
Making Your Calculator Work for You
As we’ve already mentioned, the calculator is a very important
part of the Math IC 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: those that
are calculatorfriendly, calculatorneutral, calculatorunfriendly,
and calculatoruseless. According to ETS, about 60 percent of the
test falls under the calculatorneutral and friendly categories.
That is, calculators are useful or necessary on 30 of the 50 questions
on SAT II Math IC. The other 20 questions are calculatorunfriendly
and useless. The trick is to be able to identify the different
types of questions when presented with them 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 these topics in this book.
CalculatorFriendly Questions
A calculator is extremely helpful and often necessary
to solve calculatorfriendly questions. Problems demanding exact
values for exponents, logarithms, or trigonometric functions will
most likely need a calculator. Computations that you can’t do easily
in your head are prime candidates. 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
testtakers wouldn’t), 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.
CalculatorNeutral Questions
You have two choices when faced with a calculatorneutral
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 of 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} =
4^{3} 2^{3}.
From the definition of logarithms, we know that if given an equation
of the form a^{x} =
b, then log_{a} b = x.
So you could type in log_{8} (4^{3} 2^{3}) 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 2^{3} =
8 and 4^{3} =
64 = 8^{2}. Put together, 4^{3} 2^{3} =
8^{2 } 8 = 8^{3}.
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.
CalculatorUnfriendly Questions
While it’s possible to answer calculatorunfriendly questions
using a calculator, it isn’t a good idea. These types of problems
often have builtin 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 take a closer look: cos^{2}(3 63°)
+ sin^{2}(3 63°) is
a trigonometric identity. More specifically, it’s a Pythagorean
identity: sin^{2}q +
cos^{2}q =
1 for any angle q. So, the expression {cos^{2}(3 63°)
+ sin^{2}(3 63°)}
^{4}/_{2} simplifies
to 1^{4 }/_{2} =
^{1}/_{2} =
.5. B is correct.
CalculatorUseless Questions
Even if you wanted to, you wouldn’t be able to use your
calculator on calculatoruseless problems. For the most part, problems
involving algebraic manipulation or problems lacking actual numerical
values would fall under this category. You should be able to easily 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,
it asks you 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 need to notice that the two
polynomials are in the format of a Difference of Two Squares: (a + b)(a – b)
= a^{2} – b^{2}.
In our case, a = x + y and b =
1. As a result, (x + y –
1)(x + y +
1) = (x + y)^{2} –
1. B is correct.
Don’t Immediately Use Your Calculator
The fact that the test contains all four of these question
types means that you shouldn’t get triggerhappy 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,
first take a brief look at each question and understand exactly
what it’s asking you to do. That short pause will save you a great
deal of time later on. For example, what if you came upon the question:

A triggerhappy 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 was
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 (like answers A and B),
won’t be equal to zero (answer C), and also won’t be
greater than 1 (answer E).
