Set Theory (Sets) Notes — Class 10 Math (SEE)

Complete bilingual study notes for Chapter 1: Sets — every concept explained step by step, with definitions, formulas, and worked examples.

1.0 Review

Students from roll no. 1 to 15 in grade 10 are surveyed about whether they like Mathematics or Science. The result is shown in a Venn diagram: students ${1, 3, 4, 11, 13}$ like Mathematics only, students ${5, 6, 7, 10, 12}$ like Science only, students ${2, 8, 15}$ like both, and students ${9, 14}$ like neither.

Let $M$ = set of students who like Mathematics, $S$ = set of students who like Science. By listing method: $M = {1, 2, 3, 4, 8, 11, 13, 15}$ , so $n (M) = 8$ . $M$ only $= {1, 3, 4, 11, 13}$ , so $n_{0} (M) = 5$ . $S = {2, 5, 6, 7, 8, 10, 12, 15}$ , so $n (S) = 8$ . $S$ only $= {5, 6, 7, 10, 12}$ , so $n_{0} (S) = 5$ . Both Mathematics and Science: $M \cap S = {2, 8, 15}$ , so $n (M \cap S) = 3$ . Neither: ${9, 14}$ , so $n (M \cup S) = 2$ . Total students surveyed $= n (U) = 15$ .

1.1 Cardinality of the Two Sets

Activity: among grade 10 students surveyed about coffee or tea — $n (C) = 15$ like coffee, $n (T) = 10$ like tea, $n (C \cap T) = 6$ like both, and $5$ dislike both. Here $n_{0} (C)$ and $n_{0} (T)$ denote coffee-only and tea-only. Since $n (C) = n_{0} (C) + n (C \cap T)$ , coffee only is $n_{0} (C) = n (C) - n (C \cap T) = 15 - 6 = 9$ . Likewise tea only $n_{0} (T) = n (T) - n (C \cap T) = 10 - 6 = 4$ . With $n (C \cup T) = 5$ (dislike both), the total class size is $n (U) = 9 + 6 + 4 + 5 = 24$ .

If $A$ and $B$ are overlapping sets: (i) $n (A \cup B) = n (A) + n (B) - n (A \cap B)$ . (ii) $n_{0} (A) = n (A) - n (A \cap B)$ . (iii) $n_{0} (B) = n (B) - n (A \cap B)$ . (iv) $n (A \cup B) = n_{0} (A) + n_{0} (B) + n (A \cap B)$ . (v) If $A, B$ are disjoint, $n (A \cup B) = n (A) + n (B)$ . (vi) If $U$ has only elements of $A$ and $B$ , then $n (U) = n (A \cup B)$ . (vii) Otherwise $n (U) = n (A \cup B) + n (A \cup B)$ . At least one: $n (A \cup B) = n (A) + n (B) - n (A \cap B) = n_{0} (A) + n_{0} (B) + n (A \cap B)$ . At most one: $n (A \cap B) = n (U) - n (A \cap B)$ . Exactly one: $n_{0} (A) + n_{0} (B) = n (A) + n (B) - 2 \times n (A \cap B)$ .

In a survey of 300 people of a community, it was found that 175 liked cricket and 150 liked football but 25 liked neither of them. Based on this, answer: a) Represent this in a Venn diagram. b) Find the number of people who like both games. c) Find the number of people who like exactly one game.

Solution: Let $C$ and $F$ denote the sets of people who like cricket and football; $U$ is the total. $n (U) = 300$ , $n (C) = 175$ , $n (F) = 150$ , $n (C \cup F) = 25$ . Let $n (C \cap F) = x$ . Then $n (U) = n_{0} (C) + n (C \cap F) + n_{0} (F) + n (C \cup F)$ : $300 = (175 - x) + x + (150 - x) + 25$ , so $300 = 350 - x$ , giving $x = 50$ . So $n (C \cap F) = 50$ people like both games. Only cricket: $n_{0} (C) = 175 - 50 = 125$ . Only football: $n_{0} (F) = 150 - 50 = 100$ . Exactly one game: $n_{0} (C) + n_{0} (F) = 125 + 100 = 225$ .

The result of a survey among 120 students of grade 10 is as follows: 30 like only Mathematics, 40 like only English, 10 like neither Mathematics nor English. a) Represent this in a Venn diagram. b) Find the number of students who like both subjects. c) Find the number of students who like at least one subject.

Solution: Let $M$ and $E$ denote sets of students who like Mathematics and English; $U$ is the total students. $n (U) = 120$ , $n_{0} (M) = 30$ , $n_{0} (E) = 40$ , $n (M \cup E) = 10$ . Let $n (M \cap E) = x$ . From the Venn diagram: $30 + x + 40 + 10 = 120$ , so $80 + x = 120$ , giving $x = 40$ . So $n (M \cap E) = 40$ students like both subjects. At least one subject: $n (M \cup E) = 30 + 40 + 40 = 110$ .

According to a survey among the SEE-appeared students from a school, 75% were interested in studying science and 55% were interested in studying staff nurse but 5% did not give any information whilst 21 students were interested in studying both science and staff nurse. Based on this, answer: a) Show this in a Venn diagram. b) Find the total number of students inquired in the survey. c) Find the number of students who were interested in studying only staff nurse.

Solution: Let $S$ and $N$ denote sets of students interested in science and staff nurse; $U$ is the total. Let $n (U) = x$ . Then $n (S) = 0.75 x$ , $n (N) = 0.55 x$ , $n (S \cap N) = 21$ , $n (S \cup N) = 0.05 x$ . From the Venn diagram: $(0.75 x - 21) + 21 + (0.55 x - 21) + 0.05 x = x$ , so $1.35 x - 21 = x$ , giving $0.35 x = 21$ , so $x = \frac{21}{0.35} = 60$ . The total number of students surveyed was $60$ . Only staff nurse $= 0.55 x - 21 = 0.55 \times 60 - 21 = 33 - 21 = 12$ .

In a survey of 300 foreign tourists visiting Nepal, it was found that the ratio of the number of tourists who visited Pokhara and Lumbini was 2:3. Among them, 90 visited both places and 60 visited neither Pokhara nor Lumbini. Based on this, answer: a) Show this in a Venn diagram. b) Determine the number of tourists who visited only one place. c) Find the number of tourists who visited at least one of the places.

Solution: Let $P$ and $L$ denote sets of tourists who visited Pokhara and Lumbini; $U$ is the total. $n (U) = 300$ , $n (P \cap L) = 90$ , $n (P \cup L) = 60$ . Let $n (P) = 2 x$ , $n (L) = 3 x$ . From the Venn diagram: $(2 x - 90) + 90 + (3 x - 90) + 60 = 300$ , so $5 x - 30 = 300$ , giving $5 x = 330$ , so $x = \frac{330}{5} = 66$ . Only Pokhara: $n_{0} (P) = 2 (66) - 90 = 42$ . Only Lumbini: $n_{0} (L) = 3 (66) - 90 = 108$ . Only one place $= 42 + 108 = 150$ . At least one place $= 300 - 60 = 240$ .

In a survey among 200 students studying in grade 10, it was found that the ratio of the number of students who like Mathematics and English was 2:3. Among them, 30% like both of them but 15% like neither Mathematics nor English. Based on this, answer: a) Represent this in a Venn diagram. b) What is the difference between the number of students who like Mathematics and the number of students who like English? Find it.

Solution: Let $M$ and $E$ denote sets of students who like Mathematics and English; $U$ is the total. $n (U) = 200$ , $n (M \cap E) = 30%$ of $200 = 60$ , $n (M \cup E) = 15%$ of $200 = 30$ . Let $n_{0} (M) = 2 x$ , $n_{0} (E) = 3 x$ . From the Venn diagram: $2 x + 60 + 3 x + 30 = 200$ , so $90 + 5 x = 200$ , giving $5 x = 110$ , so $x = \frac{110}{5} = 22$ . So $n (M) = 2 x + 60 = 2 (22) + 60 = 104$ and $n (E) = 3 x + 60 = 3 (22) + 60 = 126$ . The difference is $n (E) - n (M) = 126 - 104 = 22$ .

1.2 Cardinality of Three Sets

For three sets $A$ , $B$ , $C$ shown in a Venn diagram (Figure 1): $A = {1, 3, 4, 6, 9}$ so $n (A) = 5$ ; $B = {5, 6, 7, 8, 9}$ so $n (B) = 5$ ; $C = {2, 4, 6, 8}$ so $n (C) = 4$ ; $A \cup B \cup C = {1, 2, 3, 4, 5, 6, 7, 8, 9}$ so $n (A \cup B \cup C) = 9$ ; and since $U$ has only these elements, $n (U) = 9$ , i.e. $n (A \cup B \cup C) = n (U)$ . In a second diagram (Figure 2) where $U$ also contains element $10$ outside all three sets, $n (A \cup B \cup C) = 8$ while $n (U) = 10$ , so $n (A \cup B \cup C) < n (U)$ because of elements outside all three sets.

In a survey of a classroom, 40 students like orange, 35 like mango and 50 like banana. Among them, 15 like orange and mango, 20 like mango and banana, 25 like orange and banana, 5 like all three fruits, and 30 like none of the fruits. Let $O$ , $M$ , $B$ denote the sets. Insert $n (O \cap M \cap B) = 5$ at the centre and $n (O \cup M \cup B) = 30$ outside. Since $n (O \cap M) = 15$ already includes the $5$ who like all three, orange-and-mango-only $n_{0} (O \cap M) = 15 - 5 = 10$ . Similarly mango-and-banana-only $n_{0} (M \cap B) = 20 - 5 = 15$ and orange-and-banana-only $n_{0} (O \cap B) = 25 - 5 = 20$ . Orange only $n_{0} (O) = 40 - (5 + 10 + 20) = 5$ ; mango only $n_{0} (M) = 35 - (5 + 10 + 15) = 5$ ; banana only $n_{0} (B) = 50 - (5 + 15 + 20) = 10$ . Total participants $n (U) = 5 + 10 + 5 + 5 + 20 + 15 + 10 + 30 = 100$ .

If $A$ , $B$ , $C$ are overlapping sets, from the Venn diagram: (a) $n_{0} (A \cap B) = n (A \cap B) - n (A \cap B \cap C)$ ; (b) $n_{0} (A \cap C) = n (A \cap C) - n (A \cap B \cap C)$ ; (c) $n_{0} (B \cap C) = n (B \cap C) - n (A \cap B \cap C)$ ; (d) $n (A) = n_{0} (A) + n_{0} (A \cap B) + n_{0} (A \cap C) + n (A \cap B \cap C)$ ; (e) $n (B) = n_{0} (B) + n_{0} (A \cap B) + n_{0} (B \cap C) + n (A \cap B \cap C)$ ; (f) $n (C) = n_{0} (C) + n_{0} (A \cap C) + n_{0} (B \cap C) + n (A \cap B \cap C)$ ; (g) $n (U) = n_{0} (A) + n_{0} (B) + n_{0} (C) + n_{0} (A \cap B) + n_{0} (B \cap C) + n_{0} (A \cap C) + n (A \cap B \cap C) + n (A \cup B \cup C)$ , i.e. $n (U) = n (A \cup B \cup C) + n (A \cup B \cup C)$ . Also, $n (A \cup B \cup C) = n (A) + n (B) + n (C) - n (A \cap B) - n (B \cap C) - n (A \cap C) + n (A \cap B \cap C)$ . If the sets are disjoint, $n (A \cup B \cup C) = n (A) + n (B) + n (C)$ .

If $n (U) = 120$ , $n (A) = 48$ , $n (B) = 51$ , $n (C) = 40$ , $n (A \cap B) = 11$ , $n (B \cap C) = 10$ , $n (A \cap C) = 9$ , and $n (A \cap B \cap C) = 4$ , then find $n (A \cup B \cup C)$ and $n (A \cup B \cup C)$ . Also show this in a Venn diagram.

Solution: We know $n (A \cup B \cup C) = n (A) + n (B) + n (C) - n (A \cap B) - n (B \cap C) - n (A \cap C) + n (A \cap B \cap C) = 48 + 51 + 40 - 11 - 10 - 9 + 4 = 113$ . Also $n (U) = n (A \cup B \cup C) + n (A \cup B \cup C)$ , so $120 = 113 + n (A \cup B \cup C)$ , giving $n (A \cup B \cup C) = 7$ . In the Venn diagram: only $A = 32$ , $A \cap B$ only $= 7$ , only $B = 34$ , all three $= 4$ , $A \cap C$ only $= 5$ , $B \cap C$ only $= 6$ , only $C = 25$ , outside $= 7$ .

Among the 180 students who participated in the SLC examination in 2071 from Nepal Madhyamik Vidhyalaya, 86 passed in Science, 80 passed Maths, and 76 passed in Nepali. Out of them, 26 passed in Science and Maths, 36 passed in Maths and Nepali, and 32 passed in Science and Nepali, but 20 did not pass any subject. a) Show the given information in a Venn diagram. b) Find the number of students who passed in all three subjects.

Solution: Let $n (U) = 180$ , $n (S) = 86$ (Science), $n (M) = 80$ (Maths), $n (N) = 76$ (Nepali), $n (S \cap M) = 26$ , $n (M \cap N) = 36$ , $n (S \cap N) = 32$ , $n (M \cup N \cup S) = 20$ . We know $n (U) = n (S) + n (M) + n (N) - n (S \cap M) - n (M \cap N) - n (S \cap N) + n (S \cap M \cap N) + n (S \cup M \cup N)$ : $180 = 86 + 80 + 76 - 26 - 36 - 32 + n (S \cap M \cap N) + 20$ , so $180 = 168 + n (S \cap M \cap N)$ , giving $n (S \cap M \cap N) = 12$ . So 12 students passed in all three subjects. From the Venn diagram: only Science $= 40$ , $S \cap M$ only $= 14$ , only Maths $= 30$ , all three $= 12$ , $S \cap N$ only $= 20$ , $M \cap N$ only $= 24$ , only Nepali $= 20$ , outside $= 20$ .

A school distributed medals for the students in different events of a competition. 36 got medals in dance, 12 in drama and 18 in music. If only 45 students got medals and 4 students got medals in all three events, then find the number of students who got medals in exactly two events.

Solution: Let $n (A)$ , $n (B)$ , $n (C)$ denote students who got medals in dance, drama, music: $n (A) = 36$ , $n (B) = 12$ , $n (C) = 18$ , $n (A \cup B \cup C) = 45$ , $n (A \cap B \cap C) = 4$ . We know $n (A \cup B \cup C) = n (A) + n (B) + n (C) - n (A \cap B) - n (B \cap C) - n (A \cap C) + n (A \cap B \cap C)$ : $45 = 36 + 12 + 18 - [n (A \cap B) + n (B \cap C) + n (A \cap C)] + 4$ , so $45 = 70 - [n (A \cap B) + n (B \cap C) + n (A \cap C)]$ , giving $n (A \cap B) + n (B \cap C) + n (A \cap C) = 25$ . Now, exactly two events $= n_{0} (A \cap B) + n_{0} (B \cap C) + n_{0} (A \cap C) = [n (A \cap B) - n (A \cap B \cap C)] + [n (B \cap C) - n (A \cap B \cap C)] + [n (A \cap C) - n (A \cap B \cap C)] = 25 - 3 (4) = 25 - 12 = 13$ . So 13 students got medals in exactly two events.