Less than seven years worth.
Let's do the math.
It shouldn't take more than six years to cover all of elementary math to the algebra level.
Of course, as we know, many students fail to learn much of what's taught during these early years.
Here's a list of topics that the typical low performer fails to learn during these six years in the typical math class:
The typical student is weak in basic math facts (like, 13–7 and 6 x 8). The student has a very poor understanding of fractions and what they are. For these students, fractions are “less than one.” The students have no scheme of how fractions relate to division and to whole numbers. Students are weak on multiplying fractions, weaker on adding or subtracting fractions that have the same denominator, and incapable of adding or subtracting fractions with unlike denominators (4/9 + 1/6).
These students do not understand how to find the perimeter or the area of figures. They can’t figure out the circumference or area of circles. They are often able to work division problems that divide by a single-digit number, but can’t work problems that divide by a two-digit number. The students know virtually nothing about the coordinate system, and nothing about analyzing lines on the coordinate system (their slope; where they intersect the Y axis), or constructing lines from equations.
These topics need to be remediated. In addition, the CAHSEE contains a few algebra-like topics not typically taught in elementary math like:
Many items on the test require students to solve sophisticated word problems. The tests have problems that require knowledge of algebraic principles (3r + 7 = r – 2), problems that assume knowledge of scientific notation (What is the scientific notation for .002591?), and problems that assume knowledge about geometry (What’s the volume of concrete needed to form the illustrated stairway?).
So, the typical low-performing seventh grader requires extensive remediation in arithmetic and needs to learn a few post-arithmetic topics. The question is: how long does all this take to teach? In many schools, the answer is, unfortunately: forever. Starting late this spring, it shouldn't take more than one school year with the new DI Algebra course.
It's been three years in the making and went through three complete field tests and subsequent revisions. According to Engelmann:
A large majority of students who completed only the first 2/3 of the program passed the math exit exam for California.
I'm not sure how the CAHSEE compares to other state exams and NAEP, but I do know that most of them have very little post arithmetic content. If they are anything like the CAHSEE, they only contain about seven years of math content and that's pretty sad.
The same criticism can be said of the NAEP.
ReplyDeleteI took the CAHSEE awhile back and passed with flying colors.
ReplyDeleteheck
Can't remember if I'd finished Saxon Algebra 1 at that point. I may have.
I definitely hadn't done any of the Algebra 2 book.
Two complaints:
ReplyDelete1) Six years to get up to Algebra is reasonable (because this is what happens in Singapore and Russia), but this is not what it taught in California. An accelerated math track (I think) gets to Algebra in the 8th grade. This is one year faster than current/typical and one year slower than Russia/Singapore. So ... I'd say it would take 7-8 years to get to Algebra in California. Yes, with an ideal curriculum (AND K-6 TEACHERS WHO UNDERSTOOD MATH WELL), we'd do it in 6. But we don't have that curriculum and we don't have those teachers across the board.
So ... 7-8 years. This is still fairly low for a 12th grade exit exam ...
Consider an analogy: If we used the same criteria for passing the medical boards, doctors wouldn't need to learn *anything* in medical school to become doctors. This would probably be a bad thing :-)
2) You don't need Algebra to pass. I think you only need to answer 55% of the questions correctly to pass, and the test contains less than 45% algebra. My guess is that solid knowledge of up through 7th grade (as currently taught) would be enough.
-Mark Roulo
Mark, that was my point. With an adequate curriculum and competent teachers, it is possible for low performers to reach the algebra level in six years. Today we have at least one adequate curriculum, but often the teachers are not competent to teach math in areas with high percentages of low perfomers. I have a post up at d-ed reckoning where Engelmann points out that they had to put the upper elementary teachers through the upper levels of the math curriculum as students before they were capable of teaching it due to their low knowledge of math. These were certified teachers.
ReplyDeleteWell, when I took the TX math exit exam (TAAS, at the time) in 97, I had my 6th grade sister try the practice exam and she passed easily. Granted, she's a gifted math brain (4.0 math major), but she hadn't been taught anything beyond 6th grade math at that point. IIRC, the test was largely arithmetic with a smattering of algebra and a smidge of geometry.
ReplyDeleteIn defense of the test, though, it was given in the spring of our sophomore year. Regular-track students would only be halfway through geometry at that point, anyway. Odd to take an "exit" exam when you've still got 2.5 years of schooling left, but I think it was so students who failed had a LOT of opportunities (8, I think) to retake it.
That summary of weaknesses Ken posted is right on target (The typical student is weak in...) I see it every day. I can't figure out why things that are so simple like distinguishing between perimeter and area seem to be an insurmountable problem for so many pupils.
ReplyDeleteRegarding algebra, the Illinois ISAT for seventh grade http://www.isbe.net/assessment/pdfs/2007_ISAT_Sample_Book_Gr_7_m.pdf has algebra problems like this one: (I suspect most are doing trial and error for this one since algebra isn't taught at all or not coherently)
Ben sold some adult tickets and some student tickets for a basketball game.
• Each adult ticket cost $5.
• Each student ticket cost $3.
Ben collected $180 for the 50 tickets he sold.
1. How many adult tickets did he sell?
2. How many student tickets did he sell?
Show all your work. Explain in words how you found your answer. Tell why you took
the steps you did to solve the problem.
This is a classic example of an "algebra-without-algebra" problem.
ReplyDeleteHere's the solution that they're looking for:
If all the tickets were student tickets, he would have collected $150. This is $30 less than what he collected. Since a student ticket is $2 less than an adult ticket, this means that 30/2 = 15 of the tickets were adult tickets. Therefore, 50 - 15 = 35 of the tickets are student tickets.