SSC Higher Math Exercise 1.1 Solution Part 1

SSC higher math exercise 1.1 solution

1. i. A set of 2n elements has in all, 4ⁿ subsets. ii.Q= \[\]\left\{\frac pq:p,q ∈ ∧,q\neq0\right\} is the set of rational numbers.

iii. a, b ∈ ∇; ] a, b[ = {x : x ∈ ∇ and a < x < b}

Which combination of these statements is correct?

a.i and ii

b.ii and iii

c.i and iii

√d.i, ii and iii

Explanation: (i) is correct. Because, we know, if the number of elements of any set is n then the number of subset of that set is 2n.

 If the number of members of any set is 2n then the number of its subsets will be 2²ⁿ = (2²)ⁿ = 4ⁿ

Answer the questions (2-4) following the information below:

For every n ∈ Ι, A_n = {n, 2n, 3n, ……… }

2. Which one of the following is the value of A1 ∩ A2?

a. A₁

b. A₂

c. A₃

d. A₄

Explanation:A₁ = {1, 2, 3, 4….}, A₂ = {2, 4, 6,…….}

∴ A₁ ∩ A₂ = {2, 4, 6,…..} = A₂

3.Which one of the following denotes the value of A₃ ∩ A₆?

a. A₂

b. A₃

c. A₄

d. A₆

Explanation: A₃= {3, 6, 9, 12,…….}

A₆ = {6, 12, 18,……}

∴ A₃ ∩ A₆ = {6, 12,…….} = A₆

4. Which one of the following is to be written down instead of A₂ ∩ A₃?a. A₃

b. A₄

c. A₅

d. A₆

Explanation: Because, A₂ = {2, 4, 6, 8, 10, 12,…}

A₃ = {3, 6, 9, 12,……}

∴ A₂ ∩ A₃ = {6, 12, ……} = A₆

Class 9 and 10 general math exercise 2.1 solution

5. Given that, U = {x: 3 ≤ x ≤ 20, x ∈ ∧},

A = {x : x is an odd number} and B = {x : x is a prime number} list the elements of the following sets:

(i) A and (ii) B

(iii) Let C = {x: x ∈ A and x ∈ B} and

(iv) D = {x : x ∈ A or x ∈ B}.

Solution: Given, U = {x : 3 ≤ x ≤20, n ∈ ∧}

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

(i) A = {x : x is an odd number}

∴ A = {3, 5, 7, 9, 11, 13, 15, 17, 19}

(ii) B = {x : x is a prime number}

= {3, 5, 7, 11, 13, 17, 19}

(iii) C = {x : x ∈ A and x ∈ B}

= {x : x ∈ A ∈ B}

Now, A ∩ B = {3, 5, 7, 9, 11, 13, 15, 17, 19} ∩ {3, 5, 7, 11, 13, 17, 19}

= {3, 5, 7, 11, 13, 17, 19}

∴ C = {3, 5, 7, 11, 13, 17, 19}

(iv) D = {x : x ∈ A or x ∈ B} = {x : x ∈ A ∪ B}

Now, A ∪ B = {3, 5, 7, 9, 11, 13, 15, 17, 19} ∪ {3, 5, 7, 11, 13, 17, 19}

= {3, 5, 7, 9, 11, 13, 15, 17, 19}

∴ D = {3, 5, 7, 9, 11, 13, 15, 17, 19}

6. The elements of the sets A and B have been shown in the Venn diagram. If n(A) = n(B), find

(a) the value of x.

(b) n(A ∪ B) and n(A ∩ B′).

Solution:

(a) From the Venn diagram, we get, n(A) = 3x + x n(B) = x + 2x + 8

According to the question, n(A) = n(B)

or,3x + x = x + 2x + 8

or,4x – 3x = 8

∴ x = 8 (Ans.)

(b) From the Venn diagram, we get,

n(A ∪ B) = 3x + x + 2x + 8

= 6x + 8

= 6 × 8 + 8 [since, x = 8]

= 48 + 8 = 56

n(A ∪ B) = 56 (Ans.)

and n(A ∩ B′) = 3x

= 3 × 8 [since, x = 8]

= 24 (Ans.)

7. If U = {x: x is a positive integer},

A = {x : x ≥ 5}⊂U and B = {x : x < 12}⊂U; find the value of n(A ∩ B) and n(A′).

Solution: Given,

U = {x : x is a positive integer}

= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, …..}

A = {x : x ≥ 5} = {5, 6, 7, 8, 9, 10, 11, 12, 13, ….}

B = {x : x < 12} = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}

∴ A ∩ B = {5, 6, 7, 8, 9, 10, 11, 12, 13……} ∩

{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}

= {5, 6, 7, 8, 9, 10, 11}

and A ′= U – A

= {1, 2, 3, 4, 5, 6, ….} – {5, 6, 7, 8, 9,…..}

= {1, 2, 3, 4}

∴n(A ∩ B) = 7 and n(A′) = 4

Ans. 7, 4

8. Let U = {x : x is an even integer},

A = {x : 3x ≥ 25}⊂U and B = {x : 5x < 12}⊂U; find n(A ∩ B) and n(A′ ∪ B′).

Solution: U = {x : x is an even integer}

= {… – 4, –2, 0, 2, 4, 6, 8, 10, 12, 14, …….}

A = {x : 3x ≥ 25} = {10, 12, 14, ……..}

B = {x : 5x < 12} = {…, – 4, – 2, 0, 2}

Here, A ∩ B = {10, 12, ……..} ∩ {…, – 4, – 2, 0, 2 = ∅

∴n(A ∩ B) = 0

Again, A ′ = U – A

= {…, – 4 – 2, 0, 2, 4, 6, 8, 10, 12, …..} – {10, 12, 14, …..}

= {…, – 4, – 2, 0, 2, 4, 6, 8}

B′ = U – B

= {…, –4, – 2, 0, 2, 4, 6, 8, ……} – {…, – 4, – 2, 0, 2}

= {4, 6, 8, 10, …….}

∴A′ ∩ B′ = {…, – 4, – 2, 0, 2, 4, 6, 8} ∩ {4, 6, 8, 10, …}

= {4, 6, 8}

∴n(A′ ∩ B′) = 3

Ans. n(A ∩ B) = 0 and n(A′ ∩ B′) = 3

9. Show that, (a) A \ A = ∅ ; (b) A \ (A \ A) = A

Solution: (a) Let, x ∈A\A

So, x ∈ A and x ∉ A

⇒ x ∈ (A ∩ A′)

⇒ x ∈ ∅

∴ A \ A ⊂ ∅

Again, ∅ ⊂ A\A

So A \A = ∅ (Shown)

(b) Let, x ∈ A \ (A \ A)

So, x ∈ A and x ∉ A \ A

⇒ x ∈ A and x ∉ ∅ [since, A \ A = ∅]

⇒ x ∈ A

∴ A \ (A \ A) ⊂ A

Again Let, x ∈ A

So, x ∈ A and x ∉ ∅

⇒ x ∈ A and x ∉ (A \ A)

⇒ x ∈ A \ (A \ A)

∴ A ⊂ A \ (A \ A)

So, A \ (A \ A) = A (Shown)

10. Show that, A × (B ∪ C) = (A × B) ∪ (A × C)

Solution: According to the definition, A × (B ∪ C)

= {(x, y) : x ∈ A, y ∈(B ∪ C)}

= {(x, y) : x ∈ A, ( y ∈ B or y ∈ C)}

= {(x, y) : (x ∈ A, y ∈ B) or (x ∈ A, y ∈ C)}

= { (x, y) : (x, y) ∈ (A × B) or (x, y) ∈ (A × C)}

= {(x, y) : (x, y) ∈(A × B) ∪ (A × C)}

= (A × B) ∪ (A × C)

∴ A × (B ∪ C) ⊂ (A × B) ∪ (A × C)

Again, (A × B) ∪ (A × C)

= { (x, y) : (x, y) ∈ (A × B) or (x, y) ∈ (A × C)}

= {(x, y) : (x ∈ A, y ∈ B) or (x ∈ A, y ∈ C)}

= {(x, y) : x ∈ A, ( y ∈ B or y ∈ C)}

= {(x, y) : x ∈ A, y ∈(B ∪ C)}

= {(x, y) : (x ,y)∈ A × (B ∪ C)}

∴ (A × B) ∪ (A × C) ⊂ A × (B ∪ C)

So (A × B) ∪ (A × C) ⊂ A × (B ∪ C) (Shown)

11. If A ⊂ B and C ⊂ D, show that, (A × C) ⊂ (B × D)

Solution: Let, (x, y) ∈ (A × C)

So, x ∈ A, y∈ C

⇒ x ∈ B, y ∈ D [Since, A ⊂ B and C ⊂ D]

⇒ (x, y) ∈ (B × D)

∴ (A × C) ⊂ (B × D) (Shown)

12. Show that the sets A = {1, 2, 3, ……., n} and

B = {1, 2, 22,………,2^{n – 1} } are equivalent.

Solution: Given, A = {1, 2, 3,……..,n}

and B = {1, 2, 22, ………,2^{n – 1}}

We exhibit a one-one correspondence between the sets A and B in the picture below:

So the sets are equivalent. (Shown)

[N.B. the correspondence between the sets can be described as A ↔ B: k ↔ 2^k–1, k ∈ A]

13. Show that, the set S = {1,4,9,25,36,…………} of square of natural numbers, is an infinite set.

Solution: Given, S = { 1, 4, 9, 16, 25, 36, ………….}

= {1², 2², 3², 4², 5², 6² …… n² ……}

Set of all natural numbers, N= {1, 2, 3 …… n ……}

Now we exhibit an one-one correspondence between the sets ô and S in the picture below:

So N and S are equivalent. Since the set of natural numbers N is an infinite set.

So we can say that, S is an infinite set.(Shown)

14. Suppose, A ∩ B = Φ and n(A) = p, n(B) = q; prove that, n(A ∪ B) = P + Q

Solution: n(A) = p, n(B) = q and A ∪ B = Φ

It is required to prove that, n(A ∪ B) = p + q

Given, A ∩ B = Φ

∴ n(A ∩ B) = 0

We know that, for any finite sets A and B,

n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

= p + q

∴ n(A ∪ B) = p + q (Proved)

15. For finite sets A, B, C, prove that,

n (A ∪ B ∪ C) = n (A) + n (B) + n (C) – n (A ∩ B)

– n(B ∩ C) – n (C ∩ A) + n (A ∩ B ∩ C).

Solution: We know that, for any finite sets A and B,

n (A∪ B) = n (A) + n(B) – n (A ∩B)

Now, n(A ∪ B ∪ C) = n[A ∪ (B ∪ C)]

[ since, A ∪ B ∪ C = A ∪ (B ∪ C), commutative law ]

= n (A) + n (B ∪ C) – n [A ∩ (B ∩ C)]

= n (A) + n (B) + n (C) – n (B ∩ C) – n [(A ∩ B) ∪ (A ∩ C)]

[since A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)]

= n (A) + n (B) + n (C) – n (B ∩ C) – n (A ∩ B) – n (A ∩ C)

+ n [(A ∩B) ∩ (A ∩ C)]

= n (A) + n (B) + n (C) – n (B ∩ C) – n (A ∩ B) – n (A ∩ C) + n (A ∩ B ∩ C) [ since (A ∩ B) ∩ (A ∩ C) = A ∩ B ∩ C ]

= n (A) + n (B) + n (C) –n (A ∩ B) – n (B ∩ C) – n (C ∩ A) + n (A ∩ B ∩ C) [since A∩C = C∩A]

∴ n (A  B  C) = n(A) + n (B) + n(C) – n(A ∩ B) – n(B ∩ C)– n(C ∩ A) + n(A ∩ B ∩ C) (Proved)

16. If A = {a, b, x} and B = {c, y} are the subsets of the universal set U = {a, b, c, x, y, z}, verify that,

(a) (i) A ⊂ B′, (ii) A ∪ B′ = B′, (iii) A′ ∩ B = B

(b) Find: (A ∩ B) ∪ (A ∩ B′)

Solution: Given, U = {a, b, c, x, y, z},

A = {a, b, x} and B = {c, y}

(a) (i) B′ = U – B

= {a, b, c, x, y, z} – {c, y}

= {a, b, x, z}

∴A ⊂ B′ (Verified)

(ii)B′ = U – B = {a, b, c, x, y, z} – {c, y} = {a, b, x, z}

and A ∪ B′ = {a, b, x} ∪ {a, b, x, z}

= {a, b, x, z}

= B′

∴A ∪ B′ = B′ (Verified)

(iii)A′ = U – A = {a, b, c, x, y, z} – {a, b, x}

= {c, y, z}

Therefore, A′ ∩ B = {c, y, z} ∩ {c, y}

= {c, y} = B

∴A′ ∩ B = B (Verified)

(b) A ∩ B = {a, b, x} ∩ {c, y}

= ∅

and B′ = U – B

= {a, b, c, x, y, z} – {c, y}

= {a, b, x, z}

∴A ∩ B′ = {a, b, x} ∩ {a, b, x, z}

= {a, b, x}

So (A ∩ B) ∪ (A ∩ B′) = ∅{a, b, x} = {a, b, x}

Ans. {a, b, x}

17. Out of 30 students of a class, 19 have taken Economics, 17 have taken Geography, 11 have taken Civics, 12 have taken Economics and Geography, 4 have taken Civics and Geography, 7 have taken Economics and Civics, while 4 have taken all three subjects. How many students have taken none of the three subjects?

Solution: Out of 30 students of a class, 19 have taken Economics, 17 have taken Geography, 11 have taken Civics, 12 have taken Economics and Geography, 4 have taken Civics and Geography, 7 have taken Economics and Civics, while 4 have taken all three subjects.

Let, E = set of the students who have taken Economics

G = set of the students who have taken Geography

C = set of the students taken Civics

U = set of all students

Therefore, n(E) = 19

n(G) = 17

n(C) = 11

n(E ∩ G) = 12

n(E ∩ C) = 7

n(G ∩ C) = 4

n(E ∩ G ∩ C) = 4.

We know that,

n(E ∪ G ∪ C) = n(E) + n(G) + n(C) – n(E ∩ G) – n(E ∩ C) – n(G ∩ C) + n(E ∩ G ∩ C)

= 19 + 17 + 11 – 12 – 7 – 4 + 4 = 28

So, number of students who have taken at least one subject = 28

The number of students who have taken none of the three subjects = n(U) – n(E ∪ G ∪ C)

= (30 – 28) = 2 (Ans.)

18. The elements of the universal set U and the subsets A, B, C have been presented in the Venn diagram.

(a) If n(A ∩ B) = n(B ∩ C), find the value of x.

(b) If n(B ∩ C′) = n(A′ ∩ C) find the value of y.

Solution:

(a) Given, n(A ∩ B) = n(B ∩ C)

or, x = 4 [From the Venn diagram]

∴ x = 4.

(b) Given, n (B ∩ C′) = n (A′ ∩ C)

or, x + 6 = 4 + y [From the Venn diagram]

or, 4 + 6 = 4 + y [since, x = 4]

or, 6 = y

∴ y = 6.

= 8 + 4 + 6 + 4 + 6 [since, x = 4, y = 6]

= 28

∴ n(U) = 28 (Ans.)

19. The elements of the sets A, B, C have been given in the Venn diagram, U = A ∪ B ∪ C

(a) Find the value of x. If n(U) = 50

(b) Find the values of n(B ∩ C′) and n(A′ ∩ B).

Solution:

(a) Here,

n(U) = 2x + x + 1 + x – 1 + 2 + 3 + 0 + x + 5

[From the Venn diagram]

∴ n (U) = 5x + 10

Given, n (U) = 50

or, 5x + 10 = 50

or, 5x = 40

∴ x = 8 (Ans.)

[N.B.: Answer in the textbook is wrong.]

(b) From the Venn diagram,

∴ n(B ∩ C′) = x + 1 + x – 1

= 2x

= 2.8 [since, x = 8]

= 16

∴ n (A′ ∩ B) = x – 1 + 0

= 8 – 1

= 7 (Ans.)

[N.B.: Answer in the textbook is wrong.]

∴ n (A ∩ B ∩ C′) = x + 1

= 8 + 1= 9

∴ n(A ∩ B ∩ C′) = 9 (Ans.)

[N.B.: Answer in the textbook is wrong.]

20. The three sets A, B,C have been given such that, A∩ B = Φ, A∩C = Φ and C⊆ B. Explain the sets by drawing Venn diagram.

Solution: According to the given information, the sets are shown in the Venn diagram:

A ∩B = Φ

Explanation: There are no common elements in the set A and the set B.

That is, A and B are non-intersecting sets.

A ∩ C = Φ

Explanation: There are no common elements in the set A and the set C.

That is, A and C are non-intersecting sets.

C ⊆ B

Explanation: There are common elements in the set C and the set B.

All the elements of the set C belong to the set B.

SSC higher math exercise 1.1 solution part 2

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SSC higher math exercise 1.1 solution part 1