As we’ve already mentioned, the calculator is a very important part of the Math IIC test. You need to have the right kind of calculator, be familiar with its operations, and above all, know how to use it intelligently.

There are four types of questions on the test: calculator-friendly, calculator-neutral, calculator-unfriendly, and calculator-useless. According to the ETS, about 60 percent of the test falls under the calculator-neutral and -friendly categories. That is, calculators are useful or necessary on 30 out of the 50 questions on the SAT II Math IIC. The other 20 questions are calculator-unfriendly and -useless. The trick is to be able to identify the different types of questions on the test. Here’s a breakdown of each of the four types, with examples. If you’re not certain about the math discussed in the examples, don’t worry. We cover all of these topics in this book.

A calculator is extremely helpful and often necessary to solve calculator-friendly questions. Problems demanding exact values for exponents, logarithms, or trigonometric functions will most likely need a calculator. Computations that you wouldn’t be able to do easily in your head are prime suspects for a calculator. Here’s an example:

This is a simple function question in which you are asked to evaluate f(x) at the value 3.4. As you will learn in the Functions chapter, all you have to do to solve this problem is plug in 3.4 for the variable x and carry out the operations in the function. But unless you know the square root and square of 3.4 off the top of your head, which most test-takers probably wouldn’t (and shouldn’t), then this problem is extremely difficult to answer without a calculator.

But with a calculator, all you need to do is take the square root of 3.4, subtract twice the square of 3.4, and then add 5. You get answer choice C, –16.28.

You have two different choices when faced with a calculator-neutral question. A calculator is useful for these types of problems, but it’s probably just as quick and easy to work the problem out by hand.

When you see the variable x as a power, you should think logarithms. A logarithm is the power to which you must raise a given number to equal another number, so in this case, we need to find the exponent x, such that 8^x = 43 23. From the definition of logarithms, we know that given an equation of the form a^x = b, log_a b = x. So you could type in log₈ (4³ 2³) on your trusty calculator and find that x = 3.

Or you could recognize that 2 and 4 are both factors of 8 and, thinking a step further, that 2³ = 8 and 4³ = 64 = 8². Put together, 4³ 2³ = 8² 8 = 8³. We come to the same answer that x = 3, and that B is the right answer.

These two processes take about the same amount of time, so choosing one over the other is more a matter of personal preference than one of strategy. If you feel quite comfortable with your calculator, then you might not want to risk the possibility of making a mental math mistake and should choose the first method. But if you’re more prone to error when working with a calculator, then you should choose the second method.

It is possible to answer calculator-unfriendly questions by using a calculator. But while it’s possible, it isn’t a good idea. These types of problems often have built-in shortcuts—if you know and understand the principle being tested, you can bypass potentially tedious computation with a few simple calculations. Here’s a problem that you could solve much more quickly and effectively without the use of a calculator:

If you didn’t take a moment to think about this problem, you might just rush into it wielding your calculator, calculating the cosine and sine functions, squaring them each and then adding them together, etc. But if you take a closer look, you’ll see that cos²(3 � 63°) + sin²(3 63°) is a trigonometric identity. More specifically, it is a Pythagorean identity: sin²q + cos²q = 1 for any angle q. So, the expression {cos²(3 63°) + sin²(3 63°)}⁴/2 simplifies down to ¹⁴/ ₂ = ¹ /₂ = .5. Answer choice B is correct.

Even if you wanted to, you wouldn’t be able to use your calculator on calculator-useless problems. For the most part, problems involving algebraic manipulation or problems lacking actual numerical values would fall under this category. You should easily be able to identify problems that can’t be solved with a calculator. Quite often, the answers for these questions will be variables rather than numbers. Take a look at the following example:

This question tests you on an algebraic topic—that is, how to find the product of two polynomials—and requires knowledge of algebraic principles rather than calculator acumen. You’re asked to manipulate variables, not produce a specific value. A calculator would be of no use here.

To solve this problem, you would have to notice that the two polynomials are in the format of a difference of two squares: (a + b)(a – b) = a² – b². In our case, a = x + y and b = 1. As a result, (x + y – 1)(x + y + 1) = (x + y)² – 1. Answer choice B is correct.

The fact that the test contains all four of these question types means that you shouldn’t get trigger-happy with your calculator. Just because you’ve got an awesome shiny hammer doesn’t mean you should try to use it to pound in thumbtacks. Using your calculator to try to answer every question on the test would be just as unhelpful.

Instead of reaching instinctively for your calculator, you should come up with a problem-solving plan for each question. Take a brief look at each question so that you understand what it’s asking you to do, and then decide whether you should use a calculator to solve the problem at all. That brief instant of time invested in making such decisions will save you a great deal of time later on. For example, what if you came upon the question:

A trigger-happy calculator user might immediately plug in 3 for x. But the student who takes a moment to think about the problem will probably see that the calculation would be much simpler if the function were simplified first. To start, factor 11 out of the denominator:

Then, factor the numerator to its simplest form:

The (x – 4) cancels out, and the function becomes f(x) = ^{(x – 1)}/ ₁₁. At this point you could shift to the calculator and calculate f(x) = ^{(3 – 1)}/₁₁ = ²/₁₁ = .182, which is answer D. If you were very comfortable with math, however, you would see that you don’t even have to work out this final calculation. ²/ ₁₁ can’t work out to any answer other than D, since you know that ² /₁₁ isn’t a negative number, won’t be equal to zero, and also won’t be greater than 1.