How the NCEE
Limits
High School Math
Includes
the
New York City Modifications
Here you will find a compact version
of the NCEE "Work Sample & Commentary" sets (WS&C sets)
for
high school math. We also cover the New York City
modifications.
In brief, the New York City authors chose to omit the NCEE set that
contains
the most genuine math, and they added four WS&C sets. Three
of
these NYC sets deal with elementary school math. The other, Phone
Service, offered an opportunity to demonstrate important 8th grade
math. Unfortunately, that opportunity was missed.
Direct Access Links to Each WS&C
Set
 How Much Gold
 Miles of Words
 Cubes
 Shopping Carts
 Grazing Area
[Not in the NYC version]
 Dicing Cheese
 Galileo's Theater
 Great Deals
[Only in the NYC version]
 Gas Tank
[Only in the NYC version]
 Phone Service
[Only in the NYC version]
 Two Fair Dice
[Only in the NYC version]
The Work Sample & Commentary Sets for
NCEE High School Math
 How Much
Gold
Can You
Carry Out?
 Miles of
Words
 Task: Determine if a
person can
speak 40,000 words during a 200 mile train trip.
 Method: Allowing for
train
stops, the
student estimated the speed of the train as 35 miles per hour and then
computed the time for the trip as 200 ÷
35.
The student multiplied twice by 60 to convert to seconds.
Finally,
the student divided 40,000 by the total number of seconds to get the
rate
of approximately 1.94 words per second. The student concluded that the
rate was too fast. A calculator was used for all calculations.
 Comments:
 This is 6th grade math at best.
 Another student might make
different assumptions
and conclude that the rate was possible. Both solutions would be
honored in NCEE math. The NCEE never ranks the quality of a
student's work. All methods are seen as equally valuable.
 Cubes
 Task: Given a
"n by
n by n"
large cube, composed of "n^{3}
small cubes", find formulas for the
number of "visible small cubes" and the number of "hidden small
cubes".
(3 sides are seen in a 3D view)
 Method: The student
developed
detailed
drawings of cubes, for n = 1 through 5. The student then numbered
all visible component small cubes and constructed a table, recording
the
total number of small cubes, hidden small cubes, and visible small
cubes
for n = 1 through 5. After studying the table, the student
noticed
that the total number of small cubes for one size large cube was equal
to the number of hidden small cubes for the next larger size large
cube.
This led to the "discovery" of the formula (n  1)^{3 }
for the number of hidden small cubes. Then, after studying
the three visible sides of the big cube and taking care not to count a
visible small cube more than once, the student observed that the number
of visible small cubes could be found by adding n^{2}, n^{2}
 n, and n^{2} 2n + 1 to get the sum 3n^{2}  3n +
1.
An NCEE author note observes that the number of visible small cubes can
also be represented as n^{3}  (n  1)^{3},
so
= n^{3}  (n  1)^{3 }= 3n^{2}  3n + 1.
This observation is followed by the NCEE statement:"these
expressions make sense if and only if n is a whole number."
 Comments:
 This offers a good
demonstration
of constructivist
inductive reasoning busywork. It's all there: drawings,
numbering,
counting, listing, and pattern recognition. We are expected to
believe
that the teacher didn't provide any hints.
 Finding the sum 3n^{2}

3n + 1 is
the most advanced algebra demonstrated anywhere in NCEE math.
 The concluding "make sense"
statement is a
bit puzzling. For any real number x, we can use the
distributive
law to write:
(x  1)^{3} =
(x 
1)(x  1)^{2} = (x  1)(x^{2}  2x + 1)
= x^{3 } 2x^{2} + x  x^{2 }+
2x  1.
= x^{3 } 3x^{2 }+ 3x  1.
So x^{3}  (x  1)^{3}
= 3x^{2}  3x + 1
 The NCEE says "it
is not clear how the student arrived at the cubic function given in the
response." They are referring to the (n 1)^{3 }formula
for
the number of hidden "small cubes." This NCEE puzzlement is a bit
surprising, since the student followed generally accepted
constructivist
reasoning methods. The student studied the table and "noticed
that the number of hidden cubes was the same number of total cubes in
the
next size cube." Since the
table
supplied 4 illustrations of this fact and found no counterexamples,
this
normally qualifies as an acceptable constructivist proof.
 Shopping
Carts
 Task: "Create
a rule that can be used to predict the length of storage space needed
given
the number of carts."
If
S denotes the length of storage space needed and N denotes the number
of
carts, also find a rule that expresses N in terms of S.
 Method: The
student used
a scale model to measure 96 cm as the length of the first cart and 28.8
cm as the amount that each new cart added to the stack extends out
beyond
the other carts. Thus 28.8 (N  1) is the total length
contributed
by all carts after the first cart. So the first desired formula
is
S = 96 + 28.8 (N  1) = 67.2 + 28.8 N. The second formula is
found
by solving for N in terms of S to get N = (S  67.2) ÷ 28.8.
 Comment:
 This is important 6th grade
math.
 The NCEE correctly classified
this
task under
"Functions and Algebra" because it involves linear
functions.
But the student wrote about "rules" and never used the term
"function."
The student didn't recognize these as linear functions. The
(extremely
important) term "function" is never used by students in NCEE math.
 There's no description of the
physical storage
space. The student and the NCEE assumed that the "length of the
storage
space" is equal to the total length of the nested stack.
But
a "real world" storage space, say alongside a wall, may have
welldefined
beginning and end points, possibly associated with side
walls.
Recognizing such an additional constraint could lead to a legitimate
high
school math problem.
 Grazing Area
[Not in the New York City version]
 Task: A region is
comprised of two
triangles and sectors of three different circles. Given certain
angles
and dimensions, find the total area.
 Method:
 Triangle 1: Side
1 =
10', Side
2 = 40', Included Angle = 135^{o}.
 The student used Side 2 as
the
base and found
the altitude by extending Side 2 to create a 45^{o} 45^{o}
90^{o} isosceles right triangle, RT1, with Side 1 = 10' as the
hypotenuse. The student applied the Pythagorean Theorem for RT1 to
determine
the altitude for Triangle 1 (as one of the equal two legs for RT1), and
then used the formula A = 1/2 ( base x altitude) to find the area
of Triangle 1.
 The student also found the
length of Side
3 [needed below to find the area of Sector 2]. This could have
been
carried out as another application of the Pythagorean Theorem for RT2,
which is the right triangle which shares the 90^{o} angle of
RT1,
one side of RT1, and Side 3 as hypotenuse. But, since angle
information
was needed for circle Sector 1, the student used a graphing
calculator.
The Tan^{1 }function was used to find the acute angle a_{1}
of Triangle 1 which is opposite the side shared by RT1 and RT2, and
then
the Sin function was used to find the length of Side 3.
 Triangle 2: Side
1 =
10', Side
2 = 20', Included Angle = 135^{o}.
 Method same as used for
Triangle
1.
 Sector 1: Radius
given as 50'.
The problem is to find the sector angle.
 The student first computed
the
other two acute
angles, a_{2 }and b_{2}, for Triangles 1 and
2:
[a_{2} = (180  (135 + a_{1}) b_{2} =
(180
 (135 + b_{1}]. This led to the angle for Sector 1 as c_{1}
= 360^{o}  (a_{2 }+ b_{2}). The student
then calculated the area of Sector 1 as (c_{1} ÷ 360) x
(pi x 50^{2}).
 Sector 2: The area was
easily calculated
after first subtracting the length of Side 3 from 50 to get the radius,
and by then recognizing that the desired sector angle equals 180^{o}
 (90^{o} + a_{1}), where a_{1} denotes
the
acute angle described above for Triangle 1.
 Sector 3: Method similar
to
the method
used for Sector 2.
 Comments:
 This is real math. The
student recalls
and applies several 7th grade math ideas:
 Finding the altitude of a
triangle.
 Finding the area of a
triangle.
 Knowing properties of
isosceles
triangles.
 Knowing properties of right
triangles.
 Applying the Pythagorean
Theorem.
 Knowing that the sum of the
angles of any
triangle is 180^{o}.
 Finding the area of a circle.
 Finding the area of a sector
of
a circle.
 Introducing the Tan^{1 }and
Sin functions
puts this example beyond the 7th grade, but not too far, since a
graphing
calculator was used. Use of these functions (called "trig
properties"
by the student) could have been avoided by slightly modifying the
information
provided.
 The NCEE comments: "The
student work is an excerpt from a longterm project that was completed
over a fourweek period. During this time, one class per week was
allocated to completion of the project. The students worked in
groups
of three or four . . ."
 This should be the work of
one
student and
shouldn't take more than 15 minutes. The 135^{o} angle
and
simple numbers make the calculations easy. Thestudent has a
graphing
calculator, there's repetition in the methods used, and it's
basically
7th grade math.
 Dicing
Cheese
 Task: Given two
conditions
involving 4 unknowns, solve for two unknowns in terms of the other two.
A
rectangular block of cheese of volume V is cut into identical small
cubes.
One layer of the small cubes exactly fills a rectangular pan of
area
A. Given that the side of a small cube is of length
L.
Find L in terms of V and A. Find N, the number of small cubes, in
terms of V and A.
 Method: V = A x L,
or L
= V ÷
A. N = V ÷ L^{3} , therefore N = V ÷
(V ÷ A)^{3 } = A^{3} ÷ V^{2}.
 Comments:
 This is 8th grade math.
The
NCEE authors
consider it difficult because the student was required "to
deal with the abstractness of this 'numberless' formulation."
 Although the details are not
explained,
the method requires operations with rational expressions. This is
a bit surprising since operations with rational numbers are not
demonstrated
anywhere in NCEE math.
 Galileo's
Theater
 Task: Design a
circular theater
with "the greatest seating capacity," given 7 constraints.
 Method: Students
designed a circular
theater by repetitively using the formula for the circumference of a
circle
and calculator arithmetic. According to the NCEE, "the
students had a week to complete the task, and then a week to revise
based
on teacher feedback. They worked in groups . . . ."
 Comment: The 7
constraints
give the
impression of a wellposed problem, with one correct
solution.
But the constraints aren't precise, and the resulting "solution"
involves many subjective decisions. The overall impression is
repetitive
busywork.
 Great
Deals [Only
in the New York City version]
 Task: "A
clothing store offers a 50% discount at the end of each week that an
item
remains unsold. Patrick . . . says " I'll wait two weeks, have
100%
off, and get it for free!" Why
is he wrong?
 Method: The student
considered
the example of a $120 item and noted that the price would be $60 at the
end of the first week, $30 at the end of the second week, and $15 at
the
end of the third week.
 Comment: The NCEE appears
satisfied
that the high school student reasoned by considering one example.
A 6th grade student should be able to argue more abstractly, minimally
recognizing that the price at the end of the second week is found by
multiplying
the original price by ^{1}/_{4}. At the
high
school level, the student should quickly recognize that P_{n} ,
the price at the end of the n^{th }week, is given by the
formula: P_{n} = (^{1}/_{2})^{n}
x P, where P denotes the original price.
 Gas Tank [Only
in the New York City version]
 Task: "Jed
bought a generator that will run for 2 hours on a liter of gas. The gas
tank on the generator is a rectangular prism with dimensions 20 cm by
15
cm by 10 cm. If Jed fills the tank with gas, how long will the
generator
run?"
 Method:
The student
multiplied 20 x 15 x 10 and then divided by 1,000 to get 3
liters.
The student then multiplied by 2 to get the answer of 6 hours.
 Comment:
This is
basic fifth grade math.
 Phone Service
[Only
in the New York City version]
 Task:
Compare
two local phone service plans. The monthly base fees are $4
for
ESP and $7 for PQR. The rate per local call is 10 cents for ESP
and
6 cents for PQR.
 Subtasks:
 Which plan is
best
for 80 calls
a month?
 Which plan is
best
for 50 calls
a month?
 "FOR EACH
TELEPHONE COMPANY,
make an InOut table (a table of values: number of calls and the
monthly
cost.)"
 "FOR EACH
TELEPHONE PLAN,
write an equation or statement or rule that expresses the relationship
between the number of calls made and the total cost."
 "On the
same
set of axes,
make a graph for the ESP Telephone Company and for the PQR Telephone
Company."
 "Based on
the
responses
above, write a letter to the consumer group with your recommendations
about
which plan a family should choose. Explain how you arrived at your
recommendation."
 Method:
 The student
calculated the 80
call case. [$12 for ESP vs. $11.80 for PQR]
 The student
calculated the 50
call case. [$9 for ESP vs. $10 for PQR]
 The student
calculated ESP and
PQR cases for all multiples of 10 between 10 and 150 and used this data
to build the table.
 The student
wrote:
 ESP "Cost
= Amount of calls per month X .10 + 4.00 or x = (A * .10) + 4.00"
 PQR "Cost
= Amount of calls per month X .06 + 4.00 or x = (A * .06) + 4.00" [Note that the
4.00
should be 7.00]
 The student
graphed the two
straight lines.
 The student
wrote
the letter,
saying "at 75 calls per
month
for both services you have to pay 10.50."
[Note that this should be 11.50, not 10.50]
 Comments:
 The NCEE noted
the
student errors,
but then stated that these mistakes didn't detract from the quality of
the student's work. Such errors, with similar accompanying
comments,
are frequently found in NCEE math. The NCEE is working hard to
get
us to believe that this is "genuine student work." But they also
have let us know that precision and accuracy aren't highly valued in
NCEE
math.
 Subtasks 4 and
5
offer the opportunity
to demonstrate important eighth grade math: finding the
coordinates
of the point of intersection of two straight lines. But the
student
provided no algebraic evidence for the answer given in the
letter.
Apparently the student "eyeballed" the graph.
 There's too
much
busywork associated
with this WS&C set. At the high school level the student
should
quickly:
 Write ESP
Cost =
4.00 + .10N,
where N = the number of calls.
 Write PQR
Cost =
7.00 + .06N,
where N = the number of calls.
 Use
properties
of equality to
find the point of intersection of these two straight lines:
4.00 + 10N =
7.00 + .06N
.04N = 3.00
N = 75 and
C = 11.50
This type of
reasoning isn't
demonstrated anywhere in NCEE math, and it's not demonstrated anywhere
in the NYC version of NCEE math.
 Consider the two phone service
plan example
on pages 223  224 of the revised NCTM Standards, (Principles and
Standards
for School Mathematics [PSSM]). See Understand
patterns,
relations, and functions in the section Grades
6  8: Algebra. Note the need
for
a table and the use of a graphing calculator to find the two
graphs.
One equation, y = 20.00 + .10x, is eventually mentioned, but not the
other,
y = .45x. Students used the graphing calculator to
determine
that the graphs intersect at about x = 57. It's not expected that
the students will solve 20.00 + .10x = .45x to find the exact
answer
of ^{400}/_{7}.
 Two Fair Dice [Only
in the New York City version]
 Task: "Mohammed
and Noreen are studying probability. Their math teacher had a pair of
fair
but unusual dice. These dice each had six sides but one had a zero (0)
on a face instead of a four (4) and the other had a zero on a face
instead
of a one (1).
a)
Find
the probability of rolling a sum of eight with these dice.
b) Which sum(s) would be most likely with this pair of dice?"
 Method:
The students
developed a table showing the sums for all 36 combinations of the two
dice.
They observed four cases for the sum 8 and five cases for the
sums
5 and 6. They wrote the answer ^{4}/_{36}
for
part a. They didn't simplify. They wrote the answer "5 or
6"
for part b.
 Comment:
This is
similar to the Two Dice Sum example
found
in NCEE elementary school math. NYC high school busywork is
familiar
elementary school busywork.
Next?
Copyright
20022011 William
G. Quirk,
Ph.D.