Sat math Essentials

RATE
TIME
= 
PART OF JOB COMPLETED
Ben
3
1
0
x
=
1 sand castle
W
ylie
2
1
0
x
=
1 sand castle
Since Ben and Wylie are both working together on one sand castle, you can set the equation equal to one:
(Ben’s rate)(time) 
(Wylie’s rate)(time) 
1 sand castle
3
1
0
x
2
1
0
x
1
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P R O B L E M S O LV I N G
–
1 6 2

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Now solve by using 60 as the LCD for 30 and 20:
3
1
0
x
2
1
0
x
1
6
2
0
x
6
3
0
x
1
6
5
0
x
1
6
5
0
x
60 
1 
60
5
x
60
x
12
Thus, it will take Ben and Wylie 12 minutes to build one sand castle.
Practice Question 
Ms. Walpole can plant nine shrubs in 90 minutes. Mr. Saum can plant 12 shrubs in 144 minutes. If Ms.
Walpole and Mr. Saum work together, how many minutes will it take them to plant two shrubs?
a.
6
1
0
1
b.
10
c.
1
1
2
1
0
d.
11
e.
2
1
4
1
0
Answer
c.
Ms. Walpole can plant 9 shrubs in 90 minutes, so her rate of work is 
90
9
m
sh
i
r
n
u
u
b
t
s
es
or 
10
1
m
sh
in
ru
u
b
tes
. Mr. Saum’s
rate of work is 
14
1
4
2
m
sh
i
r
n
u
u
b
t
s
es
or 
12
1
m
sh
in
ru
u
b
tes
.
To solve this problem, making a chart will help:
RATE
TIME
=
PART OF JOB COMPLETED
Ms. 
W
alpole
1
1
0
x
=
1 shrub
Mr. Saum
1
1
2
x
=
1 shrub
Because both Ms. Walpole and Mr. Saum are working together on two shrubs, you can set the equation
equal to two:
(Ms. Walpole’s rate)(time) 
(Mr. Saum’s rate)(time) 
2 shrubs
1
1
0
x
1
1
2
x
2
Now solve by using 60 as the LCD for 10 and 12:
1
1
0
x
1
1
2
x
2
6
6
0
x
6
5
0
x
2
1
6
1
0
x
2
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P R O B L E M S O LV I N G
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1 6 3

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1
6
1
0
x
60 
2 
60
11
x
120
x
1
1
2
1
0
Thus, it will take Ms. Walpole and Mr. Saum 
1
1
2
1
0
minutes to plant two shrubs.
S p e c i a l S y m b o l s P r o b l e m s
Some SAT questions invent an operation symbol that you won’t recognize. Don’t let these symbols confuse you.
These questions simply require you to make a substitution based on information the question provides. Be sure
to pay attention to the placement of the variables and operations being performed.
Example
Given 
p
◊
q
(
p 
q
4)
2
, find the value of 2 
◊
3.
Fill in the formula with 2 replacing 
p
and 3 replacing 
q
.
(
p 
q
4)
2
(2 
3 
4)
2
(6 
4)
2
(10)
2
100
So, 2 
◊
3 
100.
Example
If
x
x
y
z
x
y
y
z
x
z
y
z
, then what is the value of
Fill in the variables according to the placement of the numbers in the triangular figure:
x
8,
y
4, and 
z
2.
8
4
8
2
8
4
4
2
8
4
2
2
1
8
4
1
4
4
1
2
4
LCD is 8.
1
8
4
2
8
8
5
8
6
Add.
9
8
8
Simplify.
4
4
9
Answer:
4
4
9
8
2
4
x
z
y
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P R O B L E M S O LV I N G
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1 6 4

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Practice Question 
The operation 
c
Ω
d
is defined by 
c
Ω
d
d
c
d 
d
c
d
. What value of
d
makes 2 
Ω
d
equal to 81?
a.
2
b.
3
c.
9
d.
20.25
e.
40.5
Answer
b.
If
c
Ω
d
d
c
d
d
c
d
, then 2 
Ω
d
d
2 
d 
d
2 
d
. Solve for 
d
when 2 
Ω
d
81:
d
2 
d 
d
2 
d
81
d
(2 
d
) 
(2 
d
)
81
d
2 
2 
d
d
81
d
4
81
d
4
81
d
2
9
d
2
9
d
3
Therefore,
d
3 when 2 
Ω
d
81.
T h e C o u n t i n g P r i n c i p l e
Some questions ask you to determine the number of outcomes possible in a given situation involving different
choices.
For example, let’s say a school is creating a new school logo. Students have to vote on one color for the back-
ground and one color for the school name. They have six colors to choose from for the background and eight col-
ors to choose from for the school name. How many possible combinations of colors are possible?
The quickest method for finding the answer is to use 
the counting principle.
Simply multiply the number
of possibilities from the first category (six background colors) by the number of possibilities from the second cat-
egory (eight school name colors):
6 
8 
48
Therefore, there are 48 possible color combinations that students have to choose from.
Remember:
When determining the number of outcomes possible when combining one out of
x
choices in
one category and one out of
y
choices in a second category, simply multiply 
x
y
.
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P R O B L E M S O LV I N G
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1 6 5

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Practice Question
At an Italian restaurant, customers can choose from one of nine different types of pasta and one of five dif-
ferent types of sauce. How many possible combinations of pasta and sauce are possible?
a.
9
5
b.
4
c.
14
d.
32
e.
45
Answer
e.
You can use the counting principle to solve this problem. The question asks you to determine the num-
ber of combinations possible when combining one out of nine types of pasta and one out of five types
of sauce. Therefore, multiply 9 
5 
45. There are 45 total combinations possible.
P e r m u t a t i o n s
Some questions ask you to determine the number of ways to arrange 
n
items in all possible groups of
r
items. For
example, you may need to determine the total number of ways to arrange the letters 
ABCD
in groups of two let-
ters. This question involves four items to be arranged in groups of two items. Another way to say this is that the
question is asking for the number of
permutations
it’s possible make of a group with two items from a group of
four items. Keep in mind when answering permutation questions that 
the order of the items matters.
In other words,
using the example, both 
AB
and 
BA
must be counted.
To solve permutation questions, you must use a special formula:
n
P
r
(
n
n
!
r
)!
P
number of permutations
n
the number of items 
r
number of items in each permutation
Let’s use the formula to answer the problem of arranging the letters 
ABCD
in groups of two letters.
the number of items (
n
) 
4
number of items in each permutation (
r
) 
2
Plug in the values into the formula:
n
P
r
(
n
n
!
r
)!
4
P
2
(4
4!
2)!
4
P
2
4
2
!
!
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4
P
2
4
3
2
2
1
1
Cancel out the 2 
1 from the numerator and denominator.
4
P
2
4 
3 
4
P
2
12
Therefore, there are 12 ways to arrange the letters 
ABCD
in groups of two:
AB
AC
AD
BA
BC
BD
CA
CB
CD
DA
DB
DC
Practice Question 
Casey has four different tickets to give away to friends for a play she is acting in. There are eight friends
who want to use the tickets. How many different ways can Casey distribute four tickets among her eight
friends?
a.
24
b.
32
c.
336
d.
1,680
e.
40,320
Answer
d.
To answer this permutation question, you must use the formula 
n
P
r
(
n
n
!
r
)!
, where 
n
the number
of friends 
8 and 
r
the number of tickets that the friends can use 
4. Plug the numbers into the
formula:
n
P
r
(
n
n
!
r
)!
8
P
4
(8
8!
4)!
8
P
4
8
4
!
!
8
P
4
Cancel out the 4 
3 
2 
1 from the numerator and denominator.
8
P
4
8 
7 
6 
5
8
P
4
1,680
Therefore, there are 1,680 permutations of friends that she can give the four different tickets to.
C o m b i n a t i o n s
Some questions ask you to determine the number of ways to arrange 
n
items in groups of
r
items without
repeated items. In other words,
the order of the items doesn’t matter.
For example, to determine the number of ways
to arrange the letters 
ABCD
in groups of two letters in which the order doesn’t matter, you would count only 
AB
,
not both 
AB
and 
BA
. These questions ask for the total number of
combinations
of items.
8 
7 
6 
5 
4 
3 
2 
1
4 
3 
2 
1