##
**Word
Problems based on Three Sets**

The representation of the three overlapping sets A, B and
C are shown in the adjoining Venn diagram. Observe the Venn diagram and learn
the representations.

n(A) = Number of elements in set A

n(B) = Number of elements in set B

n(C) = Number of elements in set C

n

_{o}(A) = Number of elements in set A only
n

_{o}(B) = Number of elements in set B only
n

_{o}(C) = Number of elements in set C only
n

_{o}(A∩B) = Number of elements common in set A and B only
n

_{o}(A∩C) = Number of elements common in set A and C only
n

_{o}(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(A∪B∪C) = 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.###
**List
of formulae based on cardinal number of sets (Three sets)**

(i) n(A∪B∪C) =
n(A) + n(B) + n(C) – n(A∩B) – n(B∩C) – n(A∩C) + n(A∩B∩C)

(ii) n(A∪B∪C)

^{c}= n(U) – n(A∪B∪C)
(iii) n

_{o}(A) = n(A) – n(A∩B) – n(A∩C) + n(A∩B∩C)
(iv) n

_{o}(B) = n(B) – n(A∩B) – n(B∩C) + n(A∩B∩C)
(v) n

_{o}(C) = n(C) – n(A∩C) – n(B∩C) + n(A∩B∩C)
(vi) n

_{o}(A∩B) = n(A∩B) – n(A∩B∩C)
(vii) n

_{o}(B∩C) = n(B∩C) – n(A∩B∩C)
(viii) n

_{o}(A∩C) = n(A∩C) – n(A∩B∩C)
(ix) Exactly
two = n

_{o}(A∩B) + n_{o}(B∩C) + n_{o}(A∩C)
(x) At
least one = n(A∪B∪C)

(xi) All
three = n(A∩B∩C)

###
**Word
Problems based on Three Sets**

####
*Workout Examples*

*Workout Examples*

*Example 1: In a survey of 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 also 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,*

*Now,*

*n(T*

*∪*

*C*

*∪*

*M) = 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 2: 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(M*

*∪*

*S*

*∪*

*H)*

*n(M) = 40%*

*n(S) = 45%*

*n(H) = 55%*

*n(M ∩ S) = 10%*

*n(S ∩ H) = 20%*

*n(H ∩ M) = 15%*

*The Venn diagram of above information is as follows:*

*We Know,*

*n(M*

*∪*

*S*

*∪*

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

*or, n(U) = 40 + 45 + 55 – 10 – 20 – 15 + n (M ∩ S ∩ H)*

*or, 100 = 95 + n (M ∩ S ∩ H)*

*or, 100 – 95 = n (M ∩ S ∩ H)*

*or, n (M ∩ S ∩ H) = 5%*

*Hence the number of students who passed in all three subjects is 5%.*

*Example 3: 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*

*n*_{o}(C) = 9

*n*_{o}(H) = 12

*n*_{o}(C ∩ H) = 8

*n*_{o}(H ∩ M) = 5

*Let, the number of students who study all subjects be ‘x’. i.e n(M ∩ S ∩ H) = x*

*Venn diagram,*

*From the Venn diagram,*

*n(H) = 8 + x + 5 + 12*

*or, 28 = x + 25*

*or, x = 3*

*∴*

*The number of students who study all the subjects, n(M ∩ S ∩ H) = 3*

*Again,*

*n*_{o}(C ∩ M) = n(C) - (9 + 8 + x)

*= 25 – (9 + 8 + 3)*

*= 25 – 20*

*= 5*

*n*_{o}(M) = n(M) – (5 +x + 5)

*= 20 – (5 + 3 + 5)*

*= 20 – 13*

*= 7*

*∴*

*The total number of students, n(U) = n(C*

*∪*

*H*

*∪*

*M) = 9 + 8 + 3 + 5 + 12 + 5 + 7*

*= 49**You can comment your questions or problems regarding the word problems based on three sets here.*

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