio 3:4 we can make a table to find the possible lengths
of each side and the corresponding area:
3:4
2
3
4
6:8=48 units squared
9:12=108 units squared
12:16=192 units squared
Since the area of each small rectangle is 108 units squared, the
sides must be 9 and 12 units, meaning x=9 and y=12. The perimeter is
P=6x+5y, since there are 6 short sides and 5 long sides around the
the large rectangle's perimeter. Plugging x and y in, the perimeter
is 6(9)+5(12), which equals 114 units.
Lucy
The perimeter of the rectangle is 114 unites
Explanation:
Since the small rectangles are all congruent, that means they have the
same area. We can divide the area of the large rectangle by 7 to find the
area of one small rectangle.
Suppose x is length of the small rectangle, and y is the width of the
small rectangle. Then 3x is length of the large rectangle, x+y is the
width of the large rectangle, and xy is the area of the small rectangle.
xy=756/7 --- here we have the area of one of the small rectangle.
xy=108
y=108/x
3x(x+y)=756 --- Length(3x) times width(x+y), which gives us area.
x(x+y)=252
x(x+108/x)=252 --- Bring y, which equals to 108/x, to the equation.
x^2+108=252
x^2=144
x=12
y=108/12
y=9
3x is the length of the large rectangle
so 3x=36
x+y is the width of the large rectangle
so x+y=9+12=21
Primeter of the rectangle is 2 (length + width). So 2(3x+x+y) is the
primeter of the large rectangle
2(3x+x+y)=2*57=114 unites
Carson
The Perimeter of the large rectangle is 114 units.
I thought that it might be easier to work with the little rectangles
rather than trying to tackle the large one. I found the area of the
little ones easily by dividing 756 (the large rectangle) by 7 (the
number of little ones). I did this, because the question said all the
small rectangles are congruent. A congruent shape is is a shape that
has the exact same: dimensions, number of sides, and angles, as
another shape. The question said all of the small rectangles were
congruent, so they'd each contain the same surface measurement. That
is why I decided to divide the large rectangle's area into 7.
756/7= 108
So each little rectangle had an area of 108 units. I then noticed in
the picture it shows that, on the horizantal middle line, it takes 4
little sides to make 3 big sides on the little rectangles. Thus, the
small rectangle's long side multiplied by 3 must be the same as the
small rectangle's short side multiplied by 4. (Lonside x 3)= (Small
Side x 4)I would use this information to try to find a factors of 108
(because all the rectangles are congruent).
With this information I began to guess and check. 27 x 4 equals 108 I
thought. However 27(units)x3(long sides) =81, and 4(units)x4 (short
sides)= 16. 81 is greater 16 so that can't work!
I went through this
factors of 108 when
9 (units) x 4(short
must be the correct
prosses several times and was trying to find
I found 12 and 9. 12(units)x 3(long sides)= 36
sides) = 36. They are both the same, and so that
answer.
Now I know that the long sides are 12 and the short sides are 9 on
the rectangle. I go back now to the large rectangle. The top has 4 of
the little triangle's short sides, so I multiply 9 x 4= 36. The sides
have one of the little triangle's short sides and little triangle's
long sides. 12+ 9= 21. Thus the dimentions are 21 x 36. To check this
I do the area (length x width). 21x 36= 756! Thats what the question
said. Now i find the perimeter.
Perimeter=(length x 2) + (width x 2)
P= (36x 2) + (21x2)
P= 114 units.
Dad's Cookies (Grades 3-5)
Dad bakes some cookies. He eats one hot out of the oven and leaves the rest on the
counter to cool. He goes outside to read.
Dave comes into the kitchen and finds the cookies. Since he is hungry, he eats half a
dozen of them.
Then Kate wanders by, feeling rather hungry as well. She eats half as many as Dave
did.
Jim and Eileen walk through next, and each of them eats one third of the remaining
cookies.
Hollis comes into the kitchen and eats half of the cookies that are left on the counter.
Last of all, Mom eats just one cookie.
Dad comes back inside, ready to pig out. "Hey!" he exclaims. "There is only one
cookie left!"
How many cookies did Dad bake in all?
Emma
Dad started with twenty-two cookies.
The only unknown variable was what Jim and Eileen ate. I
backtracked from what Mom and Hollis ate to find the possible
solutions. I figured out that Jim and Eileen each ate four, after
some trial and error.So;
Dad ate one,
Dave ate six
Kate ate three,
Jim and Eileen each ate four,
Hollis ate two
and Mom of course, ate one.
Added with the remaining cookie these equal twenty-two
cookies.Yum!
Pre-Service Teacher 1 Reply
Greetings, Emma!
Good problem-solving. The correct answer is 22. You are also right that
except for the cookies Jim and Eileen ate, it's possible to use the steps
of the problem to figure out all of the variables without trial and error.
Even though you solved the problem correctly, I have a couple of suggestions
that would make your answer even more complete. Can you tell me about the
trial and error that you did in order to find out that Jim and Eileen each
ate four cookies? Also, could you show me how you then found out the number
of cookies that everyone else ate? That way it would be easier for
me to see exactly how you solved the problem.
I look forward to getting your revisions!
-Pre-Service Teacher 1
Kristin
There where 22 cookies that dad baked in all.
What I did was I first read through all of the clues. In the
first three clues it said that dad ate one, Dave ate six, and Kate
ate three. But in the fourth and fifth clue it said that Jim and
Eileen ate one third of the remaining cookies, and Hollis had half of
what Jim and Eileen had. So now it is time to work the problem out
backwards. The last two clues said that mom ate one and when dad came
back there was only one left. So if 1plus 1 equals 2 and Hollis ate
half of what was left on the counter than that means that Hollis ate
2.
If Hollis ate 2 and he ate half of what Jim and Eileen did than
that means that Jim and Eileen ate 4 cookies each which means that
Jim and Eileen ate 8 cookies together. Then you add up all your clues
together which is 1+6+3+8+2+1+1=22. That means that dad baked 22
cookies in all.
Pre-Service Teacher 2’s Reply
Dear Kristin,
Thanks for submitting your answer to the mathforum. Your solution
taught me something I had never noticed before, that the number of cookies
Hollis ate had to be half of the number of cookies that Jim and Eileen each
ate. No one else I've mentored used that fact, and so I would really like
it if you could explain how you discovered it.
Also, even though I know how to do the problem, we really want submissions
to the mathforum to be explained so completely that even someone who was
really confused could understand it. How would you explain how many cookies
Hollis ate to that imaginary confused person?
I look forward to seeing your revisions.
Thanks,
Pre-Service Teacher 2
Hannah’s Work
From: Hannah
Date: Nov 14 2005 10:10PM
measure of <FAB = 90-x
If ABCD is a rhombus, then AD is parallel to BC, and when two
parallel lines are intersected by a transversal, the alternate
interior angles are equal in measure. Therefore, the measure of
<DBC is equal to x, then the measure of <BDA must also be x since
they are alternate interior angles.
From: Hannah
Date: Nov 14 2005 10:10PM
[The student looked at answer 1]
From: Hannah
Date: Nov 14 2005 10:22PM
measure of <FAB = 2x-90
If ABCD is a rhombus, then AD is parallel to BC, and when two
parallel lines are intersected by a transversal, the alternate
interior angles are equal in measure. Therefore, the measure of
<DBC is equal to x, then the measure of <BDA must also be x since
they are alternate interior angles. The measure of <BDA + measure
of <DBC= measure of <BAD. And since it is given that <FAD is a
right angle and right angles have 90 degrees, then to find the
measure of FAB you subtract 90 from the measure of <BAD
Date: Dec 6 2005 11:13PM
Hi Hannah,
Thanks for submitting your solution. Sorry for my late reply.
You were on the right track for most of your solution. I have a couple of
suggestions for you.
Try to break up your solution instead of writing in a paragraph. This is one
case where a paragraph is not needed :D.
>measure of <FAB = 2x-90
>
>If ABCD is a rhombus, then AD is parallel to BC, and when two
>parallel lines are intersected by a transversal, the alternate
>interior angles are equal in measure. Therefore, the measure of
><DBC is equal to x, then the measure of <BDA must also be x since
>they are alternate interior angles.
Beautiful to this point.
> The measure of <BDA + measure of <DBC= measure of <BAD.
The statement above is incorrect. Remember that the 3 angles of a triangle
must add to 180 degrees. Also, because ABCD is a rhombus then DA and BA are
equal which means triangle ABD is isosceles and angle <BDA and <DBA are
equal.
There is not a theorem/postulate that says that
The measure of <BDA + measure of <DBC= measure of <BAD.
If <BDA and
<DBC are both equal to anything other than
>And since it is given that <FAD is a
>right angle and right angles have 90 degrees, then to find the
>measure of FAB you subtract 90 from the measure of <BAD
This part was also correct. Just the middle part needed to be corrected.
I look forward to future submissions.
Problem Solving
Communication
Interpretation: Practitioner
Strategy:
Apprentice
Accuracy:
Apprentice
Completeness:
Clarity:
Reflection:
Practitioner
Practitioner
Apprentice
(for an explanation of scores see: http://mathforum.org/pow/scoring.html)
-- Karen, for the Geometry Problem of the Week
From: Hannah
Date: Feb 27 2006 11:50PM
The tree is 202.44 years old.
First, I had to find the circumference of the tree. There are
12 inches in one foot, so if their hands are 2 feet apart, than they
are 24 inches apart. To find the circumference of the tree, I added
up Tom and Susan's armspand and the 24 inch gap.
72.5 + 62.5 + 24 = 159 inches
Now, I found the diameter of the tree. I knew that the equation
for the circumference of a circle was c= pi x d. So I used my
circumference to solve for d (the diameter).
159/3.14= 50.61 inches.
If the diameter is 50.61 inches, then, it might seem like you
would just multiply that number by 8. But since the rings
themselves are circles, that would be counting the same ring twice.
So instead, you must find the radius, or half the diameter and
divide that by 1/8 inch to find the number of rings.
50.61/2= 25.3 inches
Since the rings are 1/8 inch apart, that means for every 1
inch, there are 8 rings. So to find the number of rings, I multiply
the radius by 8.
25.3 x 8= 202.44 rings
This means that the tree is about 202.44 years old.
I'm pretty sure I did this right because this seems like a
reasonable number for a tree of this size. I used the right
equation for the circumference and I know that there are 12 inches
in one foot, so I know I found the circumference correctly.
From: Hannah
Date: Feb 27 2006 11:50PM
[The student looked at answer 1]
Date: Mar 15 2006 4:44PM
Hi Hannah,
Very Nicely Done!!!
[student’s solution deleted]
Summary:
Problem Solving
Interpretation: Practitioner
Strategy:
Practitioner
Accuracy:
Practitioner
Communication
Completeness:
Clarity:
Reflection:
Practitioner
Practitioner
Practitioner
(for an explanation of scores see: http://mathforum.org/pow/scoring.html)
-- Karen, for the Geometry Problem of the Week