Mathematics Part I Solutions Solutions for Class 9 Maths Chapter 1 Sets are provided here with simple step-by-step explanations. These solutions for Sets are extremely popular among Class 9 students for Maths Sets Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part I Solutions Book of Class 9 Maths Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Part I Solutions Solutions. All Mathematics Part I Solutions Solutions for class Class 9 Maths are prepared by experts and are 100% accurate.

#### Page No 3:

#### Question 1:

Write the following sets in roster form.

(i) Set of even numbers

(ii) Set of even prime numbers from 1 to 50

(iii) Set of negative integers

(iv) Seven basic sounds of a sargam (sur)

#### Answer:

(i) *A* = {..., $-$6, $-$4, 0, 2, 4, 6, ...}

(ii) *B* = {2}

(iii) *C* = {..., $-$5, $-$4, $-$3, $-$2, $-$1}

(iv) *D* = {SA, RE, GA, MA, PA, DHA, NI}

#### Page No 3:

#### Question 2:

Write the following symbolic statements in words.

(i) $\frac{4}{3}\in $ Q (ii) $-2\notin $N (iii) P = {p | p is an odd number}

#### Answer:

(i) $\frac{4}{3}\in $ Q

$\frac{4}{3}$ belongs to set Q.

(ii) $-2\notin $ N

$-2$ does not belongs to set N.

(iii) P = {p | p is an odd number}

P is a set of p such that p is an odd number.

#### Page No 4:

#### Question 3:

Write any two sets by listing method and by rule method.

#### Answer:

(i) *A* = {6, 7, 8, 9}

*A* = {*x* | *x* $\in $ N and 5 < *x* < 10}

(ii) *B* = {a, e, i, o, u}

*B* = {*y* | *y* is vowel of English alphabet}

#### Page No 4:

#### Question 4:

(i) All months in the indian solar year.

#### Answer:

*A*= {January, February, March, April, May, June, July, August, September, October, November, December}

*B*= {C, O, M, P, L, E, M, E, N, T}

(iii)

*C*= {Nose, Tongue, Skin, Ear, Eyes}

(iv)

*D*= {2, 3, 5, 7, 11, 13, 17, 19}

(v)

*E*= {Asia, Africa, Europe, North America, South America, Australia, Antartica}

#### Page No 4:

#### Question 5:

#### Answer:

*x*|

*x*=

*n*

^{2},

*n*$\in $ N and $1\le n\le 10$}

*x*|

*x*= 6

*y*,

*y*$\in $ N and $1\le y\le 8$}

(iii) C = Set of letters in the word 'SMILE'

(v) X = Set of letters in the word 'ate'

#### Page No 6:

#### Question 1:

#### Answer:

*x*| 3

*x*$-$ 1 = 2} = {1};

*x*|

*x*is a natural number but

*x*is neither prime nor composite} = {1}; and

C = {

*x*|

*x*$\in $ N,

*x*< 2} = {1}

So, A = B = C

#### Page No 6:

#### Question 2:

#### Answer:

Since, A = Even prime numbers = {2}; and

B = {*x* | 7*x* $-$ 1 = 13} = {2}

So, A = B

hence, A and B are equal sets.

#### Page No 6:

#### Question 3:

( iii ) C = { x | 5 x - 2 = 0, x$\in $N}

#### Answer:

*a*|

*a*is a natural number smaller than zero.} = {}

So, A is an empty set.

*x*|

*x*

^{2}= 0} = {0}

So, B is not an empty set.

(iii) C = {

*x*| 5

*x*$-$ 2 = 0,

*x*$\in $N} = {}

So, C is an empty set.

#### Page No 7:

#### Question 4:

(ii) B = { y | y < -1, y is an integer}

(iv) Set of people from your village.

(vi) Set of whole numbers

(vii) Set of rational number

#### Answer:

*x*|

*x*< 10,

*x*is a natural number} = {1, 2, 3, 4, 5, 6, 7, 8, 9}

So, A is a finite set.

(ii) B = {

*y*|

*y*< $-$1,

*y*is an integer} = {..., $-$5, $-$4, $-$3, $-$2}

So, B is an infinite set.

(iii) C = Set of students of class 9 from your school.

Since, the number of elements of set C is countable number.

So, C is a finite set.

(iv) D = Set of people from your village.

Since, the number of elements of set D is countable number.

So, D is a finite set.

(v) E = Set of apparatus in laboratory

So, E is a finite set.

(vi) F = Set of whole numbers = {0,1, 2, 3, 4, ...}

So, F is an infinite set.

(vii) G = Set of rational number

So, G is an infinite set.

#### Page No 11:

#### Question 1:

#### Answer:

So, C $\subseteq $ B is false.

(ii) Since, $b\in A\mathrm{but}b\notin D$

So, A $\subseteq $ D is false.

(iii) Since, $a\in D\mathrm{but}a\notin B$

So, D $\subseteq $ B is false.

(iv) Since, all elements of set D is in set A.

So, D $\subseteq $ A is true.

(v) Since, $f\in B\mathrm{but}f\notin A$

So, B $\subseteq $ A is false.

(vi) Since, all elements of set C is in set A.

So, C $\subseteq $ A

#### Page No 11:

#### Question 2:

( i ) X = { x | x$\in $N, and 7 < x < 15}

#### Answer:

U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20};

X = {

*x*|

*x*$\in $ N, and 7 <

*x*< 15} = {8, 9 ,10, 11, 12, 13, 14}; and

Y = {

*y*|

*y*$\in $N,

*y*is prime number from 1 to 20} = {2, 3, 5, 7, 11, 13, 17, 19}

#### Page No 11:

#### Question 3:

(i) show the sets U, P and P ¢ by Venn diagram.

(ii) Verify (P ¢ ) ¢ = P

#### Answer:

We have,

U = {1, 2, 3, 7, 8, 9, 10, 11, 12} and P = {1, 3, 7, 10}

P' = U $-$ P = {2, 4, 8, 9, 11, 12}

(i)

(ii) (P')' = U $-$ P' = {1, 3, 7, 10} = P

#### Page No 11:

#### Question 4:

A = {1, 3, 2, 7} then write any three subsets of A.

#### Answer:

We have,

A = {1, 3, 2, 7}

The three subsets of A are: {1}, {2} and {1, 2, 3}.

Disclaimer: There are 2^{4} = 16 subsets possible i.e. {}, {1}, {2}, {3}, {7}, {1, 2}, {1, 3}, {1, 7}, {2, 3}, {2, 7}, {3, 7}, {1, 2, 3}, {2, 3, 7}, {1, 3, 7}, {1, 2, 7} and {1, 2, 3, 7}.

#### Page No 11:

#### Question 5:

#### Answer:

So, P $\subset $ H $\subset $ B

Similarly,

Indore is a city in Madhya Pradesh, and Madhya Pradesh is a state in India.

So, I $\subset $ M $\subset $ B

#### Page No 11:

#### Question 6:

C = set of multiples of 12

#### Answer:

A = set of multiples of 5 = {5, 10, 15, ...}; B = set of multiples of 7 = {7, 14, 21, ...}; and C = set of multiples of 12 = {12, 24, 36, ...}

The universal set for the sets A, B and C can be set of natural numbers, i.e. N.

Disclaimer: The Universal set for the sets A, B and C can be set of whole numbers, integers, rational number of real numbers and so on.

P = set of integers which are multiples of 4 = {4, 8, 12, ...}; and T = set of all even square numbers = {4, 16, 36, 64, 100, ...}

So, P can be the Universal set for the sets P and Q.

Disclaimer: Here also, the Universal set can be vary from the ablove.

#### Page No 11:

#### Question 7:

Let all the students of a class is an Universal set. Let set A be the students who secure 50% or more marks in Maths. Then write the complement of set A.1

#### Answer:

We have,

U = set of all the students of a class and A = set of the students who secure 50% or more marks in Maths

So, the complement of set A, A' = U $-$ A = set of the students who secure less than 50% marks in Maths.

#### Page No 16:

#### Question 1:

If n (A) = 15, n (A$\cup $B ) = 29, n (A $\cap $ B) = 7 then n (B) = ?

#### Answer:

We have,

n (A) = 15, n (A$\cup $B ) = 29, n (A $\cap $ B) = 7

Since, n (B) = n (A$\cup $B ) + n (A $\cap $ B) $-$ n (A) = $29+7-15$

So, n (B) = 21

#### Page No 16:

#### Question 2:

In a hostel there are 125 students, out of which 80 drink tea, 60 drink coffee and 20 drink tea and coffee both. Find the number of students who do not drink tea or coffee.

#### Answer:

Let A be the set of students who drink tea, and B be the set of setudents who drink coffee.

We have,

n(U) = 125, n(A) = 80, n(B) = 60, $\mathrm{n}(\mathrm{A}\cap \mathrm{B})=20$

Since,

$\mathrm{n}(\mathrm{A}\cup \mathrm{B})=\mathrm{n}\left(\mathrm{A}\right)+\mathrm{n}\left(\mathrm{B}\right)-\mathrm{n}(\mathrm{A}\cap \mathrm{B})=80+60-20$

$\mathrm{n}(\mathrm{A}\cup \mathrm{B})=120$

So, the number of students who do not drink tea or coffee = 125 $-$ 120 = 5

#### Page No 16:

#### Question 3:

In a competitive exam 50 students passed in English. 60 students passed in Mathematics. 40 students passed in both the subjects. None of them fail in both the subjects. Find the number of students who passed at least in one of the subjects ?

#### Answer:

Let A be the set of students who passed in English, and B be the set of students who passed in Mathematics.

We have,

n(A) = 50, n(B) = 60,

Since,

So, the number of students who passed at least in one of the subjects is 70.

#### Page No 16:

#### Question 4:

A survey was conducted to know the hobby of 220 students of class IX. Out of which 130 students informed about their hobby as rock climbing and 180 students informed about their hobby as sky watching. There are 110 students who follow both the hobbies. Then how many students do not have any of the two hobbies ? How many of them follow the hobby of rock climbing only ? How many students follow the hobby of sky watching only ?

#### Answer:

Let A be the set of students who follow the hobby of rock climbing, and B be the set of students who follow the hobby of sky watching.

We have, n (A) = 130, n (B) = 180, n (A $\cap $ B) = 110, n (U) = 220

Since,

n (A $\cup $ B) = n (A) + n (B) $-$ n (A $\cap $ B) = 130 + 180 $-$ 110 = 200

So, the number of students who do not have any of the two hobbies = n (U) $-$ n (A $\cup $ B) = 220 $-$ 200 = 20

Also, the number of students who follow the hobby of rock climbing only = n (A) $-$ n (A $\cap $ B) = 130 $-$ 110 = 20

And, the number of students who follow the hobby of sky watching only = n (B) $-$ n (A $\cap $ B) = 180 $-$ 110 = 50

#### Page No 16:

#### Question 5:

(v) A' (vi) B' (vii) (A$\cup $B) '

#### Answer:

*m*,

*n*,

*x*,

*y*,

*z*}

(ii) B = {

*m*,

*n*,

*p*,

*q*,

*r*}

(iii) A$\cup $B = {

*m*,

*n*,

*x*,

*y*,

*z*,

*p*,

*q*,

*r*}

(iv) U = {

*m*,

*n*,

*x*,

*s*,

*t*,

*y*,

*z*,

*p*,

*q*,

*r*}

(v) A' = {

*m*,

*n*,

*s*,

*t*,

*p*,

*q*,

*r*}

(vi) B' = {

*m*,

*n*,

*x*,

*s*,

*t*,

*y*,

*z\*}

(vii) (A$\cup $B)' = {

*s*,

*t*}

#### Page No 16:

#### Question 1:

(iii) P = {1, 2, ........., 10}, What type of set P is ?

(A) Null set (B) Infinite set (C) Finite set (D) None of these

(vi) Which of the following sets are empty sets ?

(B) set of even prime numbers.

*x*|

*x*$\in $ I, $-$1 <

*x*< 1}

#### Answer:

M = {1, 3, 5}, N = {2, 4, 6}

M$\cap $N = $\varphi $ = Empty set

*x*|

*x*is an odd natural number, 1 <

*x*$\le $5} = {3, 5}

(iii) Since, the elements of set P = {1, 2, ..., 10} is finite.

So, set P is a finite set.

Hence, the correct option is (C).

Since, {1, 2, 3} $\cup $ {1, 2, 4} = {1, 2, 3, 4} $\ne $ M $\cup $ N = {1, 2, 3, 4, 5, 6};

{2, 5, 6} $\cup $ {1, 2, 4} = {1, 2, 4, 5, 6} $\ne $ M $\cup $ N = {1, 2, 3, 4, 5, 6}; and

{4, 5, 6} $\cup $ {1, 2, 4} = {1, 2, 4, 5, 6} $\ne $ M $\cup $ N = {1, 2, 3, 4, 5, 6}

Now, P $\cap $ (P $\cup $ M) = P $\cap $ M = M (Since, P $\cup $ M = M; P$\subseteq $ M)

So, the correct option is (B).

(vi) Since,

the set of intersecting points of parallel lines = {};

the set of even prime numbers= {2};

P = {

*x*|

*x*$\in $ I, $-$1 <

*x*< 1} = {0}

So, the correct option is (A).

#### Page No 17:

#### Question 2:

(i) Which of the following collections is a set ?

(ii) Which of the following set represent N $\cap $ W?

(iii) P = {

*x*|

*x*is a letter of the word ' indian'} then which one of the following is set P in listing form ?

(iv) If T = {1, 2, 3, 4, 5} and M = {3, 4, 7, 8} then T$\cup $M = ?

(C) {1, 2, 3, 4, 5, 7, 8} (D) {3, 4}

#### Answer:

(A) Since, the colours of the rainbow are well defined such as Violet, Indigo, Blue, Green, Yellow, Orange and Red.

So, the collection of colours of the rainbow is a set.

(B) Since, the tall trees in the school campus are not well defined.

So, the collection of the tall trees in the school campus is not a set.

(C) Since, the rich people in the village is not well defined.

So, the colection of the rich people in the village is not a set.

(D) Since, the easy examples in the book is not well defined.

So, the collection of easy examples in the book is not a set.

(ii) Since, N $\cap $ W = {1, 2, 3, 4, ...}

So, the correct option is (A).

*x*|

*x*is a letter of the word 'indian'}

So, P = {i, n, d, i, a, n} = {i, n, d, a}

So, T$\cup $M = {1, 2, 3, 4, 5, 7, 8}

Hence, the correct option is (C).

#### Page No 17:

#### Question 3:

Out of 100 persons in a group, 72 persons speak English and 43 persons speak French. Each one out of 100 persons speak at least one language. Then how many speak only English ? How many speak only French ? How many of them speak English and French both ?

#### Answer:

Let A be the set of persons speaking English and B be the set of persons speaking French.

So, n (A) = 72; n (B) = 43; $\mathrm{n}(\mathrm{A}\cup \mathrm{B})=100$

Now,

$n\left(A\right)+n\left(B\right)=n(A\cup B)+n(A\cap B)$

$\Rightarrow \mathrm{n}(\mathrm{A}\cap \mathrm{B})=72+43-100\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{n}(\mathrm{A}\cap \mathrm{B})=15\phantom{\rule{0ex}{0ex}}\mathrm{So},\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{person}\mathrm{who}\mathrm{speak}\mathrm{French}\mathrm{and}\mathrm{English}\mathrm{both}\mathrm{is}15.$

Also,

$\mathrm{n}\left(\mathrm{A}\right)=\mathrm{n}(\mathrm{A}-\mathrm{B})+\mathrm{n}(\mathrm{A}\cap \mathrm{B})$

$\Rightarrow \mathrm{n}(\mathrm{A}-\mathrm{B})=72-15\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{n}(\mathrm{A}-\mathrm{B})=57\phantom{\rule{0ex}{0ex}}\mathrm{So},\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{person}\mathrm{who}\mathrm{speak}\mathrm{only}\mathrm{English}\mathrm{is}57.$

And,

$\mathrm{n}\left(\mathrm{B}\right)=\mathrm{n}(\mathrm{B}-\mathrm{A})+\mathrm{n}(\mathrm{A}\cap \mathrm{B})\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{n}(\mathrm{B}-\mathrm{A})=43-15\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{n}(\mathrm{B}-\mathrm{A})=28\phantom{\rule{0ex}{0ex}}\mathrm{So},\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{person}\mathrm{who}\mathrm{speak}\mathrm{only}\mathrm{French}\mathrm{is}28.$

#### Page No 17:

#### Question 4:

70 trees were planted by Parth and 90 trees were planted by Pradnya on the occasion of Tree Plantation Week. Out of these; 25 trees were planted by both of them together. How many trees were planted by Parth or Pradnya ?

#### Answer:

Let A be the set of tress planted by Parth and B be the set of trees planted by Pradnya.

So, n (A) = 70; n (B) = 90; $n(A\cap B)=25$

Now,

$n\left(A\right)+n\left(B\right)=n(A\cup B)+n(A\cap B)$

$n(A\cup B)=70+90-25$

݅݅݅$n(A\cup B)=135$

Hence, the number of trees planted by Parth or Pradnya is 135.

#### Page No 17:

#### Question 5:

If n (A) = 20, n (B) = 28 and n (A$\cup $B) = 36 then n (A $\cap $ B) = ?

#### Answer:

We have,

n (A) = 20, n (B) = 28 and n (A$\cup $B) = 36

Since, n (A $\cap $ B ) = n (A) + n (B) $-$ n (A $\cup $ B) = $20+28-36$

**$\therefore $ **n (A $\cap $ B ) = 12

#### Page No 17:

#### Question 6:

In a class, 8 students out of 28 have a dog as their pet animal at home, 6 students have a cat as their pet animal. 10 students have dog and cat both, then how many students do not have a dog or cat as their pet animal at home ?

#### Answer:

We have,

Total number of students = 28;

Students have a dog as their pet = 8;

Students have a cat as their pet = 6; and

Students have cat and dog both = 10

Solving using venn diagram, we get:

So, the number of students that do not have a dog or a cat as their pet is 4.

#### Page No 17:

#### Question 7:

(i) A ={3, 4, 5, 7} B ={1, 4, 8}

*a*,

*b*,

*c*,

*e*,

*f*} Q ={

*l*,

*m*,

*n*,

*e*,

*b*}

*x*|

*x*is a prime number between 80 and 100}

*y*|

*y*is an odd number between 90 and 100 }

#### Answer:

*a*,

*b*,

*c*,

*e*,

*f*} Q ={

*l*,

*m*,

*n*,

*e*,

*b*}

*x*|

*x*is a prime number between 80 and 100} = {83, 89, 97};

*y*|

*y*is an odd number between 90 and 100 } = {91, 93, 95, 97, 99}

#### Page No 18:

#### Question 8:

X = set of all quadrilaterals. Y = set of all rhombuses.

#### Answer:

Since, all squares are rectangle, all rectangles are parallelogram, all parallelograms are quadrilateral; and all squares are rhombus, all rhombus are parallelogram, all parallelograms are quadrilateral.

So, the subset relations are:

S < V < T < X and S < Y < T < X

#### Page No 18:

#### Question 9:

If M is any set, then write M $\cup \mathrm{\varphi}$ and M$\cap \mathrm{\varphi}$ .

#### Answer:

If M is any set, then

$\mathrm{M}\cup \mathrm{\varphi}=\mathrm{M}$

and,

$\mathrm{M}\cap \mathrm{\varphi}=\mathrm{\varphi}$

#### Page No 18:

#### Question 10:

Observe the Venn diagram and write the given sets

U, A, B, A $\cup $ B, A $\cap $ B

#### Answer:

(i) U = {1, 2, 3, 4, 5, 7, 8, 9, 10, 11, 13}

(ii) A = {1, 2, 3, 5, 7}

(iii) B = {1, 5, 8, 9, 10}

(iv) A $\cup $ B = {1, 2, 3, 5, 7, 8, 9, 10}

(v) $A\cap B=\left\{1,5\right\}$

#### Page No 18:

#### Question 11:

#### Answer:

We have,

n (A) = 7, n (B) = 13 and n (A $\cap $ B) = 4

Since, n (A $\cup $ B) = n (A) + n (B) $-$ n (A $\cap $ B) = 7 + 13 $-$ 4

**$\therefore $ **n (A $\cup $ B) = 16

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