## Maharashtra State Board Class 9 Maths Solutions Chapter 1 Sets Practice Set 1.3

Question 1.

If A = {a, b, c, d, e}, B = {c, d, e, f}, C = {b, d}, D = {a, e}, then which of the following statements are true and which are false?

i. C ⊆ 3

ii. A ⊆ D

iii. D ⊆ B

iv. D ⊆ A

V. B ⊆ A

vi. C ⊆ A

Ans:

i. C = {b, d}, B = {c, d, e ,f}

C ⊆ B

False

Since, all the elements of C are not present in B.

ii. A = {a, b, c, d, e}, D = {a, e}

A ⊆ D

False

Since, all the elements of A are not present in D.

iii. D = {a, e}, B = {c, d, e, f}

D ⊆ B

False

Since, all the elements of D are not present in B.

iv. D = {a, e}, A = {a, b, c, d, e}

D ⊆ A

True

Since, all the elements of D are present in A.

v. B = {c, d, e, f}, A = {a, b, c, d, e}

B ⊆ A

False

Since, all the elements of B are not present in A.

vi. C = {b, d}, A= {a, b, c, d, e}

C ⊆A

True

Since, all the elements of C are present in A.

Question 2.

Take the set of natural numbers from 1 to 20 as universal set and show set X and Y using Venn diagram. [2 Marks each]

i. X= {x |x ∈ N, and 7 < x < 15}

ii. Y = { y | y ∈ N, y is a prime number from 1 to 20}

Answer:

i. U = {1, 2, 3, 4, …….., 18, 19, 20}

x = {x | x ∈ N, and 7 < x < 15}

∴ x = {8, 9, 10, 11, 12, 13, 14}

ii. U = {1, 2, 3, 4, …… ,18, 19, 20}

Y = { y | y ∈ N, y is a prime number from 1 to 20}

∴ Y = {2, 3, 5, 7, 11, 13, 17, 19}

Question 3.

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

i. show the sets U, P and P’ by Venn diagram.

ii. Verify (P’)’ = P

Solution:

i. Here, U = {1,2, 3, 7, 8,9, 10, 11, 12} P = {1, 3, 7, 10}

∴ P’ = {2, 8, 9, 11, 12}

II. Here, U = {1, 2, 3, 7, 8, 9, 10, 11, 12}

P = {1, 3, 7, 10} ….(i)

∴ P’= {2, 8, 9, 11, 12}

Also, (P’)’ = {1,3,7, 10} …(ii)

∴ (P’)’ = P … [From (i) and (ii)]

Question 4.

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

Solution:

Three subsets of A:

i. B = {3}

ii. C = {2, 1}

iii. D= {1, 2, 7}

[Note: The above problem has many solutions. Students may write solutions other than the ones given]

Question 5.

i. Write the subset relation between the sets.

P is the set of all residents in Pune.

M is the set of all residents in Madhya Pradesh.

I is the set of all residents in Indore.

B is the set of all residents in India.

H is the set of all residents in Maharashtra.

ii. Which set can be the universal set for above sets ?

Solution:

i.

a. The residents of Pune are residents of India.

∴ P ⊆ B

b. The residents of Pune are residents of Maharashtra.

∴ P ⊆ H

c. The residents of Madhya Pradesh are residents of India.

∴ M ⊆ B

d. The residents of Indore are residents of India.

∴ I ⊆ B

e. The residents of Indore are residents of Madhya Pradesh.

∴ I ⊆ M

f. The residents of Maharashtra are residents of India.

∴ H ⊆B

ii. The residents of Pune, Madhya Pradesh, Indore and Maharashtra are all residents of India.

∴ B can be the Universal set for the above sets.

Question 6.

Which set of numbers could be the universal set for the sets given below?

i. A = set of multiples of 5,

B = set of multiples of 7,

C = set of multiples of 12

ii. P = set of integers which are multiples of 4.

T = set of all even square numbers.

Answer:

i. A = set of multiples of 5

∴ A = {5, 10, 15, …}

B = set of multiples of 7

∴ B = {7, 14, 21,…}

C = set of multiples of 12

∴ C = {12, 24, 36, …}

Now, set of natural numbers, whole numbers, integers, rational numbers are as follows:

N = {1, 2, 3, …}, W = {0, 1, 2, 3, …}

I = {…,-3, -2, -1, 0, 1, 2, 3, …}

Q = { \(\frac { p }{ q }\) | p,q ∈ I,q ≠ 0}

Since, set A, B and C are the subsets of sets N, W , I and Q.

∴ For set A, B and C we can take any one of the set from N, W, I or Q as universal set.

ii. P = set of integers which are multiples of 4.

P = {4, 8, 12,…}

T = set of all even square numbers T = {2^{2}, 4^{2}, 6^{2}, …]

Since, set P and T are the subsets of sets N, W, I and Q.

∴ For set P and T we can take any one of the set from N, W, I or Q as universal set.

Question 7.

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

Answer:

Here, U = all the students of a class.

A = Students who secured 50% or more marks in Maths.

∴ A’= Students who secured less than 50% marks in Maths.

Question 1.

If A = {1, 3, 4, 7, 8}, then write all possible subsets of A.

i. e. P = {1, 3}, T = {4, 7, 8}, V = {1, 4, 8}, S = {1, 4, 7, 8}

In this way many subsets can be written. Write five more subsets of set A. (Textbook pg. no, 8)

Answer:

B = { },

E = {4},

C = {1, 4},

D = {3, 4, 7},

F = {3, 4, 7,8}

Question 2.

Some sets are given below.

A ={…,-4, -2, 0, 2, 4, 6,…}

B = {1, 2, 3,…}

C = {…,-12, -6, 0, 6, 12, 18, }

D = {…, -8, -4, 0, 4, 8,…}

I = {…,-3, -2, -1, 0, 1, 2, 3, 4, }

Discuss and decide which of the following statements are true.

a. A is a subset of sets B, C and D.

b. B is a subset of all the sets which are given above. (Textbook pg. no. 9)

Solution:

a. All elements of set A are not present in set B, C and D.

∴ A ⊆ B,

∴ A ⊆ C,

∴ A ⊆ D

∴ Statement (a) is false.

b. All elements of set B are not present in set A, C and D.

∴ B ⊆ A,

∴ B ⊆ C,

∴ B ⊆ D

∴ Statement (b) is false.

Question 3.

Suppose U = {1, 3, 9, 11, 13, 18, 19}, and B = {3, 9, 11, 13}. Find (B’)’ and draw the inference. (Textbook pg. no. 10)

Solution:

U = {1, 3, 9, 11, 13, 18, 19},

B= {3, 9, 11, 13} ….(i)

∴ B’= {1, 18, 19}

(B’)’= {3, 9, 11, 13} ….(ii)

∴ (B’)’ = B … [From (i) and (ii)]

∴ Complement of a complement is the given set itself.