Samacheer Kalvi Class 9 Maths Solutions Chapter 1 Set Language Exercise 1.6

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 1 Set Language Ex 1.6

Question 1. (i) If n(A) = 25, n(B) = 40, n(AUB) = 50 and n(B') = 25, find n(A∩B) and n(U).
Answer:
Given:
\( n(A) = 25 \)
\( n(B) = 40 \)
\( n(A \cup B) = 50 \)
\( n(B') = 25 \)

We know the formula for the union of two sets:
\( n(A \cup B) = n(A) + n(B) - n(A \cap B) \)
Substitute the given values into the formula:
\( 50 = 25 + 40 - n(A \cap B) \)
\( 50 = 65 - n(A \cap B) \)
Now, we need to find \( n(A \cap B) \). Rearrange the equation:
\( n(A \cap B) = 65 - 50 \)
\( n(A \cap B) = 15 \)

Next, we need to find \( n(U) \). We know that the total number of elements in the universal set \( U \) is the sum of elements in set \( B \) and its complement \( B' \):
\( n(U) = n(B) + n(B') \)
Substitute the given values:
\( n(U) = 40 + 25 \)
\( n(U) = 65 \)

So, we have found both values:
\( n(A \cap B) = 15 \) and \( n(U) = 65 \). These steps show how to calculate the intersection and the universal set using set theory formulas.
In simple words: We used a special formula to find the number of items that are in both set A and set B. Then, we added the items in set B and the items not in set B to get the total number of items in the whole group.

🎯 Exam Tip: Remember the fundamental formula for the union of two sets: \( n(A \cup B) = n(A) + n(B) - n(A \cap B) \). Also, know that \( n(U) = n(X) + n(X') \) for any set \( X \).

Question 1. (ii) If n(A) = 300, n(AUB) = 500, n(A∩B) = 50 and n(B') = 350, find n(B) and n(U).
Answer:
Given:
\( n(A) = 300 \)
\( n(A \cup B) = 500 \)
\( n(A \cap B) = 50 \)
\( n(B') = 350 \)

First, we will find \( n(B) \) using the formula for the union of two sets:
\( n(A \cup B) = n(A) + n(B) - n(A \cap B) \)
Substitute the given values into the formula:
\( 500 = 300 + n(B) - 50 \)
Simplify the right side of the equation:
\( 500 = (300 - 50) + n(B) \)
\( 500 = 250 + n(B) \)
To find \( n(B) \), subtract 250 from both sides:
\( n(B) = 500 - 250 \)
\( n(B) = 250 \)

Next, we will find \( n(U) \) using the relationship between a set and its complement:
\( n(U) = n(B) + n(B') \)
Substitute the value of \( n(B) \) we just found and the given \( n(B') \):
\( n(U) = 250 + 350 \)
\( n(U) = 600 \)

So, the number of elements in set \( B \) is 250, and the total number of elements in the universal set \( U \) is 600.
In simple words: We used a formula that links the number of items in two sets, their union, and their intersection to find the number of items in set B. Then, we added the number of items in set B and the items not in B to find the total items in the universal set.

🎯 Exam Tip: Always write down the given information and the formula you are using before substituting values to avoid errors. Check your arithmetic carefully for calculations involving subtraction and addition.

Question 2. If U = {x : x ∈ N, x ≤ 10}, A = { 2, 3, 4, 8, 10} and B = {1, 2, 5, 8, 10}, then verify that n(AUB) = n(A) + n(B) – n(A∩B)
Answer:
First, let's list the elements of the universal set \( U \) and the given sets \( A \) and \( B \):
\( U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} \)
\( A = \{2, 3, 4, 8, 10\} \)
\( B = \{1, 2, 5, 8, 10\} \)

Now, let's find the number of elements in each set:
\( n(U) = 10 \)
\( n(A) = 5 \) (There are 5 elements in set A)
\( n(B) = 5 \) (There are 5 elements in set B)

Next, we find the union of sets \( A \) and \( B \), which includes all unique elements from both sets:
\( A \cup B = \{2, 3, 4, 8, 10\} \cup \{1, 2, 5, 8, 10\} \)
\( A \cup B = \{1, 2, 3, 4, 5, 8, 10\} \)
The number of elements in the union is:
\( n(A \cup B) = 7 \) ........(1)

Now, we find the intersection of sets \( A \) and \( B \), which includes elements common to both sets:
\( A \cap B = \{2, 3, 4, 8, 10\} \cap \{1, 2, 5, 8, 10\} \)
\( A \cap B = \{2, 8, 10\} \)
The number of elements in the intersection is:
\( n(A \cap B) = 3 \)

Finally, we verify the formula \( n(A \cup B) = n(A) + n(B) - n(A \cap B) \):
Substitute the counts we found:
\( n(A) + n(B) - n(A \cap B) = 5 + 5 - 3 \)
\( = 10 - 3 \)
\( = 7 \) ........(2)

From (1) and (2), we see that \( n(A \cup B) = 7 \) and \( n(A) + n(B) - n(A \cap B) = 7 \).
Since both sides are equal to 7, the formula is verified. This formula helps avoid double-counting common elements when finding the total elements in a union.
In simple words: We listed all items in each set and then counted them. We found the items that are in either set (union) and the items that are in both sets (intersection). We then used a formula to check if our counts were correct, and they matched.

🎯 Exam Tip: When verifying set formulas, always clearly list the elements of each set, then calculate \( n(A) \), \( n(B) \), \( n(A \cup B) \), and \( n(A \cap B) \) step-by-step. Double-check your element listings to avoid errors.

Question 3. Verify n(AUBUC) = n(A) + n(B) + n(C) – n(A∩B) – n(B∩C) – n(A∩C) + n(A∩B∩C) for the following sets.
(i) A = {a, c, e, f, h}, B = {c, d, e, f} and C = {a, b, c, f}
Answer:
Given sets:
\( A = \{a, c, e, f, h\} \)
\( B = \{c, d, e, f\} \)
\( C = \{a, b, c, f\} \)

First, find the number of elements in each set:
\( n(A) = 5 \)
\( n(B) = 4 \)
\( n(C) = 4 \)

Next, find the intersections of two sets:
\( A \cap B = \{a, c, e, f, h\} \cap \{c, d, e, f\} = \{c, e, f\} \)
\( n(A \cap B) = 3 \)

\( B \cap C = \{c, d, e, f\} \cap \{a, b, c, f\} = \{c, f\} \)
\( n(B \cap C) = 2 \)

\( A \cap C = \{a, c, e, f, h\} \cap \{a, b, c, f\} = \{a, c, f\} \)
\( n(A \cap C) = 3 \)

Now, find the intersection of all three sets:
\( A \cap B \cap C = \{a, c, e, f, h\} \cap \{c, d, e, f\} \cap \{a, b, c, f\} = \{c, f\} \)
\( n(A \cap B \cap C) = 2 \)

Next, find the union of all three sets:
\( A \cup B \cup C = \{a, c, e, f, h\} \cup \{c, d, e, f\} \cup \{a, b, c, f\} \)
\( A \cup B \cup C = \{a, b, c, d, e, f, h\} \)
The number of elements in the union is:
\( n(A \cup B \cup C) = 7 \) ........(1)

Finally, let's check the formula: \( n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(A \cap C) + n(A \cap B \cap C) \)
Substitute all the calculated values:
\( n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(A \cap C) + n(A \cap B \cap C) \)
\( = 5 + 4 + 4 - 3 - 2 - 3 + 2 \)
\( = 13 - 8 + 2 \)
\( = 5 + 2 \)
\( = 7 \) ........(2)

From (1) and (2), both sides of the equation are equal to 7. So, the formula is verified for these sets. This formula is useful for counting elements in a union of three sets without overcounting.
In simple words: We counted items in each set and in their overlaps (two sets at a time, and all three together). Then we counted items in the combined set (union). We put these counts into a special formula and found that both sides matched, proving the formula works.

🎯 Exam Tip: When working with three sets, it's crucial to systematically list all single sets, two-set intersections, and the three-set intersection. Use a Venn diagram sketch to visualize the regions and ensure all common elements are correctly identified and counted.

Question 3. (ii) A= {1, 3, 5}, B = {2, 3, 5, 6}, C = {1, 5, 6, 7}
Answer:
Given sets:
\( A = \{1, 3, 5\} \)
\( B = \{2, 3, 5, 6\} \)
\( C = \{1, 5, 6, 7\} \)

First, find the number of elements in each set:
\( n(A) = 3 \)
\( n(B) = 4 \)
\( n(C) = 4 \)

Next, find the intersections of two sets:
\( A \cap B = \{1, 3, 5\} \cap \{2, 3, 5, 6\} = \{3, 5\} \)
\( n(A \cap B) = 2 \)

\( B \cap C = \{2, 3, 5, 6\} \cap \{1, 5, 6, 7\} = \{5, 6\} \)
\( n(B \cap C) = 2 \)

\( A \cap C = \{1, 3, 5\} \cap \{1, 5, 6, 7\} = \{1, 5\} \)
\( n(A \cap C) = 2 \)

Now, find the intersection of all three sets:
\( A \cap B \cap C = \{1, 3, 5\} \cap \{2, 3, 5, 6\} \cap \{1, 5, 6, 7\} = \{5\} \)
\( n(A \cap B \cap C) = 1 \)

Next, find the union of all three sets:
\( A \cup B \cup C = \{1, 3, 5\} \cup \{2, 3, 5, 6\} \cup \{1, 5, 6, 7\} \)
\( A \cup B \cup C = \{1, 2, 3, 5, 6, 7\} \)
The number of elements in the union is:
\( n(A \cup B \cup C) = 6 \) ........(1)

Finally, let's check the formula: \( n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(A \cap C) + n(A \cap B \cap C) \)
Substitute all the calculated values:
\( n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(A \cap C) + n(A \cap B \cap C) \)
\( = 3 + 4 + 4 - 2 - 2 - 2 + 1 \)
\( = 11 - 6 + 1 \)
\( = 5 + 1 \)
\( = 6 \) ........(2)

From (1) and (2), both sides of the equation are equal to 6. So, the formula is verified for these sets as well. This demonstrates the consistency of the Principle of Inclusion-Exclusion for three sets.
In simple words: We repeated the process from the previous part with new sets. We counted all the elements, their overlaps, and their union. By plugging these numbers into the formula, we again found that both sides were equal, confirming the formula.

🎯 Exam Tip: Pay close attention to distinguishing between elements in the sets and the *number* of elements. A common mistake is to miscount the elements in the intersections or union, so be very systematic.

Question 4. In a colony, 25 students take part in music, 30 students take part in drama and 8 students take part in both music and drama. If each student takes part in at least one of these activities, find:
(i) The number of students who take part in only music.
(ii) The number of students who take part in only drama.
(iii) The total number of students in the class.
Answer:
Let \( M \) be the set of students who take part in Music.
Let \( D \) be the set of students who take part in Drama.

Given:
Number of students in Music, \( n(M) = 25 \)
Number of students in Drama, \( n(D) = 30 \)
Number of students in both Music and Drama, \( n(M \cap D) = 8 \)

We can represent this information using a Venn diagram:

From the Venn diagram or using formulas:

(i) **The number of students who take part in only music:**
This is found by subtracting the students who do both from the total students in music:
\( n(\text{only } M) = n(M) - n(M \cap D) \)
\( n(\text{only } M) = 25 - 8 \)
\( n(\text{only } M) = 17 \)

(ii) **The number of students who take part in only drama:**
Similarly, this is found by subtracting the students who do both from the total students in drama:
\( n(\text{only } D) = n(D) - n(M \cap D) \)
\( n(\text{only } D) = 30 - 8 \)
\( n(\text{only } D) = 22 \)

(iii) **The total number of students in the class:**
Since each student takes part in at least one activity, the total number of students in the class is the union of students in Music and Drama. We can add students who do only music, only drama, and both:
\( n(M \cup D) = n(\text{only } M) + n(\text{only } D) + n(M \cap D) \)
\( n(M \cup D) = 17 + 22 + 8 \)
\( n(M \cup D) = 47 \)
Alternatively, using the formula: \( n(M \cup D) = n(M) + n(D) - n(M \cap D) \)
\( n(M \cup D) = 25 + 30 - 8 \)
\( n(M \cup D) = 55 - 8 \)
\( n(M \cup D) = 47 \)

Therefore, 17 students participate only in music, 22 students participate only in drama, and there are 47 students in total in the class. This problem illustrates how set theory helps categorize and count participants in activities.
In simple words: We started with how many people like music, how many like drama, and how many like both. To find those who like *only* music or *only* drama, we took out the 'both' group. Then we added these three groups (only music, only drama, and both) to find the total number of students.

🎯 Exam Tip: Always draw a Venn diagram for problems involving "only A", "only B", or "both" to visualize the different regions and ensure you subtract the intersection correctly to find the "only" parts.

Question 5. In a party of 45 people, each one likes Tea or Coffee or both. 35 people like tea and 20 people like coffee. Find the number of people who
(i) like both Tea and Coffee.
(ii) do not like Tea.
(iii) do not like Coffee.
Answer:
Let \( T \) be the set of people who like Tea.
Let \( C \) be the set of people who like Coffee.

Given:
Total number of people in the party, \( n(T \cup C) = 45 \) (since everyone likes at least one)
Number of people who like Tea, \( n(T) = 35 \)
Number of people who like Coffee, \( n(C) = 20 \)

Let \( x \) be the number of people who like both Tea and Coffee, i.e., \( n(T \cap C) = x \).

We can represent this information using a Venn diagram. The regions would be:
Only Tea: \( n(T) - n(T \cap C) = 35 - x \)
Only Coffee: \( n(C) - n(T \cap C) = 20 - x \)
Both Tea and Coffee: \( x \)

The sum of these regions equals the total number of people:
\( (35 - x) + x + (20 - x) = 45 \)
\( 35 + 20 - x = 45 \)
\( 55 - x = 45 \)
To find \( x \), rearrange the equation:
\( x = 55 - 45 \)
\( x = 10 \)

So, \( n(T \cap C) = 10 \).

The Venn diagram with values:


Now we can answer the questions:
(i) **People who like both Tea and Coffee:**
This is \( x \), which we found to be 10.
\( n(T \cap C) = 10 \)

(ii) **People who do not like Tea:**
This means people who like only Coffee. From our Venn diagram calculations:
\( n(\text{only } C) = n(C) - n(T \cap C) = 20 - 10 = 10 \)
Alternatively, people who do not like Tea are those who are in the set C but not T, or those in the universal set who are not in T. Since everyone likes at least one, these are simply those who like only coffee.

(iii) **People who do not like Coffee:**
This means people who like only Tea. From our Venn diagram calculations:
\( n(\text{only } T) = n(T) - n(T \cap C) = 35 - 10 = 25 \)
Alternatively, people who do not like Coffee are those who are in the set T but not C, or those in the universal set who are not in C. Since everyone likes at least one, these are simply those who like only tea.

So, 10 people like both tea and coffee, 10 people like only coffee (do not like tea), and 25 people like only tea (do not like coffee). This problem uses the Principle of Inclusion-Exclusion to find overlaps and specific preferences within a group.
In simple words: We knew the total number of people and how many liked tea and coffee separately. We used a special counting trick (and a picture) to figure out how many liked both. Then, we easily found out how many liked just tea or just coffee by subtracting the 'both' group.

🎯 Exam Tip: When given the total number of people and the counts for individual preferences, if "everyone likes at least one", then the total number represents the union of the sets. This is a common setup for finding the intersection.

Question 6. In an examination 50% of the students passed in mathematics and 70% of students passed in science while 10% students failed in both subjects. 300 students passed in both the subjects. Find the total number of students who appeared in the examination, if they took examination in only two subjects.
Answer:
Let \( M \) represent students who passed in Mathematics.
Let \( S \) represent students who passed in Science.

Given percentages of students passed:
\( n(M) = 50\% \)
\( n(S) = 70\% \)

Given percentage of students who failed in both subjects:
Students failed in both means they are outside the union of \( M \) and \( S \).
\( n(M \cup S)' = 10\% \)

This means the percentage of students who passed in at least one subject (Mathematics or Science) is:
\( n(M \cup S) = 100\% - n(M \cup S)' \)
\( n(M \cup S) = 100\% - 10\% = 90\% \)

Now, we use the formula for the union of two sets:
\( n(M \cup S) = n(M) + n(S) - n(M \cap S) \)
Substitute the known percentages:
\( 90\% = 50\% + 70\% - n(M \cap S) \)
\( 90\% = 120\% - n(M \cap S) \)
Rearrange to find the percentage of students who passed in both subjects:
\( n(M \cap S) = 120\% - 90\% \)
\( n(M \cap S) = 30\% \)

We are given that 300 students passed in both subjects. This means that \( 30\% \) of the total students is equal to 300.
Let \( T \) be the total number of students who appeared in the examination.
\( 30\% \text{ of } T = 300 \)
\( 0.30 \times T = 300 \)
To find \( T \), divide 300 by 0.30:
\( T = \frac{300}{0.30} \)
\( T = 1000 \)

Therefore, the total number of students who appeared in the examination is 1000. This calculation uses percentages to find the overlap and then scales up to the total number of students.

In simple words: We knew how many students passed in Maths, how many in Science, and how many failed in both. First, we figured out the percentage of students who passed at least one subject. Using a formula, we then found the percentage of students who passed *both* subjects. Since we knew the actual number for "both", we could work backwards to find the total number of students.

🎯 Exam Tip: When dealing with percentages, convert "failed in both" to "passed in at least one" by subtracting from 100% to effectively use the union formula. Always clearly define your sets and state your percentages or numbers.

Question 7. A and B are two sets such that n(A – B) = 32 + x, n(B – A) = 5x and n(A∩B) = x. Illustrate the information by means of a venn diagram. Given that n(A) = n(B). Calculate the value of x.
Answer:
Given:
\( n(A - B) = 32 + x \) (Number of elements only in A)
\( n(B - A) = 5x \) (Number of elements only in B)
\( n(A \cap B) = x \) (Number of elements in the intersection of A and B)

We can illustrate this information using a Venn diagram:

We know that the number of elements in a set A, \( n(A) \), is the sum of elements only in A and elements in the intersection:
\( n(A) = n(A - B) + n(A \cap B) \)
\( n(A) = (32 + x) + x \)
\( n(A) = 32 + 2x \)

Similarly, for set B:
\( n(B) = n(B - A) + n(A \cap B) \)
\( n(B) = 5x + x \)
\( n(B) = 6x \)

We are given that \( n(A) = n(B) \). So, we can set the two expressions equal to each other:
\( 32 + 2x = 6x \)
Now, solve for \( x \):
Subtract \( 2x \) from both sides:
\( 32 = 6x - 2x \)
\( 32 = 4x \)
Divide by 4:
\( x = \frac{32}{4} \)
\( x = 8 \)

The value of \( x \) is 8. This method allows us to solve for unknown quantities in set problems by using the relationships between different parts of the sets.
In simple words: We used a Venn diagram to show the parts of two sets: items only in A, items only in B, and items in both. We were told that set A and set B have the same number of items. By writing out the math for this, we could find the value of the unknown number 'x'.

🎯 Exam Tip: When \( n(A) = n(B) \) is given, it means the *total* number of elements in each circle of the Venn diagram is equal. Use this equality to form an equation and solve for any unknowns, remembering \( n(A) = n(A-B) + n(A \cap B) \).

Question 8. Out of 500 car owners investigated, 400 owned car A and 200 owned car B, 50 owned both A and B cars. Is this data correct?
Answer:
Let \( A \) be the set of people who own car A.
Let \( B \) be the set of people who own car B.

Given:
Total number of car owners investigated, \( n(U) = 500 \)
Number of people who own car A, \( n(A) = 400 \)
Number of people who own car B, \( n(B) = 200 \)
Number of people who own both car A and car B, \( n(A \cap B) = 50 \)

To check if the data is correct, we can use the Principle of Inclusion-Exclusion for the union of two sets:
\( n(A \cup B) = n(A) + n(B) - n(A \cap B) \)
Substitute the given values into the formula:
\( n(A \cup B) = 400 + 200 - 50 \)
\( n(A \cup B) = 600 - 50 \)
\( n(A \cup B) = 550 \)

This means that 550 people own at least one of the cars. However, the total number of car owners investigated (universal set) is given as 500. It is impossible for the number of people who own at least one car to be greater than the total number of people investigated.
Since \( n(A \cup B) = 550 \) is greater than the total number of car owners \( n(U) = 500 \), the given data is not correct. This shows that the provided numbers contain an internal inconsistency.
In simple words: We added up the number of people who own car A and car B, then subtracted those who own both to find how many people own at least one car. This number turned out to be more than the total number of car owners surveyed, which is impossible. So, the original information given must be wrong.

🎯 Exam Tip: Always compare the calculated union of sets \( n(A \cup B) \) with the universal set \( n(U) \). The number of elements in the union of any subsets cannot exceed the total number of elements in the universal set being considered.

Question 9. In a colony, 275 families buy Tamil newspaper, 150 families buy English newspaper, 45 families buy Hindi newspaper, 125 families buy Tamil and English newspapers, 17 families buy English and Hindi newspapers, 5 families buy Tamil and Hindi newspapers and 3 families buy all the three newspapers. If each family buy atleast one of these newspapers then find
(i) Number of families buy only one newspaper
(ii) Number of families buy atleast two newspapers
(iii) Total number of families in the colony.
Answer:
Let \( T \) represent families buying Tamil newspaper.
Let \( E \) represent families buying English newspaper.
Let \( H \) represent families buying Hindi newspaper.

Given:
\( n(T) = 275 \)
\( n(E) = 150 \)
\( n(H) = 45 \)
\( n(T \cap E) = 125 \)
\( n(E \cap H) = 17 \)
\( n(T \cap H) = 5 \)
\( n(T \cap E \cap H) = 3 \)

We can use a Venn diagram to represent this information and find the numbers in each unique region.

First, fill in the center (intersection of all three):
\( n(T \cap E \cap H) = 3 \)

Next, find the numbers for two-set intersections *only*:
Only \( T \) and \( E \): \( n(T \cap E) - n(T \cap E \cap H) = 125 - 3 = 122 \)
Only \( E \) and \( H \): \( n(E \cap H) - n(T \cap E \cap H) = 17 - 3 = 14 \)
Only \( T \) and \( H \): \( n(T \cap H) - n(T \cap E \cap H) = 5 - 3 = 2 \)

Then, find the numbers for each set *only* (not overlapping with others):
Only \( T \): \( n(T) - [n(\text{only } T \cap E) + n(\text{only } T \cap H) + n(T \cap E \cap H)] \)
\( n(\text{only } T) = 275 - (122 + 2 + 3) = 275 - 127 = 148 \)

Only \( E \): \( n(E) - [n(\text{only } T \cap E) + n(\text{only } E \cap H) + n(T \cap E \cap H)] \)
\( n(\text{only } E) = 150 - (122 + 14 + 3) = 150 - 139 = 11 \)

Only \( H \): \( n(H) - [n(\text{only } T \cap H) + n(\text{only } E \cap H) + n(T \cap E \cap H)] \)
\( n(\text{only } H) = 45 - (2 + 14 + 3) = 45 - 19 = 26 \)



Now we can answer the questions:
(i) **Number of families who buy only one newspaper:**
This is the sum of families buying only Tamil, only English, and only Hindi.
\( n(\text{only one}) = n(\text{only } T) + n(\text{only } E) + n(\text{only } H) \)
\( n(\text{only one}) = 148 + 11 + 26 \)
\( n(\text{only one}) = 185 \)

(ii) **Number of families who buy at least two newspapers:**
This is the sum of families who buy only two newspapers (T&E, E&H, T&H) and those who buy all three.
\( n(\text{at least two}) = n(\text{only } T \cap E) + n(\text{only } E \cap H) + n(\text{only } T \cap H) + n(T \cap E \cap H) \)
\( n(\text{at least two}) = 122 + 14 + 2 + 3 \)
\( n(\text{at least two}) = 141 \)

(iii) **Total number of families in the colony:**
Since each family buys at least one newspaper, the total number of families is the union of the three sets \( n(T \cup E \cup H) \). This is the sum of all distinct regions in the Venn diagram.
\( n(T \cup E \cup H) = n(\text{only } T) + n(\text{only } E) + n(\text{only } H) + n(\text{only } T \cap E) + n(\text{only } E \cap H) + n(\text{only } T \cap H) + n(T \cap E \cap H) \)
\( n(T \cup E \cup H) = 148 + 11 + 26 + 122 + 14 + 2 + 3 \)
\( n(T \cup E \cup H) = 326 \)

Alternatively, using the formula for the union of three sets:
\( n(T \cup E \cup H) = n(T) + n(E) + n(H) - n(T \cap E) - n(E \cap H) - n(T \cap H) + n(T \cap E \cap H) \)
\( n(T \cup E \cup H) = 275 + 150 + 45 - 125 - 17 - 5 + 3 \)
\( n(T \cup E \cup H) = 470 - 147 + 3 \)
\( n(T \cup E \cup H) = 323 + 3 \)
\( n(T \cup E \cup H) = 326 \)

Thus, 185 families buy only one newspaper, 141 families buy at least two newspapers, and there are 326 families in the colony. This detailed breakdown helps understand the various preferences within the community.
In simple words: We organized all the information about newspaper readers into a Venn diagram. We first found how many families read only one type of newspaper, then how many read two or more. Finally, we added up all these different groups to find the total number of families in the colony.

🎯 Exam Tip: For three-set problems, always start by filling the innermost intersection (all three sets) in your Venn diagram. Then work outwards to fill the two-set intersections, and finally the "only" regions for each individual set. This systematic approach prevents errors.

Question 10. A survey of 1000 farmers found that 600 grew paddy, 350 grew ragi, 280 grew corn, 120 grew paddy and ragi, 100 grew ragi and corn, 80 grew paddy and corn. If each farmer grew atleast any one of the above three, then find the number of farmers who grew all the three.
Answer:
Let \( P \) represent farmers who grew Paddy.
Let \( R \) represent farmers who grew Ragi.
Let \( C \) represent farmers who grew Corn.

Given:
Total number of farmers surveyed, \( n(P \cup R \cup C) = 1000 \) (since each grew at least one)
\( n(P) = 600 \)
\( n(R) = 350 \)
\( n(C) = 280 \)
\( n(P \cap R) = 120 \)
\( n(R \cap C) = 100 \)
\( n(P \cap C) = 80 \)

Let \( x \) be the number of farmers who grew all three crops, i.e., \( n(P \cap R \cap C) = x \).

We use the Principle of Inclusion-Exclusion for the union of three sets:
\( n(P \cup R \cup C) = n(P) + n(R) + n(C) - n(P \cap R) - n(R \cap C) - n(P \cap C) + n(P \cap R \cap C) \)

Substitute the given values into the formula:
\( 1000 = 600 + 350 + 280 - 120 - 100 - 80 + x \)
First, sum the individual set sizes:
\( 600 + 350 + 280 = 1230 \)
Next, sum the two-set intersections:
\( 120 + 100 + 80 = 300 \)

Now substitute these sums back into the formula:
\( 1000 = 1230 - 300 + x \)
\( 1000 = 930 + x \)

To find \( x \), subtract 930 from both sides:
\( x = 1000 - 930 \)
\( x = 70 \)

Therefore, the number of farmers who grew all three crops (paddy, ragi, and corn) is 70. This application of set theory helps determine the overlap in preferences or activities when total group sizes and partial overlaps are known.
In simple words: We had the total number of farmers and how many grew each crop, plus how many grew two specific crops. We used a special formula for three groups to find the missing piece: the number of farmers who grew all three crops. We plugged in all the numbers we knew and solved for the unknown.

🎯 Exam Tip: When given the union of three sets and individual/pairwise intersections, the Principle of Inclusion-Exclusion formula is essential. Systematically sum the single sets, subtract the sums of two-set intersections, and then add the three-set intersection to solve for the unknown.

Question 11. In the adjacent diagram, if n(U) = 125, y is two times of x and z is 10 more than x, then find the value of x, y and z.
Answer:
From the given Venn diagram, the numbers in each distinct region are:
5, x, 4, y, 3, 6, 17, z

The total number of elements in the universal set \( U \) is the sum of all these regions:
\( n(U) = 5 + x + 4 + y + 3 + 6 + 17 + z \)
Given \( n(U) = 125 \). So:
\( 125 = 5 + x + 4 + y + 3 + 6 + 17 + z \)
Combine the constant numbers:
\( 5 + 4 + 3 + 6 + 17 = 35 \)
So, the equation becomes:
\( 125 = 35 + x + y + z \)

We are also given relationships between \( x, y, \) and \( z \):
1. \( y = 2x \)
2. \( z = x + 10 \)

Substitute these expressions for \( y \) and \( z \) into the total sum equation:
\( 125 = 35 + x + (2x) + (x + 10) \)
Combine like terms (all the \( x \) terms and constant numbers):
\( 125 = 35 + 10 + x + 2x + x \)
\( 125 = 45 + 4x \)
Now, solve for \( x \):
Subtract 45 from both sides:
\( 125 - 45 = 4x \)
\( 80 = 4x \)
Divide by 4:
\( x = \frac{80}{4} \)
\( x = 20 \)

Now that we have the value of \( x \), we can find \( y \) and \( z \):
For \( y \): \( y = 2x = 2 \times 20 = 40 \)
For \( z \): \( z = x + 10 = 20 + 10 = 30 \)

So, the values are \( x = 20, y = 40, \) and \( z = 30 \). This problem shows how to use information from a Venn diagram and algebraic relationships to solve for unknown quantities.

In simple words: We added up all the numbers and letters shown in the Venn diagram, which should equal the total number given for the whole set. We then used the clues that 'y' is twice 'x' and 'z' is 'x' plus ten. By replacing 'y' and 'z' with expressions involving 'x', we could solve the equation to find 'x', and then easily find 'y' and 'z'.

🎯 Exam Tip: When working with Venn diagrams with variables, ensure you sum *all* distinct regions within the universal set to form your primary equation. Carefully substitute any given relationships between variables to reduce the equation to a single unknown.

Question 12. Each student in a class of 35 plays atleast one game among chess, carrom and table tennis. 22 play chess, 21 play carrom, 15 play table tennis, 10 play chess and table tennis, 8 play carrom and table tennis and 6 play all the three games. Find the number of students who play (i) chess and carrom but not table tennis (ii) only chess (iii) only carrom (Hint: Use Venn diagram)
Answer:
Let \( C \) represent students who play Chess.
Let \( A \) represent students who play Carrom.
Let \( T \) represent students who play Table Tennis.

Given:
Total students, \( n(C \cup A \cup T) = 35 \) (since each student plays at least one game)
\( n(C) = 22 \)
\( n(A) = 21 \)
\( n(T) = 15 \)
\( n(C \cap T) = 10 \)
\( n(A \cap T) = 8 \)
\( n(C \cap A \cap T) = 6 \)

Let \( x \) be the number of students who play Chess and Carrom but not Table Tennis, i.e., \( n(\text{only } C \cap A) = x \).
We are asked to find \( x \), \( n(\text{only } C) \), and \( n(\text{only } A) \).

We'll fill the Venn diagram step-by-step:
1. **Intersection of all three:** \( n(C \cap A \cap T) = 6 \). Put 6 in the center region.

2. **Intersections of two sets (only that pair):**
* Only Chess and Table Tennis: \( n(C \cap T) - n(C \cap A \cap T) = 10 - 6 = 4 \)
* Only Carrom and Table Tennis: \( n(A \cap T) - n(C \cap A \cap T) = 8 - 6 = 2 \)
* Only Chess and Carrom (this is \( x \)): \( n(C \cap A) - n(C \cap A \cap T) = x \). The source shows the value 5 for this region, implying \( n(C \cap A) = 5 + 6 = 11 \). Let's use \( x \) as given in the question and solve for it.

3. **Only individual sets:**
* Only Chess: \( n(C) - (n(\text{only } C \cap T) + n(\text{only } C \cap A) + n(C \cap A \cap T)) \)
\( n(\text{only } C) = 22 - (4 + x + 6) = 22 - (10 + x) = 12 - x \)
* Only Carrom: \( n(A) - (n(\text{only } C \cap A) + n(\text{only } A \cap T) + n(C \cap A \cap T)) \)
\( n(\text{only } A) = 21 - (x + 2 + 6) = 21 - (8 + x) = 13 - x \)
* Only Table Tennis: \( n(T) - (n(\text{only } C \cap T) + n(\text{only } A \cap T) + n(C \cap A \cap T)) \)
\( n(\text{only } T) = 15 - (4 + 2 + 6) = 15 - 12 = 3 \)

Now, the sum of all these distinct regions must equal the total number of students \( n(C \cup A \cup T) \):
\( n(\text{only } C) + n(\text{only } A) + n(\text{only } T) + n(\text{only } C \cap T) + n(\text{only } A \cap T) + n(\text{only } C \cap A) + n(C \cap A \cap T) = 35 \)
\( (12 - x) + (13 - x) + 3 + 4 + 2 + x + 6 = 35 \)
Combine constant numbers:
\( 12 + 13 + 3 + 4 + 2 + 6 = 40 \)
Combine \( x \) terms:
\( -x - x + x = -x \)
So, the equation becomes:
\( 40 - x = 35 \)
Solve for \( x \):
\( x = 40 - 35 \)
\( x = 5 \)

Now we can find the required values:
(i) **Students who play Chess and Carrom but not Table Tennis:**
This is the value of \( x \), which is 5.

(ii) **Students who play only Chess:**
\( n(\text{only } C) = 12 - x = 12 - 5 = 7 \)

(iii) **Students who play only Carrom:**
\( n(\text{only } A) = 13 - x = 13 - 5 = 8 \)

Here's the Venn diagram with the calculated values:

The students who play chess and carrom but not table tennis are 5, only chess are 7, and only carrom are 8. This problem demonstrates a practical application of Venn diagrams to categorize and count participants in multiple activities.
In simple words: We used a Venn diagram to sort students into groups based on the games they play. We started by filling in the number of students who play all three games. Then, using the given numbers, we figured out how many students play only two games, and finally how many play only one game. We added everything up to find an unknown number and then solved the questions.

🎯 Exam Tip: When given partial information, especially involving "only X and Y" vs. "X and Y", be careful. \( n(X \cap Y) \) includes the three-set intersection, while "only X and Y" excludes it. Draw the Venn diagram and label each distinct region carefully with expressions involving the unknown \( x \).

Question 13. In a class of 50 students, each one come to school by bus or by bicycle or on foot. 25 by bus, 20 by bicycle, 30 on foot and 10 students by all the three. Now how many students come to school exactly by two modes of transport?
Answer:
Let \( B \) represent students who come by Bus.
Let \( C \) represent students who come by Bicycle.
Let \( D \) represent students who come by Foot (Drive).

Given:
Total number of students, \( n(B \cup C \cup D) = 50 \) (since each uses at least one mode)
\( n(B) = 25 \)
\( n(C) = 20 \)
\( n(D) = 30 \)
\( n(B \cap C \cap D) = 10 \)

Let \( x, y, \) and \( z \) be the number of students who come exactly by two modes of transport, i.e.,
\( x = n(\text{only } B \cap C) \)
\( y = n(\text{only } C \cap D) \)
\( z = n(\text{only } B \cap D) \)

We can represent this information using a Venn diagram and label the regions.

**Venn Diagram Regions:**
* Center (all three): 10
* Only \( B \cap C \): \( x \)
* Only \( C \cap D \): \( y \)
* Only \( B \cap D \): \( z \)

* Only Bus: \( n(B) - (x + z + 10) = 25 - x - z - 10 = 15 - x - z \)
* Only Bicycle: \( n(C) - (x + y + 10) = 20 - x - y - 10 = 10 - x - y \)
* Only Foot: \( n(D) - (y + z + 10) = 30 - y - z - 10 = 20 - y - z \)


The total number of students in the class is the sum of all these distinct regions:
\( n(B \cup C \cup D) = (15 - x - z) + (10 - x - y) + (20 - y - z) + x + y + z + 10 = 50 \)

Let's combine the constant terms:
\( 15 + 10 + 20 + 10 = 55 \)

Now, let's combine the \( x, y, \) and \( z \) terms:
\( -x - x + x = -x \)
\( -z - z + z = -z \)
\( -y - y + y = -y \)

So the equation becomes:
\( 55 - x - y - z = 50 \)
We want to find the total number of students who come exactly by two modes of transport, which is \( x + y + z \).
Rearrange the equation to solve for \( x + y + z \):
\( x + y + z = 55 - 50 \)
\( x + y + z = 5 \)

Therefore, the number of students who come to school exactly by two modes of transport is 5. This problem uses a Venn diagram to set up an equation that allows us to find the combined value of partially overlapping groups.
In simple words: We knew the total number of students and how many used each transport method, plus how many used all three. We drew a Venn diagram and labeled the unknown parts (those who use exactly two methods) as x, y, and z. By adding up all the parts of the diagram and setting it equal to the total students, we found that x + y + z, which is the number of students using exactly two methods, equals 5.

🎯 Exam Tip: When asked for the sum of "exactly two" overlaps, define these regions with distinct variables (like \( x, y, z \)). Then, write the expressions for the "only one" regions and the "all three" region. Sum all these distinct regions and equate to the total union to find the required sum.