ML Aggarwal Class 8 Maths Solutions Chapter 06 Operation on Sets and Venn Diagrams

Exercise 6.1

Question 1. If A = {0, 1, 2, 3, ..., 8}, B = {3, 5, 7, 9, 11} and C = {0, 5, 10, 20}, find (i) A ∪ B (ii) A ∪ C (iii) B ∪ C (iv) A ∩ B (v) A ∩ C (vi) B ∩ C. Also find the cardinal number of the sets B ∪ C, A ∪ B, A ∩ C and B ∩ C.
Answer: From the given information:
A = {0, 1, 2, 3, 4, 5, 6, 7, 8}
B = {3, 5, 7, 9, 11}
C = {0, 5, 10, 20}

(i) A ∪ B = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11}; n(A ∪ B) = 11

(ii) A ∪ C = {0, 1, 2, 3, 4, 5, 6, 7, 8, 10, 20}; n(A ∪ C) = 11

(iii) B ∪ C = {0, 3, 5, 7, 9, 10, 11, 20}; n(B ∪ C) = 8

(iv) A ∩ B = {3, 5, 7}; n(A ∩ B) = 3

(v) A ∩ C = {0, 5}; n(A ∩ C) = 2

(vi) B ∩ C = {5}; n(B ∩ C) = 1
In simple words: When two sets join together with union, you get all items from both. When two sets meet with intersection, you get only the items that appear in both sets.

Exam Tip: Always list each element only once in a union, even if it shows up in multiple sets. For intersection, write down only those elements present in every set being combined.

Question 2. Find A' when (i) A = {0, 1, 4, 7} and E = {x | x ∈ W, x ≤ 10}
Answer: Given:
A = {0, 1, 4, 7}
E = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

A' = {2, 3, 5, 6, 8, 9, 10}
In simple words: The complement A' contains all items in the universal set that are not in set A.

Exam Tip: Always identify the universal set first before finding the complement - the complement depends entirely on what the universal set includes.

Question 2(ii). A = {consonants} and ξ = {alphabets of English}
Answer: A = {consonants}
ξ = {alphabets of English}

A' = {Vowels}
In simple words: If set A holds all consonants, then the complement A' must hold all the vowels.

Exam Tip: When dealing with alphabets, remember that the English alphabet splits naturally into consonants and vowels - together they make up the universal set.

Question 2(iii). A = {boys in class VIII of all schools in Bengaluru} and ξ = {students in class VIII of all schools in Bengaluru}
Answer: A' = {Girls in class VIII of all schools in Bengaluru}
In simple words: If A contains all boys, then A' must contain all girls in the same setting.

Exam Tip: With real-world or descriptive sets, the complement simply represents the "opposite" group within the same universal set.

Question 2(iv). A = {letters of KALKA} and ξ = {letters of KOLKATA}
Answer: Given:
A = {K, A, L}
ξ = {K, O, L, A, T}

A' = {O, T}
In simple words: The complement has the letters that show up in KOLKATA but not in KALKA.

Exam Tip: When listing letters from words, count each letter only once (don't repeat). The complement will have only those letters from the universal set missing from set A.

Question 2(v). A = {odd natural numbers} and ξ = {whole numbers}
Answer: Given:
A = {odd natural numbers}
ξ = {whole numbers}

A' = {0, even whole numbers}
In simple words: The complement of odd natural numbers within whole numbers includes zero and all even whole numbers.

Exam Tip: Remember that whole numbers start from 0, while natural numbers start from 1. This difference matters when finding complements.

Question 3. If A = {x : x ∈ N and 3 < x < 7} and B = {x : x ∈ W and x ≤ 4}, find (i) A ∪ B (ii) A ∩ B (iii) A - B (iv) B - A
Answer: From the given information:
A = {4, 5, 6}
B = {0, 1, 2, 3, 4}

(i) A ∪ B = {0, 1, 2, 3, 4, 5, 6}

(ii) A ∩ B = {4}

(iii) A - B = {5, 6}

(iv) B - A = {0, 1, 2, 3}
In simple words: Union merges both sets. Intersection finds common items. Set difference gives items in the first set but not the second.

Exam Tip: Pay close attention to inequality signs: strict inequalities (< and >) exclude the boundary values, while non-strict ones (≤ and ≥) include them.

Question 4. If P = {x : x ∈ W and x < 6} and Q = {x : x ∈ N and 4 ≤ x ≤ 9}, find (i) P ∪ Q (ii) P ∩ Q (iii) P - Q (iv) Q - P. Is P ∪ Q a proper superset of P ∩ Q?
Answer: From the given information:
P = {0, 1, 2, 3, 4, 5}
Q = {4, 5, 6, 7, 8, 9}

(i) P ∪ Q = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

(ii) P ∩ Q = {4, 5}

(iii) P - Q = {0, 1, 2, 3}

(iv) Q - P = {6, 7, 8, 9}

Yes, P ∪ Q is a proper superset of P ∩ Q because P ∪ Q has more elements than P ∩ Q, and every element of P ∩ Q is in P ∪ Q.
In simple words: A proper superset means one set holds everything another set has, plus extra items.

Exam Tip: A proper superset requires both inclusion (all elements of one set are in the other) and the presence of at least one additional element not in the smaller set.

Question 5. If A = (letters of word INTEGRITY) and B = (letters of word RECKONING), find (i) A ∪ B (ii) A ∩ B (iii) A - B (iv) B - A. Also verify that: (a) n(A ∪ B) = n(A) + n(B) - n(A ∩ B) (b) n(A - B) = n(A ∪ B) - n(B) = n(A) - n(A ∪ B) (c) n(B - A) = n(A ∪ B) - n(A) = n(B) - n(A ∩ B) (d) n(A ∪ B) = n(A - B) + n(B - A) + n(A ∩ B)
Answer: From the given information:
A = {I, N, T, E, G, R, Y}; n(A) = 7
B = {R, E, C, K, O, N, I, G}; n(B) = 8

(i) A ∪ B = {I, N, T, E, G, R, Y, C, K, O}; n(A ∪ B) = 10

(ii) A ∩ B = {I, N, E, G, R}; n(A ∩ B) = 5

(iii) A - B = {T, Y}; n(A - B) = 2

(iv) B - A = {C, K, O}; n(B - A) = 3

Verification:
(a) n(A ∪ B) = 10; n(A) + n(B) - n(A ∩ B) = 7 + 8 - 5 = 10 ✓
(b) n(A - B) = 2; n(A ∪ B) - n(B) = 10 - 8 = 2; n(A) - n(A ∩ B) = 7 - 5 = 2 ✓
(c) n(B - A) = 3; n(A ∪ B) - n(A) = 10 - 7 = 3; n(B) - n(A ∩ B) = 8 - 5 = 3 ✓
(d) n(A - B) + n(B - A) + n(A ∩ B) = 2 + 3 + 5 = 10 = n(A ∪ B) ✓
In simple words: These formulas show how the different parts of two sets relate - union includes everything, intersection is the overlap, and differences are the unique portions.

Exam Tip: When counting letters from words, list each distinct letter only once. These cardinal number formulas are fundamental - memorize them for solving complex set problems quickly.

Question 6. If ξ = {natural numbers between 10 and 40}, A = {multiples of 5} and B = {multiples of 6}, then (i) find A ∪ B and A ∩ B (ii) verify that n(A ∪ B) = n(A) + n(B) - n(A ∩ B)
Answer: From the given information:
ξ = {11, 12, 13, ..., 39}
A = {15, 20, 25, 30, 35}
B = {12, 18, 24, 30, 36}

(i) A ∪ B = {12, 15, 18, 20, 24, 25, 30, 35, 36}
A ∩ B = {30}

(ii) Verification:
n(A ∪ B) = 9
n(A) + n(B) - n(A ∩ B) = 5 + 5 - 1 = 9 ✓
In simple words: When you join the two groups, you get 9 numbers total. The formula works because adding the sizes of both sets counts the shared number twice, so we subtract it once.

Exam Tip: Always identify the universal set constraints first. A multiple is any number that divides evenly by the given divisor - make sure all your listed elements actually satisfy this.

Question 7. If ξ = {1, 2, 3, ..., 9}, A = {1, 2, 3, 4, 6, 7, 8} and B = {4, 6, 8}, then find (i) A' (ii) B' (iii) A ∪ B (iv) A ∩ B (v) A - B (vi) B - A (vii) (A ∩ B)' (viii) A' ∪ B'. Also verify that: (a) (A ∩ B)' = A' ∪ B' (b) n(A) + n(A') = n(ξ) (c) n(A ∩ B) + n((A ∩ B)') = n(ξ) (d) n(A - B) + n(B - A) + n(A ∩ B) = n(A ∪ B)
Answer: From the given information:
ξ = {1, 2, 3, 4, 5, 6, 7, 8, 9}
A = {1, 2, 3, 4, 6, 7, 8}
B = {4, 6, 8}

(i) A' = {5, 9}
(ii) B' = {1, 2, 3, 5, 7, 9}
(iii) A ∪ B = {1, 2, 3, 4, 6, 7, 8}
(iv) A ∩ B = {4, 6, 8}
(v) A - B = {1, 2, 3, 7}
(vi) B - A = { } (empty set)
(vii) (A ∩ B)' = {1, 2, 3, 5, 7, 9}
(viii) A' ∪ B' = {1, 2, 3, 5, 7, 9}

Verification:
(a) (A ∩ B)' = {1, 2, 3, 5, 7, 9} and A' ∪ B' = {1, 2, 3, 5, 7, 9}; Therefore (A ∩ B)' = A' ∪ B' ✓
(b) n(A) + n(A') = 7 + 2 = 9 = n(ξ) ✓
(c) n(A ∩ B) + n((A ∩ B)') = 3 + 6 = 9 = n(ξ) ✓
(d) n(A - B) + n(B - A) + n(A ∩ B) = 4 + 0 + 3 = 7 = n(A ∪ B) ✓
In simple words: A set and its complement always add up to the universal set. When one set is fully inside another, the difference is empty.

Exam Tip: De Morgan's Laws (shown in part a) are extremely useful: the complement of an intersection equals the union of the complements. Practice identifying when these laws apply.

Question 8. If ξ = {x : x ∈ W, x ≤ 10}, A = {x : x ≥ 5} and B = {x : 3 ≤ x < 8}, then verify that: (i) (A ∪ B)' = A' ∩ B' (ii) (A ∩ B)' = A' ∪ B' (iii) A - B = A ∩ B' (iv) B - A = B ∩ A'
Answer: From the given information:
ξ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
A = {5, 6, 7, 8, 9, 10}
B = {3, 4, 5, 6, 7}

(i) Verification of (A ∪ B)' = A' ∩ B'
LHS: (A ∪ B) = {3, 4, 5, 6, 7, 8, 9, 10}; (A ∪ B)' = {0, 1, 2}
RHS: A' = {0, 1, 2, 3, 4}; B' = {0, 1, 2, 8, 9, 10}; A' ∩ B' = {0, 1, 2}
Therefore, (A ∪ B)' = A' ∩ B' ✓

(ii) Verification of (A ∩ B)' = A' ∪ B'
LHS: (A ∩ B) = {5, 6, 7}; (A ∩ B)' = {0, 1, 2, 3, 4, 8, 9, 10}
RHS: A' ∪ B' = {0, 1, 2, 3, 4, 8, 9, 10}
Therefore, (A ∩ B)' = A' ∪ B' ✓

(iii) Verification of A - B = A ∩ B'
LHS: A - B = {8, 9, 10}
RHS: A ∩ B' = {8, 9, 10}
Therefore, A - B = A ∩ B' ✓

(iv) Verification of B - A = B ∩ A'
LHS: B - A = {3, 4}
RHS: B ∩ A' = {3, 4}
Therefore, B - A = B ∩ A' ✓
In simple words: These rules show that you can swap between different ways of writing set operations - difference equals intersection with a complement.

Exam Tip: De Morgan's Laws and the relationship between difference and complement are critical identities. Understanding these connections helps solve complex set problems without listing all elements.

Question 9. If n(A) = 20, n(B) = 16 and n(A ∪ B) = 30, find n(A ∩ B).
Answer: Given:
n(A) = 20
n(B) = 16
n(A ∪ B) = 30

Using the formula: n(A ∪ B) = n(A) + n(B) - n(A ∩ B)
30 = 20 + 16 - n(A ∩ B)
30 = 36 - n(A ∩ B)
n(A ∩ B) = 6
In simple words: The intersection size equals the sum of both set sizes minus the union size.

Exam Tip: This formula rearrangement is one of the most useful in set theory - if you know three of the four values, you can always find the fourth.

Question 10. If n(ξ) = 20 and n(A') = 7, then find n(A).
Answer: Given:
n(ξ) = 20
n(A') = 7

Using the formula: n(A') = n(ξ) - n(A)
7 = 20 - n(A)
n(A) = 13
In simple words: A set and its complement make up the universal set, so the two must add to the total.

Exam Tip: Always remember that n(A) + n(A') = n(ξ) - this relationship lets you move between a set and its complement easily.

Question 11. If n(ξ) = 40, n(A) = 20, n(B') = 16 and n(A ∪ B) = 32, then find n(B) and n(A ∩ B).
Answer: Given:
n(ξ) = 40
n(A) = 20
n(B') = 16
n(A ∪ B) = 32

Step 1: Find n(B)
n(B') = n(ξ) - n(B)
16 = 40 - n(B)
n(B) = 24

Step 2: Find n(A ∩ B)
n(A ∪ B) = n(A) + n(B) - n(A ∩ B)
32 = 20 + 24 - n(A ∩ B)
32 = 44 - n(A ∩ B)
n(A ∩ B) = 12
In simple words: Start by finding n(B) using the complement relationship, then use the union formula to get the intersection.

Exam Tip: Multi-step problems like this require you to use one formula's result as input for another - work systematically through each step without skipping.

Question 12. If n(ξ) = 32, n(A) = 20, n(B) = 16 and n((A ∪ B)') = 4, find (i) n(A ∪ B) (ii) n(A ∩ B) (iii) n(A - B)
Answer: Given:
n(ξ) = 32
n(A) = 20
n(B) = 16
n((A ∪ B)') = 4

(i) n(A ∪ B) = n(ξ) - n((A ∪ B)')
n(A ∪ B) = 32 - 4 = 28

(ii) n(A ∩ B)
n(A ∪ B) = n(A) + n(B) - n(A ∩ B)
28 = 20 + 16 - n(A ∩ B)
28 = 36 - n(A ∩ B)
n(A ∩ B) = 8

(iii) n(A - B) = n(A) - n(A ∩ B)
n(A - B) = 20 - 8 = 12
In simple words: First recover the union from its complement. Then use the union formula to find the intersection. Finally, subtract the intersection from set A to get the difference.

Exam Tip: The complement of a union is everything NOT in either set - this lets you recover the union size from its complement using the universal set.

Question 13. If n(ξ) = 40, n(A') = 15, n(B) = 12 and n((A ∩ B)') = 32, find (i) n(A) (ii) n(B') (iii) n(A ∩ B) (iv) n(A ∪ B) (v) n(A - B) (vi) n(B - A)
Answer: Given:
n(ξ) = 40
n(A') = 15
n(B) = 12
n((A ∩ B)') = 32

(i) n(A) = n(ξ) - n(A') = 40 - 15 = 25

(ii) n(B') = n(ξ) - n(B) = 40 - 12 = 28

(iii) n(A ∩ B) = n(ξ) - n((A ∩ B)') = 40 - 32 = 8

(iv) n(A ∪ B) = n(A) + n(B) - n(A ∩ B) = 25 + 12 - 8 = 29

(v) n(A - B) = n(A) - n(A ∩ B) = 25 - 8 = 17

(vi) n(B - A) = n(B) - n(A ∩ B) = 12 - 8 = 4
In simple words: Use complement relationships to find the basic sizes, then apply the union and difference formulas to get the remaining values.

Exam Tip: These multi-part problems test your knowledge of all the key formulas - write them all down at the start so you know which formula applies to which unknown.

Question 14. If n(A - B) = 12, n(B - A) = 16 and n(A ∩ B) = 5, find (i) n(A) (ii) n(B) (iii) n(A ∪ B)
Answer: Given:
n(A - B) = 12
n(B - A) = 16
n(A ∩ B) = 5

(i) n(A) = n(A - B) + n(A ∩ B) = 12 + 5 = 17

(ii) n(B) = n(B - A) + n(A ∩ B) = 16 + 5 = 21

(iii) n(A ∪ B) = n(A) + n(B) - n(A ∩ B) = 17 + 21 - 5 = 33
In simple words: Each set breaks into two parts: the unique portion and the overlap. Add them to get the total size. The union is the sum of both set sizes minus the overlap.

Exam Tip: Remember that every set divides into three regions in a Venn diagram: elements only in A, elements only in B, and elements in both. Using these formulas correctly means understanding this three-part structure.

Exercise 6.2

Question 1. From the adjoining Venn diagram, find the following sets: (i) A (ii) B (iii) ξ (iv) A' (v) B' (vi) A ∪ B (vii) A ∩ B (viii) (A ∪ B)' (ix) (A ∩ B)'
Answer: From the Venn diagram:

(i) A = {0, 5, 7, 8, 9, 11}

(ii) B = {2, 5, 6, 8}

(iii) ξ = {0, 1, 2, 4, 5, 6, 7, 8, 9, 11, 12}

(iv) A' = ξ - A = {1, 2, 4, 6, 12}

(v) B' = ξ - B = {0, 1, 4, 7, 9, 11, 12}

(vi) A ∪ B = {0, 2, 5, 6, 7, 8, 9, 11}

(vii) A ∩ B = {5, 8}

(viii) (A ∪ B)' = {1, 4, 12}

(ix) (A ∩ B)' = {0, 1, 2, 4, 6, 7, 9, 11, 12}
In simple words: Read the Venn diagram carefully - elements in the left circle form A, elements in the right form B, and elements in the overlap belong to both.

Exam Tip: Always identify three regions in a two-set Venn diagram: left only, right only, and middle (overlap). Practice extracting sets from diagrams systematically by region.

Question 2. From the adjoining Venn diagram, find the following sets:
Answer: [This question references a Venn diagram that appears in the source but requires diagram interpretation. Please provide the diagram to generate the specific answer.]
In simple words: Identify each region of the diagram and list the elements accordingly.

Exam Tip: When reading Venn diagrams, mark each region clearly before extracting sets to avoid missing elements.

Question 1. From the Venn diagram, find:
(i) P
(ii) Q
(iii) ξ
(iv) P'
(v) Q'
(vi) P ∪ Q
(vii) P ∩ Q
(viii) (P ∪ Q)'
(ix) (P ∩ Q)'
Answer: Looking at the Venn diagram provided:
(i) P = {a, b, d, e, f, g, h, i}
(ii) Q = {b, d, e}
(iii) ξ = {a, b, c, d, e, f, g, h, i, j}
(iv) P' = {c, j}
(v) Q' = {a, c, f, g, h, i, j}
(vi) P ∪ Q = {a, b, d, e, f, g, h, i}
(vii) P ∩ Q = {b, d, e}
(viii) (P ∪ Q)' = {c, j}
(ix) (P ∩ Q)' = {a, c, f, g, h, i, j}
In simple words: A set shows all elements it holds. The union combines both sets. The intersection finds what both share. A complement finds what is left out.

Exam Tip: Always identify which elements belong inside each region - only in P, only in Q, in both, or outside both circles.

Question 2. From the adjoining Venn diagram, find the following sets:
(i) ξ
(ii) A ∩ B
(iii) A ∩ B ∩ C
(iv) C'
(v) A - C
(vi) B - C
(vii) C - B
(viii) (A ∪ B)'
(ix) (A ∪ B ∪ C)'
Answer: Using the Venn diagram shown:
(i) ξ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
(ii) A ∩ B = {0, 5, 8}
(iii) A ∩ B ∩ C = {0, 5}
(iv) C' = {2, 7, 8, 9, 10, 11, 12} (because C' = ξ - C)
(v) A - C = {8, 10}
(vi) B - C = {7, 8, 11}
(vii) C - B = {3, 4, 6}
(viii) (A ∪ B)' = {2, 4, 6, 9, 12}
(ix) (A ∪ B ∪ C)' = {2, 9, 12}
In simple words: Read the diagram carefully to find where each number sits. Numbers in the overlap belong to multiple sets. Numbers outside all circles are in the complement.

Exam Tip: Watch the regions carefully - the centre (all three sets overlap) is different from where only two circles overlap.

Question 3. Draw Venn diagrams to show the relationship between the following pairs of sets:
(i) A = {x | x ϵ N, x = 2n, n ≤ 5} and B = {x | x ϵ W, x = 4n, n < 5}
Answer: First, list the elements of each set.

For set A: Since n ≤ 5 and x = 2n, we get x = 2(1), 2(2), 2(3), 2(4), 2(5) = 2, 4, 6, 8, 10.
So A = {2, 4, 6, 8, 10}.

For set B: Since n < 5 and x = 4n, we get x = 4(0), 4(1), 4(2), 4(3), 4(4) = 0, 4, 8, 12, 16.
So B = {4, 8, 12, 16}.

The intersection A ∩ B = {4, 8}, showing both sets share these two elements. The sets overlap - they are not separate, but they also don't contain each other completely.

A Venn diagram would show two overlapping circles, with 4 and 8 in the shared region in the middle.
In simple words: First, work out what numbers belong to each set by plugging in the values. Then draw two circles that cross, putting shared numbers in the overlap.

Exam Tip: Always compute the sets from the given conditions before drawing - guessing the diagram leads to errors.

Question 4. A = {prime factors of 42} and B = {prime factors of 60}
Answer: Find the prime factors of each number.

Prime factors of 42: 42 = 2 × 21 = 2 × 3 × 7.
So A = {2, 3, 7}.

Prime factors of 60: 60 = 2 × 30 = 2 × 2 × 15 = 2 × 2 × 3 × 5 = 4 × 3 × 5, giving distinct prime factors 2, 3, 5.
So B = {2, 3, 5}.

The intersection A ∩ B = {2, 3}. Both sets have exactly these two primes in common. The sets overlap, as each contains some factors the other does not (A has 7, B has 5).

The Venn diagram shows two overlapping circles, with 7 in the A-only region, 2 and 3 in the shared centre, and 5 in the B-only region.
In simple words: Break down each number into its prime factors. Circle numbers that appear in both sets - put them in the middle of your diagram.

Exam Tip: Prime factorization must be complete - list only the prime numbers, not composite factors.

Question 5. P = {x | x ϵ W, x < 10} and Q = {prime factors of 210}
Answer: List the elements of both sets.

For set P: All whole numbers less than 10 are P = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}.

For set Q: Prime factors of 210. We factor: 210 = 2 × 105 = 2 × 3 × 35 = 2 × 3 × 5 × 7.
So Q = {2, 3, 5, 7}.

The intersection P ∩ Q = {2, 3, 5, 7}. Notice that Q is completely contained inside P, since all primes in Q are less than 10. This means Q is a subset of P.

The Venn diagram shows the Q circle drawn entirely inside the P circle, indicating the subset relationship.
In simple words: Q holds only prime numbers. P holds all whole numbers from 0 to 9. Every prime in Q fits into P, so Q sits fully inside P.

Exam Tip: When one set is a subset of another, draw the smaller circle completely inside the larger one - no parts sticking out.

Question 6. Draw a Venn diagram to illustrate the following information: n(A) = 22, n(B) = 18 and n(A ∩ B) = 5. Hence find: (i) n(A ∪ B) (ii) n(A - B) (iii) n(B - A)
Answer: Start by finding how many elements are in each region of the diagram.

We know: n(A) = 22, n(B) = 18, n(A ∩ B) = 5.

Elements only in A: n(A) - n(A ∩ B) = 22 - 5 = 17.
Elements only in B: n(B) - n(A ∩ B) = 18 - 5 = 13.

Now we can answer each part:

(i) n(A ∪ B) = elements only in A + elements in both + elements only in B = 17 + 5 + 13 = 35.

(ii) n(A - B) = elements only in A = 17.

(iii) n(B - A) = elements only in B = 13.

The Venn diagram shows two overlapping circles with 17 in the left region (A only), 5 in the middle (both), and 13 in the right region (B only).
In simple words: Subtract the overlap from each set to get what each set has alone. Then add all three regions to get the total.

Exam Tip: Always split each set into two parts: the overlap and the "only" region - this makes counting much clearer.

Question 7. Draw a Venn diagram to illustrate the following information: n(A) = 25, n(B) = 16, n(A ∩ B) = 6 and n((A ∪ B)') = 5. Hence find: (i) n(A ∪ B) (ii) n(ξ) (iii) n(A - B) (iv) n(B - A)
Answer: Begin by calculating the regions within the Venn diagram.

Given: n(A) = 25, n(B) = 16, n(A ∩ B) = 6, n((A ∪ B)') = 5.

Elements only in A: n(A) - n(A ∩ B) = 25 - 6 = 19.
Elements only in B: n(B) - n(A ∩ B) = 16 - 6 = 10.

Now solve each part:

(i) n(A ∪ B) = 19 + 6 + 10 = 35.

(ii) n(ξ) = n(A ∪ B) + n((A ∪ B)') = 35 + 5 = 40. The universal set contains the union of both sets plus those outside both sets.

(iii) n(A - B) = 19.

(iv) n(B - A) = 10.

The Venn diagram shows two overlapping circles within a rectangle. The rectangle represents the universal set. The left region has 19, the middle has 6, the right has 10, and outside both circles (but inside the rectangle) is 5.
In simple words: The complement tells you how many are outside both circles. Add this to the union to find the whole set.

Exam Tip: When the universal set is given or can be found, draw a rectangle around both circles to show everything included.

Question 8. Given n(ξ) = 25, n(A') = 7, n(B) = 10 and B ⊂ A. Draw a Venn diagram to illustrate this information. Hence, find the cardinal number of the set A - B.
Answer: Use the given information to determine set sizes and relationships.

From the given data:
n(ξ) = 25
n(A') = 7, which means n(A) = n(ξ) - n(A') = 25 - 7 = 18.
n(B) = 10
B ⊂ A means B is a subset of A, so every element of B is in A.

Since B ⊂ A and n(B) = 10, all 10 elements of B are inside A.
Elements in A but not in B: n(A - B) = n(A) - n(B) = 18 - 10 = 8.

The Venn diagram shows set B as a smaller circle completely inside set A, which sits inside the universal set rectangle. The region inside A but outside B shows 8 elements, and the inner circle B shows 10 elements. Outside A altogether shows 7 elements.

Therefore, the cardinal number of the set A - B = 8.
In simple words: If B fits completely inside A, then A - B is just the part of A that B doesn't fill up.

Exam Tip: When one set is a subset of another, draw it inside - this makes the "A - B" region very clear to identify.

Question 9. In a group of 50 boys, 20 play only cricket, 12 play only football and 5 boys play both the games. Draw a Venn diagram and find the number of boys who play (i) at least one of the two games cricket or football. (ii) neither cricket nor football.
Answer: Set up the problem using the given numbers in the correct regions.

Let P = boys who play cricket, Q = boys who play football, ξ = all boys.

From the question:
Boys playing only cricket = 20
Boys playing only football = 12
Boys playing both = 5
Total boys = 50

(i) At least one of the two games means the union P ∪ Q.
n(P ∪ Q) = only cricket + both + only football = 20 + 5 + 12 = 37 boys.

(ii) Neither cricket nor football = Total - those playing at least one
= 50 - 37 = 13 boys.

The Venn diagram shows two overlapping circles for P and Q within a rectangle for all 50 boys. The left circle has 20, the overlap has 5, the right circle has 12, and outside both circles (but in the rectangle) is 13.
In simple words: Add up all the parts that involve at least one game. Then subtract from the total to find those who play neither game.

Exam Tip: "At least one" means the union - add all overlapping and non-overlapping parts of both circles. "Neither" is everything outside both circles.

Question 10. In a group of 40 students, 26 students like orange but not banana, while 32 students like oranges. If all the students like at least one of the two fruits, find the number of students who like (i) both orange and banana (ii) only banana. Draw a Venn diagram to represent the data.
Answer: Work through the given information step by step to find the unknown regions.

Let P = students who like orange, Q = students who like banana, ξ = all students.

Given:
n(ξ) = 40
n(P - Q) = 26 (orange only)
n(P) = 32 (total who like orange)
n(P ∪ Q) = 40 (all like at least one fruit)

(i) Find n(P ∩ Q) - students who like both.
n(P ∩ Q) = n(P) - n(P - Q) = 32 - 26 = 6 students.

(ii) Find n(Q - P) - students who like only banana.
Since all 40 students like at least one fruit:
n(Q - P) = n(ξ) - n(P) = 40 - 32 = 8 students.

We can verify: orange only (26) + both (6) + banana only (8) = 40. ✓

The Venn diagram shows two overlapping circles. The P-only region has 26, the overlap has 6, and the Q-only region has 8. There is no space outside both circles since all students like at least one fruit.
In simple words: Subtract those with orange only from total oranges to find the overlap. Subtract total oranges from all students to find banana only.

Exam Tip: When told "all like at least one," the complement is empty - every person must be in at least one circle.

Question 11. In a group of 60 persons, 45 speak Bengali, 28 speak English and all the persons speak at least one language. Find how many people speak both Bengali and English. Draw a Venn diagram.
Answer: Apply the principle of counting with overlapping sets to find the intersection.

Let P = persons who speak Bengali, Q = persons who speak English, ξ = all persons.

Given:
n(ξ) = 60
n(P) = 45
n(Q) = 28
n(P ∪ Q) = 60 (all speak at least one language)

Using the formula: n(P ∩ Q) = n(P) + n(Q) - n(P ∪ Q)
n(P ∩ Q) = 45 + 28 - 60 = 13 persons.

Therefore, 13 people speak both Bengali and English.

The Venn diagram shows two overlapping circles within a rectangle. The left region (Bengali only) has 45 - 13 = 32 people. The overlap (both languages) has 13 people. The right region (English only) has 28 - 13 = 15 people. There is no space outside both circles since everyone speaks at least one language.

We can check: 32 + 13 + 15 = 60. ✓
In simple words: When you add up two groups, you count the overlap twice. So subtract it once to get the real total.

Exam Tip: Remember the formula n(P ∪ Q) = n(P) + n(Q) - n(P ∩ Q) - it's essential for solving these "at least one" problems with two overlapping sets.