If ξ = x ∣ x ∈ N , x ≤ 12 , A = x : x ≥ 7 . Then find A ' .

Given, ξ = x ∣ x ∈ N , x ≤ 12 Here, x belongs to natural number and value of x is less than equal to 12 . So, ξ = 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 . Also, we have A = x : x ≥ 7 Here, value of x is greater than equal to 7 . So, A = 7 , 8 , 9 , 10 , 11 , 12 A ' represents the numbers that are not present in A . A ' = ξ - A = 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 - 7 , 8 , 9 , 10 , 11 , 12 = 1 , 2 , 3 , 4 , 5 , 6 Hence, A ' = 1 , 2 , 3 , 4 , 5 , 6 .

If ξ = x ∣ x ∈ N , x ≤ 12 , B = x : 4 < x < 10 . Then find B ' .

Given, ξ = x ∣ x ∈ N , x ≤ 12 Here, x belongs to natural number and value of x is less than equal to 12 . So, ξ = 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 . Also, we have B = x : 4 < x < 10 Here, value of x is greater than 4 and less than 10 . So, B = 5 , 6 , 7 , 8 , 9 B ' represents the numbers that are not present in B . B ' = ξ - B = 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 - 5 , 6 , 7 , 8 , 9 = 1 , 2 , 3 , 4 , 10 , 11 , 12 Hence, B ' = 1 , 2 , 3 , 4 , 10 , 11 , 12 .

If ξ = x ∣ x ∈ N , x ≤ 12 , A = x : x ≥ 7 and B = x : 4 < x < 10 . Then find A ∪ B .

Given, ξ = x ∣ x ∈ N , x ≤ 12 Here, x belongs to natural number and value of x is less than equal to 12 . So, ξ = 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 . Also, we have A = x : x ≥ 7 Here, value of x is greater than equal to 7 . So, A = 7 , 8 , 9 , 10 , 11 , 12 Also, B = x : 4 < x < 10 So, B = 5 , 6 , 7 , 8 , 9 Now, A ∪ B represent all the elements that are present in both sets. A ∪ B = 7 , 8 , 9 , 10 , 11 , 12 ∪ 5 , 6 , 7 , 8 , 9 = 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 Hence, A ∪ B = 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 .

If ξ = x ∣ x ∈ N , x ≤ 12 , A = x : x ≥ 7 and B = x : 4 < x < 10 . Then find A ∩ B .

Given, ξ = x ∣ x ∈ N , x ≤ 12 Here, x belongs to natural number and value of x is less than equal to 12 . So, ξ = 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 . Also, we have A = x : x ≥ 7 Here, value of x is greater than equal to 7 . So, A = 7 , 8 , 9 , 10 , 11 , 12 Also, B = x : 4 < x < 10 So, B = 5 , 6 , 7 , 8 , 9 Now, A ∩ B represent all the common elements that are present in both sets. A ∩ B = 7 , 8 , 9 , 10 , 11 , 12 ∩ 5 , 6 , 7 , 8 , 9 = 7 , 8 , 9 Hence, A ∩ B = 7 , 8 , 9 .

If ξ = x ∣ x ∈ N , x ≤ 12 , A = x : x ≥ 7 and B = x : 4 < x < 10 . Then find A - B .

Given, ξ = x ∣ x ∈ N , x ≤ 12 Here, x belongs to natural number and value of x is less than equal to 12 . So, ξ = 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 . Also, we have A = x : x ≥ 7 Here, value of x is greater than equal to 7 . So, A = 7 , 8 , 9 , 10 , 11 , 12 Also, B = x : 4 < x < 10 So, B = 5 , 6 , 7 , 8 , 9 Now, for A - B , we need to subtract the elements of B from the elements of A , then we need to write the rest of the elements of A . Thus, A - B = 7 , 8 , 9 , 10 , 11 , 12 - 5 , 6 , 7 , 8 , 9 = 10 , 11 , 12 Hence, A - B is 10 , 11 , 12 .

If ξ = x ∣ x ∈ N , x ≤ 12 , A = x : x ≥ 7 and B = x : 4 < x < 10 . Then find B - A .

Given, ξ = x ∣ x ∈ N , x ≤ 12 Here, x belongs to natural number and value of x is less than equal to 12 . So, ξ = 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 . Also, we have A = x : x ≥ 7 Here, value of x is greater than equal to 7 . So, A = 7 , 8 , 9 , 10 , 11 , 12 Also, B = x : 4 < x < 10 So, B = 5 , 6 , 7 , 8 , 9 Now, for B - A , we need to subtract the elements of A from the elements of B , then we need to write the rest of the elements of B . Thus, B - A = 5 , 6 , 7 , 8 , 9 - 7 , 8 , 9 , 10 , 11 , 12 = 5 , 6 Hence, B - A is 5 , 6 .

If ξ = x ∣ x ∈ N , x ≤ 12 , A = x : x ≥ 7 and B = x : 4 < x < 10 , then find A ∪ B ' .

Given, ξ = x ∣ x ∈ N , x ≤ 12 Here, x belongs to natural number and value of x is less than equal to 12 . So, ξ = 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 . Also, we have A = x : x ≥ 7 Here, value of x is greater than equal to 7 . So, A = 7 , 8 , 9 , 10 , 11 , 12 Also, B = x : 4 < x < 10 So, B = 5 , 6 , 7 , 8 , 9 Now, A ∪ B represent all the elements that are present in both sets. A ∪ B = 7 , 8 , 9 , 10 , 11 , 12 ∪ 5 , 6 , 7 , 8 , 9 = 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 Also, A ∪ B ' represents the numbers that are not present in A ∪ B . Thus, A ∪ B ' = ξ - A ∪ B = 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 - 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 = 1 , 2 , 3 , 4 Hence, A ∪ B ' is 1 , 2 , 3 , 4 .

If ξ = x ∣ x ∈ N , x ≤ 12 , A = x : x ≥ 7 and B = x : 4 < x < 10 . then find A ' ∩ B '

Given, ξ = x ∣ x ∈ N , x ≤ 12 Here, x belongs to natural number and value of x is less than equal to 12 . So, ξ = 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 . Also, we have A = x : x ≥ 7 Here, value of x is greater than equal to 7 . So, A = 7 , 8 , 9 , 10 , 11 , 12 Also, B = x : 4 < x < 10 So, B = 5 , 6 , 7 , 8 , 9 Now, A ' represents the numbers that are not present in A . Thus, A ' = ξ - A = 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 - 7 , 8 , 9 , 10 , 11 , 12 = 1 , 2 , 3 , 4 , 5 , 6 B ' = ξ - B = 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 - 5 , 6 , 7 , 8 , 9 = 1 , 2 , 3 , 4 , 10 , 11 , 12 Now, A ' ∩ B ' represent all the common elements that are present in both sets. So, A ∪ B ' = 1 , 2 , 3 , 4 , 5 , 6 ∩ 1 , 2 , 3 , 4 , 10 , 11 , 12 = 1 , 2 , 3 , 4 Hence, A ' ∩ B ' = 1 , 2 , 3 , 4 .

E M B I B E

ICSE Understanding Mathematics Class 8>Chapter 6 - Operations on Sets; Venn Diagrams>Check Your Progress

M L Aggarwal Solutions for Chapter: Operations on Sets; Venn Diagrams, Exercise 3: Check Your Progress

Author:M L Aggarwal

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Chapter: Operations on Sets; Venn Diagrams, Exercise 3: Check Your Progress

Attempt the practice questions on Chapter 6: Operations on Sets; Venn Diagrams, Exercise 3: Check Your Progress with hints and solutions to strengthen your understanding. ICSE Understanding Mathematics Class 8 solutions are prepared by Experienced Embibe Experts.