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

1. How Much Gold
2. Miles of Words
3. Cubes
4. Shopping Carts
5. Grazing Area        [Not in the NYC version]
6. Dicing Cheese
7. Galileo's Theater
8. Great Deals         [Only in the NYC version]
9. Gas Tank            [Only in the NYC version]
10. Phone Service       [Only in the NYC version]
11. Two Fair Dice       [Only in the NYC version]

The Work Sample & Commentary Sets for NCEE High School Math

1. How Much Gold Can You Carry Out?
• Task: Move gold, using one trip per minute.  Move one pound the first trip, and then continue, always moving half of the amount moved the previous trip.  Find the amounts for 15, 20, and 60 minutes.  Convert to dollars at rate of $350 per ounce.
• Method:
• A calculator was used to compute data for the first 26 trips.  The computation details are not explained. Here are 4 rows in the student's 26-row table:
 
Trip  Oz.of gold          Total Oz.   Total $ Earned
  5    1
 15   .0009765625          31.999023   11,199.652
 25   .00000095367832      31.999999   11,200
 26   .0000047683716 (sic) 32          11,200


The student wrote "I will never be able to get more than 2 lbs. of gold or $11,200 worth."
 

• Although the student never mentioned powers of 2, the NCEE contributed this comment:
 
"finding the quantity of gold that can be carried out amounts to finding this sum: 1 + 1/2 + 1/4 + 1/8 + 1/16 +....   If this is approached as a practical problem, all that is required is to compute these terms until they get too small to be practical, then add them up.  'Too small to be practical' might be interpreted as 'too small to be represented on a calculator'.  On a calculator with an 8-digit display, this happens after about 25 terms of this series. That is,  1 ÷  2²⁴ comes out as 0.0000001 on the calculator, while 1 ÷  2²⁵ comes out as 0.0000000.  When all the terms up to 1 ÷  2²⁴   are added up, the sum comes to 1.9999999, but when all the terms up to 1 ÷  2²⁵  are added, the sum comes to 2.0000000."
• Comments:
1. The student used a calculator to repeatedly divide by 2, add trip ounces to total ounces, and multiple each total by 350.   There's also the constructivist busywork related to building the 26-row table.   But that table was necessary to "keep track" of the previously computed total ounce sums.
2. The final 32, without a decimal point, and student's concluding statement both combine to give the impression that the student believes that  that 32 ounces will be retrieved by the end of the 26th trip.  The NCEE's comments reinforces this impression.   But exactly 32 ounces will never be achieved, regardless of the number of trips.  This important fact is never mentioned by anyone.
• Why is this important?   The fundamental idea of limit is involved.  This is the key underlying concept that justifies the methods of calculus.  But limits aren't mentioned anywhere in NCEE math.   More generally, calculus isn't mentioned in NCEE math.  Constructivist math educators don't see the relevance to everyday life.
3. The idea of applying general knowledge to solve a special case problem is missing here and throughout NCEE math.   Neither the student nor the NCEE mentions the general formula for the sum of a finite geometric series:
• a₁ + a₁r + a₁r² + a₁r³ + . . . + a₁r^(n-1) =  a₁[(1 - rⁿ) ÷ (1 - r)]
• For this case, where  a₁  = 1 and  r = ¹/₂  , the sum is 2 [1 - (¹/₂)ⁿ].
• Note that this formula involves operations with fractions.   But adding, subtracting, multiplying or dividing fractions isn't  demonstrated anywhere in NCEE math.
2. 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:
1. This is 6th grade math at best.
2. 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.
3. Cubes
• Task:  Given a "n by n by n" large cube, composed of  "n³ small cubes", find formulas for the number of "visible small cubes" and the number of "hidden small cubes".  (3 sides are seen in a 3-D 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  n², n²  - n, and n² -2n + 1 to get the sum 3n² - 3n + 1.  An NCEE author note observes that the number of visible small cubes can also be represented as  n³ - (n - 1)³, so  = n³ - (n - 1)³ = 3n² - 3n + 1.  This observation is followed by the NCEE statement:"these expressions make sense if and only if n is a whole number."
• Comments:
1. 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.
2. Finding the sum 3n² - 3n + 1 is the most advanced algebra demonstrated anywhere in NCEE math.
3. 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)(x² - 2x + 1)
                                         =  x³ - 2x² + x  -  x² + 2x - 1.
                                         = x³ - 3x² + 3x  - 1.
 So x³ - (x - 1)³  = 3x² - 3x + 1
4. 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.
4. 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:
1. This is important 6th grade math.
2. 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.
3. 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 well-defined beginning and end points, possibly associated with side walls.   Recognizing such an additional constraint could lead to a legitimate high school math problem.
5. 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:
1. 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 a₁ 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.
2. Triangle 2:  Side 1 = 10', Side 2 = 20', Included Angle = 135^o.
• Method same as used for Triangle 1.
3. Sector 1:  Radius given as 50'.  The problem is to find the sector angle.
• The student first computed the other two acute angles, a₂ and b₂, for Triangles 1 and 2:   [a₂ = (180 - (135 + a₁)  b₂ = (180 - (135 + b₁].  This led to the angle for Sector 1 as c₁ = 360^o - (a₂ + b₂).  The student then calculated the area of Sector 1 as (c₁ ÷ 360) x (pi x 50²).
4. 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  + a₁), where a₁ denotes the acute angle described above for Triangle 1.
• Sector 3: Method similar to the method used for Sector 2.
• Comments:
1. 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.
2. 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.
3. The NCEE comments: "The student work is an excerpt from a long-term project that was completed over a four-week 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.
6. 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 ÷ L³ , therefore N = V ÷ (V ÷ A)³  = A³ ÷ V².
• Comments:
1. 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."
2. 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.
7. 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 well-posed 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.
8. 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.
9. 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.
10. 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:
1. Which plan is best for 80 calls a month?
2. Which plan is best for 50 calls a month?
3. "FOR EACH TELEPHONE COMPANY, make an In-Out table (a table of values: number of calls and the monthly cost.)"
4. "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."
5. "On the same set of axes, make a graph for the ESP Telephone Company and for the PQR Telephone Company."
6. "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:
1. The student calculated the 80 call case.  [$12 for ESP vs. $11.80 for PQR]
2. The student calculated the 50 call case.  [$9 for ESP vs. $10 for PQR]
3. The student calculated ESP and PQR cases for all multiples of 10 between 10 and 150 and used this data to build the table.
4. 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]
5. The student graphed the two straight lines.
6. 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:
1. 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.
2. 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.
3. 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.
4. 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.
11.  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.

Chapter 1:  A Summary View of NCEE Math
Chapter 2:  How the NCEE Limits Elementary School Math
Chapter 3:  How the NCEE Limits Middle School Math

Understanding NCTM Math "Standards" and NCTM "Math Reform"  [Home Page]