
Making Your Calculator Work for You
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: calculatorfriendly,
calculatorneutral, calculatorunfriendly, and calculatoruseless.
According to the ETS, about 60 percent of the test falls under the
calculatorneutral 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 calculatorunfriendly 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.
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 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
testtakers 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.
CalculatorNeutral Questions
You have two different 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 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 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
It is possible to answer calculatorunfriendly questions
by using a calculator. But while it’s possible, 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 if you take a closer look, you’ll see that cos^{2}(3 � 63°)
+ sin^{2}(3 63°) is
a trigonometric identity. More specifically, it is 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
down to ^{14}/
_{ 2} = ^{1}
/_{2} = .5. Answer
choice 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 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)
= 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. Answer choice 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,
you should come up with a problemsolving 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 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 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.
