Sets

ICSE Class 6 Mathematics · Solutions for Class 6 RS Aggarwal ICSE Foundation Mathematics · 53 solved questions

Exercise
Marks
1Exercise 7(A)

Which of the following collections are sets? If any collection is not a set, give reasons.

(i) All persons on earth

(ii) All male teachers of your school

(iii) All leaders of India

(iv) All healthy people of your town

(v) All fat boys of your locality

(vi) All planets in our solar system

(vii) All difficult problems in your book on Mathematics

(viii) All 3-digit natural numbers

(ix) All books in your school library

Show SolutionHide Solution↓

(i) It is a set as it is well defined. For any given person, we can definitely tell whether that person is on earth or not.

(ii) It is a set as it is well defined. The male teachers of your school can be clearly identified.

(iii) It is not a set because the given collection is not well defined—people may differ on whether a particular person is a leader of India or not.

(iv) It is not a set because the given collection is not well defined—people may differ on whether a person of the town is healthy or not.

(v) It is not a set because the given collection is not well defined—people may differ on whether a boy of the locality is fat or not.

(vi) It is a set as it is well defined. The planets in our solar system can be clearly listed.

(vii) It is not a set because the given collection is not well defined—people may differ on whether a problem is difficult or not.

(viii) It is a set as it is well defined. The 3-digit natural numbers are 100, 101, 102, ......, 999.

(ix) It is a set as it is well defined. The books in your school library can be clearly identified.

2(i)Exercise 7(A)

Describe the following set in Roster form:

A = {x : x is a natural number, 5 < x < 14}

Show SolutionHide Solution↓

The natural numbers x such that 5 < x < 14 are 6, 7, 8, 9, 10, 11, 12 and 13.

Hence, A = {6, 7, 8, 9, 10, 11, 12, 13}.

2(ii)Exercise 7(A)

Describe the following set in Roster form:

B = {x : x is a factor of 18}

Show SolutionHide Solution↓

The factors of 18 are 1, 2, 3, 6, 9 and 18.

Hence, B = {1, 2, 3, 6, 9, 18}.

2(iii)Exercise 7(A)

Describe the following set in Roster form:

C = {x : x is a prime factor of 330}

Show SolutionHide Solution↓

330 = 2 × 3 × 5 × 11

So, the prime factors of 330 are 2, 3, 5 and 11.

Hence, C = {2, 3, 5, 11}.

2(iv)Exercise 7(A)

Describe the following set in Roster form:

D = {x : x is a 2-digit number, the sum of whose digits is 8}

Show SolutionHide Solution↓

The 2-digit numbers whose digits add up to 8 are 17, 26, 35, 44, 53, 62, 71 and 80.

Hence, D = {17, 26, 35, 44, 53, 62, 71, 80}.

2(v)Exercise 7(A)

Describe the following set in Roster form:

E = {x : x is a natural number, x is a multiple of 7, x < 60}

Show SolutionHide Solution↓

The multiples of 7 that are less than 60 are 7, 14, 21, 28, 35, 42, 49 and 56.

Hence, E = {7, 14, 21, 28, 35, 42, 49, 56}.

2(vi)Exercise 7(A)

Describe the following set in Roster form:

F = {x : x is a natural number, x is neither prime nor composite}

Show SolutionHide Solution↓

The only natural number that is neither prime nor composite is 1.

Hence, F = {1}.

2(vii)Exercise 7(A)

Describe the following set in Roster form:

G = {x : x is an even prime number}

Show SolutionHide Solution↓

The only even prime number is 2.

Hence, G = {2}.

2(viii)Exercise 7(A)

Describe the following set in Roster form:

H = {x : x is an odd natural number, 50 < x < 60}

Show SolutionHide Solution↓

The odd natural numbers x such that 50 < x < 60 are 51, 53, 55, 57 and 59.

Hence, H = {51, 53, 55, 57, 59}.

3(i)Exercise 7(A)

Describe the following set in set-builder form:

A = {16, 18, 20, 22, 24, 26, 28}

Show SolutionHide Solution↓

The elements 16, 18, 20, 22, 24, 26, 28 are even natural numbers lying between 14 and 30.

Hence, A = {x : x is an even natural number, 14 < x < 30}.

3(ii)Exercise 7(A)

Describe the following set in set-builder form:

B = {23, 25, 27, 29, 31, 33, 35}

Show SolutionHide Solution↓

The elements 23, 25, 27, 29, 31, 33, 35 are odd natural numbers lying between 22 and 36.

Hence, B = {x : x is an odd natural number, 22 < x < 36}.

3(iii)Exercise 7(A)

Describe the following set in set-builder form:

C = {1, 3, 5, 9, 15, 45}

Show SolutionHide Solution↓

The elements 1, 3, 5, 9, 15, 45 are the factors of 45.

Hence, C = {x : x is a natural number, x is a factor of 45}.

3(iv)Exercise 7(A)

Describe the following set in set-builder form:

D = {15, 24, 33, 42, 51, 60}

Show SolutionHide Solution↓

The elements 15, 24, 33, 42, 51, 60 are 2-digit numbers, the sum of whose digits is 6.

Hence, D = {x : x is a 2-digit number, the sum of whose digits is 6}.

3(v)Exercise 7(A)

Describe the following set in set-builder form:

E = {6, 12, 18, 24, 30, 36, 42, 48}

Show SolutionHide Solution↓

The elements 6, 12, 18, 24, 30, 36, 42, 48 are multiples of 6 that are less than 50.

Hence, E = {x : x is a natural number, x is a multiple of 6, x < 50}.

3(vi)Exercise 7(A)

Describe the following set in set-builder form:

F = {41, 43, 47, 53, 59, 61, 67, 71, 73}

Show SolutionHide Solution↓

The elements 41, 43, 47, 53, 59, 61, 67, 71, 73 are prime numbers lying between 40 and 74.

Hence, F = {x : x is a prime number, 40 < x < 74}.

3(vii)Exercise 7(A)

Describe the following set in set-builder form:

G = {0, 1, 4, 9, 16, 25, 36, 49}

Show SolutionHide Solution↓

The elements 0, 1, 4, 9, 16, 25, 36, 49 are the squares of the whole numbers 0, 1, 2, 3, 4, 5, 6 and 7.

So, each element is of the form n2, where n is a whole number and 0 ≤ n ≤ 7.

Hence, G = {x : x = n2, where n is a whole number, 0 ≤ n ≤ 7}.

4Exercise 7(A)

Let A = {5, 7, 11, 13, 17, 19, 23, 29}.

Insert the appropriate symbol ∈ or ∉ in the blank spaces:

(i) 7 ......... A

(ii) 9 ......... A

(iii) 3 ......... A

(iv) 15 ......... A

(v) 17 ......... A

(vi) 25 ......... A

(vii) 11 ......... A

(viii) 31 ......... A

Show SolutionHide Solution↓

(i) 7 is a member of A.

Hence, 7 ∈ A.

(ii) 9 is not a member of A.

Hence, 9 ∉ A.

(iii) 3 is not a member of A.

Hence, 3 ∉ A.

(iv) 15 is not a member of A.

Hence, 15 ∉ A.

(v) 17 is a member of A.

Hence, 17 ∈ A.

(vi) 25 is not a member of A.

Hence, 25 ∉ A.

(vii) 11 is a member of A.

Hence, 11 ∈ A.

(viii) 31 is not a member of A.

Hence, 31 ∉ A.

5Exercise 7(A)

Let B = {2, 4, 6, 8, 10, ......, 74, 76, 78, 80}.

Insert the appropriate symbol ∈ or ∉ in the blank spaces:

(i) 26 ......... B

(ii) 31 ......... B

(iii) 70 ......... B

(iv) 15 ......... B

(v) 60 ......... B

(vi) 53 ......... B

(vii) 47 ......... B

(viii) 34 ......... B

Show SolutionHide Solution↓

B is the set of all even numbers from 2 to 80.

(i) 26 is an even number lying between 2 and 80.

Hence, 26 ∈ B.

(ii) 31 is an odd number.

Hence, 31 ∉ B.

(iii) 70 is an even number lying between 2 and 80.

Hence, 70 ∈ B.

(iv) 15 is an odd number.

Hence, 15 ∉ B.

(v) 60 is an even number lying between 2 and 80.

Hence, 60 ∈ B.

(vi) 53 is an odd number.

Hence, 53 ∉ B.

(vii) 47 is an odd number.

Hence, 47 ∉ B.

(viii) 34 is an even number lying between 2 and 80.

Hence, 34 ∈ B.

6Exercise 7(A)

State whether the given statement is true or false:

(i) 0 ∈ {x : x is a whole number, x < 10}

(ii) -5 ∈ {x : x is an integer, -4 < x < 4}

(iii) -2 ∈ {x : x is an integer, x2 = 4}

(iv) 2 ∈ {x : x is a prime number, 3 < x < 6}

(v) 83 ∈ {x : x is a prime number, x < 100}

Show SolutionHide Solution↓

(i) The whole numbers less than 10 are 0, 1, 2, ......, 9. Since 0 is one of them, 0 is a member of this set.

True

(ii) The integers x such that -4 < x < 4 are -3, -2, -1, 0, 1, 2, 3. Since -5 is not one of them, -5 is not a member of this set.

False

(iii) x2 = 4 ⇒ x = 2 or x = -2. So the set is {-2, 2}. Since -2 is one of them, -2 is a member of this set.

True

(iv) The prime numbers x such that 3 < x < 6 give only x = 5. So the set is {5}. Since 2 is not a member of this set, the given statement is false.

False

(v) 83 is a prime number and 83 < 100, so 83 is a member of this set.

True

1(i)Exercise 7(B)

Write the following set in Roster form and write its cardinal number:

A = {x : x ∈ W, x < 6}

Show SolutionHide Solution↓

The whole numbers less than 6 are 0, 1, 2, 3, 4 and 5.

The number of distinct elements in a finite set A is called its cardinal number, denoted by n(A).

Here the number of distinct elements = 6

Hence, A = {0, 1, 2, 3, 4, 5} and n(A) = 6.

1(ii)Exercise 7(B)

Write the following set in Roster form and write its cardinal number:

B = {x : x ∈ N is a multiple of 4, x < 30}

Show SolutionHide Solution↓

The multiples of 4 that are less than 30 are 4, 8, 12, 16, 20, 24 and 28.

The number of distinct elements in a finite set B is called its cardinal number, denoted by n(B).

Here the number of distinct elements = 7

Hence, B = {4, 8, 12, 16, 20, 24, 28} and n(B) = 7.

1(iii)Exercise 7(B)

Write the following set in Roster form and write its cardinal number:

C = {x : x ∈ N, x is a perfect square, x < 30}

Show SolutionHide Solution↓

The perfect squares less than 30 are 1, 4, 9, 16 and 25.

The number of distinct elements in a finite set C is called its cardinal number, denoted by n(C).

Here the number of distinct elements = 5

Hence, C = {1, 4, 9, 16, 25} and n(C) = 5.

1(iv)Exercise 7(B)

Write the following set in Roster form and write its cardinal number:

D = {x : x ∈ N, x is a composite number, 10 < x < 18}

Show SolutionHide Solution↓

The composite numbers x such that 10 < x < 18 are 12, 14, 15 and 16.

The number of distinct elements in a finite set D is called its cardinal number, denoted by n(D).

Here the number of distinct elements = 4

Hence, D = {12, 14, 15, 16} and n(D) = 4.

1(v)Exercise 7(B)

Write the following set in Roster form and write its cardinal number:

E = {x : x is an odd prime number, x < 18}

Show SolutionHide Solution↓

The odd prime numbers less than 18 are 3, 5, 7, 11, 13 and 17.

The number of distinct elements in a finite set E is called its cardinal number, denoted by n(E).

Here the number of distinct elements = 6

Hence, E = {3, 5, 7, 11, 13, 17} and n(E) = 6.

1(vi)Exercise 7(B)

Write the following set in Roster form and write its cardinal number:

F = {x : x ∈ N, 5 < x < 6}

Show SolutionHide Solution↓

There is no natural number lying between 5 and 6. So, F is an empty set.

The number of distinct elements in a finite set F is called its cardinal number, denoted by n(F).

Here the number of distinct elements = 0

Hence, F = { } and n(F) = 0.

1(vii)Exercise 7(B)

Write the following set in Roster form and write its cardinal number:

G = {x : x is a letter in the word 'ALLAHABAD'}.

Show SolutionHide Solution↓

The letters in the word 'ALLAHABAD' are A, L, L, A, H, A, B, A, D. Writing each letter only once, we get A, L, H, B, D.

The number of distinct elements in a finite set G is called its cardinal number, denoted by n(G).

Here the number of distinct elements = 5

Hence, G = {A, L, H, B, D} and n(G) = 5.

2Exercise 7(B)

Let :

A = {x : x is a letter in the word 'INCREASE'}

B = {x : x is a vowel in the word 'INCREASE'}

and C = {x : x is a consonant in the word 'INCREASE'}

(i) Write A, B and C in Roster form.

(ii) Find n(A), n(B) and n(C).

Show SolutionHide Solution↓

(i) The letters in the word 'INCREASE' are I, N, C, R, E, A, S, E. Writing each letter only once, we get I, N, C, R, E, A, S.

So, A = {I, N, C, R, E, A, S}.

The vowels among these letters are I, E, A.

So, B = {I, E, A}.

The consonants among these letters are N, C, R, S.

So, C = {N, C, R, S}.

Hence, A = {I, N, C, R, E, A, S}, B = {I, E, A} and C = {N, C, R, S}.

(ii) A has 7 elements, B has 3 elements and C has 4 elements.

Hence, n(A) = 7, n(B) = 3 and n(C) = 4.

3Exercise 7(B)

Which of the following sets are equal?

(i) A = {x : x ∈ N, 1 < x < 3} and B = {x : x is prime, x is even}.

(ii) C = {x : x ∈ W, x < 8} and D = {x : x ∈ N, x < 9}.

(iii) E = {x : x ∈ N, x is odd, x < 6} and F = {x : x ∈ N, x is a factor of 15, x < 15}.

(iv) G = {x : x is a prime factor of 12} and H = {x : x is prime, x < 4}.

(v) φ and {0}

Show SolutionHide Solution↓

(i) A = {x : x ∈ N, 1 < x < 3}. The only natural number lying between 1 and 3 is 2. So, A = {2}.

B = {x : x is prime, x is even}. The only even prime number is 2. So, B = {2}.

The elements of A and B are exactly the same.

Hence, A = B.

(ii) C = {x : x ∈ W, x < 8} = {0, 1, 2, 3, 4, 5, 6, 7}.

D = {x : x ∈ N, x < 9} = {1, 2, 3, 4, 5, 6, 7, 8}.

C contains 0 but not 8, while D contains 8 but not 0. So the elements are not the same.

Hence, C ≠ D.

(iii) E = {x : x ∈ N, x is odd, x < 6} = {1, 3, 5}.

F = {x : x ∈ N, x is a factor of 15, x < 15}. The factors of 15 are 1, 3, 5 and 15. Those less than 15 are 1, 3 and 5. So, F = {1, 3, 5}.

The elements of E and F are exactly the same.

Hence, E = F.

(iv) G = {x : x is a prime factor of 12}. Since 12 = 22 × 3, the prime factors of 12 are 2 and 3. So, G = {2, 3}.

H = {x : x is prime, x < 4}. The prime numbers less than 4 are 2 and 3. So, H = {2, 3}.

The elements of G and H are exactly the same.

Hence, G = H.

(v) φ is the empty set, containing no element at all. But {0} is a singleton set, containing one element, namely 0.

Hence, φ ≠ {0}.

4Exercise 7(B)

Which of the following are empty sets?

(i) A = {x : x ∈ N, x + 3 = 3}

(ii) B = {x : x ∈ W, x + 5 = 5}

(iii) C = {x : x ∈ N, x is even, x is divisible by 3} [Hint : C = {6, 12, 18, .......}].

(iv) D = {x : x is a prime number, x < 2}

(v) E = {x : x ∈ N, x < 1}

(vi) F = {x : x ∈ N, 2x = 1}

(vii) G = {0}

Show SolutionHide Solution↓

(i) Solving,

⇒ x + 3 = 3

⇒ x = 0.

But 0 is not a natural number, so no natural number satisfies the condition.

Hence, A is an empty set.

(ii) Solving,

⇒ x + 5 = 5

⇒ x = 0.

Since 0 is a whole number, B = {0}, which has one element.

Hence, B is not an empty set.

(iii) An even number that is divisible by 3 is a multiple of 6. So, C = {6, 12, 18, ......}, which has elements.

Hence, C is not an empty set.

(iv) The smallest prime number is 2, so there is no prime number less than 2.

Hence, D is an empty set.

(v) The smallest natural number is 1, so there is no natural number less than 1.

Hence, E is an empty set.

(vi) Solving,

⇒ 2x = 1

⇒ x = $\frac{1}{2}$.

But $\frac{1}{2}$ is not a natural number, so no natural number satisfies the condition.

Hence, F is an empty set.

(vii) G = {0} has one element, namely 0.

Hence, G is not an empty set.

Hence, the empty sets are A, D, E and F.

5Exercise 7(B)

Classify the following into finite and infinite sets:

(i) A = {x : x ∈ N, x < 100}

(ii) B = {x : x ∈ I, x < 100}

(iii) C = {x : x ∈ N, x > 100}

(iv) D = {x : x ∈ I, x > -1}

(v) E = {x : x ∈ N, x is a multiple of 3}

(vi) F = {x : x ∈ N, x is a factor of 1000}

(vii) G = {x : x is a point on an arc of a circle}

(viii) H = {x : x is a diameter of a circle}

Show SolutionHide Solution↓

(i) A = {x : x ∈ N, x < 100} = {1, 2, 3, ......, 99}.

The counting of its elements comes to an end.

Hence, A is a finite set.

(ii) B = {x : x ∈ I, x < 100} = {......, -3, -2, -1, 0, 1, ......, 99}.

The listing extends endlessly towards the negative side.

Hence, B is an infinite set.

(iii) C = {x : x ∈ N, x > 100} = {101, 102, 103, ......}.

The listing never comes to an end.

Hence, C is an infinite set.

(iv) D = {x : x ∈ I, x > -1} = {0, 1, 2, 3, ......}.

The listing never comes to an end.

Hence, D is an infinite set.

(v) E = {x : x ∈ N, x is a multiple of 3} = {3, 6, 9, 12, ......}.

The listing never comes to an end.

Hence, E is an infinite set.

(vi) A number has only a limited number of factors. So the factors of 1000 can be counted and the counting comes to an end.

Hence, F is a finite set.

(vii) An arc of a circle contains an unlimited (uncountable) number of points.

Hence, G is an infinite set.

(viii) A circle has an unlimited number of diameters.

Hence, H is an infinite set.

6Exercise 7(B)

Find whether the given statement is true or false:

(i) If P and Q are two sets such that n(P) = n(Q), then P = Q.

(ii) Two equivalent sets are always equal.

(iii) Two equal sets are always equivalent.

(iv) Let A = {x : x is a letter in the word 'ENGINEER'}. Then, n(A) = 5.

(v) Let B = {x : x is a consonant in the word 'MANAGEMENT'}. Then, n(B) = 4.

(vi) Let C = {x : x ∈ N, x is neither prime nor composite}. Then, n(C) = 1.

(vii) n(φ) = 0.

(viii) {x : x ∈ I, x is neither positive nor negative} = φ.

Show SolutionHide Solution↓

(i) Two sets with the same cardinal number are equivalent, but they need not be equal. For example, {1, 2} and {3, 4} have the same cardinal number but are not equal.

False

(ii) Equivalent sets have the same cardinal number, but their elements may be different. So they need not be equal.

False

(iii) Equal sets have exactly the same elements, so they must have the same cardinal number. Hence, they are always equivalent.

True

(iv) The letters in the word 'ENGINEER' are E, N, G, I, N, E, E, R. Writing each letter only once, we get E, N, G, I, R. So, A = {E, N, G, I, R} and n(A) = 5.

True

(v) The letters in the word 'MANAGEMENT' are M, A, N, A, G, E, M, E, N, T. The consonants among these are M, N, G, M, N, T. Writing each letter only once, we get M, N, G, T. So, B = {M, N, G, T} and n(B) = 4.

True

(vi) The only natural number that is neither prime nor composite is 1. So, C = {1} and n(C) = 1.

True

(vii) The empty set has no element at all, so its cardinal number is 0 i.e. n(φ) = 0.

True

(viii) The integer 0 is neither positive nor negative. So the set is {0}, which has one element and is not the empty set.

False

7Exercise 7(B)

Let A = {2, 4, 6, 8, 10} and B = {x : x ∈ W, x < 5}.

(i) Find n(A) and n(B).

(ii) Are A and B equivalent sets?

Show SolutionHide Solution↓

A = {2, 4, 6, 8, 10}

B = {x : x ∈ W, x < 5} = {0, 1, 2, 3, 4}

(i) A has 5 elements and B has 5 elements.

Hence, n(A) = 5 and n(B) = 5.

(ii) Since n(A) = n(B) = 5, the two sets have the same cardinal number.

Hence, A and B are equivalent sets.

11 Mark · Exercise 7(C) — Multiple Choice Questions

A set is a

Show SolutionHide Solution↓

A set is a well-defined collection of objects i.e. a collection for which it is possible to tell beyond doubt whether a given object belongs to it or not.

Hence, option 4 is the correct option.

21 Mark · Exercise 7(C) — Multiple Choice Questions

In Roster form, we have φ = ?

Show SolutionHide Solution↓

The empty set has no element at all. In roster form, it is written as { }.

Hence, option 1 is the correct option.

31 Mark · Exercise 7(C) — Multiple Choice Questions

n(φ) = ?

Show SolutionHide Solution↓

The empty set has no element, so its cardinal number is 0 i.e. n(φ) = 0.

Hence, option 3 is the correct option.

41 Mark · Exercise 7(C) — Multiple Choice Questions

A set having only one element is called

Show SolutionHide Solution↓

A set containing exactly one element is called a singleton set.

Hence, option 3 is the correct option.

51 Mark · Exercise 7(C) — Multiple Choice Questions

If A = {x : x is a letter in the word ENGAGEMENT}, then n(A) = ?

Show SolutionHide Solution↓

The letters in the word ENGAGEMENT are E, N, G, A, G, E, M, E, N, T. Writing each letter only once, we get E, N, G, A, M, T.

So, A = {E, N, G, A, M, T}, which has 6 elements i.e. n(A) = 6.

Hence, option 4 is the correct option.

61 Mark · Exercise 7(C) — Multiple Choice Questions

Which of the following is an empty set?

Show SolutionHide Solution↓

Checking all options:

  1. {x : x ∈ W, x + 2 = 2} ⇒ x = 0. Since 0 ∈ W, the set is {0} — not empty.

  2. {x : x is a prime number, x < 3} ⇒ the only prime number less than 3 is 2, so the set is {2} — not empty.

  3. {x : x ∈ I, x + 5 = 3} ⇒ x = -2. Since -2 ∈ I, the set is {-2} — not empty.

  4. {x : x ∈ N, x + 8 = 8} ⇒ x = 0. Since 0 ∉ N, no natural number satisfies this — an empty set.

Hence, option 4 is the correct option.

71 Mark · Exercise 7(C) — Multiple Choice Questions

Which of the following is a finite set?

Show SolutionHide Solution↓

Checking all options:

  1. {x : x ∈ I, x < 3} = {......, -1, 0, 1, 2} — infinite.

  2. {x : x ∈ N, x > 50} = {51, 52, 53, ......} — infinite.

  3. {x : x ∈ I, x > -10} = {-9, -8, ......, 0, 1, 2, ......} — infinite.

  4. {x : x ∈ N, 1 < x < 50} = {2, 3, 4, ......, 49} — finite, as the counting comes to an end.

Hence, option 4 is the correct option.

81 Mark · Exercise 7(C) — Multiple Choice Questions

Which of the following is an infinite set?

Show SolutionHide Solution↓

Checking all options:

  1. {x : x ∈ N, x < 500} = {1, 2, 3, ......, 499} — finite.

  2. {x : x ∈ N, x > 500} = {501, 502, 503, ......} — infinite, as the counting never comes to an end.

  3. {x : x ∈ N, 50 < x < 500} = {51, 52, ......, 499} — finite.

  4. {x : x ∈ I, -5000 < x < 5000} = {-4999, ......, 0, ......, 4999} — finite.

Hence, option 2 is the correct option.

91 Mark · Exercise 7(C) — Multiple Choice Questions

If A = {1, 2, 3} and B = {4, 5, 6}, then A and B are

Show SolutionHide Solution↓

A = {1, 2, 3} and B = {4, 5, 6}, so n(A) = 3 and n(B) = 3.

The two sets have the same cardinal number but different elements. So they are equivalent (not equal).

Hence, option 2 is the correct option.

101 Mark · Exercise 7(C) — Multiple Choice Questions

If A = {x : x ∈ N and x2 < 30}, then n(A) = ?

Show SolutionHide Solution↓

We take natural numbers x for which x2 < 30.

For x = 1, x2 = 1 < 30 ✓

For x = 2, x2 = 4 < 30 ✓

For x = 3, x2 = 9 < 30 ✓

For x = 4, x2 = 16 < 30 ✓

For x = 5, x2 = 25 < 30 ✓

For x = 6, x2 = 36 > 30 ✗

So, A = {1, 2, 3, 4, 5}, which has 5 elements i.e. n(A) = 5.

Hence, option 3 is the correct option.

1Exercise 7(C) — Mental Maths

Fill in the blanks:

(i) If x is not an element of A, we write ___________

(ii) If number of elements in A is 5, we write ___________

(iii) The members of a set are called its ___________

(iv) If A is a non-empty set, we write, ___________

(v) If A = φ, then n(A) = ___________

(vi) The collection of all persons on earth is a ___________ set.

Show SolutionHide Solution↓

(i) If x is not an element of A, we write x ∉ A.

(ii) If number of elements in A is 5, we write n(A) = 5.

(iii) The members of a set are called its elements (or members or points).

(iv) If A is a non-empty set, we write, A ≠ φ.

(v) If A = φ, then n(A) = 0.

(vi) The collection of all persons on earth is a finite set.

2Exercise 7(C) — Mental Maths

Write (T) for true and (F) for false for each of the following statements:

(i) If A = set of prime numbers, then 9 ∈ A. ___________

(ii) If A = set of even integers, then -8 ∈ A. ___________

(iii) If M = N, then n(M) = n(N). ___________

(iv) If n(A) = n(B), then A = B. ___________

(v) {1, 1, 2, 3} = {1, 2, 3}. ___________

(vi) Collection of tasty fruits is a set. ___________

Show SolutionHide Solution↓

(i) 9 = 3 × 3 is a composite number, not a prime number. So, 9 ∉ A.

(F)

(ii) -8 is an even integer, so it is a member of the set of even integers.

(T)

(iii) If M = N, both sets have exactly the same elements, so they have the same cardinal number.

(T)

(iv) Two sets with the same cardinal number are equivalent, but need not be equal.

(F)

(v) The repetition of elements in a set is meaningless, so {1, 1, 2, 3} = {1, 2, 3}.

(T)

(vi) The word 'tasty' is subjective, so the collection is not well defined and hence not a set.

(F)

1Exercise 7(C) — Case Study Based Questions

Consider the sets given below:

A = {x : x is an integer, $-\frac{1}{2}$ < x < $\frac{9}{2}$}

B = {x : x is a point in the interior of a circle}

C = {x : x is a perfect square, x < 0}

D = {x : x is a month of the year not having 31 days}

E = {x : x is prime, x < 3}

Look at the given sets and answer the questions given below:

Show SolutionHide Solution↓

1. Writing each set in roster form:

A = {x : x is an integer, $-\frac{1}{2} \lt \text{ x } \lt \frac{9}{2}$} = {0, 1, 2, 3, 4}, so n(A) = 5 (finite).

B = {x : x is a point in the interior of a circle} contains an unlimited number of points, so it is an infinite set.

C = {x : x is a perfect square, x < 0} = { }, since no perfect square is negative, so n(C) = 0 (finite).

E = {x : x is prime, x < 3} = {2}, so n(E) = 1 (finite).

Only B is an infinite set.

Hence, option (b) is the correct option.

2. As, n(A) = 5, n(C) = 0, n(E) = 1 and B is infinite.

The set with the minimum number of elements is C, as it is an empty set with n(C) = 0.

Hence, option (b) is the correct option.

3. E = {2} contains exactly one element, so it is a singleton set.

(A has 5 elements, B has infinitely many elements and C is the empty set.)

Hence, option (a) is the correct option.

4. The months of the year not having 31 days are February, April, June, September and November. So, D = {February, April, June, September, November} and n(D) = 5.

Since n(A) = 5, n(C) = 0, n(E) = 1 and B is infinite.

For a pair of equivalent sets, the two sets must have the same cardinal number.

Since n(D) = n(A) = 5, the sets D and A are equivalent.

Hence, option (b) is the correct option.

1Exercise 7(C) — Assertion-Reason Questions

Assertion (A) : The set of all prime numbers less than 30 is {2, 3, 5, 7, 11, 13, 17, 19, 23, 29}

Reason (R) : Every repeated element in a set is taken only once.

Show SolutionHide Solution↓

The prime numbers less than 30 are 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29. So, the given set is correct.

Thus, Assertion (A) is true.

Every repeated element in a set is indeed taken only once.

Thus, Reason (R) is true.

However, listing the distinct prime numbers less than 30 does not involve any repetition, so Reason (R) does not explain Assertion (A).

Hence, option 2 is the correct option.

2Exercise 7(C) — Assertion-Reason Questions

Assertion (A) : If A = {1, 2, 3, 4, 5, 6} and B = {a, b, c, d, e, f}, then A and B are equivalent sets.

Reason (R) : Two sets A and B are said to be equivalent if every element of A is in B and every element of B is in A.

Show SolutionHide Solution↓

A = {1, 2, 3, 4, 5, 6} and B = {a, b, c, d, e, f}, so n(A) = n(B) = 6. Since they have the same cardinal number, A and B are equivalent sets.

Thus, Assertion (A) is true.

The condition "every element of A is in B and every element of B is in A" is the definition of equal sets, not equivalent sets. Two sets are equivalent when n(A) = n(B).

Thus, Reason (R) is false.

Hence, option 3 is the correct option.

1Exercise 7(C) — Competency Focused Questions

Which of the following is a singleton set?

Show SolutionHide Solution↓

Checking all options:

  1. {x | x2 = 9, x ∈ Z} ⇒ x = 3 or x = -3, so the set is {-3, 3} — two elements.

  2. {x | x is a prime number and x is even} ⇒ the only even prime is 2, so the set is {2} — one element (singleton).

  3. {x | x = 7 - 7, x ∈ N} ⇒ x = 0, but 0 ∉ N, so the set is { } — empty.

  4. {x | x is a vowel in the word "prime"} ⇒ the vowels are i and e, so the set is {i, e} — two elements.

Hence, option 2 is the correct option.

2Exercise 7(C) — Competency Focused Questions

Which of the following sets is an empty set?

Show SolutionHide Solution↓

Checking all options:

  1. {x | x is a positive integer and x < 0} ⇒ no positive integer is less than 0, so the set is { } — empty.

  2. {x | x2 = 4, x ∈ Z} ⇒ x = 2 or x = -2, so the set is {-2, 2} — not empty.

  3. {x | x is a letter in the word "apple"} = {a, p, l, e} — not empty.

  4. {x | x is a natural number and x is odd} = {1, 3, 5, ......} — not empty.

Hence, option 1 is the correct option.

3Exercise 7(C) — Competency Focused Questions

If A = {x : x ∈ N and x is an odd prime number less than 17}, then the cardinal number of A is:

Show SolutionHide Solution↓

The prime numbers less than 17 are 2, 3, 5, 7, 11 and 13. Out of these, the odd prime numbers are 3, 5, 7, 11 and 13.

So, A = {3, 5, 7, 11, 13}, which has 5 elements i.e. n(A) = 5.

Hence, option 3 is the correct option.

4Exercise 7(C) — Competency Focused Questions

Two sets are defined as

A = {x | x is an even prime numbers less than 100}

B = {x | x is an odd composite number less than 100}

Which of the following numbers is common in both?

Show SolutionHide Solution↓

A = {x | x is an even prime numbers less than 100}. The only even prime number is 2, so A = {2}.

B = {x | x is an odd composite number less than 100} = {9, 15, 21, 25, 27, ......}, which contains only odd numbers.

Since A contains only the even number 2 and B contains only odd numbers, there is no element common to both. (Also, 29, 53 and 97 are all prime numbers, so none of them belongs to A or B.)

Hence, option 4 is the correct option.

5Exercise 7(C) — Competency Focused Questions

Two sets are defined as A = {1, 2, 3} and B = {3, 2, 1}. What can be concluded?

Show SolutionHide Solution↓

The order in which the elements of a set are listed is immaterial. Both A and B contain exactly the same elements 1, 2 and 3.

So, A = B, because they contain the same elements.

Hence, option 2 is the correct option.

6Exercise 7(C) — Competency Focused Questions

Which of the following sets is finite?

Show SolutionHide Solution↓

Checking all options:

  1. {x | x > 0, x ∈ N} = {1, 2, 3, ......} — infinite.

  2. {x | x2 < 100, x ∈ N}. Here x2 < 100 ⇒ x < 10, so x = 1, 2, 3, ......, 9. The set is {1, 2, 3, 4, 5, 6, 7, 8, 9} — finite.

  3. {x | x is a prime number} = {2, 3, 5, 7, 11, ......} — infinite.

  4. {x | x ≥ 0, x ∈ Z} = {0, 1, 2, 3, ......} — infinite.

Hence, option 2 is the correct option.

Press ↓ to reveal a solution and move to the next question. Back to Solutions for Class 6 RS Aggarwal ICSE Foundation Mathematics.