Part 2

Question 5

a) A survey was carried out among 900 people of a community. According to the survey, 525 read Madhupark, 450 read Yubamanch but 75 didn't read either of the newspapers. Using the information, answer the following questions:

1. Show the information in a Venn-diagram.
2. Find the number of people who read both the newspapers.
3. Find the number of people who read only one newspaper.

Solution:

i) Venn Diagram:

ii) Finding the number of people who read both newspapers:

Given information:

• Total people: $n(U) = 900$
• People who read Madhupark: $n(M) = 525$
• People who read Yubamanch: $n(Y) = 450$
• People who read neither: $n(\overline{M \cup Y}) = 75$

First, find the number of people who read at least one newspaper:

$n(M \cup Y) = n(U) - n(\overline{M \cup Y}) = 900 - 75 = 825$

Using the inclusion-exclusion principle:

$n(M \cup Y) = n(M) + n(Y) - n(M \cap Y)$

Substitute the known values:

$825 = 525 + 450 - n(M \cap Y)$

Simplify:

$825 = 975 - n(M \cap Y)$

Solve for $n(M \cap Y)$:

$n(M \cap Y) = 975 - 825 = 150$

Therefore, 150 people read both newspapers.

iii) Finding the number of people who read only one newspaper:

People who read only Madhupark:

$n_o(M) = n(M) - n(M \cap Y) = 525 - 150 = 375$

People who read only Yubamanch:

$n_o(Y) = n(Y) - n(M \cap Y) = 450 - 150 = 300$

Total people who read only one newspaper:

$n_o(M) + n_o(Y) = 375 + 300 = 675$

Verification:

$n_o(M) + n_o(Y) + n(M \cap Y) + n(\overline{M \cup Y}) = 375 + 300 + 150 + 75 = 900$ ✓

b) According to a survey among 150 people, 90 like modern songs, 70 like folk songs but 30 do not like either of the songs. Using the information, answer the following questions:

1. Show the information in a Venn-diagram.
2. Find the number of people who like both the songs.
3. Find the number of people who like only modern songs.

Solution:

i) Venn Diagram:

ii) Finding the number of people who like both songs:

Given information:

• Total people: $n(U) = 150$
• People who like modern songs: $n(M) = 90$
• People who like folk songs: $n(F) = 70$
• People who like neither: $n(\overline{M \cup F}) = 30$

First, find the number of people who like at least one song:

$n(M \cup F) = n(U) - n(\overline{M \cup F}) = 150 - 30 = 120$

Using the inclusion-exclusion principle:

$n(M \cup F) = n(M) + n(F) - n(M \cap F)$

Substitute the known values:

$120 = 90 + 70 - n(M \cap F)$

Simplify:

$120 = 160 - n(M \cap F)$

Solve for $n(M \cap F)$:

$n(M \cap F) = 160 - 120 = 40$

Therefore, 40 people like both songs.

iii) Finding the number of people who like only modern songs:

People who like only modern songs:

$n_o(M) = n(M) - n(M \cap F) = 90 - 40 = 50$

Therefore, 50 people like only modern songs.

Verification:

Let's verify our answer:

• People who like only modern songs: $n_o(M) = 50$
• People who like only folk songs: $n_o(F) = n(F) - n(M \cap F) = 70 - 40 = 30$
• People who like both songs: $n(M \cap F) = 40$
• People who like neither: $n(\overline{M \cup F}) = 30$

$n_o(M) + n_o(F) + n(M \cap F) + n(\overline{M \cup F}) = 50 + 30 + 40 + 30 = 150$ ✓

c) According to a survey among 360 players, 210 liked to play volleyball, 180 liked to play football but 30 liked to play neither of the games. Using the information, answer the following questions:

1. Show the information in a Venn-diagram.
2. Find the number of players who like to play both the games.
3. Find the number of people who like to play only one game.

Solution:

i) Venn Diagram:

ii) Finding the number of players who like both games:

Given information:

• Total players: $n(U) = 360$
• Players who like volleyball: $n(V) = 210$
• Players who like football: $n(F) = 180$
• Players who like neither: $n(\overline{V \cup F}) = 30$

First, find the number of players who like at least one game:

$n(V \cup F) = n(U) - n(\overline{V \cup F}) = 360 - 30 = 330$

Using the inclusion-exclusion principle:

$n(V \cup F) = n(V) + n(F) - n(V \cap F)$

Substitute the known values:

$330 = 210 + 180 - n(V \cap F)$

Simplify:

$330 = 390 - n(V \cap F)$

Solve for $n(V \cap F)$:

$n(V \cap F) = 390 - 330 = 60$

Therefore, 60 players like both games.

iii) Finding the number of players who like only one game:

Players who like only volleyball:

$n_o(V) = n(V) - n(V \cap F) = 210 - 60 = 150$

Players who like only football:

$n_o(F) = n(F) - n(V \cap F) = 180 - 60 = 120$

Total players who like only one game:

$n_o(V) + n_o(F) = 150 + 120 = 270$

Therefore, 270 players like only one game.

Verification:

Let's verify our answer:

• Players who like only volleyball: $n_o(V) = 150$
• Players who like only football: $n_o(F) = 120$
• Players who like both games: $n(V \cap F) = 60$
• Players who like neither: $n(\overline{V \cup F}) = 30$

$n_o(V) + n_o(F) + n(V \cap F) + n(\overline{V \cup F}) = 150 + 120 + 60 + 30 = 360$ ✓

Question 6

a) Out of the students who participated in an examination, 70% passed English, 60% passed Mathematics but 20% failed both the subjects and 550 students passed both the subjects. Based on the information, answer the following questions:

1. Show the above information in a Venn-diagram.
2. Find the total number of students participated in the examination.
3. Find how many students passed English only.

Solution:

ii) Finding the total number of students:

Let the total number of students be $n$.

Given information:

• Students who passed English: $n(E) = 70\% \text{ of } n = 0.7n$
• Students who passed Mathematics: $n(M) = 60\% \text{ of } n = 0.6n$
• Students who failed both: $n(\overline{E \cup M}) = 20\% \text{ of } n = 0.2n$
• Students who passed both: $n(E \cap M) = 550$

Since 20% failed both subjects, 80% passed at least one subject:

$n(E \cup M) = 80\% \text{ of } n = 0.8n$

i) Venn Diagram:

Using the inclusion-exclusion principle:

$n(E \cup M) = n(E) + n(M) - n(E \cap M)$

Substitute the known values:

$0.8n = 0.7n + 0.6n - 550$

Simplify:

$0.8n = 1.3n - 550$

Rearrange:

$550 = 1.3n - 0.8n$ $550 = 0.5n$

Solve for $n$:

$n = \frac{550}{0.5} = 1100$

Therefore, 1100 students participated in the examination.

iii) Finding students who passed English only:

Now that we know $n = 1100$:

• Students who passed English: $n(E) = 0.7 \times 1100 = 770$
• Students who passed both: $n(E \cap M) = 550$

Students who passed English only:

$n_o(E) = n(E) - n(E \cap M) = 770 - 550 = 220$

Therefore, 220 students passed English only.

Verification:

Let's verify our answer:

• Students who passed English only: $n_o(E) = 220$
• Students who passed Mathematics only: $n_o(M) = 0.6 \times 1100 - 550 = 660 - 550 = 110$
• Students who passed both: $n(E \cap M) = 550$
• Students who failed both: $n(\overline{E \cup M}) = 0.2 \times 1100 = 220$

$n_o(E) + n_o(M) + n(E \cap M) + n(\overline{E \cup M}) = 220 + 110 + 550 + 220 = 1100$ ✓

b) According to a survey of students who have appeared the examination of grade 10, 60% are interested to study Science, 70% are interested to study Management but 10% rejected in the interest to study both Science and Management whilst 400 students are interested to study both Science and Management. Based on this information, answer the following questions:

1. Show the above information in a Venn-diagram.
2. Find how many students were participated in the survey.
3. Find the number of students who are interested to study Science only.

Solution:

ii) Finding the total number of students in the survey:

Let the total number of students be $n$.

Given information:

• Students interested in Science: $n(S) = 60\% \text{ of } n = 0.6n$
• Students interested in Management: $n(M) = 70\% \text{ of } n = 0.7n$
• Students who rejected both: $n(\overline{S \cup M}) = 10\% \text{ of } n = 0.1n$
• Students interested in both: $n(S \cap M) = 400$

Since 10% rejected both subjects, 90% are interested in at least one subject:

$n(S \cup M) = 90\% \text{ of } n = 0.9n$

i) Venn Diagram:

Using the inclusion-exclusion principle:

$n(S \cup M) = n(S) + n(M) - n(S \cap M)$

Substitute the known values:

$0.9n = 0.6n + 0.7n - 400$

Simplify:

$0.9n = 1.3n - 400$

Rearrange:

$400 = 1.3n - 0.9n$ $400 = 0.4n$

Solve for $n$:

$n = \frac{400}{0.4} = 1000$

Therefore, 1000 students participated in the survey.

iii) Finding students interested in Science only:

Now that we know $n = 1000$:

• Students interested in Science: $n(S) = 0.6 \times 1000 = 600$
• Students interested in both: $n(S \cap M) = 400$

Students interested in Science only:

$n_o(S) = n(S) - n(S \cap M) = 600 - 400 = 200$

Therefore, 200 students are interested to study Science only.

Verification:

Let's verify our answer:

• Students interested in Science only: $n_o(S) = 200$
• Students interested in Management only: $n_o(M) = 0.7 \times 1000 - 400 = 700 - 400 = 300$
• Students interested in both: $n(S \cap M) = 400$
• Students who rejected both: $n(\overline{S \cup M}) = 0.1 \times 1000 = 100$

$n_o(S) + n_o(M) + n(S \cap M) + n(\overline{S \cup M}) = 200 + 300 + 400 + 100 = 1000$ ✓

c) In a survey among people of a community, 65% ride motorcycle, 35% ride scooter but 20% ride both whereas 200 people ride both motorcycle and scooter. Using this information, answer the following questions:

1. Show the above information in a Venn-diagram.
2. Find how many people were participated in the survey.
3. Find the number of people who ride motorcycles only.

Solution:

ii) Finding the total number of people in the survey:

Let the total number of people be $n$.

Given information:

• People who ride motorcycle: $n(M) = 65\% \text{ of } n = 0.65n$
• People who ride scooter: $n(S) = 35\% \text{ of } n = 0.35n$
• People who ride both (percentage): $n(M \cap S) = 20\% \text{ of } n = 0.2n$
• People who ride both (actual number): $n(M \cap S) = 200$

Since we have both the percentage and actual number for people who ride both:

$0.2n = 200$

i) Venn Diagram:

Solve for $n$:

$n = \frac{200}{0.2} = 1000$

Therefore, 1000 people participated in the survey.

iii) Finding people who ride motorcycles only:

Now that we know $n = 1000$:

• People who ride motorcycle: $n(M) = 0.65 \times 1000 = 650$
• People who ride both: $n(M \cap S) = 200$

People who ride motorcycles only:

$n_o(M) = n(M) - n(M \cap S) = 650 - 200 = 450$

Therefore, 450 people ride motorcycles only.

Verification:

Let's verify our answer:

• People who ride motorcycles only: $n_o(M) = 450$
• People who ride scooters only: $n_o(S) = 0.35 \times 1000 - 200 = 350 - 200 = 150$
• People who ride both: $n(M \cap S) = 200$
• People who ride neither: $n(\overline{M \cup S}) = 1000 - (450 + 150 + 200) = 200$

$n_o(M) + n_o(S) + n(M \cap S) + n(\overline{M \cup S}) = 450 + 150 + 200 + 200 = 1000$ ✓

Additional verification with percentages:

• People who ride at least one vehicle: $\frac{800}{1000} = 80\%$
• People who ride neither: $\frac{200}{1000} = 20\%$

Question 7

a) Among 95 people of a community, it was surveyed that the ratio of the number of people who drink tea and coffee is 4:5, whereas 10 people drink tea only but 15 do not drink either tea or coffee. Based on the information, answer the following questions:

1. Show the information in a Venn-diagram.
2. Find the number of people who drinks exactly one of tea or coffee.
3. Find the number of people who drinks at least one; either tea or coffee.

Solution:

Finding the values first:

Given information:

• Total people: $n(U) = 95$
• Ratio of tea drinkers to coffee drinkers: $n(T) : n(C) = 4 : 5$
• People who drink tea only: $n_o(T) = 10$
• People who drink neither: $n(\overline{T \cup C}) = 15$

Let $n(T) = 4k$ and $n(C) = 5k$ for some constant $k$.

i) Venn Diagram:

Since people who drink tea only = total tea drinkers - people who drink both:

$n_o(T) = n(T) - n(T \cap C)$ $10 = 4k - n(T \cap C)$ $n(T \cap C) = 4k - 10$

Using the universal set formula:

$n(U) = n_o(T) + n_o(C) + n(T \cap C) + n(\overline{T \cup C})$

We know $n_o(C) = n(C) - n(T \cap C) = 5k - (4k - 10) = k + 10$

Substitute into the universal set formula:

$95 = 10 + (k + 10) + (4k - 10) + 15$ $95 = 25 + 5k$ $70 = 5k$ $k = 14$

Therefore:

• $n(T) = 4k = 4(14) = 56$
• $n(C) = 5k = 5(14) = 70$
• $n(T \cap C) = 4k - 10 = 56 - 10 = 46$
• $n_o(C) = k + 10 = 14 + 10 = 24$

ii) Finding people who drink exactly one of tea or coffee:

People who drink exactly one beverage:

$n_o(T) + n_o(C) = 10 + 24 = 34$

Therefore, 34 people drink exactly one of tea or coffee.

iii) Finding people who drink at least one beverage:

People who drink at least one beverage:

$n(T \cup C) = n(U) - n(\overline{T \cup C}) = 95 - 15 = 80$

Alternatively:

$n(T \cup C) = n_o(T) + n_o(C) + n(T \cap C) = 10 + 24 + 46 = 80$

Therefore, 80 people drink at least one beverage.

Verification:

Let's verify our answer:

• People who drink tea only: $n_o(T) = 10$
• People who drink coffee only: $n_o(C) = 24$
• People who drink both: $n(T \cap C) = 46$
• People who drink neither: $n(\overline{T \cup C}) = 15$

$n_o(T) + n_o(C) + n(T \cap C) + n(\overline{T \cup C}) = 10 + 24 + 46 + 15 = 95$ ✓

Ratio verification:

$n(T) : n(C) = 56 : 70 = 4 : 5$ ✓

Additional information:

• Total mobile users: $n(M) = n_o(M) + n(T \cap C) = 56 + 46 = 102$
• People who use at least one beverage: $n(T \cup C) = 80$
• Percentage using at least one beverage: $\frac{80}{95} \approx 84.21\%$

b) In a survey of 64 students of a class, the ratio of number of students who like milk only and curd only is 2:1 whereas 16 like both. Based on this information, answer the following questions:

1. Show the above information in a Venn-diagram.
2. Find the number of students who like milk.
3. Find the number of students who like only one kind of drink.

Solution:

Finding the values first:

Given information:

• Total students: $n(U) = 64$
• Ratio of milk only to curd only: $n_o(M) : n_o(C) = 2 : 1$
• Students who like both: $n(M \cap C) = 16$

Let $n_o(M) = 2k$ and $n_o(C) = k$ for some constant $k$.

i) Venn Diagram:

Assuming all students like at least one drink (since no information about "neither" is given):

$n(U) = n_o(M) + n_o(C) + n(M \cap C) + n(\overline{M \cup C})$

With $n(\overline{M \cup C}) = 0$:

$64 = 2k + k + 16 + 0$
$64 = 3k + 16$
$48 = 3k$
$k = 16$

Therefore:

• Students who like milk only: $n_o(M) = 2k = 2(16) = 32$
• Students who like curd only: $n_o(C) = k = 16$
• Students who like both: $n(M \cap C) = 16$
• Students who like neither: $n(\overline{M \cup C}) = 0$

ii) Finding the number of students who like milk:

Total students who like milk:

$n(M) = n_o(M) + n(M \cap C) = 32 + 16 = 48$

Therefore, 48 students like milk.

iii) Finding students who like only one kind of drink:

Students who like only one kind of drink:

$n_o(M) + n_o(C) = 32 + 16 = 48$

Therefore, 48 students like only one kind of drink.

Verification:

Let's verify our answer:

• Students who like milk only: $n_o(M) = 32$
• Students who like curd only: $n_o(C) = 16$
• Students who like both: $n(M \cap C) = 16$
• Students who like neither: $n(\overline{M \cup C}) = 0$

$n_o(M) + n_o(C) + n(M \cap C) + n(\overline{M \cup C}) = 32 + 16 + 16 + 0 = 64$ ✓

Ratio verification:

$n_o(M) : n_o(C) = 32 : 16 = 2 : 1$ ✓

Additional information:

• Total students who like curd: $n(C) = n_o(C) + n(M \cap C) = 16 + 16 = 32$
• Students who like at least one drink: $n(M \cup C) = 64 - 0 = 64$

c) In a conference of 320 participants, it was surveyed that 60 participants only sing and 100 only dance. If the number of people who do not do both is three times the number of people who do both. With the help of this information, answer the following questions:

1. Show the above information in a Venn-diagram.
2. Find how many people do not do both genres.
3. Find the number of people who do one genre at most.

Solution:

Finding the values first:

Given information:

• Total participants: $n(U) = 320$
• People who only sing: $n_o(S) = 60$
• People who only dance: $n_o(D) = 100$
• People who do not do both = 3 × (people who do both)

Let $n(S \cap D) = x$ (people who do both activities).

Then the number of people who do not do both = $3x$.

i) Venn Diagram:

Note: "People who do not do both" means people who do neither activity.

So $n(\overline{S \cup D}) = 3x$

Using the universal set formula:

$n(U) = n_o(S) + n_o(D) + n(S \cap D) + n(\overline{S \cup D})$

Substitute the known values:

$320 = 60 + 100 + x + 3x$
$320 = 160 + 4x$
$160 = 4x$
$x = 40$

Therefore:

• People who do both: $n(S \cap D) = 40$
• People who do neither: $n(\overline{S \cup D}) = 3x = 3(40) = 120$

ii) Finding people who do not do both genres:

People who do not do both genres means people who do neither activity:

$n(\overline{S \cup D}) = 120$

Therefore, 120 people do not do both genres.

iii) Finding people who do one genre at most:

People who do one genre at most means people who do:

• Only singing, OR
• Only dancing, OR
• Neither activity

This excludes people who do both activities.

$\text{People who do one genre at most} = n_o(S) + n_o(D) + n(\overline{S \cup D})$
$= 60 + 100 + 120 = 280$

Alternatively:

$\text{People who do one genre at most} = n(U) - n(S \cap D) = 320 - 40 = 280$

Therefore, 280 people do one genre at most.

Verification:

Let's verify our answer:

• People who only sing: $n_o(S) = 60$
• People who only dance: $n_o(D) = 100$
• People who do both: $n(S \cap D) = 40$
• People who do neither: $n(\overline{S \cup D}) = 120$

$n_o(S) + n_o(D) + n(S \cap D) + n(\overline{S \cup D}) = 60 + 100 + 40 + 120 = 320$ ✓

Relationship verification:

$n(\overline{S \cup D}) = 3 \times n(S \cap D)$ $120 = 3 \times 40 = 120$ ✓

Additional information:

• Total singers: $n(S) = n_o(S) + n(S \cap D) = 60 + 40 = 100$
• Total dancers: $n(D) = n_o(D) + n(S \cap D) = 100 + 40 = 140$
• People who do at least one activity: $n(S \cup D) = 320 - 120 = 200$