The Math sections include two types of questions: Multiple-choice and Grid-ins.
They are designed to test your ability to solve problems, not to test your mathematical knowledge.
The questions in each sub-section are listed in ascending order of difficulty.
So, if a section begins with 8 multiple-choice questions followed by 10 grid-ins,
then Question 1 will be the easiest multiple-choice question and Question 8 will be the hardest.
Then Question 9 will be the easiest grid-in question and Question 18 will be the hardest.
There will be two 25-minute math sections and one 20-minute section. The sections can appear
anywhere in the test.
LEVEL
OF DIFFICULTY of the SAT EXAM
The mathematical skills tested on the SAT are very basic: only first
year algebra, geometry (no proofs), and a few basic concepts from second year algebra.
However, this does not mean that the math section is easy. The medium of basic mathematics is
chosen so that everyone taking the SAT will be on a fairly even playing
field. This way students who are concentrating in math and science
don't have an undue advantage over students who are concentrating in English
and humanities. Although the questions require only basic mathematics
and all have simple solutions, it can require considerable ingenuity to
find the simple solution. If you have taken a course in calculus
or another advanced math course, don't assume that you will find the math
section easy. Other than increasing your mathematical maturity, little
you learned in calculus will help on the SAT.
As mentioned
above, every SAT math problem has a simple solution, but finding
that simple solution may not be easy. The intent of the math
section is to test how skilled you are at finding the simple
solutions. The premise is that if you spend a lot of time
working out long solutions you will not finish as much of
the test as students who spot the short, simple solutions.
So if you find yourself performing long calculations or applying
advanced mathematics--stop. You're heading in the wrong direction.
Tackle
the math problems in the order given, and don't worry if you
fail to reach the last few questions. It's better to work
accurately than quickly.
SUBSTITUTION
Substitution is a very useful technique for solving SAT math
problems. It often reduces hard problems to routine ones.
In the substitution method, we choose numbers that have the
properties given in the problem and plug them into the answer-choices.
Example:
If n is
an odd integer, which one of the following is an even integer?
(A) 3n
+ 2 (B) n/4 (C)
2n + 3 (D) n(n + 3) (E)
nn
We are
told that n is an odd integer. So choose an odd integer for
n, say, 1 and substitute it into each answer-choice. In Choice
(A), 3(1) + 2 = 5, which is not an even integer. So eliminate
(A). Next, n/4 = 1/4 is not an even integer--eliminate (B).
Next, 2n + 3 = 2(1) + 3 = 5 is not an even integer--eliminate
(C). Next, n(n + 3) = 1(1 + 3) = 4 is even and hence the answer
is possibly (D). Finally, in Choice (E), the nn = 1(1) = 1,
which is not even--eliminate (E). The answer is (D).
When using
the substitution method, be sure to check every answer-choice
because the number you choose may work for more than one answer-choice.
If this does occur, then choose another number and plug it
in, and so on, until you have eliminated all but the answer.
This may sound like a lot of computing, but the calculations
can usually be done in a few seconds.
Sometimes
instead of making up numbers to substitute into the problem,
we can use the actual answer-choices. This is called Plugging
In. It is a very effective technique but not as common as
Substitution.
Example:
The digits of a three-digit number add up to 18. If the ten's
digit is twice the hundred's digit and the hundred's digit
is 1/3 the unit's digit, what is the number?
(A) 246
(B) 369 (C)
531 (D) 855 (E)
893
First,
check to see which of the answer-choices has a sum of digits
equal to 18. For choice (A), 2 + 4 + 6 = 12. Eliminate. For
choice (B), 3 + 6 + 9 = 18. This may be the answer. For choice
(C), 5 + 3 + 1 = 9. Eliminate. For choice (D), 8 + 5 + 5 =
18. This too may be the answer. For choice (E), 8 + 9 + 3
= 20. Eliminate. Now, in choice (D), the ten's digit is not
twice the hundred's digit, 5 does not equal 2(8). Eliminate.
Hence, by process of elimination, the answer is (B). Note
that we did not need the fact that the hundred's digit is
1/3 the unit's digit.
DEFINED
FUNCTIONS
Defined
functions are very common on the SAT, and most students struggle
with them. Yet once you get used to them, defined functions
can be some of the easiest problems on the test. In this type
of problem, you will be given a symbol and a property that
defines the symbol.
Example:
Define x # y by the equation x # y = xy - y. Then 2 # 3 =
(A) 1
(B) 3 (C)
12 (D) 15 (E)
18
From the
above definition, we know that x # y = xy - y. So all we have
to do is replace x with 2 and y with 3 in the definition:
2 # 3 = 2(3) - 3 = 3. Hence, the answer is (B).
NUMBER
THEORY
This broad
category is a popular source for SAT questions. At first,
students often struggle with these problems since they have
forgotten many of the basic properties of arithmetic. So before
we begin solving these problems, let's review some of these
basic properties.
• "The
remainder is r when p is divided by q" means p = qz + r; the
integer z is called the quotient. For instance, "The remainder
is 1 when 7 is divided by 3" means 7 = 3(2) + 1.
Example:
When the integer n is divided by 2, the quotient is u
and the remainder is 1. When the integer n is divided by 5,
the quotient is v and the remainder is 3. Which one of the
following must be true?
(A) 2u
+ 5v = 4
(B) 2u - 5v = 2
(C) 4u + 5v = 2
(D) 4u - 5v = 2
(E) 3u - 5v = 2
Translating
"When the integer n is divided by 2, the quotient is u and
the remainder is 1" into an equation gives n = 2 u + 1. Translating
"When the integer n is divided by 5, the quotient is v and
the remainder is 3" into an equation gives n = 5v + 3. Since
both expressions equal n, we can set them equal to each other:
2u + 1 = 5v + 3. Rearranging and then combining like terms
yields 2u - 5v = 2. The answer is (B).
• A number
n is even if the remainder is zero when n is divided by 2:
n = 2z + 0, or n = 2z.
• A number
n is odd if the remainder is one when n is divided by 2: n
= 2z + 1.
• The
following properties for odd and even numbers are very useful--you
should memorize them:
even x
even = even
odd x odd = odd
even x odd = even
even +
even = even
odd + odd = even
even + odd = odd
• Consecutive
integers are written as x, x + 1, x + 2, . . .
• Consecutive
even or odd integers are written as , x + 2, x + 4, . . .
• The
integer zero is neither positive nor negative, but it is even:
0 = 2(0).
• A prime
number is an integer that is divisible only by itself and
1.
The prime
numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41,
. . .
• A number
is divisible by 3 if the sum of its digits is divisible by
3.
For example,
135 is divisible by 3 because the sum of its digits (1 + 3
+ 5 = 9) is divisible by 3.
• The
absolute value of a number, | |, is always positive. In other
words, the absolute value symbol eliminates negative signs.
For example,
| -7 | = 7. Caution, the absolute value symbol acts only on
what is inside the symbol, | |. For example, -| -7 | = -(+7)
= -7. Here, only the negative sign inside the absolute value
symbol is eliminated.
Example:
If a, b, and c are consecutive integers and a < b <
c, which of the following must be true?
I. b -
c = 1
II. abc/3 is an integer.
III. a + b + c is even.
(A) I
only (B) II only (C) III only (D) I and II only (E) II and
III only
Let x,
x + 1, x + 2 stand for the consecutive integers a, b, and
c, in that order. Plugging this into Statement I yields b
- c = (x + 1) - (x + 2) = -1. Hence, Statement I is false.
As to
Statement II, since a, b, and c are three consecutive integers,
one of them must be divisible by 3. Hence, abc/3 is an integer,
and Statement II is true.
As to
Statement III, suppose a is even, b is odd, and c is even.
Then a + b is odd since even + odd = odd. Hence, a + b + c
= (a + b) + c = (odd) + even = odd. Thus, Statement III is
not necessarily true. The answer is (B).
GEOMETRY
One-fourth
of the math problems on the SAT involve geometry. (There are
no proofs.) Fortunately, the figures on the SAT are usually
drawn to scale. Hence, you can check your work and in some
cases even solve a problem by "eyeballing" the drawing.
Following
are some of the basic properties of geometry. You probably
know many of them. Memorize any that you do not know.
1. There
are 180 degrees in a straight angle.
2. Two
angles are supplementary if their angle sum is 180 degrees.
3. Two
angles are complementary if their angle sum is 90 degrees.
4. Perpendicular
lines meet at right angles.
5. A triangle
with two equal sides is called isosceles. The angles opposite
the equal sides are called the base angles.
6. The
altitude to the base of an isosceles or equilateral triangle
bisects the base and bisects the vertex angle.
7. The
angle sum of a triangle is 180 degrees.
8. In
an equilateral triangle all three sides are equal, and each
angle is 60 degrees.
9. The
area of a triangle is bh/2, where b is the base and h is the
height.
10. In
a triangle, the longer side is opposite the larger angle,
and vice versa.
11. Two
triangles are similar (same shape and usually different size)
if their corresponding angles are equal. If two triangles
are similar, their corresponding sides are proportional.
12. Two
triangles are congruent (identical) if they have the same
size and shape.
13. In
a triangle, an exterior angle is equal to the sum of its remote
interior angles and is therefore greater than either of them.
14. Opposite
sides of a parallelogram are both parallel and congruent.
15. The
diagonals of a parallelogram bisect each other.
16. If
w is the width and l is the length of a rectangle, then its
area is A = lw and its perimeter is P=2w + 2l.
17. The
volume of a rectangular solid (a box) is the product of the
length, width, and height. The surface area is the sum of
the area of the six faces.
18. If
the length, width, and height of a rectangular solid (a box)
are the same, it is a cube. Its volume is the cube of one
of its sides, and its surface area is the sum of the areas
of the six faces.
19. A
tangent line to a circle intersects the circle at only one
point. The radius of the circle is perpendicular to the tangent
line at the point of tangency.
20. An
angle inscribed in a semicircle is a right angle.
| Example:
In the figure to the right, what is the value of x?
(A) 30
(B) 32
(C) 35
(D) 40
(E) 47 |
 |
Since 2x + 60 is an exterior angle, it is equal to the sum
of the remote interior angles. That is, 2x + 60 = x + 90.
Solving for x gives x = 30. The answer is (A).
Most geometry problems on the SAT require straightforward
calculations. However, some problems measure your insight
into the basic rules of geometry. For this type of problem,
you should step back and take a "birds-eye" view of the problem.
The following example will illustrate.
| Example:
In the figure to the right, O is both the center of
the circle with radius 2 and a vertex of the square
OPRS. What is the length of diagonal PS?
(A) 1/2
(B) 1
(C) 4
(D) 2
(E) 2/3 |
 |
The diagonals
of a square are equal. Hence, line segment OR (not shown)
is equal to SP. Now, OR is a radius of the circle and therefore
OR = 2. Hence, SP = 2 as well, and the answer is (D).
COORDINATE
GEOMETRY
Distance
Formula:
The distance
between points (x, y) and (a, b) is given by the following
formula:
| Example:
In the figure to the right, the circle is centered at
the origin and passes through point P. Which of the
following points does it also pass through?
(A) (3, 3)
(B) 
(C) (2, 6)
(D) (1.5, 1.3)
(E) (-3, 4) |
 |
Since
the circle is centered at the origin and passes through the
point (0,-3), the radius of the circle is 3. Now, if any other
point is on the circle, the distance from that point to the
center of the circle (the radius) must also be 3. Look at
choice (B). Using the distance formula to calculate the distance
between 
and (0, 0) (the origin) yields
Hence, 
is on the circle, and the answer is (B).
Midpoint
Formula:
The midpoint
M between points (x, y) and (a, b) is given by
M
= ([x + a]/2, [y + b]/2)
In other
words, to find the midpoint, simply average the corresponding
coordinates of the two points.
| Example:
In the figure to the right, polygon PQRO is a square
and T is the midpoint of side QR. What are the coordinates
of T ?
(A) (1, 1)
(B) (1, 2)
(C) (1.5, 1.5)
(D) (2, 1)
(E) (2, 3) |
 |
Since
point R is on the x-axis, its y-coordinate is 0. Further,
since PQRO is a square and the x-coordinate of Q is 2, the
x-coordinate of R is also 2. Since T is the midpoint of side
QR, the midpoint formula yields T = ([2 + 2]/2, [2 + 0]/2)
= (4/2, 2/2) = (2, 1). The answer is (D).
Slope
Formula:
The slope
of a line measures the inclination of the line. By definition,
it is the ratio of the vertical change to the horizontal change.
The vertical change is called the rise, and the horizontal
change is called the run. Thus, the slope is the rise over
the run. Given the two points (x, y) and (a, b) the slope
is
M
= (y - b)/(x - a)
| Example:
In the figure to the right, what is the slope of line
passing through the two points?
(A) 1/4
(B) 1
(C) 1/2
(D) 3/2
(E) 2 |
 |
The slope formula yields m = (4 - 2)/(5 - 1) = 2/4 = 1/2.
The answer is (C).
Slope-Intercept Form:
Multiplying both sides of the equation m = (y -b)/(x - a)
by x-a yields
y - b = m(x - a)
Now, if the line passes through the y-axis at (0, b), then
the equation becomes
y - b = m(x - 0)
y - b = mx
y = mx + b
This is
called the slope-intercept form of the equation of a line,
where m is the slope and b is the y-intercept. This form is
convenient because it displays the two most important bits
of information about a line: its slope and its y-intercept.
INEQUALITIES
Inequalities
are manipulated algebraically the same way as equations with
one exception:
Multiplying
or dividing both sides of an inequality by a negative number
reverses the inequality. That is, if x > y and c < 0, then
cx < cy.
Example:
For which values of x is 4x + 3 > 6x - 8?
As with
equations, our goal is to isolate x on one side:
Subtracting
6x from both sides yields -2x + 3 > -8
Subtracting
3 from both sides yields -2x > -11
Dividing
both sides by -2 and reversing the inequality yields x <
11/2
Positive
& Negative Numbers
A number
greater than 0 is positive. On the number line, positive numbers
are to the right of 0. A number less than 0 is negative. On
the number line, negative numbers are to the left of 0. Zero
is the only number that is neither positive nor negative;
it divides the two sets of numbers. On the number line, numbers
increase to the right and decrease to the left.
The expression
x > y means that x is greater than y. In other words, x is
to the right of y on the number line.
We usually
have no trouble determining which of two numbers is larger
when both are positive or one is positive and the other negative
(e.g., 5 > 2 and 3.1 > -2). However, we sometimes hesitate
when both numbers are negative (e.g., -2 > -4.5). When in
doubt, think of the number line: if one number is to the right
of the number, then it is larger.
Miscellaneous
Properties of Positive and Negative Numbers
1. The
product (quotient) of positive numbers is positive.
2. The product (quotient) of a positive number and a negative
number is negative.
3. The product (quotient) of an even number of negative numbers
is positive.
4. The product (quotient) of an odd number of negative numbers
is negative.
5. The sum of negative numbers is negative.
6. A number raised to an even exponent is greater than or
equal to zero.
Absolute
Value
The absolute
value of a number is its distance on the number line from
0. Since distance is a positive number, absolute value of
a number is positive. Two vertical bars denote the absolute
value of a number: | x |. For example, | 3 | = 3 and | -3
| = 3.
Students
rarely struggle with the absolute value of numbers: if the
number is negative, simply make it positive; and if it is
already positive, leave it as is. For example, since -2.4
is negative, | -2.4 | = 2.4 and since 5.01 is positive | 5.01
| = 5.01.
Further,
students rarely struggle with the absolute value of positive
variables: if the variable is positive, simply drop the absolute
value symbol. For example, if x > 0, then | x | = x.
However,
negative variables can cause students much consternation.
If x is negative, then | x | = -x. This often confuses students
because the absolute value is positive but the -x appears
to be negative. It is actually positive--it is the negative
of a negative number, which is positive. To see this more
clearly let x = -k, where k is a positive number. Then x is
a negative number. So | x | = -x = -(-k) = k. Since k is positive
so is -x. Another way to view this is | x | = -x = (-1)x =
(-1)(a negative number) = a positive number.
Transitive
Property
If
x < y and y < z, then x < z.
FRACTIONS
I.
To compare two fractions, cross-multiply. The larger number
will be on the same side as the larger fraction.
Example:
|
Column
A |
|
Column
B |
|
9/10 |
|
10/11 |
Cross-multiplying
gives (9)(11) versus (10)(10), which reduces to 99 versus
100. Now, 100 is greater than 99. Hence, 10/11 is greater
than 9/10.
III.
To solve a fractional equation, multiply both sides by the
LCD (lowest common denominator) to clear fractions.
Example:
If (x + 3)/(x - 3) = y, what is the value of x in terms of
y?
(A) 3
- y (B) 3/y (C)
(2 + y)/(y - 2) (D) (-3y -3)/(1 -
y) (E) 3y/2
First,
multiply both sides of the equation by x - 3: (x - 3)(x +
3)/(x - 3) = (x - 3)y
Cancel
the (x - 3's) on the left side of the equation: x + 3 = (x
- 3)y
Distribute
the y: x + 3 = xy - 3y
Subtract
xy and 3 from both sides: x - xy = -3y - 3
Factor
out the x on the left side of the equation: x(1 - y) = -3y
- 3
Finally,
divide both sides of the equation by 1 - y: x = (-3y -3)/(1
- y)
Hence,
the answer is (D).
IV.
When dividing a fraction by a whole number (or vice versa),
you must keep track of the main division bar.
Example:
a/(b/c) = a(c/b) = ac/b. But (a/b)/c = (a/b)(1/c) = a/(bc).
V.
Two fractions can be added quickly by cross-multiplying: a/b
+ c/d = (ad + bc)/bd
Example:
1/2 - 3/4 =
(A) -5/4
(B) -2/3 (C)
-1/4 (D) 1/2 (E)
2/3
Cross
multiplying the expression 1/2 - 3/4 yields [1(4) - 2(3)]/2(4)
= (4 - 6)/8 = -2/8 = -1/4. Hence, the answer is (C).
VI.
To find a common denominator of a set of fractions, simply
double the largest denominator until all the other denominators
divide into it evenly.
VII.
Fractions often behave in unusual ways: Squaring a fraction
makes it smaller, and taking the square root of a fraction
makes it larger. (Caution: This is true only for
proper fractions, that is, fractions between 0 and 1.)
Example:
1/3 squared equals 1/9 and 1/9 is less than 1/3. Also the
square root of 1/4 is 1/2 and 1/2 is greater than 1/4.
AVERAGES
Problems
involving averages are very common on the SAT. They can be
classified into four major categories as follows.
I.
The average of N numbers is their sum divided by N, that is,
average = sum/N.
Example: What is the average of x, 2x, and 6?
By the definition of an average, we get (x + 2x + 6)/3 = (3x
+ 6)/3 = 3(x + 2)/3 = x + 2.
II.
Weighted average: The average between two sets of numbers
is closer to the set with more numbers.
Example:
If on a test three people answered 90% of the questions
correctly and two people answered 80% correctly, then the
average for the group is not 85% but rather [3(90) + 2(80)]/5
= 430/5 = 86. Here, 90 has a weight of 3--it occurs 3 times.
Whereas 80 has a weight of 2--it occurs 2 times. So the average
is closer to 90 than to 80 as we have just calculated.
III.
Using an average to find a number.
Sometimes
you will be asked to find a number by using a given average.
An example will illustrate.
Example:
If the average of five numbers is -10, and the sum of
three of the numbers is 16, then what is the average of the
other two numbers?
(A) -33
(B) -1 (C)
5 (D) 20 (E)
25
Let the
five numbers be a, b, c, d, e. Then their average is (a +
b + c + d + e)/5 = -10. Now three of the numbers have a sum
of 16, say, a + b + c = 16. So substitute 16 for a + b + c
in the average above: (16 + d + e)/5 = -10. Solving this equation
for d + e gives d + e = -66. Finally, dividing by 2 (to form
the average) gives (d + e)/2 = -33. Hence, the answer is (A).
IV.
Average Speed = Total Distance/Total Time
Although
the formula for average speed is simple, few people solve
these problems correctly because most fail to find both the
total distance and the total time.
Example:
In traveling from city A to city B, John drove for 1 hour
at 50 mph and for 3 hours at 60 mph. What was his average
speed for the whole trip?
(A) 50
(B) 53 1/2 (C)
55 (D) 56 (E)
57 1/2
The total
distance is 1(50) + 3(60) = 230. And the total time is 4 hours.
Hence, Average Speed = Total Distance/Total Time = 230/4 =
57 1/2. The answer is (E). Note, the answer is not the mere
average of 50 and 60. Rather the average is closer to 60 because
he traveled longer at 60 mph (3 hrs) than at 50 mph (1 hr).
RATIO
& PROPORTION
Ratio
A ratio
is simply a fraction. Both of the following notations express
the ratio of x to y: x:y, x/y. A ratio compares two numbers.
Just as you cannot compare apples and oranges, so too must
the numbers you are comparing have the same units. For example,
you cannot form the ratio of 2 feet to 4 yards because the
two numbers are expressed in different units--feet vs. yards.
It is quite common for the SAT to ask for the ratio of two
numbers with different units. Before you form any ratio, make
sure the two numbers are expressed in the same units.
Example: What is the ratio of 2 feet to 4 yards?
The ratio cannot be formed until the numbers are expressed in the
same units. Let's turn the yards into feet. Since there are 3 feet in a
yard, 4 yards = 4 x 3 feet = 12 feet. Forming the ratio yields (2 feet)/(12
feet) = 1/6 or 1:6.
Proportion
A proportion
is simply an equality between two ratios (fractions). For
example, the ratio of x to y is equal to the ratio of 3 to
2 is translated as x/y = 3/2. Two variables are directly proportional
if one is a constant multiple of the other:
y = kx,
where k is a constant.
The above
equation shows that as x increases (or decreases) so does
y. This simple concept has numerous applications in mathematics.
For example, in constant velocity problems, distance is directly
proportional to time: d = vt, where v is a constant. Note,
sometimes the word directly is suppressed.
Example:
If the ratio of y to x is equal to 3 and the sum of y
and x is 80, what is the value of y?
(A) -10
(B) -2 (C)
5 (D) 20 (E)
60
Translating
"the ratio of y to x is equal to 3" into an equation yields:
y/x = 3
Translating
"the sum of y and x is 80" into an equation yields: y + x
= 80
Solving
the first equation for y gives: y = 3x.
Substituting
this into the second equation yields
3x + x
= 80
4x = 80
x = 20
Hence,
y = 3x = 3(20) = 60. The answer is (E).
In many
word problems, as one quantity increases (decreases), another
quantity also increases (decreases). This type of problem
can be solved by setting up a direct proportion.
Example:
If Biff can shape 3 surfboards in 50 minutes, how many
surfboards can he shape in 5 hours?
(A) 16
(B) 17 (C)
18 (D) 19 (E)
20
As time
increases so does the number of shaped surfboards. Hence,
we set up a direct proportion. First, convert 5 hours into
minutes: 5 hours = 5 x 60 minutes = 300 minutes. Next, let
x be the number of surfboards shaped in 5 hours. Finally,
forming the proportion yields
3/50 =
x/300
3(300)/50 = x
18 =x
The answer
is (C).
If one
quantity increases (or decreases) while another quantity decreases
(or increases), the quantities are said to be inversely proportional.
The statement "y is inversely proportional to x" is written
as
y = k/x,
where k is a constant.
Multiplying
both sides of y = k/x by x yields
yx = k
Hence,
in an inverse proportion, the product of the two quantities
is constant. Therefore, instead of setting ratios equal, we
set products equal.
In many
word problems, as one quantity increases (decreases), another
quantity decreases (increases). This type of problem can be
solved by setting up a product of terms.
Example:
If 7 workers can assemble a car in 8 hours, how long would
it take 12 workers to assemble the same car?
(A) 3hrs
(B) 3 1/2hrs (C)
4 2/3hrs (D) 5hrs (E)
6 1/3hrs
As the
number of workers increases, the amount time required to assemble
the car decreases. Hence, we set the products of the terms
equal. Let x be the time it takes the 12 workers to assemble
the car. Forming the equation yields
7(8) =
12x
56/12 = x
4 2/3 = x
The answer
is (C).
To
summarize: if one quantity increases (decreases) as another
quantity also increases (decreases), set ratios equal. If
one quantity increases (decreases) as another quantity decreases
(increases), set products equal.
EXPONENTS
& ROOTS
Exponents
There
are five rules that govern the behavior of exponents:
Problems
involving these five rules are common on the SAT, and they
are often listed as hard problems. However, the process of
solving these problems is quite mechanical: simply apply the
five rules until they can no longer be applied.
Roots
There
are only two rules for roots that you need to know for the
SAT:
FACTORING
To factor
an algebraic expression is to rewrite it as a product of two
or more expressions, called factors. In general, any expression
on the SAT that can be factored should be factored, and any
expression that can be unfactored (multiplied out) should
be unfactored.
Distributive
Rule
The most
basic type of factoring involves the distributive rule:
ax
+ ay = a(x + y)
For example,
3h + 3k = 3(h + k), and 5xy + 45x = 5xy + 9(5x) = 5x(y + 9).
The distributive rule can be generalized to any number of
terms. For three terms, it looks like ax + ay + az = a(x +
y + z). For example, 2x + 4y + 8 = 2x + 2(2y) + 2(4) = 2(x
+ 2y + 4).
Example:
If x - y = 9, then (x - y/3) - (y - x/3) =
(A) -4
(B) -3 (C)
0 (D) 12 (E)
27
(x - y/3)
- (y - x/3) =
x - y/3 - y + x/3 =
4x/3 - 4y/3 =
4(x - y)/3 =
4(9)/3 =
12
The answer
is (D).
Difference
of Squares
One of
the most important formulas on the SAT is the difference of
squares:
Example:
If x does not equal -2, then 
(A) 2(x
- 2) (B) 2(x - 4) (C)
8(x + 2) (D) x - 2 (E)
x + 4
In most
algebraic expressions involving multiplication or division,
you won't actually multiply or divide, rather you will factor
and cancel, as in this problem.
2(x -
2)
The answer
is (A).
Perfect
Square Trinomials
Like the
difference of squares formula, perfect square trinomial formulas
are very common on the SAT.
For example, 
.
ALGEBRAIC
EXPRESSIONS
A mathematical
expression that contains a variable is called an algebraic
expression. Some examples of algebraic expressions are 3x
- 2y, 2z/y. Two algebraic expressions are called like terms
if both the variable parts and the exponents are identical.
That is, the only parts of the expressions that can differ
are the coefficients. For example, x + y and -7(x + y) are
like terms. However, x - y and 2 - y are not like terms.
Adding
& Subtracting Algebraic Expressions
Only like
terms may be added or subtracted. To add or subtract like
terms, merely add or subtract their coefficients:
You may
add or multiply algebraic expressions in any order. This is
called the commutative property:
x
+ y = y + x
xy
= yx
For example,
-2x + 5x = 5x + (-2x) = (5 - 2)x = -3x and (x - y)(-3) = (-3)(x
- y) = (-3)x - (-3)y = -3x + 3y.
Caution:
the commutative property does not apply to division or subtraction.
When adding
or multiplying algebraic expressions, you may regroup the
terms. This is called the associative property:
x
+ (y + z) = (x + y) + z
x(yz)
= (xy)z
Notice
in these formulas that the variables have not been moved,
only the way they are grouped has changed: on the left side
of the formulas the last two variables are grouped together,
and on the right side of the formulas the first two variables
are grouped together.
For example,
(x -2x) + 5x = (x + [-2x]) + 5x = x + (-2x + 5x) = x + 3x
= 4x and 24x = 2x(12x) = 2x(3x4x) = (2x3x)4x = 6x4x = 24x
Caution:
the associative property doesn't apply to division or subtraction.
Notice
in the first example that we changed the subtraction into
negative addition: (x - 2x) = (x + [- 2x]). This allowed us
to apply the associative property over addition.
Parentheses
When simplifying
expressions with nested parentheses, work from the inner most
parentheses out:
5x + (y
- (2x - 3x)) = 5x + (y - (-x)) = 5x + (y + x) = 6x + y
Sometimes
when an expression involves several pairs of parentheses,
one or more pairs are written as brackets. This makes the
expression easier to read:
2x(1 -[y
+ 2(3 - y)]) =
2x(1 -[y + 6 - 2y]) = 2x(1 -[-y + 6]) =
2x(1 + y - 6) =
2x(y - 5) =
2xy - 10x
Order
of Operations: (PEMDAS)
When simplifying
algebraic expressions, perform operations within parentheses
first and then exponents and then multiplication and then
division and then addition and then subtraction. This can
be remembered by the mnemonic:
PEMDAS
Please
Excuse My Dear Aunt Sally
GRAPHS
Questions
involving graphs rarely involve any significant calculating.
Usually, the solution is merely a matter of interpreting the
graph.
1. During
which year was the company's earnings 10 percent of its sales?
(A) 85
(B) 86 (C)
87 (D) 88 (E)
90
Reading
from the graph, we see that in 1985 the company's earnings
were $8 million and its sales were $80 million. This gives
8/80 = 1/10 = 10/100 = 10%. The answer is (A).
2. During
what two-year period did the company's earnings increase the
greatest?
(A) 85-87
(B) 86-87 (C)
86-88 (D) 87-89 (E)
88-90
Reading
from the graph, we see that the company's earnings increased
from $5 million in 1986 to $10 million in 1987, and then to
$12 million in 1988. The two-year increase from '86 to '88
was $7 million--clearly the largest on the graph. The answer
is (C).
3. During
the years 1986 through 1988, what were the average earnings
per year?
(A) 6
million (B) 7.5 million (C)
9 million (D) 10 million (E)
27 million
The graph
yields the following information:
| Year |
Earnings |
| 1986 |
$5
million |
| 1987 |
$10
million |
| 1988 |
$12
million |
Forming
the average yields (5 + 10 + 12)/3 = 27/3 = 9. The answer
is (C).
4. If
Consolidated Conglomerate's earnings are less than or equal
to 10 percent of sales during a year, then the stockholders
must take a dividend cut at the end of the year. In how many
years did the stockholders of Consolidated Conglomerate suffer
a dividend cut?
(A) None
(B) One (C)
Two (D) Three (E)
Four
Calculating
10 percent of the sales for each year yields
| Year |
10%
of Sales (millions) |
Earnings
(millions) |
| 85 |
.10
x 80 = 8 |
8 |
| 86 |
.10
x 70 = 7 |
5 |
| 87 |
.10
x 50 = 5 |
10 |
| 88 |
10
x 80 = 8 |
12 |
| 89 |
.10
x 90 = 9 |
11 |
| 90 |
.10
x 100 = 10 |
8 |
Comparing
the right columns shows that earnings were 10 percent or less
of sales in 1985, 1986, and 1990. The answer is (D).
WORD
PROBLEMS
Although
exact steps for solving word problems cannot be given, the
following guidelines will help:
(1) First,
choose a variable to stand for the least unknown quantity,
and then try to write the other unknown quantities in terms
of that variable.
For example,
suppose we are given that Sue's age is 5 years less than twice
Jane's and the sum of their ages is 16. Then Jane's age would
be the least unknown, and we let x = Jane's age. Expressing
Sue's age in terms of x gives Sue's age = 2x - 5.
(2) Second,
write an equation that involves the expressions in Step 1.
Most (though not all) word problems pivot on the fact that
two quantities in the problem are equal. Deciding which two
quantities should be set equal is usually the hardest part
in solving a word problem since it can require considerable
ingenuity to discover which expressions are equal.
For the
example above, we would get (2x - 5) + x = 16.
(3) Third,
solve the equation in Step 2 and interpret the result.
For the
example above, we would get by adding the x's: 3x - 5 = 16.
Then adding 5 to both sides gives 3x = 21. Finally, dividing
by 3 gives x = 7. Hence, Jane is 7 years old and Sue is 2x
- 5 = 2(7) - 5 = 9 years old.
Motion
Problems
Virtually,
all motion problems involve the formula Distance = Rate x
Time, or
D = R
x T
Example
: Scott starts jogging from point X to point Y. A half-hour
later his friend Garrett who jogs 1 mile per hour slower than
twice Scott's rate starts from the same point and follows
the same path. If Garrett overtakes Scott in 2 hours, how
many miles will Garrett have covered?
(A) 2
1/5 (B) 3 1/3 (C)
4 (D) 6 (E)
6 2/3
Following
Guideline 1, we let r = Scott's rate. Then 2r - 1 = Garrett's
rate. Turning to Guideline 2, we look for two quantities that
are equal to each other. When Garrett overtakes Scott, they
will have traveled the same distance. Now, from the formula
D = R x T, Scott's distance is D = r x 2 1/2 and Garrett's
distance is D = (2r - 1)2 = 4r - 2. Setting these expressions
equal to each other gives 4r - 2 = r x 2 1/2. Solving this
equation for r gives r = 4/3. Hence, Garrett will have traveled
D = 4r - 2 = 4(4/3) - 2 = 3 1/3 miles. The answer is (B).
Work
Problems
The formula
for work problems is Work = Rate x Time, or W = R x T. The
amount of work done is usually 1 unit. Hence, the formula
becomes 1 = R x T. Solving this for R gives R = 1/T.
Example
: If Johnny can mow the lawn in 30 minutes and with the
help of his brother, Bobby, they can mow the lawn 20 minutes,
how long would take Bobby working alone to mow the lawn?
(A) 1/2
hour (B) 3/4 hour (C)
1 hour (D) 3/2 hours (E)
2 hours
Let r
= 1/t be Bobby's rate. Now, the rate at which they work together
is merely the sum of their rates:
Total
Rate = Johnny's Rate + Bobby's Rate
1/20 =
1/30 + 1/t
1/20 - 1/30 = 1/t
(30 - 20)/(30)(20) = 1/t
1/60 = 1/t
t = 60
Hence,
working alone, Bobby can do the job in 1 hour. The answer
is (C).
Mixture
Problems
The key
to these problems is that the combined total of the concentrations
in the two parts must be the same as the whole mixture.
Example
: How many ounces of a solution that is 30 percent salt
must be added to a 50-ounce solution that is 10 percent salt
so that the resulting solution is 20 percent salt?
(A) 20
(B) 30 (C)
40 (D) 50 (E)
60
Let x
be the ounces of the 30 percent solution. Then 30%x is the
amount of salt in that solution. The final solution will be
50 + x ounces, and its concentration of salt will be 20%(50
+ x). The original amount of salt in the solution is 10%(50).
Now, the concentration of salt in the original solution plus
the concentration of salt in the added solution must equal
the concentration of salt in the resulting solution: 10%(50)
+ 30%x = 20%(50 + x). Multiply this equation by 100 to clear
the percent symbol and then solving for x yields x = 50. The
answer is (D).
Coin
Problems
The key
to these problems is to keep the quantity of coins distinct
from the value of the coins. An example will illustrate.
Example
: Laura has 20 coins consisting of quarters and dimes.
If she has a total of $3.05, how many dimes does she have?
(A) 3
(B) 7 (C)
10 (D) 13 (E)
16
Let D
stand for the number of dimes, and let Q stand for the number
of quarters. Since the total number of coins in 20, we get
D + Q = 20, or Q = 20 - D. Now, each dime is worth 10 cents,
so the value of the dimes is 10D. Similarly, the value of
the quarters is 25Q = 25(20 - D). Summarizing this information
in a table yields
|
Dimes |
Quarters |
Total |
| Number |
D |
20
- D |
20 |
| Value |
10D |
25(20
- D) |
305 |
Notice
that the total value entry in the table was converted from
$3.05 to 305 cents. Adding up the value of the dimes and the
quarters yields the following equation:
10D +
25(20 - D) = 305
10D + 500 - 25D = 305
-15D = -195
D = 13
Hence,
there are 13 dimes, and the answer is (D).
Age
Problems
Typically,
in these problems, we start by letting x be a person's current
age and then the person's age a years ago will be x - a and
the person's age a years in future will be x + a. An example
will illustrate.
Example
: John is 20 years older than Steve. In 10 years, Steve's
age will be half that of John's. What is Steve's age?
(A) 2
(B) 8 (C)
10 (D) 20 (E)
25
Steve's
age is the most unknown quantity. So we let x = Steve's age
and then x + 20 is John's age. Ten years from now, Steve and
John's ages will be x + 10 and x + 30, respectively. Summarizing
this information in a table yields
|
Age
now |
Age
in 10 years |
| Steve |
x |
x
+ 10 |
| John |
x
+ 20 |
x
+ 30 |
Since
"in 10 years, Steve's age will be half that of John's," we
get
(x + 30)/2
= x + 10
x + 30 = 2(x + 10)
x + 30 = 2x + 20
x = 10
Hence,
Steve is 10 years old, and the answer is (C).
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