Mathematics Illuminated is a set of 13 free lessons on topics such as prime numbers, infinity, and game theory.  Each lesson includes video, text, and interactive flash activities.  The intended audience is “adult learners and high school math teachers,” but we think the lessons will be beneficial to high school students as well.  Enjoy!

Here’s  a question from the May 2006 SAT exam:

19. A container in the shape of a right circular cylinder has an inside base radius of 4 inches and an inside height of 9 inches.  This cylinder is completely filled with water. All of the water is then poured into a second right circular cylinder with a larger inside base radius of 9 inches. What must be the minimum inside height, in inches, of the second cylinder.

(A) 4/3

(B) 16/9

(C) 9/4

(D) 4

(E) 6

Don’t worry about the terminology — it’s not important that you know what a right circular cylinder is. It is, however, important to remember that any time you’re asked about a cylinder, you’re most likely being asked about volume. It’s not necessary to even memorize the volume formula since it’s given to you on the reference information portion of the section (right below the directions).  However, we at Stylus suggest you memorize the formulas anyway because it will save you the time of flipping back and forth.  Cylinder volume is fairly easy to memorize if you already know the formula for the area of a circle.  Think of a cylinder as being a circle with the added dimension of height.  Thus, the formula for volume of a cylinder is the same as that of a circle, but with height added: V = πr²h.

Since this question appears toward the end of the section, we can bet on needing to do at least two separate calculations.  One way the makers of the SAT love to make geometry equations more complex is by asking you to solve one equation and plug that solution into a second (or even a third) equation to solve for a different variable — to plug and plug again.  This equation is no different.  Eventually, we’ll need to solve for the height of the second cylinder, but before we can do that, we need to solve for the volume of the original cylinder.

Since the first cylinder is completely full, we only need to plug the given numbers into the formula to find the original volume of water: V = π(4)²(9) = π(16)(9).  Don’t bother solving this equation yet — doing so won’t give us the answer we’re looking for, and any time you avoid using a calculator to do messy calculations, you save yourself time.

Onto the second equation: If the water is poured into the second cylinder, it needs to have at least the volume of the original cylinder.  So we need to use the volume formula once again, this time using the volume from equation 1 and solving for the height of cylinder 2.

V = πr²h

π(16)(9) = π(9)²h

Divide both sides by 9π to get

16 = 9h (see why I told you not to bother with the calculator?) and then divide by 9 again to get

16/9 = h, choice B.

President Obama announced on Wednesday a partnership between federal agencies and public universities to train thousands more mathematics and science teachers each year, part of the administration’s effort to make American students more competitive globally in science, technology, engineering, and math.

Leaders of 121 public universities have pledged to increase the total number of science and math teachers they prepare every year to 10,000 by 2015, up from the 7,500 teachers who graduate annually now.

Forty-one institutions, including California’s two university systems and the University of Maryland system, said they would double the number of science and math teachers they trained each year by 2015.

According to Libby Smith in “Universities Pledge to Train More Math and Science Teachers by 2015,” a recent article in the Chronicle for Higher Education, this plan is part of the Obama administration’s “Educate to Innovate” campaign to improve math and science education.  Whether training more teachers will improve the quality of math and science education in the U.S. remains to be seen.

Many urban school districts, including New York City, failed to show significant gains on the most recent round of federal math tests.  This has led critics to question whether the progress shown on state tests in districts like New York City is real, or just a result of easier tests.  According to John Hechinger of the Wall Street Journal,

“New York City failed to show significant progress on the federal test since 2007, even though its state tests showed major math gains. Joel Klein, New York City’s schools chancellor, noted that since 2003, the city had shown dramatic gains on the federal test. ‘We’re outperforming the rest of New York state and the nation,’ in terms of educational gains, Mr. Klein said. But, he added, the flat 2009 scores suggested New York state’s tests ‘have to be harder’ and its curriculum standards more rigorous.”

Continue reading more about this topic here.

Here is a question that appeared at the end of the grid-ins section of the May 2006 SAT examination.

18. In the integer 3,589 the digits are all different and increase from left to right. How many integers between 4,000 and 5,000 have digits that are all different and increase from left to right?

When confronted with this problem, many students will decide they don’t know how to approach it mathematically and, as a result, skip the problem altogether. However, there is really no reason not to attempt this problem; because it’s a grid in, you won’t be penalized for guessing the wrong answer. Although knowledge of combinations might help you solve this problem, it can also be solved by simply listing the possibilities in an organized manner.

Since the integers must be between 4,000 and 5,000, the first digit of the number must be 4. Because the integer must increase from left to right, the remaining digits must be either 5, 6, 7, 8, or 9. List possibilities one by one, starting with the smallest possibility, and moving to the largest.

4567 4568 4569
4578 4579
4589
4678 4679
4689
4789

As you can see, there are 10 integers altogether.

It is very important to be organized and systematic when listing possibilities for a problem like this. Otherwise, you are apt to miss one or two. Begin by listing all numbers that start with 456, then that start with 457, then 458, then move on to all that start with 46, and so on. Work slowly and meticulously; rushing will likewise lead to errors.

Although this problem can be solved mathematically, it would involve more advanced permutations than high school students are typically expected to know. The test-makers do not intend for you to solve this mathematically, but for all interested parties, the mathematical solution would be (5!)/(3!)(5-3!).

A City Journal article discusses the effect of the movement toward a “student-centered” math classroom, which began in the 1970s and 1980s when the pedagogical approach was being retooled in other subjects as well.  This new way of teaching mathematics focused on an  approach with little regard for sequence or arithmetic skills.  Instead, students were asked to find their own tools to master problem-solving, rather than following a model provided by the teacher.  In her article, Sandra Stotsky suggests that this shift in educational values is responsible for low mathematics performance in the United States.

The statistics on U.S. math performance are grim. American eighth-graders ranked 25th out of 30 countries in mathematics achievement on the 2006 Programme for International Student Assessment (PISA), which claims to assess application of the mathematical knowledge and skills needed in adult life through problem-solving test items. We do better on the Trends in International Mathematics and Science Study (TIMSS), whose test items are related to the content of school mathematics curricula. (Differences in participating countries aren’t significant.) But according to Mark Schneider, a former commissioner of education statistics at the Department of Education, the United States lags behind too many countries in “overall mathematics performance and in the performance of our best students.” And achievement gaps between different student groups within the United States, Schneider says, are “about the same size or even bigger than the gap between the United States and the top-performing countries in TIMSS.”

As part of his education-reform plan, President Obama wants to “make math and science education a top priority” and ensure that children have access to strong math and science curricula “at all grade levels.” But the president’s worthy aims won’t be reached so long as assessment experts, technology salesmen, and math educators—the professors, usually with education degrees, who teach prospective teachers of math from K–12—dominate the development of the content of school curricula and determine the pedagogy used, into which they’ve brought theories lacking any evidence of success and that emphasize political and social ends, not mastery of mathematics.

Continue reading here.

Here is a tricky SAT question from the October 2006 exam.

In an election, 2.8 million votes were cast and each vote was for Candidate I or Candidate II.  Candidate I received 28,000 more votes than Candidate II.  What percent of the 2.8 million votes were cast for Candidate I?

(A) 50.05%

(B) 50.1%

(C) 50.5%

(D) 51%

(E) 55%

Begin by asking yourself a) what do I already know? and b) what do I need to know?  In this case, we already know the number of total votes cast = 2.8 million, and we know that Candidate I received 28,000 more votes than Candidate II.  We need to know the percent of votes cast for Candidate I.  The percent of votes cast for Candidate I is the number of votes cast for him/her divided by the total number of votes cast.  We already know the total number of votes cast, so we only need to know the number of votes cast for Candidate I.

Can we write an equation to figure out the number of votes cast for Candidate I?  Let’s call that number x.

Let x = the number of votes cast for Candidate I

x – 28,000 = the number of votes cast for Candidate II (since candidate I received 28,000 more).

x + (x – 28,000) = 2,800,000

2x – 28,000 = 2,800,000

2x = 2,828,000

x = 1,414,000

Now we only need to figure out what percent of the total 1,414,000 represents.  Remember that percent is part/whole, so the percent of votes Candidate I receives is 1,414,000/2,800,000, which is 0.505, or 50.5%

Don’t make the mistake of oversimplifying this problem (as many students did) and figuring out that 28,000 votes is 1% of the total.  This is true, but it does not mean that Candidate I received 51% of the votes.  If you are performing only one step to solve a problem near the end of the section, you are definitely oversimplifying it.

According to a recent Wall Street Journal article, “U.S. Math  Scores Hit a Wall,” there has been little to no improvement in math scores achieved by fourth and eighth grade students, despite the educational changes imposed by the No Child Left Behind plan.

Fewer than four of 10 fourth- and eighth-graders are proficient in mathematics, according to a highly regarded federal test given in early 2009, adding to recent evidence that the U.S. drive to become more economically competitive by overhauling public education may be falling short.

The National Assessment of Educational Progress — often called the “nation’s report card” — found fourth-graders had made no learning gains since the last time the NAEP math test was given, in 2007. Previously, fourth-graders had made scoring gains on every NAEP math test given since 1990.

Significant scoring gaps between white students and their Hispanic and African-American peers also haven’t changed much in recent years, the test results showed.

The responses to the results have been mixed: some educators caution against reading too much into the scores, while others see a need for more drastic change.  The article continues here.