Here they are: Rate formula Chapter 9, Lesson 4: Distance or work = rate × time Average arithmetic mean formulas Chapter 9 Lesson 2: Average = Sum = average × number of things Slope for
Trang 1SAT Practice 4: Simplifying Problems
1. If 3y − 4z = 6 and 2y + z = 10, then y − 5z =
(A) 12
(B) 6
(C) 4
(D) 0
(E) −4
2. If a + b = 3, a + c = 5, and b + c = −6, then
a + b + c =
(A) 4
(B) 2
(C) 1
(D) 0
(E) −1
3. If and , what is the
value of ?
(A)
(B)
(C)
(D) 2
(E) 8
5
3
3
5
1
2
b
c
a c c
+ = 5
a b
b
+ = 3
Questions 4 and 5 pertain to the following
definition:
For all non-zero real numbers k, let .
4. Which of the following is equivalent to
^3 − ^2?
(A) (B) (C) ^1 (D) (E) ^2
5. Other than 0, what is the only value of k for which ^^k is undefined?
^6 5
∧5 6
∧1 6
∧k= −
k
1
2 3 4 5 7 8 9 6
1 0 2 3 4 5 7 8 9 6
1 0 2 3 4 5 7 8 9 6
1 0 2 3 4 5 7 8 9 6
Trang 24 D Begin by simplifying ^3 − ^2 by substitution:
But be careful not to pick (A) because
Notice that the choices must be evaluated first be-fore we can see which one equals 1/6 Notice that choice (D) is
5 1 Begin by simplifying ^^k by substitution:
Yikes! That doesn’t look simple! But think about
it: why is it that k can’t be 0? Because division
by 0 is undefined, and k is in a denominator But
notice that is also in a denominator, so can’t be 0, either!
Solving:
Add :
Multiply by k: k≠ 1 Then check by noticing that ^^1 is undefined:
^^1 = ^(1 − 1/1) = ^0 = 1 − 1/0, and 1/0 is undefined
1−1
k
1
k
1− ≠1 0
k
1−1k
1−1
k
∧∧ =∧⎛⎝⎜ − ⎞⎠⎟ = −
−
k
k
k
∧6 5 1 5 6 1 6.= − =
∧1 6 1 6= − = −5
∧1 6
∧3−∧2= −(1 1 3)− −(1 1 2)=2 3 1 2 1 6− =
Concept Review 4
1 −2n − (6 − 5n) − n = −2n − 6 + 5n − n = 2n − 6
2
3
4
5
6 Substitute y = 1 − x into 2y + x = 5 to get
2(1 − x) + x = 5
x x6 = ×x x3=x4
2
6 2
4 2
10
2
x x x
x
x x
x x
x
x x x
2
4
4 2
x x
x x
x x
2
2
x
x x
x x
x x
x x
−
−
( ) ( + )
+
( ) ( − ) = ( ( −−−1) )
Distribute: 2 − 2x + x = 5
Simplify: 2 − x = 5
Multiply by −1: x= −3
7 3 + m + n = n2+ m2
Subtract n and m: 3 = (n2− n) + (m2− m)
Substitute:
Substitute for <4>: = <(1 − 4)2>
Simplify: = <(−3)2>
Simplify: = <9>
Substitute for <9>: = (1 − 9)2
n2 n m2 m
6
3 6
1 2
−
SAT Practice 4
1 E Subtract the equations: 3y − 4z = 6
2 C Add the equations:
(a + b) + (a + c) + (b + c) = 3 + 5 + −6
Simplify: 2a + 2b + 2c = 2
Divide by 2: a + b + c = 1
3 D Start by simplifying the expressions:
Substituting into the original equations gives
Subtract 1:
Divide the fraction:
Simplify:
Simplify: b
c= 2
a c
b a
× = 2
a cc a b
=4 2
a b
a c
a
b
a c
+ =1 3 and + =1 5
a b b
a b
b b
a b
a c b
a c
c c
a c
+ = + = + + = + = +
1
1
y z
y z
Trang 3Lesson 5: Connecting to Knowledge
Know What You Need
Some SAT math questions require you to use special formulas or know the definitions of special terms Fortunately, you won’t need to memorize very many formulas (none of that trig stuff, for instance), and some of the most important ones are given to you right on the test!
Reference Information
Every SAT math section gives you this reference information Check it out and use it when you need it.
The arc of a circle measures 360°
Every straight angle measures 180°
The sum of the measures of the angles in a triangle is 180°
Memorize the Key Formulas They DON’T
Give You
It’s awfully nice of the SAT to give you those
formu-las, but those are not quite all you’ll need
Fortu-nately, we can fit the other key formulas on a single
page Here they are:
Rate formula (Chapter 9, Lesson 4):
Distance (or work) = rate × time
Average (arithmetic mean) formulas (Chapter 9
Lesson 2):
Average =
Sum = average × number of things
Slope formula (Chapter 10, Lesson 4):
Slope =
Midpoint formula (Chapter 10, Lesson 4):
Midpoint =
Percent change formula (Chapter 7, Lesson 5):
Percent change = final starting
starting
x1 x2 y1 y2
⎛
⎝⎜
⎞
⎠⎟
,
rise run
y y
x x
= 2−− 1
2 1
sum number of things
Memorize the Key Definitions
You’ll also want to memorize the definitions of some key terms that show up often:
Mode = the number that appears the most
fre-quently in a set Remember that mode and most both begin with mo (Chapter 9, Lesson 2).
Median = the “middle number” of a set of
numbers when they are listed in order If there are an even number of numbers, the median
is the average of the two middle numbers
(Chapter 9, Lesson 2)
Remainder = the whole number left over when
one whole number has been divided into an-other whole number a whole number of times
(Chapter 7, Lesson 7)
Absolute value = the distance a number is from
0 on the number line (Chapter 8, Lesson 6)
Prime number = an integer greater than 1 that
is divisible only by itself and 1 (Chapter 7,
Lesson 7)
Factor = a number or expression that is part of a product (Product = result of a multiplication.)
r
A = πr2
C = 2 πr
ᐉ
w
h
w h
r
a
2x
x
s
3
2
30°
60°
45 °
45 °
Trang 4Concept Review 5: Connecting to Knowledge
Write out each formula, theorem, definition, or property
1 The Pythagorean theorem
2 The zero product property
3 The parallel lines theorem
4 The rate formula
5 The average (arithmetic mean) formula
6 The definition of the median
7 The definition of the mode
8 The circumference formula
9 The circle area formula
10 The triangle area formula
Trang 5SAT Practice 5: Connecting to Knowledge
1. If x is the average (arithmetic mean) of k and 10,
and y is the average (arithmetic mean) of k and
4, what is the average of x and y, in terms of k?
(D) 7k (E) 14k
2. If, on average, x cars pass a certain point on a
highway in y hours, then, at this rate, how many
cars should be expected to pass the same point
in z hours?
Note: Figure not drawn to scale
3. A straight 8-foot board is resting on a rectangular
box that is 3 feet high, as shown in the diagram
above Both the box and the board are resting on a
horizontal surface, and one end of the board rests
on the ground 4 feet from the edge of the box If h
represents the height, in feet, of the other end of
the board from the top of the box, what is h?
x yz
xz
y
z xy
xy z
k+ 7 2
k+14 2
k+14
4
Key formula(s):
Key formula(s):
Key formula(s):
3 feet
4 feet
8 feet
h feet
.
1
2
3
4
5
7
8
6
1
0
2
3
4
5
7
8
6
1
0
2
3
4
5
7
8
6
1 0 2 3 4 5 7 8 6
Trang 63 1.8 Key formula: The Pythagorean theorem:
c2= a2+ b2 Key theorem: In similar triangles, cor-responding sides are proportional Notice that the figure has two right triangles, and they are similar The hypotenuse of the bottom triangle is 5 because
32+ 42 = 52 Therefore, the hypotenuse of the top triangle is 8 − 5 = 3 Since the two triangles are similar, the corresponding sides are proportional:
Cross-multiply: 5h = 9
Divide by 5: h = 1.8
3
5=3h
Concept Review 5
1 The Pythagorean theorem: In a right triangle, if
c is the length of the hypotenuse and a and b are
the lengths of the two legs, then c2 = a2 + b2
(Chapter 10, Lesson 3)
2 The zero product property: If a set of numbers has
a product of zero, then at least one of the numbers
is zero Conversely, if zero is multiplied by any
number, the result is zero (Chapter 8, Lesson 5)
3 The parallel lines theorem: If a line cuts through
two parallel lines, then all acute angles formed
are congruent, all obtuse angles formed are
con-gruent, and any acute angle is supplementary to
any obtuse angle (Chapter 10, Lesson 1)
4 The rate formula: distance (or work) = rate × time
(Chapter 9, Lesson 4)
5 The average (arithmetic mean) formula:
Average = sum ÷ number of things (Chapter 9,
Lesson 2)
6 The definition of the median: The “middle num-ber” of a set of numbers when they are listed in order If there are an odd number of numbers, the median is the “middle number,” and if there are
an even number of numbers, it is the average of the two middle numbers (Chapter 9, Lesson 2)
7 The definition of the mode: The number that ap-pears the most frequently in a set (Chapter 9, Lesson 2)
8 The circumference formula: Circumference = 2πr
(Chapter 10, Lesson 5)
9 The circle area formula: Area = πr2(Chapter 10, Lesson 5)
10 The triangle area formula: Area = base × height/2
(Chapter 10, Lesson 5)
SAT Practice 5
1 C Key formula: Average = sum ÷ number of things.
So if x is the average of k and 10, then
And if y is the average of k and 4, then y=
The average of x and y, then, is
2 D Key formulas: Number of cars = rate × time, and
Since x is the number of cars and y is the time in hours, the rate is x/y cars per
hour Using the first formula, then, the number of
cars that would pass in z hours is
You should notice, too, that simply plugging in
values for x, y, and z can make the problem easier
to think about Say, for instance, that x= 10 cars
pass every y = 2 hours In z = 4 hours, then, it
should be clear that 20 cars should pass by
Plug-ging these numbers into the choices, you will see
that (D) is the only one that gives an answer of 20
x y z xz
y
cars per hour hours
rate number of cars
time
k k
k k k k
10
2
4 2
2
10 4
4 4
4
7 2
k+ 4 2
x= +k210
3 feet
4 feet
8 feet
h feet
3 feet
5 feet
Trang 7Keep Your Options Open
There are often many good ways to solve an
SAT math problem Consider different
strate-gies This gives you a way to check your work
If two different methods give you the same
an-swer, you’re probably right!
Numerical Analysis—Plugging In
Let’s come back to the problem we saw in Lesson 4:
If 3x2+ 5x + y = 8 and x≠ 0, then what is the value
of ?
Back in Lesson 4 we solved this using substitution,
an algebraic method Now we’ll use a numerical method.
Notice that the equation contains two unknowns This
means that we can probably find solutions by guessing
and checking Notice that the equation works if x = 0 and
y = 8 But—darn it—the problem says x≠ 0! No worries—
notice that x = 1 and y = 0 also work (Check and see.)
Now all we have to do is plug those numbers in for x
whole different approach!
“Plugging in” works in two common situations:
when you have more unknowns than equations
and when the answer choices contain
un-knowns Always check that your numbers
sat-isfy the conditions of the problem Then solve
the problem numerically, and write down the
answer If the answer choices contain
un-knowns, plug the values into every choice and
eliminate those that don’t give the right answer
If more than one choice gives the right answer,
plug in again with different numbers
If 3m = mn + 1, then what is the value of m in terms
of n?
(A) n + 1
(B) n− 2
(C)
(D)
(E)
Because the choices contain unknowns, you can
plug in Pick a simple number for m to start, such as 1.
Plugging into the equation gives 3 = n + 1, which has the
2
3+ n
1
3+ n
1
3− n
16 2
16 2 0
16
−
−
y
x x
( )
16 2
−
+
y
x x
solution n = 2 Now notice that the question asks for m,
which is 1 Write that down and circle it Now substitute
2 for n in the choices and see what you get:
(A) 3 (B) 0 (C) 1 (D) 1/5 (E) 2/5 Only (C) gives the right answer
Algebraic Analysis
You can also solve the problem above algebraically:
3m = mn + 1 Subtract mn: 3m − mn = 1
Factor: m(3 − n) = 1
Divide by (3 − n): m =
Testing the Choices
Some SAT math questions can be solved just by
“testing” the choices Since numerical choices are usually given in order, start by testing choice (C) If (C) is too big, then (D) and (E) are too big, also, leaving you with just (A) and (B) This means that you have only one more test to
do, at most, until you find the answer
If 3(2)n+1 – 3(2)n = 24, what is the value of n?
(A) 2 (B) 3 (C) 4 (D) 5 (E) 6
Here you can take an algebraic or a numerical ap-proach That is, you can solve the equation for n or you
can “test” the choices to see if they work For this les-son, we’ll try the “testing” strategy Since the choices are given in ascending order, we’ll start with the
mid-dle number, (4) Substituting 4 for n gives us 3(2)5– 3(2)4on the left side, which equals 48, not 24 (It’s okay
to use your calculator!) Since that doesn’t work, we
can eliminate choice (C) But since it’s clearly too big,
we can also rule out choices (D) and (E) That’s why
we start with (C)—even if it doesn’t work, we still narrow down our choices as much as possible Now just test ei-ther (A) or (B) Notice that (B) gives us 3(2)4– 3(2)3, which equals 24, the right answer
Now try solving the problem algebraically, and see
if it’s any easier!
1
3− n
Trang 8Concept Review 6: Finding Alternatives
1 When can a multiple-choice problem be solved by just “testing the choices”?
2 When solving by testing the choices, why is it often best to start with choice (C)?
3 When testing the choices, when is it not necessarily best to start with choice (C)?
4 When can you simplify a multiple-choice question by plugging in values?
5 What are the four steps to solving by plugging in values?
6 Why is it best to understand more than one way to solve a problem?
Trang 9SAT Practice 6: Finding Alternatives
Try to find at least two different ways of solving each
of the following problems, and check that both
meth-ods give you the same answer
1. If m = 2x − 5 and n = x + 7, which of the
follow-ing expresses x in terms of m and n?
(A)
(B) m − n+2
(C)
(D) m − n+12
(E) 2(m − n+12)
2. Three squares have sides with lengths a, b, and c.
If b is 20% greater than a and c is 25% greater
than b, then by what percent is the area of the
largest square greater than the area of the
small-est square?
(A) 20%
(B) 50%
(C) 75%
(D) 125%
(E) 225%
3. Jim and Ellen together weigh 290 pounds Ellen
and Ria together weigh 230 pounds All three
to-gether weigh 400 pounds What is Ellen’s
weight?
(A) 110 lbs
(B) 120 lbs
(C) 130 lbs
(D) 140 lbs
(E) 170 lbs
m n− +12
2
m n− + 2
2
4. A painter used one-fourth of her paint on one room and one-third of her paint on a second room If she had 10 gallons of paint left after painting the second room, how many gallons did she have when she began?
(A) 19 (B) 24 (C) 28 (D) 30 (E) 50
5. If and 4r = 7t, what is the value of s in terms of t?
(A) 35t
(B) (C) 35t− 4 (D) 31t
(E) 70t
35 4
t
r=5s
Trang 10test choice (B) next If Ellen weighs 120 pounds, then Jim weighs 290 − 120 = 170 pounds and Ria weighs 230 − 120 = 110 pounds In total, they weigh 120 + 170 + 110 = 400 pounds Bingo! The answer is (B)
Method 2: Use algebra Let e= Ellen’s weight,
j = Jim’s weight, and r = Ria’s weight Translate
the problem into equations: e + j = 290
e + r = 230
e + j + r = 400
Add first two equations: 2e + j + r = 520
Subtract third equation: −(e + j + r = 400)
e= 120
4 Method 1: Use algebra Let x be the number of
gal-lons of paint that she starts with Translate the
problem into an equation: x − (1/4)x − (1/3)x = 10
Multiply by 12/5: x= 24 The answer is (B)
Method 2: Test the choices Look at the problem carefully She uses one-fourth of her paint, then one-third of her paint, and is left with 10 gallons
of paint, a whole number This suggests that she
started with a quantity that is divisible by 3 and 4 Since 24 is divisible by 3 and 4, it’s a good choice
to test One-fourth of 24 is 6, and one-third of 24
is 8 This means she would be left with
24 − 6 − 8 = 10 Bingo!
5 Method 1: Plug in Let s = 35, so r = 35/5 = 7 and
t= 4 (Check that they “fit.”) Since the question
asks for the value of s, write down 35 and circle it.
Plugging these values into the choices gives (A) 140 (B) 35 (C) 136 (D) 124 (E) 280 The answer is (B)
Method 2: Use algebra Solve the first equation
for s: s = 5r Then solve the second equation for r:
r = (7/4)t Then substitute: s = 5(7/4)t = 35t/4.
Concept Review 6
1 There are many situations in which this is
possi-ble, but perhaps the most common is where you’re
asked to find the solution of an equation, and the
choices are ordinary numbers
2 Because the answer choices are usually presented
in numerical order If choice (C) doesn’t work, you
may be able to tell whether it is too big or too
small, and thereby eliminate two other answers as
well This way, you will only need to “test” one
more choice to get the answer
3 When it is not easy to tell whether the choice is
“too big or too small,” or when there is no pattern
to the choices
4 Usually, when the answer choices contain un-knowns or represent ratios of unun-knowns; also when the problem contains more unknowns than equations
5 (1) Check that the values satisfy any given equa-tions or other restricequa-tions, (2) write down the values you are plugging in for each unknown,
(3) solve the problem numerically and write down this number, and (4) plug in the values to every
choice and eliminate those that don’t give the right answer If more than one choice gives the right answer, plug in different numbers
6 Because the two methods can provide a “check” against one another: if they both give the same an-swer, you are almost certainly right!
SAT Practice 6
1 Method 1: The problem asks you to solve for x in
terms of m and n Notice that every choice
con-tains the expression m − n.
By substitution: m − n = (2x − 5) − (x + 7)
Simplify: m − n = x − 12
Add 12: m − n + 12 = x
So the answer is (D)
Method 2: Just plug in simple values for the
un-knowns If x = 1, then m = (2)(1) − 5 = −3 and n =
(1) + 7 = 8 Since the problem asks for x, write
down its value, 1, and circle it Then plug in m=
−3 and n = 8 to every choice, and simplify:
(A) −4.5 (B) −9 (C) 0.5 (D) 1 (E) 2
So the answer is (D)
2 Method 1: Plug in numbers Let a be 100 If b is 20%
greater than a, then b = 120 If c is 25% greater than
b, then c= 150 The area of the largest square, then,
is (150)2 = 22,500, and the area of the smallest
square is (100)2= 10,000 The percent difference is
(22,500 − 10,000)/10,000 = 1.25 = 125% (D)
Method 2: Use algebra b = 1.2a and
c = (1.25)(1.2a) = 1.5a So the area of the smallest
square is a2and the area of the largest square is
(1.5a)2= 2.25a2 Since 2.25a2− a2= 1.25a2, the area
of the bigger square is 125% larger
3 Method 1: Test the choices, starting with (C)
If Ellen weighs 130 pounds, then Jim weighs
290 − 130 = 160 pounds and Ria weighs
230 − 130 = 100 pounds All together, their weight
would be 130 + 160 + 100 = 390 pounds Close,
but too small This means that our guess for
Ellen’s weight is too big (because increasing
Ellen’s weight decreases both Jim and Ria’s
weight by the same amount, for a net decrease).
This lets us eliminate choices (C), (D), and (E)
Since our guess wasn’t far off, it makes sense to