### 1. Sets

A set is a group of distinct objects, called elements or members. Sets are used to look at the descriptions and interactions of the elements.

Elements are typically listed in {brackets}. The best way to visualize a set is with a Venn diagram. Use overlapping circles to organize the elements of a set.

The set Animals contains the subset Mammals.

In the set of Mammals, Dogs and Cats are subsets.

Lizards are animals outside of the set Mammals.

The set S = {1, 3, 5, 7, 9} is the odd integers less than 10.

The set P = {2, 3, 5, 7} is the prime numbers less than 10.

The intersection is composed of the numbers that are in both sets {3, 5, 7}.

### 2. Number of Elements

Counting problems are one type of set questions. These questions can ask about the number of elements in the set or subsets.

## Example

After school, 20 students play soccer, 10 play basketball, and 7 play both. How many students play basketball, soccer or both?

(A) 20

(B) 22

(C) 23

(D) 25

(E) 29

### Solution

*Method 1*

Draw a diagram. First put the overlap into the diagram. Then calculate the number in each subset.

For only playing basketball, 10 − 7 = 3. For only playing soccer, 20 − 7 = 13.

So the total number of players is basketball only + soccer only + both = 3 + 13 + 7 = 23.

The correct answer is option (C).

*Method 2*

Another method to find the total number in a set is to add the number in each subset and subtract the number in the intersection.

subset + subset − intersection = total elements in set

20 + 10 − 7 = 23

The correct answer is option (C).

## Example

Out of 50 people at a buffet dinner, 14 chose ice cream for dessert, 25 chose chocolate cake, and 5 chose both.

(a) How many people had dessert?

(b) How many people chose not to have dessert?

(b) What percent of the people at the dinner had only cake?

### Solution

Draw a diagram. First put the overlap number into the diagram. The 5 people who had both are included in both of the 2 sets. Then calculate the number in each subset.

For the number of people who only had ice cream, subtract the people who had both.

14 − 5 = 9

For the number of people who only had cake, 25 − 5 = 20.

(a) The number of people who had dessert is 20 + 5 + 9 = 34.

(b) Out of the 50 people, 50 − 34 = 16 people did not have dessert.

(c) The diagram shows 20 people had only cake, out of the 50 people at the dinner.

So it is 20/50 = 40/100 = 40%.

## Example

Collins did a survey in his class about what pets the students owned. The most popular pet was a dog, with 22 students having dogs. There were 16 students with cats. Nine students have other types of pets. For students with more than one pet, 6 have a dog and a cat, 4 have a dog and another type of pet, and 3 have a cat with another type of pet. Two students have dogs, cats and another pet. If the class has 40 students, what percentage of students have no pets?

### Solution

Draw a diagram.

First put the overlap number into the diagram. Calculate the number in each subset by subtracting the numbers already in the intersection.

all 3 pets = 2 people

cat and dog = 6 – 2 = 4

dog and other = 4 – 2 = 2

cat and other = 3 – 2 = 1

cat only = 16 – 4 – 1 – 2 = 9

dog only = 22 – 4 – 2 – 2 = 14

other type only = 9 – 2 – 1 – 2 = 4

The total number of students in the circles is

9 + 4 + 14 + 1 + 2 + 2 + 4 = 36

If there are 40 students in the class, 40 – 36 = 4 students have no pets.

4 out of 40 = 4/40 = 1/10 = 10/100.

10% of the students have no pets.

## Example

There are 36 vehicles in a parking lot. Ten of the vehicles are white and 22 are trucks. There are 14 cars that are not white. How many of the trucks are white?

### Solution

*Method 1*

Draw a diagram. You are looking for the intersection of trucks and white.

The total number of vehicles is 36.

There are 14 that are neither trucks nor white. This number goes outside the circles.

The total number of trucks is 22.

The total number of white vehicles is 10.

So there are 36 vehicles = 14 neither + (22 trucks + 10 white – intersection)

36 = 14 + (22 + 10 –* x*)

22 = 32 – *x*

-10 = –*x*

So the intersection of trucks and white is 10. There are 10 white trucks.

*Method 2*

Use a table. The bottom row and the right column must each add to 36.

Start with the given information.

**white**

**not white**

**truck**

total trucks = 22

**car**

14 cars

total trucks = 22

total white = 10

total vehicles = 36

Fill in 2 more numbers using the totals for each column and row.

## Example

Students were surveyed about their summer plans. In all, 175 said they were planning to work, 120 said they planned to study and 105 said they were going to relax. No students said they planned to do all 3. If 150 students said they planned to do just 2, how many students plan to do just 1?

(A) 100

(B) 150

(C) 200

(D) 250

(E) 400

### Solution

Draw a diagram. You are looking for the areas that are not intersections of the plans.

Notice that the intersections are each counted twice:

The intersection of work and relax is included in the total count of work (175) and in the total count of relax (105).

So the answer is the number of students in each set minus twice the sum of the intersections.

175 + 120 + 105 − 2(150) = 400 − 300 = 100

The correct answer is option (A).

## Example

In a display of jack-o’-lanterns, 20% looked friendly but had no nose. Half of the jack-o’-lanterns with noses looked friendly. If 60% of the jack-o’-lanterns in the display looked friendly, what percentage had noses?

### Solution

*Method 1*

You are looking for the percentage of jack-o’-lanterns with noses. Use *n*% to represent the percent of jack-o’-lanterns with noses.

The first step is to draw a diagram. The sets are jack-o’-lanterns with noses and jack-o’-lanterns that look friendly.

The total in the 2 sets is 100% of the jack-o’-lanterns.

20% looked friendly but had no nose, so 20% is the friendly set without the intersection.

Half of the jack-o’-lanterns with noses did not look friendly. So 0.5*n*% represents the percentage of jack-o’-lanterns that did not look friendly and the percentage that did look friendly.

So the percentage of jack-o’-lanterns who looked friendly is 20% + 0.5*n*% = 60%.

0.5*n*% = 40%

*n*% = 80%

80% of the jack-o’-lanterns had noses.

*Method 2*

Use a table. The bottom row and the right column must each add to 100%.

Start with the given information. Use *n*% to represent the percent of jack-o’-lanterns with noses.

Nose | No nose | ||

Friendly | 0.5n% | 20% | total friendly = 60% |

Not | 0.5n% | ||

total nose = n% | total = 100% |

Using just the first row of the table:

0.5*n*% + 20% = 60%

*n*% = 80%

80% of the jack-o’-lanterns had noses.

## 800score Tip: Count Inclusively

When doing counting problems, always be sure to count the first and last of the elements.

ExampleA fence consists of fence posts that are 1 foot wide and rails connecting each post that are 9 feet long. How many rails and how many fence posts would be needed for an 80 foot fence?

### Solution

Each pair of post and rail covers 10 feet. So the fence needs 8 pairs of post and rail. But the trick here is to include the additional post that is needed at the end. The fence needs 8 rails and 9 posts.

## Example

(a) How many elements are in the set of integers from 5 to 20?

(b) What is the sum of the set of integers from 5 to 20?

### Solution

(a) You might think you just subtract. Last − first = 15 − 5 = 10. But make a list.

{5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15} = 11 elements

Note: The formula for the number of consecutive integers in a sequence is (last − first) + 1, or L − F + 1.

(b) The first thing to notice is the sums of the pairings of the first and last elements of the set.

(first + last) = 5 + 15 = 20

(second + second to last) = 6 + 14 = 20

To use this pairing, you need to know the number of elements. For this set, there are 11 elements.

The sum will be (11 / 2)(20) = 5.5 × 20 = 110.

### 3. Using Sets

Some Examples above use information about the set to ask an additional question. Using sets as the basis for questions about probability, averages, and combinations/permutations is a second type of SAT question that uses sets.

## Example

Using the set

A= {1, 2, 3, 4}, how many arrangements can there be if 2 and 4 are not adjacent?

### Solution

This is a question about permutations.

For a set with this few elements, a list is a simple method.

1, 2, 3, 4

1, 4, 3, 2

2, 3, 4, 1

2, 1, 4, 3

2, 1, 3, 4

2, 3, 1, 4

3, 2, 1, 4

3, 4, 1, 2

4, 1, 2, 3

4, 3, 2, 1

4, 3, 1, 2

4, 1, 3, 2

There are 12 arrangements.

## Example

Set

M= {20, 70, 10, k, 20, 90}. The arithmetic mean is 45. What is the mode of the set?(A) 20

(B) 30

(C) 40

(D) 45

(E) 60

### Solution

*Arithmetic mean: *sum of elements/number of elements

*Mode:* the element that appears the most times in a set.

To find the value of* k*, use the arithmetic mean.

45 = 20 + 70 + 10 + *k* + 20 + 90/6

45 = 210 + *k*/6

45 × 6 = 210 + *k*

*k* = 270 – 210 = 60

With *k* = 60, rewrite the elements of the set in numerical order.

{10, 20, 20, 60, 70, 90}

The mode is 20, since it is the only element that appears more than once.

The correct answer is option (A).

Notice that the value of *k* is given as an answer option. A SAT trick is to offer an answer option that is a number from a previous step in finding the real answer. The median is also given as an answer option. Since the number of elements is even, the median is the mean of the 2 center elements: (20 + 60)/2 = 40.

### 4. Range of Values

Range questions also refer to the elements in a set. When a question asks for a possible range, be sure to check both the lowest and highest possible values.

## Example

An opened box contains 3 to 5 bottles of mushrooms. Each bottle contains 30 to 40 mushrooms. If 10% of the mushrooms are flawed, what is the range of the possible number of flawed mushrooms in the box?

### Solution

First find the lowest value.

Three bottles and 30 mushrooms is the fewest number of mushrooms in the box.

3 × 30 × 10% = 90 × 0.1 = 9 At least 9 flawed mushrooms.

Then find the greatest value.

Five bottles and 40 mushrooms is the greatest number of mushrooms in the box.

5 × 40 × 10% = 200 × 0.1 = 20 At most 20 flawed mushrooms.

There are between 9 and 20 flawed mushrooms in the bottles in the box.

## Example

Set

A= {1, 2, 3, 4} and setB= {integers > 5}.

What is the range of values for the product of elementsab?

### Solution

Find the minimum values and multiply.

The minimum value in set *A* is 1 and the minimum value in set *B* is 6, so the minimum for *ab* is 1 × 6 = 6.

Find the maximum values and multiply.

The maximum value in set *A* is 4. Set *B* has no maximum, so *ab* has no maximum.

The range values for the product of Set *A* and Set *B* is *ab* > 6.