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
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
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(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) no(A)
= n(A) – n(A∩B) – n(A∩C) + n(A∩B∩C)
(iv) no(B)
= n(B) – n(A∩B) – n(B∩C) + n(A∩B∩C)
(v) no(C)
= n(C) – n(A∩C) – n(B∩C) + n(A∩B∩C)
(vi) no(A∩B)
= n(A∩B) – n(A∩B∩C)
(vii) no(B∩C)
= n(B∩C) – n(A∩B∩C)
(viii) no(A∩C)
= n(A∩C) – n(A∩B∩C)
(ix) Exactly
two = no(A∩B) + no(B∩C) + no(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
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
no(C) = 9
no(H) = 12
no(C ∩ H) = 8
no(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,
no(C ∩ M) = n(C)
- (9 + 8 + x)
= 25 –
(9 + 8 + 3)
= 25 – 20
= 5
no(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
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regarding the word problems based on three sets here.
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