Later in the chapter we’ll present a step method that you should employ on
every Problem Solving question you face. However, since different problems call
for different approaches, one of the steps, “Plan the Attack,” is open-ended and
calls for you to choose the most effective approach to the problem at hand. So
before we get to the step method itself, we’ll first demonstrate a standard
approach, as well as a few alternative approaches that may come in handy in
particular situations. Each approach discussed in this section represents a
different way to use the math concepts you reviewed in the previous chapter.
We’ll cover the following:
- Standard Applications of Math Concepts
- Alternative Approaches for Special Cases
Standard Applications of Math Concepts
There’s no need to make things more complicated than they need to be;
some questions require nothing more than straightforward applications of the
concepts you learned in chapter 2. This doesn’t necessarily mean that such
questions will be easy, since some of the concepts themselves can be
complex, and the test makers occasionally complicate matters by sprinkling
traps among the choices. Easier questions often require the application of a
single concept, while harder questions may involve multiple concepts. Some
may even require you to draw your own diagram when none is given. Regardless
of the difficulty level, the standard application approach is the same:
Scope out the situation, decide on what concept or concepts are being
tested, and then use what you know about those concepts to answer the
question before looking at the choices. If you’ve done your work well, the
answer you get will be among the choices on the screen, and you’ll click it
and move on.
Let’s look at a few examples spanning various difficulty levels. We’ll
take a look at single-concept questions based around one particular math
concept, and multiple-concept questions that require you to make use of
numerous bits of math knowledge to arrive at the answer. As you’ll see,
we’ve bolded all our Math 101 concepts as they come up to make it easier for
you to navigate through our explanations.
Here’s an example of the most basic kind of Problem Solving
question you’ll see:
||What is the value of x
if 3x – 27 = 33?
The math concept in play here is equations with one
variable, something you likely remember from junior high school.
There’s nothing to do here but apply the concept: First isolate the
variable by adding 27 to both sides to get 3x = 60, and
then divide both sides by 3 to get x = 20, choice
C. No doubt the test makers include 2 among the answer
choices to trap people who accidentally subtracted 27 from both sides,
yielding 3x = 6 and x = 2. 11, choice
B, is what you get if you divide one number in the
problem (33) by another (3), and 27, choice D, appears in
the problem itself. Assuming you didn’t fall for any of these traps,
there’s not much to it: Just apply a single, fairly basic concept
directly to the problem to pick up the point.
Not all single-concept questions are necessarily so
straightforward, however, especially as you get on in the section. Try
this one on for size:
||At a local golf club, 75 members
attend weekday lessons, 12 members attend weekend
lessons, and 4 members attend both weekday and
weekend lessons. If 10 members of the club do not
attend any lessons, how many members are in the
There’s only one concept in play here, but if you don’t know it,
you’re in for a very tough time. You need the formula for group
problems with two groups:
group 1 + group 2 – both + neither = total.
This is a formula you probably didn’t learn in junior high or
high school, or most likely forgot even if you did. You probably won’t
see a question like this early in the section, but if you’re doing well,
the CAT’s going to challenge you and start spitting out questions from
the harder end of the question pool. In any case, it’s really only
testing whether you’ve done your homework and memorized the formula.
If you did, then you’d be in great shape, since the math itself is
not particularly difficult: If we let group 1 be the 75
members who attend weekday lessons and let group 2 be
the 12 members who attend weekend lessons, we get: 75 + 12 – 4 + 10 =
total. Solving for total gives 93, choice
D. Notice how choice B, 75, is a number
contained in the problem, while choice E, 101, is what you
get if you mistakenly add 4 instead of subtract it.
Some Problem Solving questions require you to pull together two or
more choice tidbits from your arsenal of essential math concepts. One of
the most common examples of a multiple-concept question involves
geometric formulas that generate equations that need to be solved
arithmetically and/or algebraically. Here’s an example:
||If AB =
BC and x = 60,
what is the length of CE in
This is a bit more involved than a typical single-concept question
because there are a number of geometry concepts you need to know and
some genuine opportunities to slip up on the arithmetic end too. It’s a
mish-mash problem to boot, involving three triangles and a rectangle, so
if you don’t know the special and exciting properties of these geometric
figures, you’re pretty much sunk right there. If you do, then you should
be able to at least formulate the correct equation for line
EC, but then you still have to crunch the numbers to
solve it. Let’s see what a solid effort on this question might look
First, you’re best off redrawing the diagram on your scratch
paper, since you wouldn’t want to keep all the information you’re going
to add to it in your head. Since AB and
BC are equal, ∠BAC and
∠BCA must be equal since the angles in a triangle
opposite from equal sides are equal (concept
1). Since the third angle labeled x equals
60°, ∠BAC and ∠BCA together must total
120° because the three angles of a triangle add up to 180°
(concept 2). Since we determined that
∠BAC and ∠BCA are equal, they both
must be 60°. Notice anything now? A triangle with three equal
angles is an equilateral triangle (concept
3). Since all three sides in an equilateral triangle are
equal (concept 4), AB =
BC = AC = 4. Since
ACDE is a rectangle, and opposite sides of a
rectangle are equal (concept 5),
AC = ED = 4. By now your sketch should
look like this:
Now that we have two sides of right triangle ECD,
we have everything we need to figure out the length of
EC, thanks to the Pythagorean theorem: x
where x and y are the sides and
z is the hypotenuse (concept 6).
Substituting 4 and 8 as the sides and EC as the
hypotenuse gives us:
42 + 82
For convenience, we’ll denote all of the ensuing
arithmetic, including simplifying the radical,
as concept 7:
Voila!—choice B. Check out the
traps: 4 (A) is a number calculated along the way; 8
(D) is a number given in the problem; and 12
(E) is what you get if you add the two known sides of
triangle ECD together.
Notice that no fewer than seven math concepts made their way into
this problem—none of them particularly earth-shattering or treacherous,
mind you, but still adding up to a medium-level challenge with plenty of
Alternative Approaches for Special Cases
The standard “do question, look for answer” approach is all well and
good in many cases, but some questions call out for alternative approaches.
When the question contains variables in the answer choices, making up
numbers and substituting them into the problem is often very effective.
Conversely, when the answer choices contain actual numbers, you may benefit
from simply plugging them into the given situation to see which one works,
instead of hacking through some difficult arithmetic or algebra. Let’s take
a look at each of these strategies, one by one.
Making Up Numbers
Which of the following problems would you rather be faced with on
- Question 1: If x apples cost
y cents, how much will z apples
cost in dollars?
- Question 2: If 5 apples cost 50 cents, how much will 10 apples
cost in dollars?
If you’re like most people, question 2 looks much easier, and you
probably wouldn’t have much trouble solving it: If you double the number
of apples, you double the number of cents. One hundred cents equals one
The difference between question 1 and question 2 is simple. We
replaced the variables in question 1 with some made-up numbers, thus
creating the easier question 2. So, if you see x, y, m,
n, or any other variables in both the question and the answer
choices, see if you can avoid using complicated algebra by making up
numbers and inserting them into the problem. You don’t want to just make
up any old numbers, however—you want numbers that will simplify the
problem. Use the following guidelines:
- Pick easy numbers. Although you could choose
582.97 as a value, you definitely wouldn’t be making the problem any
easier. Stick to relatively small, whole numbers whenever possible.
- Avoid 0, 1, and any numbers used in the problem.
The numbers 0 and 1 have unique properties that may skew the results
when used for this technique, so don’t substitute either of those
into the problem. (We’ll give you the exact opposite advice in the
Quantitative Comparisons chapter, since in those questions the
special properties of 0 and 1 come in handy.) Also, since the test
makers sometimes use numbers from the question to construct
distractors, you may get yourself into trouble by selecting those as
well. If, for example, the problem contains the expression
3a + 5, don’t use 3 or 5. You shouldn’t have
any trouble avoiding the few numbers used in the question itself,
just for good measure.
- Choose different numbers for different variables.
For example, if the problem contains the variables
m and n, you wouldn’t want to
choose 2 for both. Instead, you might choose 2 for
m and 3 for n.
- Pay attention to units. If a problem involves a
change in units (such as minutes to hours, pennies to dollars, feet
to yards, and so on), choose a number that works well for both
units. For example, 120 would be a good choice for a vari- able
representing minutes, since 120 minutes is easily converted into 2
- Obey the rules of the problem. Occasionally, the
problem may include specific requirements for variables. For
example, if the problem says that x must be
negative, you can’t make up a positive value for x.
- Save dependent variables for last. If the value
of one variable is determined by the value of one or more other
variables, make up numbers for those other variables first. That
will automatically determine the value of the variable that depends
on the value of the others. For example, if the problem states that
a = b + c,
a is dependent on b and
c. Choose values for b and
c first, and the value of a
will then simply emerge as the sum of b and
Once you’ve selected your values, an actual number will emerge
when you work the problem out with the numbers you’ve selected. All you
need to do then is check which answer choice contains an expression that
yields the same value when you make the same substitutions. This will
make more sense in the context of an example, so let’s apply the
strategy to the following question.
||A gear makes r
rotations in m minutes. If it
rotates at a constant speed, how many rotations will
the gear make in h hours?
Sure, you could crunch through this algebraically, and if that
floats your boat, great. However, if you’re among those who get a
headache from just looking at questions like this, making up numbers may
be just the way to go. Here’s how.
The variables in this problem are r, m, and h. We can make up
whatever values we’d like for these, as long as the values we choose
make it easy to work the problem. For r, the number of rotations, let’s
choose something small, like 3. For m, the number of minutes, we should
choose a value that will make it easy to convert to hours: 120 works
well, since 120 minutes is the same as 2 hours. Finally, for h, we
should choose something small again. Remember that we need to choose a
different value for each variable, so let’s use 4. Now that we have our
numbers, simply plug them into the situation:
A gear makes 3 rotations in
120 minutes. If it rotates at a constant speed, how
many rotations will the gear make in 4 hours?
Okay, much better—that’s something we can sink our teeth into. 120
minutes is the same as 2 hours, during which time the gear rotates 3
times. If it rotates 3 times in 2 hours, how many times will it rotate
in 4 hours? That’s just twice as much time, so it will make twice as
many rotations: 2 × 3 = 6, and so 6 is what we get when we substitute
our values into the problem. Now we have to find the answer choice
that’s equal to 6 when the same values are substituted for its
variables. Just work your way down the list, using 3 for
r, 120 for m, and 4 for
That’s not 6, the answer we
seek, so move on.
Yup—this is exactly what we’re
looking for, so B
is correct. If you’re sure of your work,
there’s no need to even continue with the choices; you’d just
and move on to the next question. For practice,
though, let’s see how the other three pan out:
Way too big.
Way too small.
Four times bigger than what
we’re after. Just what we thought: B
is the only choice
that matches the number we derived from our made-up numbers, so B
gets the point.
You’ll get more practice with this strategy as we go forward.
Let’s now move on to our other specialty technique, an exercise in role
reversal that we call . . .
When the question includes an equation (or a word problem that can
be translated into an equation), and the answer choices contain
relatively simple numbers, then it may be possible to plug the choices
into the equation to see which one works. Working backward from the
choices in this manner may help you avoid setting up or solving
complicated equations and can save you time as well because of a neat
wrinkle of this technique: Since the choices in math questions are
usually written in either ascending or descending order, you can start
with the middle choice, choice C, and either get the answer
immediately or at least eliminate three choices for the price of one.
Let’s say the answer choices are in ascending order. If you start
by plugging in C, then even if that choice doesn’t work,
you can use the outcome to determine whether you need to plug in a
smaller or larger number. If you need a smaller number, then D
and E are out of the question, and you can go right
to test choice A or B. If instead you need a
larger number, chop A and B and try D
or E. Notice another nice feature: When you plug in
for the second time, that choice will either work or leave only one
choice standing. If you follow this alternative approach, you shouldn’t
ever have to check more than two choices.
As always, math strategies make the most sense in the context of
examples, so we’ll demonstrate using the following question.
||A classroom contains 31 chairs,
each of which has either a cushion or a hard back.
If the room has five more cushion chairs than
hard-backed chairs, how many cushion chairs does it
Now if you happen to be an algebra whiz, you’d go ahead and use
the information to set up a pair of simultaneous equations to solve the
problem. However, you may find it easier to work backward instead. Since
the choices are in ascending order, we’ll start with the middle one and
pretend it’s correct. If it really is correct, then
plugging it into the problem’s scenario will cause all the numbers to
work out, so let’s see if it does.
The question is looking for the number of cushion chairs, which
for the moment we’re assuming to be 16. We can bounce that number off
the information in the beginning of the second sentence (5 more cushion
chairs than hard chairs) to determine that with 16 cushion chairs, there
would have to be 11 hard chairs. Now all we have to do is check whether
this scenario matches the information in the first sentence. Would that
give us 31 chairs total? Nope: 16 + 11 = 27, so the numbers don’t jibe.
That tells us three things: Choice C isn’t correct,
choice B isn’t correct, and choice A isn’t
correct. We can knock out A and B along
with C because they’re both smaller than
C, and if the number in C isn’t big enough to
get us to our required 31 chairs, A and B
ain’t gonna cut it either.
Now let’s try D—if it works, it’s correct, and if it
doesn’t, we can select E without even trying it out: 18
cushion chairs means 18 – 5 = 13 hard chairs and 18 + 13 = 31 chairs
total. That matches the information in the question, so D
is correct. Note that you could have worked the numbers the other
way: If there are 31 chairs total, and we assume there are 18 cushion
chairs, then there would have to be 13 hard chairs. That matches the
information in the second sentence that requires 5 more cushion chairs
than hard ones. Either way you slice it, the number 18 fits the bill
when plugged back into the situation, and we didn’t have to bother with
creating and solving simultaneous equations.
Use your judgment as to when to work backward. If there are
numbers in the answer choices, then consider it, but don’t do it if the
numbers are unwieldy, such as complex fractions. One of the skills that
the best math test takers possess is the ability to determine the most
effective way to work through the problems. We’ve shown you standard
applications and a few powerful alternatives. Now let’s assimilate what
you’ve learned into the general step method you’ll use for all Problem