# SETS

1.
i) If X and Y are disjoint sets then the value of n(X∩Y) will be maximum. So,
Maximum value of n(X∩Y) = n(X) + n(Y) = 6 + 9 = 15
Again, If X is the subset of Y then the value of n(X∩Y) will be minimum.
So, Minimum value of n(X∩Y) = n(X) + n(Y) - n(X∪Y)

2.
Here,
U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19}
A= {4, 8, 12, 16}
B = {1, 3, 5, 7, 9, 11, 13, 15, 19}
Illustrating the given sets in Venn-diagram. Now, from venn-diagram.
i) A∩B = {4, 8, 12, 16} ∩ {1, 3, 5, 7, 9, 11, 13, 15, 19} = 0
∴ n(A∩B) = 0

ii) A∪B = {4, 8, 12, 16} ∪ {1, 3, 5, 7, 9, 11, 13, 15, 19} = {1, 3, 4, 5, 7, 8, 9, 11, 12, 13 , 15, 16, 19}
∴ $\overline {\left( {{\rm{AB}}} \right)}$ = U - (A∪B) = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19} - {1, 3, 4, 5, 7, 8, 9, 11, 12, 13, 15, 16, 19} = {2, 6, 10, 14 , 18}
∴ n$\overline {\left( {{\rm{AB}}} \right)}$ = 5

iii) B∪A = {1, 3, 4, 5, 7, 8, 9, 11, 12, 13, 15, 16, 19}
∴ n(B∪A) = 14

iv) (A - B) = {4, 8, 12, 16} - {1, 3, 5, 7, 9, 11, 13, 15, 19} ={ 4, 8, 12, 16}
∴ n(A - B) = 4

v) U - (A∩B) = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19} - 0
∴ n[U - (A∩B)] = 19

vi)B = {1, 3, 5, 7, 9, 11, 13, 15, 19}
∴ n(B) = 10
n(A∩B) = 0
n(B - A) = n(B) - n(A∩B) = 10 - 0 = 10

3. Let the number of students in class X be 100, and M and S represent the set of students who passed in math and science respectively.
Then,
n(M) = 75
n(S) = 65
n(M∪S) = 100
n(M∩S) = ?
Now,
n(M∪S) = n(M) + n(S) - n(M∩S)
or, 100 = 75 + 65 - n(M∩S)
or, n(M∩S) = 140 -100
∴ n(M∩S) = 40
So, the total percentage of students who passed in both subject is 40%

4. Let the set representing Japanese tourists who have already visited India and Pakistan be I and P respectively . Now,
n(U) = 500
n(I) = $\frac{{40}}{{100}}{\rm{*}}500$ =200
n(P) = $\frac{{30}}{{100}}{\rm{*}}500$ =150
n(I∩P)= $\frac{{10}}{{100}}{\rm{*}}500$ = 50
Number of tourists who have not visited either of the country = We know,
n(I∩P) = n(I) + n(P) - n(I∩P) = 200 +150 -50 = 300
Now,
n$\overline {\left( {{\rm{IP}}} \right)}$= n(U) - n(IP) = 500 - 300 = 200
Therefore, 200 of the tourists haven't visited either of the country.

5.
Let the sets representing people speaking Nepali and Newari be A and B respectively.
Venn diagram showing the given information. Here, n(U) = n(A) +n(B) + n(A∩B) + n$\overline {\left( {{\rm{AB}}} \right)}$
Or, 800 = 400 + 130 + n(A∩B) + 50
Or, 800 = 580 + n(A∩B)
∴ n(A∩B) = 800 - 580 = 220
Therefore, 220 people can speak both languages.

6. Let the sets representing students who like milk and yogurt be M and Y respectively.
n(U) = 40
n(Y) = 30
n(M) = 25
Number of students not liking both milk and yogurt = n$\overline {\left( {{\rm{MY}}} \right)}$= 5
We know,
n(M∪Y) = n(M) + n(Y) - n(M∩Y)
Or, n(M∪Y) = 25 + 30 - n(M∩Y)
=55 - n(M∪Y)
Again,
n$\overline {\left( {{\rm{MY}}} \right)}$ = n(U) - n(M∪Y)
or, 5 = 40 - ( 55 - n(M∪Y)
or, 5 = 40 - 55 + n(M∪Y)
or n(M∪Y) = 5 + 15
or, n(M∪Y) = 20
Therefore 20 students like both Milk and Yogurt
i) number of students who like milk only = n(M) - n(M∪Y) = 25 - 20 =5
ii) number of students who don't like yogurt = n(U) - n(Y) = 40 - 30 = 10
iii) number of students who like either milk or yogurt = [n(Y) - n(M∪Y)] + [n(M) - n(M∪Y)] = (30 -20) + ( 25-20) = 10 + 5
= 15

7.
Let E and M be the students who passed English and Maths respectively. Let U be the total set of students .
Then, Let, total students, n(U) be x
So,
n(E) = ${\rm{x*}}\frac{{80}}{{100}} = \frac{{80{\rm{x}}}}{{100}}$
n(M) = ${\rm{x*}}\frac{{85}}{{100}} = \frac{{85{\rm{x}}}}{{100}}$
n(E∩M) = ${\rm{x*}}\frac{{75}}{{100}} = \frac{{75{\rm{x}}}}{{100}}$
n$\overline {\left( {{\rm{EM}}} \right)}$ = 45
n(M∩E) = n(U) - n$\overline {\left( {{\rm{EM}}} \right)}$ = x - 45

We know,
N(E∪M) = n(E) +n(M) -n$\overline {\left( {{\rm{EM}}} \right)}$
Or, ${\rm{x}} - 45 = \frac{{80{\rm{x}}}}{{100}} + \frac{{85{\rm{x}}}}{{100}} - \frac{{75{\rm{x}}}}{{100 }}$
Or, 100x - 4500 = 90x
Or, 10x = 4500
∴x = 450
Hence the total number of students is 450.

8. Let E and M be the set of students who took English and Maths tuition respectively.
Here,
n(U) = $115000{\rm{*}}\frac{{20}}{{100}} = 23000$
No. of students who study English, n(E) = $23000{\rm{*}}\frac{{40}}{{100}} = 9200$
No. of students who study Maths, n(M) = $23000{\rm{*}}35/100 = 8050$
No. of students who study tuition in other subjects, n $\overline {\left( {{\rm{EM}}} \right)} = 23000 {\rm{*}}\frac{5}{{100}}$
No. of students who study tuition for both subject, n(E∩M) = ?
We know,
n(E∪M) = n(U) - n $\overline {\left( {{\rm{EM}}} \right)} =$ = 23000- 1150 = 21850
Also, n(E∪M) = n(E) + n(M) - n(E∩M)
Or, 21850 = 9200 + 8050 - n(E∩M)
Or, n(E∩M) = 21850 - 17250
∴ n(E∩M) = 4600
Illustrating the above information in a venn-diagram. 9.
Let D and W be the set of people who study daily newspaper and weekly newspaper respectively.
Then,
n(U) = 1500
n(w) = 1200
n(D) = 800
n(D∩W) = 600
n $\overline {\left( {{\rm{DW}}} \right)}$ = ?

i) Now, Illustrating the given information in the Venn-diagram. ii) From Venn diagram,
n(D∪W) = (800 - 600) + 600 + (1200 - 600) = 1400
Now,
n $\overline {\left( {{\rm{DW}}} \right)}$ = n(U) - n(D∪W) = 1500 - 1400 = 100
∴ Number of people who did not read any study any type of newspaper is 100.

10.
Let T and C denote the set of students who drank tea and coffee repectively.
Now, Using the given information in the Venn-diagram. Here,
n(T) = 88
n(C) = 26
n(U) = 120
n $\overline {\left( {{\rm{TC}}} \right)}$ =17
n $\overline {\left( {{\rm{TC}}} \right)}$ = ?
We know,
n(U) - n(T∪C) = n $\overline {\left( {{\rm{TC}}} \right)}$
or, 120 - n(T∪C) = 17
or, n(T∪C) = 120 - 17
∴ n(T∪C) = 103
Again,
n(T∪C) = n(T) + n(C) - n(T∩C)
or, 103 = 88 + 26 - n(T∩C)
or, n(T∩C) = 88+26 - 103
∴ n(T∩C) = 11
Hence, number of student who drank both tea and coffee = 11

11.
Here, Let the sets representing teachers speaking Newari and Bhojpuri language be N and B respectively.
So, n(U) = 20
n(N) = 15
N(B) = 8
Also, the number of teacher who can't speak either of the language = n($\overline {{\rm{NUB}}}$) = 3
Let the number of teachers who can speak both the language = n(N∩B) = x

i)
No. of teachers who could speak at least one of the language = n(N∩B)
n(N∪B) = n(U) - n$(\overline {{\rm{NUB}}}$)
= 20 - 3
= 17

ii)
We know,
N(NB) = n(N) + n(B) - n(N∩B)
= 15 + 8 - x
= 23 - x
Also,
n(N∪B) = 23 - x
or, 17 = 23 - x
or x = 23 - 17
∴ x = 6
Hence, 6 teachers could speak both language.

iii)
n(N) = n(N) - n(N∩B)
= 15 - 6
= 6
n(B) = n(B) - n(A∩B)
= 8 - 6
= 2

12.
For filling the diagram no(L∩N) = n(L∩N) - n(L∩N∩M) = 11 - 6 = 5
n(N∩M) = n(N∩M) - n(L∩N∩M) = 9 - 6 = 3
n(L∩M) = n(L∩M) - n(L∩N∩M) = 7 - 6 = 1
Also,
n(L) = n(L) - n(L∩M) - no(L∩N) - n(L∩M∩N)
= 14 -1 - 5 - 6 = 2
n(M) = n(M) - no(L∩M) - no(N∩M) - n(L∩M∩N)
= 13 - 1 - 3 -6 = 3
n(N) = n(N) - no(N∩M) - no(N∩M) - n(L∩M∩N)
= 22 - 3 - 5 -6 = 8
So, now, from venn-diagram,
n(U) = 2 + 3 + 8 + 1 + 5 + 3 + 6 + 4 = 32

13. Let the readers reading all types of magazine (i.e. D∩W∩M) be x
We know,
n(D∪W∪M) = n(D) + n(W) + n(M) - n(D∩ M) - n(M∩W) - n(D∩M) + n(D∩W∩M)
From the venn-diagram.
100% = 60% + 30% + 40% - 10% - 10% - 15% + x
or, 100% = 95% + x
or, x = 100% - 95%
∴ x = 5%

14. Here, let the sets representing students who passed in science, health and math be S, H and M respectively .
So,

i)
n(M∪S∪H) = 100%
n(H) = 55%
n(S) = 45%
n(M) = 40%
n(M∩S) = 10%
n(M∩H) = 15%
n(S∩H) = 20%
n(M∩S∩H) = x% (let)
We know,
n(M∪S∪H) = n(M) + n(S) + n(H) - n(M∩S) - n(M∩H) - n(S∩H) + n(M∩S∩H)
or 100% = 40% + 45% + 55% - 10% - 15% - 20% + x
or, x = 100% - 95%
∴ x = 5%

ii)

15.
Let the sets representing students who passed in Science, Maths and English be S, M and E respectively .
So,
n(U) = 1350
n(S) = 600
n(M) = 700
n(E) = 350
n(S∩M) = 200
n(S∩E) = 150
n(M∩F) = 100
n($\overline {{\rm{S}}\mathop \cup {\rm{M}}\mathop \cup {\rm{E}}}$) = 50
Let students passing in all three subjects, n(S∩M∩E), be x.
So,
n(S∪M∪E) = n(S) + n(M) + n(E) - n(S∩M) - n(S∩E) - n(M∩E) + n(S∩M∩E)
or, n(U) - n($\overline {{\rm{S}}\mathop \cup {\rm{M}}\mathop \cup {\rm{E}}}$) = 600 + 700 + 350 - 200 - 100 - 15- + x
or, 1350 - 50 = x
∴ x = 100
Hence, 100 students passed in all three subjects.

ii)
In venn-diagram. 16. Let the sets representing families who believe in Hinduism, Buddhism and other religion be H, B and O respectively.
So,
n(U) = 140
n(H) = 70
n (B) = 45
n (H∩B) = 17
n (H∩O) = 18
n(B∩O) = 16
n($\overline {{\rm{H}}\mathop \cup {\rm{B}}\mathop \cup {\rm{O}}}$) = 6
Let the families seeing no difference in any religion (i.e. n (H∩B ∩O)) be x
So, n(H∪B∪O) = n(U) - n($\overline {{\rm{H}}\mathop \cup {\rm{B}}\mathop \cup {\rm{O}}}$)
= 140 - 6 = 134
Also,
n(H∪B∪O) = n(H) + n(B) + n(O) - n(H∩B) - n(H∩O) - n(B∩O) + n(H∩B∩O)
or, 134 = 70 + 60 + 45 -17 - 18 - 16 + x
or, x = 134 - 124
∴ x = 10
Hence, 10 families see no difference among any of the religions.

17.
Let the sets representing people liking Tea, Coffee and Milk be T, C and M respectively. So,
n(T) = 60
n(C) = 45
n(M) = 30
n(C∩T) = 25
n(T∩M) = 20
n(C∩) = 15
n(T∩C∩M) = 10
no(T∩C) = n(T∩C) - n(T∩C∩M) = 25 - 10 = 15
no(T∩M) = n(T∩M) - n(T∩C∩M) = 20 - 10 = 10
no(C∩M) = n(C∩M) - n(T∩C∩M) = 15 - 10 = 5
In venn-diagram By venn-diagram
n(T∪C∪M) = 35+ 15 + 10 + 10 + 5 + 15 = 85
∴ 85 people were involved in survey.

18.
Let the sets representing consumer of food A, B and C be A, B and C respectively.
Now,
n(A) = 60%
n(B) = 50%
n(C) = 50%
n(A∩B) = 30%
n(B∩C) = 20%
n(C∩A) = 30%
n(A∩B∩C) = 10%

i)
n(A∪B∪C) = n(A) + n(B) + n(C) - n(A∩B) - n(B∩C) - n(C∩A) + n(A∩B∩C)
n(A∪B∪C) = 60% + 50% + 50% - 30% - 20% - 20% + 10% = 90%
We know,
n($\overline {{\rm{A}}\mathop \cup {\rm{B}}\mathop \cup {\rm{C}}}$) = N(U) - n(A∪B∪C)
= 100% - 90%
= 10%
∴ 10% of consumers didn't like any of the food.

ii) Illustrating the above information in Venn diagram. Go Top