CBSE Class 11 Maths Sets Ex 1.6: Practical Problems on Union and Intersection

Understanding the Core Formulas for Counting Elements

Exercise 1.6 primarily deals with determining the number of elements in the union or intersection of two or three finite sets. To tackle these problems effectively, we rely on a set of fundamental formulas derived from the Inclusion-Exclusion Principle. This principle helps us avoid double-counting elements that belong to multiple sets.

For Two Sets A and B:

The most important formula states that the number of elements in the union of two sets A and B, denoted as n(A U B), is given by:

n(A U B) = n(A) + n(B) - n(A ∩ B)

• n(A) is the number of elements in set A.
• n(B) is the number of elements in set B.
• n(A ∩ B) is the number of elements common to both A and B (their intersection). We subtract n(A ∩ B) because when we add n(A) and n(B), the elements in the intersection are counted twice. Subtracting them once corrects this overcounting.

Special Case: Disjoint Sets

If A and B are disjoint sets (meaning they have no elements in common, so A ∩ B = Ø), then n(A ∩ B) = 0. In this case, the formula simplifies to:

n(A U B) = n(A) + n(B)

For Three Sets A, B, and C:

Extending this logic for three finite sets A, B, and C, the number of elements in their union n(A U B U C) is given by:

n(A U B U C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(A ∩ C) - n(B ∩ C) + n(A ∩ B ∩ C)

This formula looks more complex, but follows the same inclusion-exclusion logic. We sum the individual set sizes, subtract the pairwise intersections (because elements in these were counted twice), and then add back the triple intersection (because elements in A ∩ B ∩ C were initially counted thrice, then subtracted thrice in the pairwise intersections, thus being excluded entirely. Adding them back once corrects this). Understanding the derivation of these formulas through Venn diagrams can greatly clarify their application.

Mastering these formulas is key to solving the practical problems in Exercise 1.6 efficiently and accurately.

Worked Examples: Applying Set Formulas

• Example 1: Two-Set Problem In a group of 65 people, 40 like cricket and 10 like both cricket and tennis. How many like tennis only and how many like tennis? Given: Total people, n(C U T) = 65 (where C = Cricket, T = Tennis) People who like cricket, n(C) = 40 * People who like both cricket and tennis, n(C ∩ T) = 10 Step 1: Find the number of people who like Tennis (n(T)). We use the formula: n(C U T) = n(C) + n(T) - n(C ∩ T) Substitute the given values: 65 = 40 + n(T) - 10 65 = 30 + n(T) n(T) = 65 - 30 n(T) = 35 Step 2: Find the number of people who like Tennis only. People who like Tennis only = n(T) - n(C ∩ T) = 35 - 10 = 25 Final Answer: 35 people like tennis, and 25 people like tennis only.
• Example 2: Three-Set Problem (Conceptual) In a survey of 100 students, 40 liked Math (M), 30 liked Physics (P), and 20 liked Chemistry (C). 10 liked Math and Physics, 8 liked Physics and Chemistry, 5 liked Math and Chemistry. All 100 students liked at least one subject. How many students liked all three subjects? Given: n(M U P U C) = 100 (since all liked at least one subject) n(M) = 40 n(P) = 30 n(C) = 20 n(M ∩ P) = 10 n(P ∩ C) = 8 * n(M ∩ C) = 5 Step 1: Apply the formula for the union of three sets. The formula is: n(M U P U C) = n(M) + n(P) + n(C) - n(M ∩ P) - n(P ∩ C) - n(M ∩ C) + n(M ∩ P ∩ C) Substitute the known values into the formula: 100 = 40 + 30 + 20 - 10 - 8 - 5 + n(M ∩ P ∩ C) Step 2: Solve for n(M ∩ P ∩ C). 100 = (40 + 30 + 20) - (10 + 8 + 5) + n(M ∩ P ∩ C) 100 = 90 - 23 + n(M ∩ P ∩ C) 100 = 67 + n(M ∩ P ∩ C) n(M ∩ P ∩ C) = 100 - 67 n(M ∩ P ∩ C) = 33 Final Answer: 33 students liked all three subjects.

Exam Tips and Common Mistakes to Avoid

Solving practical problems on sets requires careful reading and interpretation. Here are some critical tips to help you ace Exercise 1.6 and avoid common pitfalls:

1. Identify 'Only' vs. 'Total': Pay close attention to keywords. 'People who like coffee' refers to n(C), which includes those who might also like tea. 'People who like only coffee' refers to n(C) - n(C ∩ T). This is a very common source of error.
2. Draw Venn Diagrams: For complex problems, especially those involving three sets, drawing a Venn diagram can be immensely helpful. It visually represents the relationships between sets and helps in correctly identifying the regions for n(A), n(A ∩ B), n(A only), n(A U B), etc.
3. Correct Formula Application: Ensure you use the correct formula for two or three sets. Memorize them or understand their derivation to recall them accurately. Don't mix them up!
4. Check for 'Neither': Sometimes, a problem might specify a total number of elements in a universal set n(U), and then give details about sets A and B. If n(A U B) is less than n(U), it means some elements belong to neither A nor B. n(Neither) = n(U) - n(A U B).
5. Step-by-Step Calculation: Break down the problem into smaller, manageable steps. Calculate one unknown at a time. This reduces the chances of arithmetic errors and helps you track your progress, making it easier to recheck if needed.

Practice Questions with Solutions

• Q: In a committee, 50 people speak French, 20 speak Spanish and 10 speak both Spanish and French. How many people speak at least one of these two languages? A: Step 1: Identify the given values. Let F be the set of people who speak French, and S be the set of people who speak Spanish. n(F) = 50 n(S) = 20 n(F ∩ S) = 10 Step 2: Use the formula for the union of two sets. The number of people who speak at least one of the two languages is n(F U S). n(F U S) = n(F) + n(S) - n(F ∩ S) n(F U S) = 50 + 20 - 10 n(F U S) = 70 - 10 n(F U S) = 60 Final answer: 60 people speak at least one of these two languages.
• Q: In a group of 70 people, 37 like coffee, 52 like tea, and each person likes at least one of the two drinks. How many people like both coffee and tea? A: Step 1: Identify the given values. Let C be the set of people who like coffee, and T be the set of people who like tea. n(C) = 37 n(T) = 52 n(C U T) = 70 (since each person likes at least one drink) Step 2: Use the formula for the union of two sets to find the intersection. n(C U T) = n(C) + n(T) - n(C ∩ T) 70 = 37 + 52 - n(C ∩ T) 70 = 89 - n(C ∩ T) n(C ∩ T) = 89 - 70 n(C ∩ T) = 19 Final answer: 19 people like both coffee and tea.
• Q: In a school, there are 20 teachers who teach Mathematics or Physics. Of these, 12 teach Mathematics and 4 teach both Physics and Mathematics. How many teach Physics? A: Step 1: Identify the given values. Let M be the set of teachers who teach Mathematics, and P be the set of teachers who teach Physics. n(M U P) = 20 n(M) = 12 n(M ∩ P) = 4 Step 2: Use the formula for the union of two sets to find n(P). n(M U P) = n(M) + n(P) - n(M ∩ P) 20 = 12 + n(P) - 4 20 = 8 + n(P) n(P) = 20 - 8 n(P) = 12 Final answer: 12 teachers teach Physics.
• Q: In a town, 800 people are surveyed. 300 read Newspaper A, 250 read Newspaper B, and 200 read Newspaper C. 100 read A and B, 70 read B and C, 80 read A and C. 50 people read all three newspapers. How many people read exactly one newspaper? A: Step 1: Find the number of people who read at least one newspaper (n(A U B U C)). Given: n(A) = 300, n(B) = 250, n(C) = 200 n(A ∩ B) = 100, n(B ∩ C) = 70, n(A ∩ C) = 80 n(A ∩ B ∩ C) = 50 Using the formula for three sets: n(A U B U C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(A ∩ C) + n(A ∩ B ∩ C) n(A U B U C) = 300 + 250 + 200 - 100 - 70 - 80 + 50 n(A U B U C) = 750 - 250 + 50 n(A U B U C) = 500 + 50 = 550 Step 2: Calculate people who read exactly one newspaper. People who read only A = n(A) - n(A ∩ B) - n(A ∩ C) + n(A ∩ B ∩ C) = 300 - 100 - 80 + 50 = 170 People who read only B = n(B) - n(A ∩ B) - n(B ∩ C) + n(A ∩ B ∩ C) = 250 - 100 - 70 + 50 = 130 People who read only C = n(C) - n(A ∩ C) - n(B ∩ C) + n(A ∩ B ∩ C) = 200 - 80 - 70 + 50 = 100 Step 3: Sum the 'only one' categories. Exactly one newspaper = (only A) + (only B) + (only C) Exactly one newspaper = 170 + 130 + 100 = 400 Final answer: 400 people read exactly one newspaper.

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