Word Problems based on Three Sets

Word Problems Based on Three Sets


Cardinality Relations of Three Sets

 

Let, U be the universal set. A, B and C are three overlapping sets which are subsets of the universal set U. 


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Look at the following Venn diagram and learn the cardinal representation for the different portions.


Venn diagram of three overlapping sets A, B and C


n(A) = Number of elements in set A

n(B) = Number of elements in set B

n(C) = Number of elements in set C

no(A) = Number of elements in set A only

no(B) = Number of elements in set B only

no(C) = Number of elements in set C only

n(A∩B) = Number of elements common in set A and B

n(A∩C) = Number of elements common in set A and C

n(B∩C) = Number of elements common in set B and C

no(A∩B) = Number of elements common in set A and B only

no(A∩C) = Number of elements common in set A and C only

no(B∩C) = Number of elements common in set B and C only

n(A∩B∩C) = Number of elements common to A, B and C

n(ABC) = Number of elements belongs to at least one set A, B or C.

n(A∩B∩C)c = Number of elements does not belong to the set A, B or C.

 

 

Formula on Cardinality Relations of Three Sets

If A, B and C are subsets of universal set U then,

 

1.        n(ABC) = n(A) + n(B) + n(C) – n(A∩B) – n(B∩C) – n(A∩C) + n(A∩B∩C)

2.        n(A∩B∩C) = n(ABC) – n(A) – n(B) – n(C) + n(A∩B) + n(B∩C) + n(A∩C)

3.        n(ABC)c = n(U) – n(ABC)

4.        no(A) = n(A) – n(A∩B) – n(A∩C) + n(A∩B∩C)

5.        no(B) = n(B) – n(A∩B) – n(B∩C) + n(A∩B∩C)

6.        no(C) = n(C) – n(A∩C) – n(B∩C) + n(A∩B∩C)

7.        no(A∩B) = n(A∩B) – n(A∩B∩C)

8.        no(B∩C) = n(B∩C) – n(A∩B∩C)

9.        no(A∩C) = n(A∩C) – n(A∩B∩C)

10.   Exactly two = no(A∩B) + no(B∩C) + no(A∩C)

11.   Exactly one = no(A) + no(B) + no(C)

12.   At least one = n(ABC)

13.   All three = n(A∩B∩C)

14.   None of them = n(ABC)c

 

 

Word Problems Based on Three Sets

 

Workout Examples

 

Example 1: P, Q and R are subsets of a universal set U. If n(U) = 390, n(P) = 210, n(Q) = 165, n(R) = 120, n(P∩Q) = 60, n(Q∩R) = 45, n(A∩R) = 54 and n(P∩Q∩R) = 24, illustrate this information in a Venn-diagram and find the following:

 

a.        no(P)

b.        no(Q)

c.        no(R)

d.        no(P∩Q)

e.        no(Q∩R)

f.         no(P∩R)

g.        n(PQR)c

 

Solution:

 

Here,

 

Venn-diagram,

 

Example 1: Venn-diagram

From the Venn-diagram above,

 

a.        no(P) = 120

b.        no(Q) = 84

c.        no(R) = 45

d.        no(P∩Q) = 36

e.        no(Q∩R) = 21

f.         no(P∩R) = 30

g.        n(PQR)c = 30

Example 2: In a survey of a group of people, 60 liked tea, 45 liked coffee, 30 liked milk, 25 liked coffee as well as tea, 20 liked tea as well as milk, 15 liked coffee as well as milk and 10 liked all three. How many people were asked this question? Solve by using Venn-diagram.

 

Solution:

 

Let C, T and M represent the set of people who liked coffee, tea and milk respectively. Then,

 

       n(T) = 60

          n(C) = 45

          n(M) = 30

          n(C∩T) = 25

n(T∩M) = 20

n(C∩M) = 15

n(C∩T∩M) = 10

 

Venn diagram,


Example 2: Venn-diagram

 

Now,

 

n(TCM)

= n(T) + n(C) + n(M) – n(C∩T) – n(T∩M) – n(C∩M) + n(C∩T∩M)

= 60 + 45 + 30 – 25 – 20 – 15 + 10

= 85

 

Hence, the required number of people = 85.

 

 

Example 3: In an examination, 40% of candidates passed in mathematics, 45% in Science, and 55% in Health. If 10% passed in Mathematics and Science, 20% in Science and Health and 15% in Health and Mathematics,

 

(i)     Illustrate the above information by drawing a Venn diagram.

(ii)  Find the pass percentage in all three subjects.

 

Solution:

 

Let M, S, and H denote the set of students who passed in Maths, Science and Health respectively.

 

       n(U) = 100% = n(MSH)

          n(M) = 40%

          n(S) = 45%

          n(H) = 55%

          n(M∩S) = 10%

n(S∩H) = 20%

n(H∩M) = 15%

 

Since, there is  no one who failed in all three subjects,

 

n(MSH) = n(U) = 100%

 

(i)        Venn-diagram,


Example 3: Venn-diagram


(ii)     n (M∩S∩H)

= n(MSH) – n(M) – n(S) – n(H) + n(M∩S) + n(S∩H) + n(M∩H)

= 100 – 40 – 45 – 55 + 10 + 20 + 15%

= 5%


5% passed in all three subjects.

Example 4: Out of 1350 candidates, 600 passed in Science, 700 in Mathematics, 350 in English and 50 failed in all three subjects. If 200 passed in Science and Mathematics, 150 in Science and English, 100 in Mathematics and English,

 

(i)     How many candidates passed in all three subjects?

(ii)  Illustrate the above information in a Venn-diagram.

 

Solution:

 

Let, the set of candidates who passed in Science, Mathematics and English be S, M, and E.

 

       n(U) = 1350

n(S) = 600

n(M) = 700

n(E) = 350

n(SME)c = 50

n(S∩M) = 200

n(M∩E) = 100

n(S∩E) = 150

 

We know,

 

n(SME)

= n(U) – n(SME)c

= 1350 – 50

= 1300

 

Now,

n(A∩B∩C)

= n(ABC) – n(A) – n(B) – n(C) + n(A∩B) + n(B∩C) + n(A∩C)

= 1300 – 600 – 700 – 350 + 200 + 100 + 150

= 100

 

Therefore,

 

(i)      100 candidates passed in all three subjects.

(ii)   Venn-diagram,

 

Example 4: Venn-diagram


 

Example 5: In a group of students, 25 study computer, 28 study Health, 20 study Mathematics, 9 study Computer only, 12 study Health only, 8 study Computer and Health only and 5 students study Health and Mathematics only.

 

(i)        Draw a Venn diagram to illustrate the above information.

(ii)     Find how many students study all the subjects.

(iii)  How many students are there altogether?

 

Solution:

 

Let C, H, and M be the set of students who study Computer, Health, and Mathematics respectively.

 

       n(C) = 25

          n(H) = 28

          n(M) = 20

          no(C) = 9

          no(H) = 12

no(C∩H) = 8

no(H∩M) = 5

 

Filling the above information in Venn-diagram, we get


Example 5: Venn-diagram 1

From the above Venn-diagram,

n(C∩H∩M) + 8 + 12 + 5 = n(H)

or,     n(C∩H∩M) + 25 = 28

or,     n(C∩H∩M) = 28 – 25

or,     n(C∩H∩M) = 3

 

Again, 

no(C∩M) + n(C∩H∩M) + 9 + 8 = n(C)

or,     no(C∩M) + 3 + 9 + 8 = 25

or,     no(C∩M) + 20 = 25

or,     no(C∩M) = 25 – 20

or,     no(C∩M) = 5

 

Again, 

no(M) + no(C∩M) + n(C∩H∩M) + 5 = n(M)

or,     no(M) + 5 + 3 + 5 = 20

or,     no(M) + 13 = 20

or,     no(M) = 20 – 13

or,     no(M) = 7

 

(i)        Venn-diagram:


Example 5: Venn-diagram 2


(ii)     n(C∩H∩M) = 3 students study all subjects.

(iii)  From the above venn-diagram,

n(CHM)

= 9 + 8 + 3 + 5 + 12 + 5 + 7

= 49


There are 49 students altogether.

Example 6: In a group of people, 20 like milk, 30 like tea, 22 like coffee, 12 like coffee only, 6 like milk and coffee only,  2 like tea and coffee only and 8 like milk and tea only. Show these information in a Venn-diagram and find:

 

(i)        How many like at least one drink?

(ii)     How many like exactly one drink?

 

Solution:

 

Let the set of people who like milk, tea, and coffee be M, T, and C.

 

       n(M) = 20

n(T) = 30

n(C) = 22

no(C) = 12

no(M∩C) = 6

no(T∩C) = 2

no(M∩T) = 8

 

Filling the above information in the Venn-diagram, we get


Example 6: Venn-diagram 1

From the above Venn-diagram,

n(M∩T∩C) + 6 + 12 + 2 = n(C)

i.e.     n(M∩T∩C) + 20 = 22

or,     n(M∩T∩C) = 22 – 20

or,     n(M∩T∩C) = 2

 

Again,

no(M) + 6 + n(M∩T∩C) + 8 = n(M)

i.e.     no(M) + 6 + 2 + 8 = 20

or,     no(M) + 16 = 20

or,     no(M) = 20 – 16

or,     no(M) = 4

 

Again,

no(T) + 2 + n(M∩T∩C) + 8 = n(T)

i.e.     no(T) + 2 + 2 + 8 = 30

or,     no(T) + 12 = 30

or,     no(T) = 30 – 12

or,     no(T) = 18

 

Venn-diagram,


Example 6: Venn-diagram 2


From the venn-diagram,

 

(i)        n(MCT)

= 4 + 6 + 12 + 2 + 18 + 8 + 2

= 52


52 like at least one drink.


(ii)     no(M) + no(C) + no(T)

= 4 + 12 +18

= 34


34 like exactly one drink.


 

If you have any questions or problems regarding the Word Problems Based on Three Sets, you can ask here, in the comment section below.


 

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30 comments:

  1. In a survey among 100 people,50 liked coffee,30 liked milk,40 liked tea,20 liked coffee only,25 liked tea only,10 liked tea and coffee and 5 liked all tea, coffee and milk.using the Venn diagram,find the number of people who liked neither of these.

    ReplyDelete
    Replies
    1. Here is the solution to your problems,
      [image src=" https://1.bp.blogspot.com/-5CpwLMYeUfI/YLxWRvfHgQI/AAAAAAAAJCQ/uvCnTcLPmUwl6pjX1mZ8zapXMMTwUBRcwCLcBGAsYHQ/s16000/Comment%2BSolution%2B1.png"/]
      Comment here if you have more problems regarding the Word Problems Based on Three Sets.

      Delete
  2. In a group of students,20 study account,21 study maths,18 study history,7 study account only,10 study math only,6 study account and maths only and 3 study math and history only.if each student study at least one of the three subject,how many students study all the subject? How many students are there altogether?

    ReplyDelete
    Replies
    1. Here is the solution to your problems,
      [image src=" https://1.bp.blogspot.com/-v_RnnYLpy68/YLxWbakaXQI/AAAAAAAAJCU/ECRW98m3xT8G2sU62ybFhPwYXUIig9XqQCLcBGAsYHQ/s16000/comment%2Bsolution%2B2.png "/]
      Comment here if you have more problems regarding the Word Problems Based on Three Sets.

      Delete
  3. In a school 20%students passed in all subjects(Maths,English and Dance),46% failed in maths,57% failed in English 42% failed in Dance,35% failed in maths and English,25% failed in English and Dance,15% failed in Dance and maths.What percentage of the studenta failed in all the three subjects ?

    ReplyDelete
    Replies
    1. Here is the solution to your problems,

      Solution:
      Let the students who failed in Math, English and Dance be M, E and D respectively.
      Passed in all subjects = 20%
      ∴ Failed in at least one subject = n(M∪E∪D) = 100 – 20% = 80%
      n(M) = 46%
      n(E) = 57%
      n(D) = 42%
      n(M∩E) = 35%
      n(E∩D) = 25%
      n(D∩M) = 15%
      n(M∩E∩D) = ?

      We know,
      n(A∩B∩C)
      = n(M∪E∪D) – n(M) – n(E) – n(D) + n(M∩E) + n(E∩D) + n(D∩M)
      = 80 – 46 – 57 – 42 + 35 + 25 + 15%
      = 10%

      ∴ 10% students failed in all three subjects.

      Comment here if you have more problems regarding the Word Problems Based on Three Sets.

      Delete
  4. there are 100 cars in a car park. 60 of them are black, 40 of them are SUV's, 20 of the SUV's are automatic, the rest are manual. 15 of the manual SUV's are not black, 10 of the manual cars are not black, nor are they an SUV, 25 of the automatic cars are not an SUV, 41 of the black cars are not an SUV. How many automatic cars are there?

    ReplyDelete
    Replies
    1. Here is the solution to your problems,
      [image src="https://1.bp.blogspot.com/-5SFg01VbSDg/YOV08di9WgI/AAAAAAAAJbk/hQh09-tgLmAHZzQPWpFXS6GrmA0rDZRgACLcBGAsYHQ/s16000/comment%2Bsolution%2B3.png"/]
      Comment here if you have more problems regarding the Word Problems Based on Three Sets.

      Delete
  5. In a class of 140 students, 80 offered mathematics, 72 offered chemistry and 68 offered physics, 40 offered mathematics and chemistry, 35 offered mathematics and physics while 30 offered chemistry and physics. How many students offered all three

    ReplyDelete
    Replies
    1. Solution: Here,
      Let the set of students who offered mathematics, Chemestry and Physics be M, C and H respectively.
      n(U) = 140 = n(M∪C∪P)
      n(M) = 80
      n(C) = 72
      n(P) = 68
      n(M∩C) = 40
      n(M∩P) = 35
      n(P∩C) = 30
      n(M∩C∩P) = ?

      We know,
      n(M∩C∩P)
      = n(M∪C∪P) – n(M) – n(C) – n(P) + n(M∩C) + n(M∩P) + n(P∩C)
      = 140 – 80 – 72 – 68 + 40 + 35 + 30
      = 25

      ∴ 25 students offered all three subjects. Ans.

      Delete
  6. In a class of 240 students, a survey was conducted, 120 students had applied for university K, 100 students applied for university A and 96 students applied for university B. 42 students applied for both K and A, 24 applied for both K and B, 36 applied for both B and A while 16 applied neither of the aforementioned universities. Find
    (a) number of students that applied for all the universities
    (b) number of students that applied for at least two of the universities
    (c) number of students that applied at most two universities
    (d) number of students that applied for K but not A

    ReplyDelete
    Replies
    1. Solution: Here,
      n(U) = 240
      n(K) = 120
      n(A) = 100
      n(B) = 96
      n(K∩A) = 42
      n(K∩B) = 24
      n(A∩B) = 36
      n(K∪A∪B)c = 16
      We know,
      n(A∪B∪C) = n(U) – n(K∪A∪B)c = 240 – 16 = 224
      i.e. n(K) + n(A) + n(B) – n(K∩A) – n(K∩B) – n(A∩B) + n(K∩A∩B) = 224
      or, 120 + 100 + 96 – 42 – 24 – 36 + n(K∩A∩B) = 224
      or, 214 + n(K∩A∩B) = 224
      or, n(K∩A∩B) = 224 – 214 = 10

      a. No. of students who applied for all the universities = n(K∩A∩B) = 10
      b. No. of students who applied at least two universities = n(K∩A∩B) + n(K∩A) + n(K∩B) + n(A∩B)
      = 10 + 42 + 24 + 36
      = 112
      c. No. of students who applied at most two universities = n(K∪A∪B) – n(K∩A∩B)
      = 224 – 10
      = 214
      d. No. of students who applied K but not A = n(K) – n(A)
      = 120 – 100
      = 20

      Delete
  7. The following analysis shows how students from various departments registered for 2018/2019 academic session from the college of social and management science in Tai Solarin University of education. 10 from economics only, 150 from economics and geography, 90 from economics but not political science, 210 from geography, 120 from political science and geography, 180 from economics. If the total of students registered for courses is 250, determine the number of students who registered for I. Political science ii. Political science and geography but not economics iii. Economics and political science but not geography

    ReplyDelete
    Replies
    1. Solution: Here,
      no(E) = 10
      n(E∩G) = 150
      n(E – P) = 90
      n(G) = 210
      n(P∩G) = 120
      n(E) = 180
      n(U) = 250
      no(E∩P)= n(E)-no(E) + n(E∩G) = 180 – (10 + 150) = 180 – 160 = 20
      no(E∩G) = n(E – G) – no(E) = 90 – 10 = 80
      n(E∩P∩G) = n(E) – (20 + 10 + 80) = 180 – 110 = 70
      no(P∩G) = n(P∩G) - n(E∩P∩G) = 120 – 70 = 50
      no(P) = n(U) – [n(G) + 10 + 20] = 250 – [210 + 10 + 20] = 250 – 230 = 20
      n(P) = 20 + 70 + 50 + 20 = 160
      (i) Political science = n(P) = 160
      (ii) Political science and geography but not economics = no(P∩G) = 50
      (iii) Economics and political science but not geography = no(E∩P)= 20

      Delete
  8. Of 497 girls reading three newspapers 26 girls read the guardian and the vanguard,50 girls read sun and the vanguard, 38 girls read the guardian and the sun,and 11 read all the three newspapers.it was found that equal numbers read the sun,the guardian and the vanguard.(a) how many read only sun? (b) how many read only tee vanguard (c) how many read only one of these papers?

    ReplyDelete
  9. Of 1000 students applying for admission into faculty of physical sciences,Nnamdi Azikiwe University,Awka. 200 applied to read statistics, 240 applied to read chemistry and 250 applied to read computer sciences.97 applied to read statistics and chemistry only,while 64 applied to read chemistry and computer sciences only, 60 applied to read statistics and computer sciences only.if 570 students do not apply to any of these departments, how many students applied (a) only into the department of chemistry? (b)Only into one of these departments? (c) the three departments?

    ReplyDelete
  10. In a survey of people 48 liked Nepali film 40 liked English film 31 like Hindi film 19 liked nepali and hindi film 13 liked Hindi and English 11 liked Nepali only and 21 people were found not interested in any film 25 liked Nepali and English film find (a) how many liked all three films (b)how many were asked question ?

    ReplyDelete
  11. In a recent pool of 500 men and 500 women,it was observed that a total of 650 were married.of those married,275 were men and 500 claimed to be happy. Out of 750 claimed to be happy. 400 were men of which 200 were married. Represent this information in a venn diagram and find
    (a)the number of married people who are happy
    (b)the number of unmarried people who are not happy

    ReplyDelete
  12. The pandemic strikes and a subset of the 1000 students begin exhibiting symptoms of COVID, including fever, cough, and loss of smell. When the medical center screened the 200 symptomatic students, they found 150 students had a fever, 25 students had a cough, 25 students had loss of smell. In addition, 20 students appeared with fever and cough, 10 students showed cough and loss of smell, 5 students had fever and loss of smell. Calculate the number of students exhibiting all three symptoms: fever, cough, and loss of smell.

    ReplyDelete
    Replies
    1. Solution: Here,
      Let the set students having symptoms Fever, Cough and Loss of smell be F, C and L.
      n(F∪C∪L) = 200
      n(F) = 150
      n(C) = 25
      n(L) = 25
      n(F∩C) = 20
      n(C∩L) = 10
      n(F∩L) = 5
      n(F∩A∩M) = ?

      n(F∩C∩L)
      = n(F∪C∪L) – n(F) – n(C) – n(L) + n(F∩C) + n(C∩L) + n(F∩L)
      = 200 – 150 – 25 – 25 + 20 + 10 + 5
      = 35

      All three symptoms = n(F∩C∩L) = 35

      Delete
  13. London School of Business has matriculated 1000 Business graduates for the fall semester in 2017. Among them, 340 students have Finance, 320 students have Accounting, and 300 students have Management as their major in maters program. Also, some graduates have more than one concentration in their graduate program. Though, 100 students have gone in for all the three majors, it was inevitable from the chronicling by the dean office that 180 students have both finance and Accounting, 190 students have both Finance and Management, and 200 students have both accounting and Management as their major. Find out the number of students who has contrary close mental application with at least one of these three areas.

    ReplyDelete
    Replies
    1. Solution: Here,
      Let the set students having Finance, Accounting and Management be F, A and M.
      n(U) = 1000
      n(F) = 340
      n(A) = 320
      n(M) = 300
      n(F∩A∩M) = 100
      n(F∩A) = 180
      n(F∩M) = 190
      n(A∩M) = 200

      n(F∪A∪M)
      = n(F) + n(A) + n(M) - n(F∩A) - n(F∩M) - n(A∩M) + n(F∩A∩M)
      = 340 + 320 + 300 – 180 – 190 – 200 + 100
      = 490

      At least one subjects = n(F∪A∪M) = 490

      Delete
  14. Please i need a solution to this question
    A newspaper agent sells three papers, tge times,punch and herald,70 customers buy the times newspaper, 60 buy the punch and 50 buy the herald, 17 buy times and punch, 15 buy punch and herald and 16 buy the herald and times while 3 customers buy all three. Determine the number of customers

    ReplyDelete
    Replies
    1. Solution: Here,
      Let the set of customers buying tge times, punch and herald be T, P and H.
      n(T) = 70
      n(P) = 60
      n(H) = 50
      n(T∩P) = 17
      n(P∩H) = 15
      n(H∩T) = 16
      n(T∩P∩H) = 3

      By using formula,

      n(T∪P∪H)
      = n(T) + n(P) + n(H) – n(T∩P) – n(P∩H) – n(H∩T) + n(T∩P∩H)
      = 70 + 60 + 50 – 17 – 15 – 16 + 3
      = 135

      ∴ No. of customers = n(T∪P∪H) = 135

      Delete
  15. In a survey, people were asked what types of movies they like. It was found the 75 liked Nepali movies, 60 liked English, 40 liked Hindi, 35 liked Nepali and English 30 liked Nepali and Hindi, 20 liked Hindi and English. 10 liked all three and 2 people were found not interested in any types of movies. (i) How many people did not like only Hindi films?

    (ii) How many people did not like only Nepali or Hindi films?

    Can I Get it's Solutions

    ReplyDelete
    Replies
    1. Solution:
      Let the set of people who like Nepali, English and Hindi be N, E and H.
      n(N) = 75
      n(E) = 60
      n(H) = 40
      n(N∩E) = 35
      n(N∩H) = 30
      n(H∩E) = 20
      n(N∩E∩H) = 10
      n(N∪E∪H)c = 2

      no(N∩E) = n(N∩E) - n(N∩E∩H) = 35 – 10 = 25
      no(H∩E) = n(H∩E) - n(N∩E∩H) = 20 – 10 = 10

      (i) No. of people who did not like only hindi film = no(N∩E) = 25
      (ii) No. of people who did not like only Nepali or Hindi films = no(N∩E) + no(H∩E) = 25+ 10 = 35

      Delete
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