MCDONALDS TRAY SAMPLE 7TH GRADE PROBLEM IS 9TH GRADE ALGEBRA Why are we raising the bar higher than even 99th percentile kids that went to MIT were not expected to meet? No 4th or 7th grader was ever expected to solve a problem like this. No adult who needs to solve a problem like this will use anything but algebra. Again, SPI Terry Bergeson (who knows basically nothing about math) trusts her "experts" (who evidently know almost nothing about math) to say that such questions are perfectly reasonable to expect of all 7th graders. ------------------------------------------------------ Feb 5, 1999. I finally picked up the McDonalds tray liners about the new state reform assessment. Here is the sample 7th grade problem. It is nearly identical, (actually slightly harder because it has a 1/2 x to it) to a "bicycle" count the wheels problem presented as a 4th grade sample at my elementary school. In a survey of pre-1990 math textbooks that I have, only a 9th grade algebra textbook gives a solution, or even asks for this sort of problem, which is "solving an equation of the form ax + bx = c". Remember that something like only 33% f the adult population has ever taken algebra, and no more than 60% of 9th graders have passed an algebra proficiency test even when it is required of all students. Evidently some of the newest NCTM inspired textbooks, which have been given terrible reviews from organizations such as Mathematically Correct, expect even 4th graders to be able to solve such problems by using guess and check using tables, or with no standardized method at all. Is it any wonder that 70% of 7th graders failed to meet the standard when this is typical? Why am I the only guy in the United States to figure out that this problem is not appropriate for 4th or 7th graders. It's not even an appropriate standard for all high school students when no more than half can be expected to master this by 12th grade based on historic levels of performance, or international standards. ----------------------------------------------------------------- Sample: The questions on his [McDonalds] trayliner are taken from the 7th grade Washigton Assessment of Student Learning state test. Lisa put some fruit in large bowl for her friends. The bowl had twice as many apples as oranges, and half as many pears as oranges. Altogether, there were 14 pieces of fruit in the bowl. * How many apples did Lisa put in the bowl? * How many oranges? * How many pears? Explain or show how you found each answer Answer: 4 orange, 8 apple, 2 pear. They do not tell how to get the answer. Evidently any creative solution would be given full credit, whether by algebra or guess and check table. It is not evident how much credit would be given for a creative solution that was wrong. -------------------------------------------------------------- Guess and check solution - evidently promoted as G4 or G4 "problem solving math" in mid 1990s NTCM standard text book. This does not appear in any K12 or G7 G8 middle school textbooks that I have. desired sum = 14 orange = x apple = 2x pear = x/2 sum 1 1 2 1/2 not integer 2 2 4 1 7 3 3 6 3/2 not integer 4 4 8 2 14 This only works with very small values of X. This would not be the correct way to solve this problem in any "real" job that requires a college degree, and jobs that do not require a degree generally do not require solving this sort of problem. --------------------------------------------------------------- Textbook solution - my pre-1990 textbooks G3-G6 elementary and G7+G8 middle school do not tell how to solve or set up an equation of this type. Only a grade 9+ algebra textbook explains exactly how to solve this problem, after telling about negative numbers, commutative property, and basically the entire algebra book except for polynomials. orange = x apple = 2x pear = x/2 set up equation x + 2x + x/2 = 14 combine common elements (1 + 2 + 1/2) x = 14 simplify (3 1/2 ) x = 14 convert to fraction (6/2 + 1/2) x = 7/2 x = 14 divide by same x = 14 * 2/7 = 14/7 * 2 = 2 * 2 = 4 substitute formula for each unknown orange x = 4 apple 2*x = 8 pear x/2 = 2 --------------------------------------------------------- Elementary Algebra Part I by Denholm Dolciani, Cunningham 1970 Houghton Mifflin: p. 321 section 11-3 Equations of Type ax + bx = c Note: problems of this sort are presented only after mastering every other concept except for adding and subtracting polynomials and geometry! You have to know distributive, commuative, negative numbers, and every other algebra concept first. Typically only 1/3 - 1/2 of the population takes Algebra I, only 1/2 pass even when the course is required of all students. Each of the equations below is of the ax + bx = c type 3k + 7k = 48 1/3 t + 1/2 t = -6 3n/2 + n/4 = 1.7 You will probably recall that we can solve an equation of this type witout much difficulty if we first combine _similar_ terms. In the equation 5/3 m + 1/2 m = -14, the terms 5/3m and 1/3 m are similar, since each contains the variable m. So we use the distributive property as follows: 5/3 m + 1/3 m = (5/3 + 1/3) m = 6/3m = 2m The solution of 5/3m + 1/3 m = -14 can now be found by applying the multiplicative property of equality. Example 1. 5/3 m + 1/3 m = -14 Given equation 2m = -14 Combine similar terms 1/2 * 2 * m = -14 * 1/2 Multiply each member by 1/2 1 * m = -14 * 1/ 2 Multiplicatin property of reciprocals m = -7 Substitution principle