Saturday, March 17, 2012

Analysis of Difficult SAT Math Questions

The Grand Challenge Equations: San Diego Super...The Grand Challenge Equations: San Diego Supercomputer Center (Photo credit: dullhunk)
by Miriam Attia, Academic Advisor, Parliament Tutors

What is it about the final problems in the SAT math section? Why is it so consistently difficult for students to prepare for them adequately? How is it we always know them when we see them, yet only the best study guides can mimic them effectively? 

What makes them such a consistently neat combination of challenging and elegant - problems that look mind-bogglingly difficult at first glance, unsolvable even, and then after some work reveal themselves to be solvable in half a minute?
Consider, for example, this ingenious problem from the end of one SAT math section.
On the number line above, the tick marks are equally spaced. Which of the lettered points represents y?
A. A
B. B
C. C
D. D
E. E.
Two variables appear on a number line, but the number line has no numbers, and it seems we have to find both the value of one variable in relation to the other and that value's location on the number line. Most students would look at this question and either panic or skip it (or both) and that's not even taking into account how mentally drained they're feeling by the time they're near the end of a math section.
If they have the time to do so, most students will solve this problem slowly and methodically, as follows:
First, we suppose each tick mark is separated by 1 from the ticks next to it, such that if we add 2 to x, it takes us 2 ticks to the right, where we find (x + y). By the same reasoning, if we add 2 to (x + y), it takes us 2 more ticks to the right, where we find (x + y)/2. Thus, we can write two equations:
  1. x + 2 = (x + y).
  1. (x + y) + 2 = (x + y)/2.
The first equation shows us that y = 2. The second equation, if we're like most students, makes us pause and seriously second-guess our method - how can we add 2 to something to get half of what we started with?? - until we realize we're talking about negative numbers. We breathe a sigh of relief and rewrite the second equation, substituting 2 for y:
(x + 2) + 2 = (x + 2)/2
Simplifying, we get x + 4 = (x + 2)/2
Multiplying both sides by 2 gives 2x + 8 = x + 2
Finally, we combine like terms and discover that x = -6. So y must be 8 ticks to the right of x, because 2 - (-6) = 8. E is the correct answer, but it took us far, far too long to find it.
What some unusual students realize about this problem, if they can muster the mental agility to zoom out for a moment, is that the answer is staring us in the face. We want to know where y is. We already know where x is. We also know where (x + y)/2 is. The average of x and y, which is also the midpoint of x and y, is defined as (x + y)/2. The letter x is 4 ticks to the left of (x + y)/2, so y must be the same distance, 4 ticks, in the other direction. We count 4 ticks to the right of (x + y)/2 and find point E.
What's fascinating about this question, to me, is the combination of its utter simplicity - that number line, slightly modified, might as well be in an arithmetic textbook in the section that defines the term “average” - and the masterful degree to which that simplicity is disguised, misdirecting not only the students but easily 99% of teachers and tutors as well.
I believe this combination of a simple rule and clever disguise is what gives the most difficult SAT math questions their instant recognizability and what makes them so difficult to mimic. To test my hypothesis, let's see if we can use that method to create a new SAT-esque math question right now.
The SAT likes to ask students about circles, so I'll start with a fact about circles that I find interesting: Suppose six identical circles are arranged in a ring, each one tangent to two others. It happens that there's just enough room in the center for a seventh circle of the same size, tangent to the other six.
We can connect the centers of the circles to the centers of their neighbors and get six equilateral triangles forming a regular hexagon.
That's the kind of interesting insight we can use as a foundation for a difficult SAT math problem. So, let's pose a challenging question related to that fact. What would such a question look like? There are many possibilities. Here's one.

In the ring of circles below, each circle has a radius of 1 and is tangent to the two on either side of it.
What is the area of the white space in the center of the ring?
A. π 
B. 6√3 – 3π
C. 6√3 – 2π 
D. 24√3 – 8π 
E. 24√3 – 4π 

Just as the number line question we saw earlier doesn't require knowledge of averages but can be solved more easily using that insight, this question doesn't require us to think of a hexagon, but it can be solved more easily if we do. Here's how I would solve it.
  1. Draw a regular hexagon whose center is at the center of the ring and whose side length is 2. The center of each circle is now also a corner of the hexagon.
  1. To find the total area of the central white space, find the white space in one sixth of the hexagon and then multiply by six. Choosing any one of the six equilateral triangles to focus on (I'll choose the top triangle), find the triangle's area and then subtract the area of the blue sections.
  1. A triangle's area is the product of its base and half its height, so draw a line showing the triangle's height...and realize that the height line has divided the equilateral triangle into two right triangles with angle patterns 30-60-90. That means the side ratios must be 1-2-√3. The short leg does in fact have a length of 1, and the hypotenuse is 2, so the height is √3. Multiply half the base (1) by the height (√3) to find the triangle's area, √3.
  2. Subtract the area of the blue sectors from the area of the triangle. Each sector is a sixth of a circle, and each circle's area is π, so the sectors' areas together must be π/3. Thus, √3-π/3 is the area of blank space in each triangle.
  3. Remembering that what happens in this triangle happens a total of six times in the figure, multiply that difference by 6 to get 6√3 – 2π., which is answer C.
If you're preparing to take the SAT, remember that the test writers appreciate interesting features of arithmetic, algebra, and geometry. They use what makes math elegant to build their most challenging problems. As you study math, keep an eye out for features like these. When you find one, keep it in a mental file of your own. It will give you a valuable edge on test day, and (perhaps more significantly) will enrich your understanding and appreciation of mathematics throughout your life.
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