If there are four times as many blue buttons as green buttons and six times as many four-holed buttons as two-holed buttons, what is the least number of buttons that could be in the box.
Trang 1Using Venn Diagrams to Keep Track of Sets
Some counting problems involve “overlapping sets,” that is, sets that contain elements that also belong in other sets In these situations, Venn diagrams are very helpful for keeping track of things
Example:
A class of 29 students sponsored two field trips: one
to a zoo and one to a museum Every student at-tended at least one of the field trips, and 10 stu-dents attended both If twice as many stustu-dents went to the zoo as went to the museum, how many students went to the zoo?
Set up a Venn diagram of the situation We repre-sent the two sets—those who went to the museum and those who went to the zoo—as two overlapping circles, because some students went to both Notice that there are three regions to consider We know that ten students are in the overlapping region, but we don’t know how many are
in the other two regions,
so let’s use algebra Let’s
say that x students are in
the first region, represent-ing those who went to the museum but not to the
zoo This means that x+
10 students must have gone to the museum altogether Since twice as many students went to the zoo, the total number in the zoo
circle must be 2(x + 10) = 2x + 20 Since 10 of these are
already accounted for in the overlapping region, there
must be 2x + 20 − 10 = 2x + 10 in the third region So
now the diagram should look like this:
The total number of students is 29, so
(x) + (10) + (2x + 10) = 29
Simplify: 3x+ 20 = 29
So the number of students who went to the zoo is 2(3) + 20 = 26
The Fundamental Counting Principle
Some SAT questions ask you to count things
Some-times it’s easy enough to just write out the things in a
list and count them by hand Other times, though,
there will be too many, and it will help to use the
Fun-damental Counting Principle
Lesson 5: Counting Problems
To use the Fundamental Counting Principle
(FCP), you have to think of the things you’re
counting as coming from a sequence of
choices The Fundamental Counting Principle
says that the number of ways an event can
hap-pen is equal to the product of the choices that
must be made to “build” the event
Example:
How many ways can five people be arranged in
a line?
You might consider calling the five people A, B, C, D,
and E, and listing the number of arrangements After a
while, though, you’ll see that this is going to take a lot
of time, because there are a lot of possibilities (Not to
mention that it’s really easy to miss some of them.)
In-stead, think of “building” the line with a sequence of
choices: first pick the first person, then pick the second
person, etc There are five choices to make, so we’ll
have to multiply five numbers Clearly, there are
five people to choose from for the first person in line
Once you do this, though, there are only four people left
for the second spot, then three for the third spot, etc By
the Fundamental Counting Principle, then, the number
of possible arrangements is 5 × 4 × 3 × 2 × 1 = 120
Example:
How many odd integers greater than 500 and less
than 1,000 have an even digit in the tens place?
This seems a lot harder than it is Again, think of
“building” the numbers in question All integers
between 500 and 1,000 have three digits, so building
the number involves choosing three digits, so we will
multiply three numbers to get our answer If each
number is between 500 and 1,000, then there are only
five choices for the first digit: 5, 6, 7, 8, or 9 If the tens
digit must be even, we have five choices again: 2, 4, 6,
8, or 0 If the entire number is odd, then we have five
choices for the last digit as well: 1, 3, 5, 7, or 9
Therefore, the total number of such integers is
5 × 5 × 5 = 125
To the Museum
To the Zoo
Museum
(but not Zoo)
Zoo
(but not Museum)
Both
10
x 2x + 10
Trang 21 What is the fundamental counting principle?
2 How many different four-letter arrangements of the letters LMNO can be made if no letter can be repeated? Answer this first by listing all of the possible arrangements, then by using the Fundamental Counting Prin-ciple, and check that the two answers agree
3 If the first digit of a 3-digit area code cannot be 0 and the second digit is either 0 or 1, then how many dif-ferent area codes are possible?
4 A baseball team has six players, each of whom can play in any of the three outfield positions: left field, cen-ter field, and right field How many possible different arrangements of these players can the team place in the outfield? (This one is a bit harder to do by listing!)
5 Among a set of 40 sophomores, 20 students take French and 27 students take Spanish If all of the students
take either French or Spanish, how many students take both French and Spanish?
6 A box contains buttons, each of which is either blue or green and has either two or four holes If there are four times as many blue buttons as green buttons and six times as many four-holed buttons as two-holed
buttons, what is the least number of buttons that could be in the box?
Concept Review 5: Counting Problems
Trang 3SAT Practice 5: Counting Problems
1. A pizzeria offers three different sizes of pizza,
two different kinds of crust, and eight different
choices for toppings How many different
one-topping pizzas are there to choose from?
(A) 13 (B) 16 (C) 24
(D) 48 (E) 60
0, 2, 4, 6, 8
2. How many different integers between 30 and
70 contain only digits from the list above?
(A) 7 (B) 10 (C) 15
(D) 20 (E) 25
3. In how many ways can you arrange four different
paintings in a line on a wall?
(A) 12 (B) 24 (C) 36
(D) 48 (E) 64
4. At Lincoln County High School, 36 students are
taking either calculus or physics or both, and 10
students are taking both calculus and physics
If there are 31 students in the calculus class,
how many students are there in the physics
class?
(A) 5 (B) 8 (C) 11
(D) 15 (E) 21
5. Dave’s stickball team has six players How
many different six-player batting lineups can
they make if Dave must bat second and either
Zack or Paul must bat first?
(A) 48 (B) 96 (C) 192
(D) 256 (E) 720
6. Maria gave David x cards, gave Tina two more
cards than she gave David, and gave Samuel five
fewer cards than she gave Tina In terms of x,
how many cards did Maria give Tina, David, and Samuel all together?
(A) 3x+ 9 (B) 3x− 1 (C) 3x− 3 (D) x− 3 (E) x− 1
7. From a collection of six paintings, four are to be chosen to hang on a wall How many different arrangements are possible if every painting is different?
(A) 24 (B) 120 (C) 360 (D) 720 (E) 1,296
8. Every marble in a jar has either a dot, a stripe,
or both The ratio of striped marbles to non-striped marbles is 3:1, and the ratio of dotted marbles to nondotted marbles is 2:3 If six mar-bles have both a dot and a stripe, how many marbles are there all together?
(A) 16 (B) 18 (C) 20 (D) 36 (E) 40
9. An ant must walk from one vertex of a cube to the “opposite” vertex (that is, the vertex that is farthest from the starting vertex) and back again to its starting position
It may only walk along the edges of the cube For the entire trip, its path must tra-verse exactly six edges, and
it may travel on the same
edge twice How many dif-ferent six-edge paths can the ant choose from?
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 4SAT Practice 5
1 D Use the FCP: 3 × 2 × 8 = 48
2 B “Build” the number: If it’s between 30 and 70,
it must be a two-digit number that begins with 4
or 6 That’s two choices The second digit can be
anything in the list, so that’s 5 choices 2 × 5 = 10
3 B Since there are four spaces, there are four
deci-sions to make, so four numbers to multiply You can
choose from four paintings for the first spot, then three
paintings for the second spot, etc 4 × 3 × 2 × 1 = 24
4 D If there are 31 students in calculus but 10 of
these are also taking physics, then 31 − 10 = 21
dents are taking only calculus If there are 36
stu-dents taking either physics or calculus, but only 31
are taking calculus, then 36 − 31 = 5 students are
taking only physics Therefore, the Venn diagram
should look like this:
As you can see, 5 + 10 = 15
students are taking physics
5 A “Build” the lineup You have six spots to fill,
and thus six decisions to make and six numbers to
multiply You only have two choices for the first
spot (Zack or Paul) and one choice for the second
spot (Dave), then you have four players left to fill
the other slots, so you have four choices for the third spot, then three for the fourth spot, etc
2 × 1 × 4 × 3 × 2 × 1 = 48
6 B David gets x cards Tina gets two more cards than David, which is x+ 2 Samuel gets five fewer
cards than Tina, which is x + 2 − 5 = x − 3 So all together, x + x + 2 + x − 3 = 3x − 1.
7 C You have six choices for the first spot, then five for the second, then four for the third and three for the fourth 6 × 5 × 4 × 3 = 360
8 E Set up the Venn diagram: Since the ratio of striped marbles
to nonstriped marbles is 3:1, x+ 6 =
3y, and since the ratio of dotted marbles to nondotted marbles is 2:3, y + 6 = 2/3x and therefore y = 2/3x − 6 Substituting, we get x +
6 = 3(2/3x − 6) or x + 6 = 2x − 18, so x = 24 Plug this back in to get y = 2/3(24) − 6 = 10 Total =
24 + 6 + 10 = 40
9 36 Draw the cube To get from any vertex to its
opposite vertex, the ant has 3 × 2 × 1 possible paths
To see why, trace a path and notice it has three choices for the first edge, then two for the second, then only one option for the third Since it must return to the opposite vertex, it has 3 × 2 × 1 differ-ent paths it can take back 3 × 2 × 1 × 3 × 2 × 1 = 36
Concept Review 5
1 The number of ways an event can happen is equal
to the product of the choices that must be made
to “build” the event
2 Try listing all the “words” that start with L, then
all that start with M, and so on:
LMNO MLNO NLMO OLMN
LMON MLON NLOM OLNM
LNMO MNLO NMLO OMLN
LNOM MNOL NMOL OMNL
LOMN MOLN NOLM ONML
LONM MONL NOML ONLM
Whew! 24 in total Annoying, but not impossible
Using the FCP makes it a lot easier: 4 × 3 × 2 × 1 =
24 That’s it!
3 There are too many possibilities to list, but the
FCP makes it easy: We have 9 choices for the first
digit, 2 choices for the second digit, and 10
choices for the last digit, and 9 × 2 × 10 = 180
4 “Build” the outfield from left to right You have
6 players to choose from for left field, but then just
5 for center field and 4 for right field 6 × 5 × 4 = 120
5 Since 20 + 27 − 40 = 7, there must be 7 students who take both This Venn diagram shows how it works out:
6 This one’s tough Say there are g green buttons If
there are four times as many blue buttons as
green buttons, then there are 4g blue buttons, so
g + 4g = 5g buttons in total So the total number of
buttons must be a multiple of 5 Similarly, if there
are n two-holed buttons, there must be 6n four-holed buttons, so the total number of buttons is n + 6n = 7n, so the total number of buttons is also a
multiple of 7 The least common multiple of 5 and
7 is 35, so there are 35 buttons: 5 two-holed and
30 four-holed, and 7 green and 28 blue
7
13 20
F S
Answer Key 5: Counting Problems
10
5 21
P C
6
D S
Trang 5Probability Geometrical Probability
Lesson 6: Probability Problems
A probability is a number between 0 and 1 that
represents the likelihood of an event An event
with a probability of 0 is impossible, and an
event with a probability of 1 is certain Most
probabilities, of course, are somewhere in
be-tween 0 and 1 For instance, the probability of
rolling a 5 on a fair die is 1⁄6 It’s best to think of
a probability as a part-to-whole ratio There
are six possible outcomes when you roll a die
(the whole), but only one of them is 5 (the
part) Thus, the probability of rolling a 5 is 1⁄6
Example:
What is the probability of rolling a sum of 5 on
two dice?
Here is a table showing all the possible sums on a roll
of two dice:
Clearly, there are four ways of getting a sum of 5 out
of a possible 36, so the probability is 4⁄36, or 1⁄9
Die 1
Die 2
1
4
2
3
6
5
An SAT question may ask you to find the prob-ability that something hits a certain region, like a dart hitting a dartboard In these situa-tions, the probability is just the ratio of the par-ticular area to the entire area
Example:
A landing target for skydivers consists of two con-centric circles The smaller circle has a radius of
3 meters, and the larger one has a radius of 6 meters
If a skydiver hits the target, what is the probability that she hits the smaller circle?
It might help to sketch the target:
If she hits the target, then she hits an area that is π(6)2= 36π square meters in area The smaller circle, though, is only π(3)2= 9π square meters in area, so the probability that she lands within the smaller region should be just 9π/36π = 1/4
3 6
Trang 61 The probability of an impossible event is .
2 The probability of an event that is certain is
3 If a jar contains 3 red marbles, 4 white marbles, and 5 blue marbles, then what is the probability of ran-domly choosing
a red marble?
a white marble?
a blue marble?
4 A jar contains 5 red marbles and 10 white marbles
What is the probability of drawing a red marble?
If 3 more red marbles are added, then what is the probability of drawing a red marble?
5 A jar contains 24 red and blue marbles If the probability of selecting a red marble at random is 3⁄8, then how many red marbles must be added so that the probability of randomly selecting a red marble becomes 1⁄2?
6 A jar contains only black, white, and red marbles The probability of choosing a white marble is 2⁄3 If there are 4 times as many red marbles as black marbles, what is the least possible number of marbles in the jar?
Concept Review 6: Probability Problems
Trang 7SAT Practice 6: Probability Problems
1. The figure above shows a spinner in the
mid-dle of a disc divided into six equal parts, each
labeled with a number What is the
probabil-ity that the spinner will land on a number that
is either even or greater than 5?
(A) (B) (C)
(D) (E)
2. A jar contains 10 blue marbles, 8 green marbles,
and 14 red marbles How many green marbles
must be added so that the probability
of choosing a green marble at random is ?
(A) 16 (B) 32 (C) 40
(D) 64 (E) 72
3. A fair six-sided die has faces bearing the numbers
1, 2, 3, 4, 5, and 6 When the die is thrown, the
numbers on the five visible faces are added What
is the probability that this sum is greater than 18?
(A) (B) (C)
(D) (E)
4. A target consists of three concentric circles,
with radii of 1 meter, 2 meters, and 3 meters
If an arrow that hits the target hits any point
on the target with equal probability, what is
the probability that an arrow that hits the
tar-get falls in the outermost region (between the
second and third circles)?
(A) (B) (C)
(D) (E) 5
9
4
9
π 9
1 3
1
9
5 6
2
3
1 2
1 3
1
6
3 4
5 6
2
3
1 2
1 3
1
6
5. The probability of a meteor shower occurring in the skies above a particular island on any given night is Independently, the proba-bility that any given night will be cloudless is What is the probability that, on any given
night, there will be a meteor shower and it will
be cloudless?
(A) (B) (C)
(D) (E)
6. A basket contains red, green, and yellow balls, all of equal size The probability of choosing a green ball at random is If there are 3 times
as many red balls as yellow balls, what is the probability of choosing a yellow ball at random?
7. A certain disease occurs in 1 person out of every 101 people A test for the disease is 100% accurate for patients with the disease and 99% accurate for patients without it That is, it gives a “false positive” 1% of the time even if the person tested doesn’t have the disease If you take this test and it returns a positive re-sult, what is the probability that you have the disease?
(A) 1 (B) 99 (C) 95 (D) 50 (E) 01
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
4 7
8 25
4 25
17 200
3 100
1 50
1 4
2 25
5 9 2
8
3 6
Trang 8SAT Practice 6
1 D Put an “x” through any number that is either
even or greater than 5, or both This gives us 8, 9,
2, and 6, which is 4 out of the 6 spaces, giving a
probability of 4⁄6, or 2⁄3
2 D If the probability of choosing a green marble is
to be 3⁄4, then 3⁄4of the marbles should be green and
1⁄4not green There are 10 blue and 14 red, for a total
of 24 “not green” marbles, and this will not change,
since you are adding only green marbles If this is 1⁄4
of the total, then there must be 4(24) = 96 marbles
total after you add the extra green marbles The jar
now contains 10 + 8 + 14 = 32 marbles, so you must
add 96 − 32 = 64 green marbles
3 B The six sides of a die add up to 1 + 2 + 3 + 4 +
5 + 6 = 21 The sum of any five faces can be greater than
18 only if the “down” face is 1 or 2 (so that the sum of
the other faces is either 21 − 1 = 20 or 21 − 2 = 19 This
is 2 possibilities out of 6 for a probability of 2⁄6, or 1⁄3
4 E Sketch the target:
You want to know the
proba-bility of the arrow hitting the
outermost ring, which is the
ratio of the area of the ring to
the entire area of the target
The area of the whole target is π(3)2= 9π The area
of the outermost ring is 9π − π(2)2(subtract the area
of the middle circle from the area of the big circle)
= 9π − 4π = 5π So the probability is 5π/9π = 5/9
5 A Consider a stretch of 100 consecutive nights
If the probability of a meteor shower is 2⁄25, then we should expect a meteor shower on (2⁄25)(100) = 8 of those nights If only 1⁄4of the nights are cloudless, though, then (1⁄4)(8) = 2 of the nights with a meteor shower, on average, should be cloudless This gives a probability of 2⁄100, or 1⁄50 Mathematically,
we can just multiply the two probabilities (as long
as they are independent) to get the joint
probabil-ity: (2⁄25)(1⁄4) =1⁄50
6 3 ⁄ 28 Call the probability of choosing a yellow ball x.
If there are three times as many red balls as yellow balls, the probability of choosing a red ball must
be 3x The probability of choosing a green ball is
4⁄7 These probabilities must have a sum of 1:
x + 3x +4⁄7= 1 Simplify: 4x+4⁄7= 1 Subtract 4⁄7: 4x=3⁄7
Divide by 4: x=3⁄28
7 D Most people would say that this probability is quite high, because the test is so reliable But intu-ition is often wrong Imagine that you test 101 peo-ple Of these, on average, one will have the disease, and 100 will not Since the test is 100% accurate for those who have the disease, that person will test positive Of the 100 who do not have the disease,
99 will test negative, but one will test positive, be-cause of the 1% “false positive” rate So of those two who test positive, only one will have the dis-ease; thus, the probability is 1⁄
Concept Review 6
1 0
2 1
3 red marble: 3⁄12, or 1⁄4
white marble: 4⁄12, or 1⁄3
blue marble: 5⁄12
4 What is the probability of drawing a red marble?
5⁄15, or 1⁄3
If 3 more red marbles are added, what is the
prob-ability of drawing a red marble? 8⁄18, or 4⁄9
5 If the jar contains 24 red and blue marbles and the
probability of selecting a red marble at random is 3⁄8,
there must be (3⁄8)(24) = 9 red marbles, and 24 − 9 = 15
blue marbles If the probability of drawing a red
marble is to be 1⁄2, there must be as many red as blue
marbles, so you must add 15 − 9 = 6 marbles
6 Let’s say that the probability of drawing a black
marble is x Since there are 4 times as many red
marbles as black marbles, the probability of
drawing a red marble must be 4x The
probabil-ity of choosing a white marble is 2⁄3 Since we are certain to pick one of these colors, the sum of all
of these probabilities must be 1: x + 4x +2/3 = 1 Simplify: 5x+2⁄3= 1 Subtract 2⁄3: 5x=1⁄3
Divide by 5: x=1⁄15
Therefore, 1⁄15of the marbles are black, 4⁄15of the marbles are red, and 2⁄3of the marbles are white The least common denominator of these fractions
is 15, which means that 15 is the least possible number of marbles In that case, there are 1 black marble, 4 red marbles, and 10 white marbles
Answer Key 6: Probability Problems
1 2 3
Trang 9CHAPTER 10
ESSENTIAL GEOMETRY SKILLS
1 Lines and Angles
2 Triangles
3 The Pythagorean Theorem
4 Coordinate Geometry
5 Areas and Perimeters
6 Similar Figures
7 Volumes and 3-D Geometry
8 Circles
358
✓
Copyright © 2008 by The McGraw-Hill Companies, Inc Click here for terms of use
Trang 10When Two Lines Cross
When two lines cross, four angles are formed
“Vertical” angles are equal, and look like this:
Don’t be fooled by diagrams that look like
ver-tical angles, but aren’t Verver-tical angles are
formed by two and only two crossed lines
When two pairs of vertical angles are formed,
four pairs of adjacent angles (side-by-side) are
also formed Adjacent angles add up to 180°:
When a Line Crosses Parallel Lines
Imagine taking two crossed lines, making a “copy” of
them, and sliding the copy down one of the lines so
that together they look like this:
This produces a pair of parallel lines crossed by a
third line
"slide"
vertical not vertical
vertical angles
When two parallel lines are crossed by another line, all acute angles are equal, and all obtuse angles are equal Also, every acute angle is sup-plementary to every obtuse angle (that is, they add upto 180
To show that two lines are parallel, use the arrow marks “>” like those in the figure in the previous column To show that two angles are equal, use the arc marks “)” like those in the figure in the previous column
Don’t be fooled by diagrams that only look as
if they have two parallel lines crossed by
an-other line Don’t assume that two lines are par-allel just because they look parpar-allel It must be
given that they are parallel.
To help yourself to see the relationships be-tween angles in parallel line systems, you might try looking for these special “letters”: Angles that make Z’s are equal:
Angles that make C’s or U’s are supplementary (they have a sum of 180°):
Angles that make F’s are equal:
x°
x°
y° y°
b°
a° a + b = 180°
c° d°
c + d = 180°
a°
a°
Lesson 1: Lines and Angles
adjacent angles