Maharashtra Board Class 9 Maths Chapter 1 Set 1.4 Algebra Standard Part 1 Sets Solutions

Question 1. If \( n(A) = 15 \), \( n(A \cup B) = 29 \), \( n(A \cap B) = 7 \), then \( n(B) = ? \)
Answer: Given:
\( n(A) = 15 \)
\( n(A \cup B) = 29 \)
\( n(A \cap B) = 7 \)
We use the standard formula for the cardinality of the union of two sets:
\( n(A \cup B) = n(A) + n(B) - n(A \cap B) \)
Substituting the given values into the formula:
\( 29 = 15 + n(B) - 7 \)
\( \implies 29 = 8 + n(B) \)
\( \implies n(B) = 29 - 8 \)
\( \implies n(B) = 21 \)
Therefore, the value of \( n(B) \) is \( 21 \). This formula helps us easily find the size of one set when the union and intersection sizes are known.
In simple words: We use a simple formula that links the total items in both groups combined to the items in each individual group minus the items they share.

🎯 Exam Tip: Always write down the general formula first before substituting values to secure step-wise marking even if you make a calculation error.

Question 2. In a hostel there are 125 students, out of which 80 drink tea, 60 drink coffee and 20 drink tea and coffee both. Find the number of students who do not drink tea or coffee.
Answer:
i. Let \( U \) be the set of students in the hostel, \( T \) be the set of students who drink tea and \( C \) be the set of students who drink coffee.
\( n(U) = 125 \), \( n(T) = 80 \), \( n(C) = 60 \),
number of students who drink Tea and Coffee = \( n(T \cap C) = 20 \)

ii. \( n(T \cup C) = n(T) + n(C) - n(T \cap C) \)
\( = 80 + 60 - 20 \)

\( \implies n(T \cup C) = 120 \)

\( \implies 120 \) students drink tea or coffee
Also, there are 125 students in the hostel.

iii. Number of students who do not drink tea or coffee = \( n(U) - n(T \cup C) \)
\( = 125 - 120 \)
\( = 5 \)

\( \implies 5 \) students do not drink tea or coffee. This simple calculation helps us understand the beverage preferences of the hostel residents.

Alternate Method:
Let \( U \) be the set of students in the hostel, \( T \) be the set of students who drink tea and \( C \) be the set of students who drink coffee.
Based on the Venn diagram representation:

• Total students in hostel \( n(U) = 125 \)
• Students who drink only tea: \( 80 - 20 = 60 \)
• Students who drink both tea and coffee: \( 20 \)
• Students who drink only coffee: \( 60 - 20 = 40 \)

From Venn diagram,
Student who drinks tea or coffee = \( n(T \cup C) = 60 + 20 + 40 = 120 \)

\( \implies \) The number of students who do not drink tea or coffee = \( n(U) - n(T \cup C) \)
\( = 125 - 120 = 5 \)

\( \implies 5 \) students do not drink tea or coffee.
In simple words: Out of 125 total students, 120 drink tea, coffee, or both. Subtracting 120 from 125 leaves 5 students who do not drink either beverage.

🎯 Exam Tip: When using the Venn diagram method, clearly show how you subtracted the intersection (20) from the individual sets to find the students who drink only tea (60) or only coffee (40).

Question 3. In a competitive exam 50 students passed in English, 60 students passed in Mathematics and 40 students passed in both the subjects. None of them failed in both the subjects. Find the number of students who passed at least in one of the subjects ?
Answer:
Let \( U \) be the set of students who appeared for the exam,
\( E \) be the set of students who passed in English and
\( M \) be the set of students who passed in Maths.
\( \therefore n(E) = 50, n(M) = 60 \)
40 students passed in both the subjects
\( \dots n(M \cap E) = 40 \)
Since, none of the students failed in both subjects, every student passed in at least one of the two subjects.
\( \therefore \) Total students = \( n(E \cup M) \)
\( = n(E) + n(M) - n(E \cap M) \)
\( = 50 + 60 - 40 \)
\( = 70 \)
\( \dots \) The number of students who passed at least in one of the subjects is 70.

Alternate Method:
Let \( U \) be the set of students who appeared for the exam,
\( E \) be the set of students who passed in English and \( M \) be the set of students who passed in Maths.
Based on the Venn diagram representation:

• Students who passed only in English: \( 50 - 40 = 10 \)
• Students who passed in both English and Maths: \( 40 \)
• Students who passed only in Maths: \( 60 - 40 = 20 \)

Since, none of the students failed in both subjects
\( \therefore \) Total student = \( n(E \cup M) \)
\( = 10 + 40 + 20 \)
\( = 70 \)
\( \therefore \) The number of students who passed at least in one of the subjects is 70.
In simple words: To find the total number of students who passed at least one exam, we add the students who passed English and those who passed Maths, then subtract the students who passed both so we do not count them twice. This gives us 70 students.

🎯 Exam Tip: Always remember the formula \( n(A \cup B) = n(A) + n(B) - n(A \cap B) \) for set union problems, as it is highly effective and prevents double-counting the intersection.

Question 4. A survey was conducted to know the hobby of 220 students of class IX. Out of which 130 students informed about their hobby as ’rock climbing and 180 students informed about their hobby as sky watching. There are 110 students who follow...

Question. ...both the hobbies. Then how many students do not have any of the two hobbies? How many of them follow the hobby of rock climbing only? How many students follow the hobby of sky watching only?
Answer:
i. Let \( U \) be the set of students of class IX,
\( R \) be the set of students who follow the hobby of rock climbing and
\( S \) be the set of students who follow the hobby of sky watching.
\( \therefore n(U) = 220 \), \( n(R) = 130 \), \( n(S) = 180 \),
110 students follow both the hobbies
\( \therefore n(R \cap S) = 110 \)

ii. \( n(R \cup S) = n(R) + n(S) - n(R \cap S) \)
\( = 130 + 180 - 110 \)
\( \therefore n(R \cup S) = 200 \)
\( \therefore \) 200 students follow the hobby of rock climbing or sky watching.

iii. Total number of students = 220.
Number of students who do not follow the hobby of rock climbing or sky watching
\( = n(U) - n(R \cup S) \)
\( = 220 - 200 \)
\( = 20 \)

iv. Number of students who follow the hobby of rock climbing only
\( = n(R) - n(R \cap S) \)
\( = 130 - 110 \)
\( = 20 \)

v. Number of students who follow the hobby of sky watching only
\( = n(S) - n(R \cap S) \)
\( = 180 - 110 \)
\( = 70 \)

Alternate Method:
Let \( U \) be the set of students of class IX,
\( R \) be the set of students who follow the hobby of rock climbing and \( S \) be the set of students who follow the hobby of sky watching. We can also represent this using a Venn diagram to find the disjoint regions directly.
In simple words: Out of 220 students, 200 students do at least one hobby, leaving 20 students with no hobbies. There are 20 students who only do rock climbing and 70 students who only do sky watching.

🎯 Exam Tip: Always define your sets clearly at the beginning of the solution. Using Venn diagrams can help you double-check your calculations quickly.

Question 4. A survey was conducted to know the hobby of 220 students of class IX. Out of which 130 students informed about their hobby as rock climbing and 180 students informed about their hobby as sky watching. There are 110 students who follow both the hobbies. Then how many students do not have any of the two hobbies? How many of them follow the hobby of rock climbing only? How many students follow the hobby of sky watching only?
Answer:
Let \( U \) be the universal set of students surveyed. \( n(U) = 220 \)
Let \( R \) be the set of students who follow the hobby of rock climbing. \( n(R) = 130 \)
Let \( S \) be the set of students who follow the hobby of sky watching. \( n(S) = 180 \)
The number of students who follow both hobbies is \( n(R \cap S) = 110 \).

Venn Diagram Representation:

• Universal Set \( U \): 220
• Set \( R \) (Rock climbing): 130
• Set \( S \) (Sky watching): 180
• Intersection \( R \cap S \) (Both hobbies): 110
• Rock climbing only: 20
• Sky watching only: 70
• Neither hobby: 20

From the Venn diagram:

(i) Students who follow the hobby of rock climbing or sky watching:
\( = n(R \cup S) \)
\( = 20 + 110 + 70 \)
\( = 200 \)

(ii) Number of students who do not follow the hobby of rock climbing or sky watching:
\( = n(U) - n(R \cup S) \)
\( = 220 - 200 \)
\( = 20 \)

(iii) Number of students who follow the hobby of rock climbing only:
\( = n(R) - n(R \cap S) \)
\( = 130 - 110 \)
\( = 20 \)

(iv) Number of students who follow the hobby of sky watching only:
\( = n(S) - n(R \cap S) \)
\( = 180 - 110 \)
\( = 70 \)
In simple words: Out of 220 students, 200 students have at least one hobby (either rock climbing, sky watching, or both), which leaves 20 students with neither hobby. There are 20 students who only do rock climbing and 70 who only do sky watching.

🎯 Exam Tip: Always define your sets clearly and use a Venn diagram to cross-verify your subtraction steps to avoid simple calculation errors.

Question 5. Observe the given Venn diagram and write the following sets:
(i) A
(ii) B
(iii) A \(\cup\) B
(iv) U
(v) A'
(vi) B'
(vii) (A \(\cup\) B)'
Answer:
From the given Venn diagram, we can list the elements of each set as follows:
(i) A = {x, y, z, m, n}
(ii) B = {p, q, r, m, n}
(iii) A \(\cup\) B = {x, y, z, m, n, p, q, r}
(iv) U = {x, y, z, m, n, p, q, r, s, t}
(v) A' = {p, q, r, s, t}
(vi) B' = {x, y, z, s, t}
(vii) (A \(\cup\) B)' = {s, t}
In simple words: Set A and B contain their respective letters, with 'm' and 'n' shared between them. The complement sets like A' contain everything in the universal box U that is not inside circle A.

🎯 Exam Tip: When writing sets, always enclose the elements in curly brackets { } and separate them with commas. Do not forget to include the elements in the intersection when listing individual sets.

Question. Find the sets from the given Venn diagram:
(ii) B
(iii) A ∪ B
(iv) U
(v) A’
(vi) B’
(vii) (A ∪ B)’
Answer:
(i) \( A = \{x, y, z, m, n\} \)
(ii) \( B = \{p, q, r, m, n\} \)
(iii) \( A \cup B = \{x, y, z, m, n, p, q, r\} \)
(iv) \( U = \{x, y, z, m, n, p, q, r, s, t\} \)
(v) \( A' = \{p, q, r, s, t\} \)
(vi) \( B' = \{x, y, z, s, t\} \)
(vii) \( (A \cup B)' = \{s, t\} \)
In simple words: We list the elements belonging to each set by looking at their regions: A and B are the circles, U is the entire box, and the prime (') symbol means everything outside that set.

🎯 Exam Tip: Remember that the universal set \( U \) contains all elements inside the rectangle, including those outside the circles, while the complement of a set contains all elements in \( U \) that are not in that set.

Question 1. Take different examples of sets and verify the above mentioned properties.
Answer:
(i) Let \( A = \{3, 5\} \) and \( B = \{3, 5, 8, 9, 10\} \).
Then, \( A \cap B = \{3, 5\} \) and \( B \cap A = \{3, 5\} \).
Thus, \( A \cap B = B \cap A = \{3, 5\} \). This verifies the commutative property of intersection.

(ii) Let \( A = \{3, 5\} \) and \( B = \{3, 5, 8, 9, 10\} \).
Since all elements of set \( A \) are present in set \( B \), we have \( A \subseteq B \).
Also, \( A \cap B = \{3, 5\} = A \).
Thus, if \( A \subseteq B \), then \( A \cap B = A \).

(iii) Let \( A = \{2, 3, 8, 10\} \) and \( B = \{3, 8\} \).
Here, \( A \cap B = \{3, 8\} = B \).
Since all elements of set \( B \) are present in set \( A \), we have \( B \subseteq A \).
Thus, if \( A \cap B = B \), then \( B \subseteq A \).

(iv) Let \( A = \{2, 3, 8, 10\} \), \( B = \{3, 8\} \), so \( A \cap B = \{3, 8\} \).
Since all elements of set \( A \cap B \) are present in both set \( A \) and set \( B \), we have:
\( A \cap B \subseteq A \) and \( A \cap B \subseteq B \).
In simple words: We can verify set properties by choosing simple numbers for our sets. For example, the intersection of two sets is the same no matter which one we write first, and if one set is completely inside another, their shared elements are just the smaller set itself.

🎯 Exam Tip: When verifying set properties, always choose small, simple sets with clear elements to make calculations easy and error-free.

Question 1. Verify the following set properties:
(v) Let \( U = \{3, 4, 6, 8\} \), \( A = \{6, 4\} \)
(vi) \( A \cap \phi \)
(vii) Let \( A = \{6, 4\} \)
Answer:
(v) \( A' = \{3, 8\} \)
\( \implies A \cap A' = \{\} = \phi \)
(vi) \( A \cap \phi = \{\} = \phi \)
(vii) \( A \cap A = \{6, 4\} \)
\( \implies A \cap A = A \)
In simple words: These examples show that the intersection of a set with its complement or with an empty set is always empty, while intersecting a set with itself results in the same set.

🎯 Exam Tip: Remember that the complement of set A contains all elements of the universal set U that are not in A, and their intersection is always empty.

Question 2. Observe the set A, B, C given by Venn diagrams and write which of these are disjoint sets. (Textbook pg. no. 12)
Answer:
Based on the Venn diagram, the elements of the sets are:
• Set A = \( \{1, 2, 3, 4, 5, 6, 7\} \)
• Set B = \( \{3, 6, 8, 9, 10, 11, 12\} \)
• Set C = \( \{10, 11, 12\} \)
Now, comparing the sets, we find:
\( A \cap C = \phi \)
\( \implies A \) and \( C \) are disjoint sets.
In simple words: Disjoint sets are sets that do not share any common elements. Since Set A and Set C have no numbers in common, they are disjoint sets.

🎯 Exam Tip: To identify disjoint sets, look for sets that have no overlapping regions in the Venn diagram. If they do not touch or overlap, their intersection is empty.

Question 3. Let the set of English alphabets be the Universal set. The letters of the word ‘LAUGH’ is one set and the letter of the word ‘CRY’ is another set. Can we say that these are two disjoint sets? Observe that intersection of these two sets is empty. (Textbook pg. no. 13)
Answer:
Let the two sets be represented as:
• Set A = \( \{L, A, U, G, H\} \)
• Set B = \( \{C, R, Y\} \)
Now, comparing the elements of both sets, we find:
\( A \cap B = \phi \)
\( \implies \) Yes, we can say that these are two disjoint sets because their intersection is empty and they share no common letters.
In simple words: Since the words 'LAUGH' and 'CRY' do not share any of the same letters, their sets have nothing in common, making them disjoint sets.

🎯 Exam Tip: Always list the elements of both sets clearly before finding their intersection to prove they are disjoint.

Question 4. Fill in the blanks with elements of that set.
\( U = \{1, 3, 5, 8, 9, 10, 11, 12, 13, 15\} \)
\( A = \{1, 11, 13\} \)
\( B = \{8, 5, 10, 11, 15\} \)
\( A' = \{ \} \)
\( B' = \{ \} \)
\( A \cap B = \{ \} \)
\( A' \cap B' = \{ \} \)
\( A \cup B = \{ \} \)
\( A' \cup B' = \{ \} \)
\( (A \cap B)' = \{ \} \)
\( (A \cup B)' = \{ \} \)
Verify: \( (A \cap B)' = A' \cup B' \), \( (A \cup B)' = A' \cap B' \) (Textbook pg. no, 18)
Answer:
\( U = \{1, 3, 5, 8, 9, 10, 11, 12, 13, 15\} \)
\( A = \{1, 11, 13\} \)
\( B = \{8, 5, 10, 11, 15\} \)
\( A' = \{3, 5, 8, 9, 10, 12, 15\} \)
\( B' = \{1, 3, 9, 12, 13\} \)
\( A \cap B = \{11\} \)
\( A' \cap B' = \{3, 9, 12\} \) …(i)
\( A \cup B = \{1, 5, 8, 10, 11, 13, 15\} \)
\( A' \cup B' = \{1, 3, 5, 8, 9, 10, 12, 13, 15\} \) …(ii)
\( (A \cap B)' = \{1, 3, 5, 8, 9, 10, 12, 13, 15\} \) …(iii)
\( (A \cup B)' = \{3, 9, 12\} \) …(iv)
From (ii) and (iii), we get:
\( (A \cap B)' = A' \cup B' \)
From (i) and (iv), we get:
\( (A \cup B)' = A' \cap B' \)
This successfully verifies De Morgan's Laws for the given sets.
In simple words: To find the complement of a set, we list all elements of the universal set \( U \) that are not in that set. De Morgan's Laws show that the complement of an intersection is the union of the complements, and vice versa.

🎯 Exam Tip: Label your equations clearly as (i), (ii), (iii), and (iv) so the examiner can easily follow your verification steps for De Morgan's Laws.

Question 5. \( A = \{1, 2, 3, 5, 7, 9, 11, 13\} \)
\( B = \{1, 2, 4, 6, 8, 12, 13\} \)
Verify the above rule for the given set A and set B. (Textbook pg. no. 14)
Answer:
The rule to be verified is: \( n(A \cup B) = n(A) + n(B) - n(A \cap B) \)
Given:
\( A = \{1, 2, 3, 5, 7, 9, 11, 13\} \)
\( \implies n(A) = 8 \)
\( B = \{1, 2, 4, 6, 8, 12, 13\} \)
\( \implies n(B) = 7 \)
Now, let us find \( A \cap B \) (elements common to both A and B):
\( A \cap B = \{1, 2, 13\} \)
\( \implies n(A \cap B) = 3 \)
Next, let us find \( A \cup B \) (all elements from both A and B without repetition):
\( A \cup B = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13\} \)
\( \implies n(A \cup B) = 12 \)
Now, let us calculate the right-hand side of the formula:
\( n(A) + n(B) - n(A \cap B) = 8 + 7 - 3 = 12 \)
Since both sides equal 12, we have verified that \( n(A \cup B) = n(A) + n(B) - n(A \cap B) \).
In simple words: This formula helps us count the total unique elements in two groups combined by adding the sizes of both groups and subtracting the elements that were counted twice.

🎯 Exam Tip: Always list the elements of \( A \cap B \) and \( A \cup B \) clearly before writing down their cardinal numbers \( n(A \cap B) \) and \( n(A \cup B) \) to avoid calculation errors.

Question 6. Verify the above rule for the given Venn diagram. (Textbook pg. no. 14)
Answer:
Venn Diagram Elements:

• Elements in Set A: 3, 9, 12, 15, 24
• Elements in Set B: 6, 12, 18, 24, 30, 36
• Common elements in both A and B (Intersection): 12, 24

From the Venn diagram, we can determine the number of elements in each set:
\( n(A) = 5 \), \( n(B) = 6 \)
\( n(A \cup B) = 9 \), \( n(A \cap B) = 2 \)
Now, \( n(A \cup B) = 9 \) ...(i)
\( n(A) + n(B) - n(A \cap B) = 5 + 6 - 2 = 9 \) ...(ii)
This verification confirms that the cardinality of the union of two sets is always equal to the sum of their individual cardinalities minus the cardinality of their intersection.
\( \implies n(A \cup B) = n(A) + n(B) - n(A \cap B) \) ...[From (i) and (ii)]
In simple words: We count the total unique items in both groups combined, which is 9. This is the same as adding the count of group A (5) and group B (6) and then subtracting the 2 items they share so we don't count them twice.

🎯 Exam Tip: Always list the elements of each set clearly from the Venn diagram before writing their cardinalities to avoid simple counting mistakes.