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.
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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.
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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.
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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.
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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
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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