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. 

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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10 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