Chapter 3 Sets (Q & A)

A set is a well-defined collection of distinct objects. This means that for any given object, we can definitively determine whether it belongs to the collection or not.

Let's analyze each option:

(A) The collection of beautiful flowers in a garden.

Beauty is subjective. What one person finds beautiful, another may not. Therefore, this collection is not well-defined.

(B) The collection of honest people in your city.

Honesty can also be subjective and difficult to quantify definitively. There isn't a universally agreed-upon criterion to classify someone as "honest" or not, making this collection not well-defined.

(C) The collection of all even numbers less than 20.

This collection is well-defined. We can clearly identify all numbers that are even and are less than 20. The even numbers less than 20 are 0, 2, 4, 6, 8, 10, 12, 14, 16, and 18. For any number, we can determine if it meets these criteria.

(D) The collection of talented singers in India.

Talent is subjective and can be judged differently by different people. There are no objective criteria to define "talented singers," making this collection not well-defined.

Therefore, the correct answer is (C).

The question asks to represent the set A in roster form, where A is defined as the set of all prime numbers less than 15.

First, let's understand what a prime number is. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.

Now, let's identify the numbers less than 15:

$1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14$.

Next, we need to determine which of these numbers are prime:

• 1 is not a prime number because it is not greater than 1.
• 2 is a prime number (divisors are 1 and 2).
• 3 is a prime number (divisors are 1 and 3).
• 4 is not a prime number (divisors are 1, 2, and 4).
• 5 is a prime number (divisors are 1 and 5).
• 6 is not a prime number (divisors are 1, 2, 3, and 6).
• 7 is a prime number (divisors are 1 and 7).
• 8 is not a prime number (divisors are 1, 2, 4, and 8).
• 9 is not a prime number (divisors are 1, 3, and 9).
• 10 is not a prime number (divisors are 1, 2, 5, and 10).
• 11 is a prime number (divisors are 1 and 11).
• 12 is not a prime number (divisors are 1, 2, 3, 4, 6, and 12).
• 13 is a prime number (divisors are 1 and 13).
• 14 is not a prime number (divisors are 1, 2, 7, and 14).

The prime numbers less than 15 are 2, 3, 5, 7, 11, and 13.

Therefore, the set A in roster form is $\{2, 3, 5, 7, 11, 13\}$.

Comparing this with the given options:

• (A) $\{1, 2, 3, 5, 7, 11, 13\}$ - Incorrect, as 1 is not prime.
• (B) $\{2, 3, 5, 7, 11, 13\}$ - Correct.
• (C) $\{1, 3, 5, 7, 11, 13\}$ - Incorrect, as 1 is not prime.
• (D) $\{2, 3, 5, 7, 9, 11, 13\}$ - Incorrect, as 9 is not prime.

The correct option is (B).

The question asks to represent the set of all integers between -3 and 4, inclusive of -3 and 4, in roster form.

An integer is a whole number (not a fractional number) that can be positive, negative, or zero.

The phrase "between -3 and 4 (inclusive of -3 and 4)" means we need to list all integers starting from -3 and ending at 4, including both -3 and 4.

Let's list the integers that fit this description:

• Starting with -3: -3
• The next integer is: -2
• The next integer is: -1
• The next integer is: 0
• The next integer is: 1
• The next integer is: 2
• The next integer is: 3
• The next integer is: 4 (since it's inclusive)

So, the set of all integers between -3 and 4, inclusive, is: $\{-3, -2, -1, 0, 1, 2, 3, 4\}$.

Now, let's compare this with the given options:

• (A) $\{-2, -1, 0, 1, 2, 3\}$ - Incorrect, as it does not include -3 and 4.
• (B) $\{-3, -2, -1, 0, 1, 2, 3\}$ - Incorrect, as it does not include 4.
• (C) $\{-3, -2, -1, 0, 1, 2, 3, 4\}$ - Correct, as it includes all integers from -3 to 4.
• (D) $\{-2, -1, 0, 1, 2, 3, 4\}$ - Incorrect, as it does not include -3.

The correct option is (C).

The given set is $\{1, 4, 9, 16, 25, \dots\}$. We need to represent this set in set-builder form.

Let's examine the elements of the set:

• $1 = 1^2$
• $4 = 2^2$
• $9 = 3^2$
• $16 = 4^2$
• $25 = 5^2$

It appears that each element in the set is the square of a natural number.

The set-builder form describes the general form of the elements and the condition they must satisfy.

Let's analyze the given options:

(A) $\{x : x \text{ is a square number}\}$

This is a valid description, as all elements are square numbers. However, set-builder form usually aims for a more precise mathematical representation.

(B) $\{x^2 : x \in \mathbb{N}\}$

Here, $x \in \mathbb{N}$ means $x$ is a natural number. The set of natural numbers is typically $\{1, 2, 3, \dots\}$. So, the elements are $1^2, 2^2, 3^2, \dots$, which are $1, 4, 9, \dots$. This accurately represents the given set.

(C) $\{n^2 : n \in \mathbb{Z}\}$

Here, $n \in \mathbb{Z}$ means $n$ is an integer. Integers include negative numbers, zero, and positive numbers. If $n$ can be negative, e.g., $n=-1$, then $n^2 = (-1)^2 = 1$. If $n=0$, then $n^2 = 0^2 = 0$. The given set starts with 1 and does not include 0. While $n^2$ for negative integers will produce the same squares as positive integers, the inclusion of $0^2 = 0$ makes this representation include an element not in the given set.

(D) $\{n^2 : n \in \mathbb{N}, n \geq 1\}$

This option is essentially the same as option (B) because the definition of natural numbers ($\mathbb{N}$) often starts from 1. If $\mathbb{N}$ is defined as $\{1, 2, 3, \dots\}$, then the condition $n \geq 1$ is redundant. However, if $\mathbb{N}$ were defined to include 0 in some contexts, then this condition would be necessary. Given the common convention for $\mathbb{N}$, both (B) and (D) are very similar.

Comparing (B) and (D), option (B) is more concise and directly states that the elements are squares of natural numbers. Option (D) explicitly adds a condition that is often implied by the definition of natural numbers starting from 1.

In most mathematical contexts, natural numbers are taken to be $\{1, 2, 3, \dots\}$. Therefore, $\{x^2 : x \in \mathbb{N}\}$ correctly generates the sequence $1^2, 2^2, 3^2, \dots$ which matches the given set.

Option (A) is a descriptive form, while (B), (C), and (D) are more formal set-builder notations. Between (B) and (D), (B) is the most standard and direct representation.

The most appropriate set-builder form is (B).

An empty set, also known as a null set, is a set that contains no elements. It is denoted by $\emptyset$ or {}. We need to find which of the given options represents a collection with no elements.

Let's analyze each option:

(A) The set of all even prime numbers.

A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. The even numbers are numbers divisible by 2. The only even prime number is 2. Therefore, this set is $\{2\}$, which is not an empty set.

(B) The set of all integers greater than 5 and less than 6.

Integers are whole numbers. We are looking for an integer $x$ such that $5 < x < 6$. There is no integer that is strictly greater than 5 and strictly less than 6. Therefore, this set contains no elements, making it an empty set.

(C) The set of all months in a year starting with the letter 'J'.

The months of the year are January, February, March, April, May, June, July, August, September, October, November, December. The months starting with the letter 'J' are January, June, and July. Therefore, this set is \{January, June, July\}, which is not an empty set.

(D) The set of all squares of odd integers.

Odd integers are integers that are not divisible by 2 (e.g., $\dots, -3, -1, 1, 3, \dots$). When we square an odd integer, we get a positive number. For example:

• $(-3)^2 = 9$
• $(-1)^2 = 1$
• $1^2 = 1$
• $3^2 = 9$

The squares of odd integers are $\{1, 9, 25, 49, \dots\}$. This set contains elements and is therefore not an empty set.

Based on the analysis, option (B) is the only example of an empty set.

A singleton set has exactly one element.

(A) $\{0\}$ contains one element (0). This is a singleton set.

(B) $\emptyset$ is the empty set, containing zero elements. Not a singleton.

(C) $\{x : x \in \mathbb{R}, x^2 = 4\}$ represents the set $\{-2, 2\}$. It has two elements. Not a singleton.

(D) $\{x : x \in \mathbb{N}, x < 2\}$ represents the set of natural numbers less than 2. Assuming $\mathbb{N} = \{1, 2, 3, \dots\}$, this set is $\{1\}$. It has one element. This is a singleton set.

Both (A) and (D) are singleton sets. However, if only one answer is correct, option (A) is the most direct and unambiguous representation of a singleton set.

The answer is (A).

Two sets are considered equal if and only if they contain exactly the same elements. The order in which the elements are listed in a set does not affect its equality.

Let's analyze the given sets:

Set $A = \{1, 2, 3\}$. The elements of set A are 1, 2, and 3.

Set $B = \{3, 1, 2\}$. The elements of set B are 3, 1, and 2.

Now, let's compare the elements of set A and set B:

• Does set A contain 1? Yes. Does set B contain 1? Yes.
• Does set A contain 2? Yes. Does set B contain 2? Yes.
• Does set A contain 3? Yes. Does set B contain 3? Yes.

Since every element in set A is also an element in set B, and every element in set B is also an element in set A, the sets contain exactly the same elements.

Let's evaluate the options:

(A) Yes, because they have the same number of elements.

While it's true that they have the same number of elements (both have 3 elements), this is not the *definition* of set equality. Sets can have the same number of elements but not be equal (e.g., $\{1, 2\}$ and $\{3, 4\}$).

(B) Yes, because they contain exactly the same elements.

This statement accurately reflects the definition of set equality. Since both sets have the elements 1, 2, and 3, they are equal.

(C) No, because the order of elements is different.

This is incorrect. The order of elements in a set does not matter for equality.

(D) No, because they are just equivalent sets.

The term "equivalent sets" in set theory usually refers to sets that have the same cardinality (the same number of elements), but not necessarily the same elements. While these sets are equivalent in terms of cardinality, the question is about equality, and they *are* equal sets, not merely equivalent.

Therefore, the correct reason why sets A and B are equal is that they contain exactly the same elements.

Two sets are called equivalent if they have the same number of elements (i.e., the same cardinality). The elements themselves do not need to be the same.

Let's analyze the given sets:

Set $A = \{a, b, c\}$.

Set $B = \{1, 2, 3\}$.

First, let's find the number of elements in each set. This is denoted by $n(A)$ and $n(B)$.

For set A, the elements are a, b, and c. So, the number of elements in A is $n(A) = 3$.

For set B, the elements are 1, 2, and 3. So, the number of elements in B is $n(B) = 3$.

Since $n(A) = 3$ and $n(B) = 3$, we have $n(A) = n(B)$.

Now let's evaluate the options:

(A) Yes, because they contain exactly the same elements.

This is incorrect. The elements of set A are letters $\{a, b, c\}$, while the elements of set B are numbers $\{1, 2, 3\}$. They do not contain exactly the same elements, so they are not equal sets.

(B) Yes, because $\text{n}(A) = \text{n}(B)$.

This statement is correct. Since both sets have 3 elements, they are equivalent sets.

(C) No, because the elements are different.

While the elements are indeed different, this is the reason why they are *not equal*, not why they are *not equivalent*. Equivalence depends on the number of elements, not the identity of the elements.

(D) No, because they are not equal sets.

It is true that they are not equal sets, but the question asks if they are equivalent. The fact that they are not equal does not automatically mean they are not equivalent. In this case, they are not equal but *are* equivalent.

Therefore, sets A and B are equivalent because they have the same number of elements.

An infinite set is a set that contains an unending number of elements. In contrast, a finite set has a countable number of elements, meaning we can count them and reach a final number.

Let's analyze each option:

(A) The set of all states in India.

India has a fixed number of states. Currently, there are 28 states. This is a countable and finite number. Therefore, this is a finite set.

(B) The set of all positive integers.

The set of positive integers is $\{1, 2, 3, 4, 5, \dots\}$. This sequence continues indefinitely. There is no largest positive integer, and we can always find a next integer by adding 1. Therefore, this set has an unending number of elements and is an infinite set.

(C) The set of all solutions to $x^2 - 4 = 0$.

To find the solutions, we solve the equation: $x^2 = 4$. Taking the square root of both sides gives $x = \pm 2$. So, the solutions are $x = 2$ and $x = -2$. This set is $\{2, -2\}$, which contains exactly two elements. Therefore, this is a finite set.

(D) The set of all days in a week.

A week has a fixed number of days, which is 7. Therefore, this is a finite set.

Based on the analysis, the set of all positive integers is an infinite set.

Let's analyze the Assertion (A) and the Reason (R) separately.

Assertion (A): The set of all rivers in India is a finite set.

A finite set is a set whose elements can be counted and the counting process terminates. While the exact definition of "river" can be debated (e.g., are very small streams rivers?), in a practical sense, the number of distinct geographical features that are recognized as rivers in India is countable and has a limit. We can list them and count them. Therefore, the assertion that the set of all rivers in India is a finite set is considered true in the context of set theory's practical application to real-world objects.

Reason (R): You can count the number of major rivers in India.

The reason states that one can count the number of *major* rivers. This is true. The term "major" implies a classification that makes them countable. If the set of major rivers is finite, it supports the idea that the set of all rivers (which would include major rivers and potentially others) is also finite, as the total number cannot exceed the number of major ones plus any minor ones.

Now, let's determine the relationship between A and R.

The reason (R) provides a justification for the assertion (A). If you can count the number of major rivers, it implies that the collection of rivers is not infinitely vast. The ability to count (even if it's only the major ones) is the basis for classifying a set as finite. Therefore, Reason (R) correctly explains why Assertion (A) is true.

Let's evaluate the options based on this understanding:

(A) Both A and R are true and R is the correct explanation of A.

This option aligns with our analysis. Both statements are true, and R provides a valid reason for A.

(B) Both A and R are true but R is not the correct explanation of A.

This is incorrect because R does explain A.

(C) A is true but R is false.

This is incorrect because R is true.

(D) A is false but R is true.

This is incorrect because A is true.

Therefore, both the assertion and the reason are true, and the reason correctly explains the assertion.

A set is a well-defined collection. The issue here is whether "most popular" subjects are well-defined from the survey results.

In a survey context, "most popular" is usually determined by objective criteria from the data (e.g., subjects with the highest number of votes). If such a criterion is applied, the collection is well-defined.

(A) Subjects are clearly defined categories. If the criteria for "most popular" are also clearly defined based on survey data, then it's a set.

(B) "Most popular" *can be* subjective in general, but in a survey, it's typically made objective.

(C) Finiteness does not guarantee a set.

(D) Assuming a standard survey interpretation, it can be determined.

Assuming "most popular" is determined by objective criteria from the survey data, the collection is well-defined and thus a set.

The answer is (A).

We need to match each set description with its correct type.

Let's analyze each description:

(i) Set of all even numbers

The set of all even numbers is $\{\dots, -4, -2, 0, 2, 4, \dots\}$. This set continues infinitely in both positive and negative directions. Therefore, it is an infinite set.

Matching: (i) - (b)

(ii) Set of all integers between 1 and 2

Integers are whole numbers. There are no integers strictly between 1 and 2. Thus, this collection has no elements. It is an empty set.

Matching: (ii) - (a)

(iii) Set of all even prime numbers

A prime number is a number greater than 1 with only two divisors: 1 and itself. An even number is divisible by 2. The only number that is both even and prime is 2. Therefore, this set is $\{2\}$. A set with exactly one element is a singleton set.

Matching: (iii) - (c)

(iv) Set of all planets in our solar system

Our solar system has a fixed and countable number of planets (currently 8). Therefore, this is a finite set.

Matching: (iv) - (d)

Combining the matches:

(i) - (b)

(ii) - (a)

(iii) - (c)

(iv) - (d)

Comparing this with the given options, option (A) matches our findings.

The correct matching is (A).

We are given the set $A = \{1, \{2\}, 3\}$. We need to determine which of the given statements is correct based on the definition of set membership ($\in$) and subset ($\subseteq$).

The elements of set A are:

• The number 1
• The set $\{2\}$
• The number 3

Let's analyze each option:

(A) $2 \in A$

This statement claims that the number 2 is an element of set A. However, the elements of A are 1, $\{2\}$, and 3. The number 2 itself is not listed as an element of A. The element $\{2\}$ is a set containing the number 2, which is different from the number 2 being an element of A.

Therefore, $2 \notin A$. This statement is false.

(B) $\{2\} \in A$

This statement claims that the set containing the number 2 (i.e., $\{2\}$) is an element of set A. Looking at the definition of A, $A = \{1, \{2\}, 3\}$, we can see that $\{2\}$ is indeed listed as one of the elements of A.

Therefore, this statement is true.

(C) $\{1\} \in A$

This statement claims that the set containing the number 1 (i.e., $\{1\}$) is an element of set A. The elements of A are 1, $\{2\}$, and 3. The element 1 is in A, but the set $\{1\}$ is not listed as an element of A.

Therefore, $\{1\} \notin A$. This statement is false.

(D) $3 \subseteq A$

The symbol $\subseteq$ denotes a subset. A set X is a subset of a set Y if every element of X is also an element of Y. This statement claims that the number 3 is a subset of set A. However, a single number cannot be a subset of a set; only sets can be subsets of other sets. To be a subset, 3 would need to be a set itself, and all its elements would need to be in A. The correct way to express the relationship of 3 with A is membership, i.e., $3 \in A$.

Therefore, this statement is false.

Based on the analysis, the only correct statement is (B).

To find the number of subsets of a set, we use the formula: If a set has $n$ elements, then it has $2^n$ subsets.

The given set is $A = \{a, b\}$.

First, we need to determine the number of elements in set A. The elements are 'a' and 'b'.

So, the number of elements in set A is $n = 2$.

Now, we can use the formula to find the number of subsets:

Number of subsets = $2^n$

Substitute $n=2$ into the formula:

Number of subsets = $2^2$

Let's list the subsets to verify:

• The empty set: $\emptyset$
• Subsets with one element: $\{a\}$, $\{b\}$
• Subsets with two elements: $\{a, b\}$

The total number of subsets is $1 + 2 + 1 = 4$.

Comparing this result with the given options:

• (A) 2
• (B) 3
• (C) 4
• (D) 8

Our calculated number of subsets is 4.

Therefore, the correct option is (C).

The number of subsets of a set with $n$ elements is $2^n$.

A proper subset of a set A is any subset of A except for the set A itself.

We are given that the number of elements in set A is $n(A) = 5$.

First, let's find the total number of subsets of A:

Total number of subsets = $2^{n(A)}$

So, set A has 32 subsets in total.

Now, to find the number of proper subsets, we subtract the set A itself from the total number of subsets.

Number of proper subsets = (Total number of subsets) - 1

Number of proper subsets = $32 - 1$

Comparing this result with the given options:

• (A) 32
• (B) 31
• (C) 16
• (D) 25

The calculated number of proper subsets is 31.

Therefore, the correct option is (B).

A set $B$ is a subset of set $A$ if all elements of $B$ are also elements of $A$. We are given set $A = \{1, 2, 3, 4\}$.

Let's examine each option to see if it meets the definition of a subset of A:

(A) $\{1, 2, 5\}$

For this to be a subset of A, all its elements (1, 2, and 5) must be in A. While 1 and 2 are in A, the element 5 is not in A. Therefore, $\{1, 2, 5\}$ is not a subset of A.

(B) $\{2, 4\}$

For this to be a subset of A, all its elements (2 and 4) must be in A. The element 2 is in A, and the element 4 is also in A. Since all elements of $\{2, 4\}$ are in A, $\{2, 4\}$ is a subset of A.

(C) $\{0\}$

For this to be a subset of A, all its elements (0) must be in A. The element 0 is not in A. Therefore, $\{0\}$ is not a subset of A.

(D) $\{1, 2, 3, 4, 5\}$

For this to be a subset of A, all its elements (1, 2, 3, 4, and 5) must be in A. While 1, 2, 3, and 4 are in A, the element 5 is not in A. Therefore, $\{1, 2, 3, 4, 5\}$ is not a subset of A. (In fact, it is a superset if it contains A, but here it contains an element not in A.)

The only option that represents a subset of A is (B).

A set $B$ is a subset of a set $A$ if every element of $B$ is also an element of $A$. We need to find which of the given options is a subset of *any* set.

Let's analyze each option:

(A) The set itself

Every set is a subset of itself. For any set $A$, it is true that $A \subseteq A$. This is because every element in $A$ is indeed an element of $A$. So, this is always a subset.

(B) $\emptyset$

The empty set, $\emptyset$, is the set that contains no elements. The condition for being a subset is that every element of the subset must be in the larger set. Since the empty set has no elements, this condition is vacuously true for any set. There are no elements in $\emptyset$ that are not in any given set $A$. Therefore, the empty set is always a subset of every set, including itself.

(C) $\{0\}$

This is a set containing only the element 0. For $\{0\}$ to be a subset of every set, 0 must be an element of every set. This is not true. For example, if we consider the set $A = \{1, 2, 3\}$, then $0 \notin A$, so $\{0\}$ is not a subset of $A$. Thus, this is not always a subset.

(D) The universal set

The universal set, usually denoted by $U$, is the set of all possible elements relevant to a particular context. By definition, every element in any set within that context belongs to the universal set. Therefore, every set is a subset of the universal set ($A \subseteq U$). However, the question asks what is a subset of *every* set. The universal set is a superset of every set, but not necessarily a subset of every set. For instance, if $A$ is the universal set, then $A \subseteq A$ is true. But if we have another set $B$ that is *not* the universal set, then the universal set is not a subset of $B$ (unless $B$ is also the universal set). So, the universal set is not always a subset of *every* set.

We have identified two options that are always subsets of every set: (A) The set itself, and (B) $\emptyset$. However, the question asks for *the* subset that is *always* a subset of every set. This phrasing usually points to the empty set as the unique answer that fits this property universally across all contexts, without needing a specific set to be defined first.

The empty set is a subset of every set. Every set is a subset of itself. When faced with such a question, the empty set is the more fundamental answer for "always a subset of every set" in a general context.

Let's consider common conventions in set theory questions. The empty set is universally acknowledged as a subset of all sets.

The most universally correct answer is the empty set.

The notation $A \subseteq B$ means that set A is a subset of set B.

By the definition of a subset, if $A \subseteq B$, then every element of set A must also be an element of set B.

Let's analyze each statement:

(A) Every element of B is an element of A.

This statement means that $B \subseteq A$. If $A \subseteq B$ and $B \subseteq A$, then $A = B$. However, $A \subseteq B$ does not necessarily imply $B \subseteq A$. For example, if $A = \{1\}$ and $B = \{1, 2\}$, then $A \subseteq B$, but $B$ is not a subset of $A$ (because $2 \in B$ but $2 \notin A$). So, this statement is not always true.

(B) Every element of A is an element of B.

This is the definition of $A \subseteq B$. If A is a subset of B, it means that all elements found in A are also present in B. This statement must be true.

(C) A and B are equal sets.

For A and B to be equal, we need both $A \subseteq B$ and $B \subseteq A$. The condition $A \subseteq B$ alone does not guarantee $B \subseteq A$. For instance, $A = \{1\}$ and $B = \{1, 2\}$ satisfy $A \subseteq B$, but $A \neq B$. So, this statement is not always true.

(D) B is an empty set.

If $A \subseteq B$, it is possible for B to be empty, but only if A is also empty. For example, if $A = \emptyset$ and $B = \emptyset$, then $A \subseteq B$ is true, and B is empty. However, $A \subseteq B$ can also be true when B is not empty. For example, if $A = \{1\}$ and $B = \{1, 2\}$, then $A \subseteq B$ is true, but B is not empty. So, this statement is not always true.

The statement that must always be true if $A \subseteq B$ is that every element of A is an element of B.

Let's analyze the Assertion (A) and the Reason (R) separately.

Assertion (A): If A has $n$ elements, the number of subsets of A is $2^n$.

This is a fundamental theorem in set theory. For a set with $n$ distinct elements, the total number of possible subsets (including the empty set and the set itself) is indeed $2^n$. This assertion is true.

Reason (R): Each element of A can either be included or excluded in a subset.

To form any subset of a given set A, we consider each element of A independently. For each element, there are two possibilities: either the element is included in the subset, or it is excluded from the subset. If set A has $n$ elements, and for each element there are 2 choices, the total number of ways to form a subset is the product of the number of choices for each element, which is $2 \times 2 \times \dots \times 2$ ($n$ times), resulting in $2^n$. This statement accurately describes the process of forming subsets and the basis for the formula. This reason is true.

Now, let's determine if Reason (R) correctly explains Assertion (A).

Reason (R) explains *why* the number of subsets is $2^n$. The fact that each of the $n$ elements has two independent choices (include or exclude) leads directly to the total number of combinations being $2^n$. Therefore, Reason (R) provides the correct justification for Assertion (A).

Evaluating the options:

(A) Both A and R are true and R is the correct explanation of A.

This option accurately reflects our analysis.

(B) Both A and R are true but R is not the correct explanation of A.

This is incorrect, as R is indeed the explanation for A.

(C) A is true but R is false.

This is incorrect, as R is true.

(D) A is false but R is true.

This is incorrect, as A is true.

Therefore, both the assertion and the reason are true, and the reason correctly explains the assertion.

Let's define the sets based on the case study:

• Set I: The set of all inspected items.
• Set D: The set of defective items in the batch.
• Set N: The set of non-defective items in the batch.

We know that a batch of 100 items was inspected. This means that all items considered in this scenario are part of the inspected batch.

Consider the relationship between Set I and the batch of items.

Since Set I is defined as "the set of all inspected items," and the entire batch of 100 items was inspected, every item in the batch is an inspected item. Thus, Set I represents all 100 items in the batch.

Now consider the relationship between the items in the batch, defective items (D), and non-defective items (N).

Any item in the batch is either defective or non-defective. This means that the set of all items in the batch (which is Set I) is composed of the union of defective items and non-defective items. Also, a defective item cannot be non-defective, and vice-versa, so D and N are disjoint sets.

Therefore, the set of all inspected items (I) contains all the defective items (D) and all the non-defective items (N).

This means that D is a part of I, and N is also a part of I.

In terms of set notation:

• Every defective item is an inspected item, so $D \subseteq I$.
• Every non-defective item is an inspected item, so $N \subseteq I$.
• Also, $I = D \cup N$ and $D \cap N = \emptyset$.

Let's evaluate the given options based on these relationships:

(A) $I \subseteq D$

This would mean that every inspected item is defective. This is incorrect, as there are also non-defective items.

(B) $D \subseteq I$

This means that every defective item is an inspected item. Since the batch of 100 items was inspected, and defective items are part of this batch, this statement is correct.

(C) $N \subseteq D$

This would mean that every non-defective item is also a defective item, which is a contradiction. Non-defective items are distinct from defective items. So, this statement is incorrect.

(D) $I \subseteq N$

This would mean that every inspected item is non-defective. This is incorrect, as there are also defective items among the inspected items.

The only correct relationship based on the given information is that the set of defective items (D) is a subset of the set of all inspected items (I).

From the previous question and the case study setup:

• Set I: The set of all inspected items (the entire batch of 100 items). This acts as our universal set for this context.
• Set D: The set of defective items.
• Set N: The set of non-defective items.

We established that $D \subseteq I$ and $N \subseteq I$.

Now let's analyze the relationships between D and N within I.

Consider statement (A): $D = N'$ (Complement of N)

The complement of set N, denoted as $N'$, consists of all elements in the universal set I that are *not* in N. In this context, $N'$ would be the set of all inspected items that are not non-defective. Since an item is either defective or non-defective, the items that are not non-defective must be the defective ones. Therefore, the complement of N with respect to I is precisely the set of defective items, D.

$N' = \{x \in I \mid x \notin N\}$. Since $I = D \cup N$ and $D \cap N = \emptyset$, if $x \in I$ and $x \notin N$, then $x$ must be in $D$. So, $N' = D$. This statement is true.

Consider statement (B): $D \cup N = I$ and $D \cap N = \emptyset$

$D \cup N = I$: This means that the union of the set of defective items and the set of non-defective items is equal to the set of all inspected items. This is true because every inspected item is either defective or non-defective, and these two categories cover all items in the batch.

$D \cap N = \emptyset$: This means that the intersection of the set of defective items and the set of non-defective items is empty. This is true because an item cannot be both defective and non-defective simultaneously; these are mutually exclusive categories.

Therefore, this statement is also true.

Consider option (C): Both (A) and (B)

Since both statement (A) and statement (B) are true based on our analysis, this option is the most comprehensive and correct answer.

Consider option (D): Neither (A) nor (B)

This is incorrect because both (A) and (B) are true.

Thus, both relationships correctly describe the relationship between sets D and N within the universal set I.

We need to match each interval notation with its corresponding set-builder notation.

(i) $(a, b]$

This interval notation represents all real numbers $x$ such that $x$ is strictly greater than $a$ and less than or equal to $b$. In set-builder notation, this is written as $\{x \in \mathbb{R} : a < x \leq b\}$. This matches with option (d).

Matching: (i) - (d)

(ii) $[a, b)$

This interval notation represents all real numbers $x$ such that $x$ is greater than or equal to $a$ and strictly less than $b$. In set-builder notation, this is written as $\{x \in \mathbb{R} : a \leq x < b\}$. This matches with option (b).

Matching: (ii) - (b)

(iii) $(-\infty, b)$

This interval notation represents all real numbers $x$ such that $x$ is strictly less than $b$. The notation $-\infty$ indicates that the interval extends infinitely to the left. In set-builder notation, this is written as $\{x \in \mathbb{R} : x < b\}$. This matches with option (a).

Matching: (iii) - (a)

(iv) $[a, \infty)$

This interval notation represents all real numbers $x$ such that $x$ is greater than or equal to $a$. The notation $\infty$ indicates that the interval extends infinitely to the right. In set-builder notation, this is written as $\{x \in \mathbb{R} : a \leq x\}$. This matches with option (c).

Matching: (iv) - (c)

Combining the matches:

(i) - (d)

(ii) - (b)

(iii) - (a)

(iv) - (c)

Let's compare this with the given options:

(A) (i)-(d), (ii)-(b), (iii)-(a), (iv)-(c)

(B) (i)-(b), (ii)-(d), (iii)-(a), (iv)-(c)

(C) (i)-(d), (ii)-(b), (iii)-(c), (iv)-(a)

(D) (i)-(c), (ii)-(a), (iii)-(d), (iv)-(b)

Our combined matches perfectly align with option (A).

Therefore, the correct option is (A).

The interval notation $[-5, 2)$ represents a set of real numbers.

The square bracket `[` before -5 indicates that -5 is included in the interval (i.e., $-5$ is greater than or equal to -5).

The parenthesis `)` after 2 indicates that 2 is not included in the interval (i.e., $2$ is strictly greater than the numbers in the interval).

Therefore, the interval $[-5, 2)$ includes all real numbers $x$ such that $x$ is greater than or equal to -5 and strictly less than 2.

In mathematical inequality notation, this is written as: $-5 \leq x < 2$.

To represent this in set-builder form, we write:

$\{x \in \mathbb{R} : -5 \leq x < 2\}$

Here, $x \in \mathbb{R}$ signifies that $x$ is a real number.

Let's compare this with the given options:

• (A) $\{x \in \mathbb{R} : -5 < x < 2\}$: This corresponds to the interval $(-5, 2)$. Incorrect.
• (B) $\{x \in \mathbb{R} : -5 \leq x \leq 2\}$: This corresponds to the interval $[-5, 2]$. Incorrect.
• (C) $\{x \in \mathbb{R} : -5 \leq x < 2\}$: This corresponds to the interval $[-5, 2)$. Correct.
• (D) $\{x \in \mathbb{R} : -5 < x \leq 2\}$: This corresponds to the interval $(-5, 2]$. Incorrect.

Thus, the correct set-builder form for the interval $[-5, 2)$ is $\{x \in \mathbb{R} : -5 \leq x < 2\}$.

The interval notation $(-\infty, \infty)$ represents a range of real numbers.

The symbol $-\infty$ indicates that the interval extends infinitely to the left, meaning it includes all numbers less than any finite value.

The symbol $\infty$ indicates that the interval extends infinitely to the right, meaning it includes all numbers greater than any finite value.

The parentheses on both ends $(-\infty, \infty)$ signify that neither $-\infty$ nor $\infty$ are included as actual numbers in the set (as they are not numbers but concepts of unboundedness).

Therefore, the interval $(-\infty, \infty)$ encompasses all possible real numbers.

Let's analyze the options:

• (A) The set of all positive real numbers: This would be represented by the interval $(0, \infty)$. Incorrect.
• (B) The set of all negative real numbers: This would be represented by the interval $(-\infty, 0)$. Incorrect.
• (C) The set of all real numbers: This perfectly matches the description of $(-\infty, \infty)$. Correct.
• (D) The empty set: The empty set has no elements, whereas $(-\infty, \infty)$ represents an infinite collection of numbers. Incorrect.

Thus, the interval $(-\infty, \infty)$ represents the set of all real numbers.

In mathematics, an interval denoted by $(a, b)$ is called an open interval.

An open interval $(a, b)$ represents all real numbers that are strictly greater than $a$ and strictly less than $b$.

This means that the endpoints $a$ and $b$ are not included in the interval.

Option (A) is incorrect because it states that the endpoints $a$ and $b$ are included.

Option (B) is incorrect because it states that the endpoints $a$ and $b$ are included.

Option (D) is incorrect because intervals represent real numbers, not just integers.

Therefore, the interval $(a, b)$ includes all numbers between $a$ and $b$, excluding $a$ and $b$.

The correct option is (C).

The given set is $\{x \in \mathbb{R} : x > 10\}$.

This set represents all real numbers $x$ that are strictly greater than 10.

In interval notation:

• Numbers strictly greater than 10 are represented using an open interval with the left endpoint as 10. This is denoted by $(10, \dots$.
• Since there is no upper bound specified, it means all numbers greater than 10 are included, extending to positive infinity. Positive infinity is always represented with an open interval. This is denoted by $\dots, \infty)$.

Combining these, the interval notation for $x > 10$ is $(10, \infty)$.

Let's analyze the given options:

(A) $[10, \infty)$ represents all real numbers greater than or equal to 10. The square bracket '[' indicates that 10 is included.

(B) $(10, \infty)$ represents all real numbers strictly greater than 10. The parenthesis '(' indicates that 10 is excluded.

(C) $(-\infty, 10]$ represents all real numbers less than or equal to 10.

(D) $(-\infty, 10)$ represents all real numbers strictly less than 10.

Therefore, the correct interval notation for the set $\{x \in \mathbb{R} : x > 10\}$ is $(10, \infty)$.

The correct option is (B).

Assertion (A): The interval $[2, 5]$ represents all real numbers $x$ such that $2 \leq x \leq 5$.

In interval notation, a square bracket '[' at an endpoint signifies that the endpoint is included in the interval. The notation $[a, b]$ denotes the set of all real numbers $x$ such that $a \leq x \leq b$. In this case, for $[2, 5]$, it means all real numbers $x$ where $2 \leq x \leq 5$. Thus, Assertion (A) is true.

Reason (R): Square brackets indicate that the endpoints are included in the interval.

This statement correctly defines the convention for using square brackets in interval notation. Square brackets signify inclusion of the endpoint, while parentheses '(' or ')' signify exclusion.

Thus, Reason (R) is true.

Now we need to determine if Reason (R) is the correct explanation for Assertion (A).

Assertion (A) states what the interval $[2, 5]$ represents. Reason (R) explains the meaning of the square brackets used in this interval notation. The reason directly supports and explains why the interval $[2, 5]$ corresponds to the inequality $2 \leq x \leq 5$. The inclusion of the endpoints as specified in the inequality is precisely due to the use of square brackets.

Therefore, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

The correct option is (A).

The given interval for the company's profit P is $(10, 50]$.

Let's break down the meaning of this interval:

• The parenthesis '(' before 10 indicates that 10 is not included in the interval. This means the profit P is strictly greater than 10 lakhs of $\textsf{₹}$ ($P > 10$).
• The square bracket ']' after 50 indicates that 50 is included in the interval. This means the profit P is less than or equal to 50 lakhs of $\textsf{₹}$ ($P \leq 50$).

Combining these, the interval $(10, 50]$ represents all profits P such that $10 < P \leq 50$ (in lakhs of $\textsf{₹}$).

Now let's evaluate each option:

(A) The minimum profit is $\textsf{₹}\,10$ lakhs.

This is incorrect because the interval starts with an open parenthesis $(10$, meaning the profit must be strictly greater than 10. There is no specific minimum value within the interval, but values can get arbitrarily close to 10.

(B) The maximum profit is $\textsf{₹}\,50$ lakhs.

This is correct because the interval ends with a closed square bracket $]$, meaning 50 is included, and no value greater than 50 is allowed. Therefore, 50 is the maximum possible profit.

(C) The profit can be exactly $\textsf{₹}\,10$ lakhs.

This is incorrect because the open parenthesis $(10$ at the beginning of the interval means that 10 is excluded.

(D) The profit cannot be $\textsf{₹}\,50$ lakhs.

This is incorrect because the closed square bracket $]$ at the end of the interval means that 50 is included.

Therefore, the only correct statement is that the maximum profit is $\textsf{₹}\,50$ lakhs.

The correct option is (B).

In mathematics, intervals are classified based on whether their endpoints are included or excluded.

The notation $(a, b)$ signifies an interval where both endpoints, $a$ and $b$, are excluded.

An interval where both endpoints are excluded is called an open interval.

Let's look at the other options to confirm:

(A) A closed interval is denoted by $[a, b]$ and includes both endpoints ($a \leq x \leq b$).

(C) A semi-closed or half-open interval is one where one endpoint is included and the other is excluded, such as $[a, b)$ or $(a, b]$.

(D) An infinite interval involves infinity, such as $(a, \infty)$ or $(-\infty, b]$. While $(a, b)$ is a type of interval, its classification is not based on whether it's infinite or not.

Therefore, the interval $(a, b)$ is also known as an open interval.

The correct option is (B).

Given the sets $A = \{1, 2, 3\}$ and $B = \{3, 4, 5\}$.

The union of two sets, denoted by $A \cup B$, is the set containing all the elements that are in either set A or set B (or both).

To find $A \cup B$, we combine all the elements from set A and set B, ensuring that each unique element is listed only once.

Elements in set A are: 1, 2, 3.

Elements in set B are: 3, 4, 5.

Combining these elements, we get: 1, 2, 3 (from A) and 3, 4, 5 (from B).

The element '3' is present in both sets. When forming the union, we list it only once.

So, $A \cup B = \{1, 2, 3, 4, 5\}$.

Comparing this result with the given options:

(A) $\{1, 2, 3\}$ is just set A.

(B) $\{3\}$ is the intersection of A and B ($A \cap B$).

(C) $\{1, 2, 3, 4, 5\}$ matches our calculated union.

(D) $\{1, 2, 4, 5\}$ is incorrect as it omits the common element 3.

Therefore, the correct answer is $\{1, 2, 3, 4, 5\}$.

The correct option is (C).

Given the sets $A = \{1, 2, 3\}$ and $B = \{3, 4, 5\}$.

The intersection of two sets, denoted by $A \cap B$, is the set containing all the elements that are common to both set A and set B.

To find $A \cap B$, we need to identify the elements that appear in both set A and set B.

Elements in set A: 1, 2, 3.

Elements in set B: 3, 4, 5.

By comparing the elements of both sets, we can see that the only element present in both A and B is 3.

Therefore, $A \cap B = \{3\}$.

Let's evaluate the given options:

(A) $\{1, 2, 3, 4, 5\}$ is the union of A and B ($A \cup B$).

(B) $\{3\}$ matches our calculated intersection.

(C) $\emptyset$ (the empty set) would be the result if there were no common elements.

(D) $\{1, 2, 4, 5\}$ is neither the union nor the intersection.

Therefore, the correct answer is $\{3\}$.

The correct option is (B).

Given the sets $A = \{1, 2, 3, 4, 5\}$ and $B = \{4, 5, 6, 7\}$.

The set difference $A - B$ (read as "A minus B" or "A set difference B") contains all the elements that are in set A but not in set B.

To find $A - B$, we take all the elements of set A and remove any elements that are also present in set B.

Elements in set A are: 1, 2, 3, 4, 5.

Elements in set B are: 4, 5, 6, 7.

We look at each element in A and check if it is in B:

• 1 is in A, but not in B. So, 1 is in $A - B$.
• 2 is in A, but not in B. So, 2 is in $A - B$.
• 3 is in A, but not in B. So, 3 is in $A - B$.
• 4 is in A, and also in B. So, 4 is removed from $A - B$.
• 5 is in A, and also in B. So, 5 is removed from $A - B$.

Therefore, the elements remaining in A after removing the elements of B are 1, 2, and 3.

So, $A - B = \{1, 2, 3\}$.

Let's check the given options:

(A) $\{1, 2, 3\}$ matches our calculated result.

(B) $\{6, 7\}$ represents the elements in B that are not in A ($B - A$).

(C) $\{4, 5\}$ represents the intersection of A and B ($A \cap B$).

(D) $\{1, 2, 3, 6, 7\}$ is incorrect; it combines elements only in A with elements only in B.

The correct answer is $\{1, 2, 3\}$.

The correct option is (A).

Given the universal set $\text{U} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ and the set $A = \{1, 3, 5, 7, 9\}$.

The complement of a set A, denoted by $A'$ (or sometimes $A^c$ or $\bar{A}$), is the set of all elements in the universal set U that are not in A.

Mathematically, $A' = \{x \in \text{U} \mid x \notin A\}$.

We need to find the elements present in U but not in A.

Elements in U are: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.

Elements in A are: 1, 3, 5, 7, 9.

Let's go through each element of U and see if it is in A:

• 1 is in U and in A.
• 2 is in U but not in A. So, 2 is in $A'$.
• 3 is in U and in A.
• 4 is in U but not in A. So, 4 is in $A'$.
• 5 is in U and in A.
• 6 is in U but not in A. So, 6 is in $A'$.
• 7 is in U and in A.
• 8 is in U but not in A. So, 8 is in $A'$.
• 9 is in U and in A.
• 10 is in U but not in A. So, 10 is in $A'$.

Therefore, the complement of A, $A'$, is the set of elements from U that are not in A.

$A' = \{2, 4, 6, 8, 10\}$.

Let's compare this with the given options:

(A) $\{2, 4, 6, 8, 10\}$ matches our result.

(B) $\{1, 3, 5, 7, 9\}$ is the set A itself.

(C) $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ is the universal set U.

(D) $\emptyset$ (the empty set) would be the complement if A was equal to U.

The correct answer is $\{2, 4, 6, 8, 10\}$.

The correct option is (A).

We need to simplify the expression $A \cap (B \cup A)$. This involves understanding the properties of set operations.

The expression involves the intersection ($\cap$) and union ($\cup$) of sets.

Let's consider the definition of union and intersection:

• $B \cup A$: This set contains all elements that are in B, or in A, or in both.
• $A \cap (B \cup A)$: This set contains all elements that are common to both set A AND the set $(B \cup A)$.

Consider an element $x$. For $x$ to be in $A \cap (B \cup A)$, two conditions must be met:

1. $x$ must be in A ($x \in A$).
2. $x$ must be in $(B \cup A)$ ($x \in (B \cup A)$).

If an element $x$ is in set A, then it is automatically also in the set $B \cup A$ (because $B \cup A$ contains all elements of A).

So, if $x \in A$, then $x \in (B \cup A)$ is always true.

This means that the condition for $x$ to be in $A \cap (B \cup A)$ simplifies to just the first condition: $x$ must be in A.

Therefore, the set $A \cap (B \cup A)$ is equivalent to the set A.

This property is known as the absorption law for sets: $A \cap (B \cup A) = A$.

Let's verify this with an example:

Let $A = \{1, 2\}$ and $B = \{2, 3\}$.

$B \cup A = \{1, 2\} \cup \{2, 3\} = \{1, 2, 3\}$.

$A \cap (B \cup A) = \{1, 2\} \cap \{1, 2, 3\}$.

The common elements are 1 and 2. So, $A \cap (B \cup A) = \{1, 2\}$, which is equal to set A.

Comparing our result with the given options:

(A) A

(B) B

(C) $\emptyset$

(D) $A \cup B$

The correct answer is A.

The correct option is (A).

We are given the following information about two sets A and B:

• The number of elements in set A, $\text{n}(A) = 20$.
• The number of elements in set B, $\text{n}(B) = 28$.
• The number of elements in the union of A and B, $\text{n}(A \cup B) = 36$.

We need to find the number of elements in the intersection of A and B, $\text{n}(A \cap B)$.

We can use the Principle of Inclusion-Exclusion for two sets, which states:

Now, we can substitute the given values into this formula:

First, calculate the sum of $\text{n}(A)$ and $\text{n}(B)$:

So, the equation becomes:

To find $\text{n}(A \cap B)$, we can rearrange the equation:

Now, perform the subtraction:

Therefore, the number of elements in the intersection of A and B is 12.

Comparing this result with the given options:

(A) 12

(B) 16

(C) 20

(D) 24

The correct answer is 12.

The correct option is (A).

De Morgan's Laws are a pair of transformation rules in set theory that relate the intersection and union of sets with complements.

There are two De Morgan's Laws for sets:

1. The complement of the intersection of two sets is the union of their complements: $$ (A \cap B)' = A' \cup B' $$
2. The complement of the union of two sets is the intersection of their complements: $$ (A \cup B)' = A' \cap B' $$

The question asks for the expression equal to $(A \cap B)'$.

According to the first De Morgan's Law, $(A \cap B)'$ is equal to $A' \cup B'$.

Let's examine the given options:

(A) $A' \cap B'$: This is the result of $(A \cup B)'$, not $(A \cap B)'$.

(B) $A' \cup B'$: This matches the first De Morgan's Law for $(A \cap B)'$.

(C) $A \cup B$: This is the union of the original sets.

(D) $A \cap B$: This is the intersection of the original sets.

Therefore, according to De Morgan's Law, $(A \cap B)'$ is equal to $A' \cup B'$.

The correct option is (B).

The provided description of the Venn diagram states: "Venn diagram showing two intersecting circles A and B, with the intersection area shaded."

In a Venn diagram, when two circles representing sets A and B intersect:

• The entire area covered by both circles represents the union ($A \cup B$).
• The region where the circles overlap, representing elements common to both sets, is the intersection ($A \cap B$).
• The part of circle A that does not overlap with circle B represents $A - B$ (elements in A but not in B).
• The part of circle B that does not overlap with circle A represents $B - A$ (elements in B but not in A).

Since the shaded region is described as the "intersection area," it represents the elements that are common to both set A and set B.

Therefore, the shaded region represents the intersection of sets A and B, which is denoted by $A \cap B$.

Let's check the options based on this understanding:

(A) $A \cup B$: This would be the total area covered by both circles.

(B) $A \cap B$: This is the region where the circles overlap, which matches the description of the shaded area.

(C) $A - B$: This would be the part of circle A that is outside of circle B.

(D) $B - A$: This would be the part of circle B that is outside of circle A.

The correct answer is $A \cap B$.

The correct option is (B).

Assertion (A): For any two sets A and B, $A \cup B = B \cup A$.

The union of two sets $A$ and $B$ is the set containing all elements that are in $A$, or in $B$, or in both. The order in which the sets are combined does not change the resulting set of elements. For example, if $A = \{1, 2\}$ and $B = \{2, 3\}$, then $A \cup B = \{1, 2, 3\}$ and $B \cup A = \{2, 3\} \cup \{1, 2\} = \{1, 2, 3\}$. Since $A \cup B$ and $B \cup A$ result in the same set, the assertion is true.

Reason (R): The union operation is commutative.

A binary operation is said to be commutative if the order of the operands does not affect the result. In set theory, the union operation ($\cup$) is commutative because, as shown above, $A \cup B = B \cup A$ for any two sets A and B. Therefore, the reason is true.

Now we assess if Reason (R) correctly explains Assertion (A).

Assertion (A) states that $A \cup B = B \cup A$. Reason (R) states that the union operation is commutative. The fact that the union operation is commutative is precisely the reason why $A \cup B$ is equal to $B \cup A$. The definition of commutativity for union is the equality stated in Assertion (A).

Thus, both the assertion and the reason are true, and the reason correctly explains the assertion.

The correct option is (A).

Let C be the set of students who play cricket, and F be the set of students who play football.

We are given the following information:

• Total number of students in the class = 100.
• Number of students who play cricket, $\text{n}(C) = 60$.
• Number of students who play football, $\text{n}(F) = 50$.
• Number of students who play both cricket and football, $\text{n}(C \cap F) = 20$.

We need to find the number of students who play only cricket.

The number of students who play only cricket is the number of students in set C minus the number of students who are in both C and F (i.e., those who play cricket and football).

Mathematically, the number of students who play only cricket is given by:

Substitute the given values into the formula:

Performing the subtraction:

We can also visualize this using a Venn diagram. Draw two intersecting circles, one for Cricket (C) and one for Football (F). The overlapping region represents students who play both.

• The intersection region ($C \cap F$) has 20 students.
• The circle C has a total of 60 students. The part of C that does not overlap with F (only cricket) is $60 - 20 = 40$.
• The circle F has a total of 50 students. The part of F that does not overlap with C (only football) is $50 - 20 = 30$.

The sum of students playing only cricket, only football, and both is $40 + 30 + 20 = 90$. This means $100 - 90 = 10$ students play neither sport.

The number of students who play only cricket is 40.

Comparing this with the given options:

(A) 60 (This is the total number of students who play cricket)

(B) 40 (This is the number of students who play only cricket)

(C) 20 (This is the number of students who play both)

(D) 80 (This is likely $60 + 20$ or an incorrect calculation)

The correct answer is 40.

The correct option is (B).

From the previous question (Q39), we have the following information:

• Total number of students in the class = 100.
• Number of students who play cricket, $\text{n}(C) = 60$.
• Number of students who play football, $\text{n}(F) = 50$.
• Number of students who play both cricket and football, $\text{n}(C \cap F) = 20$.
• Number of students who play only cricket = $\text{n}(C) - \text{n}(C \cap F) = 60 - 20 = 40$.
• Number of students who play only football = $\text{n}(F) - \text{n}(C \cap F) = 50 - 20 = 30$.

The number of students who play at least one of the sports (cricket or football or both) is given by the union of the sets C and F, $\text{n}(C \cup F)$.

Using the Principle of Inclusion-Exclusion:

Substitute the given values:

Calculate the union:

This means 90 students play at least one of the sports.

The number of students who play neither cricket nor football is the total number of students minus the number of students who play at least one of the sports.

Number of students playing neither = Total students - $\text{n}(C \cup F)$

Substitute the values:

Performing the subtraction:

Alternatively, using the components from the Venn diagram breakdown:

Students playing only cricket = 40

Students playing only football = 30

Students playing both = 20

Total playing at least one sport = $40 + 30 + 20 = 90$.

Students playing neither = Total students - (Only Cricket + Only Football + Both)

Students playing neither = $100 - (40 + 30 + 20) = 100 - 90 = 10$.

The number of students who play neither cricket nor football is 10.

Comparing this with the given options:

(A) 10

(B) 20

(C) 30

(D) 40

The correct answer is 10.

The correct option is (A).

Let's analyze each set operation and match it with its corresponding property:

(i) $A \cup A = A$

This property states that the union of a set with itself is the set itself. This is known as the Idempotent Law for union.

Therefore, (i) matches with (b) Idempotent Law.

(ii) $A \cup \emptyset = A$

This property states that the union of any set with the empty set ($\emptyset$) is the set itself. The empty set acts as an identity element for the union operation. This is known as the Identity Law for union.

Therefore, (ii) matches with (c) Identity Law.

(iii) $(A \cup B) \cup C = A \cup (B \cup C)$

This property states that the order in which unions are performed does not matter when combining three or more sets. This is known as the Associative Law for union.

Therefore, (iii) matches with (a) Associative Law.

(iv) $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$

This property shows how the intersection operation distributes over the union operation. This is known as the Distributive Law (specifically, intersection over union).

Therefore, (iv) matches with (d) Distributive Law.

Matching the items:

• (i) - (b)
• (ii) - (c)
• (iii) - (a)
• (iv) - (d)

Comparing this combination with the given options:

(A) (i)-(b), (ii)-(c), (iii)-(a), (iv)-(d)

(B) (i)-(a), (ii)-(c), (iii)-(b), (iv)-(d)

(C) (i)-(b), (ii)-(a), (iii)-(c), (iv)-(d)

(D) (i)-(d), (ii)-(c), (iii)-(a), (iv)-(b)

The correct option is (A).

We are given the following information:

• The number of elements in set A, $\text{n}(A) = 10$.
• The number of elements in set B, $\text{n}(B) = 15$.
• Sets A and B are disjoint.

Two sets are said to be disjoint if they have no elements in common. This means their intersection is the empty set:

From this, it follows that the number of elements in the intersection is zero:

We need to find the number of elements in the union of A and B, $\text{n}(A \cup B)$.

The general formula for the union of two sets is:

Since A and B are disjoint, $\text{n}(A \cap B) = 0$. Substituting this into the formula:

This simplifies to:

Now, substitute the given values for $\text{n}(A)$ and $\text{n}(B)$:

Performing the addition:

Therefore, the number of elements in the union of A and B is 25.

Let's check the options:

(A) 10

(B) 15

(C) 25

(D) 5

The correct answer is 25.

The correct option is (C).

Let's analyze each statement based on set theory properties, where $\text{U}$ is the universal set and $A'$ is the complement of set A.

(A) $A \cup A' = \text{U}$

The union of a set and its complement includes all elements that are either in A or not in A. By definition, this covers all elements within the universal set. Thus, this statement is correct.

(B) $A \cap A' = \emptyset$

The intersection of a set and its complement contains elements that are common to both. An element cannot be both in set A and not in set A simultaneously. Therefore, the intersection is always the empty set. Thus, this statement is correct.

(C) $(\text{U})' = \emptyset$

The complement of the universal set ($\text{U}'$) consists of all elements in the universal set that are not in the universal set. There are no such elements. Therefore, the complement of the universal set is the empty set. Thus, this statement is correct.

(D) $\emptyset' = \emptyset$

The complement of the empty set ($\emptyset'$) consists of all elements in the universal set that are not in the empty set. Since the empty set contains no elements, all elements of the universal set are not in the empty set. Therefore, the complement of the empty set is the universal set itself ($\text{U}$), not the empty set. Thus, this statement is incorrect.

The question asks for the incorrect statement.

The incorrect statement is (D).

The power set of a set S, denoted by $\text{P}(S)$ or $2^S$, is the set of all subsets of S, including the empty set and the set S itself.

We are asked to find the power set of the empty set, $\text{P}(\emptyset)$.

The empty set, $\emptyset$, is a set that contains no elements.

The subsets of the empty set are the sets that contain only elements from the empty set. Since the empty set has no elements, the only possible subset is the empty set itself.

Therefore, the set of all subsets of $\emptyset$ is $\{\emptyset\}$.

The power set $\text{P}(\emptyset)$ is the set containing this single subset:

Let's analyze the options:

(A) $\emptyset$: This is the empty set itself, not the set containing the subset.

(B) $\{0\}$: This is a set containing the number 0. It is not related to the power set of the empty set.

(C) $\{\emptyset\}$: This is a set containing the empty set as its only element. This correctly represents the power set of the empty set.

(D) $\{ \{\emptyset\} \}$: This is a set containing a set which itself contains the empty set. This is incorrect.

The correct answer is $\{\emptyset\}$.

The correct option is (C).

We are given the universal set $\text{U} = \{1, 2, 3, 4, 5\}$, set $A = \{1, 2, 3\}$, and set $B = \{3, 4\}$.

We need to find $(A \cup B)'$. This involves two steps:

1. Find the union of A and B ($A \cup B$).
2. Find the complement of the resulting union $(A \cup B)'$.

Step 1: Find the union of A and B ($A \cup B$).

The union of two sets contains all the elements that are in either set, or in both.

$A = \{1, 2, 3\}$

$B = \{3, 4\}$

$A \cup B = \{1, 2, 3\} \cup \{3, 4\}$

Combining the elements and listing unique ones:

Step 2: Find the complement of $(A \cup B)$, denoted by $(A \cup B)'$.

The complement of a set consists of all elements in the universal set $\text{U}$ that are not in the set.

We have $\text{U} = \{1, 2, 3, 4, 5\}$ and $A \cup B = \{1, 2, 3, 4\}$.

To find $(A \cup B)'$, we look for elements in $\text{U}$ that are not in $A \cup B$.

• 1 is in $\text{U}$ but also in $A \cup B$.
• 2 is in $\text{U}$ but also in $A \cup B$.
• 3 is in $\text{U}$ but also in $A \cup B$.
• 4 is in $\text{U}$ but also in $A \cup B$.
• 5 is in $\text{U}$ but not in $A \cup B$.

Therefore, the only element not in $A \cup B$ but in $\text{U}$ is 5.

Let's check the given options:

(A) $\{1, 2, 3, 4\}$ (This is $A \cup B$ itself)

(B) $\{1, 2, 3, 4, 5\}$ (This is the universal set $\text{U}$)

(C) $\{5\}$ (This matches our calculated result)

(D) $\emptyset$ (This would be the complement if $A \cup B$ was equal to $\text{U}$)

The correct answer is $\{5\}$.

The correct option is (C).

The given interval is $(-3, 5]$.

This notation represents all real numbers $x$ such that $-3 < x \leq 5$.

The parenthesis $(-3$ indicates that the number -3 is not included in the interval. So, the numbers in the interval are strictly greater than -3.

The square bracket $5]$ indicates that the number 5 is included in the interval. So, the numbers in the interval are less than or equal to 5.

We are looking for the largest integer that is contained within this interval.

Let's list the integers that satisfy the condition $-3 < x \leq 5$:

• Integers strictly greater than -3 are -2, -1, 0, 1, 2, 3, 4, 5, ...
• Integers less than or equal to 5 are ..., 2, 3, 4, 5.

The integers that satisfy both conditions are: -2, -1, 0, 1, 2, 3, 4, 5.

Among these integers, the largest one is 5.

Let's review the options:

(A) 5: This is an integer and it is in the interval $(-3, 5]$. It is the largest integer in the interval.

(B) 4: This is an integer in the interval, but 5 is larger.

(C) -2: This is an integer in the interval, but it is the smallest integer, not the largest.

(D) 6: This integer is not in the interval because the interval only goes up to 5 (inclusive).

The largest integer contained in the interval $(-3, 5]$ is 5.

The correct option is (A).

Let $S = \{a, b, c, d\}$. We need to determine which of the given options is a proper subset of S.

A set $A$ is a subset of a set $S$ (denoted $A \subseteq S$) if every element of $A$ is also an element of $S$.

A set $A$ is a proper subset of a set $S$ (denoted $A \subset S$) if $A$ is a subset of $S$ and $A$ is not equal to $S$. In other words, $A$ must be a subset of $S$, and $A$ must contain fewer elements than $S$, or at least not all the same elements.

Let's examine each option:

(A) $\{a, b, c, d\}$

This set contains all the same elements as S. Therefore, it is a subset of S, but it is not a *proper* subset because it is equal to S.

(B) $\{a, b, c, d, e\}$

This set contains an element 'e' which is not in S. Therefore, it is not a subset of S at all.

(C) $\{a, b\}$

Every element in $\{a, b\}$ (which are 'a' and 'b') is also an element of $S = \{a, b, c, d\}$. So, $\{a, b\}$ is a subset of S. Furthermore, $\{a, b\}$ contains fewer elements than S and is not equal to S. Therefore, $\{a, b\}$ is a proper subset of S.

(D) $\emptyset$

The empty set $\emptyset$ contains no elements. By definition, the empty set is a subset of every set. Since $\emptyset$ does not contain all the elements of S (it contains none), it is also a proper subset of S (as long as S is not empty, which it is not).

The question asks for *a* proper subset. Both (C) and (D) are proper subsets. However, in multiple-choice questions, we typically choose the option that is most representative or the one explicitly intended. Option (C) is a non-empty proper subset, while (D) is the empty set, which is always a proper subset of any non-empty set. Typically, when asked for "a proper subset," a non-empty one is often implied if available.

Let's re-read the question: "Which of the following is *a* proper subset". Both (C) and (D) fit this. However, if we consider typical question construction, there might be a preference. Often, when asked for *a* proper subset, the intention is to identify a non-trivial proper subset if one exists.

Both (C) and (D) are mathematically correct as proper subsets.

Let's assume the question expects a non-trivial proper subset if available, making (C) a strong candidate. If only one correct answer is allowed, and both are technically correct, it's a matter of interpretation or common convention in the context where the question is posed.

In most contexts, when given options like these, if a non-empty proper subset is an option, it is often preferred. However, mathematically, $\emptyset$ is also a proper subset.

Given the standard understanding and the options provided, option (C) is a clear example of a proper subset that is not the empty set.

Let's go with the most direct example of a non-empty proper subset.

The correct answer is $\{a, b\}$.

The correct option is (C).

The roster form of a set lists all the distinct elements of the set within curly braces.

The word is "MATHEMATICS". Let's identify all the letters in this word:

M, A, T, H, E, M, A, T, I, C, S

To write this in roster form, we need to list each distinct letter only once. We ignore any repeated letters.

Let's go through the letters and keep track of the unique ones:

• M: (unique) -> M
• A: (unique) -> M, A
• T: (unique) -> M, A, T
• H: (unique) -> M, A, T, H
• E: (unique) -> M, A, T, H, E
• M: (already listed)
• A: (already listed)
• T: (already listed)
• I: (unique) -> M, A, T, H, E, I
• C: (unique) -> M, A, T, H, E, I, C
• S: (unique) -> M, A, T, H, E, I, C, S

So, the set of distinct letters in "MATHEMATICS" in roster form is $\{M, A, T, H, E, I, C, S\}$.

Now let's compare this with the given options:

(A) $\{M, A, T, H, E, M, A, T, I, C, S\}$: This lists all letters including duplicates, which is not the correct roster form of a set.

(B) $\{M, A, T, H, E, I, C, S\}$: This lists all the distinct letters exactly once. This is the correct roster form.

(C) $\{M, A, T, H, E, I, C, S, T\}$: This includes the distinct letters but also includes 'T' twice, which is incorrect for roster form.

(D) $\{M, A, T, H, E, M, I, C, S\}$: This lists the first 'M' but omits the second 'M' while still listing other letters. It's also incorrect due to the duplicate 'M' and missing 'A' and 'T' that were duplicated. It omits the first 'A' and 'T' from the unique list and includes the second 'M' from the original word.

The correct roster form lists each distinct element exactly once.

The correct option is (B).

We are given that for two sets A and B, $A \cup B = A$.

Let's analyze the meaning of the union of two sets. $A \cup B$ is the set containing all elements that are in A, or in B, or in both.

The condition $A \cup B = A$ means that adding all the elements of B to set A does not introduce any new elements to A. In other words, all the elements of B are already present in A.

If every element of B is also an element of A, then B is a subset of A. This is denoted as $B \subseteq A$.

Let's verify this with an example:

Suppose $A = \{1, 2, 3\}$ and $B = \{2, 3\}$.

$A \cup B = \{1, 2, 3\} \cup \{2, 3\} = \{1, 2, 3\}$.

Here, $A \cup B = A$ is true. And indeed, $B = \{2, 3\}$ is a subset of $A = \{1, 2, 3\}$, so $B \subseteq A$.

Now let's consider the given options:

(A) $A \subset B$: This would mean all elements of A are in B, which contradicts $A \cup B = A$ unless both A and B are empty or equal.

(B) $B \subset A$: This means all elements of B are in A. This is consistent with $A \cup B = A$.

(C) $A = B$: This is a special case of $B \subset A$ (and $A \subset B$), but it's not necessarily true. For example, if $A = \{1, 2, 3\}$ and $B = \{2, 3\}$, then $B \subset A$ but $A \neq B$. So, $A=B$ is not always true.

(D) $A \cap B = \emptyset$: This means A and B are disjoint. If they were disjoint, $A \cup B$ would contain all elements of both A and B, and if B had any elements, $A \cup B$ would be larger than A. So, this is generally false.

The condition $A \cup B = A$ implies that B must be a subset of A ($B \subseteq A$).

The correct option is (B).

We need to find the number of elements in the power set of $\text{P}(\emptyset)$.

First, let's recall what $\text{P}(\emptyset)$ is.

The power set of a set S, denoted $\text{P}(S)$, is the set of all subsets of S. The number of elements in the power set of a set S with $n$ elements is $2^n$.

The empty set $\emptyset$ has 0 elements. So, $n=0$ for the empty set.

The power set of the empty set is $\text{P}(\emptyset)$. The number of elements in $\text{P}(\emptyset)$ is $2^0 = 1$.

What is that one element? It is the empty set itself, because the empty set is a subset of every set, including the empty set. So, $\text{P}(\emptyset) = \{\emptyset\}$.

Now, we need to find the number of elements in the power set of $\text{P}(\emptyset)$. This is $\text{P}(\text{P}(\emptyset))$.

We know that $\text{P}(\emptyset) = \{\emptyset\}$. Let's call this set S for simplicity, so $S = \{\emptyset\}$.

Now we need to find the power set of S, which is $\text{P}(S) = \text{P}(\{\emptyset\})$.

The set S = $\{\emptyset\}$ has only one element, which is the empty set itself. So, $n=1$ for the set S.

The number of elements in the power set of S is $2^n$, where $n$ is the number of elements in S.

Number of elements in $\text{P}(S) = 2^1 = 2$.

What are these two elements? The subsets of $S = \{\emptyset\}$ are:

1. The empty set: $\emptyset$
2. The set S itself: $\{\emptyset\}$

So, $\text{P}(\text{P}(\emptyset)) = \text{P}(\{\emptyset\}) = \{\emptyset, \{\emptyset\}\}$.

The number of elements in this power set is 2.

Let's check the options:

(A) 0

(B) 1 (This is the number of elements in $\text{P}(\emptyset)$)

(C) 2 (This is the number of elements in $\text{P}(\text{P}(\emptyset))$)

(D) 4

The correct answer is 2.

The correct option is (C).

We need to find the solutions to the quadratic equation $x^2 - 5x + 6 = 0$.

We can solve this equation by factoring, using the quadratic formula, or completing the square. Factoring is often the simplest method if it's possible.

We look for two numbers that multiply to 6 and add up to -5.

The pairs of numbers that multiply to 6 are:

• 1 and 6
• -1 and -6
• 2 and 3
• -2 and -3

Now, let's check which pair adds up to -5:

• $1 + 6 = 7$
• $-1 + (-6) = -7$
• $2 + 3 = 5$
• $-2 + (-3) = -5$

The pair of numbers that satisfies both conditions is -2 and -3.

We can use these numbers to factor the quadratic equation:

For the product of two factors to be zero, at least one of the factors must be zero. So, we set each factor equal to zero:

Case 1: $x - 2 = 0$

Adding 2 to both sides, we get $x = 2$.

Case 2: $x - 3 = 0$

Adding 3 to both sides, we get $x = 3$.

The solutions to the equation are $x = 2$ and $x = 3$.

The set of solutions in roster form is $\{2, 3\}$.

Let's check the options:

(A) $\{2, 3\}$: This matches our calculated set of solutions.

(B) $\{-2, -3\}$: These are the numbers we found that multiply to +6 and add to -5, but they are the roots of $x^2 + 5x + 6 = 0$, not $x^2 - 5x + 6 = 0$.

(C) $\{2, -3\}$: This is an incorrect combination of roots.

(D) $\{-2, 3\}$: This is an incorrect combination of roots.

The correct answer is $\{2, 3\}$.

The correct option is (A).

We are given the set $A = \{1, 2, 3, 4, 5\}$.

We need to find $\text{n}(\text{P}(A))$, which means the number of elements in the power set of A.

The number of elements in a set A is denoted by $\text{n}(A)$. In this case:

The power set of a set A, denoted by $\text{P}(A)$, is the set of all possible subsets of A. If a set A has $n$ elements, then its power set $\text{P}(A)$ has $2^n$ elements.

In this problem, $n = \text{n}(A) = 5$.

Therefore, the number of elements in the power set of A is:

Substituting the value of $n$:

Now, we calculate $2^5$:

$2^1 = 2$

$2^2 = 4$

$2^3 = 8$

$2^4 = 16$

$2^5 = 32$

So, the number of elements in the power set of A is 32.

Let's check the given options:

(A) 5 (This is $\text{n}(A)$)

(B) 10

(C) 25

(D) 32 (This matches our calculated result)

The correct answer is 32.

The correct option is (D).

We need to simplify the expression $A \cap (A \cup B)'$.

We can use set properties, particularly De Morgan's Laws and distributive laws, to simplify this expression.

First, let's apply De Morgan's Law to $(A \cup B)'$. De Morgan's Law states that $(A \cup B)' = A' \cap B'$.

So, the expression becomes: $A \cap (A' \cap B')$.

Now we can use the associative property of intersection to group the terms: $(A \cap A') \cap B'$.

We know that the intersection of any set with its complement is the empty set, i.e., $A \cap A' = \emptyset$.

Substituting this into the expression: $\emptyset \cap B'$.

The intersection of the empty set with any other set is always the empty set. This is because the empty set has no elements, so there can be no common elements between it and any other set.

Therefore, $\emptyset \cap B' = \emptyset$.

The expression $A \cap (A \cup B)'$ simplifies to $\emptyset$.

Let's check with an example:

Let $A = \{1, 2\}$ and $B = \{2, 3\}$.

$A \cup B = \{1, 2, 3\}$.

Assume a universal set $\text{U} = \{1, 2, 3, 4\}$.

$(A \cup B)' = \{1, 2, 3\}' = \{4\}$.

$A \cap (A \cup B)' = \{1, 2\} \cap \{4\}$.

The intersection of $\{1, 2\}$ and $\{4\}$ is $\emptyset$, because there are no common elements.

Let's check the options:

(A) A

(B) B

(C) $\emptyset$ (This matches our result)

(D) $A \cap B$ (In our example, $A \cap B = \{2\}$)

The correct answer is $\emptyset$.

The correct option is (C).

The English alphabet consists of a specific, limited number of letters.

Let's list them out (though not strictly necessary for the definition): A, B, C, ..., X, Y, Z.

There are exactly 26 letters in the English alphabet.

Now, let's consider the definitions of the given set types:

• Finite set: A set that has a limited number of elements. The number of elements can be counted, and it is a non-negative integer.
• Infinite set: A set that has an unlimited number of elements. It is not possible to count all the elements in an infinite set.
• Empty set: A set containing no elements. It is denoted by $\emptyset$ or \{\}.
• Singleton set: A set containing exactly one element.

Since the English alphabet has exactly 26 letters, which is a specific and countable number, the set of all letters in the English alphabet is a finite set.

Let's evaluate the options:

(A) finite: This correctly describes a set with a countable number of elements.

(B) infinite: This is incorrect because the alphabet has a fixed, limited number of letters.

(C) empty: This is incorrect because the alphabet contains 26 letters, not zero.

(D) singleton: This is incorrect because the alphabet contains 26 letters, not just one.

The correct answer is finite.

The correct option is (A).

We are asked to find the notation that represents the set of all real numbers greater than or equal to 7.

In set notation, this can be written as $\{x \in \mathbb{R} \mid x \geq 7\}$.

We need to translate this into interval notation:

• The condition "$x \geq 7$" means that $x$ is greater than or equal to 7.
• The "greater than or equal to" part means that the number 7 is included in the set.
• When a number is included in an interval, we use a square bracket '[' next to it.
• The condition "$x \geq 7$" implies that $x$ can be any number from 7 upwards, extending to positive infinity.
• Infinity ($\infty$) is not a number, so it is always represented with a parenthesis ')'.

Therefore, the interval notation for numbers greater than or equal to 7 is $[7, \infty)$.

Let's analyze the given options:

(A) $(7, \infty)$: This represents numbers strictly greater than 7 (7 is excluded).

(B) $[7, \infty)$: This represents numbers greater than or equal to 7 (7 is included).

(C) $(-\infty, 7)$: This represents numbers strictly less than 7.

(D) $(-\infty, 7]$: This represents numbers less than or equal to 7.

The notation that correctly represents the set of all real numbers greater than or equal to 7 is $[7, \infty)$.

The correct option is (B).

We are given the following information:

• Universal set $\text{U}$, $\text{n}(\text{U}) = 50$.
• Set A, $\text{n}(A) = 20$.
• Set B, $\text{n}(B) = 30$.
• Intersection of A and B, $\text{n}(A \cap B) = 10$.

We need to find $\text{n}(A' \cap B')$.

We can use De Morgan's Law, which states that $(A \cup B)' = A' \cap B'$.

Therefore, to find $\text{n}(A' \cap B')$, we need to find $\text{n}((A \cup B)')$.

The complement of a set $(A \cup B)'$ is given by $\text{n}(\text{U}) - \text{n}(A \cup B)$.

First, let's find $\text{n}(A \cup B)$ using the Principle of Inclusion-Exclusion:

Substitute the given values into the formula:

Calculate the union:

Now we can find $\text{n}(A' \cap B')$, which is equal to $\text{n}((A \cup B)')$:

Substitute the values:

Calculate the result:

Therefore, the number of elements in $A' \cap B'$ is 10.

Let's check the options:

(A) 10

(B) 20

(C) 30

(D) 40

The correct answer is 10.

The correct option is (A).

The union of two sets, denoted by $A \cup B$, is the set that contains all the elements belonging to set A, or to set B, or to both.

In a Venn diagram with two sets A and B represented by circles, the union $A \cup B$ is visually represented by shading all regions that are part of either circle A, or circle B, or the overlapping region between them.

Let's examine each option based on the descriptions provided:

(A) Option A: "Venn diagram showing two intersecting circles A and B. Only circle A is shaded." This represents set A ($A$).

(B) Option B: "Venn diagram showing two intersecting circles A and B. Only the intersection area is shaded." This represents the intersection of A and B ($A \cap B$).

(C) Option C: "Venn diagram showing two intersecting circles A and B. Both circles are shaded." This means all the area within circle A and all the area within circle B is shaded. This visually corresponds to the union of the two sets ($A \cup B$).

(D) Option D: "Venn diagram showing two intersecting circles A and B. Only the area outside both circles is shaded within the universal set rectangle." This represents the complement of the union of A and B ($(A \cup B)'$).

Therefore, the Venn diagram that represents $A \cup B$ is the one where both circles (representing A and B) are shaded.

The correct option is (C).

The set A is defined as $\{x : x \text{ is an integer and } -2 \leq x < 2\}$.

This means we need to find all integers $x$ that satisfy two conditions:

1. $x$ is an integer.
2. $-2 \leq x < 2$.

The condition $-2 \leq x < 2$ means that $x$ is greater than or equal to -2 AND $x$ is strictly less than 2.

Let's list the integers that satisfy this condition:

• The integers that are greater than or equal to -2 are: -2, -1, 0, 1, 2, 3, ...
• The integers that are strictly less than 2 are: ..., -2, -1, 0, 1.

We need the integers that are in both of these lists. These are the integers $x$ such that $-2 \leq x < 2$.

The integers are: -2, -1, 0, 1.

Therefore, the set A in roster form is $\{-2, -1, 0, 1\}$.

Let's check the given options:

(A) $\{-2, -1, 0, 1, 2\}$: This includes 2, but the condition is $x < 2$, so 2 should not be included.

(B) $\{-1, 0, 1\}$: This excludes -2, but the condition is $-2 \leq x$, so -2 should be included.

(C) $\{-2, -1, 0, 1\}$: This correctly includes all integers from -2 up to, but not including, 2.

(D) $\{-1, 0, 1, 2\}$: This excludes -2 and includes 2, both of which are incorrect based on the given condition.

The correct set in roster form is $\{-2, -1, 0, 1\}$.

The correct option is (C).

Sets can be described in two primary ways: roster form and set-builder form.

Roster Form: This form involves listing all the elements of the set, enclosed in curly braces, and separated by commas. For example, the set of vowels in the English alphabet in roster form is $\{a, e, i, o, u\}$.

Set-Builder Form: This form does not list all the elements. Instead, it describes the elements by stating a property or a rule that all elements in the set must satisfy. The general format is $\{x \mid P(x)\}$, where $x$ is a variable representing an element, and $P(x)$ is the property or rule that $x$ must satisfy.

Let's analyze the given options in relation to set-builder form:

(A) Listing all elements: This is a characteristic of the roster form, not set-builder form.

(B) Using a rule to describe the elements: This is the fundamental characteristic of set-builder form. It defines the set based on a shared property.

(C) Enclosing elements in curly braces: Both roster form and set-builder form use curly braces to denote a set. So, this is a characteristic of both, but not unique to set-builder form and not its primary defining characteristic.

(D) Listing elements separated by commas: This is a characteristic of the roster form, where elements are explicitly listed and separated by commas.

Therefore, the primary characteristic of the set-builder form is using a rule to describe the elements.

The correct option is (B).

For a collection of objects to be considered a set, it must satisfy the "well-defined" criterion. This means that for any given object, it must be possible to definitively determine whether that object belongs to the collection or not.

Assertion (A): The set of "smart students" in a class is not a set.

The term "smart students" is ambiguous. What one person considers smart, another might not. There is no clear, objective criterion given to determine if a student is "smart." For example, is smartness based on grades, creativity, problem-solving skills, or something else? Without a precise definition, we cannot definitively say whether a particular student is included in this collection or not. Therefore, the collection of "smart students" is not well-defined, and hence, it is not a set. The assertion is true.

Reason (R): "Smartness" is subjective and not well-defined.

As discussed above, the concept of "smartness" in the context of students is subjective. It depends on individual perception and criteria, and there isn't a universally agreed-upon objective standard. Because it's subjective, it's not well-defined in a mathematical sense required for forming a set. The reason is true.

Now, we check if Reason (R) correctly explains Assertion (A).

The reason why the collection of "smart students" is not a set (Assertion A) is precisely because the term "smartness" is subjective and not well-defined (Reason R). The lack of a clear, objective criterion is the reason for its failure to meet the definition of a set.

Therefore, both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A).

The correct option is (A).

We are given two sets: $A = \{1, 2, 3\}$ and $B = \{4, 5\}$.

We need to find the intersection of A and B, denoted by $A \cap B$.

The intersection of two sets is the set containing all elements that are common to both sets.

To find $A \cap B$, we look for elements that are present in set A AND also present in set B.

Elements in set A are: 1, 2, 3.

Elements in set B are: 4, 5.

Comparing the elements of both sets, we can see that there are no elements that appear in both set A and set B.

When there are no common elements between two sets, their intersection is the empty set, denoted by $\emptyset$ or \{\}.

Therefore, $A \cap B = \emptyset$.

Let's check the given options:

(A) $\{1, 2, 3, 4, 5\}$: This is the union of A and B ($A \cup B$).

(B) $\emptyset$: This represents the empty set, which is the correct intersection.

(C) $\{1, 2, 3\}$: This is set A.

(D) $\{4, 5\}$: This is set B.

The correct answer is $\emptyset$.

The correct option is (B).

The set difference $A - B$ is defined as the set of all elements that are in set A but not in set B.

We can express this definition using intersection and complements.

An element $x$ is in $A - B$ if and only if $x \in A$ AND $x \notin B$.

The condition "$x \notin B$" is equivalent to "$x$ is in the complement of B", which is denoted by $B'$.

So, an element $x$ is in $A - B$ if and only if $x \in A$ AND $x \in B'$.

This means that $A - B$ is the set of elements that are in A AND in $B'$. This is precisely the definition of the intersection of A and $B'$, denoted by $A \cap B'$.

Therefore, $A - B = A \cap B'$.

Let's verify with an example:

Let $A = \{1, 2, 3\}$ and $B = \{3, 4\}$.

$A - B$ would be the elements in A that are not in B. So, $A - B = \{1, 2\}$.

Now let's check $A \cap B'$:

Assume a universal set $\text{U} = \{1, 2, 3, 4, 5\}$.

$B' = \text{U} - B = \{1, 2, 3, 4, 5\} - \{3, 4\} = \{1, 2, 5\}$.

$A \cap B' = \{1, 2, 3\} \cap \{1, 2, 5\}$.

The common elements are 1 and 2. So, $A \cap B' = \{1, 2\}$.

Since $A - B = \{1, 2\}$ and $A \cap B' = \{1, 2\}$, the equivalence holds.

Let's look at the options:

(A) $A \cap B$: This is the intersection, containing only common elements.

(B) $A \cup B$: This is the union, containing all elements from both sets.

(C) $A \cap B'$: This is equivalent to $A - B$, as we've shown.

(D) $A' \cap B$: This is equivalent to $B - A$.

The correct answer is $A \cap B'$.

The correct option is (C).

The empty set is a set that contains no elements. Let's examine each option to see if it represents the empty set.

(A) $\emptyset$

This is the standard mathematical symbol for the empty set. So, this is a valid representation.

(B) \{\}

This notation, curly braces with nothing inside, also represents a set with no elements, which is the empty set. So, this is a valid representation.

(C) $\{0\}$

This notation represents a set that contains exactly one element, which is the number 0. Since it contains an element, it is not the empty set. So, this is NOT a valid representation of the empty set.

(D) $\{x : x \in \mathbb{N}, x < 1\}$

This is a set described in set-builder form. It represents all numbers $x$ such that $x$ is a natural number ($\mathbb{N}$), and $x$ is less than 1. The natural numbers are typically considered to be $\{1, 2, 3, \dots\}$. The only natural number less than 1 would be none. If $\mathbb{N}$ is considered to include 0, then the set would be $\{0\}$, which is not empty. However, if $\mathbb{N}$ means $\{1, 2, 3, \ldots\}$, then there are no natural numbers less than 1. Assuming the common definition of $\mathbb{N}$ as $\{1, 2, 3, \ldots\}$, this set represents no elements, making it the empty set. If $\mathbb{N}$ includes 0 (sometimes denoted $\mathbb{N}_0$), then the set would be $\{0\}$, which is not empty. Given the other options, and the typical context of such questions, it's likely that $\mathbb{N}$ here implies positive integers. If $\mathbb{N}$ starts from 1, then this is indeed the empty set.

Let's assume the most common interpretation of $\mathbb{N}$ as positive integers $\{1, 2, 3, ...\}$. In this case, there are no natural numbers less than 1. So, this set-builder notation correctly describes the empty set.

The question asks for the representation that is NOT valid for the empty set.

Option (C) $\{0\}$ is definitively not the empty set, as it contains the element 0.

The correct answer is $\{0\}$.

The correct option is (C).

We are given that A and B are disjoint sets.

By definition, two sets are disjoint if they have no elements in common. In other words, their intersection is the empty set.

Mathematically, if A and B are disjoint, then:

The number of elements in the empty set is 0.

Therefore, the number of elements in the intersection of A and B is:

The information that $\text{n}(A) = 3$ and $\text{n}(B) = 4$ is relevant if we were asked to find $\text{n}(A \cup B)$, which would be $\text{n}(A) + \text{n}(B) = 3 + 4 = 7$ since they are disjoint. However, the question specifically asks for $\text{n}(A \cap B)$.

The correct answer is 0.

Let's check the options:

(A) 0

(B) 3

(C) 4

(D) 7

The correct option is (A).

The interval notation $[2, 8)$ describes a set of real numbers.

Let's break down the notation:

• The square bracket '[' before 2 indicates that the number 2 is included in the interval.
• The parenthesis ')' before 8 indicates that the number 8 is not included in the interval.

So, the interval $[2, 8)$ represents all real numbers $x$ that are greater than or equal to 2 AND strictly less than 8.

In mathematical inequality notation, this is written as:

Now, let's compare this with the given options:

(A) $2 < x < 8$: This represents the open interval $(2, 8)$, where both endpoints are excluded.

(B) $2 \leq x \leq 8$: This represents the closed interval $[2, 8]$, where both endpoints are included.

(C) $2 < x \leq 8$: This represents a half-open interval $(2, 8]$, where 2 is excluded and 8 is included.

(D) $2 \leq x < 8$: This represents a half-open interval $[2, 8)$, where 2 is included and 8 is excluded, matching our requirement.

Therefore, the interval $[2, 8)$ represents the set of real numbers $x$ such that $2 \leq x < 8$.

The correct option is (D).

We are given the universal set $\text{U} = \{a, b, c, d, e, f\}$, set $A = \{a, b, c\}$, and set $B = \{c, d, e\}$.

We need to find $A' \cup B'$.

First, let's find the complements of A and B:

Complement of A ($A'$): This is the set of elements in $\text{U}$ that are not in A.

$A' = \text{U} - A = \{a, b, c, d, e, f\} - \{a, b, c\} = \{d, e, f\}$.

Complement of B ($B'$): This is the set of elements in $\text{U}$ that are not in B.

$B' = \text{U} - B = \{a, b, c, d, e, f\} - \{c, d, e\} = \{a, b, f\}$.

Now, we need to find the union of $A'$ and $B'$, which is $A' \cup B'$:

$A' \cup B' = \{d, e, f\} \cup \{a, b, f\}$

The union contains all elements that are in $A'$, or in $B'$, or in both.

Combining the elements and listing unique ones:

Alternatively, we can use De Morgan's Law: $A' \cup B' = (A \cap B)'$.

First, find the intersection of A and B: $A \cap B = \{a, b, c\} \cap \{c, d, e\} = \{c\}$.

Now, find the complement of the intersection: $(A \cap B)' = \{c\}' = \text{U} - \{c\} = \{a, b, d, e, f\}$.

Both methods yield the same result.

Let's check the given options:

(A) $\{a, b, c, d, e\}$: Incorrect. It omits 'f' but includes 'c'.

(B) $\{f\}$: Incorrect. This is only the element 'f'.

(C) $\{a, b, d, e, f\}$: This matches our calculated result.

(D) $\{c\}$: This is the intersection $A \cap B$.

The correct answer is $\{a, b, d, e, f\}$.

The correct option is (C).

De Morgan's Laws are fundamental rules in set theory that relate the operations of union, intersection, and complement.

There are two De Morgan's Laws:

1. The complement of the union of two sets is the intersection of their complements: $$ (A \cup B)' = A' \cap B' $$
2. The complement of the intersection of two sets is the union of their complements: $$ (A \cap B)' = A' \cup B' $$

The question asks for the expression equivalent to $(A \cup B)'$.

According to the first De Morgan's Law listed above, $(A \cup B)'$ is equal to $A' \cap B'$.

Let's examine the given options:

(A) $A' \cup B'$: This is equivalent to $(A \cap B)'$ by the second De Morgan's Law.

(B) $A' \cap B'$: This is equivalent to $(A \cup B)'$ by the first De Morgan's Law.

(C) $A \cap B$: This is the intersection of the original sets.

(D) $A \cup B$: This is the union of the original sets.

Therefore, the correct representation of $(A \cup B)'$ according to De Morgan's Law is $A' \cap B'$.

The correct option is (B).

We are given that $A \subseteq B$. This means that A is a subset of B.

By the definition of a subset, if $A \subseteq B$, then every element of A is also an element of B.

We need to find $A \cap B$, which is the intersection of sets A and B. The intersection contains all elements that are common to both A and B.

Since every element in A is also an element in B, all elements of A are common to both A and B.

Therefore, the intersection of A and B will contain exactly the same elements as set A.

So, $A \cap B = A$.

Let's verify with an example:

Let $A = \{1, 2\}$ and $B = \{1, 2, 3\}$. Here, $A \subseteq B$ is true.

$A \cap B = \{1, 2\} \cap \{1, 2, 3\}$.

The common elements are 1 and 2. So, $A \cap B = \{1, 2\}$, which is equal to set A.

Let's look at the options:

(A) A: This matches our derived result.

(B) B: This would be true if $B \subseteq A$, not the other way around.

(C) $\emptyset$: This would be true if A and B were disjoint sets.

(D) U: This is the universal set, which is generally not equal to the intersection unless A and B together cover the entire universal set and are subsets of each other.

The correct answer is A.

The correct option is (A).

We need to find the cardinality of the power set of the set $\{1, 2\}$. The cardinality of a set is the number of elements in the set.

Let $S = \{1, 2\}$.

The number of elements in S is $\text{n}(S) = 2$.

The power set of S, denoted by $\text{P}(S)$, is the set of all possible subsets of S.

The formula for the number of elements in the power set of a set with $n$ elements is $2^n$.

In this case, $n = \text{n}(S) = 2$.

So, the number of elements in the power set of S is:

Calculating $2^2$:

$2^2 = 2 \times 2 = 4$.

Therefore, the cardinality of the set $\text{P}(\{1, 2\})$ is 4.

For completeness, the subsets of $\{1, 2\}$ are:

• The empty set: $\emptyset$
• The set containing 1: $\{1\}$
• The set containing 2: $\{2\}$
• The set containing both 1 and 2: $\{1, 2\}$

So, $\text{P}(\{1, 2\}) = \{\emptyset, \{1\}, \{2\}, \{1, 2\}\}$. The number of elements in this power set is indeed 4.

Let's check the options:

(A) 2 (This is the number of elements in the original set)

(B) 3

(C) 4 (This matches our calculated cardinality)

(D) 8

The correct answer is 4.

The correct option is (C).

We need to find the intersection of sets A and B. First, let's determine the elements of each set.

Set A: $A = \{x : x^2 = 9, x \in \mathbb{Z}\}$

This definition means set A consists of all integers ($x \in \mathbb{Z}$) whose square is 9.

To find these integers, we solve the equation $x^2 = 9$:

Taking the square root of both sides, we get $x = \pm\sqrt{9}$.

So, $x = 3$ or $x = -3$. Both 3 and -3 are integers.

Therefore, in roster form, $A = \{-3, 3\}$.

Set B: $B = \{x : x \text{ is an integer and } -4 < x < 4\}$

This definition means set B consists of all integers $x$ that are strictly greater than -4 AND strictly less than 4.

Let's list these integers:

• Integers greater than -4: -3, -2, -1, 0, 1, 2, 3, 4, ...
• Integers less than 4: ..., -2, -1, 0, 1, 2, 3.

The integers that satisfy both conditions are: -3, -2, -1, 0, 1, 2, 3.

Therefore, in roster form, $B = \{-3, -2, -1, 0, 1, 2, 3\}$.

Now we need to find the intersection of A and B, which is $A \cap B$.

$A \cap B$ consists of all elements that are common to both set A and set B.

$A = \{-3, 3\}$

$B = \{-3, -2, -1, 0, 1, 2, 3\}$

Comparing the elements of A and B, we see that both -3 and 3 are present in both sets.

Therefore, $A \cap B = \{-3, 3\}$.

Let's check the options:

(A) $\{-3, 3\}$: This matches our calculated intersection.

(B) $\{-3, -2, -1, 0, 1, 2, 3\}$: This is set B.

(C) $\{3\}$: This is only one of the common elements.

(D) $\emptyset$: This would be the case if there were no common elements.

The correct answer is $\{-3, 3\}$.

The correct option is (A).

We are given two intervals: $A = [1, 5]$ and $B = (3, 7]$. We need to find their intersection, $A \cap B$.

Interval A: $[1, 5]$ means all real numbers $x$ such that $1 \leq x \leq 5$. The square brackets indicate that both 1 and 5 are included.

Interval B: $(3, 7]$ means all real numbers $x$ such that $3 < x \leq 7$. The parenthesis indicates 3 is excluded, and the square bracket indicates 7 is included.

To find the intersection ($A \cap B$), we need to find the numbers that are common to both intervals. This means the numbers must satisfy both conditions:

$1 \leq x \leq 5$ AND $3 < x \leq 7$

Let's combine these inequalities:

For the lower bound: $x$ must be greater than or equal to 1 ($x \geq 1$) AND greater than 3 ($x > 3$). The stricter condition is $x > 3$.

For the upper bound: $x$ must be less than or equal to 5 ($x \leq 5$) AND less than or equal to 7 ($x \leq 7$). The stricter condition is $x \leq 5$.

Combining these results, the numbers common to both intervals must satisfy $3 < x \leq 5$.

Now, let's write this in interval notation:

Since $x > 3$, the lower bound is 3 and it is excluded, so we use a parenthesis: $(3$.

Since $x \leq 5$, the upper bound is 5 and it is included, so we use a square bracket: $5]$.

Therefore, $A \cap B = (3, 5]$.

Let's check the options:

(A) $(3, 5]$: This matches our result.

(B) $[3, 5]$: This would mean $3 \leq x \leq 5$, which is incorrect because 3 is excluded from interval B.

(C) $(1, 7]$: This represents the union of the intervals excluding 1 but including 7.

(D) $[1, 7]$: This represents the union of the intervals including both 1 and 7.

The correct answer is $(3, 5]$.

The correct option is (A).

We are given two intervals: $A = [1, 5]$ and $B = (3, 7]$. We need to find their union, $A \cup B$.

Interval A: $[1, 5]$ means all real numbers $x$ such that $1 \leq x \leq 5$. The square brackets indicate that both endpoints, 1 and 5, are included.

Interval B: $(3, 7]$ means all real numbers $x$ such that $3 < x \leq 7$. The parenthesis indicates 3 is excluded, and the square bracket indicates 7 is included.

To find the union ($A \cup B$), we combine all the numbers that belong to either interval A or interval B (or both).

Let's consider the range of numbers covered by these intervals:

• Interval A starts at 1 (inclusive) and ends at 5 (inclusive).
• Interval B starts at 3 (exclusive) and ends at 7 (inclusive).

The union will start at the smallest number included in either interval and end at the largest number included in either interval, covering all numbers in between.

The smallest number involved is 1, which is included in A (due to the square bracket '[').

The largest number involved is 7, which is included in B (due to the square bracket ']').

So, the union starts at 1 (inclusive) and goes up to 7 (inclusive). This covers all numbers from 1 to 7.

Combining the intervals: We need to cover all numbers from the start of A (1, inclusive) up to the end of B (7, inclusive). The overlap from 3 (exclusive) to 5 (inclusive) is automatically covered.

Thus, the union is all numbers $x$ such that $1 \leq x \leq 7$.

In interval notation, this is written as $[1, 7]$.

Let's check the options:

(A) $(3, 5]$: This is the intersection $A \cap B$, not the union.

(B) $[3, 5]$: This is incorrect.

(C) $(1, 7]$: This would mean $1 < x \leq 7$, which is incorrect because 1 is included in A.

(D) $[1, 7]$: This means $1 \leq x \leq 7$, which correctly represents the union of the two intervals.

The correct answer is $[1, 7]$.

The correct option is (D).

Let's analyze each statement to determine its truthfulness.

(A) Every set is a subset of itself.

This statement is true. By the definition of a subset, if all elements of set A are also elements of set A, then A is a subset of A. This is universally true for all sets.

(B) The empty set is a subset of every set.

This statement is true. The empty set ($\emptyset$) contains no elements. The condition for being a subset is that every element in the first set must also be in the second set. Since the empty set has no elements, this condition is vacuously true for any set.

(C) If A is a proper subset of B, then $A \subseteq B$.

This statement is true. The definition of a proper subset ($A \subset B$) requires two conditions: 1) A must be a subset of B ($A \subseteq B$), and 2) A must not be equal to B. Therefore, if A is a proper subset of B, it inherently means that A is also a subset of B.

(D) If A is a proper subset of B, then $\text{n}(A) = \text{n}(B)$.

This statement is false. The definition of a proper subset ($A \subset B$) requires that A is a subset of B ($A \subseteq B$) AND that A is not equal to B ($A \neq B$). If A is not equal to B, and A is a subset of B, it means that B must contain at least one element that is not in A. In terms of cardinality (number of elements), this implies that the number of elements in A must be strictly less than the number of elements in B. That is, $\text{n}(A) < \text{n}(B)$ for finite sets. The statement claims $\text{n}(A) = \text{n}(B)$, which contradicts the definition of a proper subset.

The question asks for the false statement.

The false statement is (D).

We are given the following information about two sets A and B:

• The number of elements in set A, $\text{n}(A) = 4$.
• The number of elements in set B, $\text{n}(B) = 5$.
• The number of elements in the intersection of A and B, $\text{n}(A \cap B) = 2$.

We need to find the number of elements in the union of A and B, $\text{n}(A \cup B)$.

We use the Principle of Inclusion-Exclusion for two sets, which states:

Now, substitute the given values into this formula:

First, perform the addition:

So the equation becomes:

Finally, perform the subtraction:

Therefore, the number of elements in the union of A and B is 7.

Let's check the options:

(A) 7

(B) 9

(C) 11

(D) 1

The correct answer is 7.

The correct option is (A).

Let's analyze each statement to determine which is always true for any set A and the universal set U.

(A) $A \cap \text{U} = \text{U}$

The intersection of a set A with the universal set U includes all elements that are common to both A and U. Since A is a subset of U, all elements of A are also in U. Therefore, the common elements are simply the elements of A. So, $A \cap \text{U} = A$. This statement claims $A \cap \text{U} = \text{U}$, which is only true if A is equal to U. Thus, this statement is not always true.

(B) $A \cup \text{U} = \text{U}$

The union of a set A with the universal set U includes all elements that are in A or in U (or both). Since A is a subset of U, all elements of A are already included in U. Therefore, combining A with U does not add any new elements to U. Thus, $A \cup \text{U} = \text{U}$. This statement is always true.

(C) $A \cap \emptyset = \text{U}$

The intersection of any set A with the empty set $\emptyset$ is always the empty set, because there are no elements common to A and the empty set. The statement claims $A \cap \emptyset = \text{U}$, which is only true if U itself is the empty set, and A is also the empty set. This is not always true.

(D) $A \cup \emptyset = \emptyset$

The union of a set A with the empty set $\emptyset$ results in the set A itself, because the empty set contains no elements to add. So, $A \cup \emptyset = A$. This statement claims $A \cup \emptyset = \emptyset$, which is only true if A is the empty set. Thus, this statement is not always true.

The statement that is always true is $A \cup \text{U} = \text{U}$.

The correct option is (B).

We need to represent the set of all natural numbers greater than 5 in set-builder form.

The set-builder form typically has the structure $\{x \mid \text{condition}\}$.

The elements we are considering are natural numbers. The set of natural numbers is usually denoted by $\mathbb{N}$. Depending on the convention, $\mathbb{N}$ can be $\{1, 2, 3, \dots\}$ or $\{0, 1, 2, 3, \dots\}$. In most contexts related to number theory and basic set theory problems, $\mathbb{N}$ refers to the positive integers $\{1, 2, 3, \dots\}$.

The condition given is "greater than 5". This means the numbers must be strictly larger than 5.

So, we need to combine these two aspects:

• The elements must be natural numbers: $x \in \mathbb{N}$.
• The condition is that they must be greater than 5: $x > 5$.

Putting this into set-builder notation, we get: $\{x \mid x \in \mathbb{N} \text{ and } x > 5\}$.

Let's examine the options:

(A) $\{x : x \in \mathbb{N}, x \geq 5\}$: This includes natural numbers greater than or equal to 5, so it includes 5 itself. The requirement is "greater than 5", which excludes 5.

(B) $\{x : x \in \mathbb{N}, x > 5\}$: This specifies natural numbers that are strictly greater than 5. This perfectly matches our requirement.

(C) $\{x : x \in \mathbb{Z}, x > 5\}$: This specifies integers ($x \in \mathbb{Z}$) greater than 5. While the integers greater than 5 are the same as the natural numbers greater than 5 in this case, the question specifically asks for "natural numbers", so using $\mathbb{N}$ is more precise.

(D) $\{x : x \in \mathbb{N}, x < 5\}$: This represents natural numbers less than 5, which is the opposite of what's required.

The correct representation is $\{x : x \in \mathbb{N}, x > 5\}$.

The correct option is (B).

We are given the following sets:

• Universal set $\text{U} = \{a, b, c, p, q, r, x, y\}$.
• Set $A = \{a, b, c\}$.
• Set $B = \{p, q, r\}$.

We need to find the complement of the intersection of A and B, which is $(A \cap B)'$.

First, let's find the intersection of A and B ($A \cap B$). The intersection contains elements that are common to both sets.

$A = \{a, b, c\}$

$B = \{p, q, r\}$

Comparing the elements of A and B, we see that there are no elements common to both sets.

Therefore, the intersection of A and B is the empty set:

Next, we need to find the complement of this intersection, $(A \cap B)'$.

The complement of a set is found by taking the universal set $\text{U}$ and removing the elements of that set.

So, $(A \cap B)' = \text{U} - (A \cap B)$.

Substituting the value of $A \cap B$ we found:

$(A \cap B)' = \text{U} - \emptyset$.

Removing the empty set from the universal set leaves the universal set unchanged.

Therefore, $(A \cap B)' = \text{U}$.

Let's verify this with the given universal set:

$\text{U} = \{a, b, c, p, q, r, x, y\}$

Since $A \cap B = \emptyset$, the complement of $A \cap B$ is all elements in U that are not in $\emptyset$. This means all elements of U.

So, $(A \cap B)' = \{a, b, c, p, q, r, x, y\}$, which is equal to $\text{U}$.

Let's check the options:

(A) $\emptyset$: This is $A \cap B$, not its complement.

(B) $\{a, b, c, p, q, r\}$: This is the union of A and B ($A \cup B$).

(C) $\{x, y\}$: This is the complement of $A \cup B$, i.e., $(A \cup B)'$.

(D) $\text{U}$: This matches our calculated result.

The correct answer is $\text{U}$.

The correct option is (D).

We are asked to identify the property that states $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$.

Let's review the basic laws of set theory:

• Commutative Laws:
• $A \cup B = B \cup A$
• $A \cap B = B \cap A$
• Associative Laws:
• $(A \cup B) \cup C = A \cup (B \cup C)$
• $(A \cap B) \cap C = A \cap (B \cap C)$
• Distributive Laws:
• $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$
• $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$
• Idempotent Laws:
• $A \cup A = A$
• $A \cap A = A$

The property given in the question, $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$, matches the second form of the Distributive Law, where the union operation distributes over the intersection operation.

Let's examine the options:

(A) Commutative Law: This deals with the order of operands not affecting the result (e.g., $A \cup B = B \cup A$).

(B) Associative Law: This deals with the grouping of operands when performing the same operation multiple times (e.g., $(A \cup B) \cup C = A \cup (B \cup C)$).

(C) Distributive Law: This deals with how one operation distributes over another (e.g., intersection over union, or union over intersection).

(D) Idempotent Law: This deals with an operation on a set with itself resulting in the same set (e.g., $A \cup A = A$).

The property $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$ is indeed the Distributive Law.

The correct option is (C).

We are given two sets:

• $A = \{x : x \text{ is an even number}\}$. This means A contains all integers that are divisible by 2. Examples: ..., -4, -2, 0, 2, 4, ...
• $B = \{x : x \text{ is an odd number}\}$. This means B contains all integers that are not divisible by 2. Examples: ..., -5, -3, -1, 1, 3, 5, ...

We need to find the intersection of A and B, which is $A \cap B$.

The intersection contains all elements that are common to both set A and set B.

An even number is an integer that can be written in the form $2k$ for some integer $k$. An odd number is an integer that can be written in the form $2k + 1$ for some integer $k$.

By definition, an integer is either even or odd, but it cannot be both simultaneously. An integer cannot be divisible by 2 and not divisible by 2 at the same time.

Therefore, there are no elements that are common to the set of even numbers (A) and the set of odd numbers (B).

The intersection of A and B is the empty set.

Now let's look at the options:

(A) The set of all integers: This is incorrect; the intersection contains no integers.

(B) The set of all real numbers: This is incorrect; the intersection contains no real numbers.

(C) A finite set: While the empty set is finite, this option doesn't specifically state what finite set it is. The empty set is a more precise answer.

(D) The empty set: This correctly describes the intersection of the set of even numbers and the set of odd numbers.

The correct answer is the empty set.

The correct option is (D).

We need to identify all months in a year that have strictly less than 30 days. We are asked to ignore leap year variations, which means we consider February to have 28 days for this purpose.

Let's list the number of days in each month:

• January: 31 days
• February: 28 days (ignoring leap year)
• March: 31 days
• April: 30 days
• May: 31 days
• June: 30 days
• July: 31 days
• August: 31 days
• September: 30 days
• October: 31 days
• November: 30 days
• December: 31 days

Now, we need to find the months with less than 30 days. Looking at the list above, the only month with less than 30 days is February (with 28 days).

So, the set of all months of a year having less than 30 days is \{February\}.

Let's check the options:

(A) \{February\}: This matches our finding.

(B) $\emptyset$: This would be correct if no month had less than 30 days.

(C) \{February, September, April, June, November\}: These are the months with 30 days, not less than 30 days.

(D) \{February, April, June, September, November\}: This is the same as (C), listing months with 30 days and also February.

The correct answer is \{February\}.

The correct option is (A).

We are given the set $A = \{1, 2, \{3, 4\}\}$. We need to find its cardinality, denoted by $\text{n}(A)$, which is the number of elements in the set.

In set notation, the curly braces \{\} define the boundaries of the set. Everything within these braces that is separated by commas is considered an element of the set.

Let's identify the elements of set A:

• The first element is the number 1.
• The second element is the number 2.
• The third element is the set $\{3, 4\}$. It's important to note that the curly braces around $\{3, 4\}$ indicate that this entire set is treated as a single element of A. The elements 3 and 4 are inside this element, but they are not directly elements of A.

So, the elements of set A are: 1, 2, and \{3, 4\}.

Counting these elements:

• Element 1: 1
• Element 2: 2
• Element 3: $\{3, 4\}$

There are exactly 3 distinct elements in set A.

Therefore, the cardinality of set A is $\text{n}(A) = 3$.

Let's check the options:

(A) 2: Incorrect, as there are 3 elements.

(B) 3: This matches our count.

(C) 4: Incorrect, as it might be counting 3 and 4 as separate elements of A.

(D) 5: Incorrect.

The correct answer is 3.

The correct option is (B).

We need to represent the set of all real numbers $x$ such that $x > -5$ and $x \leq 0$ using interval notation.

Let's analyze the conditions:

• $x > -5$: This means $x$ is strictly greater than -5. The number -5 is not included.
• $x \leq 0$: This means $x$ is less than or equal to 0. The number 0 is included.

In interval notation:

• "Strictly greater than -5" is represented by a parenthesis '('. So, we start with $(-5$.
• "Less than or equal to 0" is represented by a square bracket ']'. So, we end with $0]$.

Combining these, the interval is $(-5, 0]$.

Let's check the options:

(A) $[-5, 0]$: This means $-5 \leq x \leq 0$. It includes -5 but excludes numbers greater than 0.

(B) $(-5, 0]$: This means $-5 < x \leq 0$. This matches our requirement.

(C) $[-5, 0)$: This means $-5 \leq x < 0$. It includes -5 but excludes 0.

(D) $(-5, 0)$: This means $-5 < x < 0$. It excludes both -5 and 0.

The correct interval notation is $(-5, 0]$.

The correct option is (B).

We are given that $A \cap B = B$.

The intersection of two sets, $A \cap B$, contains all elements that are common to both set A and set B.

The condition $A \cap B = B$ means that the set of elements common to both A and B is exactly the set B itself.

This implies that all elements of B must also be elements of A. If an element is in B, it must also be in A for it to be in their intersection.

Therefore, B must be a subset of A. This is denoted as $B \subseteq A$.

Let's verify with an example:

Let $A = \{1, 2, 3\}$ and $B = \{1, 2\}$.

$A \cap B = \{1, 2\} \cap \{1, 2, 3\} = \{1, 2\}$.

In this case, $A \cap B = B$ is true. And indeed, $B = \{1, 2\}$ is a subset of $A = \{1, 2, 3\}$, so $B \subseteq A$.

Now let's consider the options:

(A) $A \subseteq B$: This would mean all elements of A are in B. If this were true, then $A \cap B$ would be A, not B (unless A and B are equal).

(B) $B \subseteq A$: This means all elements of B are in A. This is consistent with $A \cap B = B$.

(C) $A \cup B = \emptyset$: This would imply both A and B are empty, which is a special case but not generally true.

(D) $A = \emptyset$: This would imply A is empty. If A were empty, then $A \cap B$ would be $\emptyset$, not B (unless B is also empty).

The condition $A \cap B = B$ implies that B is a subset of A.

The correct option is (B).

Set: A set is a collection of distinct and well-defined objects. These objects are called elements or members of the set.

Examples of well-defined collections:

1. The collection of all vowels in the English alphabet. This is well-defined because it is clear which letters are vowels (a, e, i, o, u).

2. The collection of all even numbers between 1 and 10. This is well-defined as we can clearly identify the numbers that are even and fall within this range (2, 4, 6, 8, 10).

Examples of collections that are not well-defined:

1. The collection of all beautiful flowers in a garden. This is not well-defined because the concept of "beautiful" is subjective and varies from person to person.

2. The collection of all good cricket players in the world. This is not well-defined as "good" is a subjective judgment and depends on various criteria which are not universally agreed upon.

The given set is defined as $A = \{x : x \text{ is an integer, } -3 < x \leq 4\}$.

This means that the set A contains all integers that are strictly greater than -3 and less than or equal to 4.

The integers that satisfy these conditions are:

-2, -1, 0, 1, 2, 3, 4

Therefore, the set A in roster form is:

$A = \{-2, -1, 0, 1, 2, 3, 4\}$

The given set is $B = \{ \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6} \}$.

We observe that each element in the set is a fraction where the numerator is one less than the denominator.

Let the general element of the set be represented by $\frac{n}{n+1}$.

For the first element, $\frac{1}{2}$, we have $n=1$. For the second element, $\frac{2}{3}$, we have $n=2$, and so on.

The values of $n$ range from 1 to 5.

Therefore, the set B in set-builder form can be written as:

$B = \{ \frac{n}{n+1} : n \text{ is an integer and } 1 \leq n \leq 5 \}$

Alternatively, we can also express it as:

$B = \{ \frac{n}{n+1} : n \in \{1, 2, 3, 4, 5\} \}$

A set is called finite if it is empty or its elements can be counted, meaning there is a natural number that is the count of its elements. A set is called infinite if it is not finite.

i) The set of lines parallel to the x-axis.

This set is infinite. We can draw infinitely many lines parallel to the x-axis, each with a different y-intercept.

ii) The set of letters in the English alphabet.

This set is finite. The English alphabet has a fixed number of letters (26).

iii) The set of numbers which are multiples of 5.

This set is infinite. We can have multiples of 5 such as 5, 10, 15, 20, and so on, extending indefinitely.

iv) The set of living persons in India.

This set is finite. Although the number is very large, it is a countable and fixed number at any given time.

An empty set is a set which contains no elements. It is denoted by $\emptyset$ or {}.

i) Set of odd natural numbers divisible by 2.

Odd natural numbers are 1, 3, 5, 7, ...

A number is divisible by 2 if it is an even number.

Since a number cannot be both odd and divisible by 2 (i.e., even) at the same time, there are no elements in this set.

Therefore, this is an example of the empty set.

ii) Set of even prime numbers greater than 2.

A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. The prime numbers are 2, 3, 5, 7, 11, ...

The only even prime number is 2.

The set asks for even prime numbers *greater than* 2. Since there are no such numbers, this set contains no elements.

Therefore, this is an example of the empty set.

iii) $\{x : x \text{ is a natural number, } x < 5 \text{ and } x > 7 \}$.

This set requires $x$ to be a natural number that is simultaneously less than 5 AND greater than 7.

There is no natural number that can satisfy both conditions. A number cannot be both less than 5 and greater than 7.

Therefore, this is an example of the empty set.

iv) $\{y : y \text{ is a point common to two parallel lines}\}$.

Parallel lines are defined as lines in a plane that do not meet; that is, two lines in a plane that do not intersect no matter how far they are extended.

By definition, two parallel lines do not have any common points.

Therefore, this is an example of the empty set.

Two sets are equal if and only if they have the same elements. This means that every element of the first set must be an element of the second set, and every element of the second set must be an element of the first set.

i) $A = \{1, 2, 3\}$, $B = \{x : x \text{ is a solution of } x^2 - 5x + 6 = 0 \}$.

First, let's find the elements of set B by solving the quadratic equation $x^2 - 5x + 6 = 0$.

We can factor the quadratic equation:

$(x - 2)(x - 3) = 0$

This gives us two possible solutions:

So, the set B in roster form is $B = \{2, 3\}$.

Now, let's compare sets A and B.

$A = \{1, 2, 3\}$

$B = \{2, 3\}$

The element '1' is in set A but not in set B. Therefore, the sets A and B are not equal.

Reason: The sets are not equal because set A contains the element 1, which is not present in set B.

ii) $A = \{x : x \text{ is a letter in the word FOLLOW}\}$, $B = \{y : y \text{ is a letter in the word FLOW}\}$.

Let's write the sets in roster form by listing the distinct letters in each word.

For the word FOLLOW, the distinct letters are F, O, L, W.

So, $A = \{F, O, L, W\}$.

For the word FLOW, the distinct letters are F, L, O, W.

So, $B = \{F, L, O, W\}$.

Now, let's compare sets A and B.

$A = \{F, O, L, W\}$

$B = \{F, L, O, W\}$

Both sets contain the same elements: F, O, L, and W. The order of elements in a set does not matter.

Therefore, the sets A and B are equal.

Reason: The sets are equal because both sets contain the same distinct letters: F, O, L, and W.

The given set is $A = \{1, 2, \{3, 4\}, 5\}$. This set contains four elements: the number 1, the number 2, the set $\{3, 4\}$, and the number 5.

i) $\{3, 4\} \in A$

The symbol '$\in$' denotes "is an element of". We need to check if the collection $\{3, 4\}$ is directly one of the elements listed within set A.

Looking at set A, we see that one of its elements is indeed the set $\{3, 4\}$.

Therefore, the statement $\{3, 4\} \in A$ is correct.

ii) $\{3, 4\} \subseteq A$

The symbol '$\subseteq$' denotes "is a subset of". For $\{3, 4\}$ to be a subset of A, every element of $\{3, 4\}$ must be an element of A. The elements of $\{3, 4\}$ are 3 and 4.

From set A, we have elements 1, 2, $\{3, 4\}$, and 5.

The number 3 is not an element of A (instead, the set $\{3, 4\}$ is an element of A). Similarly, the number 4 is not an element of A.

Therefore, the statement $\{3, 4\} \subseteq A$ is incorrect.

iii) $\{ \{3, 4\} \} \subseteq A$

This statement checks if the set containing the single element $\{3, 4\}$ is a subset of A. This means that the element $\{3, 4\}$ must be an element of A.

As we identified in part (i), $\{3, 4\}$ is indeed an element of A.

So, the set $\{ \{3, 4\} \}$ has one element, which is $\{3, 4\}$, and this element is present in A.

Therefore, the statement $\{ \{3, 4\} \} \subseteq A$ is correct.

A subset of a set is a set containing only elements that are also in the original set. The empty set is a subset of every set, and every set is a subset of itself.

Given the set $A = \{a, b\}$.

The subsets of A are:

1. The empty set: $\emptyset$ (or {})

2. Subsets with one element:

$\{a\}$

$\{b\}$

3. Subsets with two elements:

$\{a, b\}$

So, all the subsets of A are: $\emptyset$, $\{a\}$, $\{b\}$, $\{a, b\}$.

To find the number of subsets of a set with $n$ elements, we use the formula $2^n$.

In this case, the set $A$ has 2 elements ($n=2$).

Number of subsets = $2^n = 2^2 = 4$.

Therefore, the set $A$ has 4 subsets.

Interval notation is a way of writing subsets of the real number line. Set-builder form describes the set using a property that its elements must satisfy.

i) $[-3, 5]$

This notation represents all real numbers $x$ such that $x$ is greater than or equal to -3 and less than or equal to 5.

In set-builder form: $\{x : x \in \mathbb{R}, -3 \leq x \leq 5\}$

ii) $(2, 7)$

This notation represents all real numbers $x$ such that $x$ is strictly greater than 2 and strictly less than 7.

In set-builder form: $\{x : x \in \mathbb{R}, 2 < x < 7\}$

iii) $[-4, 0)$

This notation represents all real numbers $x$ such that $x$ is greater than or equal to -4 and strictly less than 0.

In set-builder form: $\{x : x \in \mathbb{R}, -4 \leq x < 0\}$

iv) $(-\infty, 3)$

This notation represents all real numbers $x$ such that $x$ is strictly less than 3. The $-\infty$ indicates that the interval extends infinitely to the left.

In set-builder form: $\{x : x \in \mathbb{R}, x < 3\}$

Interval notation is a way of writing subsets of the real number line. We convert the set-builder notation into interval notation based on the inequalities.

i) $\{x : x \in \mathbb{R}, -5 < x \leq 1\}$

The condition $-5 < x \leq 1$ means $x$ is strictly greater than -5 and less than or equal to 1. This corresponds to an open interval on the left and a closed interval on the right.

Interval form: $(-5, 1]$

ii) $\{x : x \in \mathbb{R}, 0 \leq x \leq 10\}$

The condition $0 \leq x \leq 10$ means $x$ is greater than or equal to 0 and less than or equal to 10. This corresponds to a closed interval on both sides.

Interval form: $[0, 10]$

iii) $\{x : x \in \mathbb{R}, x \geq -2\}$

The condition $x \geq -2$ means $x$ is greater than or equal to -2 and extends infinitely to the right. This corresponds to a closed interval on the left and extends to positive infinity.

Interval form: $[-2, \infty)$

iv) $\{x : x \in \mathbb{R}, x < 8\}$

The condition $x < 8$ means $x$ is strictly less than 8 and extends infinitely to the left. This corresponds to an open interval on the right and extends from negative infinity.

Interval form: $(-\infty, 8)$

Given the universal set $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$, set $A = \{1, 2, 3, 4\}$, and set $B = \{2, 4, 6, 8\}$.

Union of sets ($A \cup B$):

The union of two sets A and B, denoted by $A \cup B$, is the set containing all elements that are in A, or in B, or in both. We list each unique element only once.

$A \cup B = \{1, 2, 3, 4\} \cup \{2, 4, 6, 8\}$

Combining the elements from both sets and removing duplicates:

$A \cup B = \{1, 2, 3, 4, 6, 8\}$

Intersection of sets ($A \cap B$):

The intersection of two sets A and B, denoted by $A \cap B$, is the set containing all elements that are common to both A and B.

$A \cap B = \{1, 2, 3, 4\} \cap \{2, 4, 6, 8\}$

The elements that are present in both set A and set B are 2 and 4.

$A \cap B = \{2, 4\}$

From Question 11, we have:

$U = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$

$A = \{1, 2, 3, 4\}$

$B = \{2, 4, 6, 8\}$

Difference of sets ($A - B$):

The difference of two sets A and B, denoted by $A - B$, is the set of all elements that are in A but not in B.

$A - B = \{x : x \in A \text{ and } x \notin B\}$

We look at the elements of A and remove any element that is also present in B.

Elements in A: 1, 2, 3, 4.

Elements in B: 2, 4, 6, 8.

The elements in A that are not in B are 1 and 3.

So, $A - B = \{1, 3\}$.

Difference of sets ($B - A$):

The difference of two sets B and A, denoted by $B - A$, is the set of all elements that are in B but not in A.

$B - A = \{x : x \in B \text{ and } x \notin A\}$

We look at the elements of B and remove any element that is also present in A.

Elements in B: 2, 4, 6, 8.

Elements in A: 1, 2, 3, 4.

The elements in B that are not in A are 6 and 8.

So, $B - A = \{6, 8\}$.

Complement of a set ($A'$):

The complement of a set A, denoted by $A'$ (or $A^c$), is the set of all elements in the universal set U that are not in A.

$A' = \{x : x \in U \text{ and } x \notin A\}$

We look at the elements of the universal set U and remove any element that is present in set A.

$U = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$

$A = \{1, 2, 3, 4\}$

The elements in U that are not in A are 5, 6, 7, 8, 9.

So, $A' = \{5, 6, 7, 8, 9\}$.

Given set $A = \{1, 3, 5, 7\}$ and set $B = \{2, 4, 6, 8\}$.

Union of sets ($A \cup B$):

The union of sets A and B is the set of all elements that are in A, or in B, or in both.

$A \cup B = \{1, 3, 5, 7\} \cup \{2, 4, 6, 8\}$

Combining all elements from both sets:

$A \cup B = \{1, 2, 3, 4, 5, 6, 7, 8\}$

Intersection of sets ($A \cap B$):

The intersection of sets A and B is the set of all elements that are common to both A and B.

$A \cap B = \{1, 3, 5, 7\} \cap \{2, 4, 6, 8\}$

We observe that there are no common elements between set A and set B.

$A \cap B = \emptyset$ (the empty set).

Kind of sets with respect to their intersection:

When the intersection of two sets is the empty set ($A \cap B = \emptyset$), the sets are called disjoint sets.

In this case, since $A \cap B = \emptyset$, sets A and B are disjoint sets.

To represent $(A \cup B)'$ using a Venn diagram, we need to understand the components:

• The universal set, typically represented by a rectangle.
• Set A, represented by a circle.
• Set B, represented by another circle, possibly overlapping with A.
• The union of A and B ($A \cup B$), which is the region encompassing all elements in A, or in B, or in both. This is the combined area of both circles.
• The complement of $(A \cup B)$, denoted as $(A \cup B)'$, which includes all elements in the universal set that are *not* in $A \cup B$.

Construction of the Venn Diagram:

1. Draw the Universal Set: Draw a rectangle. This rectangle represents the universal set, U, containing all possible elements under consideration.

2. Draw Sets A and B: Draw two circles inside the rectangle. Label one circle 'A' and the other 'B'. These circles can overlap, indicating that there might be elements common to both sets.

3. Identify the Union ($A \cup B$): The region representing $A \cup B$ is the area covered by circle A, circle B, and the overlapping region between them. This is the entire area encompassed by both circles.

4. Represent the Complement ($(A \cup B)'$): The complement of the union means all elements in the universal set (the rectangle) that are *outside* of the union of A and B. Therefore, you need to shade the region of the rectangle that is *not* covered by either circle A or circle B.

Visual Description:

The Venn diagram will show a rectangle (U). Inside this rectangle, there will be two circles, A and B, which may overlap. The shaded region representing $(A \cup B)'$ will be the entire area within the rectangle, excluding all parts of circle A and all parts of circle B. In simpler terms, it's the area outside of both circles but within the universal set's boundary.

To represent the set $\mathbf{A \cap (B \cup C)}$ using a Venn diagram, we follow these steps:

First, draw a rectangle to represent the universal set. Inside this rectangle, draw three overlapping circles representing sets A, B, and C.

Identify the region corresponding to the union of sets B and C, which is $\mathbf{B \cup C}$. This region includes all elements that are in B, or in C, or in both B and C. Visually, this is the area covered by the circles B and C combined.

Next, consider the set A, which is represented by the circle A.

The expression $\mathbf{A \cap (B \cup C)}$ represents the intersection of set A and the region $(B \cup C)$. This includes all elements that are common to both set A and the region $(B \cup C)$.

Therefore, the region representing $\mathbf{A \cap (B \cup C)}$ in the Venn diagram is the area where the circle A overlaps with the combined area of circles B and C ($B \cup C$).

In the diagram, shade the portion of circle A that is also within either circle B or circle C (or both). This shaded area is the visual representation of $\mathbf{A \cap (B \cup C)}$.

Given:

$|A| = 20$

$|B| = 25$

$|A \cup B| = 40$

To Find:

The value of $|A \cap B|$.

Solution:

We use the formula relating the sizes of the union and intersection of two sets:

$|A \cup B| = |A| + |B| - |A \cap B|$

We are given the values for $|A|$, $|B|$, and $|A \cup B|$. We can rearrange the formula to solve for $|A \cap B|$.

Rearranging the formula, we get:

$|A \cap B| = |A| + |B| - |A \cup B|$

Substitute the given values into this equation:

$|A \cap B| = 20 + 25 - 40$

Perform the addition and subtraction:

$|A \cap B| = 45 - 40$

$|A \cap B| = 5$

Thus, the number of elements in the intersection of sets A and B is 5.

Final Answer:

$|A \cap B| = 5$

Given:

Total number of people in the group = 60

Number of people who like tea = 35

Number of people who like coffee = 40

Each person likes at least one of the drinks.

To Find:

The number of people who like both tea and coffee.

Solution:

Let T be the set of people who like tea, and C be the set of people who like coffee.

We are given:

$|T| = 35$

$|C| = 40$

Since each person likes at least one drink, the total number of people in the group is the number of people who like tea or coffee (or both). This is represented by the union of the two sets.

$|T \cup C| = 60$

We need to find the number of people who like both tea and coffee, which is the intersection of the two sets, $|T \cap C|$.

We use the principle of inclusion-exclusion for two sets:

$|T \cup C| = |T| + |C| - |T \cap C|$

Substitute the given values into the formula:

Simplify the right side of the equation:

Now, we solve for $|T \cap C|$. Add $|T \cap C|$ to both sides and subtract 60 from both sides:

Perform the subtraction:

$|T \cap C| = 15$

Therefore, 15 people like both tea and coffee.

Final Answer:

The number of people who like both tea and coffee is 15.

Given that $\mathbf{A \subseteq B}$. This means that every element in set A is also an element in set B.

About $A \cap B$:

The intersection of A and B, denoted by $\mathbf{A \cap B}$, consists of all elements that are common to both set A and set B.

Since every element of A is also an element of B (because $A \subseteq B$), the elements common to both sets are exactly the elements of set A.

Therefore, if $A \subseteq B$, then $\mathbf{A \cap B = A}$.

About $A \cup B$:

The union of A and B, denoted by $\mathbf{A \cup B}$, consists of all elements that are in set A or in set B (or in both).

Since every element in A is already in B, the union of A and B will include all the elements of B and all the elements of A. As A's elements are already in B, including A's elements doesn't add anything new to B.

Therefore, if $A \subseteq B$, then $\mathbf{A \cup B = B}$.

Conclusion:

If $A \subseteq B$, then:

$\mathbf{A \cap B = A}$

$\mathbf{A \cup B = B}$

Given:

$U$: The set of all students in a school.

$A$: The set of students who play cricket.

$B$: The set of students who play football.

Description of the set $A' \cap B'$ in words:

The set $A'$ represents the complement of set A with respect to the universal set U. In this context, $A'$ is the set of all students in the school who are not in set A, meaning the students who do not play cricket.

The set $B'$ represents the complement of set B with respect to the universal set U. In this context, $B'$ is the set of all students in the school who are not in set B, meaning the students who do not play football.

The intersection $A' \cap B'$ consists of the elements that are common to both $A'$ and $B'$. Therefore, $A' \cap B'$ is the set of all students who are in $A'$ and in $B'$.

Putting this together, the set $A' \cap B'$ describes the set of all students in the school who do not play cricket and do not play football.

Alternatively, using De Morgan's Law, $A' \cap B' = (A \cup B)'$. The set $A \cup B$ is the set of students who play cricket or football (or both). Therefore, $(A \cup B)'$ is the set of students who are not in $A \cup B$, meaning the students who play neither cricket nor football.

Thus, the set $A' \cap B'$ can be described as:

The set of students in the school who play neither cricket nor football.

The power set of a set $S$, denoted by $\mathcal{P}(S)$, is the set of all subsets of $S$, including the empty set and the set $S$ itself.

The given set is $S = \{a, b, c\}$.

To find the power set, we list all possible subsets of $S$. These subsets can have 0, 1, 2, or 3 elements.

Subsets with 0 elements: The empty set, $\emptyset$.

Subsets with 1 element: $\{a\}, \{b\}, \{c\}$.

Subsets with 2 elements: $\{a, b\}, \{a, c\}, \{b, c\}$.

Subsets with 3 elements: $\{a, b, c\}$.

Combining all these subsets, we get the power set $\mathcal{P}(S)$.

$\mathcal{P}(S) = \{\emptyset, \{a\}, \{b\}, \{c\}, \{a, b\}, \{a, c\}, \{b, c\}, \{a, b, c\}\}$.

To find the number of elements in the power set of a set with $n$ elements, we use the formula $2^n$.

In this case, the set $S$ has $n = 3$ elements.

The number of elements in the power set of $S$ is $2^3$.

$2^3 = 2 \times 2 \times 2 = 8$.

The power set of $S = \{a, b, c\}$ is $\mathcal{P}(S) = \{\emptyset, \{a\}, \{b\}, \{c\}, \{a, b\}, \{a, c\}, \{b, c\}, \{a, b, c\}\}$.

There are 8 elements in the power set of $S$.

Given the sets $A = \{x : x^2 - 9 = 0\}$ and $B = \{x : (x-3)(x+1) = 0\}$.

First, let's find the elements of set A by solving the equation $x^2 - 9 = 0$.

$x^2 - 9 = 0$

$x^2 = 9$

Taking the square root of both sides:

$x = \pm\sqrt{9}$

$x = 3$ or $x = -3$.

So, the elements of set A are the solutions to this equation.

$A = \{-3, 3\}$.

Next, let's find the elements of set B by solving the equation $(x-3)(x+1) = 0$.

For the product of two factors to be zero, at least one of the factors must be zero.

So, either $x - 3 = 0$ or $x + 1 = 0$.

If $x - 3 = 0$, then $x = 3$.

If $x + 1 = 0$, then $x = -1$.

So, the elements of set B are the solutions to this equation.

$B = \{-1, 3\}$.

Now, we need to find the union of sets A and B, $A \cup B$. The union of two sets is the set of all elements which are in either set, or in both sets.

$A \cup B = \{-3, 3\} \cup \{-1, 3\}$

$A \cup B = \{-3, -1, 3\}$.

Finally, we need to find the intersection of sets A and B, $A \cap B$. The intersection of two sets is the set of all elements which are common to both sets.

$A \cap B = \{-3, 3\} \cap \{-1, 3\}$

The element common to both sets is 3.

$A \cap B = \{3\}$.

Thus, $A \cup B = \{-3, -1, 3\}$ and $A \cap B = \{3\}$.

Given set $A = \{x : x \in \mathbb{Z}, x^2 \leq 4\}$ and set $B = \{-2, -1, 0, 1, 2\}$.

First, let's determine the elements of set A. Set A consists of integers ($x \in \mathbb{Z}$) such that the square of the integer is less than or equal to 4 ($x^2 \leq 4$).

We need to find all integers $x$ satisfying $x^2 \leq 4$.

This inequality $x^2 \leq 4$ is equivalent to $-\sqrt{4} \leq x \leq \sqrt{4}$, which simplifies to $-2 \leq x \leq 2$.

The integers that satisfy the condition $-2 \leq x \leq 2$ are $-2, -1, 0, 1, 2$.

Therefore, the elements of set A are:

$A = \{-2, -1, 0, 1, 2\}$.

Now, we compare the elements of set A with the elements of set B.

$A = \{-2, -1, 0, 1, 2\}$

$B = \{-2, -1, 0, 1, 2\}$

By comparing the elements, we see that every element in set A is also in set B, and every element in set B is also in set A.

Two sets are considered equal if and only if they have exactly the same elements.

Since set A and set B contain precisely the same elements, they are equal sets.

Thus, the sets A and B are equal because $A = \{-2, -1, 0, 1, 2\}$ and $B = \{-2, -1, 0, 1, 2\}$, showing that $A \subseteq B$ and $B \subseteq A$.

Given the sets $A = \{1, 2, 3\}$ and $B = \{2, 3, 4\}$.

We need to find $(A \setminus B) \cup (B \setminus A)$.

First, let's find $A \setminus B$. The set $A \setminus B$ consists of elements that are in set A but not in set B.

Elements in A: 1, 2, 3

Elements in B: 2, 3, 4

Comparing the two sets, the elements in A that are not in B are only 1.

$A \setminus B = \{1\}$.

Next, let's find $B \setminus A$. The set $B \setminus A$ consists of elements that are in set B but not in set A.

Elements in B: 2, 3, 4

Elements in A: 1, 2, 3

Comparing the two sets, the elements in B that are not in A are only 4.

$B \setminus A = \{4\}$.

Finally, we find the union of $A \setminus B$ and $B \setminus A$. The union of two sets contains all elements that are in either set.

$(A \setminus B) \cup (B \setminus A) = \{1\} \cup \{4\}$.

$(A \setminus B) \cup (B \setminus A) = \{1, 4\}$.

Thus, $(A \setminus B) \cup (B \setminus A) = \mathbf{\{1, 4\}}$.

Given the universal set $U = \{1, 2, 3, 4, 5\}$, set $A = \{1, 3, 5\}$, and set $B = \{2, 4\}$.

First, we find $A \cup B$. The union of sets A and B contains all elements that are in A or B or both.

$A \cup B = \{1, 3, 5\} \cup \{2, 4\}$

$A \cup B = \{1, 2, 3, 4, 5\}$.

Next, we find $(A \cup B)'$. The complement of $(A \cup B)$ is the set of elements in the universal set $U$ that are not in $A \cup B$.

$(A \cup B)' = U \setminus (A \cup B)$

$(A \cup B)' = \{1, 2, 3, 4, 5\} \setminus \{1, 2, 3, 4, 5\}$

$(A \cup B)' = \emptyset$ (the empty set).

Now, we find $A'$. The complement of A is the set of elements in the universal set $U$ that are not in A.

$A' = U \setminus A$

$A' = \{1, 2, 3, 4, 5\} \setminus \{1, 3, 5\}$

$A' = \{2, 4\}$.

Next, we find $B'$. The complement of B is the set of elements in the universal set $U$ that are not in B.

$B' = U \setminus B$

$B' = \{1, 2, 3, 4, 5\} \setminus \{2, 4\}$

$B' = \{1, 3, 5\}$.

Finally, we find $A' \cap B'$. The intersection of $A'$ and $B'$ contains elements common to both $A'$ and $B'$.

$A' \cap B' = \{2, 4\} \cap \{1, 3, 5\}$

$A' \cap B' = \emptyset$.

We found that $(A \cup B)' = \emptyset$ and $A' \cap B' = \emptyset$.

Observation: We observe that $(A \cup B)' = A' \cap B'$. This illustrates one of De Morgan's Laws for sets.

Two sets are said to be disjoint if they have no common elements.

In other words, the intersection of two disjoint sets is the empty set.

If A and B are two sets, they are disjoint if and only if $A \cap B = \emptyset$.

Example:

Let set $P = \{1, 3, 5, 7\}$ be the set of the first four positive odd numbers.

Let set $Q = \{2, 4, 6, 8\}$ be the set of the first four positive even numbers.

To check if sets P and Q are disjoint, we find their intersection, $P \cap Q$.

$P \cap Q = \{1, 3, 5, 7\} \cap \{2, 4, 6, 8\}$

We look for elements that are present in both set P and set Q.

There are no elements that are common to both sets P and Q.

Therefore, the intersection of P and Q is the empty set.

$P \cap Q = \emptyset$.

Since their intersection is the empty set, the sets P and Q are disjoint sets.

Given that $A$ is a subset of $B$, which is denoted by $A \subseteq B$. This means that every element in set A is also an element in set B.

The universal set is $U$. The complement of a set X, denoted by $X'$, is the set of all elements in $U$ that are not in X. Mathematically, $X' = U \setminus X = \{x \in U \mid x \notin X\}$.

We want to find the relationship between $A'$ and $B'$.

Consider an element $x$.

If $x \in B'$, by definition of complement, $x \in U$ and $x \notin B$.

Since $A \subseteq B$, if an element is not in B, it cannot be in A either (because if it were in A, it would also have to be in B). So, if $x \notin B$, then $x \notin A$.

Therefore, if $x \in B'$, then $x \in U$ and $x \notin A$.

By the definition of complement, this means $x \in A'$.

Thus, if an element is in $B'$, it must also be in $A'$. This shows that $B'$ is a subset of $A'$.

Relationship: If $A \subseteq B$, then $B' \subseteq A'$.

In words, the complement of the larger set (B) is a subset of the complement of the smaller set (A).

Venn Diagram:

Let the rectangle represent the universal set $U$. Draw a circle for set B inside $U$, and a circle for set A inside set B (since $A \subseteq B$).

$A'$ is the region outside of the circle for A within the rectangle U.

$B'$ is the region outside of the circle for B within the rectangle U.

Visually, the region outside B is clearly contained within the region outside A.

Example Diagram (text description):

[Universal Set U]

[Set B]

[Set A]

The area outside B (B') is part of the area outside A (A').

Given the set $S = \{ \phi, \{1\}, \{2\}, \{1, 2\} \}$.

A power set is the set of all subsets of a given set. Let the given set be $A$. Then its power set is $\mathcal{P}(A)$.

The number of elements in the power set of a set with $n$ elements is $2^n$.

In the given set $S$, there are 4 elements: $\phi$, $\{1\}$, $\{2\}$, and $\{1, 2\}$.

If $S$ is a power set of some set $A$, then the number of elements in $S$ must be $2^{|A|}$, where $|A|$ is the number of elements in $A$.

We have $|S| = 4$. So, we are looking for a set $A$ such that $2^{|A|} = 4$.

Solving for $|A|$, we get $|A| = 2$ since $2^2 = 4$.

So, if $S$ is a power set of some set, that set must contain exactly 2 elements.

The elements of a power set are the subsets of the original set. The elements of $S$ are $\phi$, $\{1\}$, $\{2\}$, and $\{1, 2\}$.

These elements are the subsets of the original set.

The elements of the original set are the elements that appear in the single-element subsets within the power set (excluding the empty set, which is always a subset). The single-element subsets in S are $\{1\}$ and $\{2\}$.

This suggests that the original set might contain the elements 1 and 2.

Let's consider the set $A = \{1, 2\}$.

The subsets of $A = \{1, 2\}$ are:

Subset with 0 elements: $\emptyset$ (or $\phi$).

Subsets with 1 element: $\{1\}, \{2\}$.

Subset with 2 elements: $\{1, 2\}$.

The power set of $A = \{1, 2\}$ is $\mathcal{P}(A) = \{\emptyset, \{1\}, \{2\}, \{1, 2\}\}$.

Comparing the given set $S$ with $\mathcal{P}(\{1, 2\})$, we see that they are identical.

$S = \{ \phi, \{1\}, \{2\}, \{1, 2\} \}$

$\mathcal{P}(\{1, 2\}) = \{\emptyset, \{1\}, \{2\}, \{1, 2\}\}$. Note that $\phi$ and $\emptyset$ both represent the empty set.

Yes, $S$ is a power set of some set.

The set whose power set is S is $\{1, 2\}$.

Justification:

The number of elements in S is 4. If S is a power set, the original set must have 2 elements ($2^2=4$).

The elements of the original set must be the elements whose singletons (sets containing only that element) are present in the power set.

The singletons in S are $\{1\}$ and $\{2\}$. This indicates the original set is $\{1, 2\}$.

We verified that the power set of $\{1, 2\}$ is indeed $\{\emptyset, \{1\}, \{2\}, \{1, 2\}\}$, which is equal to S.

Given the sets $A = \{x : x \in \mathbb{N}, x < 10\}$ and $B = \{x : x \text{ is a prime number less than } 15\}$.

First, let's list the elements of set A.

Set A consists of natural numbers ($x \in \mathbb{N}$) that are less than 10 ($x < 10$).

The natural numbers are $\{1, 2, 3, 4, 5, ...\}$.

The natural numbers less than 10 are 1, 2, 3, 4, 5, 6, 7, 8, 9.

So, $A = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$.

Next, let's list the elements of set B.

Set B consists of prime numbers that are less than 15.

A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.

The prime numbers less than 15 are 2, 3, 5, 7, 11, 13.

So, $B = \{2, 3, 5, 7, 11, 13\}$.

We need to find $A \cap B$, which is the intersection of sets A and B. The intersection of two sets is the set containing all elements that are common to both sets.

$A = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$

$B = \{2, 3, 5, 7, 11, 13\}$

The elements that are present in both set A and set B are 2, 3, 5, and 7.

So, $A \cap B = \{2, 3, 5, 7\}$.

Thus, the intersection of sets A and B is $\mathbf{\{2, 3, 5, 7\}}$.

Given that $U$ is the universal set and $A$ is a set within $U$.

We need to simplify the expression $(A')'$.

The complement of a set $A$, denoted by $A'$, is the set of all elements in the universal set $U$ that are not in $A$.

Mathematically, $A' = \{x \in U \mid x \notin A\}$.

Now, $(A')'$ is the complement of the set $A'$. This means it is the set of all elements in the universal set $U$ that are not in $A'$.

Mathematically, $(A')' = \{x \in U \mid x \notin A'\}$.

Let's consider an element $x$.

An element $x$ is in $(A')'$ if and only if $x$ is in $U$ and $x$ is not in $A'$.

$x \in (A')' \Leftrightarrow x \in U \text{ and } x \notin A'$.

According to the definition of $A'$, an element $x$ is in $A'$ if and only if $x$ is in $U$ and $x$ is not in $A$.

$x \in A' \Leftrightarrow x \in U \text{ and } x \notin A$.

Therefore, an element $x$ is not in $A'$ if and only if it is not the case that ($x \in U$ and $x \notin A$).

Assuming $A$ is a subset of $U$, $x \in U$ is usually implicit when discussing elements of $A'$. So, $x \in A' \Leftrightarrow x \notin A$.

Thus, $x \notin A' \Leftrightarrow \neg(x \notin A) \Leftrightarrow x \in A$.

So, $x \in (A')' \Leftrightarrow x \in U \text{ and } x \in A$.

Since $A \subseteq U$, if $x \in A$, then $x \in U$ is automatically true.

Therefore, $x \in (A')' \Leftrightarrow x \in A$.

This shows that the elements of $(A')'$ are exactly the elements of $A$.

Hence, $(A')' = A$.

This is often referred to as the Involution Law or the Double Complement Law in set theory.

The simplified form of $(A')'$ is A.

We will use Venn diagrams to illustrate the regions corresponding to the Left Hand Side (LHS) and the Right Hand Side (RHS) of the given equation and show that they are the same.

Left Hand Side (LHS): $A \cup (B \cap C)$

To represent $A \cup (B \cap C)$ using a Venn diagram, we first need to identify the region representing $B \cap C$, and then take the union with set A.

Step 1: Represent $B \cap C$.

In a Venn diagram with three overlapping circles representing sets A, B, and C within a universal set U, the set $B \cap C$ is represented by the region common to both circle B and circle C. This is the overlapping area between B and C.

(Imagine shading the intersection area of B and C)

Step 2: Represent $A \cup (B \cap C)$.

The set $A \cup (B \cap C)$ is the union of set A and the region representing $B \cap C$. This is represented by shading the entire region corresponding to circle A along with the region representing the intersection of B and C (shaded in Step 1).

The final shaded region for the LHS includes all of set A, and the area where B and C overlap.

Right Hand Side (RHS): $(A \cup B) \cap (A \cup C)$

To represent $(A \cup B) \cap (A \cup C)$ using a Venn diagram, we first need to identify the regions representing $A \cup B$ and $A \cup C$, and then find their intersection.

Step 1: Represent $A \cup B$.

The set $A \cup B$ is the union of sets A and B. This is represented by shading the entire region corresponding to circle A and the entire region corresponding to circle B.

(Imagine shading all of circle A and all of circle B)

Step 2: Represent $A \cup C$.

The set $A \cup C$ is the union of sets A and C. This is represented by shading the entire region corresponding to circle A and the entire region corresponding to circle C.

(Imagine shading all of circle A and all of circle C)

Step 3: Represent $(A \cup B) \cap (A \cup C)$.

The set $(A \cup B) \cap (A \cup C)$ is the intersection of the regions representing $A \cup B$ (from Step 1) and $A \cup C$ (from Step 2). This is represented by shading only the region that is common to both the shaded area from Step 1 and the shaded area from Step 2.

The final shaded region for the RHS is the area where the shaded region for $A \cup B$ overlaps with the shaded region for $A \cup C$. This results in the region that includes all of set A, and the area where B and C overlap.

Observation:

By comparing the final shaded region representing $A \cup (B \cap C)$ (LHS) and the final shaded region representing $(A \cup B) \cap (A \cup C)$ (RHS), we observe that both shaded regions are exactly the same.

This visual representation using Venn diagrams demonstrates that $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$. This property is known as the Distributive Law of Union over Intersection in set theory.

Given:

$|A| = 30$ (The number of elements in set A is 30)

$|B| = 40$ (The number of elements in set B is 40)

$|A \cap B| = 15$ (The number of elements in the intersection of A and B is 15)

To Find:

$|A \cup B|$ (The number of elements in the union of A and B)

Solution:

We can find the number of elements in the union of two sets A and B using the Inclusion-Exclusion Principle, which is given by the formula:

$|A \cup B| = |A| + |B| - |A \cap B|$

Substitute the given values into the formula:

$|A \cup B| = 30 + 40 - 15$

$|A \cup B| = 70 - 15$

$|A \cup B| = 55$

Thus, the number of elements in the union of sets A and B is 55.

$|A \cup B| = \mathbf{55}$.

Given the interval $[-2, 5)$.

The interval notation $[-2, 5)$ represents all real numbers $x$ such that $x$ is greater than or equal to -2 and less than 5.

The square bracket `[` at -2 indicates that -2 is included in the interval.

The round bracket `)` at 5 indicates that 5 is excluded from the interval.

In terms of inequalities, this can be written as:

$-2 \leq x < 5$

The elements in the interval are real numbers, so we specify that $x \in \mathbb{R}$.

The set-builder form is written as $\{x \mid \text{condition(s) on } x\}$.

Combining the conditions, the set-builder form for the interval $[-2, 5)$ is:

$\{x \mid x \in \mathbb{R}, -2 \leq x < 5\}$

Thus, the set-builder form of the interval $[-2, 5)$ is $\{x \mid x \in \mathbb{R}, -2 \leq x < 5\}$.

Given the set $A = \{a, \{b\}\}$.

We need to list all the subsets of set A.

First, identify the elements of set A. Set A has two elements:

Element 1: $a$

Element 2: $\{b\}$

So, the number of elements in A is $|A| = 2$.

The number of subsets of a set with $n$ elements is $2^n$. In this case, $n=2$, so the number of subsets is $2^2 = 4$.

Now, let's list all the possible subsets of A:

1. The empty set ($\emptyset$ or $\phi$). The empty set is a subset of every set.

Subset: $\emptyset$

2. Subsets containing one element from A.

Take the first element 'a': $\{a\}$

Take the second element '{b}': $\{\{b\}\}$ (Note: the element itself is $\{b\}$, so the subset containing this element is a set with $\{b\}$ inside it).

3. Subsets containing all elements from A.

This is the set A itself: $\{a, \{b\}\}$.

Listing all the subsets together, we get:

The subsets of $A = \{a, \{b\}\}$ are $\emptyset$, $\{a\}$, $\{\{b\}\}$, and $\{a, \{b\}\}$.

Thus, the list of all subsets of $A$ is $\{\emptyset, \{a\}, \{\{b\}\}, \{a, \{b\}\}\}$.

Given the universal set $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$.

Set $A = \{x : x \text{ is prime}\}$ within $U$.

Set $B = \{x : x \text{ is even}\}$ within $U$.

First, let's identify the elements of set A.

The prime numbers in the universal set U are the numbers greater than 1 that have no positive divisors other than 1 and themselves.

The prime numbers in $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ are 2, 3, 5, and 7.

So, $A = \{2, 3, 5, 7\}$.

Next, let's identify the elements of set B.

The even numbers in the universal set U are the numbers divisible by 2.

The even numbers in $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ are 2, 4, 6, 8, and 10.

So, $B = \{2, 4, 6, 8, 10\}$.

We need to find $A'$. The complement of A ($A'$) is the set of elements in the universal set $U$ that are not in A.

$A' = U \setminus A$

$A' = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} \setminus \{2, 3, 5, 7\}$

$A' = \{1, 4, 6, 8, 9, 10\}$.

Next, we need to find $A \cap B$. The intersection of A and B ($A \cap B$) is the set of elements that are common to both set A and set B.

$A = \{2, 3, 5, 7\}$

$B = \{2, 4, 6, 8, 10\}$

The elements common to both A and B are those numbers that appear in both lists.

The only common element is 2.

$A \cap B = \{2\}$.

Thus, $A' = \mathbf{\{1, 4, 6, 8, 9, 10\}}$ and $A \cap B = \mathbf{\{2\}}$.

Let $U$ be the set of all people surveyed.

Let $A$ be the set of people who like product A.

Let $B$ be the set of people who like product B.

Given:

Total number of people surveyed, $|U| = 50$.

Number of people who like product A, $|A| = 30$.

Number of people who like product B, $|B| = 25$.

Number of people who like both product A and product B, $|A \cap B| = 10$.

To Find:

The number of people who like neither product A nor product B. This is the number of elements in the complement of the union of A and B, i.e., $|(A \cup B)'|$.

Solution:

We know that the number of elements in the complement of a set is the total number of elements minus the number of elements in the set.

$|(A \cup B)'| = |U| - |A \cup B|$

To find $|(A \cup B)'|$, we first need to find $|A \cup B|$.

We can use the Inclusion-Exclusion Principle to find the number of elements in the union of two sets A and B:

$|A \cup B| = |A| + |B| - |A \cap B|$

Substitute the given values into the formula for the union:

$|A \cup B| = 30 + 25 - 10$

$|A \cup B| = 55 - 10$

$|A \cup B| = 45$

So, 45 people like at least one of the products (A or B or both).

Now, substitute the values of $|U|$ and $|A \cup B|$ into the formula for the complement:

$|(A \cup B)'| = |U| - |A \cup B|$

$|(A \cup B)'| = 50 - 45$

$|(A \cup B)'| = 5$

Therefore, the number of people who like neither product A nor product B is 5.

The number of people who like neither product A nor product B is 5.

Two sets A and B are said to be comparable if either $A \subseteq B$ (A is a subset of B) or $B \subseteq A$ (B is a subset of A).

Given set $A = \{x : x^2 - 5x = 0\}$.

To find the elements of set A, we solve the equation $x^2 - 5x = 0$.

Factor out x:

$x(x - 5) = 0$

This gives two possible solutions:

$x = 0$ or $x - 5 = 0$, which means $x = 5$.

So, the elements of set A are the solutions to this equation.

$A = \{0, 5\}$.

Given set $B = \{x : x \text{ is a positive integer less than 6}\}$.

A positive integer is an integer greater than 0. The positive integers less than 6 are 1, 2, 3, 4, 5.

So, $B = \{1, 2, 3, 4, 5\}$.

Now, let's check if A and B are comparable.

Is $A \subseteq B$? This means every element in A must be in B.

The elements of A are 0 and 5.

The element 0 is in A, but 0 is not in B (since B contains only positive integers). Therefore, $A$ is not a subset of $B$ ($A \not\subseteq B$).

Is $B \subseteq A$? This means every element in B must be in A.

The elements of B are 1, 2, 3, 4, 5.

The element 1 is in B, but 1 is not in A. The element 2 is in B, but 2 is not in A, and so on. Therefore, $B$ is not a subset of $A$ ($B \not\subseteq A$).

Since neither $A \subseteq B$ nor $B \subseteq A$ is true, the sets A and B are not comparable.

Thus, the sets A and B are not comparable.

The given interval is $[2, 8)$.

This interval notation represents the set of all real numbers $x$ such that $x$ is greater than or equal to 2 and strictly less than 8.

In inequality form, this is written as $2 \leq x < 8$.

To represent this interval on a number line, follow these steps:

1. Draw a horizontal line to represent the number line.

2. Mark some points on the number line, including 0, 2, and 8, for reference.

3. At the point corresponding to the number 2, draw a solid circle (a filled dot). The square bracket `[` in the interval notation indicates that 2 is included in the interval, corresponding to the $\leq$ inequality.

4. At the point corresponding to the number 8, draw an open circle (an unfilled dot). The round bracket `)` in the interval notation indicates that 8 is excluded from the interval, corresponding to the $<$ inequality.

5. Draw a thick line or shade the region on the number line between the solid circle at 2 and the open circle at 8. This shaded region represents all the real numbers $x$ that are in the interval $[2, 8)$.

Visual representation (description):

A number line extending horizontally. Points like 0, 2, and 8 are marked.

At the point 2, there is a filled circle.

At the point 8, there is an unfilled circle.

The segment of the number line between the filled circle at 2 and the unfilled circle at 8 is shaded or drawn with a thicker line.

We need to write the set of all vowels in the English alphabet in roster form.

The vowels in the English alphabet are 'a', 'e', 'i', 'o', and 'u'.

Roster form is a way of representing a set by listing all its elements within curly braces $\{ \}$, with the elements separated by commas.

Let the set of all vowels in the English alphabet be denoted by V.

Listing the elements in curly braces:

$V = \{a, e, i, o, u\}$

Thus, the set of all vowels in the English alphabet in roster form is $\mathbf{\{a, e, i, o, u\}}$.

A set is called a finite set if it is the empty set or if it can be put into one-to-one correspondence with the set $\{1, 2, 3, ..., n\}$ for some positive integer $n$. In simpler terms, a finite set has a definite and countable number of elements.

Example of a Finite Set:

Let $D$ be the set of days in a week.

In roster form, $D = \{\text{Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday}\}$.

The number of elements in this set is $|D| = 7$, which is a definite number. Therefore, $D$ is a finite set.

A set is called an infinite set if it is not a finite set. In simpler terms, an infinite set has an uncountable number of elements.

Example of an Infinite Set:

Let $\mathbb{N}$ be the set of natural numbers.

In roster form, $\mathbb{N} = \{1, 2, 3, 4, ...\}$.

The elements of this set continue indefinitely. There is no last element, and we cannot count all the elements. Therefore, $\mathbb{N}$ is an infinite set.

Given the sets $A = \{1, 2, 3\}$ and $B = \{1, 2, 3, 4, 5\}$.

Relationship between A and B:

To determine the relationship between sets A and B, we compare their elements.

Set A contains the elements 1, 2, and 3.

Set B contains the elements 1, 2, 3, 4, and 5.

We observe that every element in set A (1, 2, and 3) is also present in set B.

However, set B contains elements (4 and 5) that are not in set A.

This indicates that A is a subset of B.

The relationship between A and B is that A is a subset of B, denoted as $A \subseteq B$.

Finding $A \cup B$:

The union of two sets $A$ and $B$, denoted by $A \cup B$, is the set of all elements that are in A, or in B, or in both.

$A \cup B = \{1, 2, 3\} \cup \{1, 2, 3, 4, 5\}$

Combining all unique elements from both sets:

$A \cup B = \{1, 2, 3, 4, 5\}$.

Finding $A \cap B$:

The intersection of two sets $A$ and $B$, denoted by $A \cap B$, is the set of all elements that are common to both A and B.

$A \cap B = \{1, 2, 3\} \cap \{1, 2, 3, 4, 5\}$

Identifying the elements present in both sets:

The common elements are 1, 2, and 3.

$A \cap B = \{1, 2, 3\}$.

Notice that since $A \subseteq B$, the union $A \cup B$ is equal to B, and the intersection $A \cap B$ is equal to A.

Therefore, the relationship is $A \subseteq B$, $A \cup B = \mathbf{\{1, 2, 3, 4, 5\}}$, and $A \cap B = \mathbf{\{1, 2, 3\}}$.

Given the universal set $U = \{a, b, c, d, e\}$.

Given set $A = \{a, b\}$.

Given set $B = \{b, c, d\}$.

We need to find $(A \cup B)'$ and $A' \cap B'$.

First, let's find the union of A and B, $A \cup B$. The union contains all elements present in A or B or both.

$A \cup B = \{a, b\} \cup \{b, c, d\}$

$A \cup B = \{a, b, c, d\}$.

Next, let's find the complement of $(A \cup B)$, denoted by $(A \cup B)'$. The complement contains all elements in the universal set $U$ that are not in $A \cup B$.

$(A \cup B)' = U \setminus (A \cup B)$

$(A \cup B)' = \{a, b, c, d, e\} \setminus \{a, b, c, d\}$

$(A \cup B)' = \{e\}$.

Now, let's find the complement of A, $A'$. The complement of A contains all elements in $U$ that are not in A.

$A' = U \setminus A$

$A' = \{a, b, c, d, e\} \setminus \{a, b\}$

$A' = \{c, d, e\}$.

Next, let's find the complement of B, $B'$. The complement of B contains all elements in $U$ that are not in B.

$B' = U \setminus B$

$B' = \{a, b, c, d, e\} \setminus \{b, c, d\}$

$B' = \{a, e\}$.

Finally, let's find the intersection of $A'$ and $B'$, denoted by $A' \cap B'$. The intersection contains all elements common to both $A'$ and $B'$.

$A' \cap B' = \{c, d, e\} \cap \{a, e\}$

$A' \cap B' = \{e\}$.

We found that $(A \cup B)' = \{e\}$ and $A' \cap B' = \{e\}$.

Thus, $(A \cup B)' = \mathbf{\{e\}}$ and $A' \cap B' = \mathbf{\{e\}}$.

To draw a Venn diagram showing that $A \subseteq B$ (A is a subset of B), we represent the universal set $U$ as a rectangle and the sets A and B as circles within the rectangle.

The condition $A \subseteq B$ means that every element in set A is also an element in set B.

In a Venn diagram, this is represented by drawing the circle for set A completely inside the circle for set B.

Steps to draw the Venn diagram:

1. Draw a rectangle to represent the universal set $U$. Label it 'U'.

2. Inside the rectangle, draw a circle to represent set B. Label it 'B'.

3. Inside the circle representing set B, draw another circle to represent set A. Ensure that the entire circle for A is within the boundaries of the circle for B.

Visual representation (description):

[Rectangle representing Universal Set U]

[Large Circle representing Set B]

[Small Circle representing Set A, drawn entirely inside the large circle B]

This diagram visually shows that all elements within set A are also within set B, illustrating the subset relationship $A \subseteq B$.

Given that the number of elements in set A is $|A| = n$.

The power set of A, denoted by $\mathcal{P}(A)$, is the set of all possible subsets of A.

Let's consider examples:

If $A = \emptyset$, then $|A|=0$. The subsets are $\{\emptyset\}$. The power set is $\mathcal{P}(A) = \{\emptyset\}$. The number of elements in $\mathcal{P}(A)$ is 1. Note that $2^0 = 1$.

If $A = \{a\}$, then $|A|=1$. The subsets are $\emptyset, \{a\}$. The power set is $\mathcal{P}(A) = \{\emptyset, \{a\}\}$. The number of elements in $\mathcal{P}(A)$ is 2. Note that $2^1 = 2$.

If $A = \{a, b\}$, then $|A|=2$. The subsets are $\emptyset, \{a\}, \{b\}, \{a, b\}$. The power set is $\mathcal{P}(A) = \{\emptyset, \{a\}, \{b\}, \{a, b\}\}$. The number of elements in $\mathcal{P}(A)$ is 4. Note that $2^2 = 4$.

If $A = \{a, b, c\}$, then $|A|=3$. The subsets are $\emptyset, \{a\}, \{b\}, \{c\}, \{a, b\}, \{a, c\}, \{b, c\}, \{a, b, c\}$. The power set is $\mathcal{P}(A) = \{\emptyset, \{a\}, \{b\}, \{c\}, \{a, b\}, \{a, c\}, \{b, c\}, \{a, b, c\}\}$. The number of elements in $\mathcal{P}(A)$ is 8. Note that $2^3 = 8$.

From these examples, we observe a pattern: if a set has $n$ elements, its power set has $2^n$ elements.

This is because for each element in the original set, there are two possibilities when forming a subset: either the element is included in the subset or it is not included. Since there are $n$ elements, and each element has 2 independent possibilities, the total number of ways to form a subset is $2 \times 2 \times ... \times 2$ (n times), which is $2^n$.

Therefore, if $|A| = n$, the number of elements in the power set of A is $2^n$. This is denoted as $|\mathcal{P}(A)| = 2^n$.

The number of elements in the power set of $A$ is $\mathbf{2^n}$.

Given set $A = \{x : x \in \mathbb{Z}, x^2 = 25\}$.

Given set $B = \{x : x \in \mathbb{Z}, x = 5 \text{ or } x = -5\}$.

Two sets are equal if and only if they contain exactly the same elements.

First, let's find the elements of set A.

Set A consists of integers ($x \in \mathbb{Z}$) such that $x^2 = 25$.

We solve the equation $x^2 = 25$ for integer values of $x$.

$x^2 = 25$

Taking the square root of both sides:

$x = \pm\sqrt{25}$

$x = 5$ or $x = -5$.

Both 5 and -5 are integers. So, the elements of set A are -5 and 5.

$A = \{-5, 5\}$.

Next, let's find the elements of set B.

Set B consists of integers ($x \in \mathbb{Z}$) such that $x = 5$ or $x = -5$.

The integers that satisfy the condition "$x = 5$ or $x = -5$" are exactly 5 and -5.

So, the elements of set B are -5 and 5.

$B = \{-5, 5\}$.

Now, we compare the elements of set A and set B.

$A = \{-5, 5\}$

$B = \{-5, 5\}$

Both sets A and B contain the same elements: -5 and 5.

Since set A and set B have exactly the same elements, they are equal sets.

Thus, the sets A and B are equal sets.

Let $C$ be the set of students who play cricket.

Let $F$ be the set of students who play football.

The problem states that every student in the class plays either cricket or football. This means the total number of students in the class is equal to the number of students who play at least one of the two sports, which is given by the number of elements in the union of the sets C and F, i.e., $|C \cup F|$.

Given:

Number of students who play cricket, $|C| = 30$.

Number of students who play football, $|F| = 25$.

Number of students who play both cricket and football, $|C \cap F| = 15$.

To Find:

The total number of students in the class, which is $|C \cup F|$.

Solution:

We can use the Inclusion-Exclusion Principle for two sets to find the number of elements in the union of sets C and F.

The formula is:

$|C \cup F| = |C| + |F| - |C \cap F|$

Substitute the given values into the formula:

$|C \cup F| = 30 + 25 - 15$

$|C \cup F| = 55 - 15$

$|C \cup F| = 40$

Since every student plays at least one of the sports, the total number of students in the class is equal to $|C \cup F|$.

Total number of students $= 40$.

Thus, there are 40 students in the class.

We are asked to write the set of all positive integers less than 10 in roster form and set-builder form.

A positive integer is an integer greater than 0. The positive integers are $\{1, 2, 3, 4, ...\}$.

We are interested in positive integers that are less than 10.

The positive integers less than 10 are 1, 2, 3, 4, 5, 6, 7, 8, 9.

Roster Form:

Roster form (or tabular form) lists all the elements of a set within curly braces $\{ \}$ and separated by commas.

The set of all positive integers less than 10 in roster form is:

$\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$.

Set-Builder Form:

Set-builder form describes the elements of a set by stating the properties that the elements must satisfy.

It is typically written in the format $\{x \mid \text{property of } x\}$.

The elements of our set are integers ($x \in \mathbb{Z}$). They are positive ($x > 0$) or belong to the set of natural numbers ($\mathbb{N}$, usually starting from 1). And they are less than 10 ($x < 10$).

Using natural numbers, we can write it as:

$\{x \mid x \in \mathbb{N}, x < 10\}$.

Using integers, we can write it as:

$\{x \mid x \in \mathbb{Z}, x > 0, x < 10\}$ or $\{x \mid x \in \mathbb{Z}, 0 < x < 10\}$.

The most common and concise set-builder form using natural numbers is:

$\{x \mid x \in \mathbb{N}, x < 10\}$.

Thus, the set of all positive integers less than 10 is:

In roster form: $\mathbf{\{1, 2, 3, 4, 5, 6, 7, 8, 9\}}$

In set-builder form: $\mathbf{\{x \mid x \in \mathbb{N}, x < 10\}}$

Given the sets $A = \{1, 2, 3, 4, 5\}$ and $B = \{4, 5, 6, 7, 8\}$.

We need to find the difference of set A and set B, $A \setminus B$. The set $A \setminus B$ consists of elements that are in set A but not in set B.

Elements in A: 1, 2, 3, 4, 5

Elements in B: 4, 5, 6, 7, 8

We look for elements that are in A but not in B.

The elements 1, 2, and 3 are in A but not in B.

The elements 4 and 5 are in both A and B, so they are not included in $A \setminus B$.

$A \setminus B = \{1, 2, 3\}$.

Next, we need to find the difference of set B and set A, $B \setminus A$. The set $B \setminus A$ consists of elements that are in set B but not in set A.

Elements in B: 4, 5, 6, 7, 8

Elements in A: 1, 2, 3, 4, 5

We look for elements that are in B but not in A.

The elements 4 and 5 are in both A and B, so they are not included in $B \setminus A$.

The elements 6, 7, and 8 are in B but not in A.

$B \setminus A = \{6, 7, 8\}$.

We have found $A \setminus B = \{1, 2, 3\}$ and $B \setminus A = \{6, 7, 8\}$.

Now, we need to determine if $A \setminus B$ and $B \setminus A$ are equal.

Two sets are equal if they have the exact same elements.

The elements of $A \setminus B$ are 1, 2, 3.

The elements of $B \setminus A$ are 6, 7, 8.

Since the elements are different, the sets are not equal.

$\{1, 2, 3\} \neq \{6, 7, 8\}$.

Thus, $A \setminus B = \mathbf{\{1, 2, 3\}}$ and $B \setminus A = \mathbf{\{6, 7, 8\}}$. They are not equal.

Given that $A$ is the set of even numbers and $B$ is the set of odd numbers. The universal set $U$ is the set that contains all the elements being considered in a particular context.

Set A = $\{x \mid x \text{ is an even number}\}$. Examples of even numbers are ..., -4, -2, 0, 2, 4, ...

Set B = $\{x \mid x \text{ is an odd number}\}$. Examples of odd numbers are ..., -3, -1, 1, 3, 5, ...

In standard mathematics, the collection of all even numbers and all odd numbers together forms the set of all integers.

An integer is a number that can be written without a fractional component. The set of integers is denoted by $\mathbb{Z}$.

$\mathbb{Z} = \{..., -3, -2, -1, 0, 1, 2, 3, ...\}$

The union of the set of even numbers and the set of odd numbers is the set of all integers:

$A \cup B = \{\text{even numbers}\} \cup \{\text{odd numbers}\} = \{\text{all integers}\}$

$A \cup B = \mathbb{Z}$.

Since the problem defines the sets A and B as subsets of some universal set $U$, and given that the union of A and B covers all types of integers (even and odd), the most appropriate universal set in this context is the set of all integers.

If the context were restricted to only positive numbers, $U$ might be the set of positive integers. However, "even numbers" and "odd numbers" typically refer to the entire set of integers unless specified otherwise.

Therefore, the universal set $U$ for $A = \{ \text{even numbers} \}$ and $B = \{ \text{odd numbers} \}$ is the set of all integers.

The universal set $U$ is the set of all integers, $\mathbb{Z}$.

Given set $A = \{x : x^2 \leq 16, x \in \mathbb{Z}\}$.

Given set $B = \{-4, -3, -2, -1, 0, 1, 2, 3, 4\}$.

Two sets are equal if and only if they contain exactly the same elements.

First, let's find the elements of set A.

Set A consists of integers ($x \in \mathbb{Z}$) such that $x^2 \leq 16$.

The inequality $x^2 \leq 16$ is equivalent to $-\sqrt{16} \leq x \leq \sqrt{16}$.

This simplifies to $-4 \leq x \leq 4$.

We need to find all integers $x$ that satisfy the condition $-4 \leq x \leq 4$.

The integers are ..., -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, ...

The integers between -4 and 4 (inclusive) are -4, -3, -2, -1, 0, 1, 2, 3, 4.

So, the elements of set A are:

$A = \{-4, -3, -2, -1, 0, 1, 2, 3, 4\}$.

Now, we compare the elements of set A with the elements of set B.

$A = \{-4, -3, -2, -1, 0, 1, 2, 3, 4\}$

$B = \{-4, -3, -2, -1, 0, 1, 2, 3, 4\}$

By comparing the elements, we see that set A and set B have the exact same elements.

Since set A and set B contain precisely the same elements, they are equal sets.

Thus, the sets A and B are equal sets.

We are asked to write the given sets, which are in set-builder form, as intervals.

For the set $\{x \in \mathbb{R} : x > -1\}$:

This set contains all real numbers $x$ such that $x$ is strictly greater than -1.

The condition $x > -1$ means that the number -1 is not included in the set, but all numbers greater than -1 are included.

On a number line, this corresponds to starting just after -1 and extending indefinitely towards positive infinity.

In interval notation, a strict inequality ($>$ or $<$) corresponds to a round bracket `(`. The interval extends to positive infinity, which is represented by $\infty$ and always uses a round bracket `)`. The lower bound is -1, and since it's not included, we use `(`. The upper bound is positive infinity.

The interval corresponding to $\{x \in \mathbb{R} : x > -1\}$ is $(-1, \infty)$.

For the set $\{x \in \mathbb{R} : x \leq 5\}$:

This set contains all real numbers $x$ such that $x$ is less than or equal to 5.

The condition $x \leq 5$ means that the number 5 is included in the set, and all numbers less than 5 are also included.

On a number line, this corresponds to starting from negative infinity and extending up to and including 5.

In interval notation, an inequality with equality ($\leq$ or $\geq$) corresponds to a square bracket `[`. The interval starts from negative infinity, which is represented by $-\infty$ and always uses a round bracket `(`. The upper bound is 5, and since it's included, we use `]`. The lower bound is negative infinity.

The interval corresponding to $\{x \in \mathbb{R} : x \leq 5\}$ is $(-\infty, 5]$.

Thus, the intervals are:

For $\{x \in \mathbb{R} : x > -1\}$, the interval is $\mathbf{(-1, \infty)}$.

For $\{x \in \mathbb{R} : x \leq 5\}$, the interval is $\mathbf{(-\infty, 5]}$.

Given:

The number of elements in the universal set $U$ is $|U| = 100$.

The number of elements in set A is $|A| = 50$.

The number of elements in set B is $|B| = 60$.

The number of elements in the intersection of A and B is $|A \cap B| = 30$.

To Find:

The number of elements in the complement of the union of A and B, $|(A \cup B)'|$.

Solution:

We know that the number of elements in the complement of a set $X$ within a universal set $U$ is given by $|X'| = |U| - |X|$.

In this case, we want to find $|(A \cup B)'|$, so we use the formula:

$|(A \cup B)'| = |U| - |A \cup B|$

To use this formula, we first need to find $|A \cup B|$, the number of elements in the union of sets A and B.

We use the Inclusion-Exclusion Principle for two sets:

$|A \cup B| = |A| + |B| - |A \cap B|$

Substitute the given values into the formula for the union:

$|A \cup B| = 50 + 60 - 30$

$|A \cup B| = 110 - 30$

$|A \cup B| = 80$

This means there are 80 elements in the union of A and B.

Now, substitute the values of $|U|$ and $|A \cup B|$ into the formula for the complement:

$|(A \cup B)'| = |U| - |A \cup B|$

$|(A \cup B)'| = 100 - 80$

$|(A \cup B)'| = 20$

Therefore, the number of elements in the complement of the union of A and B is 20.

$|(A \cup B)'| = \mathbf{20}$.

Let $U$ be the set of all students surveyed.

Let $M$ be the set of students who studied Mathematics.

Let $P$ be the set of students who studied Physics.

Let $C$ be the set of students who studied Chemistry.

Given:

$|U| = 100$

$|M| = 28$

$|P| = 30$

$|C| = 24$

$|M \cap P| = 8$

$|M \cap C| = 10$

$|P \cap C| = 5$

$|M \cap P \cap C| = 3$

To Find:

i) $|P \text{ only}|$

ii) $|C \text{ only}|$

iii) $|M \text{ only}|$

iv) $|(M \cup P \cup C)'|$

v) Number of students who studied exactly one subject.

vi) Number of students who studied exactly two subjects.

Solution:

We can calculate the number of students in each disjoint region of the Venn diagram.

Number of students who studied all three subjects:

$|M \cap P \cap C| = 3$

Number of students who studied Mathematics and Physics only (not Chemistry):

$|(M \cap P) \text{ only}| = |M \cap P| - |M \cap P \cap C| = 8 - 3 = 5$

Number of students who studied Mathematics and Chemistry only (not Physics):

$|(M \cap C) \text{ only}| = |M \cap C| - |M \cap P \cap C| = 10 - 3 = 7$

Number of students who studied Physics and Chemistry only (not Mathematics):

$|(P \cap C) \text{ only}| = |P \cap C| - |M \cap P \cap C| = 5 - 3 = 2$

Number of students who studied Mathematics only:

$|M \text{ only}| = |M| - |(M \cap P) \text{ only}| - |(M \cap C) \text{ only}| - |M \cap P \cap C| = 28 - 5 - 7 - 3 = 28 - 15 = 13$

So, i) Physics only = $\mathbf{20}$

Number of students who studied Physics only:

$|P \text{ only}| = |P| - |(M \cap P) \text{ only}| - |(P \cap C) \text{ only}| - |M \cap P \cap C| = 30 - 5 - 2 - 3 = 30 - 10 = 20$

So, ii) Chemistry only = $\mathbf{12}$

Number of students who studied Chemistry only:

$|C \text{ only}| = |C| - |(M \cap C) \text{ only}| - |(P \cap C) \text{ only}| - |M \cap P \cap C| = 24 - 7 - 2 - 3 = 24 - 12 = 12$

So, iii) Mathematics only = $\mathbf{13}$

Number of students who studied at least one subject is the union $|M \cup P \cup C|$.

Using the Inclusion-Exclusion Principle for three sets:

$|M \cup P \cup C| = |M| + |P| + |C| - |M \cap P| - |M \cap C| - |P \cap C| + |M \cap P \cap C|$

$|M \cup P \cup C| = 28 + 30 + 24 - 8 - 10 - 5 + 3$

$|M \cup P \cup C| = 82 - 23 + 3$

$|M \cup P \cup C| = 59 + 3 = 62$

Alternatively, summing the disjoint regions:

$|M \cup P \cup C| = |M \text{ only}| + |P \text{ only}| + |C \text{ only}| + |(M \cap P) \text{ only}| + |(M \cap C) \text{ only}| + |(P \cap C) \text{ only}| + |M \cap P \cap C|$

$|M \cup P \cup C| = 13 + 20 + 12 + 5 + 7 + 2 + 3 = 62$

Number of students who studied none of the subjects:

This is the complement of the union $(M \cup P \cup C)'$.

$|(M \cup P \cup C)'| = |U| - |M \cup P \cup C| = 100 - 62 = 38$

So, iv) None of the subjects = $\mathbf{38}$

Number of students who studied exactly one subject:

This is the sum of the numbers in the regions for Mathematics only, Physics only, and Chemistry only.

Exactly one subject $= |M \text{ only}| + |P \text{ only}| + |C \text{ only}| = 13 + 20 + 12 = 45$

So, v) Exactly one of the subjects = $\mathbf{45}$

Number of students who studied exactly two subjects:

This is the sum of the numbers in the regions for Mathematics and Physics only, Mathematics and Chemistry only, and Physics and Chemistry only.

Exactly two subjects $= |(M \cap P) \text{ only}| + |(M \cap C) \text{ only}| + |(P \cap C) \text{ only}| = 5 + 7 + 2 = 14$

So, vi) Exactly two of the subjects = $\mathbf{14}$

To Prove:

$A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$

This property is known as the Distributive Law of Union over Intersection.

Proof:

To prove that two sets are equal, we need to show that each set is a subset of the other.

We will show that $A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C)$ and $(A \cup B) \cap (A \cup C) \subseteq A \cup (B \cap C)$.

Part 1: Show that $A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C)$.

Let $x$ be an arbitrary element such that $x \in A \cup (B \cap C)$.

By the definition of union, this means $x \in A$ or $x \in (B \cap C)$.

$x \in A \cup (B \cap C) \implies x \in A \text{ or } x \in (B \cap C)$

Case 1: Suppose $x \in A$.

If $x \in A$, then by the definition of union, $x$ is in the union of A with any other set.

So, if $x \in A$, then $x \in A \cup B$.

Also, if $x \in A$, then $x \in A \cup C$.

Since $x \in A \cup B$ and $x \in A \cup C$, by the definition of intersection, $x \in (A \cup B) \cap (A \cup C)$.

Case 2: Suppose $x \in (B \cap C)$.

If $x \in (B \cap C)$, then by the definition of intersection, $x \in B$ and $x \in C$.

$x \in (B \cap C) \implies x \in B \text{ and } x \in C$

If $x \in B$, then by the definition of union, $x \in A \cup B$.

If $x \in C$, then by the definition of union, $x \in A \cup C$.

Since $x \in A \cup B$ and $x \in A \cup C$, by the definition of intersection, $x \in (A \cup B) \cap (A \cup C)$.

In both cases (whether $x \in A$ or $x \in (B \cap C)$), we have shown that $x \in (A \cup B) \cap (A \cup C)$.

Therefore, $A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C)$.

Part 2: Show that $(A \cup B) \cap (A \cup C) \subseteq A \cup (B \cap C)$.

Let $x$ be an arbitrary element such that $x \in (A \cup B) \cap (A \cup C)$.

By the definition of intersection, this means $x \in (A \cup B)$ and $x \in (A \cup C)$.

$x \in (A \cup B) \cap (A \cup C) \implies x \in (A \cup B) \text{ and } x \in (A \cup C)$

By the definition of union:

$x \in (A \cup B) \implies x \in A \text{ or } x \in B$

$x \in (A \cup C) \implies x \in A \text{ or } x \in C$

So, we have that $x \in A \text{ or } x \in B$, AND $x \in A \text{ or } x \in C$.

Consider the logical statement: $(x \in A \text{ or } x \in B) \text{ and } (x \in A \text{ or } x \in C)$.

Using the distributive property of logical disjunction over conjunction (which states that $P \lor (Q \land R) \Leftrightarrow (P \lor Q) \land (P \lor R)$), let $P = (x \in A)$, $Q = (x \in B)$, and $R = (x \in C)$.

The statement $(P \lor Q) \land (P \lor R)$ is logically equivalent to $P \lor (Q \land R)$.

So, $(x \in A \text{ or } x \in B) \text{ and } (x \in A \text{ or } x \in C)$ is equivalent to $x \in A \text{ or } (x \in B \text{ and } x \in C)$.

By the definition of intersection, $x \in B \text{ and } x \in C$ is equivalent to $x \in (B \cap C)$.

So, we have $x \in A \text{ or } x \in (B \cap C)$.

By the definition of union, $x \in A \text{ or } x \in (B \cap C)$ is equivalent to $x \in A \cup (B \cap C)$.

Therefore, if $x \in (A \cup B) \cap (A \cup C)$, then $x \in A \cup (B \cap C)$.

This shows that $(A \cup B) \cap (A \cup C) \subseteq A \cup (B \cap C)$.

Conclusion:

Since $A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C)$ (from Part 1) and $(A \cup B) \cap (A \cup C) \subseteq A \cup (B \cap C)$ (from Part 2), we conclude that the two sets are equal.

$A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$.

De Morgan's Laws are fundamental in set theory and logic, providing relationships between set operations (union, intersection, complement) and logical operations (OR, AND, NOT). We can prove these laws using Venn diagrams.

Given:

Two sets, A and B, within a universal set U.

To Prove:

i) $(A \cup B)' = A' \cap B'$

ii) $(A \cap B)' = A' \cup B'$

Proof using Venn Diagrams:

Part i) Proving $(A \cup B)' = A' \cap B'$

Left-Hand Side: $(A \cup B)'$

First, consider the union of sets A and B, denoted as $A \cup B$. This region includes all elements that are in A, or in B, or in both.

The complement of $(A \cup B)$, denoted as $(A \cup B)'$, represents all elements in the universal set U that are *not* in $A \cup B$. This shaded region would be everything outside of the combined area of A and B.

Right-Hand Side: $A' \cap B'$

Now, let's consider the complements of A and B separately.

$A'$ (complement of A) represents all elements in U that are not in A.

$B'$ (complement of B) represents all elements in U that are not in B.

The intersection of $A'$ and $B'$, denoted as $A' \cap B'$, represents the elements that are common to both $A'$ and $B'$. This means these elements are neither in A nor in B.

When we shade $A'$ and $B'$ on a Venn diagram, the overlapping region is precisely the area outside of both A and B.

Conclusion for Part i):

By observing the Venn diagrams for $(A \cup B)'$ and $A' \cap B'$, we see that both represent the same region in the universal set U – the region that contains elements not present in either A or B. Therefore, $(A \cup B)' = A' \cap B'$.

Part ii) Proving $(A \cap B)' = A' \cup B'$

Left-Hand Side: $(A \cap B)'$

First, consider the intersection of sets A and B, denoted as $A \cap B$. This region includes all elements that are common to both A and B.

The complement of $(A \cap B)$, denoted as $(A \cap B)'$, represents all elements in the universal set U that are *not* in $A \cap B$. This shaded region would be everything outside the common overlap of A and B.

Right-Hand Side: $A' \cup B'$

As discussed before:

$A'$ represents all elements in U that are not in A.

$B'$ represents all elements in U that are not in B.

The union of $A'$ and $B'$, denoted as $A' \cup B'$, represents all elements that are in $A'$, or in $B'$, or in both. This means it includes elements that are not in A, or not in B, or not in both.

When we shade $A'$ and $B'$ on a Venn diagram, the union of these shaded regions covers all areas except for the intersection of A and B.

Conclusion for Part ii):

By observing the Venn diagrams for $(A \cap B)'$ and $A' \cup B'$, we see that both represent the same region in the universal set U – the region that excludes the intersection of A and B. Therefore, $(A \cap B)' = A' \cup B'$.

These proofs visually demonstrate the validity of De Morgan's Laws using Venn diagrams.

Let U be the set of all people in the town. Let A be the set of people who read Newspaper A, B be the set of people who read Newspaper B, and C be the set of people who read Newspaper C.

We are given the following information:

Total number of people in the town, $|U| = 10,000$

Number of people who read Newspaper A, $|A| = 5,500$

Number of people who read Newspaper B, $|B| = 3,000$

Number of people who read Newspaper C, $|C| = 2,500$

Number of people who read A and B, $|A \cap B| = 1,000$

Number of people who read B and C, $|B \cap C| = 800$

Number of people who read A and C, $|A \cap C| = 1,200$

Number of people who read all three newspapers, $|A \cap B \cap C| = 300$

i) Number of people who read at least one newspaper:

This is given by the union of the three sets, $|A \cup B \cup C|$. We use the Principle of Inclusion-Exclusion for three sets:

Substituting the given values:

So, 8,300 people read at least one newspaper.

ii) Number of people who read exactly two newspapers:

To find this, we first calculate the number of people who read exactly A and B, exactly B and C, and exactly A and C, and then sum them up.

Number of people who read only A and B = $|A \cap B| - |A \cap B \cap C|$

Number of people who read only B and C = $|B \cap C| - |A \cap B \cap C|$

Number of people who read only A and C = $|A \cap C| - |A \cap B \cap C|$

Number of people who read exactly two newspapers = (Only A and B) + (Only B and C) + (Only A and C)

So, 2,100 people read exactly two newspapers.

iii) Number of people who read no newspaper:

This is the complement of reading at least one newspaper with respect to the total population.

Number of people who read no newspaper = $|U| - |A \cup B \cup C|$

So, 1,700 people read no newspaper.

iv) Number of people who read only Newspaper A:

This can be found by subtracting the number of people who read A and B (but not C), A and C (but not B), and A, B, and C from the total number of people who read A.

Number of people who read only A = $|A| - (\text{Only A and B}) - (\text{Only A and C}) - |A \cap B \cap C|$

Alternatively, using set notation:

Only A = $|A| - |A \cap B| - |A \cap C| + |A \cap B \cap C|$ (This formula accounts for subtracting the overlaps correctly)

So, 3,600 people read only Newspaper A.

Summary of results:

i) At least one newspaper: 8,300

ii) Exactly two newspapers: 2,100

iii) No newspaper: 1,700

iv) Only Newspaper A: 3,600

We need to prove the distributive property of intersection over union for sets:

$A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$

We will prove this by showing that each side is a subset of the other.

Part 1: Proving $A \cap (B \cup C) \subseteq (A \cap B) \cup (A \cap C)$

Let $x$ be an arbitrary element such that $x \in A \cap (B \cup C)$.

By the definition of intersection, this means:

By the definition of union, $x \in (B \cup C)$ implies:

So, we have two cases:

Case 1: $x \in A$ and $x \in B$

If $x \in A$ and $x \in B$, then by the definition of intersection, $x \in (A \cap B)$.

Since $x \in (A \cap B)$, it follows that $x \in (A \cap B) \cup (A \cap C)$ because if an element belongs to one set in a union, it belongs to the union itself.

Case 2: $x \in A$ and $x \in C$

If $x \in A$ and $x \in C$, then by the definition of intersection, $x \in (A \cap C)$.

Since $x \in (A \cap C)$, it follows that $x \in (A \cap B) \cup (A \cap C)$ because if an element belongs to one set in a union, it belongs to the union itself.

In both cases, we have shown that if $x \in A \cap (B \cup C)$, then $x \in (A \cap B) \cup (A \cap C)$.

Therefore, $A \cap (B \cup C) \subseteq (A \cap B) \cup (A \cap C)$.

Part 2: Proving $(A \cap B) \cup (A \cap C) \subseteq A \cap (B \cup C)$

Let $y$ be an arbitrary element such that $y \in (A \cap B) \cup (A \cap C)$.

By the definition of union, this means:

Now we consider these two possibilities:

Possibility 1: $y \in (A \cap B)$

If $y \in (A \cap B)$, then by the definition of intersection:

If $y \in B$, then by the definition of union, $y \in (B \cup C)$.

Since $y \in A$ and $y \in (B \cup C)$, by the definition of intersection, $y \in A \cap (B \cup C)$.

Possibility 2: $y \in (A \cap C)$

If $y \in (A \cap C)$, then by the definition of intersection:

If $y \in C$, then by the definition of union, $y \in (B \cup C)$.

Since $y \in A$ and $y \in (B \cup C)$, by the definition of intersection, $y \in A \cap (B \cup C)$.

In both possibilities, we have shown that if $y \in (A \cap B) \cup (A \cap C)$, then $y \in A \cap (B \cup C)$.

Therefore, $(A \cap B) \cup (A \cap C) \subseteq A \cap (B \cup C)$.

Conclusion:

Since we have shown that $A \cap (B \cup C) \subseteq (A \cap B) \cup (A \cap C)$ and $(A \cap B) \cup (A \cap C) \subseteq A \cap (B \cup C)$, we can conclude that:

This proves the distributive property of intersection over union.

Given Sets:

Set $A = \{x \in \mathbb{R} : -5 \leq x < 2\}$

Set $B = \{x \in \mathbb{R} : 1 < x \leq 6\}$

Interval Notation:

Set $A$ as an interval is $[-5, 2)$.

Set $B$ as an interval is $(1, 6]$.

Representing on a Number Line:

We will draw a number line and mark the intervals for A and B.

For $A = [-5, 2)$: Mark -5 with a closed circle (inclusive) and 2 with an open circle (exclusive). Shade the region between them.

For $B = (1, 6]$: Mark 1 with an open circle (exclusive) and 6 with a closed circle (inclusive). Shade the region between them.

Finding $A \cup B$:

The union of two sets contains all elements that are in either set A or set B (or both).

Looking at the number line, the combined region starts from -5 (inclusive) and goes up to 6 (inclusive), covering all numbers in between.

$A \cup B = \{x \in \mathbb{R} : -5 \leq x \leq 6\}$

As an interval, $A \cup B = [-5, 6]$.

Finding $A \cap B$:

The intersection of two sets contains all elements that are common to both set A and set B.

On the number line, the overlapping region between A and B starts from where B begins (1, exclusive) and ends where A ends (2, exclusive).

$A \cap B = \{x \in \mathbb{R} : 1 < x < 2\}$

As an interval, $A \cap B = (1, 2)$.

Finding $A \setminus B$:

The set difference $A \setminus B$ contains all elements that are in set A but not in set B.

This means we take the interval for A and remove any part that overlaps with B.

Set A is $[-5, 2)$. Set B is $(1, 6]$. The part of A that is also in B is $(1, 2)$.

So, $A \setminus B$ will be the part of $[-5, 2)$ that is not in $(1, 2)$. This is the interval from -5 up to, but not including, 1.

$A \setminus B = \{x \in \mathbb{R} : -5 \leq x \leq 1\}$

As an interval, $A \setminus B = [-5, 1]$.

Finding $B \setminus A$:

The set difference $B \setminus A$ contains all elements that are in set B but not in set A.

This means we take the interval for B and remove any part that overlaps with A.

Set B is $(1, 6]$. Set A is $[-5, 2)$. The part of B that is also in A is $(1, 2)$.

So, $B \setminus A$ will be the part of $(1, 6]$ that is not in $(1, 2)$. This is the interval from 2 (exclusive, because it's the upper bound of the removed part from B) up to 6 (inclusive).

$B \setminus A = \{x \in \mathbb{R} : 2 \leq x \leq 6\}$

As an interval, $B \setminus A = [2, 6]$.

Summary of Intervals:

$A = [-5, 2)$

$B = (1, 6]$

$A \cup B = [-5, 6]$

$A \cap B = (1, 2)$

$A \setminus B = [-5, 1]$

$B \setminus A = [2, 6]$

Representation on Number Line:

(Please imagine a number line for this description. The description below outlines how to draw it.)

1. Draw a horizontal line representing the number line.
2. Mark important points: -5, 1, 2, 6.
3. For set A ($[-5, 2)$): Place a closed circle at -5 and an open circle at 2. Shade the line segment connecting them.
4. For set B ($(1, 6]$): Place an open circle at 1 and a closed circle at 6. Shade the line segment connecting them.
5. For $A \cup B$ ($[-5, 6]$): Place a closed circle at -5 and a closed circle at 6. Shade the entire line segment between them.
6. For $A \cap B$ ($(1, 2)$): Place an open circle at 1 and an open circle at 2. Shade the line segment between them.
7. For $A \setminus B$ ($[-5, 1]$): Place a closed circle at -5 and a closed circle at 1. Shade the line segment between them.
8. For $B \setminus A$ ($[2, 6]$): Place a closed circle at 2 and a closed circle at 6. Shade the line segment between them.

To Show: $A \cup B = (A \setminus B) \cup (B \setminus A) \cup (A \cap B)$, and the union is disjoint.

Proof:

Part 1: Showing $A \cup B \subseteq (A \setminus B) \cup (B \setminus A) \cup (A \cap B)$

Let $x \in A \cup B$. This means that $x \in A$ or $x \in B$ (or both).

We can consider three cases:

1. Case 1: $x \in A$ and $x \notin B$.

   In this case, $x \in A \setminus B$. Therefore, $x \in (A \setminus B) \cup (B \setminus A) \cup (A \cap B)$.

2. Case 2: $x \in B$ and $x \notin A$.

   In this case, $x \in B \setminus A$. Therefore, $x \in (A \setminus B) \cup (B \setminus A) \cup (A \cap B)$.

3. Case 3: $x \in A$ and $x \in B$.

   In this case, $x \in A \cap B$. Therefore, $x \in (A \setminus B) \cup (B \setminus A) \cup (A \cap B)$.

Since any element in $A \cup B$ must fall into one of these three cases, we can conclude that $A \cup B \subseteq (A \setminus B) \cup (B \setminus A) \cup (A \cap B)$.

Part 2: Showing $(A \setminus B) \cup (B \setminus A) \cup (A \cap B) \subseteq A \cup B$

Let $y \in (A \setminus B) \cup (B \setminus A) \cup (A \cap B)$. This means that $y$ is in at least one of the sets $A \setminus B$, $B \setminus A$, or $A \cap B$.

1. If $y \in A \setminus B$, then $y \in A$ and $y \notin B$. Since $y \in A$, it follows that $y \in A \cup B$.
2. If $y \in B \setminus A$, then $y \in B$ and $y \notin A$. Since $y \in B$, it follows that $y \in A \cup B$.
3. If $y \in A \cap B$, then $y \in A$ and $y \in B$. Since $y \in A$ (or $y \in B$), it follows that $y \in A \cup B$.

In all cases, if $y$ is in $(A \setminus B) \cup (B \setminus A) \cup (A \cap B)$, then $y \in A \cup B$.

Therefore, $(A \setminus B) \cup (B \setminus A) \cup (A \cap B) \subseteq A \cup B$.

From Part 1 and Part 2, we have shown that $A \cup B = (A \setminus B) \cup (B \setminus A) \cup (A \cap B)$.

Part 3: Showing the union is disjoint

We need to show that the pairwise intersections of these three sets are empty:

1. $(A \setminus B) \cap (B \setminus A) = \emptyset$

Let $x \in (A \setminus B) \cap (B \setminus A)$.

This means $x \in A \setminus B$ AND $x \in B \setminus A$.

If $x \in A \setminus B$, then $x \in A$ and $x \notin B$.

If $x \in B \setminus A$, then $x \in B$ and $x \notin A$.

We have a contradiction: $x \notin B$ and $x \in B$. Also, $x \notin A$ and $x \in A$.

Thus, there is no such element $x$, so $(A \setminus B) \cap (B \setminus A) = \emptyset$.

2. $(A \setminus B) \cap (A \cap B) = \emptyset$

Let $x \in (A \setminus B) \cap (A \cap B)$.

This means $x \in A \setminus B$ AND $x \in A \cap B$.

If $x \in A \setminus B$, then $x \in A$ and $x \notin B$.

If $x \in A \cap B$, then $x \in A$ and $x \in B$.

We have a contradiction: $x \notin B$ and $x \in B$.

Thus, there is no such element $x$, so $(A \setminus B) \cap (A \cap B) = \emptyset$.

3. $(B \setminus A) \cap (A \cap B) = \emptyset$

Let $x \in (B \setminus A) \cap (A \cap B)$.

This means $x \in B \setminus A$ AND $x \in A \cap B$.

If $x \in B \setminus A$, then $x \in B$ and $x \notin A$.

If $x \in A \cap B$, then $x \in A$ and $x \in B$.

We have a contradiction: $x \notin A$ and $x \in A$.

Thus, there is no such element $x$, so $(B \setminus A) \cap (A \cap B) = \emptyset$.

Since all pairwise intersections are empty, the union $(A \setminus B) \cup (B \setminus A) \cup (A \cap B)$ is disjoint.

Conclusion:

We have shown that $A \cup B = (A \setminus B) \cup (B \setminus A) \cup (A \cap B)$ and that the sets $A \setminus B$, $B \setminus A$, and $A \cap B$ are mutually disjoint.

Given:

Total number of students enrolled, $|U| = 150$.

Number of students who opted for Physics, $|P| = 60$.

Number of students who opted for Chemistry, $|C| = 70$.

Number of students who opted for Mathematics, $|M| = 80$.

Number of students who opted for Physics and Chemistry, $|P \cap C| = 30$.

Number of students who opted for Chemistry and Mathematics, $|C \cap M| = 35$.

Number of students who opted for Physics and Mathematics, $|P \cap M| = 40$.

Number of students who opted for all three subjects, $|P \cap C \cap M| = 20$.

Using the Principle of Inclusion-Exclusion or a Venn Diagram, we can find the required values. Let's use the formulas derived from the Principle of Inclusion-Exclusion for clarity and then confirm with Venn diagram logic.

Number of students in exactly two subjects:

Number of students in Physics and Chemistry only = $|P \cap C| - |P \cap C \cap M| = 30 - 20 = 10$.

Number of students in Chemistry and Mathematics only = $|C \cap M| - |P \cap C \cap M| = 35 - 20 = 15$.

Number of students in Physics and Mathematics only = $|P \cap M| - |P \cap C \cap M| = 40 - 20 = 20$.

Number of students in exactly one subject:

Number of students in Physics only = $|P| - (|P \cap C| - |P \cap C \cap M|) - (|P \cap M| - |P \cap C \cap M|) - |P \cap C \cap M|$

$= |P| - |P \cap C| - |P \cap M| + |P \cap C \cap M|$

$= 60 - 30 - 40 + 20 = 10$.

Number of students in Chemistry only = $|C| - (|P \cap C| - |P \cap C \cap M|) - (|C \cap M| - |P \cap C \cap M|) - |P \cap C \cap M|$

$= |C| - |P \cap C| - |C \cap M| + |P \cap C \cap M|$

$= 70 - 30 - 35 + 20 = 25$.

Number of students in Mathematics only = $|M| - (|C \cap M| - |P \cap C \cap M|) - (|P \cap M| - |P \cap C \cap M|) - |P \cap C \cap M|$

$= |M| - |C \cap M| - |P \cap M| + |P \cap C \cap M|$

$= 80 - 35 - 40 + 20 = 25$.

Now we can answer the specific questions:

i) Physics but not Chemistry.

This includes students who opted for Physics only, and students who opted for Physics and Mathematics only (but not Chemistry).

Number of students in Physics but not Chemistry = (Physics only) + (Physics and Mathematics only)

$= ( |P| - (|P \cap C| - |P \cap C \cap M|) - (|P \cap M| - |P \cap C \cap M|) - |P \cap C \cap M| ) + ( |P \cap M| - |P \cap C \cap M| )$

$= 10 + 20 = 30$.

Alternatively, this can be calculated as $|P| - |P \cap C| = 60 - 30 = 30$. This formula works because $|P|$ includes everyone in P, and subtracting $|P \cap C|$ removes everyone who opted for Chemistry along with Physics, leaving only those who opted for Physics but not Chemistry.

Answer: 30 students.

ii) Exactly two subjects.

This is the sum of students in Physics and Chemistry only, Chemistry and Mathematics only, and Physics and Mathematics only.

Number of students in exactly two subjects = (Physics and Chemistry only) + (Chemistry and Mathematics only) + (Physics and Mathematics only)

= 10 + 15 + 20 = 45.

Answer: 45 students.

iii) At least one subject.

This is the total number of students who opted for Physics, Chemistry, or Mathematics, which can be found using the Principle of Inclusion-Exclusion:

$|P \cup C \cup M| = |P| + |C| + |M| - |P \cap C| - |C \cap M| - |P \cap M| + |P \cap C \cap M|$

$|P \cup C \cup M| = 60 + 70 + 80 - 30 - 35 - 40 + 20$

$|P \cup C \cup M| = 210 - 105 + 20 = 105 + 20 = 125$.

Alternatively, sum of students in exactly one subject, exactly two subjects, and exactly three subjects:

= (Physics only) + (Chemistry only) + (Mathematics only) + (Exactly two subjects) + (All three subjects)

= 10 + 25 + 25 + 45 + 20 = 125.

Answer: 125 students.

iv) None of the subjects.

This is the total number of students minus the number of students who opted for at least one subject.

Number of students who opted for none of the subjects = $|U| - |P \cup C \cup M|$

= 150 - 125 = 25.

Answer: 25 students.

Venn Diagram Representation:

*(To fully represent this with a Venn diagram, you would draw three overlapping circles for P, C, and M within a rectangle representing U. The numbers for each region would be filled in as calculated above:*

• Center (P $\cap$ C $\cap$ M): 20
• P $\cap$ C only: 10
• C $\cap$ M only: 15
• P $\cap$ M only: 20
• P only: 10
• C only: 25
• M only: 25
• Outside the circles (None): 25

The sum of all regions within the circles is $20 + 10 + 15 + 20 + 10 + 25 + 25 = 125$. The total outside is $150 - 125 = 25$.

Given Universal Set:

$U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$

Listing the elements of sets A, B, and C:

Set $A$ = {x | x is a prime number}. The prime numbers in U are 2, 3, 5, 7.

So, $A = \{2, 3, 5, 7\}$.

Set $B$ = {x | x is an even number}. The even numbers in U are 2, 4, 6, 8, 10.

So, $B = \{2, 4, 6, 8, 10\}$.

Set $C$ = {x | x is a multiple of 3}. The multiples of 3 in U are 3, 6, 9.

So, $C = \{3, 6, 9\}$.

Now, let's find the required sets:

First, find $B \cup C$:

$B \cup C$ = {elements in B or C or both}

$B \cup C = \{2, 4, 6, 8, 10\} \cup \{3, 6, 9\}$

$B \cup C = \{2, 3, 4, 6, 8, 9, 10\}$

i) $A \cap (B \cup C)$

$A \cap (B \cup C)$ = {elements common to A and $(B \cup C)$}

$A \cap (B \cup C) = \{2, 3, 5, 7\} \cap \{2, 3, 4, 6, 8, 9, 10\}$

$A \cap (B \cup C) = \{2, 3\}$

Next, find $A \cap B$ and $A \cap C$:

$A \cap B$ = {elements common to A and B}

$A \cap B = \{2, 3, 5, 7\} \cap \{2, 4, 6, 8, 10\}$

$A \cap B = \{2\}$

$A \cap C$ = {elements common to A and C}

$A \cap C = \{2, 3, 5, 7\} \cap \{3, 6, 9\}$

$A \cap C = \{3\}$

ii) $(A \cap B) \cup (A \cap C)$

$(A \cap B) \cup (A \cap C)$ = {elements in $(A \cap B)$ or $(A \cap C)$ or both}

$(A \cap B) \cup (A \cap C) = \{2\} \cup \{3\}$

$(A \cap B) \cup (A \cap C) = \{2, 3\}$

Verification of $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$

We found that $A \cap (B \cup C) = \{2, 3\}$ and $(A \cap B) \cup (A \cap C) = \{2, 3\}$.

Thus, $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$ is verified. This is the distributive property of intersection over union.

Now, let's find the remaining sets:

iii) $A \setminus (B \cap C)$

First, find $B \cap C$: {elements common to B and C}

$B \cap C = \{2, 4, 6, 8, 10\} \cap \{3, 6, 9\}$

$B \cap C = \{6\}$

$A \setminus (B \cap C)$ = {elements in A that are not in $(B \cap C)$}

$A \setminus (B \cap C) = \{2, 3, 5, 7\} \setminus \{6\}$

$A \setminus (B \cap C) = \{2, 3, 5, 7\}$

Next, find $A \setminus B$ and $A \setminus C$:

$A \setminus B$ = {elements in A that are not in B}

$A \setminus B = \{2, 3, 5, 7\} \setminus \{2, 4, 6, 8, 10\}$

$A \setminus B = \{3, 5, 7\}$

$A \setminus C$ = {elements in A that are not in C}

$A \setminus C = \{2, 3, 5, 7\} \setminus \{3, 6, 9\}$

$A \setminus C = \{2, 5, 7\}$

iv) $(A \setminus B) \cup (A \setminus C)$

$(A \setminus B) \cup (A \setminus C)$ = {elements in $(A \setminus B)$ or $(A \setminus C)$ or both}

$(A \setminus B) \cup (A \setminus C) = \{3, 5, 7\} \cup \{2, 5, 7\}$

$(A \setminus B) \cup (A \setminus C) = \{2, 3, 5, 7\}$

Verification of $A \setminus (B \cap C) = (A \setminus B) \cup (A \setminus C)$

We found that $A \setminus (B \cap C) = \{2, 3, 5, 7\}$ and $(A \setminus B) \cup (A \setminus C) = \{2, 3, 5, 7\}$.

Thus, $A \setminus (B \cap C) = (A \setminus B) \cup (A \setminus C)$ is verified. This is De Morgan's Law for sets, applied to the difference operation.

Summary of Results:

$A = \{2, 3, 5, 7\}$

$B = \{2, 4, 6, 8, 10\}$

$C = \{3, 6, 9\}$

i) $A \cap (B \cup C) = \{2, 3\}$

ii) $(A \cap B) \cup (A \cap C) = \{2, 3\}$

iii) $A \setminus (B \cap C) = \{2, 3, 5, 7\}$

iv) $(A \setminus B) \cup (A \setminus C) = \{2, 3, 5, 7\}$

Part 1: Prove that $A \setminus B = A \cap B'$

We need to show that every element in $A \setminus B$ is in $A \cap B'$, and every element in $A \cap B'$ is in $A \setminus B$.

Step 1: Show $A \setminus B \subseteq A \cap B'$

Let $x \in A \setminus B$. By the definition of set difference, this means $x \in A$ and $x \notin B$.

If $x \notin B$, then by the definition of a complement, $x \in B'$.

So, we have $x \in A$ and $x \in B'$. By the definition of intersection, this means $x \in A \cap B'$.

Therefore, $A \setminus B \subseteq A \cap B'$.

Step 2: Show $A \cap B' \subseteq A \setminus B$

Let $x \in A \cap B'$. By the definition of intersection, this means $x \in A$ and $x \in B'$.

If $x \in B'$, then by the definition of a complement, $x \notin B$.

So, we have $x \in A$ and $x \notin B$. By the definition of set difference, this means $x \in A \setminus B$.

Therefore, $A \cap B' \subseteq A \setminus B$.

From Step 1 and Step 2, we have proven that $A \setminus B = A \cap B'$.

Part 2: Prove that $(A \setminus B) \cup (B \setminus A) = (A \cup B) \setminus (A \cap B)$

We need to show that every element in $(A \setminus B) \cup (B \setminus A)$ is in $(A \cup B) \setminus (A \cap B)$, and vice versa.

Step 1: Show $(A \setminus B) \cup (B \setminus A) \subseteq (A \cup B) \setminus (A \cap B)$

Let $x \in (A \setminus B) \cup (B \setminus A)$. This means $x \in A \setminus B$ or $x \in B \setminus A$.

Case 1: $x \in A \setminus B$. This means $x \in A$ and $x \notin B$.

Since $x \in A$, it follows that $x \in A \cup B$.

Since $x \notin B$, it cannot be in $A \cap B$ (because for an element to be in $A \cap B$, it must be in both A and B). So, $x \notin (A \cap B)$.

Since $x \in A \cup B$ and $x \notin (A \cap B)$, it means $x \in (A \cup B) \setminus (A \cap B)$.

Case 2: $x \in B \setminus A$. This means $x \in B$ and $x \notin A$.

Since $x \in B$, it follows that $x \in A \cup B$.

Since $x \notin A$, it cannot be in $A \cap B$. So, $x \notin (A \cap B)$.

Since $x \in A \cup B$ and $x \notin (A \cap B)$, it means $x \in (A \cup B) \setminus (A \cap B)$.

In both cases, if $x \in (A \setminus B) \cup (B \setminus A)$, then $x \in (A \cup B) \setminus (A \cap B)$.

Therefore, $(A \setminus B) \cup (B \setminus A) \subseteq (A \cup B) \setminus (A \cap B)$.

Step 2: Show $(A \cup B) \setminus (A \cap B) \subseteq (A \setminus B) \cup (B \setminus A)$

Let $x \in (A \cup B) \setminus (A \cap B)$. This means $x \in (A \cup B)$ and $x \notin (A \cap B)$.

From $x \in (A \cup B)$, we know that $x \in A$ or $x \in B$ (or both).

From $x \notin (A \cap B)$, we know that it is NOT the case that ($x \in A$ AND $x \in B$).

Now consider the possibilities:

Possibility 1: $x \in A$ and $x \notin B$.

In this case, $x \in A \setminus B$. Since $A \setminus B \subseteq (A \setminus B) \cup (B \setminus A)$, we have $x \in (A \setminus B) \cup (B \setminus A)$.

Possibility 2: $x \notin A$ and $x \in B$.

In this case, $x \in B \setminus A$. Since $B \setminus A \subseteq (A \setminus B) \cup (B \setminus A)$, we have $x \in (A \setminus B) \cup (B \setminus A)$.

We cannot have the case where $x \in A$ and $x \in B$, because that would imply $x \in A \cap B$, which contradicts our initial condition that $x \notin (A \cap B)$.

Therefore, if $x \in (A \cup B) \setminus (A \cap B)$, then $x$ must be in either $A \setminus B$ or $B \setminus A$. This means $x \in (A \setminus B) \cup (B \setminus A)$.

Thus, $(A \cup B) \setminus (A \cap B) \subseteq (A \setminus B) \cup (B \setminus A)$.

From Step 1 and Step 2, we have proven that $(A \setminus B) \cup (B \setminus A) = (A \cup B) \setminus (A \cap B)$.

Given:

Total number of persons, $|U| = 80$.

Number of persons who like Hindi, $|H| = 40$.

Number of persons who like English, $|E| = 50$.

Number of persons who like neither Hindi nor English, $|(H \cup E)'| = 10$.

To find:

i) Number of persons who like both Hindi and English, $|H \cap E|$.

ii) Number of persons who like only Hindi, $|H \setminus E|$.

iii) Number of persons who like only English, $|E \setminus H|$.

Solution:

First, let's find the number of persons who like at least one of the subjects (Hindi or English).

The number of persons who like at least one subject is the total number of persons minus those who like neither.

$|H \cup E| = |U| - |(H \cup E)'|$

$|H \cup E| = 80 - 10 = 70$.

Now, we use the Principle of Inclusion-Exclusion for two sets:

$|H \cup E| = |H| + |E| - |H \cap E|$

We can rearrange this formula to find $|H \cap E|$:

$|H \cap E| = |H| + |E| - |H \cup E|$

$|H \cap E| = 40 + 50 - 70$

$|H \cap E| = 90 - 70 = 20$.

i) How many like both Hindi and English?

The number of persons who like both Hindi and English is $|H \cap E|$.

Answer: 20 persons.

ii) How many like only Hindi?

The number of persons who like only Hindi is the total number of persons who like Hindi minus those who like both Hindi and English.

$|H \setminus E| = |H| - |H \cap E|$

$|H \setminus E| = 40 - 20 = 20$.

Answer: 20 persons.

iii) How many like only English?

The number of persons who like only English is the total number of persons who like English minus those who like both Hindi and English.

$|E \setminus H| = |E| - |H \cap E|$

$|E \setminus H| = 50 - 20 = 30$.

Answer: 30 persons.

Verification:

Total persons = (Only Hindi) + (Only English) + (Both Hindi and English) + (Neither)

$80 = 20 + 30 + 20 + 10$

$80 = 80$. The results are consistent.

Given Universal Set: $\mathbb{R}$ (The set of all real numbers).

Given Sets:

Set $A = \{x \in \mathbb{R} : x \geq 3\}$. In interval notation, $A = [3, \infty)$.

Set $B = \{x \in \mathbb{R} : x < 7\}$. In interval notation, $B = (-\infty, 7)$.

Finding the required sets in interval notation:

1. $A \cup B$

$A \cup B$ is the set of all real numbers that are in A or in B (or both).

Since A starts at 3 and goes to infinity, and B starts from negative infinity and goes up to 7, their union covers all real numbers.

$A \cup B = [3, \infty) \cup (-\infty, 7)$

$A \cup B = (-\infty, \infty) = \mathbb{R}$.

2. $A \cap B$

$A \cap B$ is the set of all real numbers that are common to both A and B.

We need numbers that are $x \geq 3$ AND $x < 7$.

$A \cap B = [3, \infty) \cap (-\infty, 7)$

$A \cap B = [3, 7)$.

3. $A \setminus B$

$A \setminus B$ is the set of all real numbers that are in A but not in B.

This means $x \geq 3$ AND $x \not< 7$ (which is equivalent to $x \geq 7$).

$A \setminus B = [3, \infty) \setminus (-\infty, 7)$

$A \setminus B = [7, \infty)$.

4. $B \setminus A$

$B \setminus A$ is the set of all real numbers that are in B but not in A.

This means $x < 7$ AND $x \not\geq 3$ (which is equivalent to $x < 3$).

$B \setminus A = (-\infty, 7) \setminus [3, \infty)$

$B \setminus A = (-\infty, 3)$.

5. $A'$

$A'$ is the complement of A with respect to the universal set $\mathbb{R}$.

If $A = \{x \in \mathbb{R} : x \geq 3\}$, then $A'$ is the set of all real numbers that are NOT greater than or equal to 3.

This means $x < 3$.

$A' = \{x \in \mathbb{R} : x < 3\}$

$A' = (-\infty, 3)$.

6. $B'$

$B'$ is the complement of B with respect to the universal set $\mathbb{R}$.

If $B = \{x \in \mathbb{R} : x < 7\}$, then $B'$ is the set of all real numbers that are NOT less than 7.

This means $x \geq 7$.

$B' = \{x \in \mathbb{R} : x \geq 7\}$

$B' = [7, \infty)$.

Summary of Results:

$A = [3, \infty)$

$B = (-\infty, 7)$

$A \cup B = (-\infty, \infty)$

$A \cap B = [3, 7)$

$A \setminus B = [7, \infty)$

$B \setminus A = (-\infty, 3)$

$A' = (-\infty, 3)$

$B' = [7, \infty)$

To Prove: $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$

We will prove this by showing that each side is a subset of the other.

Part 1: Show $A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C)$

Let $x \in A \cup (B \cap C)$.

By the definition of union, this means $x \in A$ or $x \in (B \cap C)$.

Case 1: $x \in A$.

If $x \in A$, then by the definition of union, $x \in A \cup B$.

Also, if $x \in A$, then by the definition of union, $x \in A \cup C$.

Since $x \in A \cup B$ and $x \in A \cup C$, by the definition of intersection, $x \in (A \cup B) \cap (A \cup C)$.

Case 2: $x \in (B \cap C)$.

By the definition of intersection, this means $x \in B$ and $x \in C$.

If $x \in B$, then by the definition of union, $x \in A \cup B$.

If $x \in C$, then by the definition of union, $x \in A \cup C$.

Since $x \in A \cup B$ and $x \in A \cup C$, by the definition of intersection, $x \in (A \cup B) \cap (A \cup C)$.

In both cases, if $x \in A \cup (B \cap C)$, then $x \in (A \cup B) \cap (A \cup C)$.

Therefore, $A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C)$.

Part 2: Show $(A \cup B) \cap (A \cup C) \subseteq A \cup (B \cap C)$

Let $x \in (A \cup B) \cap (A \cup C)$.

By the definition of intersection, this means $x \in (A \cup B)$ and $x \in (A \cup C)$.

This implies that ($x \in A$ or $x \in B$) AND ($x \in A$ or $x \in C$).

We can use the distributive property of "or" over "and" for logical statements:

$(P \lor Q) \land (P \lor R) \equiv P \lor (Q \land R)$

Here, $P$ is $x \in A$, $Q$ is $x \in B$, and $R$ is $x \in C$.

So, the statement becomes: $x \in A$ or ($x \in B$ and $x \in C$).

By the definition of intersection, ($x \in B$ and $x \in C$) means $x \in (B \cap C)$.

So, the statement is $x \in A$ or $x \in (B \cap C)$.

By the definition of union, this means $x \in A \cup (B \cap C)$.

Therefore, $(A \cup B) \cap (A \cup C) \subseteq A \cup (B \cap C)$.

Conclusion:

Since we have shown that $A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C)$ and $(A \cup B) \cap (A \cup C) \subseteq A \cup (B \cap C)$, we can conclude that $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$.

Let:

$N(A)$ be the percentage of people who watch news channel A.

$N(B)$ be the percentage of people who watch news channel B.

$N(A \cap B)$ be the percentage of people who watch both channels.

$N(A \cup B)$ be the percentage of people who watch at least one channel.

$N(\text{neither A nor B})$ be the percentage of people who watch neither channel.

Given:

$N(A) = 63\%$.

$N(B) = 76\%$.

$N(A \cap B) = x\%$.

Using the Principle of Inclusion-Exclusion:

$N(A \cup B) = N(A) + N(B) - N(A \cap B)$

$N(A \cup B) = 63\% + 76\% - x\%$

$N(A \cup B) = 139\% - x\%$.

We know that the total percentage of people must be $100\%$. The percentage of people who watch at least one channel ($N(A \cup B)$) cannot be more than $100\%$. Also, the percentage of people who watch neither channel is $100\% - N(A \cup B)$.

From $N(A \cup B) = 139\% - x\%$, we have:

$N(A \cup B) \leq 100\%$.

$139\% - x\% \leq 100\%$.

$139\% - 100\% \leq x\%$

$39\% \leq x\%$.

Also, the percentage of people watching both channels ($x\%$) cannot be more than the percentage of people watching either channel individually.

$x\% \leq N(A) \implies x\% \leq 63\%$.

$x\% \leq N(B) \implies x\% \leq 76\%$.

So, $x\%$ must be between $39\%$ and $63\%$.

To find a specific value for $x$, we need to consider the minimum overlap required for the percentages to be consistent.

Consider the percentage of people who watch at least one channel. This cannot exceed 100%.

So, $N(A \cup B) \leq 100$.

Using the formula $N(A \cup B) = N(A) + N(B) - N(A \cap B)$: $N(A \cup B) = 63 + 76 - x = 139 - x$.

Since $N(A \cup B) \leq 100$, we have $139 - x \leq 100$.

This implies $139 - 100 \leq x$, so $x \geq 39$.

Also, the number of people who watch both channels cannot be more than the number of people watching either channel.

So, $x \leq N(A)$ and $x \leq N(B)$.

$x \leq 63$ and $x \leq 76$. Therefore, $x \leq 63$.

The question implies a specific value for $x$. The minimum overlap occurs when $N(A \cup B)$ is maximized (i.e., 100%).

If $N(A \cup B) = 100\%$, then:

$100\% = 63\% + 76\% - x\%$

$100\% = 139\% - x\%$

$x\% = 139\% - 100\%$

$x\% = 39\%$.

This means the minimum percentage of people who watch both channels is $39\%$. If $x$ is exactly $39\%$, then everyone watches at least one channel ($N(A \cup B) = 100\%$) and $100\% - 100\% = 0\%$ watch neither.

Value of $x$:

The problem formulation suggests finding a single value for $x$. In such problems, the minimum overlap is usually implied or the situation where everyone watches at least one channel is assumed if not stated otherwise.

Assuming that at least everyone watches one channel (i.e., $N(A \cup B) = 100\%$), we find $x = 39$.

So, $x = 39\%$.

Percentage of people who watch exactly one channel:

Percentage of people who watch only channel A = $N(A) - N(A \cap B) = 63\% - 39\% = 24\%$.

Percentage of people who watch only channel B = $N(B) - N(A \cap B) = 76\% - 39\% = 37\%$.

Percentage of people who watch exactly one channel = (Only A) + (Only B)

= $24\% + 37\% = 61\%$.

Let's check the total:

$N(\text{only A}) + N(\text{only B}) + N(A \cap B) + N(\text{neither})$

$24\% + 37\% + 39\% + 0\% = 100\%$. This confirms our assumption that $N(A \cup B)=100\%$.

Therefore:

The value of $x$ is $39\%$.

The percentage of people who watch exactly one channel is $61\%$.

To Prove: $P(A \cap B) = P(A) \cap P(B)$

Here, $P(X)$ denotes the power set of set X, which is the set of all subsets of X.

We will prove this by showing that each side is a subset of the other.

Part 1: Show $P(A \cap B) \subseteq P(A) \cap P(B)$

Let $S$ be an arbitrary element of $P(A \cap B)$.

By the definition of a power set, $S$ is a subset of $(A \cap B)$.

$S \subseteq (A \cap B)$

By the definition of intersection, this means that every element of $S$ is also an element of $A$, AND every element of $S$ is also an element of $B$.

So, for any $x \in S$, we have $x \in A$ and $x \in B$.

If every element of $S$ is an element of $A$, then $S$ is a subset of $A$.

$S \subseteq A$

If every element of $S$ is an element of $B$, then $S$ is a subset of $B$.

$S \subseteq B$

Since $S$ is a subset of $A$, $S \in P(A)$ by definition of the power set.

Since $S$ is a subset of $B$, $S \in P(B)$ by definition of the power set.

If $S \in P(A)$ and $S \in P(B)$, then by the definition of intersection, $S \in P(A) \cap P(B)$.

Since $S$ was an arbitrary element of $P(A \cap B)$, we have shown that $P(A \cap B) \subseteq P(A) \cap P(B)$.

Part 2: Show $P(A) \cap P(B) \subseteq P(A \cap B)$

Let $S$ be an arbitrary element of $P(A) \cap P(B)$.

By the definition of intersection, this means $S \in P(A)$ and $S \in P(B)$.

Since $S \in P(A)$, $S$ is a subset of $A$ ($S \subseteq A$).

Since $S \in P(B)$, $S$ is a subset of $B$ ($S \subseteq B$).

If $S$ is a subset of $A$, it means every element of $S$ is in $A$.

If $S$ is a subset of $B$, it means every element of $S$ is in $B$.

So, for any element $x \in S$, we have $x \in A$ AND $x \in B$.

By the definition of intersection, this means $x \in (A \cap B)$.

Since every element of $S$ is an element of $(A \cap B)$, by the definition of a subset, $S$ is a subset of $(A \cap B)$.

$S \subseteq (A \cap B)$

Since $S$ is a subset of $(A \cap B)$, by the definition of the power set, $S \in P(A \cap B)$.

Since $S$ was an arbitrary element of $P(A) \cap P(B)$, we have shown that $P(A) \cap P(B) \subseteq P(A \cap B)$.

Conclusion:

Since we have shown that $P(A \cap B) \subseteq P(A) \cap P(B)$ and $P(A) \cap P(B) \subseteq P(A \cap B)$, we can conclude that $P(A \cap B) = P(A) \cap P(B)$.

Example:

Let $A = \{1, 2\}$ and $B = \{2, 3\}$.

First, find $A \cap B$: $A \cap B = \{1, 2\} \cap \{2, 3\} = \{2\}$.

Now, find the power set of $A \cap B$: $P(A \cap B) = P(\{2\})$. The subsets of $\{2\}$ are the empty set and $\{2\}$. So, $P(A \cap B) = \{\emptyset, \{2\}\}$.

Next, find the power set of A:

$P(A) = P(\{1, 2\})$. The subsets of $\{1, 2\}$ are $\emptyset$, $\{1\}$, $\{2\}$, and $\{1, 2\}$. So, $P(A) = \{\emptyset, \{1\}, \{2\}, \{1, 2\}\}$.

Next, find the power set of B:

$P(B) = P(\{2, 3\})$. The subsets of $\{2, 3\}$ are $\emptyset$, $\{2\}$, $\{3\}$, and $\{2, 3\}$. So, $P(B) = \{\emptyset, \{2\}, \{3\}, \{2, 3\}\}$.

Now, find the intersection of $P(A)$ and $P(B)$: $P(A) \cap P(B) = \{\emptyset, \{1\}, \{2\}, \{1, 2\}\} \cap \{\emptyset, \{2\}, \{3\}, \{2, 3\}\}$. The common elements are $\emptyset$ and $\{2\}$. So, $P(A) \cap P(B) = \{\emptyset, \{2\}\}$.

Comparing the results:

$P(A \cap B) = \{\emptyset, \{2\}\}$

$P(A) \cap P(B) = \{\emptyset, \{2\}\}$

Thus, $P(A \cap B) = P(A) \cap P(B)$ is illustrated by this example.

Let:

$H$ be the set of people who speak Hindi.

$E$ be the set of people who speak English.

Given:

Number of people who speak Hindi, $|H| = 50$.

Number of people who speak English, $|E| = 20$.

Number of people who speak both Hindi and English, $|H \cap E| = 10$.

To find:

The total number of people in the committee, given that everyone speaks at least one of the two languages. This is $|H \cup E|$.

Solution:

We can use the Principle of Inclusion-Exclusion to find the total number of people in the committee.

The formula for the union of two sets is:

$|H \cup E| = |H| + |E| - |H \cap E|$

Substitute the given values into the formula:

$|H \cup E| = 50 + 20 - 10$

$|H \cup E| = 70 - 10$

$|H \cup E| = 60$.

Since everyone speaks at least one of the two languages, the total number of people in the committee is equal to the number of people who speak Hindi or English or both.

Answer: There are 60 people in the committee.

To Prove: $A \cap B = A \setminus (A \setminus B)$

We will prove this by showing that each side is a subset of the other.

Part 1: Show $A \cap B \subseteq A \setminus (A \setminus B)$

Let $x \in A \cap B$. By the definition of intersection, this means $x \in A$ and $x \in B$.

We want to show that $x \in A \setminus (A \setminus B)$. This means we need to show that $x \in A$ and $x \notin (A \setminus B)$.

From our initial assumption, we already know that $x \in A$. So, we only need to show that $x \notin (A \setminus B)$.

For $x$ to be in $A \setminus B$, it must be the case that $x \in A$ and $x \notin B$.

However, we know that $x \in B$ (from $x \in A \cap B$). Since $x \in B$, it cannot be true that $x \notin B$.

Therefore, $x$ cannot be in $A \setminus B$. This means $x \notin (A \setminus B)$.

Since $x \in A$ and $x \notin (A \setminus B)$, by the definition of set difference, $x \in A \setminus (A \setminus B)$.

Thus, $A \cap B \subseteq A \setminus (A \setminus B)$.

Part 2: Show $A \setminus (A \setminus B) \subseteq A \cap B$

Let $x \in A \setminus (A \setminus B)$. By the definition of set difference, this means $x \in A$ and $x \notin (A \setminus B)$.

We want to show that $x \in A \cap B$. This means we need to show that $x \in A$ and $x \in B$.

From our initial assumption, we already know that $x \in A$. So, we only need to show that $x \in B$.

We know that $x \notin (A \setminus B)$. By the definition of set difference, this means it is NOT true that ($x \in A$ and $x \notin B$).

Using De Morgan's law for logic, $\neg (P \land Q) \equiv \neg P \lor \neg Q$.

So, $\neg (x \in A \land x \notin B)$ is equivalent to $\neg (x \in A) \lor \neg (x \notin B)$.

This means $x \notin A$ or $x \in B$.

We also know that $x \in A$ (from our initial assumption $x \in A \setminus (A \setminus B)$).

We have the statements: ($x \notin A$ or $x \in B$) AND ($x \in A$).

Using the distributive law of logic: $(P \lor Q) \land R \equiv (P \land R) \lor (Q \land R)$.

Let $P$ be $x \notin A$, $Q$ be $x \in B$, and $R$ be $x \in A$.

So, $(x \notin A \lor x \in B) \land (x \in A) \equiv (x \notin A \land x \in A) \lor (x \in B \land x \in A)$.

The statement $(x \notin A \land x \in A)$ is a contradiction, so it is false.

Therefore, the statement simplifies to $(False) \lor (x \in B \land x \in A)$, which is equivalent to $(x \in B \land x \in A)$.

This means $x \in A$ and $x \in B$. By the definition of intersection, $x \in A \cap B$.

Thus, $A \setminus (A \setminus B) \subseteq A \cap B$.

Conclusion:

Since we have shown that $A \cap B \subseteq A \setminus (A \setminus B)$ and $A \setminus (A \setminus B) \subseteq A \cap B$, we can conclude that $A \cap B = A \setminus (A \setminus B)$.

Let:

$E$ be the set of students who study Economics.

$C$ be the set of students who study Commerce.

Given:

Total number of students, $|U| = 60$.

Number of students who study Economics, $|E| = 20$.

Number of students who study Commerce, $|C| = 25$.

Number of students who study both Economics and Commerce, $|E \cap C| = 15$.

To find:

Number of students who study neither Economics nor Commerce.

Solution:

First, we need to find the number of students who study at least one of the subjects (Economics or Commerce). We can use the Principle of Inclusion-Exclusion for this:

$|E \cup C| = |E| + |C| - |E \cap C|$

Substitute the given values into the formula:

$|E \cup C| = 20 + 25 - 15$

$|E \cup C| = 45 - 15$

$|E \cup C| = 30$.

This means 30 students study at least one of the two subjects.

Now, to find the number of students who study neither Economics nor Commerce, we subtract the number of students who study at least one subject from the total number of students:

Number of students who study neither = $|U| - |E \cup C|$

Number of students who study neither = $60 - 30$

Number of students who study neither = $30$.

Answer: 30 students study neither Economics nor Commerce.

To Prove: $A \cup (A \cap B) = A$

We will prove this using the properties of set operations.

We start with the left-hand side of the equation:

$A \cup (A \cap B)$

We know the property of intersection: $A \cap B \subseteq A$.

This means that any element that is in the intersection of A and B is also in A.

Now consider the union of a set A with a subset of A. If we take the union of a set with any of its subsets, the result is always the original set.

This property is called the Idempotent Law of Union, which states that for any set X, $X \cup X = X$. However, here we have $A \cup (\text{a subset of } A)$.

Let's use another property: For any sets X and Y, if $Y \subseteq X$, then $X \cup Y = X$.

In our case, we have $Y = A \cap B$ and $X = A$. We already established that $A \cap B \subseteq A$.

Therefore, applying this property:

$A \cup (A \cap B) = A$

Alternatively, using algebraic proof:

$A \cup (A \cap B)$

Using the distributive law of union over intersection: $X \cup (Y \cap Z) = (X \cup Y) \cap (X \cup Z)$.

Applying this, we get:

$A \cup (A \cap B) = (A \cup A) \cap (A \cup B)$

Now, we use the Idempotent Law of Union, which states that $A \cup A = A$.

So, $(A \cup A) \cap (A \cup B) = A \cap (A \cup B)$

Now, we use the property of intersection with union, which states that $X \cap (X \cup Y) = X$. This is called the Absorption Law.

Applying this, where $X=A$ and $Y=B$, we get:

$A \cap (A \cup B) = A$

Both methods lead to the same conclusion.

Conclusion:

Thus, $A \cup (A \cap B) = A$ has been proven using the properties of set operations.

Given:

Universal Set, $U = \{1, 2, 3, 4, 5, 6\}$.

Set $A = \{1, 2\}$.

Set $B = \{2, 3, 4\}$.

We need to verify De Morgan's Law: $(A \cap B)' = A' \cup B'$.

Step 1: Find the elements of the Left-Hand Side (LHS): $(A \cap B)'$

First, find the intersection of A and B ($A \cap B$):

$A \cap B = \{x \mid x \in A \text{ and } x \in B\}$

$A \cap B = \{1, 2\} \cap \{2, 3, 4\}$

$A \cap B = \{2\}$

Next, find the complement of $(A \cap B)$, denoted as $(A \cap B)'$. The complement consists of all elements in the universal set U that are not in $(A \cap B)$.

$(A \cap B)' = U \setminus (A \cap B)$

$(A \cap B)' = \{1, 2, 3, 4, 5, 6\} \setminus \{2\}$

$(A \cap B)' = \{1, 3, 4, 5, 6\}$

Step 2: Find the elements of the Right-Hand Side (RHS): $A' \cup B'$

First, find the complement of A ($A'$):

$A' = U \setminus A$

$A' = \{1, 2, 3, 4, 5, 6\} \setminus \{1, 2\}$

$A' = \{3, 4, 5, 6\}$

Next, find the complement of B ($B'$):

$B' = U \setminus B$

$B' = \{1, 2, 3, 4, 5, 6\} \setminus \{2, 3, 4\}$

$B' = \{1, 5, 6\}$

Now, find the union of $A'$ and $B'$ ($A' \cup B'$):

$A' \cup B' = \{x \mid x \in A' \text{ or } x \in B'\}$

$A' \cup B' = \{3, 4, 5, 6\} \cup \{1, 5, 6\}$

$A' \cup B' = \{1, 3, 4, 5, 6\}$

Step 3: Compare LHS and RHS

LHS: $(A \cap B)' = \{1, 3, 4, 5, 6\}$

RHS: $A' \cup B' = \{1, 3, 4, 5, 6\}$

Since LHS = RHS, De Morgan's Law $(A \cap B)' = A' \cup B'$ is verified.

Given:

Number of elements in set A, $|A| = 2$.

Number of elements in set B, $|B| = 3$.

We need to find the minimum and maximum possible values for $|A \cup B|$ and $|A \cap B|$.

Understanding the relationships:

The Principle of Inclusion-Exclusion states: $|A \cup B| = |A| + |B| - |A \cap B|$.

This can be rewritten as: $|A \cap B| = |A| + |B| - |A \cup B|$.

Consider the intersection $|A \cap B|$:

The intersection $A \cap B$ contains elements that are common to both A and B. The number of common elements cannot be more than the size of the smaller set, and it cannot be less than zero.

Maximum value of $|A \cap B|$:

The maximum number of common elements occurs when the smaller set is a subset of the larger set. In this case, the smaller set is A ($|A|=2$).

Maximum $|A \cap B| = \min(|A|, |B|) = \min(2, 3) = 2$.

This happens when A is a subset of B.

Example for Maximum Intersection:

Let $A = \{1, 2\}$ and $B = \{1, 2, 3\}$.

Here, $A \cap B = \{1, 2\}$, so $|A \cap B| = 2$.

Minimum value of $|A \cap B|$:

The minimum number of common elements occurs when the sets have as little overlap as possible. The number of common elements cannot be negative.

Using the formula $|A \cap B| = |A| + |B| - |A \cup B|$.

The maximum value of $|A \cup B|$ is when the sets are disjoint as much as possible. However, the total number of elements in the union cannot exceed the sum of the sizes of the sets plus the universal set size (if it was defined). Since the universal set is not specified, we consider the case where the union is minimized in terms of overlap.

If the sets have no overlap, $|A \cup B| = |A| + |B| = 2 + 3 = 5$. In this case, $|A \cap B| = 0$.

Minimum $|A \cap B| = \max(0, |A| + |B| - |U|)$, where $|U|$ is the size of the universal set. If $|U|$ is not specified, we assume it's large enough for the sets to potentially be disjoint.

In the absence of a universal set constraint, the minimum possible intersection is 0 if the sets can be disjoint.

Minimum $|A \cap B| = 0$.

This happens when the sets have no common elements.

Example for Minimum Intersection:

Let $A = \{1, 2\}$ and $B = \{3, 4, 5\}$.

Here, $A \cap B = \emptyset$, so $|A \cap B| = 0$.

Consider the union $|A \cup B|$:

Using $|A \cup B| = |A| + |B| - |A \cap B|$.

Minimum value of $|A \cup B|$:

The minimum value of $|A \cup B|$ occurs when $|A \cap B|$ is at its maximum.

Minimum $|A \cup B| = |A| + |B| - \text{Maximum } |A \cap B|$

Minimum $|A \cup B| = 2 + 3 - 2 = 3$.

This happens when A is a subset of B.

Example for Minimum Union:

Let $A = \{1, 2\}$ and $B = \{1, 2, 3\}$.

Here, $A \cup B = \{1, 2, 3\}$, so $|A \cup B| = 3$.

Maximum value of $|A \cup B|$:

The maximum value of $|A \cup B|$ occurs when $|A \cap B|$ is at its minimum.

Maximum $|A \cup B| = |A| + |B| - \text{Minimum } |A \cap B|$

Maximum $|A \cup B| = 2 + 3 - 0 = 5$.

This happens when the sets are disjoint.

Example for Maximum Union:

Let $A = \{1, 2\}$ and $B = \{3, 4, 5\}$.

Here, $A \cup B = \{1, 2, 3, 4, 5\}$, so $|A \cup B| = 5$.

Summary:

Minimum value of $|A \cup B| = 3$.

Maximum value of $|A \cup B| = 5$.

Minimum value of $|A \cap B| = 0$.

Maximum value of $|A \cap B| = 2$.

To Prove: $A \setminus (B \cup C) = (A \setminus B) \cap (A \setminus C)$

We will prove this using an element-wise proof (algebraic proof).

Part 1: Show $A \setminus (B \cup C) \subseteq (A \setminus B) \cap (A \setminus C)$

Let $x \in A \setminus (B \cup C)$.

By the definition of set difference, this means $x \in A$ and $x \notin (B \cup C)$.

Since $x \notin (B \cup C)$, by the definition of union, it is not true that ($x \in B$ or $x \in C$).

Using De Morgan's law for logic, $\neg (P \lor Q) \equiv \neg P \land \neg Q$.

So, $x \notin (B \cup C)$ means $x \notin B$ and $x \notin C$.

Now we have: $x \in A$ and ($x \notin B$ and $x \notin C$).

Let's rearrange this:

($x \in A$ and $x \notin B$) and ($x \in A$ and $x \notin C$).

By the definition of set difference:

($x \in A$ and $x \notin B$) means $x \in A \setminus B$.

($x \in A$ and $x \notin C$) means $x \in A \setminus C$.

So, we have $x \in A \setminus B$ and $x \in A \setminus C$. By the definition of intersection, $x \in (A \setminus B) \cap (A \setminus C)$.

Therefore, $A \setminus (B \cup C) \subseteq (A \setminus B) \cap (A \setminus C)$.

Part 2: Show $(A \setminus B) \cap (A \setminus C) \subseteq A \setminus (B \cup C)$

Let $x \in (A \setminus B) \cap (A \setminus C)$.

By the definition of intersection, this means $x \in (A \setminus B)$ and $x \in (A \setminus C)$.

If $x \in A \setminus B$, then $x \in A$ and $x \notin B$.

If $x \in A \setminus C$, then $x \in A$ and $x \notin C$.

Combining these conditions, we have $x \in A$, $x \notin B$, and $x \notin C$.

Since $x \notin B$ and $x \notin C$, by the definition of union, it is not true that ($x \in B$ or $x \in C$).

This means $x \notin (B \cup C)$.

So, we have $x \in A$ and $x \notin (B \cup C)$.

By the definition of set difference, this means $x \in A \setminus (B \cup C)$.

Therefore, $(A \setminus B) \cap (A \setminus C) \subseteq A \setminus (B \cup C)$.

Conclusion:

Since we have shown that $A \setminus (B \cup C) \subseteq (A \setminus B) \cap (A \setminus C)$ and $(A \setminus B) \cap (A \setminus C) \subseteq A \setminus (B \cup C)$, we can conclude that $A \setminus (B \cup C) = (A \setminus B) \cap (A \setminus C)$.

Let:

$C$ be the set of students who played chess.

$R$ be the set of students who played carrom.

$T$ be the set of students who played table tennis.

Given:

Number of students who played chess, $|C| = 30$.

Number of students who played carrom, $|R| = 25$.

Number of students who played table tennis, $|T| = 20$.

Number of students who played chess and carrom, $|C \cap R| = 10$.

Number of students who played carrom and table tennis, $|R \cap T| = 8$.

Number of students who played chess and table tennis, $|C \cap T| = 12$.

Everyone played at least one game, which means $|C \cup R \cup T|$ is the total number of students.

To find:

i) The total number of students surveyed, $|C \cup R \cup T|$.

ii) The number of students who played only chess, $|C \setminus (R \cup T)|$.

Solution:

Part i) Total number of students surveyed:

We use the Principle of Inclusion-Exclusion for three sets:

$|C \cup R \cup T| = |C| + |R| + |T| - (|C \cap R| + |R \cap T| + |C \cap T|) + |C \cap R \cap T|$

We are missing the number of students who played all three games, $|C \cap R \cap T|$. Let's denote this as $x$.

So, $|C \cup R \cup T| = 30 + 25 + 20 - (10 + 8 + 12) + x$

$|C \cup R \cup T| = 75 - 30 + x$

$|C \cup R \cup T| = 45 + x$.

To find $x$, we can use the information about the number of students playing exactly one game, exactly two games, and exactly three games. However, we don't have direct information for exactly one or exactly two games.

Let's re-evaluate the given information. We need to find $x = |C \cap R \cap T|$.

The number of students who played chess and carrom only is $|C \cap R| - |C \cap R \cap T| = 10 - x$.

The number of students who played carrom and table tennis only is $|R \cap T| - |C \cap R \cap T| = 8 - x$.

The number of students who played chess and table tennis only is $|C \cap T| - |C \cap R \cap T| = 12 - x$.

The number of students who played only chess is $|C| - (\text{chess and carrom only}) - (\text{chess and table tennis only}) - (\text{all three})$

$|C|_{\text{only}} = |C| - (10 - x) - (12 - x) - x$

$|C|_{\text{only}} = 30 - 10 + x - 12 + x - x = 8 + x$.

Similarly:

$|R|_{\text{only}} = |R| - (10 - x) - (8 - x) - x = 25 - 10 + x - 8 + x - x = 7 + x$.

$|T|_{\text{only}} = |T| - (8 - x) - (12 - x) - x = 20 - 8 + x - 12 + x - x = 0 + x = x$.

The sum of all these disjoint regions must equal the total number of students:

$|C \cup R \cup T| = |C|_{\text{only}} + |R|_{\text{only}} + |T|_{\text{only}} + (|C \cap R|_{\text{only}}) + (|R \cap T|_{\text{only}}) + (|C \cap T|_{\text{only}}) + |C \cap R \cap T|$

$|C \cup R \cup T| = (8 + x) + (7 + x) + x + (10 - x) + (8 - x) + (12 - x) + x$

$|C \cup R \cup T| = 8 + x + 7 + x + x + 10 - x + 8 - x + 12 - x + x$

$|C \cup R \cup T| = (8 + 7 + 10 + 8 + 12) + (x + x + x - x - x - x + x)$

$|C \cup R \cup T| = 45 + x$.

This equation is consistent with what we got from the Inclusion-Exclusion Principle directly. However, we need a value for $x$. This type of problem usually implies that the number of people in each "only" category must be non-negative.

Consider the constraints on $x$: $10 - x \geq 0 \implies x \leq 10$ $8 - x \geq 0 \implies x \leq 8$ $12 - x \geq 0 \implies x \leq 12$ $8 + x \geq 0 \implies x \geq -8$ (always true for counts) $7 + x \geq 0 \implies x \geq -7$ (always true for counts) $x \geq 0$

Combining these, we get $0 \leq x \leq 8$.

Without further information or a Venn diagram that implies a specific value for $x$, we cannot determine a unique value for $x$ or the total number of students. However, problems of this nature often provide enough information for a unique solution. Let's check if there's any implicit information we missed or if a common value for $x$ is assumed.

A common structure in these problems is that the sum of students in specific categories allows us to deduce the triple intersection. If the question implies a unique answer, there might be a typical value for the "all three" category that makes the numbers work out neatly.

Let's re-read carefully: "Everyone played at least one game." This is already incorporated in the total being $|C \cup R \cup T|$.

Let's assume there's a missing piece of information or that a standard value for $x$ is implied by the problem setter. If we assume the number of students playing all three games, say $x$, is a specific value that makes the problem solvable without additional constraints, we would need more context.

However, if we assume a typical problem structure, sometimes the sum of the pairwise intersections minus the sum of single sets plus the total can give clues, but that's not directly applicable here.

Let's consider if there's a way to deduce $x$ from the fact that the "only" categories must be non-negative. The tightest constraint is $x \leq 8$.

Let's revisit the structure of such problems. If the question is solvable with the given information, it implies $x$ has a specific value. A common scenario is when the sum of those playing only one game, only two games, and all three games adds up to the total.

If we assume a specific value for $x$ (e.g., a value that makes the "only" categories sensible), let's try $x=5$ as a plausible value that is within $0 \leq x \leq 8$.

If $x=5$:

$|C \cap R \cap T| = 5$.

Chess and Carrom only: $10 - 5 = 5$.

Carrom and Table Tennis only: $8 - 5 = 3$.

Chess and Table Tennis only: $12 - 5 = 7$.

Only Chess: $30 - (5 + 7 + 5) = 30 - 17 = 13$.

Only Carrom: $25 - (5 + 3 + 5) = 25 - 13 = 12$.

Only Table Tennis: $20 - (3 + 7 + 5) = 20 - 15 = 5$.

Total students = $13 + 12 + 5 + 5 + 3 + 7 + 5 = 50$.

But the total is supposed to be derived, and this leads to 50 if $x=5$. If the total surveyed students were indeed 50, then $x=5$. If the total is not given, and "everyone played at least one game" implies the total is $|C \cup R \cup T|$, then we must be able to find $x$.

Let's re-examine the formula $|C \cup R \cup T| = 45 + x$. If the total number of students surveyed is not explicitly given but implied to be the union, there must be a way to find $x$.

There might be an assumption about the number of students playing all three games when not explicitly stated. Often, these problems are constructed such that the "only" values are positive integers. The constraint $x \leq 8$ is the most restrictive upper bound.

Let's reconsider the wording. "Find the total number of students surveyed" implies this is something to be calculated. If $x$ were a variable, the total would be $45+x$. This suggests that $x$ should be deducible.

Is it possible that some information is missing, or that there's a property I'm overlooking?

Let's think about the "only" categories from the counts directly:

$|C| = |C|_{\text{only}} + |C \cap R|_{\text{only}} + |C \cap T|_{\text{only}} + |C \cap R \cap T|$

$30 = |C|_{\text{only}} + (10-x) + (12-x) + x$

$30 = |C|_{\text{only}} + 22 - x \implies |C|_{\text{only}} = 8 + x$. (This matches our previous calculation).

Let's assume for the sake of providing an answer that there is a specific value for $x$ intended. A common pattern is that the problem is set up so that all categories are non-negative integers. The tightest constraint is $x \leq 8$.

If we consider a Venn Diagram, the segments would be:

Chess only: $8+x$ Carrom only: $7+x$ Table Tennis only: $x$ Chess & Carrom only: $10-x$ Carrom & Table Tennis only: $8-x$ Chess & Table Tennis only: $12-x$ Chess, Carrom & Table Tennis: $x$

Total surveyed = $(8+x) + (7+x) + x + (10-x) + (8-x) + (12-x) + x = 45 + x$.

It's possible that the question intends for us to use a specific value of $x$ that is the largest possible value consistent with the data, which is $x=8$. Let's see if this leads to a sensible answer.

If $x=8$:

$|C \cap R \cap T| = 8$.

Chess and Carrom only: $10 - 8 = 2$.

Carrom and Table Tennis only: $8 - 8 = 0$.

Chess and Table Tennis only: $12 - 8 = 4$.

Only Chess: $8 + 8 = 16$.

Only Carrom: $7 + 8 = 15$.

Only Table Tennis: $8$.

Total students = $16 + 15 + 8 + 2 + 0 + 4 + 8 = 53$.

So, if $x=8$, the total number of students surveyed is 53, and the number who played only chess is 16.

This seems like a plausible interpretation given the constraints. The phrasing "Find the total number of students surveyed" implies a unique answer, suggesting $x$ must be determined by the non-negativity constraints.

Part i) Total number of students surveyed:

Assuming $x = 8$ (the maximum possible value for the triple intersection to keep all categories non-negative).

Total students = $45 + x = 45 + 8 = 53$.

Part ii) Number of students who played only chess:

Number of students who played only chess = $|C|_{\text{only}} = 8 + x$.

With $x=8$, the number of students who played only chess = $8 + 8 = 16$.

Answer:

The total number of students surveyed is 53.

The number of students who played only chess is 16.

To Prove: $A \subseteq B$ if and only if $A \cap B = A$.

This is a biconditional statement, so we need to prove two implications:

1. If $A \subseteq B$, then $A \cap B = A$.
2. If $A \cap B = A$, then $A \subseteq B$.

Part 1: Prove that if $A \subseteq B$, then $A \cap B = A$.

Assume that $A \subseteq B$. This means that every element in set A is also in set B.

We need to show that $A \cap B = A$. To do this, we will show that $A \cap B \subseteq A$ and $A \subseteq A \cap B$.

1a: Show $A \cap B \subseteq A$.

Let $x \in A \cap B$. By the definition of intersection, this means $x \in A$ and $x \in B$.

Since $x \in A$, it follows directly that $A \cap B \subseteq A$.

1b: Show $A \subseteq A \cap B$.

Let $y \in A$. Since we assumed $A \subseteq B$, every element in A is also in B. Therefore, $y \in B$.

Since $y \in A$ and $y \in B$, by the definition of intersection, $y \in A \cap B$.

Therefore, $A \subseteq A \cap B$.

Since $A \cap B \subseteq A$ and $A \subseteq A \cap B$, we can conclude that $A \cap B = A$.

Part 2: Prove that if $A \cap B = A$, then $A \subseteq B$.

Assume that $A \cap B = A$. This means that the intersection of A and B results in set A itself.

We need to show that $A \subseteq B$. This means we need to show that every element in A is also in B.

Let $z \in A$. Since we assumed $A \cap B = A$, it means that any element in A must also be in the intersection of A and B.

So, if $z \in A$, then $z \in A \cap B$.

By the definition of intersection, if $z \in A \cap B$, then $z \in A$ and $z \in B$.

From this, we can see that $z \in B$.

Since every element $z$ in A is also in B, by the definition of subset, $A \subseteq B$.

Conclusion:

We have proven both implications:

• If $A \subseteq B$, then $A \cap B = A$.
• If $A \cap B = A$, then $A \subseteq B$.

Therefore, $A \subseteq B$ if and only if $A \cap B = A$.

Let:

$N(X)$ be the number of people who liked Product X.

$N(Y)$ be the number of people who liked Product Y.

$N(Z)$ be the number of people who liked Product Z.

$N(X \cap Y)$ be the number of people who liked X and Y.

$N(X \cap Z)$ be the number of people who liked X and Z.

$N(Y \cap Z)$ be the number of people who liked Y and Z.

$N(X \cap Y \cap Z)$ be the number of people who liked all three products.

$N(U)$ be the total number of people surveyed.

Given:

$N(U) = 500$.

$N(X) = 280$.

$N(Y) = 230$.

$N(Z) = 150$.

$N(X \cap Y) = 100$.

$N(X \cap Z) = 60$.

$N(Y \cap Z) = 50$.

$N(X \cap Y \cap Z) = 20$.

We will use a Venn diagram approach to calculate the number of people in each specific region.

1. Number of people who liked all three products:

$N(X \cap Y \cap Z) = 20$.

2. Number of people who liked exactly two products:

Number who liked X and Y only = $N(X \cap Y) - N(X \cap Y \cap Z) = 100 - 20 = 80$.

Number who liked X and Z only = $N(X \cap Z) - N(X \cap Y \cap Z) = 60 - 20 = 40$.

Number who liked Y and Z only = $N(Y \cap Z) - N(X \cap Y \cap Z) = 50 - 20 = 30$.

3. Number of people who liked exactly one product:

Number who liked only X = $N(X) - (\text{X and Y only}) - (\text{X and Z only}) - (\text{X, Y, and Z})$

Number who liked only X = $N(X) - (N(X \cap Y) - N(X \cap Y \cap Z)) - (N(X \cap Z) - N(X \cap Y \cap Z)) - N(X \cap Y \cap Z)$

Number who liked only X = $280 - 80 - 40 - 20 = 140$.

Number who liked only Y = $N(Y) - (\text{X and Y only}) - (\text{Y and Z only}) - (\text{X, Y, and Z})$

Number who liked only Y = $230 - 80 - 30 - 20 = 100$.

Number who liked only Z = $N(Z) - (\text{X and Z only}) - (\text{Y and Z only}) - (\text{X, Y, and Z})$

Number who liked only Z = $150 - 40 - 30 - 20 = 60$.

Now, let's answer the specific questions:

i) Only Product X?

The number of people who liked only Product X is 140.

Answer: 140 people.

ii) Exactly one product?

This is the sum of those who liked only X, only Y, and only Z.

Exactly one product = (Only X) + (Only Y) + (Only Z)

Exactly one product = $140 + 100 + 60 = 300$.

Answer: 300 people.

iii) At least one product?

This is the total number of people who liked X or Y or Z, found using the Principle of Inclusion-Exclusion:

$N(X \cup Y \cup Z) = N(X) + N(Y) + N(Z) - N(X \cap Y) - N(X \cap Z) - N(Y \cap Z) + N(X \cap Y \cap Z)$

$N(X \cup Y \cup Z) = 280 + 230 + 150 - 100 - 60 - 50 + 20$

$N(X \cup Y \cup Z) = 660 - 210 + 20$

$N(X \cup Y \cup Z) = 450 + 20 = 470$.

Alternatively, sum of all regions within the circles:

At least one product = (Only X) + (Only Y) + (Only Z) + (X and Y only) + (X and Z only) + (Y and Z only) + (All three)

= $140 + 100 + 60 + 80 + 40 + 30 + 20 = 470$.

Answer: 470 people.

iv) None of the products?

This is the total number of people surveyed minus the number of people who liked at least one product.

None of the products = $N(U) - N(X \cup Y \cup Z)$

None of the products = $500 - 470 = 30$.

Answer: 30 people.

To Prove: $A \setminus (B \cap C) = (A \setminus B) \cup (A \setminus C)$

We will prove this using an element-wise proof (algebraic proof).

Part 1: Show $A \setminus (B \cap C) \subseteq (A \setminus B) \cup (A \setminus C)$

Let $x \in A \setminus (B \cap C)$.

By the definition of set difference, this means $x \in A$ and $x \notin (B \cap C)$.

Since $x \notin (B \cap C)$, by the definition of intersection, it is not true that ($x \in B$ and $x \in C$).

Using De Morgan's law for logic, $\neg (P \land Q) \equiv \neg P \lor \neg Q$.

So, $x \notin (B \cap C)$ means $x \notin B$ or $x \notin C$.

Now we have: $x \in A$ and ($x \notin B$ or $x \notin C$).

We can distribute the $x \in A$ condition using the distributive law of logic: $(P \land (Q \lor R)) \equiv (P \land Q) \lor (P \land R)$.

So, ($x \in A$ and $x \notin B$) or ($x \in A$ and $x \notin C$).

By the definition of set difference:

($x \in A$ and $x \notin B$) means $x \in A \setminus B$.

($x \in A$ and $x \notin C$) means $x \in A \setminus C$.

So, we have $x \in A \setminus B$ or $x \in A \setminus C$. By the definition of union, $x \in (A \setminus B) \cup (A \setminus C)$.

Therefore, $A \setminus (B \cap C) \subseteq (A \setminus B) \cup (A \setminus C)$.

Part 2: Show $(A \setminus B) \cup (A \setminus C) \subseteq A \setminus (B \cap C)$

Let $x \in (A \setminus B) \cup (A \setminus C)$.

By the definition of union, this means $x \in (A \setminus B)$ or $x \in (A \setminus C)$.

Case 1: $x \in A \setminus B$. This means $x \in A$ and $x \notin B$.

Case 2: $x \in A \setminus C$. This means $x \in A$ and $x \notin C$.

If $x \in A \setminus B$, then $x \in A$. Since $x \notin B$, it cannot be in $B \cap C$. Therefore, $x \notin (B \cap C)$. Since $x \in A$ and $x \notin (B \cap C)$, $x \in A \setminus (B \cap C)$.

If $x \in A \setminus C$, then $x \in A$. Since $x \notin C$, it cannot be in $B \cap C$. Therefore, $x \notin (B \cap C)$. Since $x \in A$ and $x \notin (B \cap C)$, $x \in A \setminus (B \cap C)$.

More formally, if $x \in (A \setminus B) \cup (A \setminus C)$, then:

$x \in A \setminus B$ OR $x \in A \setminus C$.

This translates to: ($x \in A$ AND $x \notin B$) OR ($x \in A$ AND $x \notin C$).

Using the distributive law of logic: $(P \land Q) \lor (P \land R) \equiv P \land (Q \lor R)$.

Let $P$ be $x \in A$, $Q$ be $x \notin B$, and $R$ be $x \notin C$.

The statement becomes: $x \in A$ AND ($x \notin B$ OR $x \notin C$).

By De Morgan's law for logic, $\neg P \lor \neg Q \equiv \neg (P \land Q)$.

So, ($x \notin B$ OR $x \notin C$) is equivalent to NOT ($x \in B$ AND $x \in C$).

This means NOT ($x \in B \cap C$).

Therefore, the statement is: $x \in A$ AND $x \notin (B \cap C)$.

By the definition of set difference, this means $x \in A \setminus (B \cap C)$.

Thus, $(A \setminus B) \cup (A \setminus C) \subseteq A \setminus (B \cap C)$.

Conclusion:

Since we have shown that $A \setminus (B \cap C) \subseteq (A \setminus B) \cup (A \setminus C)$ and $(A \setminus B) \cup (A \setminus C) \subseteq A \setminus (B \cap C)$, we can conclude that $A \setminus (B \cap C) = (A \setminus B) \cup (A \setminus C)$.

To Prove: $|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|$

This is the Principle of Inclusion-Exclusion for three sets. We will prove this by considering the number of times each element in $A \cup B \cup C$ is counted on the right-hand side.

Let $x$ be an element in $A \cup B \cup C$. We consider the cases based on how many of the sets A, B, and C contain $x$.

Case 1: $x$ belongs to exactly one set (e.g., only in A).

If $x \in A$ only, then:

• It is counted once in $|A|$.
• It is not counted in $|B|$, $|C|$, $|A \cap B|$, $|A \cap C|$, $|B \cap C|$, or $|A \cap B \cap C|$.

So, $x$ is counted $1 - 0 - 0 - 0 - 0 - 0 + 0 = 1$ time on the right-hand side.

Case 2: $x$ belongs to exactly two sets (e.g., in A and B, but not C).

If $x \in A$ and $x \in B$ only, then:

• It is counted once in $|A|$.
• It is counted once in $|B|$.
• It is not counted in $|C|$.
• It is counted once in $|A \cap B|$.
• It is not counted in $|A \cap C|$.
• It is not counted in $|B \cap C|$.
• It is not counted in $|A \cap B \cap C|$.

So, $x$ is counted $1 + 1 - 0 - 1 - 0 - 0 + 0 = 1$ time on the right-hand side.

Case 3: $x$ belongs to all three sets (in A, B, and C).

If $x \in A$, $x \in B$, and $x \in C$, then:

• It is counted once in $|A|$.
• It is counted once in $|B|$.
• It is counted once in $|C|$.
• It is counted once in $|A \cap B|$.
• It is counted once in $|A \cap C|$.
• It is counted once in $|B \cap C|$.
• It is counted once in $|A \cap B \cap C|$.

So, $x$ is counted $1 + 1 + 1 - 1 - 1 - 1 + 1 = 1$ time on the right-hand side.

In all possible cases, every element in $A \cup B \cup C$ is counted exactly once on the right-hand side of the equation.

Therefore, the formula correctly represents the number of elements in the union of three sets.

Conclusion:

The Principle of Inclusion-Exclusion for three sets is proven.

Given:

$U$ = Set of all students in a college.

$A$ = Set of students who got admission in Science stream.

$B$ = Set of students who got admission in Commerce stream.

$C$ = Set of students who got admission in Humanities stream.

Description of the sets in words:

i) $A \cap B'$

$B'$ represents the complement of set B, which are the students who did NOT get admission in the Commerce stream.

$A \cap B'$ represents the intersection of students in the Science stream and students not in the Commerce stream.

Description: The set of students who got admission in the Science stream but not in the Commerce stream.

ii) $A \cup B \cup C$

$A \cup B \cup C$ represents the union of the three sets, meaning students who are in set A, or set B, or set C, or any combination of these.

Description: The set of students who got admission in the Science stream, or the Commerce stream, or the Humanities stream, or any combination of these streams. In other words, it's the set of students who got admission in at least one of these three streams.

iii) $(A \cup B)' \cap C'$

First, let's understand $(A \cup B)'$. This represents the complement of the union of A and B. $(A \cup B)$ is the set of students who got admission in Science OR Commerce (or both). So, $(A \cup B)'$ is the set of students who did NOT get admission in Science OR Commerce. This means students who are in neither Science nor Commerce.

$C'$ represents the set of students who did NOT get admission in the Humanities stream.

The intersection of these two sets, $(A \cup B)' \cap C'$, represents students who are in neither Science nor Commerce, AND also not in Humanities.

Description: The set of students who got admission in neither the Science stream nor the Commerce stream, and also not in the Humanities stream. This means students who did not get admission in any of the three streams (Science, Commerce, or Humanities).

iv) $(A \cap B) \setminus C$

$A \cap B$ represents the set of students who got admission in both the Science stream and the Commerce stream.

$C$ represents the set of students who got admission in the Humanities stream.

$(A \cap B) \setminus C$ represents the set difference, meaning students who are in $(A \cap B)$ but not in $C$.

Description: The set of students who got admission in both Science and Commerce streams, but did not get admission in the Humanities stream.