Operations on Sets — Multiple Choice Questions

If A = {x | x is a colour of rainbow} and B = {white, red, green}, then A ∩ B is

1. B

2. {green}

3. {red}

4. {green, red}

Answer

A = {x | x is a colour of rainbow}

A = {violet, indigo, blue, green, yellow, orange, red}

B = {white, red, green}

A ∩ B - contains all the common elements in set A and B.

A ∩ B = {violet, indigo, blue, green, yellow, orange, red} ∩ {white, red, green}

A ∩ B = {green, red}

Hence, option 4 is the correct option.

If P = {-1, 0, 1, 2, 5} and Q = {3, 5, 7}, then P ∪ Q is

1. {5}

2. {-1, 0, 1, 2, 3, 7}

3. {-1, 0, 1, 2, 3, 5, 7}

4. none of these

Answer

P ∪ Q - contains all the elements in set P and Q.

P ∪ Q = {-1, 0, 1, 2, 5} ∪ {3, 5, 7}

P ∪ Q = {-1, 0, 1, 2, 3, 5, 7}

Hence, option 3 is the correct option.

If A and B are two sets, then A - B is defined as

1. {x | x ∈ A or x ∈ B}

2. {x | x ∈ A and x ∈ B}

3. {x | x ∈ A and x ∉ B}

4. {x | x ∈ B and x ∉ A}

Answer

A - B is the set consisting of all those elements which belong to A but do not belong to B.

∴ A - B = {x | x ∈ A and x ∉ B}

Hence, option 3 is the correct option.

If A is any set, then A ∩ φ is

1. A

2. φ

3. ξ

4. none of these

Answer

As the empty set φ has no elements, there can be no element common to A and φ.

∴ A ∩ φ = φ

Hence, option 2 is the correct option.

A ∩ ξ is same as

1. A

2. φ

3. A′

4. ξ

Answer

As every element of A belongs to the universal set ξ, the elements common to A and ξ are all the elements of A.

∴ A ∩ ξ = A

Hence, option 1 is the correct option.

If ξ = W and A = {x | x ∈ W and x ≤ 10}, then A′ is

1. φ

2. {x | x ∈ W and 0 ≤ x ≤ 10}

3. {x | x ∈ W and x ≥ 10}

4. {x | x ∈ W and x ≥ 11}

Answer

ξ = W = {0, 1, 2, 3, ...}

A = {x | x ∈ W and x ≤ 10}

A = {0, 1, 2, ..., 10}

A′ - contains all the elements in universal set which are not in A.

A′ = ξ - A

A′ = {0, 1, 2, 3, ...} - {0, 1, 2, ..., 10}

A′ = {11, 12, 13, ...}

∴ A′ = {x | x ∈ W and x ≥ 11}

Hence, option 4 is the correct option.

If ξ = {x | x ∈ N, x ≤ 12}, A = {x | x is a multiple of 3} and B = {x | x is a multiple of 4}, then A ∩ B is

1. φ

2. {0}

3. {12}

4. {0, 12}

Answer

ξ = {x | x ∈ N, x ≤ 12}

ξ = {1, 2, 3, ..., 12}

A = {x | x is a multiple of 3}

A = {3, 6, 9, 12}

B = {x | x is a multiple of 4}

B = {4, 8, 12}

A ∩ B - contains all the common elements in set A and B.

A ∩ B = {3, 6, 9, 12} ∩ {4, 8, 12}

A ∩ B = {12}

Hence, option 3 is the correct option.

If ξ = {all digits in our number system}, A = {x | x is prime} and B = {x | x is even}, then B - A is

1. {4, 6, 8}

2. {0, 4, 6, 8}

3. {3, 5, 7}

4. {2}

Answer

ξ = {all digits in our number system}

ξ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

A = {x | x is prime}

A = {2, 3, 5, 7}

B = {x | x is even}

B = {0, 2, 4, 6, 8}

B - A - contains all the elements in set B but not in A.

B - A = {0, 2, 4, 6, 8} - {2, 3, 5, 7}

B - A = {0, 4, 6, 8}

Hence, option 2 is the correct option.

If A and B are two sets such that n(A) = 22, n(B) = 18 and n(A ∪ B) = 35, then n(A ∩ B)

1. 4

2. 5

3. 15

4. 75

Answer

Given:

n(A) = 22

n(B) = 18

n(A ∪ B) = 35

∵ n(A ∪ B) = n(A) + n(B) - n(A ∩ B)

Putting the values, we get

⇒ 35 = 22 + 18 - n(A ∩ B)

⇒ n(A ∩ B) = 40 - 35

⇒ n(A ∩ B) = 5

Hence, option 2 is the correct option.