MATH: CONTENT
60 questions in 60 minutes
Topics – arithmetic to trigonometry
Calculators are permitted but not required
(cannot use TI-89 or TI-92)
Questions come in two formulas:
straightforward “plug and chug” calculation
questions and word problems
THIS IS A
THINKING AND
REASONING TEST!
MATH: CONTENT
Questions: 1st 20 = easy, 2nd 20 = medium, 3rd 20 =
difficult
Questions arranged with most (but not all) of easy
problems in first twenty ?’s and most(but not all) of
difficult problems in last twenty ?’s
 Pre Algebra – 14
 Elementary Algebra – 10
 Intermediate Algebra - 9
 Plane Geometry – 14
 Coordinate Geometry – 9
 Trigonometry - 4
MATH: STRATEGIES
Now, Later, Never = bank as many points as you can
on easy/medium questions before tackling difficult ones
Use your personal order of difficulty to decide which
questions you find easiest = NOW
Do tougher questions LATER, if at all. Circle the
questions where you are unsure of your answer or
think you should come back to it.
Unless shooting for over a 30, there will be some
questions you NEVER do. Use a letter of the day on
these questions.
MATH: STRATEGIES
Do less problems to get more
points = 40 ish, 45 ish, 50 ish
Slow down on easy questions
Use process of elimination to save
time and help you to guess on
harder questions
MATH: STRATEGIES
Ball parking – helps you avoid careless
mistakes on questions in your personal
order of difficulty and make educated
guesses on more difficult questions
Draw a picture if one isn’t given.
38) Jamal sees a tree on the shore directly across a lake and wonders what the
distance is across the lake to the tree. He turns 90 degrees to the right and
walks in a straight line for 100 meters. Jamal turns to face the tree and finds
the angle between his line of sight and his path measures 25 degrees. Which
of the following is the closest distance, in meters, from Jamal’s initial position
to the tree?
F. 42
G. 47
H. 91
J. 213
K. 238
Using ballparking, we can eliminate answers. In this
problem, simply look at the picture!
We want to know the distance from Jamal’s initial
position to the tree. This is the leg of the right triangle.
The side length 100 is opposite the larger acute angle in
the right triangle, so the missing side is shorter.
So, the answer can’t be J or K.
Actual solution: (Use SOHCAHTOA)
tan(25) = x/100
So, x = 100tan(25) = 100(0.47) = 47.
MATH: STRATEGIES
Plugging in – plug in values to
help you solve the problem –
pick 2, 5, 10 generally.
Plugging in the answer
Be careful with your
calculator!
PLUGGING IN: (Look at #29)
29) If a = 2c and b = 6c, which of the following
relationships holds between a and b for each
Nonzero value of c?
A)
B)
C)
D)
E)
a = 3b
a = 2b
a=b
a = (1/6)b
a = (1/3)b
Plug in a nice value for either a, b, or c to help
solve the problem. In this case, let’s pick a
value for a. I’ll choose a = 10 (because this will
make c a nice number). If a = 10 and a = 2c,
then c = 5. Since b = 6c, b = 30.
If a = 10 and b = 30, the b is triple a. The
answer is b = 3a (which isn’t a choice), so divide
both sides by 3. Thus, b/3 = a. So, a = (1/3)b.
Try #26 and #30 using the plugging in strategy.
EXPLANATIONS FOR #26 AND #30
26) Which of the following is equivalent to (x+2)0 whenever x isn’t -2?
Pick any number for x other than -2. Use your calculator if you desire.
Don’t forget parenthesis. See what you get. It should be 1.
30) Since you are given a range of values to plug in for F, plug in the
smallest value in the range (59) and the biggest value in the range (68) and
see what you get. Plugging in 59 should yield 15, and plugging in 68 should
yield 20. Thus, the answer is J. This is an example of being careful with
your calculator!!
PLUGGING IN THE ANSWER (PITA)
10) What is the least common
denominator for adding the fractions
4/35, 1/56, and 3/16?
F) 80
H) 1,960
K) 31,360
G) 560
J) 4,480
USING THE STRATEGY
Since you are looking for the least common
denominator, start with the smallest choice
(80) and see if each denominator divides
into it.
Normally, we start with the middle value
when we plug in the answer. HOWEVER, we
aren’t going to here. WHY?
The answer is G.
PLUGGING IN THE ANSWER #2
5) What is the largest value of x for
which there exists a real value of y such
that x2 + y2 = 256?
A)
16
B)
128
C)
240
D)
256
E)
512
PLUGGING IN THE ANSWER #2
Use the choices. I generally like to start with the middle
choice.
If x = 240, then we have 2402 + y2 = 256. This equals 57600 +
y2 = 256.So, y2 = -57344. So, y is not a real number. Thus, 240
was too big of a choice for x. So, we should use a smaller
choice. Next, use either 16 or 128.
If x = 128 and we follow the same procedure, we again get y2
equal to a negative, which we know isn’t possible.
If x = 16, we get y2 equal to 0, which means y = 0. The answer
is A.
BE CAREFUL WITH YOUR
CALCULATOR!!
18) |5(-4) + 3(6)| = ?
F) -2
J) 19
G) 2
K) 38
H) 10
Since the given expression is in absolute value bars, you
need the absolute value command on your calculator.
Hit MATH, go over to NUM, and choose abs.
18) |5(-4) + 3(6)| = ?
F) -2
J) 19
G) 2
K) 38
H) 10
Try question #12 on
your calculator.
BE CAREFUL ON THIS ONE.
Make sure BOTH the numerator and denominator
are in parenthesis. Otherwise, your calculator won’t
understand your problem correctly.
Then, if the answer isn’t a whole number (like here),
turn your answer into a fraction on your calculator.
Hit MATH and choose option #1 and hit enter. On
the main screen, hit enter again.
If a geometry question is given WITHOUT a
picture, draw one!!
25) The sides of one triangle are 12 inches, 14
inches, and 15 inches long, respectively. In a
second triangle similar to the first, the shortest
side is 8 inches long. To the nearest tenth of an
inch, what is the length of the longest side of the
second triangle?
15
12
8
x
14
This picture tells us that 8/12 = x/15. (If you don’t
remember how to solve an equation like this, plug in the
answers.)
Multiply both sides by 15 and we get x = 10.
WORD PROBLEMS: THE BASIC
APPROACH
1.
2.
3.
Know the question. Read the whole problem
before calculating anything, and underline the
actual question.
Let the answers help. Look for clues on how
to solve and ways to use POE
Break the problem into bite-sized pieces.
Watch out for trick phrasing.
KNOW THE QUESTION #37
A data set has 15 elements. The 15 elements in a
second data set are obtained by multiplying each
element in the first data set by 10. The 15 elements in
third data set are obtained by decreasing each element
of the second data set by 20. The median of the third
data set is 50. What is the median of the first data
set?
LET THE ANSWERS HELP #37
In this problem, I’m not sure the answers do
you a lot of good, unless you utilize a PITA
strategy. However, either PITA or ballparking
can be beneficial if you are given a diagram or
a problem that immediately allows you to rule
out some answers.
BITE SIZED PIECES #37
Part 1: A data set has 15 elements.
Part 2: The 15 elements in a second data set are
obtained by multiplying each element in the first data
set by 10.
Part 3: The 15 elements in third data set are obtained
by decreasing each element of the second data set by
20.
RESULT: The median of the third data set is 50.
What is the median of the first data set?
SOLUTION #37
Since we are talking about the median, only the
middle piece of data matters.
Doing any operation to a data set keeps the data in
the same order.
Let the first data set have median “x”.
In set 2, this piece becomes 10x.
In set 3, the piece becomes 10x – 20, which is also
known as 50. If 10x – 20 = 50, 10x = 70, so x = 7.
KNOW THE QUESTION #40
Each edge of a cube is 4 inches long. Each edge of a
second cube is triple the length of each edge of the
first cube. The volume of the second cube is how
many cubic inches bigger than the volume of the
first cube?
LET THE ANSWERS HELP #40
If you recognize that the volume of the first
cube is 4 x 4 x 4 = 64, and you recognize that
answers G and H and answers J and K are 64
units apart, you can actually rule out H and K
as possibilities. This is because H and K are
distractors and are provided if you don’t read
the question carefully.
BITE SIZED PIECES #40
#1) The first cube has volume 4 x 4 x 4 = 64.
#2) The second cube has edges that are triple in
length of the first cube. So, the second cube is 12 x
12 x 12 = 1728.
#3) The difference in their volumes is 1728 – 64 =
1664.
MATH: STRATEGIES
SLOW DOWN! Accuracy is more important than
speed.
You don’t get points for showing your work – use it
when you need to. Watch out for careless mistakes.
Be careful of the trap answers – just because you came
up with a number that is a choice doesn’t mean you got
that right answer.