OP Malhotra Class 11 Maths Solutions Chapter 1 Sets Exercise 1 (D)

Question 1. If n (\( \xi \)) = 80, n (A) = 48, n (B) = 40 and n (A \(\cap\) B) = 25, draw a Venn diagram to find :
(i) n (A \(\cup\) B)
(ii) n (A \(\cup\) B)'
(iii) n (A - B)
(iv) n (B – A)
(v) n (A \(\cap\) B')
(vi) n (A' \(\cap\) B)
Answer:
Given:
n (\( \xi \)) = 80
n (A) = 48
n (B) = 40
n (A \(\cap\) B) = 25

(i) We use the formula for the union of two sets:
n(A \(\cup\) B) = n (A) + n (B) – n (A \(\cap\) B)
\( = 48 + 40 – 25 \)
\( = 88 – 25 \)
\( = 63 \)

(ii) The complement of a set's union is found by subtracting it from the universal set:
n(A \(\cup\) B)' = n(\( \xi \)) – n(A \(\cup\) B)
\( = 80 - 63 \)
\( = 17 \)

(iii) To find elements only in A, we subtract the intersection from A:
n(A – B) = n (A) – n(A \(\cap\) B)
\( = 48 - 25 \)
\( = 23 \)

(iv) Similarly, for elements only in B, we subtract the intersection from B:
n(B – A) = n (B) – n(A \(\cap\) B)
\( = 40 - 25 \)
\( = 15 \)

(v) The number of elements in A but not in B is the same as n(A – B):
n (A \(\cap\) B') = n (A – B)
\( = 23 \)

(vi) The number of elements in B but not in A is the same as n(B – A):
n(A' \(\cap\) B) = n (B – A)
\( = 15 \)

Venn diagram illustration:
The universal set is \( \xi \). The two circles represent sets A and B. The numbers inside show how many elements are in each part of the sets.
The numbers are placed in their respective regions: 23 in A only, 25 in the intersection (A \(\cap\) B), 15 in B only, and 17 outside both A and B.

In simple words: We calculated how many items are in set A or B, outside both, only in A, or only in B, using the given total numbers. The Venn diagram helps to see these parts visually.

🎯 Exam Tip: Always draw the Venn diagram first and fill in the intersection (A \(\cap\) B) value, then the 'only A' and 'only B' parts, and finally the 'neither' part. This helps visualize and verify your calculations.

Question 2. If \( \xi \) = {x | x \( \in \) N, x < 10}, A = {x | x is a prime number, x < 10}, B = {x | x is an even number, x < 10} draw a Venn diagram to find :
(i) n (A \(\cup\) B)
(ii) n (A \(\cap\) B)
(iii) n (A \(\cup\) B)'
(iv) n (A \(\cap\) B')
(v) n (A' \(\cap\) B)
Answer:
First, let's list the elements of each set:
\( \xi \) = {x | x \( \in \) N, x < 10} = {1, 2, 3, 4, 5, 6, 7, 8, 9}
A = {x | x is a prime number, x < 10} = {2, 3, 5, 7}
B = {x | x is an even number, x < 10} = {2, 4, 6, 8}

Now, let's find the number of elements in each set:
n(\( \xi \)) = 9
n(A) = 4
n(B) = 4

Next, find the intersection of A and B:
A \(\cap\) B = {2, 3, 5, 7} \(\cap\) {2, 4, 6, 8} = {2}
So, n(A \(\cap\) B) = 1

Now we can find the required values:
(i) n(A \(\cup\) B) = n(A) + n(B) – n(A \(\cap\) B)
\( = 4 + 4 - 1 \)
\( = 7 \)

(ii) n (A \(\cap\) B) = 1 (as calculated above)

(iii) n (A \(\cup\) B)' = n (\( \xi \)) – n(A \(\cup\) B)
\( = 9 - 7 \)
\( = 2 \)

(iv) n(A \(\cap\) B') = n(A) – n(A \(\cap\) B)
\( = 4 - 1 \)
\( = 3 \)

(v) n (A' \(\cap\) B) = n (B) – n(A \(\cap\) B)
\( = 4 - 1 \)
\( = 3 \)

Venn diagram illustration:
The universal set \( \xi \) contains numbers from 1 to 9. Set A has prime numbers, and Set B has even numbers less than 10. The number 2 is in both sets. The diagram shows all these elements in their correct regions.

In simple words: We first listed all numbers in each set up to 9. Then we counted how many numbers were in each set, in their shared part, and in their combined total. We also showed these numbers in a Venn diagram.

🎯 Exam Tip: When defining sets from rules like 'prime number less than 10', always list the elements explicitly first. This helps avoid errors when calculating intersections and unions.

Question 3. Given n (\( \xi \)) = 40, n (A') = 12, n (B) = 15 and B \( \subset \) A. Draw a Venn diagram to illustrate this information. Hence find n (A – B).
Answer:
Given:
n (\( \xi \)) = 40
n (A') = 12 (number of elements not in A)
n (B) = 15
B \( \subset \) A (B is a subset of A, meaning all elements of B are also in A)

First, let's find the number of elements in A:
n(A) = n(\( \xi \)) - n(A')
\( = 40 - 12 \)
\( = 28 \)

Since B \( \subset \) A, all elements of B are inside A. This means the intersection of A and B is simply B itself.
So, n(A \(\cap\) B) = n(B) = 15.

Now, let's find n(A - B), which represents elements in A but not in B:
n(A - B) = n(A) - n(A \(\cap\) B)
\( = 28 - 15 \)
\( = 13 \)

Venn diagram illustration:
The diagram shows a universal set \( \xi \). Set B is completely inside Set A. The number 13 represents elements only in A (A - B). The number 15 represents elements in B (which is also A \(\cap\) B). The number 12 represents elements outside A (A').

In simple words: Since all of B is part of A, the elements only in A are found by taking A's total and removing B's total. We drew a Venn diagram where the circle for B is completely inside the circle for A, showing this relationship clearly.

🎯 Exam Tip: When one set is a subset of another (B \( \subset \) A), always remember that n(A \(\cap\) B) = n(B). This simplifies calculations significantly and is crucial for drawing the Venn diagram correctly.

Question 4. The Venn diagram shows :
\( \xi \) = {pupils in class 8}
A = {pupils who play cricket}
B = {pupils who play basketball}
How many pupils:
(i) are in class 8 ?
(ii) play cricket
(iii) play basketball ?
(iv) play both cricket and basketball
(v) play neither cricket nor basketball ?
Answer:
Let's analyze the given Venn diagram first:

From the diagram, we have:
Pupils who play only cricket = 15
Pupils who play both cricket and basketball = 8
Pupils who play only basketball = 9
Pupils who play neither game = 13

Now, let's answer the questions:
(i) The number of pupils in class 8 is the total of all pupils in the universal set:
n(\( \xi \)) = 15 + 8 + 9 + 13
\( = 45 \)

(ii) The number of pupils who play cricket (Set A) includes those who play only cricket and those who play both:
n(A) = 15 + 8
\( = 23 \)

(iii) The number of pupils who play basketball (Set B) includes those who play only basketball and those who play both:
n(B) = 8 + 9
\( = 17 \)

(iv) The number of pupils who play both cricket and basketball is found in the intersection of Set A and Set B:
n(A \(\cap\) B) = 8

(v) The number of pupils who play neither cricket nor basketball are those outside both circles:
n(A' \(\cap\) B') = n(A \(\cup\) B)'
\( = \) n(\( \xi \)) – n(A \(\cup\) B)
First, find n(A \(\cup\) B) = 15 + 8 + 9 = 32
\( = 45 – 32 \)
\( = 13 \)
This also matches the number directly shown outside the circles in the diagram.
In simple words: We used the numbers shown in the Venn diagram to count students in different groups. We added up all numbers to find the total class size. Then, we counted those who play only one sport, both, or neither.

🎯 Exam Tip: Always identify the values for each distinct region in the Venn diagram (A only, B only, A \(\cap\) B, and neither A nor B) before answering questions. This makes all calculations straightforward.

Question 5. In the Venn diagram
\( \xi \) = {people at a function}
A = {those who asked for tea}
B = {those who watched the ballet}
Write down the number who :
(i) asked for tea
(ii) asked for tea and watched the ballet
(iii) neither asked for tea nor watched the ballet
(iv) attended the function
Answer:
Let's refer to the given Venn diagram:

From the diagram, we have:
People who asked for tea only = 59
People who asked for tea AND watched ballet = 23
People who watched ballet only = 47
People who neither asked for tea NOR watched ballet = 25

Now, let's answer the questions:
(i) The number of people who asked for tea (Set A) includes those who asked only for tea and those who asked for tea and watched the ballet:
n(A) = 59 + 23
\( = 82 \)

(ii) The number of people who asked for tea and watched the ballet is the intersection of Set A and Set B:
n (A \(\cap\) B) = 23

(iii) The number of people who neither asked for tea nor watched the ballet are those outside both circles:
n(A' \(\cap\) B') = n(A \(\cup\) B)'
\( = 25 \)

(iv) The number of people who attended the function (total people in \( \xi \)) is the sum of all parts in the diagram:
n(\( \xi \)) = 59 + 23 + 47 + 25
\( = 154 \)
In simple words: We used the numbers from the Venn diagram. We added the numbers in the "tea only" and "tea and ballet" parts for the total tea-drinkers. We directly found the number for "both" and "neither". The total number of people was found by adding all the numbers in the diagram.

🎯 Exam Tip: Always sum up all the distinct regions in a Venn diagram to find the total number of elements in the universal set if it's not given directly. Pay close attention to keywords like "only", "both", and "neither".

Question 6. In a group of 30 people, 18 play squash and 19 play tennis. How many play both games, provided everyone plays at least one game ?
Answer:
Let S be the set of people who play squash.
Let T be the set of people who play tennis.
We are given:
Total number of people in the group, n(S \(\cup\) T) = 30 (since everyone plays at least one game)
Number of people who play squash, n(S) = 18
Number of people who play tennis, n(T) = 19

We need to find the number of people who play both games, n(S \(\cap\) T).
Using the formula for the union of two sets:
n(S \(\cup\) T) = n(S) + n(T) – n(S \(\cap\) T)
Now, we can rearrange the formula to find the intersection:
n(S \(\cap\) T) = n(S) + n(T) – n(S \(\cup\) T)
\( = 18 + 19 - 30 \)
\( = 37 - 30 \)
\( = 7 \)
So, 7 people play both squash and tennis. This means there is an overlap in the groups, as expected.
In simple words: We know how many people play squash, how many play tennis, and the total number of people. We subtracted the total from the sum of squash and tennis players to find how many play both, because those who play both were counted twice.

🎯 Exam Tip: When given the total number of elements in the union and the sizes of two sets, always use the formula n(A \(\cup\) B) = n(A) + n(B) – n(A \(\cap\) B) to find the intersection. Remember that "everyone plays at least one game" means the total group size is the union of the sets.

Question 7. In a class of 50 students, 22 like History, 25 like Geography and 10 like both subjects. Draw a Venn diagram and find the number of students who
(i) do not like History
(ii) do not like Geography
(iii) like neither History nor Geography
Answer:
Let H be the set of students who like History.
Let G be the set of students who like Geography.
Let \( \xi \) be the set of all students in the class.
We are given:
n(\( \xi \)) = 50
n(H) = 22
n(G) = 25
n(H \(\cap\) G) = 10 (students who like both History and Geography)

First, let's find the number of students who like only History and only Geography:
Only History = n(H) - n(H \(\cap\) G) = 22 - 10 = 12
Only Geography = n(G) - n(H \(\cap\) G) = 25 - 10 = 15

Next, let's find the number of students who like at least one subject (union of H and G):
n(H \(\cup\) G) = n(H) + n(G) - n(H \(\cap\) G)
\( = 22 + 25 - 10 \)
\( = 47 - 10 \)
\( = 37 \)

Now, let's answer the questions:
(i) The number of students who do not like History is the total number of students minus those who like History:
n(H') = n(\( \xi \)) – n(H)
\( = 50 - 22 \)
\( = 28 \)

(ii) The number of students who do not like Geography is the total number of students minus those who like Geography:
n(G') = n(\( \xi \)) – n(G)
\( = 50 - 25 \)
\( = 25 \)

(iii) The number of students who like neither History nor Geography is the total number of students minus those who like at least one subject:
n(H' \(\cap\) G') = n(H \(\cup\) G)' = n(\( \xi \)) – n(H \(\cup\) G)
\( = 50 - 37 \)
\( = 13 \)

Venn diagram illustration:
The diagram shows the distribution of students. 12 students like only History, 10 like both, 15 like only Geography, and 13 like neither. These add up to the total of 50 students.

In simple words: We first found out how many students like only one subject, then how many like at least one. Using these numbers, we calculated how many students do not like History, do not like Geography, and like neither. The diagram visually breaks down the class into these groups.

🎯 Exam Tip: To avoid confusion, always calculate the 'only A', 'only B', and 'A \(\cap\) B' values first and draw your Venn diagram. This makes it easier to answer questions about complements and 'neither' categories.

Question 8. 2000 candidates appear in a written test in Mathematics and General Awareness for a Government job. 1800 passed in at least one subject. If 1200 passed in Mathematics and 1500 in General Awareness find :
(i) how many passed in both the subjects ?
(ii) how many passed in Mathematics only ?
(iii) how many failed in General Awareness ?
Answer:
Let \( \xi \) be the set of all candidates.
Let M be the set of candidates who passed in Mathematics.
Let G be the set of candidates who passed in General Awareness.
We are given:
Total candidates, n(\( \xi \)) = 2000
Candidates who passed in at least one subject, n(M \(\cup\) G) = 1800
Candidates who passed in Mathematics, n(M) = 1200
Candidates who passed in General Awareness, n(G) = 1500

(i) To find how many passed in both subjects (M \(\cap\) G), we use the union formula:
n(M \(\cup\) G) = n(M) + n(G) – n(M \(\cap\) G)
Rearranging to find n(M \(\cap\) G):
n(M \(\cap\) G) = n(M) + n(G) – n(M \(\cup\) G)
\( = 1200 + 1500 - 1800 \)
\( = 2700 - 1800 \)
\( = 900 \)
So, 900 candidates passed in both subjects.

(ii) To find how many passed in Mathematics only, we subtract those who passed both from the total who passed Mathematics:
n(M \(\cap\) G') = n(M) – n(M \(\cap\) G)
\( = 1200 - 900 \)
\( = 300 \)
So, 300 candidates passed in Mathematics only.

(iii) To find how many failed in General Awareness, we find the complement of G:
n(G') = n(\( \xi \)) – n(G)
\( = 2000 - 1500 \)
\( = 500 \)
So, 500 candidates failed in General Awareness. This number includes those who passed Math only, and those who failed both subjects.
In simple words: We used the numbers of students who passed each subject and the total who passed at least one. We found the overlap (passed both), then those who passed only one, and finally, those who failed a specific subject by subtracting passes from the total candidates.

🎯 Exam Tip: "Passed in at least one subject" directly refers to the union of the sets (A \(\cup\) B). Always distinguish between those who passed *only* one subject and those who passed *a specific* subject, which might include passing both.

Question 9. In a group of 80 people, 40 like Indian food, 36 like Chinese food and 27 do not like any kind of these foods. Draw Venn diagram to find :
(i) how many like both kind of food ?
(ii) how many like only the Indian food ?
(iii) how many like only the Chinese food ?
Answer:
Let A be the set of people who like Indian food.
Let B be the set of people who like Chinese food.
Let \( \xi \) be the set of all people in the group.
We are given:
n(\( \xi \)) = 80
n(A) = 40
n(B) = 36
Number of people who do not like any kind of food = n(A' \(\cap\) B') = 27

From n(A' \(\cap\) B') = n(A \(\cup\) B)', we know:
n(A \(\cup\) B)' = n(\( \xi \)) – n(A \(\cup\) B)
\( 27 = 80 – n(A \(\cup\) B) \)
\( \implies \) n(A \(\cup\) B) = 80 - 27
\( = 53 \)

Now let's find the required values:
(i) To find how many like both kinds of food (A \(\cap\) B):
n(A \(\cup\) B) = n(A) + n(B) – n(A \(\cap\) B)
\( 53 = 40 + 36 – n(A \(\cap\) B) \)
\( 53 = 76 – n(A \(\cap\) B) \)
\( \implies \) n(A \(\cap\) B) = 76 - 53
\( = 23 \)
So, 23 people like both Indian and Chinese food.

(ii) To find how many like only Indian food (A \(\cap\) B'):
n(A \(\cap\) B') = n(A) – n(A \(\cap\) B)
\( = 40 - 23 \)
\( = 17 \)
So, 17 people like only Indian food.

(iii) To find how many like only Chinese food (A' \(\cap\) B):
n(A' \(\cap\) B) = n(B) – n(A \(\cap\) B)
\( = 36 - 23 \)
\( = 13 \)
So, 13 people like only Chinese food.

Venn diagram illustration:
The diagram shows 17 people like only Indian food, 23 like both, 13 like only Chinese food, and 27 like neither. These sum up to the total of 80 people.

In simple words: We used the total number of people and those who liked no food to find how many liked at least one food. Then, using the numbers for Indian and Chinese food, we calculated the overlap (both), and then those who liked only Indian or only Chinese.

🎯 Exam Tip: When given the number of elements *not* in any set (n(A' \(\cap\) B')), use this to calculate n(A \(\cup\) B) first, as it's the complement of the universal set minus those who like neither. This is a common starting point for these problems.

Question 10. In a group of people, two-seventh speak Bengali only and three-seventh speak Hindi only. If 20 people speak none of these languages and 80 speak both, find using Venn diagram the total number of people in the group.
Answer:
Let the total number of people in the group be \( x \). So, n(\( \xi \)) = \( x \).
Let B be the set of people who speak Bengali.
Let H be the set of people who speak Hindi.

We are given:
Number of people who speak Bengali only = n(B \(\cap\) H') = \( \frac{2}{7}x \)
Number of people who speak Hindi only = n(H \(\cap\) B') = \( \frac{3}{7}x \)
Number of people who speak both languages = n(B \(\cap\) H) = 80
Number of people who speak none of these languages = n(B' \(\cap\) H') = 20

Venn diagram illustration:
The diagram shows the distribution of people in fractions and absolute numbers. The sum of all regions must equal the total group size \( x \).

From the Venn diagram, the sum of all distinct regions must equal the total number of people \( x \):
(People who speak Bengali only) + (People who speak both) + (People who speak Hindi only) + (People who speak none) = Total people
\( \frac{2}{7}x + 80 + \frac{3}{7}x + 20 = x \)
Combine the fractions and constants:
\( \frac{2x + 3x}{7} + 100 = x \)
\( \frac{5x}{7} + 100 = x \)
To solve for \( x \), subtract \( \frac{5x}{7} \) from both sides:
\( 100 = x - \frac{5x}{7} \)
\( 100 = \frac{7x - 5x}{7} \)
\( 100 = \frac{2x}{7} \)
Now, multiply both sides by 7:
\( 100 \times 7 = 2x \)
\( 700 = 2x \)
Finally, divide by 2:
\( x = \frac{700}{2} \)
\( x = 350 \)
Thus, the total number of people in the group is 350. The Venn diagram helps set up the equation to solve for the total.
In simple words: We added up all the parts of the group from the Venn diagram – those who speak only Bengali, only Hindi, both, and neither. This sum equals the total number of people. We then solved the equation to find that the total group size is 350 people.

🎯 Exam Tip: When parts of the population are given as fractions of the total, represent the total as \( x \) and set up an equation where the sum of all distinct regions in the Venn diagram equals \( x \). Solve this equation to find the total number of people.

Question 11. In a class of 150 students, the following results were obtained in a certain exami-nation ; 45 students failed in Maths, 50 students failed in Physics, 48 students failed in Chemistry, 30 students failed in both Maths and Physics, 32 failed in Physics and Chemistry. 35 failed in both Maths and Chemistry, 25 failed in all the three subjects. Draw a Venn diagram corresponding to this data and find the number of students who have failed in at least one subject.
Answer:
Let \( \xi \) be the set of all students in the class.
Let M be the set of students who failed in Maths.
Let P be the set of students who failed in Physics.
Let C be the set of students who failed in Chemistry.

We are given:
n(\( \xi \)) = 150
n(M) = 45
n(P) = 50
n(C) = 48
n(M \(\cap\) P) = 30
n(P \(\cap\) C) = 32
n(M \(\cap\) C) = 35
n(M \(\cap\) P \(\cap\) C) = 25 (students who failed in all three subjects)

First, let's determine the number of students in each specific region of the Venn diagram for three sets. We start from the innermost intersection:
1. Failed in all three: n(M \(\cap\) P \(\cap\) C) = 25
2. Failed in M and P only: n(M \(\cap\) P) - n(M \(\cap\) P \(\cap\) C) = 30 - 25 = 5
3. Failed in P and C only: n(P \(\cap\) C) - n(M \(\cap\) P \(\cap\) C) = 32 - 25 = 7
4. Failed in M and C only: n(M \(\cap\) C) - n(M \(\cap\) P \(\cap\) C) = 35 - 25 = 10
5. Failed in M only: n(M) - (failed in M \(\cap\) P only + failed in M \(\cap\) C only + failed in all three)
\( = 45 - (5 + 10 + 25) = 45 - 40 = 5 \)
6. Failed in P only: n(P) - (failed in M \(\cap\) P only + failed in P \(\cap\) C only + failed in all three)
\( = 50 - (5 + 7 + 25) = 50 - 37 = 13 \)
7. Failed in C only: n(C) - (failed in P \(\cap\) C only + failed in M \(\cap\) C only + failed in all three)
\( = 48 - (7 + 10 + 25) = 48 - 42 = 6 \)

Venn diagram illustration:
The diagram visually represents the numbers of students failing in various combinations of subjects. Each region has a specific count, and their sum forms the total who failed in at least one subject.

The number of students who have failed in at least one subject is the union of all three sets, n(M \(\cup\) P \(\cup\) C):
n(M \(\cup\) P \(\cup\) C) = (M only) + (P only) + (C only) + (M \(\cap\) P only) + (P \(\cap\) C only) + (M \(\cap\) C only) + (M \(\cap\) P \(\cap\) C)
\( = 5 + 13 + 6 + 5 + 7 + 10 + 25 \)
\( = 71 \)

Alternatively, using the principle of inclusion-exclusion for three sets:
n(M \(\cup\) P \(\cup\) C) = n(M) + n(P) + n(C) - n(M \(\cap\) P) - n(P \(\cap\) C) - n(M \(\cap\) C) + n(M \(\cap\) P \(\cap\) C)
\( = 45 + 50 + 48 - 30 - 32 - 35 + 25 \)
\( = 143 - 97 + 25 \)
\( = 46 + 25 \)
\( = 71 \)
So, 71 students failed in at least one subject. This means 71 students will need extra help or retesting. The number of students who passed all subjects (or failed none) would be \( 150 - 71 = 79 \). This is shown in the diagram outside the circles.
In simple words: We used all the given numbers to fill in each part of the three-circle Venn diagram, starting from the middle. Then, we added up all the numbers inside the circles to find the total number of students who failed at least one subject.

🎯 Exam Tip: For three-set Venn diagrams, always calculate the elements in the intersection of all three (A \(\cap\) B \(\cap\) C) first. Then calculate elements in two sets *only* (e.g., A \(\cap\) B only) by subtracting the three-set intersection. Work outwards from the center to fill the diagram correctly.

Question 12. A firm has 40 workers working in the factory premises, 30 working in its office and 20 working in both the places. How many workers are there in the firm ? How many are working in the (i) factory (ii) office alone?
Answer:
Let F be the set of workers working in the factory premises.
Let O be the set of workers working in the office.
We are given:
n(F) = 40 (total factory workers)
n(O) = 30 (total office workers)
n(F \(\cap\) O) = 20 (workers in both factory and office)

First, let's find the total number of workers in the firm. This is the union of F and O, as it represents all workers whether in factory, office, or both:
Total workers = n(F \(\cup\) O) = n(F) + n(O) – n(F \(\cap\) O)
\( = 40 + 30 - 20 \)
\( = 70 - 20 \)
\( = 50 \)
There are 50 workers in the firm.

Now, let's find workers in (i) factory alone and (ii) office alone:
(i) Workers working in factory alone = n(F \(\cap\) O') = n(F) – n(F \(\cap\) O)
\( = 40 - 20 \)
\( = 20 \)

(ii) Workers working in office alone = n(O \(\cap\) F') = n(O) – n(F \(\cap\) O)
\( = 30 - 20 \)
\( = 10 \)

Venn diagram illustration:
The diagram shows 20 workers are in the factory only, 20 workers are in both the factory and the office, and 10 workers are in the office only. These sums up to the total of 50 workers in the firm.

In simple words: We used the number of workers in the factory, in the office, and in both places. We found the total number of workers by adding the factory workers and office workers, then subtracting those counted twice (who work in both places). Then, we found how many work only in the factory or only in the office.

🎯 Exam Tip: The phrase "total number of workers in the firm" implies the union of all relevant sets (F \(\cup\) O), unless specified that some workers are not in either category. "Alone" questions refer to the region of one set that does not overlap with the other.