Let \(A\), \(B\) and \(C\) be sets. Then show that
\[ A \cap (B \cup C) = (A \cap B) \cup (A \cap C). \]
To prove: \(A \cap (B \cup C) = (A \cap B) \cup (A \cap C)\).
Proof:
(1) Show that \(A \cap (B \cup C) \subseteq (A \cap B) \cup (A \cap C)\)
Let \(x \in A \cap (B \cup C)\). Then:
Since \(x \in B \cup C\), \(x\) is in either \(B\) or \(C\).
Case 1: If \(x \in B\), then \(x \in A\) and \(x \in B\), so \(x \in A \cap B\).
Case 2: If \(x \in C\), then \(x \in A\) and \(x \in C\), so \(x \in A \cap C\).
Thus in either case, \(x \in (A \cap B) \cup (A \cap C)\).
(2) Show that \((A \cap B) \cup (A \cap C) \subseteq A \cap (B \cup C)\)
Let \(x \in (A \cap B) \cup (A \cap C)\). Then:
In both cases, \(x \in A\). Also:
If \(x \in B\), then \(x \in B \cup C\).
If \(x \in C\), then \(x \in B \cup C\).
Thus \(x \in A\) and \(x \in B \cup C\), which means:
\[ x \in A \cap (B \cup C). \]
Hence, both sides are equal.
\[ A \cap (B \cup C) = (A \cap B) \cup (A \cap C). \]
Out of 100 students; 15 passed in English, 12 passed in Mathematics, 8 in Science, 6 in English and Mathematics, 7 in Mathematics and Science, 4 in English and Science, and 4 in all the three. Find how many passed:
Given:
1. English and Mathematics but not Science
\[ (E \cap M) - S = 6 - 4 = 2 \]
2. Mathematics and Science but not English
\[ (M \cap S) - E = 7 - 4 = 3 \]
3. Mathematics only
\[ M - (E \cap M) - (M \cap S) + (E \cap M \cap S) \]
\[ = 12 - 6 - 7 + 4 = 3 \]
4. More than one subject only
Sum of students in exactly two-subject combinations:
\[ (E \cap M - E \cap M \cap S) + (M \cap S - E \cap M \cap S) + (E \cap S - E \cap M \cap S) \]
\[ = (6-4) + (7-4) + (4-4) = 2 + 3 + 0 = 5 \]
Thus the number of students who passed more than one subject only = \(9\).
In a class of 60 students, 25 students play cricket and 20 students play tennis, and 10 students play both the games. Find the number of students who play neither.
Given:
Using the formula:
\[ |C \cup T| = |C| + |T| - |C \cap T| \]
\[ = 25 + 20 - 10 = 35 \]
Number who play neither:
\[ 60 - 35 = 25 \]
Thus, 25 students play neither game.
In a survey of 200 students of a school, it was found that 120 study Mathematics, 90 study Physics and 70 study Chemistry, 40 study Mathematics and Physics, 30 study Physics and Chemistry, 50 study Chemistry and Mathematics and 20 none of these subjects. Find the number of students who study all the three subjects.
Given:
Students studying at least one subject:
\[ 200 - 20 = 180 \]
Using the three-set union formula:
\[ |M \cup P \cup C| = M + P + C - (M \cap P) - (P \cap C) - (C \cap M) + (M \cap P \cap C) \]
Substituting:
\[ 180 = 120 + 90 + 70 - 40 - 30 - 50 + x \]
\[ 180 = 160 + x \]
\[ x = 20 \]
Thus, 20 students study all three subjects.
In a town of 10,000 families it was found that 40% families buy newspaper A, 20% families buy newspaper B, 10% families buy newspaper C, 5% families buy A and B, 3% buy B and C and 4% buy A and C. If 2% families buy all the three newspapers, find:
Convert percentages to numbers:
Total families = 10,000
1. Families buying A only
\[ |A| - (|A \cap B| + |A \cap C|) + |A \cap B \cap C| \]
\[ = 4000 - (500 + 400) + 200 = 3300 \]
2. Families buying none
Use union formula:
\[ |A \cup B \cup C| = A + B + C - (AB + BC + CA) + ABC \]
\[ = 4000 + 2000 + 1000 - (500 + 300 + 400) + 200 \]
\[ = 6000 - 1200 + 200 = 5000 \]
Thus families buying none:
\[ 10000 - 5000 = 4000 \]
In a group of 50 students, the number of students studying French, English, Sanskrit were found to be as follows:
French = 17, English = 13, Sanskrit = 15
French and English = 9, English and Sanskrit = 4, French and Sanskrit = 5
English, French and Sanskrit = 3
Find the number of students who study:
Given:
Compute the exact two-language intersections:
\(F \cap E \text{ only} = 9 - 3 = 6\)
\(E \cap S \text{ only} = 4 - 3 = 1\)
\(F \cap S \text{ only} = 5 - 3 = 2\)
1. French only
\[ 17 - (6 + 2 + 3) = 6 \]
2. English only
\[ 13 - (6 + 1 + 3) = 3 \]
3. Sanskrit only
\[ 15 - (2 + 1 + 3) = 9 \]
4. English and Sanskrit but not French
\[ 1 \]
5. French and Sanskrit but not English
\[ 2 \]
6. French and English but not Sanskrit
\[ 6 \]
7. At least one language
Sum all regions:
\[ 6 + 3 + 9 + 6 + 1 + 2 + 3 = 30 \]
8. None
\[ 50 - 30 = 20 \]