problem solving involving sets grade 7 ppt

Mathematics 7 Quarter 1 – Module 2: Problems Involving Sets

This module was designed and written with you in mind. It is here to help you master your skills in solving mathematical problems involving sets. The scope of this module permits it to be used in many different learning situations. The language used recognizes the diverse vocabulary level of students. The lessons are arranged to follow the standard sequence of the course. But the order in which you read them can be changed to correspond with the textbook you are now using.

The module is all about Solving Problems Involving Sets.

After going through this module, you are expected to:

1. solve problems involving sets using Venn diagram;

2. apply set operations to solve a variety of word problems.

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PROBLEMS INVOLVING SETS

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Math 7 Module 2: Problems Involving Sets

• MATHEMATICS Quarter 4 – Module 5 Measures of Central Tendency
• MATHEMATICS Quarter 4 – Module 3 Frequency Distribution
• MATHEMATICS Quarter 4 – Module 6 Measures of Variability
• MATHEMATICS Quarter 4 – Module 7 Statistical Data Interpretation
• MATHEMATICS Quarter 4 – Module 2 Gathering Statistical Data

WORD PROBLEMS ON SETS AND VENN DIAGRAMS

Basic stuff.

To understand, how to solve Venn diagram word problems with 3 circles, we have to know the following basic stuff.  

u ----> union (or)

n ----> intersection (and)

Addition Theorem on Sets

Theorem 1 :

n(AuB) = n(A) + n(B) - n(AnB)

Theorem 2 :

=n(A) + n(B) + n(C) - n(AnB) - n(BnC) - n(AnC) + n(AnBnC)

Explanation :

Let us come to know about the following terms in details.

n(AuB) = Total number of elements related to any of the two events A & B.

n(AuBuC) = Total number of elements related to any of the three events A, B & C.

n(A) = Total number of elements related to A

n(B) = Total number of elements related to B

n(C) = Total number of elements related to C

For  three events A, B & C, we have

n(A) - [n(AnB) + n(AnC) - n(AnBnC)] :

Total number of elements related to A only

n(B) - [n(AnB) + n(BnC) - n(AnBnC)] :

Total number of elements related to B only

n(C) - [n(BnC) + n(AnC) + n(AnBnC)] :

Total number of elements related to C only

Total number of elements related to both A & B

n(AnB) - n(AnBnC) :

Total number of elements related to both (A & B) only

Total number of elements related to both B & C

n(BnC) - n(AnBnC) :

Total number of elements related to both (B & C) only

Total number of elements related to both A & C

n(AnC) - n(AnBnC) :

Total number of elements related to both (A & C) only

For  two events A & B, we have

n(A) - n(AnB) :

n(B) - n(AnB) :

Solved Problems

Problem 1 :

In a survey of university students, 64 had taken mathematics course, 94 had taken chemistry course, 58 had taken physics course, 28 had taken mathematics and physics, 26 had taken mathematics and chemistry, 22 had taken chemistry and physics course, and 14 had taken all the three courses. Find how many had taken one course only.

Let M, C, P represent sets of students who had taken mathematics, chemistry and physics respectively.

From the given information, we have

n(M) = 64, n(C) = 94, n(P) = 58,

n(MnP) = 28, n(MnC) = 26, n(CnP) = 22

n(MnCnP) = 14

From the basic stuff, we have

Number of students who had taken only Math

= n(M) - [n(MnP) + n(MnC) - n(MnCnP)]

= 64 - [28 + 26 - 14]

Number of students who had taken only Chemistry :

= n(C) - [n(MnC) + n(CnP) - n(MnCnP)]

= 94 - [26+22-14]

Number of students who had taken only Physics :

= n(P) - [n(MnP) + n(CnP) - n(MnCnP)]

= 58 - [28 + 22 - 14]

Total n umber of students who had taken only one course :

= 24 + 60 + 22

Hence, the total number of students who had taken only one course is 106.

Alternative Method (Using venn diagram) :

Venn diagram related to the information given in the question:

From the venn diagram above, we have

Number of students who had taken only math = 24

Number of students who had taken only chemistry = 60

Number of students who had taken only physics = 22

Total  Number of students who had taken only one course :

Problem 2 :

In a group of students, 65 play foot ball, 45 play hockey, 42 play cricket, 20 play foot ball and hockey, 25 play foot ball and cricket, 15 play hockey and cricket and 8 play all the three games. Find the total number of students in the group  (Assume that each student in the group plays at least one game).

Let F, H and C represent the set of students who play foot ball, hockey and cricket respectively.

n(F) = 65, n(H) = 45, n(C) = 42,

n(FnH) = 20, n(FnC) = 25, n(HnC) = 15

n(FnHnC) = 8

Total number of students in the group is  n(FuHuC).

n(FuHuC) is equal to

= n(F) + n(H) + n(C) - n(FnH) - n(FnC) - n(HnC) + n(FnHnC)

n(FuHuC) = 65 + 45 + 42 -20 - 25 - 15 + 8

n(FuHuC) = 100

Hence, the total number of students in the group is 100.

Alternative Method (Using Venn diagram) :

Venn diagram related to the information given in the question :

Total number of students in the group :

= 28 + 12 + 18 + 7 + 10 + 17 + 8

So, the total number of students in the group is 100.

Problem 3 :

In a college, 60 students enrolled in chemistry,40 in physics, 30 in biology, 15 in chemistry and physics,10 in physics and biology, 5 in biology and chemistry. No one enrolled in all the three. Find how many are enrolled in at least one of the subjects.

Let C, P and B represents the subjects Chemistry, Physics  and Biology respectively.

Number of students enrolled in Chemistry :

Number of students enrolled in Physics :

Number of students enrolled in Biology :

Number of students enrolled in Chemistry and Physics :

n(CnP) = 15

Number of students enrolled in Physics and Biology :

n(PnB) = 10

Number of students enrolled in Biology and Chemistry :

No one enrolled in all the three.  So, we have

n(CnPnB) = 0

The above information can be put in a Venn diagram as shown below.

From, the above Venn diagram, number of students enrolled in at least one of the subjects :

= 40 + 15 + 15 + 15 + 5 + 10 + 0

So, the number of students  enrolled in at least one of the subjects is 100.

Problem 4 :

In a town 85% of the people speak Tamil, 40% speak English and 20% speak Hindi. Also 32% speak Tamil and English, 13% speak Tamil and Hindi and 10% speak English and Hindi, find the percentage of people who can speak all the three languages.

Let T, E and H represent the people who speak Tamil, English and Hindi respectively.

Percentage of people who speak Tamil :

Percentage of people who speak English :

Percentage of people who speak Hindi :

n(H)  =  20

Percentage of people who speak English and Tamil :

n(TnE) = 32

Percentage of people who speak Tamil and Hindi :

n(TnH) = 13

Percentage of people who speak English and Hindi :

n(EnH) = 10

Let x be the percentage of people who speak all the three language.

From the above Venn diagram, we can have 

100 = 40 + x + 32 – x + x + 13 – x + 10 – x – 2 + x – 3 + x

100 = 40 + 32 + 13 + 10 – 2 – 3 + x 

100 = 95 – 5 + x

100 = 90 + x

x = 100 - 90

x = 10% 

So, the percentage of people who speak all the three languages is 10%.

Problem 5 :

An advertising agency finds that, of its 170 clients, 115 use Television, 110 use Radio and 130 use Magazines. Also 85 use Television and Magazines, 75 use Television and Radio, 95 use Radio and Magazines, 70 use all the three. Draw Venn diagram to represent these data. Find 

(i) how many use only Radio?

(ii) how many use only Television?

(iii) how many use Television and Magazine but not radio?

Let T, R and M represent the people who use Television, Radio and Magazines respectively.

Number of people who use Television :

Number of people who use Radio :

Number of people who use Magazine :

Number of people who use Television and Magazines

n (TnM) = 85

Number of people who use Television and Radio :

n(TnR) = 75

Number of people who use Radio and Magazine :

n(RnM) = 95

Number of people who use all the three :

n(TnRnM) = 70

From the above Venn diagram, we have

(i) Number of people who use only Radio is 10.

(ii) Number of people who use only Television is 25.

(iii) Number of people who use Television and Magazine but not radio is 15.

Problem 6 :

In a class of 60 students, 40 students like math, 36 like science, 24 like both the subjects. Find the number of students who like

(i) Math only, (ii) Science only  (iii) Either Math or Science (iv) Neither Math nor science.

Let M and S represent the set of students who like math and science respectively.

From the information given in the question, we have

n(M) = 40, n(S) = 36, n(MnS) = 24

Answer (i) :

Number  of students who like math only :

= n(M) - n(MnS)

Answer (ii) :

Number  of students who like science only :

= n(S) - n(MnS)

=   12

Answer (iii) :

Number  of students who like either math or science :

= n(M or S) 

= n(MuS) 

= n(M) + n(S) - n(MnS)

= 40 + 36 - 24

Answer (iv) :

Total n umber  students who like Math or Science subjects :

n(MuS) = 52

Number  of students who like neither math nor science

Problem 7 :

At a certain conference of 100 people there are 29 Indian women and 23 Indian men. Out of these Indian people 4 are doctors and 24 are either men or doctors. There are no foreign doctors. Find the number of women doctors attending the conference.

Let M and D represent the set of Indian men and Doctors respectively.

n(M) = 23, n(D) = 4, n(MuD) = 24

n(MuD) = n(M) + n(D) - n(MnD)

24 = 23 + 4 - n(MnD)

n(MnD) = 3 

n(Indian Men and Doctors) = 3

So, out of the 4 Indian doctors,  there are 3 men.

And the remaining 1 is Indian women doctor.

So, the number women doctors attending the conference is 1.

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Word Problems on Sets

Word problems on sets are solved here to get the basic ideas how to use the  properties of union and intersection of sets.

Solved basic word problems on sets:

1. Let A and B be two finite sets such that n(A) = 20, n(B) = 28 and n(A ∪ B) = 36, find n(A ∩ B).

Solution:  Using the formula n(A ∪ B) = n(A) + n(B) - n(A ∩ B).  then n(A ∩ B) = n(A) + n(B) - n(A ∪ B)                       = 20 + 28 - 36                       = 48 - 36                       = 12 

2.  If n(A - B) = 18, n(A ∪ B) = 70 and n(A ∩ B) = 25, then find n(B).

Solution:  Using the formula n(A∪B) = n(A - B) + n(A ∩ B) + n(B - A)                                   70 = 18 + 25 + n(B - A)                                   70 = 43 + n(B - A)                           n(B - A) = 70 - 43                           n(B - A) = 27  Now n(B) = n(A ∩ B) + n(B - A)                 = 25 + 27                 = 52 

Different types on word problems on sets:

3.  In a group of 60 people, 27 like cold drinks and 42 like hot drinks and each person likes at least one of the two drinks. How many like both coffee and tea? 

Solution:  Let A = Set of people who like cold drinks.       B = Set of people who like hot drinks.  Given   (A ∪ B) = 60            n(A) = 27       n(B) = 42 then;

n(A ∩ B) = n(A) + n(B) - n(A ∪ B)              = 27 + 42 - 60              = 69 - 60 = 9              = 9  Therefore, 9 people like both tea and coffee. 

4.  There are 35 students in art class and 57 students in dance class. Find the number of students who are either in art class or in dance class.

•  When two classes meet at different hours and 12 students are enrolled in both activities.  •  When two classes meet at the same hour.  Solution:  n(A) = 35,       n(B) = 57,       n(A ∩ B) = 12  (Let A be the set of students in art class.  B be the set of students in dance class.)  (i) When 2 classes meet at different hours n(A ∪ B) = n(A) + n(B) - n(A ∩ B)                                                                            = 35 + 57 - 12                                                                            = 92 - 12                                                                            = 80  (ii) When two classes meet at the same hour, A∩B = ∅ n (A ∪ B) = n(A) + n(B) - n(A ∩ B)                                                                                                = n(A) + n(B)                                                                                                = 35 + 57                                                                                                = 92

Further concept to solve word problems on sets:

5. In a group of 100 persons, 72 people can speak English and 43 can speak French. How many can speak English only? How many can speak French only and how many can speak both English and French?

Solution: Let A be the set of people who speak English. B be the set of people who speak French. A - B be the set of people who speak English and not French. B - A be the set of people who speak French and not English. A ∩ B be the set of people who speak both French and English. Given, n(A) = 72       n(B) = 43       n(A ∪ B) = 100 Now, n(A ∩ B) = n(A) + n(B) - n(A ∪ B)                      = 72 + 43 - 100                      = 115 - 100                      = 15 Therefore, Number of persons who speak both French and English = 15 n(A) = n(A - B) + n(A ∩ B) ⇒ n(A - B) = n(A) - n(A ∩ B)                 = 72 - 15                 = 57 and n(B - A) = n(B) - n(A ∩ B)                    = 43 - 15                    = 28 Therefore, Number of people speaking English only = 57 Number of people speaking French only = 28

Word problems on sets using the different properties (Union & Intersection):

6. In a competition, a school awarded medals in different categories. 36 medals in dance, 12 medals in dramatics and 18 medals in music. If these medals went to a total of 45 persons and only 4 persons got medals in all the three categories, how many received medals in exactly two of these categories?

Solution: Let A = set of persons who got medals in dance. B = set of persons who got medals in dramatics. C = set of persons who got medals in music. Given, n(A) = 36                              n(B) = 12       n(C) = 18 n(A ∪ B ∪ C) = 45       n(A ∩ B ∩ C) = 4 We know that number of elements belonging to exactly two of the three sets A, B, C = n(A ∩ B) + n(B ∩ C) + n(A ∩ C) - 3n(A ∩ B ∩ C) = n(A ∩ B) + n(B ∩ C) + n(A ∩ C) - 3 × 4       ……..(i) n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(A ∩ C) + n(A ∩ B ∩ C) Therefore, n(A ∩ B) + n(B ∩ C) + n(A ∩ C) = n(A) + n(B) + n(C) + n(A ∩ B ∩ C) - n(A ∪ B ∪ C) From (i) required number = n(A) + n(B) + n(C) + n(A ∩ B ∩ C) - n(A ∪ B ∪ C) - 12 = 36 + 12 + 18 + 4 - 45 - 12 = 70 - 57 = 13

Apply set operations to solve the word problems on sets:

7. Each student in a class of 40 plays at least one indoor game chess, carrom and scrabble. 18 play chess, 20 play scrabble and 27 play carrom. 7 play chess and scrabble, 12 play scrabble and carrom and 4 play chess, carrom and scrabble. Find the number of students who play (i) chess and carrom. (ii) chess, carrom but not scrabble.

Solution: Let A be the set of students who play chess B be the set of students who play scrabble C be the set of students who play carrom Therefore, We are given n(A ∪ B ∪ C) = 40, n(A) = 18,         n(B) = 20         n(C) = 27, n(A ∩ B) = 7,     n(C ∩ B) = 12    n(A ∩ B ∩ C) = 4 We have n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) + n(A ∩ B ∩ C) Therefore, 40 = 18 + 20 + 27 - 7 - 12 - n(C ∩ A) + 4 40 = 69 – 19 - n(C ∩ A) 40 = 50 - n(C ∩ A) n(C ∩ A) = 50 - 40 n(C ∩ A) = 10 Therefore, Number of students who play chess and carrom are 10. Also, number of students who play chess, carrom and not scrabble. = n(C ∩ A) - n(A ∩ B ∩ C) = 10 – 4 = 6

Therefore, we learned how to solve different types of word problems on sets without using Venn diagram.

● Set Theory

● Sets Theory

● Representation of a Set

● Types of Sets

● Finite Sets and Infinite Sets

● Power Set

● Problems on Union of Sets

● Problems on Intersection of Sets

● Difference of two Sets

● Complement of a Set

● Problems on Complement of a Set

● Problems on Operation on Sets

● Word Problems on Sets

● Venn Diagrams in Different Situations

● Relationship in Sets using Venn Diagram

● Union of Sets using Venn Diagram

● Intersection of Sets using Venn Diagram

● Disjoint of Sets using Venn Diagram

● Difference of Sets using Venn Diagram

● Examples on Venn Diagram

8th Grade Math Practice From Word Problems on Sets to HOME PAGE

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How To Solve a Word Problem Using Venn Diagrams - PowerPoint PPT Presentation

How To Solve a Word Problem Using Venn Diagrams

... skateboarding, one set for bicycling, and one set for college student chat rooms. ... all the students who joined the bicycling chat room or n(bicycling) ... – powerpoint ppt presentation.

• Suppose that a group of 200 students are surveyed and ask which chatrooms they have joined. There are three chatrooms in our survey one for skateboarding, one for bicycling, and one for college students.
• 90 students joined the room for skateboarding
• 50 students joined the room for bicycling
• 70 students joined the room for college students
• 15 students joined rooms for skateboarding and college students
• 12 students joined rooms for bicycling and college students
• 25 students joined rooms for skateboarding and bicycling
• 10 students joined all three rooms.
• 1.) How many students joined the room for skateboarding OR bicycling?
• 2.) How many students did not join any of these three rooms?
• 3.) How many students joined the bicycling AND skateboarding rooms BUT NOT the room for college students?
• 4.) How many students joined EXACTLY 1 of these rooms?
• 5.) How many students joined AT MOST 2 of these rooms?
• This problem seems too difficult to solve! But it isnt. You just need to use a Venn Diagram to represent the relationship between the three chat rooms and the answers to all 5 of these questions will be perfectly clear.
• For this type of problem we fill in our regions for our Venn Diagram. We use the information given to fill in the number of students in each region. We start from the bottom and work our way to the top.
• First we draw our Venn Diagram representing three sets, one set for Skateboarding, one set for Bicycling, and one set for College Student chat rooms. Then we label our regions and start putting in the number of elements or students for each region. 10 students joined all three rooms means these 10 students are in all three circles - the intersection of circles for skateboarding, bicycling, and college students. This is represented by region 5 - we need to put in a 10 in region 5.
• Our next piece of information is
• What do these 25 students represent? They are the the set of students who joined the Skateboarding chatroom AND the Bicycling chatroom or n(Skateboarding n Bicycling). These 25 students are in both the Skateboarding circle AND the Bicycling circle. This intersection is represented by regions 2 5 - we already have 10 students in region 5 so the number of students to put in region 2 25-10 or 15.
• What do these 12 students represent? They are the the students who joined the Bicycling chatroom AND the College Students chatroom or n(Bicycling n College Students). These 12 students are in both the Bicycling circle AND the College Student circle. This intersection is represented by regions 5 6 - we already have 10 students in region 5 so the number of students to put in region 6 12-10 or 2.
• What do these 15 students represent? They represent the students who joined the Skateboarding chatroom AND the students who joined the College Students chatroom or
• n(Skateboarding n College Students). These 15 students are in both the Skateboarding circle AND the College Student circle. This intersection is represented by regions 4 5 - we already have 10 students in region 5 so the number of students to put in region 4 15-10 or 5.
• What do these 70 students represent? They represent ALL the students who joined the College Students chat room or n(College Students). This is represented by regions 4 5 6 7 - the sum of the students in these four regions 70. We already have 5 students in region 4, 10 students in region 5, and 2 students in region 6 so the number of students to put in region 7 70-(1052) 53.
• What do these 50 students represent? They represent ALL the students who joined the Bicycling chat room or n(Bicycling). This is represented by regions 2 3 5 6 - the sum of the students in these four regions 50. We already have 15 students in region 2, 10 students in region 5, and 2 students in region 6 so the number of students to put in region 3 50-(15102) 23.
• What do these 90 students represent? They represent ALL the students who joined the Skateboarding chatroom or n(Skateboarding). This is represented by regions 1 2 4 5 - the sum of the students in these four regions 90. We already have 15 students in region 2, 5 students in region 4, and 10 students in region 5 so the number of students to put in region 1 90-(15510) 60.
• We have now filled in regions 1 thru 7, we only have to fill in region 8. What students does region 8 represent? Region 8 represents the students surveyed who did not join any of the three chatrooms in our survey. These students are not in any of the circles that represent our sets. The sum of ALL 8 regions must add up to all the students we surveyed - our universe - or 200 students. Therefore, the number of students in region 8 200-(601523510253)32.
• Now that we have all our regions filled in we can answer our five questions.
• How many students joined the room for skateboarding OR bicycling?
• This is asking us for the union of the skateboarding and bicycling chatrooms or n(Skateboarding U Bicycling). For union we bring all the members of the skateboarding and bicycling chatrooms together. We want all the regions in the skateboarding and bicycling circle (remember dont count regions more than once). So the students we are interested in are in regions 1 2 3 4 5 6 or
• 60 15 23 5 10 2 115 students
• How many students did not join any of these three rooms?
• The students who did not join any groups are not in ANY of the three circles. These students are in region 8. So the number of students who did not join any of these three chatrooms 32 students
• How many students joined the bicycling AND skateboarding rooms BUT NOT the room for college students?
• We want the students who joined bicycling AND skateboarding - this represents the intersection of the skateboarding and bicycling sets - regions 2 5. The next part of the question tells us that we do not want students in the College Students set, so we do not want region 5 since this region while it is in the intersection of skateboarding and bicycling it is also in the college students circle, we only want region 2. The answer is 15 students.
• How many students joined EXACTLY 1 of these rooms?
• The students who joined EXACTLY 1 of the rooms will be in regions that are only in 1 circle. These regions include 1, 3, and 7, we will add all the students in these regions to get our answer.
• 60 23 53 136 students
• Students who joined exactly 2 chatrooms will be in regions in exactly 2 circles, these regions are 2,4, and 6.
• Students who joined exactly 3 chatrooms will be in regions in exactly 3 circles, this region is 5.
• How many students joined AT MOST 2 of these rooms?
• AT MOST 2 means the number of rooms we want a student to join is LESS THAN OR EQUAL TO 2. ( 2 rooms). We want regions that are in 2 circles, 1 circle, or none of the circles. These include regions 1,2,3,4,6,7,and 8. We add up all these regions to get our answer.
• 601523525332 190 students
• NOTE If the question had asked for the number of students who joined AT LEAST 2 of these rooms we would be interested in regions that are in 2 OR MORE circles (at least mean means 2). We would want to add together the students in regions 2,4,5,and 6.

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