Solved Examples on Set Operations

For ACT Students
The ACT is a timed exam...60 questions for 60 minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no negative penalty for a wrong answer.

For WASSCE Students
Any question labeled WASCCE is a question for the WASCCE General Mathematics
Any question labeled WASSCE-FM is a question for the WASSCE Further Mathematics/Elective Mathematics

For JAMB Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

Corequisites: Probability

Solve all questions
Show all work.

(1.) WASSCE A = {2, 4, 6, 8}, B = {2, 3, 7, 9} and C = {x: 3 < x < 9} are subsets of the universal set, $\mu$ = {2, 3, 4, 5, 6, 7, 8, 9}
Find:
(a.) $A \cap (B' \cap C')$
(b.) $(A \cup B) \cap (B \cup C)$

$ A = \{2, 4, 6, 8\} \\[3ex] B = \{2, 3, 7, 9\} \\[3ex] C = \{x: 3 \lt x \lt 9\} \\[3ex] \rightarrow C = \{4, 5, 6, 7, 8\} \\[3ex] \mu = \{2, 3, 4, 5, 6, 7, 8, 9\} \\[3ex] (a.) \\[3ex] B' = \{4, 5, 6, 8\} \\[3ex] C' = \{2, 3, 9\} \\[3ex] B' \cap C' = \phi \\[3ex] A \cap (B' \cap C') = A \cap \phi = \phi \\[3ex] (b.) \\[3ex] A \cup B = \{2, 3, 4, 6, 7, 8, 9\} \\[3ex] B \cup C = \{2, 3, 4, 5, 6, 7, 8, 9\} \\[3ex] (A \cup B) \cap (B \cup C) = \{2, 3, 4, 6, 7, 8, 9\} $

(2.) WASSCE Given that $\mu$ = {1, 2, 3, 4, 5, 6, 7, 8}, X = {1, 3, 5, 7} and Y = {1, 5, 8},
find:
(i) $X' \cap Y$
(ii) $(X' \cup Y)'$

$ (i) \\[3ex] X = \{1, 3, 5, 7\} \\[3ex] X' = \{2, 4, 6, 8\} \\[3ex] X' \cap Y = \{8\} \\[3ex] (ii) \\[3ex] X' \cup Y = \{1, 2, 4, 5, 6, 8\} \\[3ex] (X' \cup Y)' = \{3, 7\} $

(3.) Sets C, D, and E are:
C = {love, peace, joy}
D = {hope, strength}
E = {favor, blessing}

Write the following Cartesian products using n-tuple notation in roster form.

$ (a.)\:\: C * D \\[3ex] (b.)\:\: D * C \\[3ex] (c.)\:\: D * E \\[3ex] (d.)\:\: C * D * E \\[3ex] (e.)\:\: D * C * E \\[3ex] $

$ (a.) \\[3ex] n(C * D) = n(C) * n(D) = 3 * 2 = 6 \\[3ex] C * D = \{love, peace, joy\} * \{hope, strength\} \\[3ex] = \{(love, hope), (love, strength), (peace, hope), (peace, strength), (joy, hope), (joy, strength)\} \\[3ex] n(C * D) = 6 \\[3ex] (b.) \\[3ex] n(D * C) = n(D) * n(C) = 2 * 3 = 6 \\[3ex] D * C = \{hope, strength\} * \{love, peace, joy\} \\[3ex] = \{(hope, love), (hope, peace), (hope, joy), (strength, love), (strength, peace), (strength, joy)\} \\[3ex] n(D * C) = 6 \\[3ex] (c.) \\[3ex] n(D * E) = n(D) * n(E) = 2 * 2 = 4 \\[3ex] D * E = \{hope, strength\} * \{favor, blessing\} \\[3ex] = \{(hope, favor), (hope, blessing), (strength, favor), (strength, blessing)\} \\[3ex] n(D * E) = 4 \\[3ex] (d.) \\[3ex] n(C * D * E) = n(C) * n(D) * n(E) = 3 * 2 * 2 = 12 \\[3ex] C * D * E = \{love, peace, joy\} * \{hope, strength\} * \{favor, blessing\} \\[3ex] = \{ \\[3ex] (love, hope, favor), (love, hope, blessing), (love, strength, favor), (love, strength, blessing), \\[3ex] (peace, hope, favor), (peace, hope, blessing), (peace, strength, favor), (peace, strength, blessing), \\[3ex] (joy, hope, favor), (joy, hope, blessing), (joy, strength, favor), (joy, strength, blessing) \\[3ex] \} \\[3ex] n(C * D * E) = 12 \\[3ex] (e.) \\[3ex] n(D * C * E) = n(D) * n(C) * n(E) = 2 * 3 * 2 = 12 \\[3ex] D * C * E = \{hope, strength\} * \{love, peace, joy\} * \{favor, blessing\} \\[3ex] = \{ \\[3ex] (hope, love, favor), (hope, love, blessing), \\[3ex] (hope, peace, favor), (hope, peace, blessing), \\[3ex] (hope, joy, favor), (hope, joy, blessing), \\[3ex] (strength, love, favor), (strength, love, blessing), \\[3ex] (strength, peace, favor), (strength, peace, blessing), \\[3ex] (strength, joy, favor), (strength, joy, blessing) \\[3ex] \} \\[3ex] n(D * C * E) = 12 \\[3ex] $

(4.) Set C = {b, o, y}
(a.) Determine the cardinality of set C
(b.) List the power set of C
(c.) What is the cardinality of the power set of C?

C = {b, o, y}

(a.) n(C) = 3

(b.) P(C) = {φ, {b}, {o}, {y}, {b, o}, {b, y}, {o, y}, {b, o, y}}

(c.) n(P(C)) = 8

(6.) JAMB If $\mu$ = {x|x is a positive integer less than 10} and P = {x|x is a prime factor of 30}, find the complement of P

$ A.\;\; \{1, 2, 4, 7, 8, 9\} \\[3ex] B.\;\; \{1, 2, 4, 6, 7, 8, 9\} \\[3ex] C.\;\; \{1, 4, 6, 7, 8, 9\} \\[3ex] D.\;\; \{1, 4, 7, 8, 9\} \\[3ex] $

$ \mu = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} \\[3ex] P = \{2, 3, 5\} \\[3ex] P' = \{1, 4, 6, 7, 8, 9\} $

(8.) WASCCE Given that X = {x: 10 ≤ x < 15} and Y = {even numbers < 18} are subsets of U = {10, 11, 12, ..., 20}, find:
(a.) $X \cap Y$
(b.) $n(X' \cap Y)$

$ U = \{10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20\} \\[3ex] X = \{10, 11, 12, 13, 14\} \\[3ex] Y = \{2, 4, 6, 8, 10, 12, 14, 16\} \\[3ex] X' = \{15, 16, 17, 18, 19, 20\} \\[3ex] (a) \\[3ex] X \cap Y = \{10, 12, 14\} \\[3ex] (b) \\[3ex] X' \cap Y = \{16\} \\[3ex] n(X' \cap Y) = 1 $

(10.) ACT The 3 statements below are true for the elements of sets A, B, C, and D
I. All elements of A are elements of B
II. All elements of C are elements of D
III. No elements of D are elements of B

Which of the following statements must be true?
F. All elements of A are elements of C
G. All elements of B are elements of D
H. All elements of C are elements of B
J. No elements of A are elements of B
K. No elements of A are elements of C

Let us analyze each option
Option J.
No elements of A are elements of B
This is incorrect because: All elements of A are elements of B

Option G.
All elements of B are elements of D
This is false.
If: No elements of D are elements of B, it means that: B does not contain any element of D

Option H.
All elements of C are elements of B
This is not correct.
Here's why it is not true:
No elements of D are elements of B
This means that: B does not contain any element of D
All elements of C are elements of D
This means that B does not contain any element of C

Option F.
All elements of A are elements of C
This is also not correct.
Here's the reason:
No elements of D are elements of B
This means that: B does not contain any element of D
All elements of C are elements of D
This means that B does not contain any element of C
All elements of A are elements of B
Because: B does not contain any element of C, no elements of A can be elements of C

Option K.
No elements of A are elements of C
This is the correct option.
Based on the conclusion from the previous reason:
Because: B does not contain any element of C, no elements of A are elements of C

(12.) JAMB If U = {x | x is a positive integer less than 10} and P = {x | x is a prime factor of 30}, find the complement of P

$ A.\;\; \{1, 2, 4, 7, 8, 9\} \\[3ex] B.\;\; \{1, 2, 4, 6, 7, 8, 9\} \\[3ex] C.\;\; \{1, 4, 6, 7, 8, 9\} \\[3ex] D.\;\; \{1, 4, 7, 8, 9\} \\[3ex] $

$ U = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} \\[3ex] P = \{2, 3, 5\} \\[3ex] P' = \{1, 4, 6, 7, 8, 9\} $

(14.) Set K = {l, o, v, e}
(a.) Determine the cardinality of set K
(b.) List the power set of K
(c.) Determine the cardinality of the power set of K.

C = {l, o, v, e}

(a.) n(K) = 4

(b.) P(K) = {{}, {l}, {o}, {v}, {e}, {l, o}, {l, v}, {l, e}, {o, v}, {o, e}, {v, e}, {l, o, v}, {l, o, e}, {l, v, e}, {o, v, e}, {l, o, v, e}}

(c.) n(P(C)) = 16

(18.) Decide whether these statements make sense (or is clearly true) or does not make sense (or is clearly false).
Explain your reasoning.

(I.) I counted an irrational number of students in my class.

(II.) The people who live in Boise form a subset of those who rent apartments in Boise.

(III.) My professor asked me to draw a Venn diagram from a categorical proposition, but I couldn't do it because the proposition was clearly false.

(IV.) I surveyed my class to find out whether students rode a bike on campus or not.
Then I made a Venn diagram with one circle (inside a rectangle) to summarize the results.

(I.) The statement does not make sense. Irrational numbers have non-repeating decimal expansion, so there could not be an irrational number of students.

(II.) The statement does not make sense because the set of people who live in Boise also includes home owners which are not part of the set of people who rent apartments in Boise.

(III.) The statement does not make sense because to draw a Venn diagram, the proposition can be either true or false.

(IV.) The statement makes sense because students will either answer with ride a bike or not ride a bike, which can be summarized using one circle in a Venn diagram.

(20.) JAMB If P = {x | x is odd, -1 < x ≤ 20} and Q = {y: y is prime, -2 < y ≤ 25}, find $P \cap Q$

$ A.\;\; \{3, 5, 7, 11, 17, 19\} \\[3ex] B.\;\; \{3, 5, 7, 11, 13, 17, 19\} \\[3ex] C.\;\; \{3, 5, 7, 11, 17, 19\} \\[3ex] D.\;\; \{2, 3, 5, 7, 11, 13, 17, 19\} \\[3ex] $

$ P = \{1, 3, 5, 7, 9, 11, 13, 15, 17, 19\} \\[3ex] Q = \{2, 3, 5, 7, 11, 13, 17, 19, 23\} \\[3ex] P \cap Q = \{3, 5, 7, 11, 13, 17, 19\} $

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(23.) JAMB If $\mu = \{1, 2, 3, 6, 7, 8, 9, 10\}$ is the universal set, $E = \{0, 4, 6, 8, 10\}$ and $F = \{x: x^2 = 2^6,\;\;x\;\;is\;\;odd\}$
Find $(E \cap F)'$, where ' means the complement of a set.

$ A.\;\; \{0\} \\[3ex] B.\;\; \mu \\[3ex] C.\;\; \{8\} \\[3ex] D.\;\; \phi \\[3ex] $

$ \mu = \{1, 2, 3, 6, 7, 8, 9, 10\} \\[3ex] E = \{0, 4, 6, 8, 10\} \\[3ex] 2^6 = 64 \\[3ex] 8^2 = 64\;\;but\;\;8\;\;is\;\;not\;\;odd \\[3ex] F = \phi \\[3ex] E \cap F = \{0, 4, 6, 8, 10\} \cap \phi = \phi...Domination\;\;Law \\[3ex] (E \cap F)' = \phi ' = \mu...Complement\;\;Law $

(24.) ACT Consider sets A, B, C, and D such that B is a subset of A, C is a subset of B, and D is a subset of C.
Whenever x is an element of B, x must be an element of:
A. A
B. D
C. A and C
D. C and D
E. A, C, and D

$ B \subseteq A \\[3ex] C \subseteq B \\[3ex] D \subseteq C \\[3ex] x \in B \;\;implies\;\;that\;\; x \in A \;\;because\;\; B \subseteq A
Option A is the correct option $

(29.) JAMB Given that for sets A and B, in a universal set E, $A \subseteq B$ then $A \cap (A \cap B)'$ is

$ A.\;\; A \\[3ex] B.\;\; \phi \\[3ex] C.\;\; B \\[3ex] D.\;\; E \\[3ex] $

$ A \subseteq B \\[3ex] means\;\;that \\[3ex] A \cap B = A ...because\;\;A\;\;is\;\;contained\;\;in,\;\;or\;\;equal\;\;to\;\;B \\[3ex] (A \cap B)' = A' \\[5ex] \implies \\[3ex] A \cap (A \cap B)' \\[3ex] = A \cap A' = \phi...Complement\;\;Law $

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