Finite Mathematics Quiz: Venn Diagrams And Set Relations

11 questions · exam conditions

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Venn Diagrams And Set RelationsQuestion 1 of 11

If $A \subseteq B$ and $B \subseteq C$ , and we know that $|A| = 12$ , $|C| = 30$ , and $|A \cup C| = 30$ , what can we conclude about $|B|$ ?

|B|

must be exactly 21 elements

|B|

can be any value with

12 \leq |B| \leq 30

|B|

must be exactly 30 elements

|B|

can be any value with

12 \leq |B| \leq 18

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Finite Mathematics Quiz

Finite Mathematics Quiz: Venn Diagrams And Set Relations

Practice Venn Diagrams And Set Relations in Finite Mathematics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Venn Diagrams And Set Relations, giving you a quick way to practice the rules, question types, and explanations that matter most for Finite Mathematics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

If $A \subseteq B$ and $B \subseteq C$ , and we know that $|A| = 12$ , $|C| = 30$ , and $|A \cup C| = 30$ , what can we conclude about $|B|$ ?

$|B|$ must be exactly 21 elements
$|B|$ can be any value with $12 \leq |B| \leq 30$ (correct answer)
$|B|$ must be exactly 30 elements
$|B|$ can be any value with $12 \leq |B| \leq 18$

Explanation: Since

A \subseteq B \subseteq C

and

|A \cup C| = |C| = 30

, this confirms that

A \subseteq C

(which we already knew from transitivity). The subset relationships give us

|A| \leq |B| \leq |C|

, so

12 \leq |B| \leq 30

. The condition

|A \cup C| = 30

doesn't provide additional constraints on

|B|

beyond what the subset relationships already give us. Choice A assumes a specific value without justification. Choice C assumes

B = C

. Choice D incorrectly restricts the upper bound.

Question 2

A Venn diagram for sets $A$ , $B$ , and $C$ is partially labeled with the cardinalities of its regions in terms of a variable $x$ . The cardinalities for the regions representing elements in only one set are $|A \text{ only}| = 10$ , $|B \text{ only}| = 15$ , and $|C \text{ only}| = 20$ . The cardinalities for the intersection regions are $|A \cap B \cap C'| = x$ , $|A \cap C \cap B'| = 2x$ , $|B \cap C \cap A'| = 3x$ , and $|A \cap B \cap C| = 5$ . If the total number of elements in set $A$ is 21, what is the total number of elements in the set $B \cup C$ ?

28
35
52 (correct answer)
63

Explanation: First, use the given total cardinality of set

A

to find the value of

x

. The set

A

consists of four disjoint regions:

|A| = |A \text{ only}| + |A \cap B \cap C'| + |A \cap C \cap B'| + |A \cap B \cap C|

. Plugging in the given values:

21 = 10 + x + 2x + 5

. This simplifies to

21 = 15 + 3x

, which gives

3x = 6

, so

x=2

. Now, we can find the cardinality of

B \cup C

. This is the sum of all disjoint regions that are part of

B

C

|B \cup C| = |B \text{ only}| + |C \text{ only}| + |A \cap B \cap C'| + |A \cap C \cap B'| + |B \cap C \cap A'| + |A \cap B \cap C|

. Substituting the values:

|B \cup C| = 15 + 20 + x + 2x + 3x + 5 = 40 + 6x

. Now substitute

x=2

into this expression:

|B \cup C| = 40 + 6(2) = 40 + 12 = 52

Question 3

A survey of 200 business owners about their online presence found the following:

120 have a Website
90 have Social Media
70 have an E-commerce platform
50 have a Website and Social Media
35 have a Website and an E-commerce platform
25 have Social Media and an E-commerce platform
15 have none of the three

Based on the survey data, how many business owners use exactly two of these three types of online platforms?

15
65 (correct answer)
95
110

Explanation: Let

W

S

, and

E

be the sets of owners with a Website, Social Media, and E-commerce platform, respectively. First, find the number of owners with at least one platform:

|W \cup S \cup E| = 200 - 15 = 185

. Next, use the Principle of Inclusion-Exclusion to find the number with all three,

|W \cap S \cap E|

|W \cup S \cup E| = |W| + |S| + |E| - (|W \cap S| + |W \cap E| + |S \cap E|) + |W \cap S \cap E|

185 = 120 + 90 + 70 - (50 + 35 + 25) + |W \cap S \cap E|

185 = 280 - 110 + |W \cap S \cap E|

, which gives

|W \cap S \cap E| = 15

. The number of owners with exactly two platforms is the sum of those with (W and S only), (W and E only), and (S and E only). Exactly W and S:

|W \cap S| - |W \cap S \cap E| = 50 - 15 = 35

. Exactly W and E:

|W \cap E| - |W \cap S \cap E| = 35 - 15 = 20

. Exactly S and E:

|S \cap E| - |W \cap S \cap E| = 25 - 15 = 10

. The total is

35 + 20 + 10 = 65

Question 4

At a technology conference of 500 attendees, 350 use cloud service X, and 280 use cloud service Y. If every attendee uses at least one of these two services, how many attendees use service X but not service Y?

70
130
150
220 (correct answer)

Explanation: Let

X

be the set of attendees using service X and

Y

be the set for service Y. We are given

|X|=350

|Y|=280

. The condition 'every attendee uses at least one' means

|X \cup Y| = 500

. We want to find the number of attendees who use X but not Y, which is

|X \setminus Y|

. We can calculate this as

|X \setminus Y| = |X| - |X \cap Y|

. First, we must find

|X \cap Y|

using the Principle of Inclusion-Exclusion:

|X \cup Y| = |X| + |Y| - |X \cap Y|

. Substituting the known values:

500 = 350 + 280 - |X \cap Y|

. This gives

500 = 630 - |X \cap Y|

, so

|X \cap Y| = 130

. Now we can find

|X \setminus Y|

|X \setminus Y| = |X| - |X \cap Y| = 350 - 130 = 220

Question 5

In a survey of smartphone users, 70% have app A installed, and 55% have app B installed. If 10% of users have neither app A nor app B installed, what percentage of users have both app A and app B installed?

15%
25%
35% (correct answer)
45%

Explanation: Let

A

be the set of users with app A and

B

be the set with app B. We are given

|A|=70\%

and

|B|=55\%

. The percentage of users with neither app is

|(A \cup B)'| = 10\%

. This means the percentage with at least one of the apps is

|A \cup B| = 100\% - 10\% = 90\%

. We can use the Principle of Inclusion-Exclusion,

|A \cup B| = |A| + |B| - |A \cap B|

, to find the percentage with both apps,

|A \cap B|

. Plugging in the values:

90\% = 70\% + 55\% - |A \cap B|

. This simplifies to

90\% = 125\% - |A \cap B|

. Solving for

|A \cap B|

gives

|A \cap B| = 125\% - 90\% = 35\%

Question 6

A company's inventory of 150 products is categorized by three attributes: Handmade (H), Organic (O), and Vegan (V).

60 products are Handmade
75 products are Organic
55 products are Vegan
30 products are Handmade and Organic
25 products are Organic and Vegan
20 products are Handmade and Vegan
12 products are Handmade, Organic, and Vegan

How many products in the inventory are Organic and Vegan, but not Handmade?

12
13 (correct answer)
22
25

Explanation: We need to find the number of products that are in the set 'Organic and Vegan' but not in the set 'Handmade'. This corresponds to the set cardinality

|(O \cap V) \setminus H|

|O \cap V \cap H'|

. This can be calculated by taking the total number of products that are Organic and Vegan,

|O \cap V|

, and subtracting the number of products that are Organic, Vegan, AND Handmade,

|O \cap V \cap H|

. From the data provided, we have

|O \cap V| = 25

and

|H \cap O \cap V| = 12

. Therefore, the number of products that are Organic and Vegan but not Handmade is

25 - 12 = 13

Question 7

Let $A$ and $B$ be subsets of a universal set $U$ . Given $|U|=100$ , $|A|=50$ , and $|B|=65$ , what is the minimum possible value of $|A \cap B'|$ ?

0 (correct answer)
15
35
50

Explanation: The set

A \cap B'

represents the elements that are in

A

but not in

B

, which is also written as

A \setminus B

. The cardinality is given by

|A \setminus B| = |A| - |A \cap B|

. To minimize

|A \setminus B|

, we must maximize the term being subtracted,

|A \cap B|

. The maximum possible size of the intersection of two sets is the size of the smaller set. In this case, max(

|A \cap B|

) = min(

|A|

|B|

) = min(50, 65) = 50

. This configuration is possible if

is a subset of

. We must check if this fits within the universal set: if

A \subseteq B

, then

|A \cup B| = |B| = 65

, which is less than

|U|=100

. So, this is a valid configuration. Therefore, the maximum possible value of

|A \cap B|

is 50. The minimum value of

|A \setminus B|

is then

50 - 50 = 0$.

Question 8

Let $A$ , $B$ , and $C$ be subsets of a universal set $U$ . If $A \setminus B \subseteq C$ , which of the following statements must be true?

$A \subseteq B \cup C$ (correct answer)
$C \subseteq A$
$A \cap C = \emptyset$
$B \cap C = \emptyset$

Explanation: The condition

A \setminus B \subseteq C

means that any element in

A

but not in

B

must also be in

C

. To check if

A \subseteq B \cup C

must be true, consider an arbitrary element

x \in A

. There are two possibilities for

x

with respect to set

B

: either

x \in B

x \notin B

. Case 1: If

x \in B

, then

x

is certainly in

B \cup C

. Case 2: If

x \notin B

, then since

x \in A

, we know

x \in A \setminus B

. By the given condition, this implies

x \in C

. If

x \in C

, then

x

is also in

B \cup C

. In both cases, any element of

A

is also an element of

B \cup C

. Therefore,

A \subseteq B \cup C

must be true.

Question 9

For two sets, $A$ and $B$ , the number of elements that belong to exactly one of the two sets is 30. If the total number of elements in $A \cup B$ is 42, how many elements are in $A \cap B$ ?

6
12 (correct answer)
15
27

Explanation: The set of elements in

A \cup B

can be partitioned into three disjoint subsets: elements in

A

only (

A \setminus B

), elements in

B

only (

B \setminus A

), and elements in both (

A \cap B

). The number of elements that belong to 'exactly one' of the two sets is the sum of the cardinalities of the first two subsets:

|A \setminus B| + |B \setminus A|

. We are given this value is 30. The total number of elements in the union is the sum of the cardinalities of these three disjoint parts:

|A \cup B| = (|A \setminus B| + |B \setminus A|) + |A \cap B|

. We are given

|A \cup B| = 42

. Substituting the known values, we get

42 = 30 + |A \cap B|

. Solving for

|A \cap B|

gives

42 - 30 = 12

Question 10

Consider sets $X$ and $Y$ where $X \subseteq Y \subseteq U$ and $|U| = 200$ . If $|(X \cup Y)^c| = 60$ and $|X^c \cap Y| = 45$ , what is $|X|$ ?

$|X| = 85$ elements based on set complement relationships
$|X| = 95$ elements based on set complement relationships (correct answer)
$|X| = 105$ elements based on set complement relationships
$|X| = 115$ elements based on set complement relationships

Explanation: Since

X \subseteq Y

, we have

X \cup Y = Y

. Therefore,

|(X \cup Y)^c| = |Y^c| = 60

, which means

|Y| = 200 - 60 = 140

. The set

X^c \cap Y

represents elements in

Y

but not in

X

. Since

X \subseteq Y

, we have

Y = X \cup (X^c \cap Y)

and these are disjoint. Therefore,

|Y| = |X| + |X^c \cap Y|

, which gives us

140 = |X| + 45

. Solving:

|X| = 95

. Choice A subtracts incorrectly. Choice C adds instead of subtracting. Choice D uses wrong intermediate values.

Question 11

Given sets $A = \{x : x^2 - 5x + 6 \leq 0\}$ and $B = \{x : |x - 4| < 2\}$ , what is $A \cap B$ ?

$\{x : 2 \leq x \leq 3\}$
$\{x : 2 < x \leq 3\}$ (correct answer)
$\{x : 2 \leq x < 6\}$
$\{x : 2 < x < 6\}$

Explanation: First, solve

x^2 - 5x + 6 \leq 0

. Factoring:

(x-2)(x-3) \leq 0

, so

A = [2,3]

. Next, solve

|x-4| < 2

, which gives

-2 < x-4 < 2

, so

2 < x < 6

, meaning

B = (2,6)

. The intersection

A \cap B = [2,3] \cap (2,6) = (2,3]

. Choice A incorrectly includes the endpoint 2. Choice C uses the wrong upper bound. Choice D excludes both correct endpoints.