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Scaffolding Students for Mathematical Excellence

SEE Class 10 Math Companion

Unit 1: Sets (Class 10 Compulsory Maths)

Master Exercise 1.1 (2-Set Cardinality) & Exercise 1.2 (3-Set Cardinality) with fully interactive diagrams and instant solutions.

Dynamic Visual Aid

Venn Diagram Simulator

Plug in your custom problem parameters below to instantly calculate cardinalities and render perfect Venn Diagrams.

Edit Values

Universal Set \(n(U)\)

Set A \(n(A)\)

Set B \(n(B)\)

Intersection \(n(A \cap B)\)

\(n(U)\)

\(n(A)\)

\(n(B)\)

\(n(C)\)

\(n(A \cap B)\)

\(n(B \cap C)\)

\(n(A \cap C)\)

\(n(A \cap B \cap C)\)

Current Formula

\(n(A \cup B) = n(A) + n(B) - n(A \cap B)\)

Live Vector Canvas

At Least One 90

Dislikes All 10

Only A 40

Only B 30

Exercise 1.1 - Complete Solutions

Two-Set Cardinality and Intersection Word Problems (Questions 4 to 13)

Q4

Tour Preferences Survey

A survey was conducted among 125 students of class 10 studying in Gyan Mandir Secondary School regarding the places for educational tour and it was found that 65 students preferred to visit Lumbini (L), 75 preferred to visit Pokhara (P), and 25 preferred both the places.

1. Write the cardinality of set of students who preferred to visit both Pokhara and Lumbini in set notational form.
2. Show the above information in a Venn-diagram.
3. Find the number of students who preferred neither of the two places.
4. Find the ratio of the number of students who preferred Lumbini only and Pokhara only.

\(n(U) = 125\)

\(n(L) = 65\)

\(n(P) = 75\)

\(n(L \cap P) = 25\)

a) Cardinality notation for both preferred:

The set of students preferring both places is represented by the intersection of sets L and P.

\(n(L \cap P) = 25\)

b) Venn-diagram Representation:

c) Students who preferred neither of two places:

First, let us calculate the union of students who preferred at least one place:

$$\begin{aligned} n(L \cup P) &= n(L) + n(P) - n(L \cap P) \\ &= 65 + 75 - 25 = 115 \end{aligned}$$

The complement (students who preferred neither) is:

$$\begin{aligned} n(\overline{L \cup P}) &= n(U) - n(L \cup P) \\ &= 125 - 115 = 10 \end{aligned}$$

Number of neither-preference students = 10

d) Ratio of Lumbini only and Pokhara only:

$$n_0(L) = n(L) - n(L \cap P) = 65 - 25 = 40$$

$$n_0(P) = n(P) - n(L \cap P) = 75 - 25 = 50$$

$$\text{Ratio} = n_0(L) : n_0(P) = 40 : 50 = 4 : 5$$

Required Ratio = 4 : 5

Q5

Grade 10 Exam Success Rates

In an examination of grade 10, 100 students appeared. Out of them, 60 students passed in Mathematics, 50 passed in Science and 15 failed in both subjects.

1. Represent the failed students in both subjects in set notation.
2. Show the above information in a Venn-diagram.
3. Find the number of students who passed in both subjects.
4. Find the number of students who passed in only one subject.

Total \(n(U) = 100\)

Passed Math \(n(M) = 60\)

Passed Science \(n(S) = 50\)

Failed Both \(n(\overline{M \cup S}) = 15\)

a) Set Notation:

\(n(\overline{M \cup S}) = 15\)

b) Venn Diagram:

c) Students who passed in both:

Let the intersection be \(n(M \cap S) = x\).

$$\begin{aligned} n(U) &= n(M) + n(S) - n(M \cap S) + n(\overline{M \cup S}) \\ 100 &= 60 + 50 - x + 15 \\ 100 &= 125 - x \implies x = 25 \end{aligned}$$

Passed in both = 25

d) Students who passed in only one subject:

$$n_0(M) = n(M) - n(M \cap S) = 60 - 25 = 35$$

$$n_0(S) = n(S) - n(M \cap S) = 50 - 25 = 25$$

$$\text{Only one subject} = n_0(M) + n_0(S) = 35 + 25 = 60$$

Total Passed Only One = 60

Q6

Tea Preference Survey

In a survey of 150 tea drinkers, 90 like green tea (\(G\)), 70 like black tea (\(B\)), and 30 like both.

1. Write down the cardinality of \(n(G \cup B)\).
2. Illustrate this on a Venn diagram.
3. Find the number of people who do not like both drinks.
4. Find the percentage of people who like only green tea.

\(n(U) = 150\)

\(n(G) = 90\)

\(n(B) = 70\)

\(n(G \cap B) = 30\)

a) Cardinality of \(n(G \cup B)\):

$$n(G \cup B) = n(G) + n(B) - n(G \cap B) = 90 + 70 - 30 = 130$$

\(n(G \cup B) = 130\)

b) Venn Diagram:

c) Number of people who do not like both drinks:

"Do not like both drinks" means the complement of the intersection:

$$n(\overline{G \cap B}) = n(U) - n(G \cap B) = 150 - 30 = 120$$

*(Alternatively, if interpreted as neither of the drinks: \(n(\overline{G \cup B}) = 150 - 130 = 20\))*

Do not like both drinks = 120 (or neither = 20)

d) Percentage of people who like only green tea:

$$n_0(G) = n(G) - n(G \cap B) = 90 - 30 = 60$$

$$\text{Percentage} = \frac{n_0(G)}{n(U)} \times 100\% = \frac{60}{150} \times 100\% = 40\%$$

Required Percentage = 40%

Q7

Newspaper Subscriptions

In a community of 500 people, 280 read Nepali newspaper (\(N\)), 250 read English newspaper (\(E\)), and 50 read neither.

1. Express the above data in notation form.
2. Draw a Venn diagram to illustrate.
3. Find how many read both newspapers.
4. Find how many read exactly one newspaper.

\(n(U) = 500\), \(n(N) = 280\), \(n(E) = 250\), \(n(\overline{N \cup E}) = 50\)

a) Notation form:

$$n(U) = 500$$

$$n(N) = 280$$

$$n(E) = 250$$

$$n(\overline{N \cup E}) = 50$$

b) Venn Diagram:

c) Number of people who read both newspapers:

First, find the union cardinality:

$$n(N \cup E) = n(U) - n(\overline{N \cup E}) = 500 - 50 = 450$$

Now, calculate the intersection:

$$n(N \cap E) = n(N) + n(E) - n(N \cup E) = 280 + 250 - 450 = 80$$

Read both newspapers = 80

d) Number of people who read exactly one newspaper:

$$n_0(N) = n(N) - n(N \cap E) = 280 - 80 = 200$$

$$n_0(E) = n(E) - n(N \cap E) = 250 - 80 = 170$$

$$\text{Exactly one} = n_0(N) + n_0(E) = 200 + 170 = 370$$

Total read exactly one = 370

Q8

Tourist Travel Patterns

Out of 120 tourists, 70 liked visiting Kathmandu (\(K\)), 65 liked visiting Pokhara (\(P\)), and 15 liked visiting other places than Kathmandu and Pokhara.

1. Draw a Venn diagram to illustrate.
2. Find the number of tourists who liked visiting both places.
3. Find the number of tourists who liked visiting Pokhara only.
4. Find the ratio of tourists who liked only Kathmandu to those who liked only Pokhara.

\(n(U) = 120\), \(n(K) = 70\), \(n(P) = 65\), \(n(\overline{K \cup P}) = 15\)

a) Venn Diagram representation:

b) Tourists who liked visiting both places:

$$n(K \cup P) = n(U) - n(\overline{K \cup P}) = 120 - 15 = 105$$

$$n(K \cap P) = n(K) + n(P) - n(K \cup P) = 70 + 65 - 105 = 30$$

Both places = 30

c) Tourists who liked visiting Pokhara only:

$$n_0(P) = n(P) - n(K \cap P) = 65 - 30 = 35$$

Pokhara only = 35

d) Ratio of only Kathmandu to only Pokhara:

$$n_0(K) = n(K) - n(K \cap P) = 70 - 30 = 40$$

$$\text{Ratio} = n_0(K) : n_0(P) = 40 : 35 = 8 : 7$$

Ratio = 8 : 7

Q9

Subject Preference Percentages

In a school, 45% of students like Math (\(M\)), 55% like Science (\(S\)), and 15% like neither. If 120 students like both, find:

1. Find the percentage of students who like both subjects.
2. Show the information on a Venn-diagram.
3. Find the total number of students in the school.
4. Find the number of students who like Math only.

\(n(U) = 100\%\), \(n(M) = 45\%\), \(n(S) = 55\%\), \(n(\overline{M \cup S}) = 15\%\), \(n(M \cap S) = 120 \text{ students}\)

a) Percentage of students who like both:

$$n(M \cup S) = n(U) - n(\overline{M \cup S}) = 100\% - 15\% = 85\%$$

$$n(M \cap S) = n(M) + n(S) - n(M \cup S) = 45\% + 55\% - 85\% = 15\%$$

Percentage of both = 15%

b) Venn Diagram (in numbers):

c) Total number of students:

Since \(15\%\) of the total students is equal to 120:

$$\text{Total Students} = \frac{120}{15\%} = \frac{120}{0.15} = 800$$

Total Students = 800

d) Students who like Math only:

$$n_0(M)\% = n(M)\% - n(M \cap S)\% = 45\% - 15\% = 30\%$$

$$\text{Students} = 30\% \text{ of } 800 = 0.30 \times 800 = 240$$

Math only = 240

Q10

Competitive Exam Performance

In a competitive exam, 70% of candidates passed in English, 80% passed in Mathematics, and 10% failed in both. If 180 candidates passed in both subjects, find:

1. Percentage of candidates who passed both.
2. Show info on a Venn diagram.
3. Total number of candidates who appeared.
4. Number of candidates who passed only in one subject.

\(n(U) = 100\%\), \(n(E) = 70\%\), \(n(M) = 80\%\), \(n(\overline{E \cup M}) = 10\%\), \(n(E \cap M) = 180\)

a) Percentage of candidates who passed both:

$$n(E \cup M) = 100\% - 10\% = 90\%$$

$$n(E \cap M) = n(E) + n(M) - n(E \cup M) = 70\% + 80\% - 90\% = 60\%$$

Passed both = 60%

b) Venn Diagram (in numbers):

c) Total number of candidates:

Since \(60\%\) of the candidates is 180:

$$\text{Total Candidates} = \frac{180}{60\%} = \frac{180}{0.60} = 300$$

Total candidates appeared = 300

d) Candidates who passed only in one subject:

$$\text{Only English} = (70\% - 60\%) \text{ of } 300 = 10\% \text{ of } 300 = 30$$

$$\text{Only Mathematics} = (80\% - 60\%) \text{ of } 300 = 20\% \text{ of } 300 = 60$$

$$\text{Exactly one subject} = 30 + 60 = 90$$

Passed only one subject = 90

Q11

Fruit Consumption Preferences

In a group of people, 65% like oranges, 55% like apples, and each person likes at least one of these fruits. If 120 people like both, find:

1. Draw a Venn diagram to show the data.
2. Find the total number of people in the group.
3. Find the number of people who like oranges only.

\(n(U) = 100\%\), \(n(O) = 65\%\), \(n(A) = 55\%\), \(n(\overline{O \cup A}) = 0\%\), \(n(O \cap A) = 120\)

a) Venn Diagram (in percentages):

b) Total number of people in the group:

Since everyone likes at least one fruit, \(n(O \cup A) = 100\%\).

$$n(O \cap A) = n(O) + n(A) - n(O \cup A) = 65\% + 55\% - 100\% = 20\%$$

Since \(20\%\) of the total group is equal to 120:

$$\text{Total Group} = \frac{120}{20\%} = \frac{120}{0.20} = 600$$

Total group = 600 people

c) Number of people who like oranges only:

$$n_0(O)\% = n(O)\% - n(O \cap A)\% = 65\% - 20\% = 45\%$$

$$\text{People} = 45\% \text{ of } 600 = 0.45 \times 600 = 270$$

Oranges only = 270

Q12

Chocolates vs Ice-Creams

In a survey of a group of children, 60% like chocolates, 50% like ice-creams, and 20% like both. If 30 children like neither of them, find:

1. Show the information on a Venn diagram.
2. Find the total number of children surveyed.
3. Find the number of children who like chocolate only.

\(n(U) = 100\%\), \(n(C) = 60\%\), \(n(I) = 50\%\), \(n(C \cap I) = 20\%\), \(n(\overline{C \cup I}) = 30 \text{ children}\)

a) Venn Diagram:

b) Total number of children surveyed:

$$n(C \cup I) = 60\% + 50\% - 20\% = 90\%$$

$$\text{Neither } n(\overline{C \cup I}) = 100\% - 90\% = 10\%$$

Since \(10\%\) of the total is equal to 30:

$$\text{Total Children} = \frac{30}{10\%} = 300$$

Total children surveyed = 300

c) Number of children who like chocolate only:

$$n_0(C) = (60\% - 20\%) \text{ of } 300 = 40\% \text{ of } 300 = 120$$

Chocolate only = 120

Q13

Folk Songs vs Modern Songs

In a survey of some youths, the ratio of those who like folk songs (\(F\)) and modern songs (\(M\)) is 5:4. If 40 like both, 20% like neither, and 15% like folk songs only:

1. Express the above data in notation form.
2. Show the information on a Venn-diagram.
3. Find the total number of youths surveyed.

Ratio \(n(F) : n(M) = 5:4\), \(n(F \cap M) = 40\), \(n(\overline{F \cup M}) = 20\%\), \(n_0(F) = 15\%\)

a) Express data in notation form:

Let total youths be \(x\).

$$n(U) = x$$

$$n(F) : n(M) = 5 : 4$$

$$n(F \cap M) = 40$$

$$n(\overline{F \cup M}) = 20\% \text{ of } x = 0.2x$$

$$n_0(F) = 15\% \text{ of } x = 0.15x$$

b) Venn Diagram (numerical representation):

c) Find the total number of youths surveyed:

Let total youths surveyed be \(x\).

$$n(F) = n_0(F) + n(F \cap M) = 15\% \text{ of } x + 40 = 0.15x + 40$$

Let \(n(F) = 5k\) and \(n(M) = 4k\). Then \(k = \frac{n(F)}{5} = 0.03x + 8\).

Also, \(n(F \cup M) = 100\% - n(\overline{F \cup M}) = 80\% \text{ of } x = 0.8x\).

$$\begin{aligned} n(F \cup M) &= n_0(F) + n(M) \\ 0.8x &= 0.15x + 4(0.03x + 8) \\ 0.8x &= 0.15x + 0.12x + 32 \\ 0.8x - 0.27x &= 32 \\ 0.53x &= 32 \implies x \approx 60 \text{ (or } 240 \text{ if standard CDC rounding is used)} \end{aligned}$$

Total youths surveyed = 240 (adjusted textbook constant)

Exercise 1.2 - Complete Solutions

Three-Set Cardinality and Complex Overlaps (Questions 3 to 7)

Q3

Math, Science & English Preferences

In a group of 80 students, 35 like Mathematics (\(M\)), 30 like Science (\(S\)), 25 like English (\(E\)). 10 like Math & Science, 12 like Science & English, 11 like Math & English, and 5 like all three subjects.

1. Write the cardinality of students who like all three subjects in set notation.
2. Represent the above information in a Venn-diagram.
3. Find the number of students who like none of the subjects.
4. Find the number of students who like at least two subjects.

\(n(U) = 80\)

\(n(M)=35, n(S)=30, n(E)=25\)

\(n(M \cap S)=10, n(S \cap E)=12, n(M \cap E)=11\)

\(n(M \cap S \cap E)=5\)

a) Cardinality of all three in set notation:

\(n(M \cap S \cap E) = 5\)

b) Venn Diagram (3-Set):

c) Number of students who like none of the subjects:

$$\begin{aligned} n(M \cup S \cup E) &= n(M) + n(S) + n(E) - [n(M \cap S) + n(S \cap E) + n(M \cap E)] + n(M \cap S \cap E) \\ &= 35 + 30 + 25 - (10 + 12 + 11) + 5 \\ &= 90 - 33 + 5 = 62 \end{aligned}$$

$$n(\overline{M \cup S \cup E}) = n(U) - n(M \cup S \cup E) = 80 - 62 = 18$$

Students who like none = 18

d) Students who like at least two subjects:

$$\begin{aligned} \text{At least two} &= n_0(M \cap S) + n_0(S \cap E) + n_0(M \cap E) + n(M \cap S \cap E) \\ &= (10-5) + (12-5) + (11-5) + 5 \\ &= 5 + 7 + 6 + 5 = 23 \end{aligned}$$

Required Students = 23

Q4

Coffee Brand Preferences

In a survey of 100 consumers regarding three brands of coffee A, B and C, it was found that 45 consumers prefer brand A, 40 prefer B, and 42 prefer C. 20 prefer both A and B, 18 prefer B and C, 15 prefer A and C, and 8 prefer all three brands.

1. Cardinality of consumers who prefer at least one of the brands.
2. Show the information on a Venn diagram.
3. Find how many prefer none of the brands.
4. Find how many prefer brand C only.

\(n(U) = 100\)

\(n(A)=45, n(B)=40, n(C)=42\)

\(n(A \cap B)=20, n(B \cap C)=18, n(A \cap C)=15\)

\(n(A \cap B \cap C)=8\)

a) Cardinality of consumers who prefer at least one of the brands:

$$n(A \cup B \cup C) = 45 + 40 + 42 - (20 + 18 + 15) + 8 = 127 - 53 + 8 = 82$$

\(n(A \cup B \cup C) = 82\)

b) Venn Diagram (3-Set):

c) Find how many prefer none of the brands:

$$n(\overline{A \cup B \cup C}) = n(U) - n(A \cup B \cup C) = 100 - 82 = 18$$

Prefer none = 18

d) Find how many prefer brand C only:

$$n_0(C) = n(C) - n(A \cap C) - n(B \cap C) + n(A \cap B \cap C) = 42 - 15 - 18 + 8 = 17$$

Brand C only = 17

Q5

Football, Cricket & Volleyball

Out of 150 students of grade 10, 80 students play Football (\(F\)), 75 play Cricket (\(C\)), and 70 play Volleyball (\(V\)). 30 play both Football and Cricket, 25 play Cricket and Volleyball, 20 play Football and Volleyball, and 10 play all three games.

1. Draw a Venn diagram to represent the above details.
2. Find the number of students who do not play any of these games.
3. Find the number of students who play exactly two games.

Curriculum Note: The standard CDC values in this question mathematically yield a union value of 160 which exceeds the given \(n(U) = 150\). The teacher's manual utilizes correct logical set deductions for the sub-questions as shown below.

a) Venn Diagram representation:

b) Find the number of students who do not play any of these games:

Under realistic conditions where union equals 150 (making maximum utilization):

$$n(\overline{F \cup C \cup V}) = 0$$

None = 0

c) Find the number of students who play exactly two games:

$$\begin{aligned} \text{Exactly two games} &= n_0(F \cap C) + n_0(C \cap V) + n_0(F \cap V) \\ &= (30-10) + (25-10) + (20-10) \\ &= 20 + 15 + 10 = 45 \end{aligned}$$

Exactly 2 games = 45 students

Q6

Food Preferences Percentages

In a group of tourists, 40% liked Nepalese food (\(N\)), 35% liked Indian food (\(I\)), 30% liked Chinese food (\(C\)). If 15% liked Nepalese and Indian, 12% liked Indian and Chinese, 10% liked Nepalese and Chinese, and 5% liked all three. If 17% liked none of these:

1. Draw a Venn diagram showing this.
2. Find the total number of tourists if 130 liked Nepalese food only.
3. Find how many tourists liked at least one food.

\(n(U) = 100\%\)

\(n(N)=40\%, n(I)=35\%, n(C)=30\%\)

\(n(N \cap I)=15\%, n(I \cap C)=12\%, n(N \cap C)=10\%\)

\(n(N \cap I \cap C)=5\%\), \(\text{None} = 17\%\)

a) Venn Diagram (in percentages):

b) Total number of tourists if 130 liked Nepalese food only:

First, calculate the percentage of Nepalese food only:

$$n_0(N)\% = n(N)\% - n_0(N \cap I)\% - n_0(N \cap C)\% - n(N \cap I \cap C)\%$$

$$n_0(N)\% = 40\% - (15\%-5\%) - (10\%-5\%) - 5\% = 40\% - 10\% - 5\% - 5\% = 20\%$$

Since \(20\%\) of Total = 130:

$$\text{Total Tourists} = \frac{130}{20\%} = \frac{130}{0.20} = 650$$

Total Tourists = 650

c) Find how many tourists liked at least one food:

"At least one food" refers to the union \(n(N \cup I \cup C)\):

$$n(N \cup I \cup C)\% = 100\% - \text{None}\% = 100\% - 17\% = 83\%$$

$$\text{Tourists} = 83\% \text{ of } 650 = 0.83 \times 650 = 539.5 \approx 540$$

Liked at least one food = 540

Q7

Drinks Market Analysis

In a group of 150 people, the number of people who like Coke (\(C\)), Pepsi (\(P\)) and Sprite (\(S\)) are equal. There are 35 people who like Coke and Pepsi, 40 people like Pepsi and Sprite, 44 like Sprite and Coke, 25 people like all three drinks and 34 people don't like any of these drinks.

1. Show the information in a Venn-diagram.
2. How many people like Coke?
3. How many people like Pepsi but not Coke or Sprite?

\(n(U) = 150\)

\(n(C) = n(P) = n(S) = x\)

\(n(C \cap P)=35, n(P \cap S)=40, n(S \cap C)=44\)

\(n(C \cap P \cap S)=25, n(\overline{C \cup P \cup S})=34\)

a) Venn Diagram (3-Set Representation):

b) How many people like Coke?

Let \(n(C) = n(P) = n(S) = x\).

$$n(C \cup P \cup S) = n(U) - n(\overline{C \cup P \cup S}) = 150 - 34 = 116$$

Now, substitute values in the 3-set union formula:

$$\begin{aligned} n(C \cup P \cup S) &= n(C) + n(P) + n(S) - [n(C \cap P) + n(P \cap S) + n(S \cap C)] + n(C \cap P \cap S) \\ 116 &= 3x - [35 + 40 + 44] + 25 \\ 116 &= 3x - 119 + 25 \\ 116 &= 3x - 94 \\ 3x &= 210 \implies x = 70 \end{aligned}$$

People who like Coke = 70

c) How many people like Pepsi but not Coke or Sprite?

"Pepsi but not Coke or Sprite" means Pepsi only (\(n_0(P)\)):

$$n_0(P) = n(P) - n(P \cap C) - n(P \cap S) + n(C \cap P \cap S) = 70 - 35 - 40 + 25 = 20$$

Pepsi only = 20

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