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Sets

This section summarizes the essential concepts and formulas for Sets, Subsets, Set Operations, and the Principle of Inclusion and Exclusion, foundational for understanding Algebra at the engineering level.

A set is a collection of distinct elements, usually denoted by capital letters. The number of elements in a set is called its cardinality and is written as $|S|$ or $n(S)$.

Common Sets

Subsets

If every element of set $A$ is also in $B$, we write:

$$ A \subseteq B $$

A proper subset is one where $A \subseteq B$ but $A \ne B$.

Complement and Relative Complement

Set Operations

Union: All elements in $A$ or $B$

$$ A \cup B $$

Intersection: All elements common to $A$ and $B$

$$ A \cap B $$

Principle of Inclusion and Exclusion

This principle helps count elements in the union of overlapping sets while avoiding overcounting.

$$ \left| \bigcup_{i=1}^{n} A_i \right| = \sum_{i=1}^{n} |A_i| - \sum_{1 \le i < j \le n} |A_i \cap A_j| + \sum_{1 \le i < j < k \le n} |A_i \cap A_j \cap A_k| - \cdots + (-1)^{n+1} \left| A_1 \cap A_2 \cap \cdots \cap A_n \right| $$
$$ \begin{aligned} |A \cup B \cup C \cup D| &= |A| + |B| + |C| + |D| \quad \text{(all singletons)} \\\\ &\quad - \left( |A \cap B| + |A \cap C| + |A \cap D| + |B \cap C| + |B \cap D| + |C \cap D| \right) \quad \text{(all pairs)} \\\\ &\quad + \left( |A \cap B \cap C| + |A \cap B \cap D| + |A \cap C \cap D| + |B \cap C \cap D| \right) \quad \text{(all triples)} \\\\ &\quad - |A \cap B \cap C \cap D| \quad \text{(all quadruples)} \end{aligned} $$

This formula is especially useful in solving problems involving 2 or more overlapping sets, common in surveys, probability, and classification tasks in engineering.

Concept Concept Concept Concept Concept Concept Concept Concept Concept

Problem:

Among the 90 students in a dormitory, 35 own an automobile, 40 own a bicycle, and 10 have both an automobile and a bicycle. Find the number of students who:
a. have an automobile or bicycle
b. do not have an automobile
c. have neither an automobile nor a bicycle
d. have an automobile or bicycle, but not both

Sets | Algebra – Problem 1: – Diagram Sets | Algebra – Problem 1: – Diagram Sets | Algebra – Problem 1: – Diagram

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Sets | Algebra – Problem 1: – Diagram Sets | Algebra – Problem 1: – Diagram Sets | Algebra – Problem 1: – Diagram Sets | Algebra – Problem 1: – Diagram

Problem:

In a consumer market, 85% of those surveyed liked at least one of three products: 1, 2, or 3. 50% of those asked liked product 1, 30% liked product 2, and 20% liked product 3. If 5% of the people in the survey liked all three of the products, what percentage of the survey participants liked more than one of the three products?
A. 10%
B. 15%
C. 25%
D. 30%

Sets | Algebra – Problem 2: – Diagram Sets | Algebra – Problem 2: – Diagram Sets | Algebra – Problem 2: – Diagram

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Sets | Algebra – Problem 2: – Diagram Sets | Algebra – Problem 2: – Diagram Sets | Algebra – Problem 2: – Diagram Sets | Algebra – Problem 2: – Diagram

Problem:

In a coffee shop that offers four kinds of drinks: Mocha, Choco, Coffee, and Vanilla. Thirty ordered Mocha. 29 ordered Choco. 23 ordered Coffee. 26 ordered Vanilla. Eighteen ordered Mocha and Choco, 9 ordered Mocha and Coffee, 15 ordered Mocha and Vanilla, 15 ordered Choco and Coffee, 14 ordered Choco and Vanilla, and 14 ordered Coffee and Vanilla. Six ordered Mocha, Choco, and Coffee; 9 ordered Mocha, Choco, and Vanilla; 7 ordered Mocha, Coffee, and Vanilla; and 9 ordered Choco, Coffee, and Vanilla. Four ordered Mocha, Choco, Coffee, and Vanilla. Assuming that each person ordered one of a kind for each purchase, determine the total number of people who ordered.
A. 50
B. 40
C. 75
D. 100

Sets | Algebra – Problem 3: – Diagram Sets | Algebra – Problem 3: – Diagram Sets | Algebra – Problem 3: – Diagram

Answer: 19

Sets | Algebra – Problem 3: – Diagram Sets | Algebra – Problem 3: – Diagram Sets | Algebra – Problem 3: – Diagram Sets | Algebra – Problem 3: – Diagram

Problem:

Consider the following data for 120 mathematics students: 65 study French, 20 study French and German, 45 study German, 25 study French and Russian, 42 study Russian, 15 study German and Russian, and 8 study all three languages.
a. Find the number of students studying at least one of the three languages Ans. 100
b. Find the number of students studying at least two languages Ans. 44
c. Find the number of students studying exactly one language Ans. 56

Sets | Algebra – Problem 4: – Diagram Sets | Algebra – Problem 4: – Diagram Sets | Algebra – Problem 4: – Diagram

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Sets | Algebra – Problem 4: – Diagram Sets | Algebra – Problem 4: – Diagram Sets | Algebra – Problem 4: – Diagram Sets | Algebra – Problem 4: – Diagram

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Sets | Algebra – Problem 5: – Diagram Sets | Algebra – Problem 5: – Diagram Sets | Algebra – Problem 5: – Diagram

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Sets | Algebra – Problem 5: – Diagram Sets | Algebra – Problem 5: – Diagram Sets | Algebra – Problem 5: – Diagram Sets | Algebra – Problem 5: – Diagram

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Sets | Algebra – Problem 6: – Diagram Sets | Algebra – Problem 6: – Diagram Sets | Algebra – Problem 6: – Diagram

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Sets | Algebra – Problem 6: – Diagram Sets | Algebra – Problem 6: – Diagram Sets | Algebra – Problem 6: – Diagram Sets | Algebra – Problem 6: – Diagram

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Sets | Algebra – Problem 7: – Diagram Sets | Algebra – Problem 7: – Diagram Sets | Algebra – Problem 7: – Diagram

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Sets | Algebra – Problem 7: – Diagram Sets | Algebra – Problem 7: – Diagram Sets | Algebra – Problem 7: – Diagram Sets | Algebra – Problem 7: – Diagram

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Sets | Algebra – Problem 8: – Diagram Sets | Algebra – Problem 8: – Diagram Sets | Algebra – Problem 8: – Diagram

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Sets | Algebra – Problem 8: – Diagram Sets | Algebra – Problem 8: – Diagram Sets | Algebra – Problem 8: – Diagram Sets | Algebra – Problem 8: – Diagram
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Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: q54

MSTE - Algebra / Sets / Engr. Janclyde Espinosa (Clidez)

Among the 90 students in a dormitory, 35 own an automobile, 40 own a bicycle, and 10 have both an automobile and a bicycle. Find the number of students who:

Have an automobile or bicycle:

  1. 65
  2. 45
  3. 50
  4. 25

Do not have an automobile:

  1. 55
  2. 50
  3. 30
  4. 45

Have neither an automobile nor a bicycle:

  1. 25
  2. 55
  3. 35
  4. 50

Have an automobile or bicycle but not both:

  1. 55
  2. 90
  3. 75
  4. 30

Part 1.

Use inclusion-exclusion for automobile $A$ and bicycle $B$:
$n(A\cup B)=n(A)+n(B)-n(A\cap B)$
$n(A\cup B)=35+40-10$
$\boxed{65}$

Part 2.

Students who do not have an automobile are the complement of the automobile owners:
$90-35=55$
$\boxed{55}$

Part 3.

Students with neither are outside $A\cup B$:
$90-65=25$
$\boxed{25}$

Part 4.

Students who have automobile or bicycle but not both are counted in exactly one set:
$n(A\cup B)-n(A\cap B)=65-10$
$\boxed{55}$

Question Bank: q55

MSTE - Algebra / Sets / Engr. Janclyde Espinosa (Clidez)

Consider the following data for 120 mathematics students. 65 study French, 20 study French and German, 45 study German, 25 study French and Russian, 42 study Russian, 15 study German and Russian, and 8 study all three languages.

Find the number of students studying at least one of the three languages.

  1. 100
  2. 65
  3. 75
  4. 45

Find the number of students studying at least 2 languages.

  1. 44
  2. 67
  3. 23
  4. 50

Find the number of students studying exactly one language.

  1. 56
  2. 57
  3. 35
  4. 38

Part 1.

Use inclusion-exclusion for French $F$, German $G$, and Russian $R$:
$n(F\cup G\cup R)=65+45+42-20-25-15+8$
$\boxed{100}$

Part 2.

Students studying at least two languages are the pair intersections, with the triple intersection counted once overall:
$(20+25+15)-2(8)=44$
$\boxed{44}$

Part 3.

Students studying exactly one language equals at least one minus at least two:
$100-44=56$
$\boxed{56}$

Question Bank: q56

MSTE - Algebra / Sets / Engr. Janclyde Espinosa (Clidez)

Ninety people at a Superbowl party were surveyed to see what they ate while watching the game. The following data were collected: 48 had nachos, 39 had wings, 35 had potato skins, 20 had both wings and potato skins, 19 had both potato skins and nachos, 22 had both wings and nachos, 10 had nachos, wings, and potato skins. How many had nothing?

Answer:

  1. 19
  2. 22
  3. 16
  4. 24
Use inclusion-exclusion:
$n(N\cup W\cup P)=48+39+35-22-19-20+10=71$
Thus the number who had nothing is:
$90-71=19$
$\boxed{19}$

Question Bank: q681

MSTE - Algebra / Sets / Engr. Janclyde Espinosa (Clidez)

At XYZ University, 40% of all students are members of both a chess club and a swim team. If 20% of members of the swim team are not members of the chess club, what percentage of all XYZ students are members of the swim team?

If function f(x) satisfies:
f(x2)=f(x)2 for all x, which of the following must be true?

  1. 50%
  2. 20%
  3. 30%
  4. 40%
Let $S$ be swim team members and $C$ chess club members. Since 20% of swim team members are not in chess, 80% of swim team members are in both:
$0.80S=40\%$ of all students
$S=40\%/0.80=50\%$
$\boxed{50\%}$

Question Bank: q682

MSTE - Algebra / Sets / Engr. Janclyde Espinosa (Clidez)

A number of light bulbs were purchased to illuminate a gym. However, only 2/3 of them were needed. After 160 light bulbs were returned, and 60% percent of their cost ($96) was reimbursed, how much money was spent on illuminating the gym, in dollars?

  1. 384
  2. 360
  3. 320
  4. 364
The 160 returned bulbs are the unused $1/3$ of the bulbs, so total purchased was:
$160/(1/3)=480$ bulbs
The refund was 60% of the returned bulbs' cost and equaled $96, so the returned bulbs originally cost:
$96/0.60=160$ dollars
Cost per bulb is $160/160=1$ dollar. Bulbs used were $2/3(480)=320$, so money spent for illumination is:
$\boxed{384}$

Question Bank: t125

MSTE - Algebra / Algebra Fundamentals / Gemini mapped Chapter 1 to 3

Given the following set of numbers: Set A = {12, 16, 17}; Set B = {$X$, 7.2, 22}. If the average of A and B are equal, what is the value of $X$?

  1. 14.6
  2. 13.2
  3. 15.8
  4. 16.2
Mean of A = $(12+16+17)/3 = 45/3 = 15$
Set mean of B = 15: $(X+7.2+22)/3 = 15$
$X = 45 - 29.2$
$\boxed{X = 15.8}$

Question Bank: t223

MSTE - Algebra / Algebra Fundamentals / Gemini mapped Chapter 1 to 3

Set $M$ and $N$ have 20 pairs where each element in $M$ is paired with one in $N$ such that the element of $M$ is always greater than its pair from $N$. Which of the following is always true?

  1. range of $M$ > range of $N$
  2. mode of $M$ > mode of $N$
  3. mean of $M$ > mean of $N$
  4. median of $M$ > median of $N$
Each paired element in $M$ is greater than its corresponding element in $N$.
Adding all 20 pair inequalities gives:
$\sum M_i>\sum N_i$
Dividing both sides by 20 gives $\bar{M}>\bar{N}$.
$\boxed{\text{mean of }M>\text{mean of }N}$

Question Bank: t273

MSTE - Algebra / Algebra Fundamentals / Gemini mapped Chapter 1 to 3

Given the following set of numbers arranged from smallest to largest: ${1, x, 4, 6, x^2}$. If the range of the set is 8, what is the average of the set?

  1. $4.3$
  2. $4.2$
  3. $4.6$
  4. $5.2$
The set is arranged from smallest to largest, so the range is $x^2-1$.
$x^2-1=8 \Rightarrow x^2=9$.
Since $x$ is between 1 and 4 in the ordered set, $x=3$.
Average $=\frac{1+3+4+6+9}{5}=4.6$.
$\boxed{4.6}$

Question Bank: w16

MSTE - Algebra / Sets / MSTE May 2019

The passengers on an excursion bus consisted of 14 married couples, 8 of whom brought no children and 6 of whom brought 3 children apiece. Counting the driver, the bus has 31 occupants. How is this possible?

  1. included among the 18 children were 8 married couples.
  2. included among the 18 children were 6 married couples.
  3. included among the 18 children were 10 married couples.
  4. included among the 18 children were 4 married couples.
14 couples $= 28$ adults; children $= 6 \times 3 = 18$; driver $= 1$.
Counting everyone separately gives $28 + 18 + 1 = 47$. To reach 31 occupants, some persons counted as “children” are also among the married couples (double counted): overlap $= 47 - 31 = 16$ persons $= 8$ couples.
$\boxed{\text{8 married couples are among the 18 children}}$