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Legal Notice
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of this book.
New SAT Math Problems
arranged by Topic and
Difficulty Level
For the Revised SAT
March 2016 and Beyond
Dr. Steve Warner
© 2015, All Rights Reserved
iii
BOOKS FROM THE GET 800 COLLECTION FOR
COLLEGE BOUND STUDENTS
28 SAT Math Lessons to Improve Your Score in One Month
Beginner Course
Intermediate Course
Advanced Course
320 SAT Math Problems Arranged by Topic and Difficulty Level
320 SAT Math Subject Test Problems Arranged by Topic and
Difficulty Level
Level 1 Test
Level 2 Test
SAT Prep Book of Advanced Math Problems
The 32 Most Effective SAT Math Strategies
SAT Prep Official Study Guide Math Companion
SAT Vocabulary Book
ACT Prep Red Book – 320 ACT Math Problems with Solutions
320 AP Calculus AB Problems Arranged by Topic and Difficulty
Level
320 AP Calculus BC Problems Arranged by Topic and Difficulty
Level
555 Math IQ Questions for Middle School Students
555 Geometry Problems for High School Students
CONNECT WITH DR. STEVE WARNER
iv
Table of Contents
Introduction: The Proper Way to Prepare
1. Using this book effectively
2. The magical mixture for success
3. Practice problems of the appropriate level
4. Practice in small amounts over a long period of time
5. Redo the problems you get wrong over and over and
over until you get them right
6. Check your answers properly
7. Guess when appropriate
8. Pace yourself
9. Attempt the right number of questions
10. Use your calculator wisely
11. Grid your answers correctly
7
7
8
9
10
Problems by Level and Topic with Fully Explained Solutions
Level 1: Heart of Algebra
Level 1: Geometry and Trig
Level 1: Passport to Advanced Math
Level 1: Problem Solving and Data
17
17
23
29
37
10
11
11
11
12
13
15
Level 2: Heart of Algebra
Level 2: Geometry and Trig
Level 2: Passport to Advanced Math
Level 2: Problem Solving and Data
42
47
54
57
Level 3: Heart of Algebra
Level 3: Geometry and Trig
Level 3: Passport to Advanced Math
Level 3: Problem Solving and Data
65
75
81
87
Level 4: Heart of Algebra
Level 4: Geometry and Trig
Level 4: Passport to Advanced Math
Level 4: Problem Solving and Data
92
98
109
118
v
Level 5: Heart of Algebra
Level 5: Geometry and Trig
Level 5: Passport to Advanced Math
Level 5: Problem Solving and Data
122
127
136
143
Supplemental Problems – Questions
Level 1: Heart of Algebra
Level 1: Geometry and Trig
Level 1: Passport to Advanced Math
Level 1: Problem Solving and Data
Level 2: Heart of Algebra
Level 2: Geometry and Trig
Level 2: Passport to Advanced Math
Level 2: Problem Solving and Data
Level 3: Heart of Algebra
Level 3: Geometry and Trig
Level 3: Passport to Advanced Math
Level 3: Problem Solving and Data
Level 4: Heart of Algebra
Level 4: Geometry and Trig
Level 4: Passport to Advanced Math
Level 4: Problem Solving and Data
Level 5: Heart of Algebra
Level 5: Geometry and Trig
Level 5: Passport to Advanced Math
Level 5: Problem Solving and Data
151
151
152
154
156
158
159
161
162
164
166
167
168
169
170
172
173
176
177
179
180
Answers to Supplemental Problems
182
About the Author
187
Books by Dr. Steve Warner
188
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I N T R O D U C T I O N
THE PROPER WAY TO PREPARE
his book is for the revised SAT beginning in March 2016. If
you are preparing for an SAT being administered before this date, then
this is not the right book for you. The PSAT being given in October 2015
will have the new format, so you can use this book to prepare for that test,
especially if you are going for a national merit scholarship.
There are many ways that a student can prepare for the SAT. But not
all preparation is created equal. I always teach my students the methods
that will give them the maximum result with the minimum amount of
effort.
The book you are now reading is self-contained. Each problem was
carefully created to ensure that you are making the most effective use of
your time while preparing for the SAT. By grouping the problems given
here by level and topic I have ensured that you can focus on the types of
problems that will be most effective to improving your score.
1. Using this book effectively



Begin studying at least three months before the SAT
Practice SAT math problems twenty minutes each day
Choose a consistent study time and location
You will retain much more of what you study if you study in short bursts
rather than if you try to tackle everything at once. So try to choose about
a twenty minute block of time that you will dedicate to SAT math each
day. Make it a habit. The results are well worth this small time
commitment.



Every time you get a question wrong, mark it off, no matter
what your mistake.
Begin each study session by first redoing problems from previous
study sessions that you have marked off.
If you get a problem wrong again, keep it marked off.
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Note that this book often emphasizes solving each problem in more than
one way. Please listen to this advice. The same question is not generally
repeated on any SAT so the important thing is learning as many
techniques as possible.
Being able to solve any specific problem is of minimal importance. The
more ways you have to solve a single problem the more prepared you will
be to tackle a problem you have never seen before, and the quicker you
will be able to solve that problem. Also, if you have multiple methods for
solving a single problem, then on the actual SAT when you “check over”
your work you will be able to redo each problem in a different way. This
will eliminate all “careless” errors on the actual exam. Note that in this
book the quickest solution to any problem will always be marked with an
asterisk (*).
2. The magical mixture for success
A combination of three components will maximize your SAT math score
with the least amount of effort.
 Learning test taking strategies that work specifically for
standardized tests.
 Practicing SAT problems for a small amount of time each day for
about three months before the SAT.
 Taking about four practice tests before test day to make sure you
are applying the strategies effectively under timed conditions.
I will discuss each of these three components in a bit more detail.
Strategy: The more SAT specific strategies that you know the better off
you will be. Throughout this book you will see many strategies being
used. Some examples of basic strategies are “plugging in answer
choices,” “taking guesses,” and “picking numbers.” Some more advanced
strategies include “trying a simple operation,” and “moving the sides of a
figure around.” Pay careful attention to as many strategies as possible and
try to internalize them. Even if you do not need to use a strategy for that
specific problem, you will certainly find it useful for other problems in
the future.
Practice: The problems given in this book, together with the problems in
the practice tests from the College Board’s Official Study Guide (2016
Edition), are more than enough to vastly improve your current SAT math
score. All you need to do is work on these problems for about ten to
twenty minutes each day over a period of three to four months and the
final result will far exceed your expectations.
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Let me further break this component into two subcomponents – topic and
level.
Topic: You want to practice each of the four general math topics
given on the SAT and improve in each independently. The four topics are
Heart of Algebra, Geometry and Trig, Passport to Advanced Math,
and Problem Solving and Data Analysis. The problem sets in this book
are broken into these four topics.
Level: You will make the best use of your time by primarily
practicing problems that are at and slightly above your current ability
level. For example, if you are struggling with Level 2 Geometry and Trig
problems, then it makes no sense at all to practice Level 5 Geometry and
Trig problems. Keep working on Level 2 until you are comfortable, and
then slowly move up to Level 3. Maybe you should never attempt those
Level 5 problems. You can get an exceptional score without them (higher
than a 700).
Tests: You want to take about four practice tests before test day to make
sure that you are implementing strategies correctly and using your time
wisely under pressure. For this task you should use “The Official SAT
Study Guide (2016 Edition).” Take one test every few weeks to make
sure that you are implementing all the strategies you have learned
correctly under timed conditions.
3. Practice problems of the appropriate level
Roughly speaking about one third of the math problems on the SAT are
easy, one third are medium, and one third are hard. If you answer two
thirds of the math questions on the SAT correctly, then your score will be
approximately a 600 (out of 800). That’s right—you can get about a 600
on the math portion of the SAT without answering a single hard question.
Keep track of your current ability level so that you know the types of
problems you should focus on. If you are currently scoring around a 400
on your practice tests, then you should be focusing primarily on Level 1,
2, and 3 problems. You can easily raise your score 100 points without
having to practice a single hard problem.
If you are currently scoring about a 500, then your primary focus should
be Level 2 and 3, but you should also do some Level 1 and 4 problems.
If you are scoring around a 600, you should be focusing on Level 2, 3,
and 4 problems, but you should do some Level 1 and 5 problems as well.
Those of you at the 700 level really need to focus on those Level 4 and 5
problems.
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If you really want to refine your studying, then you should keep track of
your ability level in each of the four major categories of problems:




Heart of Algebra
Geometry and Trig
Passport to Advanced Math
Problem Solving and Data Analysis
For example, many students have trouble with very easy Geometry and
Trig problems, even though they can do more difficult algebra problems.
This type of student may want to focus on Level 1, 2, and 3 Geometry
and Trig questions, but Level 3 and 4 Heart of Algebra questions.
4. Practice in small amounts over a long period of time
Ideally you want to practice doing SAT math problems ten to twenty
minutes each day beginning at least 3 months before the exam. You will
retain much more of what you study if you study in short bursts than if
you try to tackle everything at once.
The only exception is on a day you do a practice test. You should do at
least four practice tests before you take the SAT. Ideally you should do
your practice tests on a Saturday or Sunday morning. At first you can do
just the math sections. The last one or two times you take a practice test
you should do the whole test in one sitting. As tedious as this is, it will
prepare you for the amount of endurance that it will take to get through
this exam.
So try to choose about a twenty minute block of time that you will
dedicate to SAT math every night. Make it a habit. The results are well
worth this small time commitment.
5. Redo the problems you get wrong over and over and
over until you get them right
If you get a problem wrong, and never attempt the problem again, then it
is extremely unlikely that you will get a similar problem correct if it
appears on the SAT.
Most students will read an explanation of the solution, or have someone
explain it to them, and then never look at the problem again. This is not
how you optimize your SAT score. To be sure that you will get a similar
problem correct on the SAT, you must get the problem correct before the
SAT—and without actually remembering the problem.
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This means that after getting a problem incorrect, you should go over and
understand why you got it wrong, wait at least a few days, then attempt
the same problem again. If you get it right you can cross it off your list of
problems to review. If you get it wrong, keep revisiting it every few days
until you get it right. Your score does not improve by getting problems
correct. Your score improves when you learn from your mistakes.
6. Check your answers properly
When you go back to check your earlier answers for careless errors do
not simply look over your work to try to catch a mistake. This is usually a
waste of time. Always redo the problem without looking at any of your
previous work. Ideally, you want to use a different method than you used
the first time.
For example, if you solved the problem by picking numbers the first time,
try to solve it algebraically the second time, or at the very least pick
different numbers. If you do not know, or are not comfortable with a
different method, then use the same method, but do the problem from the
beginning and do not look at your original solution. If your two answers
do not match up, then you know that this a problem you need to spend a
little more time on to figure out where your error is.
This may seem time consuming, but that’s okay. It is better to spend more
time checking over a few problems than to rush through a lot of problems
and repeat the same mistakes.
7. Take a guess whenever you cannot solve a problem
There is no guessing penalty on the SAT. Whenever you do not know
how to solve a problem take a guess. Ideally you should eliminate as
many answer choices as possible before taking your guess, but if you
have no idea whatsoever do not waste time overthinking. Simply put
down an answer and move on. You should certainly mark it off and come
back to it later if you have time.
8. Pace yourself
Do not waste your time on a question that is too hard or will take too
long. After you’ve been working on a question for about 30 to 45 seconds
you need to make a decision. If you understand the question and think
that you can get the answer in another 30 seconds or so, continue to work
on the problem. If you still do not know how to do the problem or you are
using a technique that is going to take a long time, mark it off and come
back to it later if you have time.
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If you do not know the correct answer, eliminate as many answer choices
as you can and take a guess. But you still want to leave open the
possibility of coming back to it later. Remember that every problem is
worth the same amount. Do not sacrifice problems that you may be able
to do by getting hung up on a problem that is too hard for you.
9. Attempt the right number of questions
Many students make the mistake of thinking that they have to attempt
every single SAT math question when they are taking the test. There is no
such rule. In fact, most students will increase their SAT score by
reducing the number of questions they attempt.
There are two math sections on the SAT – one where a calculator is
allowed and one where a calculator is not allowed. The calculator section
has 30 multiple choice (mc) questions and 8 free response (grid in)
questions. The non-calculator section has 15 multiple choice (mc)
questions and 5 free response (grid in) questions.
You should first make sure that you know what you got on your last SAT
practice test, actual SAT, or actual PSAT (whichever you took last).
What follows is a general goal you should go for when taking the exam.
Score
< 330
330 – 370
380 – 430
440 – 490
500 – 550
560 – 620
630 – 800
MC
Grid In
MC
Grid In
(Calculator
Allowed)
(Calculator
Allowed)
(Calculator
Not Allowed)
(Calculator
Not Allowed)
10/30
15/30
18/30
21/30
24/30
27/30
30/30
3/8
4/8
5/8
6/8
6/8
7/8
8/8
4/15
6/15
8/15
9/15
11/15
13/15
15/15
1/5
2/5
2/5
3/5
4/5
4/5
5/5
For example, a student with a current score of 450 should attempt 21
multiple choice questions and 6 grid ins from the section where a
calculator is allowed, and 9 multiple choice questions and 3 grid in
questions from the section where a calculator is not allowed.
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This is just a general guideline. Of course it can be fine-tuned. As a
simple example, if you are particularly strong at Algebra problems, but
very weak at Geometry and Trig problems, then you may want to try
every Algebra problem no matter where it appears, and you may want to
reduce the number of Geometry and Trig problems you attempt.
Remember that there is no guessing penalty on the SAT, so you should
not leave any questions blank. This does not mean you should attempt
every question. It means that if you are running out of time make sure
you fill in answers for all the questions you did not have time to attempt.
10. Use your calculator wisely.



Use a TI-84 or comparable calculator if possible when practicing
and during the SAT.
Make sure that your calculator has fresh batteries on test day.
You may have to switch between DEGREE and RADIAN modes
during the test. If you are using a TI-84 (or equivalent) calculator
press the MODE button and scroll down to the third line when
necessary to switch between modes.
Below are the most important things you should practice on your
graphing calculator.


Practice entering complicated computations in a single step.
Know when to insert parentheses:
 Around numerators of fractions
 Around denominators of fractions
 Around exponents
 Whenever you actually see parentheses in the expression
Examples:
We will substitute a 5 in for x in each of the following examples.
Expression
7x  3
2 x  11
(3x  8) 2 x9


Calculator computation
(7*5 + 3)/(2*5 – 11)
(3*5 – 8)^(2*5 – 9)
Clear the screen before using it in a new problem. The big screen
allows you to check over your computations easily.
Press the ANS button (2ND (-) ) to use your last answer in the
next computation.
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





Press 2ND ENTER to bring up your last computation for editing.
This is especially useful when you are plugging in answer
choices, or guessing and checking.
You can press 2ND ENTER over and over again to cycle
backwards through all the computations you have ever done.
Know where the √ , 𝜋, and ^ buttons are so you can reach them
quickly.
Change a decimal to a fraction by pressing MATH ENTER
ENTER.
Press the MATH button - in the first menu that appears you can
take cube roots and nth roots for any n. Scroll right to NUM and
you have lcm( and gcd(. Scroll right to PRB and you have nPr,
nCr, and ! to compute permutations, combinations and factorials
very quickly.
Know how to use the SIN, COS and TAN buttons as well as
SIN-1, COS-1 and TAN-1.
You may find the following graphing tools useful.




Press the Y= button to enter a function, and then hit ZOOM 6 to
graph it in a standard window.
Practice using the WINDOW button to adjust the viewing
window of your graph.
Practice using the TRACE button to move along the graph and
look at some of the points plotted.
Pressing 2ND TRACE (which is really CALC) will bring up a
menu of useful items. For example selecting ZERO will tell you
where the graph hits the x-axis, or equivalently where the
function is zero. Selecting MINIMUM or MAXIMUM can find
the vertex of a parabola. Selecting INTERSECT will find the
point of intersection of 2 graphs.
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11. Grid your answers correctly
The computer only grades what you have marked in
the bubbles. The space above the bubbles is just for
your convenience, and to help you do your bubbling
correctly.
Never mark more than one circle in a column or the
problem will automatically be marked wrong. You do
not need to use all four columns. If you don’t use a
column just leave it blank.
The symbols that you can grid in are the digits 0
through 9, a decimal point, and a division symbol for
fractions. Note that there is no negative symbol. So
answers to grid-ins cannot be negative. Also, there are
only four slots, so you can’t get an answer such as 52,326.
Sometimes there is more than one correct answer to a grid-in question.
Simply choose one of them to grid-in. Never try to fit more than one
answer into the grid.
If your answer is a whole number such as 2451 or a decimal that only
requires four or less slots such as 2.36, then simply enter the number
starting at any column. The two examples just written must be started in
the first column, but the number 16 can be entered starting in column 1, 2
or 3.
Note that there is no zero in column 1, so if your answer is 0 it must be
gridded into column 2, 3 or 4.
Fractions can be gridded in any form as long as there are enough slots.
The fraction 2/100 must be reduced to 1/50 simply because the first
representation won’t fit in the grid.
Fractions can also be converted to decimals before being gridded in. If a
decimal cannot fit in the grid, then you can simply truncate it to fit. But
you must use every slot in this case. For example, the decimal
.167777777… can be gridded as .167, but .16 or .17 would both be
marked wrong.
Instead of truncating decimals you can also round them. For example, the
decimal above could be gridded as .168. Truncating is preferred because
there is no thinking involved and you are less likely to make a careless
error.
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Here are three ways to grid in the number 8/9.
1
Never grid-in mixed numerals. If your answer is 24, and you grid in the
1
mixed numeral 24, then this will be read as 21/4 and will be marked
wrong. You must either grid in the decimal 2.25 or the improper fraction
9/4.
𝟏
Here are two ways to grid in the mixed numeral 1𝟐 correctly.
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PROBLEMS BY LEVEL AND TOPIC WITH
FULLY EXPLAINED SOLUTIONS
Note: An asterisk (*) before a question indicates that a calculator is
required. An asterisk (*) before a solution indicates that the quickest
solution is being given.
LEVEL 1: HEART OF ALGEBRA
1.
Which of the following expressions
5𝑎 + 10𝑏 + 15𝑐 ?
(A)
(B)
(C)
(D)
is
equivalent
to
5(𝑎 + 2𝑏 + 3𝑐)
5(𝑎 + 2𝑏 + 15𝑐)
5(𝑎 + 10𝑏 + 15𝑐)
5(𝑎 + 2𝑏) + 3𝑐
Solution by picking numbers: Let’s choose values for a, b, and c, say
a = 2, b = 3, c = 4. Then
5a + 10b + 15c = 5(2) + 10(3) + 15(4) = 10 + 30 + 60 = 100.
Put a nice big dark circle around 100 so you can find it easier later. We
now substitute a = 2, b = 3, c = 4 into each answer choice:
(A) 5(2 + 2 ∙ 3 + 3 ∙ 4) = 100
(B) 5(2 + 2 ∙ 3 + 15 ∙ 4) = 340
(C) 5(2 + 10 ∙ 3 + 15 ∙ 4) = 460
(D) 5(2 + 2 ∙ 3) + 3 ∙ 4 = 52
Since (B), (C), and (D) each came out incorrect, the answer is choice (A).
Important note: (A) is not the correct answer simply because it is equal
to 100. It is correct because all three of the other choices are not 100.
You absolutely must check all four choices!
Remark: All of the above computations can be done in a single step with
your calculator (if a calculator is allowed for this problem).
Notes about picking numbers: (1) Observe that we picked a different
number for each variable. We are less likely to get more than one answer
choice to come out to the correct answer this way.
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(2) We picked numbers that were simple, but not too simple. The number
2 is usually a good choice to start, if it is allowed. We then also picked 3
and 4 so that the numbers would be distinct (see note (1)).
(3) When using the strategy of picking numbers it is very important that
we check every answer choice. It is possible for more than one choice to
come out to the correct answer. We would then need to pick new
numbers to try to eliminate all but one choice.
* Algebraic solution: We simply factor out a 5 to get
5a + 10b + 15c = 5(a + 2b + 3c)
This is choice (A).
Remarks: (1) If you have trouble seeing why the right hand side is the
same as what we started with on the left, try working backwards and
multiplying instead of factoring. In other words we have
5(a + 2b + 3c) = 5a + 10b + 15c
Note how the distributive property is being used here. Each term in
parentheses is multiplied by the 5.
In general, the distributive property says that if 𝑥, 𝑦, and 𝑧 are real
numbers, then
𝑥(𝑦 + 𝑧) = 𝑥𝑦 + 𝑥𝑧.
This property easily extends to expressions with more than two terms.
For example,
𝑥(𝑦 + 𝑧 + 𝑤) = 𝑥𝑦 + 𝑥𝑧 + 𝑥𝑤.
(2) We can also solve this problem by starting with the answer choices
and multiplying (as we did in Remark (1)) until we get 5a + 10b + 15c.
2.
Joseph joins a gym that charges $79.99 per month plus tax for a
premium membership. A tax of 6% is applied to the monthly
fee. Joseph is also charged a one-time initiation fee of $95 as
soon as he joins. There is no contract so that Joseph can cancel
at any time without having to pay a penalty. Which of the
following represents Joseph’s total charge, in dollars, if he keeps
his membership for 𝑡 months?
(A)
(B)
(C)
(D)
1.06(79.99 + 95)𝑡
1.06(79.99𝑡 + 95)
1.06(79.99𝑡) + 95
(79.99 + .06𝑡) + 95
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Solution by picking a number: (We will be using a calculator for this
solution)
Let’s choose a value for 𝑡, say 𝑡 = 2, so that Joseph keeps his gym
membership for 2 months.
Now 6% of 79.99 is 4.80 (to the nearest cent). So each month of
membership, including tax, is 79.99 + 4.80 = 84.79 dollars. It follows
that 2 months of membership, with tax, is 2 ⋅ 84.79 = 169.58 dollars.
When we add the initiation fee we get 169.58 + 95 = 𝟐𝟔𝟒. 𝟓𝟖 dollars.
Put a nice big, dark circle around the number 264.58 so you can find it
easily later. We now substitute 𝑡 = 2 into each answer choice and use our
calculator:
(A)
(B)
(C)
(D)
1.06(79.99 + 95)*2 ≈ 370.98
1.06(79.99*2 + 95) ≈ 270.28
1.06(79.99*2) + 95 ≈ 264.58
(79.99 + .06*2) + 95 = 175.11
Since choices (A), (B), and (D) came out incorrect, we can eliminate
them. Therefore the answer is choice (C).
Important note: (C) is not the correct answer simply because it came
out to 264.58. It is correct because all three of the other choices did not
come out correct.
* Algebraic solution: Since the monthly membership fee is 79.99
dollars, and the tax is 6%, the total monthly fee, with tax, is 1.06(79.99)
dollars per month. It follows that the total monthly fee for 𝑡 months is
1.06(79.99𝑡). Finally, we add in the one-time initiation fee to get
1.06(79.99𝑡) + 95, choice (C).
Notes: (1) 6% can be written either as the decimal .06 or the fraction
6
.
100
To change a percent to a decimal, simply divide by 100, or equivalently,
move the decimal point two places to the left, adding in zeros if
necessary. Note that an integer has a “hidden” decimal point right after
the number. In other words, 6 can be written as 6., so when we move the
decimal point two places to the left we get .06 (we had to add in a zero as
a placeholder).
To change a percent to a fraction, simply place the number in front of the
percent symbol (%) over 100.
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(2) Since the tax is 6%, it follows that the tax for $79.99 is .06(79.99) or
6
(79.99) dollars.
100
It follows that the total monthly fee, including tax, is
79.99 + .06(79.99) dollars.
We can use the distributive property to simplify this expression as
follows:
79.99 + .06(79.99) = 1(79.99) + .06(79.99) = 1.06(79.99)
(3) See problem 1 for more information on the distributive property.
(4) In note (2) we saw that one way to get the total monthly fee, including
tax, is to add the amount of tax to the untaxed amount. A quicker way is
to simply multiply the monthly fee by 1.06. A justification for why this
works is given in the last line of note (2).
(5) If you need to pay a certain dollar amount more than once, simply
multiply by the number of times you need to pay.
For example, if you need to pay 100 dollars five times, then the final
result is that you pay 100 ⋅ 5 = 500 dollars. More generally, if you need
to pay 100 dollars 𝑡 times, then the final result is that you pay 100𝑡
dollars.
In this problem we want to pay the monthly fee 𝑡 times. Since the
monthly fee is 1.06(79.99), the final result is 1.06(79.99)𝑡, or
equivalently 1.06(79.99𝑡)
(6) Don’t forget to add on the one-time initiation fee to 1.06(79.99𝑡) to
get 1.06(79.99𝑡) + 95 dollars.
3.
A high school has a $1000 budget to buy calculators. Each
scientific calculator will cost the school $12.97 and each
graphing calculator will cost the school $73.89. Which of the
following inequalities represents the possible number of
scientific calculators 𝑆 and graphing calculators 𝐺 that the
school can purchase while staying within their specified budget?
(A) 12.97𝑆 + 73.89𝐺 > 1000
(B) 12.97𝑆 + 73.89𝐺 ≤ 1000
(C)
(D)
12.97
73.89
+ 𝐺
𝑆
12.97
73.89
+ 𝐺
𝑆
> 1000
≤ 1000
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* Algebraic solution: The total cost, in dollars, for 𝑆 scientific
calculators is 12.97𝑆, and the total cost, in dollars, for 𝐺 graphing
calculators is 73.89𝐺.
It follows that the total cost, in dollars, for 𝑆 scientific calculators and 𝐺
graphing calculators is 12.97𝑆 + 73.89𝐺.
To stay within the school’s budget, we need this total cost to be less than
or equal to 1000 dollars.
So the answer is 12.97𝑆 + 73.89𝐺 ≤ 1000, choice (B).
Notes: (1) When using the symbols “<” and “>”, the symbol always
points to the smaller number (and similarly for “≤” and “≥”).
(2) To stay within the specified budget means that the total must not
exceed $1000. Some equivalent ways to say this are as follows:

the total must not be greater than $1000.

the total must be less than or equal to $1000.

the total 𝑇 must satisfy 𝑇 ≤ 1000.
(3) If the school were to spend exactly $1000, they would still be within
their budget. This is why the solution has “≤” instead of “<.”
4.
27
If − < 2 − 5𝑥 < −
10
20𝑥 − 8.
13
,
5
then give one possible value of
* Solution by trying a simple operation: Observe that
20𝑥 − 8 = 4(5𝑥 − 2) = −4(2 − 5𝑥).
So we have
(−4) (−
13
)
5
< −4(2 − 5𝑥) < (−4)(−
27
)
10
or equivalently
52
5
< 20𝑥 − 8 <
54
5
So we can grid in 𝟓𝟑/𝟓.
Notes: (1) The simple operation we used here was multiplication by −4.
We simply multiplied each of the three parts of the given inequality by
−4, noting that the inequalities reverse because we are multiplying by a
negative number.
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27
(2) Take careful note of how − 10 and −
multiplied by the negative number −4.
13
5
changed positions when we
(3) If we are allowed to use a calculator for this problem we could
27
13
multiply each of − 10 and − 5 by −4 in our calculator to get
(−4) (−
27
)
10
= 10.8 and (−4) (−
13
)
5
= 10.4
So we can grid in 𝟏𝟎. 𝟓, 𝟏𝟎. 𝟔, or 𝟏𝟎. 𝟕.
(4) We actually do not need to worry too much about the inequalities
27
13
reversing in this problem. We can simply multiply each of − 10 and − 5
by −4, and then choose a number between the two numbers that we get.
5.
The expression 3(5𝑥 + 8) − 4(3𝑥 − 2) is simplified to the form
𝑎𝑥 + 𝑏. What is the value of ab ?
* Algebraic solution:
3(5x + 8) – 4(3x – 2) = 15x + 24 – 12x + 8 = 3x + 32.
So 𝑎 = 3, 𝑏 = 32, and therefore 𝑎𝑏 = 3 ⋅ 32 = 𝟗𝟔.
Note: Make sure you are using the distributive property correctly here.
For example 3(5x + 8) = 15x + 24. A common mistake would be to write
3(5x + 8) = 15x + 8.
Also, −4(3𝑥 − 2) = −12𝑥 + 8. A common mistake would be to write
−4(3𝑥 − 2) = −12𝑥 − 2.
See problem 1 for more information on the distributive property.
6.
If 𝑥 + 7𝑦 = 15 and 𝑥 + 3𝑦 = 7, what is the value of 𝑥 + 5𝑦?
* Solution by trying a simple operation: We add the two equations
𝑥 + 7𝑦 = 15
𝑥 + 3𝑦 = 7
2𝑥 + 10𝑦 = 22
Now observe that 2𝑥 + 10𝑦 = 2(𝑥 + 5𝑦). So 𝑥 + 5𝑦 =
22
2
= 𝟏𝟏.
Notes: (1) We can also finish the problem by dividing each term of
2𝑥 + 10𝑦 = 22 by 2.
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2𝑥
10𝑦
We have 2 = 𝑥, 2 = 5𝑦, and
equivalently 𝑥 + 5𝑦 = 11.
22
2
= 11. So we get
2𝑥
2
+
10𝑦
2
=
22
,
2
or
(2) Although I do not recommend this for this problem, we could solve
the system of equations for 𝑥 and 𝑦, and then substitute those values in
for 𝑥 and 𝑦 in the expression 𝑥 + 5𝑦.
See problem 73 for several different ways to do this.
LEVEL 1: GEOMETRY AND TRIG
7.
Given right triangle ∆𝑃𝑄𝑅 below, what is the length of ̅̅̅̅
𝑃𝑄 ?
(A)
(B)
(C)
(D)
√2
√5
5
7
* Solution using Pythagorean triples: We use the Pythagorean triple
5, 12, 13 to see that PQ = 5, choice (C).
Note: The most common Pythagorean triples are 3, 4, 5 and 5, 12, 13.
Two others that may come up are 8, 15, 17 and 7, 24, 25.
Solution by the Pythagorean Theorem: By the Pythagorean Theorem,
we have 132 = (PQ)2 + 122. So 169 = (PQ)2 + 144. Subtracting 144 from
each side of this equation yields 25 = (PQ)2, or PQ = 5, choice (C).
Remarks: (1) The Pythagorean Theorem says that if a right triangle has
legs of lengths a and b, and a hypotenuse of length c, then c2 = a2 + b2.
(2) Be careful in this problem: the length of the hypotenuse is 13. So we
replace c by 13 in the Pythagorean Theorem.
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(3) The equation x2 = 25 would normally have two solutions: x = 5 and
x = –5. But the length of a side of a triangle cannot be negative, so we
reject –5.
What is the radius of a circle whose circumference is 𝜋?
8.
(A)
1
2
(B) 1
(C)
𝜋
2
(D) 𝜋
Solution by plugging in answer choices: The circumference of a circle
𝜋
is C = 2𝜋r. Let’s start with choice (C) as our first guess. If r = 2 , then
𝜋
C = 2𝜋(2 ) = π2. Since this is too big we can eliminate choices (C) and
(D).
Let’s try choice (B) next. If r = 1, then C = 2𝜋(1) = 2π, still too big.
1
The answer must therefore be choice (A). Let’s verify this. If r = 2, then
1
C = 2𝜋(2) = 𝜋. So the answer is indeed choice (A).
Note: When plugging in answer choices, it’s always a good idea to start
with choice (B) or (C) unless there is a specific reason not to. In this
problem, eliminating choice (C) allowed us to eliminate choice (D) as
well, possibly saving us from having to do one extra computation.
* Algebraic solution: We use the circumference formula C = 2𝜋r, and
substitute 𝜋 in for C.
C = 2𝜋r
𝜋 = 2𝜋r
𝜋
=r
2𝜋
1
=r
2
This is choice (A).
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9.
5
In ∆𝑃𝑄𝑅 above, tan 𝑅 = 12. What is the length of side PR ?
(A)
(B)
(C)
(D)
* Since tan 𝑅 =
11
13
15
16
OPP
,
ADJ
5
we have 12 =
OPP
.
ADJ
Since the adjacent side is 12, the
opposite side must be 5. So we have the following picture.
We now find PR by using the Pythagorean Theorem, or better yet,
recognizing the Pythagorean triple 5, 12, 13.
So 𝑃𝑅 = 13, choice (B).
Remarks: (1) If you don’t remember the Pythagorean triple 5, 12, 13,
you can use the Pythagorean Theorem.
In this problem we have 𝑐 2 = 52 + 122 = 169. So 𝑐 = 13.
(2) See problem 7 for more information about Pythagorean triples and the
Pythagorean Theorem.
(3) The equation 𝑐 2 = 169 would normally have two solutions: 𝑐 = 13
and 𝑐 = −13. But the length of a side of a triangle cannot be negative, so
we reject –13.
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Here is a quick lesson in right triangle trigonometry for those of you
that have forgotten.
Let’s begin by focusing on angle A in the following picture:
Note that the hypotenuse is ALWAYS the side opposite the right angle.
The other two sides of the right triangle, called the legs, depend on which
angle is chosen. In this picture we chose to focus on angle A. Therefore
the opposite side is BC, and the adjacent side is AC.
Now you should simply memorize how to compute the six trig functions:
OPP
HYP
sin A = HYP
csc A = OPP
ADJ
cos A = HYP
tan A =
sec A =
OPP
ADJ
HYP
ADJ
ADJ
cot A = OPP
Here are a couple of tips to help you remember these:
(1) Many students find it helpful to use the word SOHCAHTOA. You
can think of the letters here as representing sin, opp, hyp, cos, adj, hyp,
tan, opp, adj.
(2) The three trig functions on the right are the reciprocals of the three
trig functions on the left. In other words, you get them by interchanging
the numerator and denominator. It’s pretty easy to remember that the
reciprocal of tangent is cotangent. For the other two, just remember that
the “s” goes with the “c” and the “c” goes with the “s.” In other words,
the reciprocal of sine is cosecant, and the reciprocal of cosine is secant.
To make sure you understand this, compute all six trig functions for each
of the angles (except the right angle) in the triangle given in this problem.
Please try this yourself before looking at the answers below.
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12
sin P = 13
5
cos P = 13
tan P =
12
5
13
5
csc P = 12
sin R = 13
13
5
cos R = 13
sec P =
5
cot P = 12
12
5
tan R = 12
csc R =
13
5
13
sec R = 12
cot R =
12
5
10. Let 𝑥 = cos 𝜃 and 𝑦 = sin 𝜃 for any real value 𝜃. Then x2 + y2 =
(A)
(B)
(C)
(D)
−1
0
1
It cannot be determined from the information given
* Solution using a Pythagorean identity:
𝑥 2 + 𝑦 2 = (cos 𝜃)2 + (sin 𝜃)2 = 1
This is choice (C).
Notes: (1) (cos 𝜃)2 is usually abbreviated as cos2 𝜃.
Similarly, (sin 𝜃)2 is usually abbreviated as sin2 𝜃.
In particular, (cos 𝜃)2 + (sin 𝜃)2 would be written as cos2 𝜃 + sin2 𝜃.
(2) One of the most important trigonometric identities is the Pythagorean
Identity which says
𝐜𝐨𝐬 𝟐 𝒙 + 𝐬𝐢𝐧𝟐 𝒙 = 𝟏.
2
11. A line with slope 3 is translated up 5 units and right 1 unit. What
is the slope of the new line?
* Any translation of a line is parallel to the original line and therefore has
the same slope. The new line therefore has a slope of 𝟐/𝟑.
Notes: (1) If we only moved some of the points on the line, then the slope
might change. But here we are moving all points on the line
simultaneously. Therefore the exact shape and orientation of the line are
preserved.
(2) We could also grid in one of the decimals .666 or .667.
(3) If the solution is not clear, it is recommended that you draw a picture.
2
Start by drawing a line with slope 3. One way to do this would be to plot
points at (0,0) and (3,2) and then draw a line through these two points.
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Now take those same two points and move them up 5 units and right 1
unit to the points (1,5) and (4,7). Draw a line through these two points.
Note that the two lines are parallel.
(4) Recall that the formula for the slope of a line is
Slope = 𝑚 =
rise
run
𝑦 −𝑦
= 𝑥2 −𝑥1
2
1
Let’s verify that the slopes of the two lines mentioned in note (3) are the
same.
2−0
2
For the line passing through (0,0) and (3,2), the slope is 3−0 = 3, and for
7−5
2
the line passing through (1,5) and (4,7), the slope is 4−1 = 3.
So we see that the two slopes are equal.
(5) Parallel lines always have the same slope.
12. In the figure above, adjacent sides meet at right angles and the
lengths given are in inches. What is the perimeter of the figure,
in inches?
* Solution by moving the sides of the figure around: Recall that to
compute the perimeter of the figure we need to add up the lengths of all 8
line segments in the figure. We “move” the two smaller vertical segments
to the right, and each of the smaller horizontal segments up or down as
shown below.
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Note that the “bold” length is equal to the “dashed” length. We get a
rectangle with length 30 and width 15. Thus, the perimeter is
(2)(30) + (2)(15) = 60 + 30 = 90.
Warning: Although lengths remain unchanged by moving line segments
around, areas will be changed. This method should not be used in
problems involving areas.
LEVEL 1: PASSPORT TO ADVANCED MATH
13. If 2𝑥 2 − 11 = 5 − 2𝑥 2 , what are all possible values of x ?
(A)
(B)
(C)
(D)
2 only
−2 only
0 only
2 and −2 only
Solution by plugging in the answer choices: According to the answer
choices we need only check 0, 2, and −2.
𝑥 = 0:
2(0)2 − 11 = 5 − 2(0)2 −11 = 5
False
𝑥 = 2:
2(2)2 − 11 = 5 − 2(2)2
−3 = −3
True
𝑥 = −2: 2(−2)2 − 11 = 5 − 2(−2)2
−3 = −3
True
So the answer is choice (D).
Notes: (1) Since all powers of 𝑥 in the given equation are even, 2 and −2
must give the same answer. So we didn’t really need to check −2.
(2) Observe that when performing the computations above, the proper
order of operations was followed. Exponentiation was done first,
followed by multiplication, and then subtraction was done last.
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For example, we have 2(2)2 − 11 = 2 ⋅ 4 − 11 = 8 − 11 = −3 and
5 − 2(2)2 = 5 − 2 ⋅ 4 = 5 − 8 = −3.
Order of Operations: Here is a quick review of order of operations.
PEMDAS
P
E
M
D
A
S
Parentheses
Exponentiation
Multiplication
Division
Addition
Subtraction
Note that multiplication and division have the same priority, and addition
and subtraction have the same priority.
* Algebraic solution: We add 2𝑥 2 to each side of the given equation to
get 4𝑥 2 − 11 = 5. We then add 11 to get 4𝑥 2 = 5 + 11 = 16. Dividing
16
each side of this last equation by 4 gives 𝑥 2 = 4 = 4. We now use the
square root property to get 𝑥 = ±2. So the answer is choice (D).
Notes: (1) The equation 𝑥 2 = 4 has two solutions: 𝑥 = 2 and 𝑥 = −2. A
common mistake is to forget about the negative solution.
(2) The square root property says that if 𝑥 2 = 𝑐, then 𝑥 = ±√𝑐.
This is different from taking the positive square root of a number. For
example, √4 = 2, whereas the equation 𝑥 2 = 4 has two solutions 𝑥 = ±2.
(3) Another way to solve the equation 𝑥 2 = 4 is to subtract 4 from each
side of the equation, and then factor the difference of two squares as
follows:
𝑥2 − 4 = 0
(𝑥 − 2)(𝑥 + 2) = 0
We now set each factor equal to 0 to get 𝑥 − 2 = 0 or 𝑥 + 2 = 0.
So 𝑥 = 2 or 𝑥 = −2.
14. A function 𝑔(𝑥) is defined as 𝑔(𝑥) = −5𝑥 2 . What is 𝑔(−2)?
(A)
(B)
(C)
(D)
−100
−20
20
50
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2
* 𝑔(−2) = −5(– 2) = −5(4) = −20, choice (B).
Notes: (1) The variable x is a placeholder. We evaluate the function 𝑔 at
a specific value by substituting that value in for x. In this question we
replaced x by –2.
(2) The exponentiation was done first, followed by the multiplication. See
the end of the solution to problem 13 for more information on order of
operations.
(3) To square a number means to multiply it by itself. So
(–2)2 = (–2)( –2) = 4.
(4) We can do the whole computation in our calculator (if a calculator is
allowed for the problem) in one step. Simply type -5(-2)^2 ENTER. The
output will be −20.
Make sure to use the minus sign and not the subtraction symbol.
Otherwise the calculator will give an error.
𝑥2 − 𝑦2 = 9
𝑥2
16
+
𝑦2
4
=1
𝑥 + 2𝑦 = 4
15. A system of three equations in two unknowns and their graphs
in the 𝑥𝑦-plane are shown above. How many solutions does the
system have?
(A)
(B)
(C)
(D)
None
Two
Four
Six
* Solution by looking at the graph: There is no point that is common to
all three graphs. So the system has no solutions, choice (A).
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Notes: (1) A solution to the system of equations is a point that satisfies
all three equations simultaneously. Graphically this means that the point
is on all three graphs. Although there are several points that are common
to two of the graphs, there are none that are common to all three.
(2) The graph of the equation 𝑥 2 − 𝑦 2 = 9 is the hyperbola in the figure
above with vertices (−3,0) and (3,0).
𝑥2
𝑦2
(3) The graph of the equation 16 + 4 = 1 is the ellipse in the figure
above with vertices (−4,0), (4, 0), (0,2), and (0, −2)
(4) The graph of the equation 𝑥 + 2𝑦 = 4 is the line in the figure above
with intercepts (4, 0) and (0,2).
(4,0) is the 𝒙-intercept of the line, and (0,2) is the 𝒚-intercept of the
line.
(5) Consider the following system of equations:
𝑥2
16
+
𝑦2
4
=1
𝑥 + 2𝑦 = 4
This system has the two solutions (0,2) and (4,0). These are the two
points common to the graphs of these two equations (the ellipse and the
line), also known as points of intersection of the two graphs.
(6) Consider the following system of equations:
𝑥2 − 𝑦2 = 9
𝑥 + 2𝑦 = 4
This system also has two solutions. These are the two points common to
the hyperbola and the line. Finding these two solutions requires solving
the system algebraically, which we won’t do here. One of these solutions
can be seen on the graph. It looks to be approximately (3.1,0.5). The
second solution does not appear on the portion of the graph that is
displayed. If we continued to graph the line and hyperbola to the left we
would see them intersect one more time.
(6) Consider the following system of equations:
𝑥2 − 𝑦2 = 9
𝑥2
16
+
𝑦2
4
=1
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This system has four solutions. These are the four points common to the
hyperbola and the ellipse. Finding these four solutions requires solving
the system algebraically, which we won’t do here. These solutions can be
seen clearly on the graph.
Algebraic solution: Observe from the graph that the points (0,2) and
(4,0) are intersection points of the line and the ellipse. In other words
they are solutions to the following system:
𝑥2
16
+
𝑦2
4
=1
𝑥 + 2𝑦 = 4
We can verify this by substituting each point into each equation.
(0,2):
𝑥2
16
+
𝑦2
4
02
= 1 ⇔ 16 +
22
4
4
= 1 ⇔4 = 1 ⇔1 = 1
𝑥 + 2𝑦 = 4 ⇔ 0 + 2(2) = 4 ⇔ 4 = 4
(4,0):
𝑥2
16
+
𝑦2
4
42
= 1 ⇔ 16 +
02
4
16
= 1 ⇔ 16 = 1 ⇔ 1 = 1
𝑥 + 2𝑦 = 4 ⇔ 4 + 2(0) = 4 ⇔ 4 = 4
When we plug each of these points into the equation for the hyperbola
however, we get the following:
(0,2):
𝑥 2 − 𝑦 2 = 9 ⇔ 02 − 22 = 9 ⇔ −4 = 9
(4,0):
𝑥 2 − 𝑦 2 = 9 ⇔ 42 − 02 = 9 ⇔ 16 = 9
Since we wound up with false equations, neither of these points are on
the hyperbola.
It follows that the system of equations has no solutions, choice (A).
Notes: (1) Although I do not recommend this for this problem, we can
solve the following system formally using the substitution method.
𝑥2
16
+
𝑦2
4
=1
𝑥 + 2𝑦 = 4
Let’s begin by solving the second equation for 𝑥 by subtracting 2𝑦 from
each side of the equation to get 𝑥 = 4 − 2𝑦.
We now replace 𝑥 by 4 − 2𝑦 in the first equation and solve for 𝑦.
𝑥2
𝑦2
+
=
16
4
(4−2𝑦)2
𝑦2
+
16
4
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We multiply each side of this last equation by 16 to get
(4 − 2𝑦)2 + 4𝑦 2 = 16
Now (4 − 2𝑦)2 = (4 − 2𝑦)(4 − 2𝑦) = 16 − 8𝑦 − 8𝑦 + 4𝑦 2 . So we
have
16 − 8𝑦 − 8𝑦 + 4𝑦 2 + 4𝑦 2 = 16
We cancel the 16 from each side and combine like terms on the left to get
−16𝑦 + 8𝑦 2 = 0
We factor −8𝑦 and note that
−16𝑦
−8𝑦
8𝑦 2
= 2 and −8𝑦 = −𝑦 to get
−8𝑦(2 − 𝑦) = 0
We now set each factor equal to zero.
−8𝑦 = 0
2−𝑦 =0
or
So we get the two solutions 𝑦 = 0 and 𝑦 = 2.
We can now substitute these 𝑦-values into either equation. Let’s use the
equation of the line since it’s simpler:
𝑦 = 0: 𝑥 + 2𝑦 = 4 ⇔ 𝑥 + 2(0) = 4 ⇔ 𝑥 = 4
𝑦 = 2: 𝑥 + 2𝑦 = 4 ⇔ 𝑥 + 2(2) = 4 ⇔ 𝑥 + 4 = 4 ⇔ 𝑥 = 0
So we see that the two points of intersection of the ellipse and the line are
(4,0) and (0,2).
(2) If we wanted to find the intersection points of the line and the
hyperbola we would solve the following system as we did in note (1):
𝑥2 − 𝑦2 = 9
𝑥 + 2𝑦 = 4
In this case however the algebra will be much messier and the solutions
do not “look very nice.”
It will never be necessary to do such messy algebra on the SAT, so we
leave this an optional exercise for the interested reader.
Similarly for the intersection points of the hyperbola and the ellipse we
would solve the following system:
𝑥2 − 𝑦2 = 9
𝑥2
16
+
𝑦2
4
=1
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Again, the algebra here is messy, and we leave this as an optional
exercise.
16. Which of the following graphs could not be the graph of a
function?
* Only choice (D) fails the vertical line test. In other words, we can draw
a vertical line that hits the graph more than once:
So the answer is choice (D).
𝑓(𝑥) = 5𝑥 + 3
𝑔(𝑥) = 𝑥 2 − 5𝑥 + 2
17. The functions 𝑓 and 𝑔 are defined above. What is the value of
𝑓(10) − 𝑔(5)?
* We have
𝑓(10) = 5(10) + 3 = 50 + 3 = 53
and
𝑔(5) = 52 − 5(5) + 2 = 25 − 25 + 2 = 2.
Therefore 𝑓(10) − 𝑔(5) = 53 − 2 = 𝟓𝟏.
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𝑥
𝑝(𝑥)
𝑞(𝑥)
𝑟(𝑥)
1
5
6
11
2
–3
7
–10
3
–4
–7
3
4
–5
–7
–2
5
–6
0
5
18. The table above gives some values of the functions 𝑝, 𝑞, and 𝑟.
At which value of 𝑥 does 𝑞(𝑥) = 𝑝(𝑥) + 𝑟(𝑥)?
Solution by guessing: The answer is an integer between 1 and 5
inclusive (these are the 𝑥-values given). So let’s start with 𝑥 = 3 as our
first guess. From the table 𝑝(3) = −4, 𝑞(3) = −7, and 𝑟(3) = 3.
Therefore 𝑝(3) + 𝑟(3) = −4 + 3 = −1. This is not equal to 𝑞(3) = −7
so that 3 is not the answer.
Let’s try 𝑥 = 4 next. From the table 𝑝(4) = −5, 𝑞(4) = −7, and
𝑟(4) = −2. So 𝑝(4) + 𝑟(4) = −5 + (−2) = −7 = 𝑞(4).
Therefore the answer is 4.
* Quick solution: We can just glance at the rows quickly and observe
that in the row corresponding to 𝑥 = 4, we have −5 + (−2) = −7. Thus,
the answer is 4.
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LEVEL 1: PROBLEM SOLVING AND DATA
Questions 19 - 21 refer to the following information.
Ten 25 year old men were asked how many hours per week they
exercise and their resting heart rate was taken in beats per
minute (BPM). The results are shown as points in the scatterplot
below, and the line of best fit is drawn.
19. How many of the men have a resting heart rate that differs by
more than 5 BPM from the resting heart rate predicted by the
line of best fit?
(A)
(B)
(C)
(D)
None
Two
Three
Four
* The points that are more than 5 BPM away from the line of best fit
occur at 1, 4, and 8 hours. So there are Three of them, choice (C).
Notes: (1) One of the two men that exercise 1 hour per week has a resting
heart rate of approximately 68 BPM. The line of best fit predicts
approximately 77 BPM. So this difference is 77 – 68 = 9 BPM.
Similarly, at 4 we have a difference of approximately 75 – 67 = 8 BPM,
and at 8 we have a difference of approximately 60 – 54 = 6 BPM.
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(2) At 5, the point below the curve corresponds to a heart rate that differs
from that predicted by the line of best fit by approximately 64 – 59 = 5
BPM. Since this is not more than 5, we do not include this point in the
count.
20. Based on the line of best fit, what is the predicted resting heart
rate for someone that exercises three and a half hours per week?
(A)
(B)
(C)
(D)
66 BPM
68 BPM
70 BPM
72 BPM
* The point (3.5, 68) seems to be on the line of best fit. So the answer is
68 BPM, choice (B).
21. Which of the following is the best interpretation of the slope of
the line of best fit in the context of this problem?
(A) The predicted number of hours that a person must
exercise to maintain a resting heart rate of 50 BPM.
(B) The predicted resting heart rate of a person that does not
exercise.
(C) The predicted decrease in resting heart rate, in BPM, for
each one hour increase in weekly exercise.
(D) The predicted increase in the number of hours of exercise
needed to increase the resting heart rate by one BPM.
* The slope of the line is the
change in predicted heart rate
.
change in hours of exercise
If we make the
denominator a 1 hour increase, then the fraction is the change in predicted
heart rate per 1 hour increase. Since the line is moving downward from
left to right, we can replace “change” in the numerator by “decrease.” So
the answer is choice (C).
*Note: Recall that the slope of a line is
Slope = 𝑚 =
rise
run
change in vertical distance
= change in horizontal distance
In this problem the change in vertical distance is the change in resting
heart rate, in BPM, and the change in horizontal distance is the change in
hours of exercise per week.
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22. The mean annual salary of an NBA player, 𝑆, can be estimated
using the equation 𝑆 = 161,400(1.169)𝑡 , where 𝑆 is measured
in thousands of dollars, and 𝑡 represents the number of years
since 1980 for 0 ≤ 𝑡 ≤ 20. Which of the following statements is
the best interpretation of 161,400 in the context of this
problem?
(A) The estimated mean annual salary, in dollars, of an NBA
player in 1980.
(B) The estimated mean annual salary, in dollars, of an NBA
player in 2000.
(C) The estimated yearly increase in the mean annual salary
of an NBA player.
(D) The estimated yearly decrease in the mean annual salary
of an NBA player.
* When 𝑡 = 0, we have
𝑆 = 161,400(1.169)0 = 161,400(1) = 161,400.
Since 𝑡 = 0 corresponds to the year 1980, it follows that 161,400 is the
estimated mean annual salary, in dollars, of an NBA player in 1980. This
is choice (A).
Notes: (1) The year 2000 corresponds with 𝑡 = 20. So the estimated
mean annual salary, in dollars, of an NBA player in 2000 would be
𝑆 = 161,400(1.169)20. This is a number much larger than 161,400 (it is
approximately 3,666,011).
(2) The function given in this problem is an exponential function. In
general, exponential functions have the form 𝑦 = 𝑎𝑏 𝑡 . Note that 𝑡 = 0
corresponds to 𝑦 = 𝑎. In other words, the initial amount is always 𝑎.
In this problem 𝑡 = 0 corresponds to the year 1980, and so the 161,400
gives the mean annual salary in 1980.
Unlike a linear function, an exponential function does not have a constant
slope. So in this problem the yearly increase or decrease in mean annual
salary cannot be described by a single number.
(3) Let’s compare this to the analogous linear function. Suppose for a
moment that the equation given instead was
𝑆 = 1.169𝑡 + 161,400
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In this case, the number 161,400 would still describe the estimated mean
annual salary, in dollars, of an NBA player in 1980.
The number 1.169 would describe the estimated yearly increase in the
mean annual salary of an NBA player.
23. A biologist was interested in the number of times a field cricket
chirps each minute on a sunny day. He randomly selected 100
field crickets from a garden, and found that the mean number of
chirps per minute was 112, and the margin of error for this
estimate was 6 chirps. The biologist would like to repeat the
procedure and attempt to reduce the margin of error. Which of
the following samples would most likely result in a smaller
margin of error for the estimated mean number of times a field
cricket chirps each minute on a sunny day?
(A) 50 randomly selected crickets from the same garden.
(B) 50 randomly selected field crickets from the same
garden.
(C) 200 randomly selected crickets from the same garden.
(D) 200 randomly selected field crickets from the same
garden.
* Increasing the sample size while keeping the population the same will
most likely decrease the margin of error. So the answer is choice (D).
Notes: (1) Decreasing the sample size will increase the margin of error.
This allows us to eliminate choices (A) and (B).
(2) The original sample consisted of only field crickets. If we were to
allow the second sample to include all crickets, then we have changed the
population. We cannot predict what impact this would have on the mean
and margin of error. This allows us to eliminate choice (C).
Technical note: In reality there is a correlation between the frequency of
cricket chirps and temperature. You can estimate the current temperature,
in degrees Fahrenheit, by counting the number of times a cricket chirps in
15 seconds and adding 37 to the result.
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24. A survey was conducted among a randomly chosen sample of
250 single men and 250 single women about whether they
owned any dogs or cats. The table below displays a summary of
the survey results.
Dogs
Only
Cats
Only
Both
Neither
Total
Men
92
14
18
126
250
Women
75
42
35
98
250
Total
167
56
53
224
500
What fraction of the people surveyed who said they own dogs
are women?
* There are 75 + 35 = 110 women who said they own dogs, and there
are a total of 167 + 53 = 220 people who said they own dogs. Therefore
110
the fraction of reported dog owners that are women is
= 𝟏/𝟐 or . 𝟓.
220
Notes: (1) There are two columns that represent people who said they
own dogs: the column labeled “Dogs Only,” and the column labeled
“Both.”
This is the end of this free sample. To
Remember that the word “Both” indicates both dog and cat ownership,
order
thisdogbook
from Amazon click the
and in particular
ownership.
(2) The noncept. image.
following
41
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