# Class 10 Mathematics Solution

# 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%

Hence, 5% readers read all types of newspaper.

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.