Solved Examples on Venn Diagrams for Two Sets

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
Draw Venn Diagrams for each question and answer as applicable
Show all work.

(1.) ACT Fifty shoppers at a pet store were asked if they owned at least 1 cat or at least 1 dog.
Data from their answers were recorded below.

Ownership	Number of shoppers
Cat(s) only	13
Dog(s) only	24
Both cat(s) and dog(s)	7

How many of these shoppers said that they owned NEITHER a cat NOR a dog?

$ A.\;\; 0 \\[3ex] B.\;\; 6 \\[3ex] C.\;\; 7 \\[3ex] D.\;\; 13 \\[3ex] E.\;\; 43 \\[3ex] $

$ Let \\[3ex] Cat = C \\[3ex] Dog = D \\[3ex] Neither\;\;a\;\;cat\;\;nor\;\;a\;\;dog = k \\[3ex] $

$ 13 + 7 + 24 + k = 50 \\[3ex] 44 + k = 50 \\[3ex] k = 50 - 44 \\[3ex] k = 6 $

(2.) GCSE In a group of 20 people
7 own a dog
3 own a cat
12 do not own a dog or a cat.
Aidan shows this information on a Venn diagram.

Make two criticisms of his Venn diagram.

Choose any two
The criticisms of his Venn diagram are:

(1.) Statement: 7 own a dog
His Venn diagram: 7 own a dog ONLY
These are two different statements

(2.) Statement: 3 own a cat
His Venn diagram: 3 own a cat ONLY
These are two different statements

(3.) Statement: 12 do not own a dog or a cat.
However, 12 is still a subset of the universal set.
In other words, 12 is still a part of the group of 20 people
His Venn diagram: 12 is outside the universal set...not part of the group of 20 people
That is incorrect

(3.) CSEC T and E are subsets of a universal set, U, such that:
U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
T = {multiples of 3}
E = {even numbers}

(i) Draw a Venn diagram to represent this information.

(ii) List the members of the set
(a) $T \cap E$
(b) $(T \cup E)'$

$ U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\} \\[3ex] T = \{3, 6, 9, 12\} \\[3ex] E = \{2, 4, 6, 8, 10, 12\} \\[3ex] (ii) \\[3ex] (a) \\[3ex] T \cap E = \{6, 12\} \\[3ex] (b) \\[3ex] T \cup E = \{2, 3, 4, 6, 8, 9, 10, 12\} \\[3ex] (T \cup E)' = \{1, 5, 7, 11\} \\[3ex] $ (i) The Venn diagram is:

(4.) JAMB In a science class of 42 students, each offers at least one of Mathematics and Physics.
If 22 students offer Physics and 28 students offer Mathematics, find how many students offer Physics only.

$ A.\;\; 6 \\[3ex] B.\;\; 8 \\[3ex] C.\;\; 12 \\[3ex] D.\;\; 14 \\[3ex] $

$ Let: \\[3ex] Physics = P \\[3ex] Mathematics = M \\[3ex] n(P) = 22 \\[3ex] n(M) = 28 \\[3ex] n(P \cap M) = k \\[3ex] n(P \cap M') = 22 - k \\[3ex] n(M \cap P') = 28 - k \\[3ex] n(\xi) = 42 \\[3ex] $

$ 22 - k + 28 - k + k = 42 \\[3ex] 50 - k = 42 \\[3ex] 50 - 42 = k \\[3ex] k = 8 \\[3ex] n(P\;\;ONLY) = n(P \cap M') \\[3ex] = 22 - k \\[3ex] = 22 - 8 \\[3ex] = 14 \\[3ex] $ 14 students offer Physics only

(5.) CSEC The Universal set, U, is given as
U = {Whole numbers from 1 to 12}

H is a subset of U, such that H = {Odd numbers between 4 and 12}
(i) List the members of the set H

J is a subset of U, such that J = {Prime numbers}
(ii) List the members of the set J

(iii) Draw a Venn diagram to represent the sets, U, H and J, showing ALL the elements in the subsets.

$ U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\} \\[3ex] (i) \\[3ex] H = \{5, 7, 9, 11\} \\[3ex] (ii) \\[3ex] J = \{2, 3, 5, 7, 11\} \\[3ex] H \cap J = \{5, 7, 11\} \\[3ex] H \cup J = \{2, 3, 5, 7, 9, 11\} \\[3ex] (H \cup J)' = \{1, 4, 6, 8, 10, 12\} \\[3ex] $ (iii) The Venn diagram is:

(6.) ACT Among a group of 20 students, 13 students are members of the Math Club, 11 students are members of the Drama Club, and 9 students are members of both clubs.
How many of the 20 students are NOT members of either club?

$ F.\;\; 4 \\[3ex] G.\;\; 5 \\[3ex] H.\;\; 6 \\[3ex] J.\;\; 11 \\[3ex] K.\;\; 13 \\[3ex] $

$ Let: \\[3ex] Math\;\;Club = M \\[3ex] Drama\;\;Club = D \\[3ex] NOT\;\;members\;\;of\;\;either\;\;club = p \\[3ex] $

$ 4 + 9 + 2 + p = 20 \\[3ex] 15 + p = 20 \\[3ex] p = 20 - 15 \\[3ex] p = 5 \\[3ex] $ 5 students are NOT members of either club.

(7.) USSCE Advance Mathematics Paper 2, 2011
If U = {Positive whole numbers ≤ 12} be the universal set.
A = {Positive primes ≤ 12}
B = {Positive factors of 12}

a) List the elements of sets A and B
b) Show the sets A, B, and U on a Venn diagram.
c) List the elements not in A (or A')

$ U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\} \\[3ex] a) \\[3ex] A = \{2, 3, 5, 7, 11\} \\[3ex] B = \{1, 2, 3, 4, 6, 12\} \\[3ex] b) \\[3ex] A \cap B = \{2, 3\} \\[3ex] A \cup B = \{1, 2, 3, 4, 5, 6, 7, 11, 12\} \\[3ex] (A \cup B)' = \{8, 9, 10\} \\[3ex] $ The Venn diagram is:

$ c) \\[3ex] A' = \{1, 4, 6, 8, 9, 10, 12\} $

(8.) CSEC A club has 160 members, some of whom play tennis (T) or cricket (C) or both.
97 play tennis, 86 play cricket and 10 play neither.
x play both tennis and cricket.

(i) Draw a Venn diagram to represent this information.
(ii) How many members play both tennis and cricket?

(i) The Venn Diagram is:

$ (ii) \\[3ex] 97 - x + 86 - x + x + 10 = 160 \\[3ex] 193 - x = 160 \\[3ex] 193 - 160 = x \\[3ex] x = 33 $

(9.) GCSE $\varepsilon$ = {2, 3, 4, 5, 6, 7, 8, 9}
P = {even numbers}
Q = {numbers divisible by 3}

(a) Complete the Venn diagram below.

(b) A number is chosen at random from the numbers 2 to 9
What is the probability that the number chosen is odd and not divisible by 3?

$ \varepsilon = \{2, 3, 4, 5, 6, 7, 8, 9\} \\[3ex] P = \{2, 4, 6, 8\} \\[3ex] Q = \{3, 6, 9\} \\[3ex] P \cap Q = \{6\} \\[3ex] P \cup Q = \{2, 3, 4, 6, 8, 9\} \\[3ex] (P \cup Q)' = \{5, 7\} \\[3ex] $ (a)
The Venn diagram is:

(b)
Let the set: R = {odd and not divisible by 3}

$ R = \{5, 7\} \\[3ex] n(R) = 2 \\[3ex] n(\varepsilon) = 8 \\[3ex] P(R) = \dfrac{n(R)}{n(\varepsilon)} = \dfrac{2}{8} = \dfrac{1}{4} $

(10.) ACT A health club surveyed 175 members about which types of equipment they had used in the past month.
Of the 175 members, 117 had used treadmills, 89 had used stationary bikes, and 53 had used both types of equipment.
Some members had used neither type of equipment.
Of the 175 members, how many had used treadmills, stationary bikes, or both?

$ A.\;\; 53 \\[3ex] B.\;\; 81 \\[3ex] C.\;\; 122 \\[3ex] D.\;\; 134 \\[3ex] E.\;\; 153 \\[3ex] $

The members that had used treadmills, stationary bikes, or both are those that:
used treadmills ONLY
used stationary bikes ONLY
used both treadmills and stationary bikes

$ Let: \\[3ex] treadmills = T \\[3ex] stationary\;\;bikes = B \\[3ex] n(\xi) = 175 \\[3ex] n(T) = 175 \\[3ex] n(B) = 89 \\[3ex] n(T \cap B) = 53 \\[3ex] $

The members that had used treadmills, stationary bikes, or both

$ = n(T \cup B) \\[3ex] = 64 + 36 + 53 \\[3ex] = 153 \\[3ex] $ 153 members had used treadmills, stationary bikes, or both

(11.) WASCCE A number of tourists were interviewed on their choice of means of travel.
Two-thirds said that they travelled by road, $\dfrac{13}{30}$ by air and $\dfrac{4}{15}$ by both air and road.
If 20 tourists did not travel by either air or road,

(i) represent the information on a Venn diagram

(ii) how many tourists
(a) were interviewed?
(b) travelled by air only?

$ Let: \\[3ex] number\;\;of\;\;people\;\;that\;\;were\;\;interviewed = n(universal\;\;set) = p \\[3ex] road = R \\[3ex] air = A \\[3ex] n(R \cap A) = \dfrac{4p}{15} \\[5ex] n(R) = \dfrac{2p}{3} \\[5ex] n(R\;\;only) = \dfrac{2p}{3} - \dfrac{4p}{15} = \dfrac{10p - 4p}{15} = \dfrac{6p}{15} = \dfrac{2p}{5} \\[5ex] n(A) = \dfrac{13p}{30} \\[5ex] n(A\;\;only) = \dfrac{13p}{30} - \dfrac{4p}{15} = \dfrac{13p - 8p}{30} = \dfrac{5p}{30} = \dfrac{p}{6} \\[5ex] $ (i) The Venn diagram is:

$ (ii) \\[3ex] (a) \\[3ex] \dfrac{2p}{5} + \dfrac{p}{6} + \dfrac{4p}{15} + 20 = p \\[5ex] LCD = 30 \\[3ex] 30\left(\dfrac{2p}{5}\right) + 30\left(\dfrac{p}{6}\right) + 30\left(\dfrac{4p}{15}\right) + 30(20) = 30(p) \\[5ex] 6(2p) + 5p + 2(4p) + 600 = 30p \\[3ex] 12p + 5p + 8p - 30p = -600 \\[3ex] -5p = -600 \\[3ex] p = \dfrac{-600}{-5} \\[5ex] p = 120 \\[3ex] $ 120 tourists were interviewed

$ (b) \\[3ex] n(A\;\;only) = \dfrac{p}{6} = \dfrac{120}{6} = 20 \\[5ex] $ 20 people travelled by air only

(12.) JAMB

From the Venn diagram above, the complement of the set $P \cap Q$ is given by

$ A.\;\; \{a, b, d, e\} \\[3ex] B.\;\; \{b, d\} \\[3ex] C.\;\; \{a, e\} \\[3ex] D.\;\; \{c\} \\[3ex] $

$ \mu = \{a, b c, d, e\} \\[3ex] P = \{b, c\} \\[3ex] Q = \{c, d\} \\[3ex] P \cap Q = \{c\} \\[3ex] (P \cap Q)' = \{a, b, d, e\} $

(13.) GCSE There are 135 passengers on a plane.
3 of the passengers in Business Class are flying for the first time.
In total, there are 15 passengers in Business Class.
$\dfrac{1}{4}$ of the passengers not in Business Class are flying for the first time.

(a.) In the Venn diagram,
$\xi$ = passengers on the plane
B = passengers in Business Class
F = passengers flying for the first time
Complete the Venn diagram.

(b.) One of the passengers is chosen at random.
Write down the probability that the passenger is in Business Class.

$ (a.) \\[3ex] n(\xi) = 135 \\[3ex] n(B \cap F) = 3 \\[3ex] n(B) = 15 \\[3ex] \implies \\[3ex] n(B\;\;ONLY) = n(B \cap F') \\[3ex] = 15 - 3 = 12 \\[3ex] n(B') = n(\xi) - n(B) \\[3ex] = 135 - 15 \\[3ex] = 120 \\[3ex] n(F \cap B') = \dfrac{1}{4} * 120 = 30 \\[5ex] n(B' \cap F') \\[3ex] = n(neither\;\;B\;\;nor\;\;F) \\[3ex] = 135 - (12 + 30 + 3) \\[3ex] = 135 - 45 \\[3ex] = 90 \\[3ex] $

$ (b.) \\[3ex] n(B) = 15 \\[3ex] n(\xi) = 135 \\[3ex] P(B) = \dfrac{n(B)}{n(\xi)} \\[5ex] = \dfrac{15}{135} \\[5ex] = \dfrac{1}{9} $

(14.) CSEC The incomplete Venn diagram below shows the number of students in a class of 28 who play football and tennis.

U = {all students in the class}
F = {students who play football}
T = {students who play tennis}

Additional information about the class is that
12 students play tennis
15 students play football
8 students play neither football nor tennis
x students play BOTH football and tennis

(i) Complete the Venn diagram above to represent the information, showing the number of students in EACH subset.

(ii) Calculate the value of x

(i)

$ (ii) \\[3ex] 15 - x + 12 - x + x + 8 = 28 \\[3ex] 35 - x = 28 \\[3ex] 35 - 28 = x \\[3ex] x = 7 $

(15.) Assume that A represents the set of all plums and B represents the set of all fruits.
What is the correct Venn diagram for the relationship between these sets?

All plums are fruits.
Hence the Venn diagram shows that the set of plums is a proper subset of the set of fruits.
Correct option is D.

(16.) WASCCE In a class of 50 students, 30 offered History, 15 offered History and Geography while 3 did not offer any of the two subjects.
(i) Represent the information on a Venn diagram.

(ii) Find the number of candidates that offered:
(a) History only
(b) Geography only

$ n(Universal\;\;Set) = n(\xi) = 50 \\[3ex] Let: \\[3ex] History = H \\[3ex] Geography = G \\[3ex] n(H) = 30 \\[3ex] n(G) = g \\[3ex] n(H \cap G) = 15 \\[3ex] n(H\;\;only) = 30 - 15 = 15 \\[3ex] n(G\;\;only) = g - 15 \\[3ex] n(neither\;\;H\;\;nor\;\;G) = 3 \\[3ex] $ (i) The Venn diagram is:

$ (ii) \\[3ex] (a) \\[3ex] n(H\;\;only) = 30 - 15 = 15 \\[3ex] $ 15 students offered only History

$ (b) \\[3ex] 15 + (g - 15) + 15 + 3 = 50 \\[3ex] 15 + g - 15 + 15 + 3 = 50 \\[3ex] 18 + g = 50 \\[3ex] g = 50 - 18 \\[3ex] g = 32 \\[3ex] n(G\;\;only) \\[3ex] = g - 15 \\[3ex] = 32 - 15 \\[3ex] = 17 \\[3ex] $ 17 students offered only Geography

(17.) What do amusement park rides and films have in common?
As the thrill and interest in 3D rides becomes more realistic, motion capture brings the computer enhanced graphics of film to amusement park rides.
Universal entertainment paired with Illumination Entertainment to produce the blockbuster film Despicable Me in 2010 and paired again for the 2013 creation of Despicable Me 2.
Illumination Entertainment is also working with Universal Studios to create a 3D ride which debuted in summer of 2012 at Universal Studios in Florida and will also be built and open in spring of 2014 in California.
A company produces work for 250 productions. 175 are films, 80 are 3D rides, and 25 are motion capture.
How many of the productions are purely for 3D rides?

$ A.\;\; 80 \\[3ex] B.\;\; 30 \\[3ex] C.\;\; 55 \\[3ex] D.\;\; Not\;\;enough\;\;information \\[3ex] $

Let:
the productions for Film = F
the productions for 3D Rides = R

$ n(\mu) = 250 \\[3ex] n(Motion\;\;Capture) = n(F \cap R) = 25 \\[3ex] n(R) = 80 \\[3ex] n(R\;\;only) = 80 - 25 = 55 \\[3ex] $ 55 productions are purely for 3D rides.

(18.) GCSE Here are five shapes, A to E

A	Parallelogram
B	Regular pentagon
C	Rhombus
D	Scalene triangle
E	Trapezium

In the Venn diagram,
$\xi$ is the set of all shapes
Q is the set of quadrilaterals
R is the set of shapes which always have rotational symmetry

Complete the Venn diagram with the letters A to E.

Quadrilateral: four-sided polygon
They are: Parallelogram (A), Rhombus (C), and Trapezium (E)

Rotational Symmetry: shape looks the same after some rotations from it's initial position
They are: Parallelogram (A) (order: 2), Regular pentagon (B) (order: 5), and Rhombus (C) (order: 2)

Order of rotational symmetry of 1 (for Trapezium and Scalene triangle) does not count as a rotational symmetry
Explain to students if they ask questions.

$ \xi = \{A, B, C, D, E\} \\[3ex] Q = \{A, C, E\} \\[3ex] R = \{A, B, C\} \\[3ex] Q \cap R = \{A, C\} \\[3ex] Q \cup R = \{A, B, C, E\} \\[3ex] (Q \cup R)' = \{D\} \\[3ex] $ The Venn diagram is:

(19.) ACT Kelly asked 120 students questions about skiing.
The results of the poll are shown in the table below.

Question	Yes	No
1. Have you skied either cross-country or downhill?	65	55
2. If you answered Yes to Question 1, did you ski downhill?	28	37
3. If you answered Yes to Question 1, did you ski cross-country?	45	20

After completing the poll, Kelly wondered how many of the students polled had skied both cross-country and downhill.
How many of the students polled indicated that they had skied both cross-country and downhill?

$ A.\;\; 73 \\[3ex] B.\;\; 65 \\[3ex] C.\;\; 47 \\[3ex] D.\;\; 18 \\[3ex] E.\;\; 8 \\[3ex] $

To solve this question, we shall not deal with the No responses

$ Let: \\[3ex] C = cross-country \\[3ex] D = downhill \\[3ex] n(C \cup D) = 65 \\[3ex] n(D) = 28 \\[3ex] n(C) = 45 \\[3ex] Both\;\;cross-country\;\;and\;\;downhill = n(C \cap D) = e \\[3ex] n(\mu) = 120 \\[3ex] $

$ 45 - e + e + 28 - e = 65 \\[3ex] 73 - e = 65 \\[3ex] 73 - 65 = e \\[3ex] e = 8 \\[3ex] $ 8 students skied both cross-country and downhill

(20.) CSEC In a survey of 30 families, the findings were that:
15 families owned dogs
12 families owned cats
x families owned BOTH dogs and cats
8 families owned NEITHER dogs NOR cats

(i.) Given that:
U = {families in the survey}
C = {families who owned cats}
D = {families who owned dogs}
Use the given information to copy and complete the Venn diagram below.

(ii.) Write an expression, in x, which represents the TOTAL number of families in the survey.

(iii.) Write an equation which may be used to solve for x

(i.) The Venn diagram is:

$ (ii.) \\[3ex] 12 - x + 15 - x + x + 8 \\[3ex] 35 - x \\[3ex] (iii.) \\[3ex] 35 - x = 30 $

Top

(21.) GCSE In a tennis tournament,
98 players took part in the singles only
34 players took part in the doubles only
twice as many players took part in the singles as took part in the doubles

How many players took part in both the singles and the doubles?
You may use the Venn diagram to help you.

Let the number of players in both the singles and the doubles = p

twice as many players took part in the singles as took part in the doubles

$ n(Singles) = 98 + p \\[3ex] n(Doubles) = 34 + p \\[3ex] \implies \\[3ex] 98 + p = 2(34 + p) \\[3ex] 98 + p = 68 + 2p \\[3ex] 68 + 2p = 98 + p \\[3ex] 2p - p = 98 - 68 \\[3ex] p = 30 \\[3ex] $ 30 players took part in both the singles and the doubles

(22.) ACT In a small high school with 20 seniors, 8 of the seniors are in soccer, 9 of the seniors are in band, and 5 of the seniors are in both.
How many seniors are in neither soccer nor band?

$ A.\;\; 15 \\[3ex] B.\;\; 12 \\[3ex] C.\;\; 11 \\[3ex] D.\;\; 8 \\[3ex] E.\;\; 3 \\[3ex] $

$ Let: \\[3ex] Soccer = C \\[3ex] Band = B \\[3ex] Neither\;\;soccer\;\;nor\;\;band = p \\[3ex] $

$ 3 + 4 + 5 + p = 20 \\[3ex] 12 + p = 20 \\[3ex] p = 20 - 12 \\[3ex] p = 8 \\[3ex] $ 8 seniors are in neither soccer nor band.

(23.) CSEC The universal set U is defined as follows:
$U = \{x:x \in N, 2 \lt x \lt 12\}$
The sets M and R are subsets of U such that
M = {odd numbers}
R = {square numbers}

(i) List the members of the subset M

(ii) List the members of the subset R

(iii) Draw a Venn Diagram that represents the relationship among the defined subsets of U

$ U = \{x:x \in N, 2 \lt x \lt 12\} \\[3ex] U = \{3, 4, 5, 6, 7, 8, 9, 10, 11\} \\[3ex] (i) \\[3ex] M = \{odd\;\;numbers\} \\[3ex] M = \{3, 5, 7, 9, 11\} \\[3ex] (ii) \\[3ex] R = \{square\;\;numbers\} \\[3ex] R = \{4, 9\} \\[3ex] (iii) \\[3ex] M \cap R = \{9\} \\[3ex] M \cup R = \{3, 4, 5, 7, 9, 11\} \\[3ex] (M \cup R)' = \{6, 8, 10\} \\[3ex] $ The Venn diagram is:

(24.) ACT After polling a class of 20 music students by a show of hands, you find that 8 students play the guitar and 9 students play the piano.
Given that information, what is the minimum number of students in this music class who play both the guitar and the piano?

$ F.\;\; 0 \\[3ex] G.\;\; 1 \\[3ex] H.\;\; 8 \\[3ex] J.\;\; 9 \\[3ex] K.\;\; 17 \\[3ex] $

$ Let: \\[3ex] guitar = G \\[3ex] piano = P \\[3ex] n(G) = 8 \\[3ex] n(P) = 9 \\[3ex] Both\;\;guitar\;\;and\;\;piano = n(G \cap P) = k \\[3ex] n(\mu) = 20 \\[3ex] $

$ 8 - k + k + 9 - k = 20 \\[3ex] 17 - k = 20 \\[3ex] 17 - 20 = k \\[3ex] k = -3 \\[3ex] $ This is not possible because the number that play both the guitar and the piano cannot be negative
That number must be non-negative (zero and positive integers only)
So, let us analyze the options.
The number of students that play ONLY the guitar cannot be negative. This eliminates Options J. and K.
Any of Options F., G., and H. can be the number of students that play both the guitar and the piano.
However, the question is asking for the minimum number
That leaves us with zero...Option F. as the correct answer

(25.) A scan of all biology and business majors at a college shows the accompanying breakdown by gender.
Let set A equals women and set B equals business majors.

$ n(Women) = 33 + 113 = 146 \\[3ex] n(Men) = 20 + 88 = 108 \\[3ex] n(Biology) = 33 + 20 = 53 \\[3ex] n(Business) = 113 + 88 = 201 \\[3ex] n(\mu) = n(Women) + n(Men) = 146 + 108 = 254 \\[3ex] OR \\[3ex] n(\mu) = n(Biology) + n(Business) = 53 + 201 = 254 \\[5ex] Women = A \\[3ex] Business = B \\[3ex] n(A \cap B) = 113 \\[3ex] n(A) = n(Women) = 146 \\[3ex] n(only\;\;A) = n(A \cap B') = 146 - 113 = 33 \\[3ex] n(B) = n(Business) = 201 \\[3ex] n(only\;\;B) = n(B \cap A') = 201 - 113 = 88 \\[3ex] n(neither\;\;A\;\;nor\;\;B) \\[3ex] = n(A \cup B)' \\[3ex] = 254 - (33 + 113 + 88) \\[3ex] = 254 - 234 \\[3ex] = 20 \\[3ex] $ The Venn diagram for the set of comedies and the set of favorable performances is:

Option C.

(26.) GCSE The Venn diagram shows information about 50 people who are in bands.

(a) How many of the people are guitar players?

(b) How many of the people are singers but not guitar players?

(c) One of the people is chosen at random.
Write down the probability that the person is not a singer and not a guitar player.

$ (a) \\[3ex] n(guitar\;\;players) = n(G) = 21 + 8 = 29 \\[3ex] $ 29 people are guitar players

$ (b) \\[3ex] n(singers\;\;but\;\;not\;\;guitar\;\;players) \\[3ex] = n(singers\;\;only) \\[3ex] = 4 \\[3ex] $ 4 people are singers but not guitar players

$ (c) \\[3ex] n(universal\;\;set) = n(\xi) = 50 \\[3ex] n(not\;\;a\;\;singer\;\;and\;\;not\;\;a\;\;guitar\;\;player) = 17 \\[3ex] P(not\;\;a\;\;singer\;\;and\;\;not\;\;a\;\;guitar\;\;player) \\[3ex] = \dfrac{n(not\;\;a\;\;singer\;\;and\;\;not\;\;a\;\;guitar\;\;player)}{n(\xi)} \\[5ex] = \dfrac{17}{50} $

(28.) CSEC On a certain day, 300 customers visited a bakery that sell bread and cakes.
70 customers bought cakes only
80 customers bought neither bread nor cakes
2x customers boought bread only
x customers bought both bread and cakes

(i) U represents the rest of customers visiting the bakery on that day, B represents the set of customers who bought bread, and C represents the set of customers who bought cake.
Copy and complete the Venn diagram to illustrate the information.

(ii) Write an expression in x to represent the TOTAL number of customers who visited the bakery on that day.

(iii) Calculate the number of customers who bought bread ONLY.

(i) The Venn diagram is:

$ (ii) \\[3ex] TOTAL\;\;number\;\;of\;\;customers\;\;that\;\;day \\[3ex] = 2x + x + 70 + 80 \\[3ex] = 3x + 150 \\[3ex] (iii) \\[3ex] 3x + 150 = 300 \\[3ex] 3x = 300 - 150 \\[3ex] 3x = 150 \\[3ex] x = \dfrac{150}{3} \\[5ex] x = 50 \\[3ex] n(B\;\;only) \\[3ex] = 2x \\[3ex] = 2(50) \\[3ex] 100 \\[3ex] $ 100 customers bought bread only

(29.) GCSE A and B are two events.
Some probabilities are shown on the Venn diagram.

Work out $P(A' \cup B)$

Please do this quickly...find the shaded area of $P(A' \cup B)$...as seen in Number (5.): Venn Diagram for Two Sets
As noted, we are concerned with those three shaded areas
But, first; let us find the probability of neither A nor B
The sum of all the probabilities is 1

$ P(A \cup B)' = P(neither) \\[3ex] 0.3 + 0.15 + 0.35 + P(neither) = 1 \\[3ex] 0.8 + P(neither) = 1 \\[3ex] P(neither) = 1 - 0.8 \\[3ex] P(neither) = 0.2 \\[3ex] $ So, we need to add these shaded areas

$ P(A' \cup B) \\[3ex] = 0.15 + 0.35 + 0.2 \\[3ex] = 0.7 $

(30.) Use the Venn diagram to answer questions (a.) through (d.)

(a.) How many women at the party are under 30?
(b.) How many men at the party are not under 30?
(c.) How many women are at the party?
(d.) How many people are at the party?

Let the set of:
men = M
people under 30 = E

(a.) Women at the party under 30 = $n(M' \cap E)$ = the people under 30 only = 11
11 women at the party are under 30

(b.) Men at the party are not under 30 = $n(M \cap E')$ = the men only = 18
18 men at the party are not under 30

(c.) Women at the party = $n(A')$ = the sum of: under the age 30 only and neither men nor under age 30 = 11 + 25 = 36
36 women were at the party.

(d.) Number of people at the party = n(Universal Set) = the sum of: men only and under age 30 only and the men under age 30 and neither men nor under the age 30 = 18 + 11 + 12 + 25 = 66
66 people were at the party.

(31.) CSEC A Universal set, U, is defined as
U = {15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25}
Sets M and N are subsets of U such that
M = {Prime Numbers} and N = {Even Numbers}

(i) Draw a Venn diagram to represent the sets M, N and U

(ii) List the elements of the set $(M \cup N)'$

$ (ii) \\[3ex] U = \{15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25\} \\[3ex] M = \{17, 19, 23\} \\[3ex] N = \{16, 18, 20, 22, 24\} \\[3ex] M \cap N = \phi \\[3ex] M \cup N = \{16, 17, 18, 19, 20, 22, 23, 24\} \\[3ex] (M \cup N)' = \{15, 21, 25\} \\[3ex] $ (i)
Sets M and N are disjoint sets

(32.) ACT All 30 students in a history class took 2 tests.
Of these students, 28 passed the first test and 25 passed the second test.
What is the maximum possible number of students who passed both tests?

$ F.\;\; 23 \\[3ex] G.\;\; 25 \\[3ex] H.\;\; 27 \\[3ex] J.\;\; 28 \\[3ex] K.\;\; 30 \\[3ex] $

$ Let: \\[3ex] first\;\;test = F \\[3ex] second\;\;test = E \\[3ex] n(F) = 28 \\[3ex] n(E) = 25 \\[3ex] n(both\;\;tests) = n(F \cap E) = p \\[3ex] n(\xi) = 30 \\[3ex] $

$ 28 - p + p + 25 - p = 30 \\[3ex] 53 - p = 30 \\[3ex] 53 - 30 = p \\[3ex] p = 23 \\[3ex] $ The number of students who passed both tests is 23
However, the maximum number of students who passed both tests is 25

Student: How is it 25?
Teacher: Because the number of students who passed either test ONLY cannot be negative
The number of students who passed the either test ONLY should be at least the number of people who passed either test
For example: the number of people who passed the second test ONLY should be at least the number of people who passed the second test.
In other words, the number of people who passed the second test ONLY cannot be negative.
Student: You used the second test as the example because 28 is greater than 25?
Teacher: That is correct.

(34.) Of the 36 theater performances that a critic reviewed, 19 were comedies.
She gave favorable reviews to 7 of the comedies and unfavorable reviews to 9 of the non-comedies.
(a.) Complete a two-way table summarizing the reviews.
(b.) Make a Venn diagram for two sets from the table in part (a) using the set of comedies and the set of favorable performances as the two sets.
(c.) How many comedies received unfavorable reviews?
(d.) How many non-comedies received favorable reviews?

Total performances = 36
Comedy = 19
Non-comedy = 36 − 19 = 17
Favorable comedy = 7
Unfavorable comedy = 19 − 7 = 12
Unfavorable non-comedy = 9
Favorable non-comedy = 17 − 9 = 8

(a.) The two-way table summarizing the reviews is:

	Favorable	Unfavorable
Comedy	7	12
Non-comedy	8	9

Let the set of:
comedy = C
favorable performances = F

$ n(\mu) = 36 \\[3ex] n(C \cap F) = 7 \\[3ex] n(C) = 19 \\[3ex] n(only\;\;C) = n(C \cap F') = 19 - 7 = 12 \\[3ex] n(F) = 7 + 8 = 15 \\[3ex] n(only\;\;F) = n(F \cap C') = 15 - 7 = 8 \\[3ex] n(neither\;\;C\;\;nor\;\;F) \\[3ex] = n(C \cup F)' \\[3ex] = 36 - (12 + 8 + 7) \\[3ex] = 36 - 27 \\[3ex] = 9 \\[3ex] $ (b.) The Venn diagram for the set of comedies and the set of favorable performances is:

Based on the table in Part (a.):
(c.) 12 comedies received unfavorable reviews.
(d.) 8 non-comedies received favorable reviews.

(35.) GCSE A school has 86 teachers.
42 are male and 44 are female.

$\dfrac{1}{3}$ of the male teachers have blue eyes.

$\dfrac{1}{4}$ of the female teachers have blue eyes.

(a) $\xi$ = teachers in the school
M = male teachers
B = teachers who have blue eyes

Complete the Venn diagram.

(b) One teacher who has blue eyes is chosen at random.
Work out the probability that the teacher is male.

$ (a) \\[3ex] Male\;\;teachers\;\;with\;\;blue\;\;eyes \\[3ex] = n(M \cap B) \\[3ex] = \dfrac{1}{3} * 42 \\[5ex] = 14 \\[3ex] Male\;\;teachers = n(M) = 42 \\[3ex] n(Male\;\;teachers\;\;only) = 42 - 14 = 28 \\[3ex] Female\;\;teachers\;\;with\;\;blue\;\;eyes \\[3ex] = \dfrac{1}{4} * 44 \\[5ex] = 11 \\[3ex] Teachers\;\;with\;\;blue\;\;eyes \\[3ex] = n(blue\;\;eyes) \\[3ex] = 14 + 11 = 25 \\[3ex] n(Blue\;\;eyes\;\;only) = 25 - 14 = 11 \\[3ex] Let\;\;n(neither\;\;male\;\;nor\;\;blue\;\;eyes) = k \\[3ex] 28 + 14 + 11 + k = 86 \\[3ex] 53 + k = 86 \\[3ex] k = 86 - 53 \\[3ex] k = 33 \\[3ex] $ The Venn Diagram is:

$ n(sample\;\;space) = n(B) = 25 \\[3ex] n(M \cap B) = 14 \\[3ex] P(M \cap B) = \dfrac{n(M \cap B)}{n(B)} = \dfrac{14}{25} $

(36.) CSEC A school has 90 students in Form 5
54 students study Physical Education
42 students study Music
6 students study neither Physical Education nor Music
x students study both Physical Education and Music

(i) Copy the Venn diagram shown below.

(ii) Show on your Venn diagram the information relating to the students in Form 5

(iii) Calculate the number of students who study BOTH Physical Education and Music.

(ii) The information on the Venn diagram is:

$ (iii) \\[3ex] 54 - x + 42 - x + x + 6 = 90 \\[3ex] 102 - x = 90 \\[3ex] 102 - 90 = x \\[3ex] x = 12 \\[3ex] $ 12 students study BOTH Physical Education and Music

(38.) One hundred and four people who grew up in either City A or City B were surveyed to determine whether they preferred country music or blues (both and neither were not acceptable responses).
Of those who grew up in City B, 44 preferred blues and 15 preferred country music.
Of those who grew up in City A, 21 preferred country music.
(a.) Complete a two-way table summarizing the reviews.
(b.) Draw a Venn diagram for the sets: City B and Country.
(c.) How many City B residents preferred blues?
(d.) How many respondents preferred blues?

Total number of residents in City A and City B = 104
City B:
Blues = 44
Country music = 15
City A:
Country music = 21
Blues = 104 − (44 + 15 + 21)
Blues = 104 − 80 = 24

(a.) The two-way table summarizing the reviews is:

	Country Music	Blues
City A	21	24
City B	15	44

Let the set of:
City B = B
Country Music = C

$ n(\mu) = 104 \\[3ex] n(B \cap C) = 15 \\[3ex] n(B) = 44 + 15 = 59 \\[3ex] n(only\;\;B) = n(B \cap C') = 59 - 15 = 44 \\[3ex] n(C) = 15 + 21 = 36 \\[3ex] n(only\;\;C) = n(C \cap B') = 36 - 15 = 21 \\[3ex] n(neither\;\;B\;\;nor\;\;C) \\[3ex] = n(B \cup C)' \\[3ex] = 104 - (44 + 21 + 15) \\[3ex] = 104 - 80 \\[3ex] = 24 \\[3ex] $ (b.) The Venn diagram for the set of City B and the set of Country Music is:

Based on the table in Part (a.):
(c.) 44 City B residents preferred blues.
(d.) Number of respondents that preferred blues = 24 + 44 = 68.

(39.) CSEC (a) Describe, using set notation only, the shaded regions in each Venn diagram below.
The first one is done for you.

(b) The following information is given.
U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
P = {prime numbers}
Q = {odd numbers}
Draw a Venn diagram to represent the information above.

(c) The Venn diagram below shows the number of elements in each region.

Determine how many elements are in EACH of the following sets:
(i) $A \cup B$
(ii) $A \cap B$
(iii) $(A \cap B)'$
(iv) $U$

(a)
(i) The shaded area is the intersection of the sets: A and B ...what they have in common
$A \cap B$

(ii) The shaded area is the union of sets A and B
$A \cup B$

(iii) We can determine it...as I explained it in the Video Number (3.) in the playlist
Notice that sets A and B are disjoint sets
So, let us make up an example:
U = {1, 2, 3, 4, 5}
A = {1, 2}
B = {3, 4}
A' = {3, 4, 5}
B' = {1, 2, 5}
5 is the only number outside sets A and B... that is shaded
$A' \cap B' = \{5\}$
Therefore, the shaded area is $A' \cap B'$

(iv)
The shaded area is Set A

$ (b) \\[3ex] U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} \\[3ex] P = \{2, 3, 5, 7\} \\[3ex] Q = \{1, 3, 5, 7, 9\} \\[3ex] P \cap Q = \{3, 5, 7\} \\[3ex] P \cup Q = \{1, 2, 3, 5, 7, 9\} \\[3ex] (P \cup Q)' = \{4, 6, 8, 10\} \\[3ex] $ The Venn Diagram is:

$ (c) \\[3ex] (i) \\[3ex] n(A \cup B) = 10 + 3 + 4 = 17 \\[3ex] (ii) \\[3ex] n(A \cap B) = 4 \\[3ex] (iii) \\[3ex] n(A \cap B)' = 10 + 3 + 8 = 21 \\[3ex] (iv) \\[3ex] n(U) = 10 + 4 + 3 + 8 = 25 $

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(42.) In a trial of a new allergy medicine, 85 people were given the medicine and 85 were given a placebo.
Of those given the medicine, 60 showed improvement in their allergy symptoms.
Of those given the placebo, 55 did not show improvement.
(a.) Construct a two-way table summarizing the reviews.
(b.) Make a Venn diagram for the sets: Medicine and Improvement.
(c.) How many people who received medicine did not improve?
(d.) How many people who received a placebo improved?

Medicine
Total = 85
Improvement = 60
No Improvement = 85 − 60 = 25
Placebo
Total = 85
No Improvement = 55
Improvement = 85 − 55 = 30

(a.) The two-way table summarizing the reviews is:

	Improvement	No Improvement
Medicine	60	25
Placebo	30	55

Let the set of:
Medicine = M
Improvement = I

$ n(\mu) = 85 + 85 = 170 \\[3ex] n(M \cap I) = 60 \\[3ex] n(M) = 85 \\[3ex] n(only\;\;M) = n(M \cap I') = 85 - 60 = 25 \\[3ex] n(I) = 60 + 30 = 90 \\[3ex] n(only\;\;I) = n(I \cap M') = 90 - 60 = 30 \\[3ex] n(neither\;\;M\;\;nor\;\;I) \\[3ex] = n(M \cup I)' \\[3ex] = 170 - (25 + 60 + 30) \\[3ex] = 170 - 115 \\[3ex] = 55 \\[3ex] $ (b.) The Venn diagram for the set of Medicine and the set of Improvement is:

Based on the table in Part (a.):
(c.) 25 people who received medicine did not improve.
(d.) 30 people who received a placebo improved.

(43.) WASSCE Out of 120 customers in a shop, 45 bought bags and shoes.
If all the customers bought either bags or shoes and 11 more customers bought shoes than bags:

(a) Illustrate this information in a diagram

(b) Find the number of customers who bought shoes

(c) Calculate the probability that a customer selected at random bought bags

$ n(\xi) = 120 \\[3ex] Let: \\[3ex] bags = B \\[3ex] shoes = H \\[3ex] n(B \cap H) = n(both) = 45 \\[3ex] n(B) = n(bags) = b \\[3ex] \rightarrow n(H) = n(shoes) = 11 + b \\[3ex] n(bags\;\;only) = b - 45 \\[3ex] n(shoes\;\;only) = (11 + b) - 45 = 11 + b - 45 = b - 34 \\[3ex] $ (a)
The Venn diagram is:

$ (b - 45) + (b - 34) + 45 = 120 \\[3ex] b - 45 + b - 34 + 45 = 120 \\[3ex] 2b = 120 + 34 \\[3ex] 2b = 154 \\[3ex] b = \dfrac{154}{2} \\[5ex] b = 77 \\[3ex] (b) \\[3ex] n(shoes) = n(H) \\[3ex] = (b - 34) + 45 \\[3ex] = b - 34 + 45 \\[3ex] = 77 - 34 + 45 \\[3ex] = 88 \\[3ex] $ 88 customers bought shoes

$ (c) \\[3ex] n(bags) = n(B) \\[3ex] = (b - 45) + 45 \\[3ex] = 77 - 45 + 45 \\[3ex] = 77 \\[3ex] n(\xi) = 120 \\[3ex] P(bags) = \dfrac{n(bags)}{n\xi} = \dfrac{77}{120} $

(44.) CSEC A universal set, U, is defined as:
U = {51, 52, 53, 54, 55, 56, 57, 58, 59}
A and B are subsets of U, such that:
A = {odd numbers}
B = {prime numbers}

(i) List the members of the set A

(ii) List the members of the set B

(iii) Draw a Venn diagram to represent the sets A, B and U

$ U = \{51, 52, 53, 54, 55, 56, 57, 58, 59\} \\[3ex] (i) \\[3ex] A = \{51, 53, 55, 57, 59\} \\[3ex] (ii) \\[3ex] B = \{53, 59\} \\[3ex] A \cap B = \{53, 59\} \\[3ex] A \cup B = \{51, 53, 55, 57, 59\} \\[3ex] (A \cup B)' = \{52, 54, 56, 58\} \\[3ex] $ The Venn diagram that represents the sets A, B and U is:

(45.) CSEC The universal set, U, is defined as the set of integers between 11 and 26
A and B are subsets of U such that:
A = {even numbers}
B = {multiples of 3}

(i) How many members are in the universal set, U?

(ii) List the members of the subset, A

(iii) List the members of the subset, B

(iv) Draw a Venn diagram to represent the relationships among A, B and U
(Assume: between means that the two endpoints are not included)

$ (i) \\[3ex] U = \{12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25\} \\[3ex] n(U) = 14 \\[3ex] (ii) \\[3ex] A = \{12, 14, 16, 18, 20, 22, 24\} \\[3ex] (iii) \\[3ex] B = \{12, 15, 18, 21, 24\} (iv) \\[3ex] A \cap B = \{12, 18, 24\} \\[3ex] A \cup B = \{12, 14, 15, 16, 18, 20, 21, 22, 24\} \\[3ex] (A \cup B)' = \{13, 17, 19, 23, 25\} \\[3ex] $ The Venn diagram is:

(46.) JAMB

The Venn diagram below shows the number of students offering Music and History in a class of 80 students.
If a student is picked at random from the class, what is the probability that he offers Music only?

$ A.\;\; 0.13 \\[3ex] B.\;\; 0.25 \\[3ex] C.\;\; 0.38 \\[3ex] D.\;\; 0.50 \\[3ex] $

$ 30 - x + 40 - x + x + 20 = 80 \\[3ex] 90 - x = 80 \\[3ex] 90 - 80 = x \\[3ex] 10 = x \\[3ex] x = 10 \\[3ex] n(Music\;\;only) \\[3ex] = 30 - x \\[3ex] = 30 - 10 \\[3ex] = 20 \\[3ex] P(Music\;\;only) = \dfrac{n(Music\;\;only)}{n(\mu)} = \dfrac{20}{80} = \dfrac{1}{4} = 0.25 $

(49.) GCSE In the Venn diagram
$\xi$ represents 31 students in a class
C is students who have a cat
D is students who have a dog

(a) One student from the class is picked at random.
Work out the probability that the student has a dog.

(b) One of the students who has a cat is picked at random.
Work out the probability that this student has a dog.

$ 6 + 2x + 5 + (x + 2) = 31 \\[3ex] 11 + 2x + x + 2 = 31 \\[3ex] 3x + 13 = 31 \\[3ex] 3x = 31 - 13 \\[3ex] 3x = 18 \\[3ex] x = \dfrac{18}{3} \\[5ex] x = 6 \\[3ex] (a) \\[3ex] n(sample\;\;space) = n(\xi) = 31 \\[3ex] n(event\;\;space) = n(D) \\[3ex] = 2x + 5 \\[3ex] = 2(6) + 5 \\[3ex] = 12 + 5 \\[3ex] = 17 \\[3ex] P(D) = \dfrac{n(D)}{n(\xi)} = \dfrac{17}{31} \\[5ex] (b) \\[3ex] n(sample\;\;space) = n(C) = 6 + 5 = 11 \\[3ex] n(event\;\;space) = n(C \cap D) = 5 \\[3ex] P(C \cap D) = \dfrac{n(C \cap D)}{n(C)} = \dfrac{5}{11} $

(50.) CSEC There are 50 students in a class.
Students in the class were given awards for Mathematics or Science.
36 students received awards in either Mathematics or Science.
6 students received awards in BOTH Mathematics and Science
2x students received awards for Mathematics only.
x students received awards for Science only.
In the Venn diagram below:
U = {all students in the class}
M = {students who received awards for Mathematics}
S = {students who received awards for Science}

(i) Copy and complete the Venn diagram to represent the information about the awards given, showing the number of students in EACH subsets.

(ii) Calculate the value of x

$ n(U) = 50 \\[3ex] n(M \cup S) = 36 \\[3ex] n(neither\;\;M\;\;nor\;\;S) = n(M \cup S)' = 50 - 36 = 14 \\[3ex] $ (i) The Venn diagram is:

$ (ii) \\[3ex] 2x + x + 6 = 36 \\[3ex] 3x = 36 - 6 \\[3ex] 3x = 30 \\[3ex] x = \dfrac{30}{3} \\[5ex] x = 10 $

(56.) CSEC The Venn diagram below shows the number of students in Form 5A who have visited Canada (C) or Dominica (D)

(i) How many students have visited Dominica ONLY?

(ii) Write an expression, in terms of x, to represent the TOTAL number of students who have visited Canada.

(iii) Given that there are 25 students in Form 5A, calculate the value of x

(iv) Hence, write down the number of students in each of the following subsets:

$C \cup D$
$C \cap D$
$(C \cup D)'$

(i) 10 students have visited Dominica ONLY

(ii) The total number of number of students who have visited Canada is: $3 + x$

$ (iii) \\[3ex] n(U) = 25 \\[3ex] 3 + 10 + x + 2x = 25 \\[3ex] 13 + 3x = 25 \\[3ex] 3x = 25 - 13 \\[3ex] 3x = 12 \\[3ex] x = \dfrac{12}{3} \\[5ex] x = 4 \\[3ex] (iv) \\[3ex] n(C \cup D) = 3 + 10 + x = 13 + 4 = 17 \\[3ex] n(C \cap D) = x = 4 \\[3ex] n(C \cup D)' = 2x = 2(4) = 8 $

(60.) GCSE Twenty people go on a trip to the seaside.
Of these 20 people

13 swim in the sea
17 go to the funfair
2 do not swim in the sea or go to the funfair

(a) Complete the Venn diagram below to show this information.

One person is chosen at random.

(b) Find the probability that this person swims in the sea and goes to the funfair.

(c) Find the probability that this person either swims in the sea or goes to the funfair, but does not do both.

(d.) One person is chosen at random from those who swim in the sea.
Find the probability that this person does not go to the funfair.

$ Let: \\[3ex] swim\;\;in\;\;sea = W \\[3ex] go\;\;to\;\;the\;\;funfair = G \\[3ex] n(swim) = n(W) = 13 \\[3ex] n(go) = n(G) = 17 \\[3ex] n(both) = n(W \cap G) = p n(swim\;\;only) = 13 - p \\[3ex] n(go\;\;only) = 17 - p \\[3ex] n(neither) = n((W \cup G)') = 2 \\[3ex] $ The Venn diagram is:

$ 13 - p + 17 - p + p + 2 = 20 \\[3ex] 32 - p = 20 \\[3ex] 32 - 20 = p \\[3ex] p = 12 \\[3ex] \implies \\[3ex] n(both) = 12 \\[3ex] n(swim\;\;only) \\[3ex] = 13 - p \\[3ex] = 13 - 12 \\[3ex] = 1 \\[3ex] n(go\;\;only) \\[3ex] = 17 - p \\[3ex] = 17 - 12 \\[3ex] = 5 \\[3ex] $ The completed Venn diagram is:

$ (b) \\[3ex] P(both) = \dfrac{n(both)}{n(\varepsilon)} \\[5ex] = \dfrac{12}{20} \\[5ex] = \dfrac{3}{5} \\[5ex] (c) \\[3ex] $ Swimming in the sea or going to the funfair without doing both are:
those who do only one: swim the sea only OR go to the funfair only

$ n(only\;\;one) \\[3ex] = 1 + 5 = 6 \\[3ex] P(only\;\;one) \\[3ex] = \dfrac{n(only\;\;one)}{n(\varepsilon)} \\[5ex] = \dfrac{6}{20} \\[5ex] = \dfrac{3}{10} \\[5ex] $ Selecting somone who swims but does not go to the funfair means that the person swims only

$ (d) \\[3ex] n(swim) = 13 \\[3ex] n(swim\;\;only) = 1 \\[3ex] P(swim\;\;only) = \dfrac{n(swim\;\;only)}{n(swim)} = \dfrac{1}{13} $

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(62.) CSEC In a survey of 39 students, it was found that
18 can ride a bicycle
15 can drive a car
x can ride a bicycle and drive a car
3x can do neither

(i) Copy and complete the Venn diagram to represent the information

(ii) Write an expression in x for the number of students in the survey.

(iii) Calculate the value of x

(i) The Venn diagram is:

$ (ii) \\[3ex] Number\;\;of\;\;students \\[3ex] = 18 - x + 15 - x + x + 3x \\[3ex] = 33 + 2x \\[3ex] = 2x + 33 \\[3ex] (iii) \\[3ex] \implies \\[3ex] 2x + 33 = 39 \\[3ex] 2x = 39 - 33 \\[3ex] 2x = 6 \\[3ex] x = \dfrac{6}{2} \\[5ex] x = 3 $

(68.) CSEC In a survey of 50 students,
23 owned cellular phones
18 owned digital cameras
x owned cellular phones and digital cameras
2x owned neither.
Let C represent the set of students in the survey who owned cellular phones, and D the set of students who owned digital cameras.

(i) Copy and complete the Venn diagram below to represent the information obtained from the survey.

(ii) Write an expression in x for the TOTAL number of students in the survey.

(iii) Calculate the value of x

(i) The Venn diagram is:

$ (ii) \\[3ex] TOTAL\;\;number\;\;of\;\;students \\[3ex] = 23 - x + 18 - x + x + 2x \\[3ex] = 41 + x \\[3ex] = x + 41 \\[3ex] (iii) \\[3ex] \implies \\[3ex] x + 41 = 50 \\[3ex] x = 50 - 41 \\[3ex] x = 9 $

(71.) GCSE $\mathcal{E}$ = {1, 2, 3, 4, 5, 6, 7, 8, 9}
A = {1, 5, 6, 8, 9}
B = {2, 6, 9}

(a) Complete the Venn diagram to represent this information.

A number is chosen at random from the universal set $\mathcal{E}$
(b) Find the probability that the number is in the set $A \cap B$

$ A \cap B = \{6, 9\} \\[3ex] A \cup B = \{1, 2, 5, 6, 8, 9\} \\[3ex] (A \cup B)' = \{3, 4, 7\} \\[3ex] $ (a) The Venn diagram is:

$ (b) \\[3ex] n(A \cap B) = 2 \\[3ex] n(\mathcal{E}) = 9 \\[3ex] P(A \cap B) = \dfrac{n(A \cap B)}{n(\mathcal{E})} = \dfrac{2}{9} $

(72.) CSEC In a class of 32 students, ALL the students study Spanish (S) and 20 of the students study French (F)

(i) Represent this information on a Venn diagram

(ii) Calculate the number of students who study Spanish (S) but NOT French (F)

(iii) Write, using set notation, the relationship between F and S

All the students stidy Spanish
This implies that the Spanish set, S is also the universal set
It also implies that the French set, F is a subset of S because every set is a subset of the universal set
In other words, F is included in S

(i) The Venn diagram is:

$ (ii) \\[3ex] n(Spanish\;\;but\;\;NOT\;\;French) \\[3ex] = n(S\;\;ONLY) \\[3ex] = 32 - 20 \\[3ex] = 12 \\[3ex] $ 12 students study Spanish but not French

$ (iii) \\[3ex] Because\;\; n(S) = n(\mu) = 32 \\[3ex] F \subseteq S $

(76.) CSEC The Venn diagram below shows the number of students who study Music and Art in a class of 35 students.
U = {students in the class}
M = {students who study Music}
A = {students who study Art}

(i) How many students study neither Art nor Music?

(ii) Calculate the value of x

(iii) Hence, state the number of students who study Music only.

(i)
4 students study neither Art nor Music

$ (ii) \\[3ex] (x + 5) + x + 8 + 4 = 35 \\[3ex] x + 5 + x + 8 + 4 = 35 \\[3ex] 2x + 17 = 35 \\[3ex] 2x = 35 - 17 \\[3ex] 2x = 18 \\[3ex] x = \dfrac{18}{2} \\[5ex] x = 9 \\[3ex] (iii) \\[3ex] n(Music\;\;ONLY) \\[3ex] = x + 5 \\[3ex] = 9 + 5 \\[3ex] = 14 \\[3ex] $ 14 students study only Music.

(77.) CSEC In a survey of 36 students, it was found that
30 play tennis,
x play volleyball ONLY,
9x play BOTH tennis and volleyball
4 play neither tennis nor volleyball.

(i) Given that:
U = {students in the survey}
V = {students who play Volleyball}
T = {students who play Tennis}
Copy and complete the Venn diagram below to show the number of students in the subsets marked y and z

(ii) (a) Write an expression in x to represent the TOTAL number of students in the survey.

(b) Write an equation in x to represent the total number of students in the survey and hence solve for x

(i) The Venn diagram is:

$ (ii) \\[3ex] (a) \\[3ex] Total\;\;number\;\;of\;\;students \\[3ex] = x + y + 9x + z \\[3ex] = x + (30 - 9x) + 9x + 4 \\[3ex] = x + 30 - 9x + 9x + 4 \\[3ex] = x + 34 \\[3ex] (b) \\[3ex] \implies \\[3ex] x + 34 = 36 \\[3ex] x = 36 - 34 \\[3ex] x = 2 $

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(83.) CSEC A surevy of the 30 students in Form 5 showed that some students used cameras (C) or mobile phones (M) to take photographs.
20 students used mobile phones
4x students used ONLY cameras
x students used BOTH mobile phones and cameras
2 students did not use either cameras or phones.

(i) Copy the Venn diagram below and complete it to show, in terms of x, the number of students in each region.

(ii) Write an expression, in terms of x, which represents the TOTAL number of students in the survey.

(iii) Determine the number of students in Form 5 who used ONLY cameras.

(i) The Venn diagram is:

$ (ii) \\[3ex] Total\;\;number\;\;of\;\;students \\[3ex] = 4x + (20 - x) + x + 2 \\[3ex] = 4x + 20 -x + x + 2 \\[3ex] = 4x + 22 \\[3ex] (iii) \\[3ex] \implies \\[3ex] 4x + 22 = 30 \\[3ex] 4x = 30 - 22 \\[3ex] 4x = 8 \\[3ex] x = \dfrac{8}{4} \\[5ex] x = 2 $

(88.) CSEC

In the diagram shown above, the Universal set, (U), represents all the students in a class.
The set M represents the students who take Music.
The set D represents the students who take Drama.
If 24 students take Music, calculate

(i) the number of students who take BOTH Music and Drama

(ii) the number of students who take Drama ONLY

$ (i) \\[3ex] n(M) = 24 \\[3ex] n(M) = 3x + x \\[3ex] \implies \\[3ex] 3x + x = 24 \\[3ex] 4x = 24 \\[3ex] x = \dfrac{24}{4} \\[5ex] x = 6 \\[3ex] $ 6 students take BOTH Music and Drama

$ (ii) \\[3ex] n(D\;\;ONLY) \\[3ex] = 7 + x \\[3ex] = 7 + 6 \\[3ex] = 13 \\[3ex] $ 13 students take only Drama

(91.) CSEC A survey was conducted among 40 tourists. The results were:
28 visited Antigua (A)
30 visited Barbados (B)
3x visited both Antigua and Barbados
x visited neither Antigua nor Barbados

(i) Copy and complete the Venn diagram below to represent the given information above.

(ii) Write an expression, in x, to represent the TOTAL number of tourists in the survey.

(iii) Calculate the value of x

(i) The Venn diagram is:

$ (ii) \\[3ex] The\;\;total\;\;number\;\;of\;\;tourists \\[3ex] = (28 - 3x) + (30 - 3x) + 3x + x \\[3ex] = 28 - 3x + 30 - 3x + 3x + x \\[3ex] = 58 - 2x \\[3ex] (iii) \\[3ex] The\;\;total\;\;number\;\;of\;\;tourists = 40 \\[3ex] \implies \\[3ex] 58 - 2x = 40 \\[3ex] 58 - 40 = 2x \\[3ex] 18 = 2x \\[3ex] 2x = 18 \\[3ex] x = \dfrac{18}{2} \\[5ex] x = 9 $

(92.) GCSE $\mathcal{E}$ = {odd numbers less than 30}
A = {3, 9, 15, 21, 27}
B = {5, 15, 25}

(a) Complete the Venn diagram to represent this information.

A number is chosen at random from the universal set, $\mathcal{E}$
(b) What is the probability that the number is in the set $A \cup B$?

$ \mathcal{E} = \{1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29\} \\[3ex] A = \{3, 9, 15, 21, 27\} \\[3ex] A' = \{1, 5, 7, 11, 13, 17, 19, 23, 25, 29\} \\[3ex] B = \{5, 15, 25\} \\[3ex] B' = \{1, 3, 7, 9, 11, 13, 17, 19, 21, 23, 27, 29\} \\[3ex] A \cap B = \{15\} \\[3ex] A \cup B = \{3, 5, 9, 15, 21, 25, 27\} \\[3ex] (A \cup B)' = \{1, 7, 11, 13, 17, 19, 23, 29\} \\[3ex] \underline{A\;\;only} \\[3ex] A \cap B' = \{3, 9, 21, 27\} \\[3ex] \underline{B\;\;only} \\[3ex] B \cap A' = \{5, 25\} \\[3ex] $ (a) The Venn diagram is:

$ (b) \\[3ex] A \cup B = \{3, 5, 9, 15, 21, 25, 27\} \\[3ex] n(A \cup B) = 7 \\[3ex] \mathcal{E} = \{1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29\} \\[3ex] n(\mathcal{E}) = 15 \\[3ex] P(A \cup B) = \dfrac{n(A \cup B)}{n(\mathcal{E})} = \dfrac{7}{15} $

(94.) CSEC The Venn diagram below shows the number of students who study History and French in a class of 30 students.
U = {students in the class}
H = {students who study History}
F = {students who study French}

(i) Write an expression, in x, in its simplest form, for the TOTAL number of students in the class.

(ii) State whether the following relationships are true or false:

$H \cup F = U$
$H \cap F' = \phi$

(iii) Determine the number of students who study BOTH History and French.

$ (i) \\[3ex] The\;\;total\;\;number\;\;of\;\;students \\[3ex] = (18 - x) + (14 - x) + x + 5 \\[3ex] = 18 - x + 14 - x + x + 5 \\[3ex] = 37 - x \\[3ex] (ii) \\[3ex] H \cup F = U \\[3ex] $ This is false because the Universal set, U, also contains those who study neither History nor French
5 students do not study History or French, hence they are outside sets H and F
But they are still part of the universal set

$ H \cap F' = \phi \\[3ex] H \cap F' = ONLY\;\;History \\[3ex] $ This is also false.
Those who study History only is not an empty set

$ (iii) \\[3ex] n(U) = 30 \\[3ex] \implies \\[3ex] 37 - x = 30 \\[3ex] 37 - 30 = x \\[3ex] x = 7 $

(98.) CSEC The Venn diagram below shows the number of students who play guitar (G) or the violin (V) in a class of 40 students.

(i) How many students play neither the guitar nor the violin?

(ii) Write an expression, in terms of x, which represents the TOTAL number of students in the class.

(iii) Write an equation which may be used to determine the total number of students in the class.

(iv) How many students play the guitar?

(i) 12 students play neither the guitar nor the violin

$ (ii) \\[3ex] \underline{Expression}:\;\;The\;\;total\;\;number\;\;of\;\;students \\[3ex] = 2x + 4 + x + 12 \\[3ex] = 3x + 16 \\[3ex] (iii) \\[3ex] The\;\;total\;\;number\;\;of\;\;students = 40 \\[3ex] \implies \\[3ex] \underline{Equation}:\;\;The\;\;total\;\;number\;\;of\;\;students \\[3ex] 3x + 16 = 40 \\[3ex] (iv) \\[3ex] 3x + 16 = 40 \\[3ex] 3x = 40 - 16 \\[3ex] 3x = 24 \\[3ex] x = \dfrac{24}{3} \\[5ex] x = 8 \\[3ex] n(G) = 2x + x \\[3ex] = 3x = 3(8) \\[3ex] = 24 \\[3ex] $ 24 students play the guitar

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(102.) CSEC The universal set $U = \{b, d, e, f, g, i, k, s, t, v, w\}$.
The Venn diagram shows $U$ and three sets, M, P, and R, which are subsets of U.

(i) State the value of $n(P \cup R)$

(ii) List the members of:
(a.) $M \cap P$
(b.) $M \cup R'$

$ U = \{b, d, e, f, g, i, k, s, t, v, w\} \\[3ex] M = \{b, d, i, k\} \\[3ex] P = \{b, d, e, f, s, v\} \\[3ex] R = \{d, e, f, g, i\} \\[3ex] (i) \\[3ex] P \cup R = \{b, d, e, f, g, i, s, v\} \\[3ex] n(P \cup R) = 8 \\[3ex] (ii) \\[3ex] (a.) \\[3ex] M \cap P = \{b, d\} \\[3ex] (b.) \\[3ex] R' = \{b, k, s, t, v, w\} \\[3ex] M \cup R' = \{b, d, i, k, s, t, v, w\} $

(103.) GCSE The Venn diagram shows some information about 150 students.
$\xi$ = 150 students
C = students who study Chemistry
P = students who study Physics

The probability that a Physics student, chosen at random, also studies Chemistry is $\dfrac{5}{12}$

One of the 150 students is chosen at random.
Work out the probability that the student does not study either Chemistry or Physics

$ n(sample\;\;space) = n(P) = x + 35...number\;\;of\;\;Physics\;\;students \\[3ex] n(P \cap C) = x \\[3ex] P(P \cap C) = \dfrac{n(P \cap C)}{n(P)} = \dfrac{5}{12} \\[5ex] \implies \\[3ex] \dfrac{x}{x + 35} = \dfrac{5}{12} \\[5ex] 12x = 5(x + 35) \\[3ex] 12x = 5x + 175 \\[3ex] 12x - 5x = 175 \\[3ex] 7x = 175 \\[3ex] x = \dfrac{175}{7} \\[5ex] x = 25 \\[3ex] 47 + 35 + x + y = 150 \\[3ex] 82 + 25 + y = 150 \\[3ex] 107 + y = 150 \\[3ex] y = 150 - 107 \\[3ex] y = 43 \\[3ex] n(neither) = n(P \cup C)' = y = 43 \\[3ex] n(sample\;\;space) = n(\xi) = 150 \\[3ex] P(P \cup C)' = \dfrac{n(P \cup C)'}{n(\xi)} \\[5ex] = \dfrac{y}{150} \\[5ex] = \dfrac{43}{150} $

(104.) ACT At Brookfield High School, 55 seniors are enrolled in the sociology class and 40 seniors are enrolled in the drawing class.
Of these seniors, 20 are enrolled in both the sociology class and the drawing class.
How many of the 120 seniors enrolled at Brookfield High School are NOT enrolled in either the sociology class or the drawing class?

$ F.\;\; 5 \\[3ex] G.\;\; 15 \\[3ex] H.\;\; 20 \\[3ex] J.\;\; 35 \\[3ex] K.\;\; 45 \\[3ex] $

$ n(\mu) = 120 \\[3ex] Let: \\[3ex] sociology\;\;class = C \\[3ex] drawing\;\;class = D \\[3ex] n(C) = 55 \\[3ex] n(D) = 40 \\[3ex] n(both\;\;C\;\;and\;\;D) = 20 \\[3ex] n(neither\;\;C\;\;nor\;\;D) = k \\[3ex] $ The Venn diagram is:

$ \implies \\[3ex] 35 + 20 + 20 + k = 120 \\[3ex] 75 + k = 120 \\[3ex] k = 120 - 75 \\[3ex] k = 45 \\[3ex] $ 45 students are NOT enrolled in either the sociology class or the drawing class

(106.) WASSCE Out of 30 candidates applying for a post, 17 have degrees, 15 diplomas and 4 neither degree nor diploma.
How many of them have both?

$ n(\xi) = 30 \\[3ex] Let: \\[3ex] degree = DE \\[3ex] diploma = DI \\[3ex] n(DE) = 17 \\[3ex] n(DI) = 15 \\[3ex] n(neither) = 4 \\[3ex] n(both) = n(DE \cap DI) = d \\[3ex] $ Let us draw the Venn diagram for this information:

$ \implies \\[3ex] (17 - d) + (15 - d) + d + 4 = 30 \\[3ex] 17 - d + 15 - d + d + 4 = 30 \\[3ex] 36 - d = 30 \\[3ex] 36 - 30 = d \\[3ex] d = 6 \\[3ex] $ 6 candidates have both degrees and diplomas