Mathematics · Chapter 01

88 solved questions

Chapter 1: Sets

Every question in this chapter, answered and explained — step-by-step solutions drawn live from the chapter library across 3 question sections.

Exercise 1.134 Exercise 1.231 Mixed Exercise23

Exercise 1.1

1Textbookex1.1-q1-a

Question 1 Present the cardinality of sets with examples and show it to your teacher.

This is an open, activity-based task: choose any two overlapping sets of real objects (e.g. students who play football and students who play basketball), list their elements, find $n (A)$ , $n (B)$ , $n (A \cap B)$ , $n (A \cup B)$ , and present this to the teacher.

2Textbookex1.1-q1-b

For two sets $A$ and $B$ , $A \subset B$ , find the values of $n (A \cup B)$ and $n (A \cap B)$ .

If $A \subset B$ (every element of $A$ is in $B$ ), then $A \cup B = B$ and $A \cap B = A$ . So $n (A \cup B) = n (B)$ and $n (A \cap B) = n (A)$ .

3Textbookex1.1-q1-c

If $A$ and $B$ are overlapping sets, state the formula for $n (A \cup B)$ .

For overlapping sets, $n (A \cup B) = n (A) + n (B) - n (A \cap B)$ .

4Textbookex1.1-q1-d

There are 12 and 8 elements in the sets $A$ and $B$ respectively. Find the minimum number of elements that would be in the set $n (A \cup B)$ .

$n (A \cup B) = n (A) + n (B) - n (A \cap B)$ . This is minimum when $n (A \cap B)$ is as large as possible; the largest possible intersection equals the size of the smaller set, i.e. $n (A \cap B) = 8$ (when $B \subset A$ ). So the minimum $n (A \cup B) = 12 + 8 - 8 = 12$ .

5Textbookex1.1-q2-a

In the given Venn diagram, 80 people are in set $M$ , 90 people are in set $E$ , and 15 people are not in both the sets. Determine the cardinality of the following sets. $n_{0} (M)$

From the given Venn diagram, the region for $M$ only shows $n_{0} (M) = 20$ .

6Textbookex1.1-q2-b

$n_{0} (E)$

From the given Venn diagram, the region for $E$ only shows $n_{0} (E) = 30$ .

7Textbookex1.1-q2-c

$n (M)$

$n (M) = n_{0} (M) + n (M \cap E) = 20 + 60 = 80$ (as given).

8Textbookex1.1-q2-d

$n (E)$

$n (E) = n_{0} (E) + n (M \cap E) = 30 + 60 = 90$ (as given).

9Textbookex1.1-q2-e

$n (M \cup E)$

$n (M \cup E) = n_{0} (M) + n (M \cap E) + n_{0} (E) = 20 + 60 + 30 = 110$ .

10Textbookex1.1-q2-f

$n (M \cap E)$

From the given Venn diagram, the overlap region shows $n (M \cap E) = 60$ .

11Textbookex1.1-q2-g

$n (M \cup E)$

From the given Venn diagram, the region outside both circles shows $n (M \cup E) = 15$ .

12Textbookex1.1-q2-h

$n (U)$

$n (U) = n (M \cup E) + n (M \cup E) = 110 + 15 = 125$ .

13Textbookex1.1-q3-a

Question 3 If $n (U) = 200$ , $n_{0} (M) = 2 x$ , $n_{0} (E) = 3 x$ , $n (M \cap E) = 60$ and $n (M \cup E) = 40$ , find the value of $x$ .

$n (U) = n_{0} (M) + n (M \cap E) + n_{0} (E) + n (M \cup E)$ : $200 = 2 x + 60 + 3 x + 40$ , so $200 = 5 x + 100$ , giving $5 x = 100$ , so $x = 20$ .

14Textbookex1.1-q3-b

If $n (U) = 350$ , $n (A) = 200$ , $n (B) = 220$ and $n (A \cap B) = 120$ , then find $n (A \cup B)$ and $n (A \cup B)$ .

$n (A \cup B) = n (A) + n (B) - n (A \cap B) = 200 + 220 - 120 = 300$ . Then $n (A \cup B) = n (U) - n (A \cup B) = 350 - 300 = 50$ .

15Textbookex1.1-q3-c

If $n (A) = 35$ and $n (A) = 25$ , then find the value of $n (U)$ .

$n (U) = n (A) + n (A) = 35 + 25 = 60$ .

16Textbookex1.1-q3-d

Out of two sets $P$ and $Q$ , there are 40 elements in $P$ , 60 elements in $(P \cup Q)$ and 10 elements in $(P \cap Q)$ . How many elements are there in $Q$ ? Find it.

$n (P \cup Q) = n (P) + n (Q) - n (P \cap Q)$ : $60 = 40 + n (Q) - 10$ , so $60 = 30 + n (Q)$ , giving $n (Q) = 30$ .

17Textbookex1.1-q4-a

Question 4 In a survey of 180 students of a school, 45 like Nepali only and 60 like English only but 15 like none of the subjects. Based on this information, answer: i) Show the above information in a Venn diagram. ii) Find the number of students who like both the subjects. iii) Find the number of students who like at least one subject.

i) Let $N$ = Nepali, $E$ = English. $n_{0} (N) = 45$ , $n_{0} (E) = 60$ , $n (N \cup E) = 15$ ; drawn as two overlapping circles inside $U$ . ii) $n (U) = n_{0} (N) + n (N \cap E) + n_{0} (E) + n (N \cup E)$ : $180 = 45 + n (N \cap E) + 60 + 15$ , so $n (N \cap E) = 180 - 120 = 60$ . iii) At least one subject $= n (N \cup E) = n (U) - n (N \cup E) = 180 - 15 = 165$ .

18Textbookex1.1-q4-b

In a survey among 1200 students of a school, 100 like Mathematics only and 200 like Science only but 700 like neither of the subjects. Based on the information, answer: i) Show the above information in a Venn diagram. ii) Find the number of students who like both the subjects. iii) Find the number of students who like at least one subject.

i) Let $M$ = Mathematics, $S$ = Science. $n_{0} (M) = 100$ , $n_{0} (S) = 200$ , $n (M \cup S) = 700$ ; drawn as two overlapping circles inside $U$ . ii) $n (U) = n_{0} (M) + n (M \cap S) + n_{0} (S) + n (M \cup S)$ : $1200 = 100 + n (M \cap S) + 200 + 700$ , so $n (M \cap S) = 1200 - 1000 = 200$ . iii) At least one subject $= n (U) - n (M \cup S) = 1200 - 700 = 500$ .

19Textbookex1.1-q4-c

In a survey among 60 students, 10 play football only and 20 play volleyball only but 12 play neither of the games. Based on the information, answer: i) Show the above information in a Venn diagram. ii) Find the number of students who play both the games. iii) Find the number of students who play at least one game.

i) Let $F$ = football, $V$ = volleyball. $n_{0} (F) = 10$ , $n_{0} (V) = 20$ , $n (F \cup V) = 12$ ; drawn as two overlapping circles inside $U$ . ii) $n (U) = n_{0} (F) + n (F \cap V) + n_{0} (V) + n (F \cup V)$ : $60 = 10 + n (F \cap V) + 20 + 12$ , so $n (F \cap V) = 60 - 42 = 18$ . iii) At least one game $= n (U) - n (F \cup V) = 60 - 12 = 48$ .

20Textbookex1.1-q5-a

Question 5 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: i) Show the information in a Venn diagram. ii) Find the number of people who read both the newspapers. iii) Find the number of people who read only one newspaper.

i) Let $M$ = Madhupark readers, $Y$ = Yubamanch readers. $n (U) = 900$ , $n (M) = 525$ , $n (Y) = 450$ , $n (M \cup Y) = 75$ . ii) $n (M \cup Y) = n (U) - n (M \cup Y) = 900 - 75 = 825$ . Also $n (M \cup Y) = n (M) + n (Y) - n (M \cap Y)$ : $825 = 525 + 450 - n (M \cap Y)$ , so $n (M \cap Y) = 975 - 825 = 150$ . iii) Only Madhupark $= n (M) - n (M \cap Y) = 525 - 150 = 375$ ; only Yubamanch $= 450 - 150 = 300$ ; only one newspaper $= 375 + 300 = 675$ .

21Textbookex1.1-q5-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: i) Show the information in a Venn diagram. ii) Find the number of people who like both the songs. iii) Find the number of people who like only modern songs.

i) Let $M o$ = modern songs, $F o$ = folk songs. $n (U) = 150$ , $n (M o) = 90$ , $n (F o) = 70$ , $n (M o \cup F o) = 30$ . ii) $n (M o \cup F o) = 150 - 30 = 120$ . $n (M o) + n (F o) - n (M o \cap F o) = 120$ : $90 + 70 - n (M o \cap F o) = 120$ , so $n (M o \cap F o) = 160 - 120 = 40$ . iii) Only modern songs $= n (M o) - n (M o \cap F o) = 90 - 40 = 50$ .

22Textbookex1.1-q5-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: i) Show the information in a Venn diagram. ii) Find the number of players who like to play both the games. iii) Find the number of people who like to play only one game.

i) Let $V$ = volleyball, $F$ = football. $n (U) = 360$ , $n (V) = 210$ , $n (F) = 180$ , $n (V \cup F) = 30$ . ii) $n (V \cup F) = 360 - 30 = 330$ . $210 + 180 - n (V \cap F) = 330$ , so $n (V \cap F) = 390 - 330 = 60$ . iii) Only one game $= n (V \cup F) - n (V \cap F) = 330 - 60 = 270$ .

23Textbookex1.1-q6-a

Question 6 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: i) Show the above information in a Venn diagram. ii) Find the total number of students participated in the examination. iii) Find how many students passed English only.

i) Let $E$ = English, $M$ = Mathematics, $n (U) = x$ (unknown). $n (E) = 0.7 x$ , $n (M) = 0.6 x$ , $n (E \cup M) = 0.2 x$ , $n (E \cap M) = 550$ . ii) $n (E \cup M) = n (U) - n (E \cup M) = x - 0.2 x = 0.8 x$ . Also $n (E \cup M) = n (E) + n (M) - n (E \cap M) = 0.7 x + 0.6 x - 550 = 1.3 x - 550$ . Equating: $0.8 x = 1.3 x - 550$ , so $550 = 0.5 x$ , giving $x = 1100$ . So $1100$ students participated. iii) English only $= n (E) - n (E \cap M) = 0.7 (1100) - 550 = 770 - 550 = 220$ .

24Textbookex1.1-q6-b

According to a survey of students who have appeared for 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: i) Show the above information in a Venn diagram. ii) Find how many students participated in the survey. iii) Find the number of students who are interested to study Science only.

i) Let $S$ = Science, $M g$ = Management, $n (U) = x$ . $n (S) = 0.6 x$ , $n (M g) = 0.7 x$ , $n (S \cup M g) = 0.1 x$ , $n (S \cap M g) = 400$ . ii) $n (S \cup M g) = x - 0.1 x = 0.9 x$ . Also $= 0.6 x + 0.7 x - 400 = 1.3 x - 400$ . Equating: $0.9 x = 1.3 x - 400$ , so $400 = 0.4 x$ , giving $x = 1000$ . iii) Science only $= 0.6 (1000) - 400 = 600 - 400 = 200$ .

25Textbookex1.1-q6-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: i) Show the above information in a Venn diagram. ii) Find how many people participated in the survey. iii) Find the number of people who ride motorcycles only.

i) Let $M t$ = motorcycle, $S c$ = scooter. Since $20%$ of the total ride both, and $200$ people ride both: $0.2 x = 200$ , giving the total $n (U) = x = 1000$ . ii) So $1000$ people participated in the survey. iii) $n (M t) = 65%$ of $1000 = 650$ . Motorcycle only $= n (M t) - n (M t \cap S c) = 650 - 200 = 450$ .

26Textbookex1.1-q7-a

Question 7 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 both tea and coffee but 15 do not drink either tea or coffee. Based on the information, answer: i) Show the information in a Venn diagram. ii) Find the number of people who drink exactly one of tea or coffee. iii) Find the number of people who drink at least one; either tea or coffee.

i) Let $T$ = tea, $C$ = coffee. $n (U) = 95$ , $n (T) : n (C) = 4 : 5$ so let $n (T) = 4 x$ , $n (C) = 5 x$ ; $n (T \cap C) = 10$ ; $n (T \cup C) = 15$ . ii) $n (U) = n_{0} (T) + n (T \cap C) + n_{0} (C) + n (T \cup C)$ , i.e. $(4 x - 10) + 10 + (5 x - 10) + 15 = 95$ , so $9 x + 5 = 95$ , giving $9 x = 90$ , so $x = 10$ . Then $n (T) = 40$ , $n (C) = 50$ , $n_{0} (T) = 40 - 10 = 30$ , $n_{0} (C) = 50 - 10 = 40$ . Exactly one $= n_{0} (T) + n_{0} (C) = 30 + 40 = 70$ . iii) At least one $= n (U) - n (T \cup C) = 95 - 15 = 80$ .

27Textbookex1.1-q7-b

In a survey of 64 students of a class, the ratio of the number of students who like milk only and curd only is 2:1 whereas 16 like both. Based on this information, answer: i) Show the above information in a Venn diagram. ii) Find the number of students who like milk. iii) Find the number of students who like only one kind of drink.

i) Let $M k$ = milk, $C d$ = curd. $n (U) = 64$ , $n_{0} (M k) : n_{0} (C d) = 2 : 1$ so let $n_{0} (M k) = 2 y$ , $n_{0} (C d) = y$ ; $n (M k \cap C d) = 16$ . ii) $n (U) = n_{0} (M k) + n (M k \cap C d) + n_{0} (C d)$ : $64 = 2 y + 16 + y$ , so $3 y = 48$ , giving $y = 16$ . So $n_{0} (M k) = 32$ , $n_{0} (C d) = 16$ . $n (M k) = n_{0} (M k) + n (M k \cap C d) = 32 + 16 = 48$ . iii) Only one kind of drink $= n_{0} (M k) + n_{0} (C d) = 32 + 16 = 48$ .

28Textbookex1.1-q7-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: i) Show the above information in a Venn diagram. ii) Find how many people do not do both genres. iii) Find the number of people who do one genre at most.

i) Let $S$ = sing, $D$ = dance. $n (U) = 320$ , $n_{0} (S) = 60$ , $n_{0} (D) = 100$ . Let both $= n (S \cap D) = b$ ; then those who don't do both $= n (U) - b = 320 - b$ , and this equals $3 b$ : $320 - b = 3 b$ , so $320 = 4 b$ , giving $b = 80$ . ii) People who don't do both $= 3 b = 3 (80) = 240$ . iii) At most one genre $= n (U) - n (S \cap D) = 320 - 80 = 240$ .

29Textbookex1.1-q8-i

According to a survey of 200 people of a community, it was found that the ratio of the number of people who use laptop only and mobile only is 2:3, among them, 30% use both but 15% does not use both the gadgets. Based on this information, answer: Show the above information in a Venn diagram.

Let $L$ = laptop, $M b$ = mobile. $n (U) = 200$ , $n_{0} (L) : n_{0} (M b) = 2 : 3$ so let $n_{0} (L) = 2 y$ , $n_{0} (M b) = 3 y$ ; $n (L \cap M b) = 30%$ of $200 = 60$ ; $n (L \cup M b) = 15%$ of $200 = 30$ . From $n (U) = n_{0} (L) + n (L \cap M b) + n_{0} (M b) + n (L \cup M b)$ : $200 = 2 y + 60 + 3 y + 30$ , so $5 y + 90 = 200$ , giving $5 y = 110$ , so $y = 22$ . Thus $n_{0} (L) = 44$ , $n_{0} (M b) = 66$ .

30Textbookex1.1-q8-ii

Find the number of people who use laptop.

$n (L) = n_{0} (L) + n (L \cap M b) = 44 + 60 = 104$ .

31Textbookex1.1-q8-iii

Find how many people use one gadget at most.

One gadget at most $= n (U) - n (L \cap M b) = 200 - 60 = 140$ (equivalently, $n_{0} (L) + n_{0} (M b) + n (L \cup M b) = 44 + 66 + 30 = 140$ ).

32Textbookex1.1-q9

Out of 300 players in a survey, one-third players play volleyball only. 60% of the remaining players play football only. But 60 players do not play both. Then, find the ratio of the number of players who play volleyball and football by using the Venn diagram.

Let $V$ = volleyball, $F$ = football. $n (U) = 300$ . Volleyball only $= \frac{1}{3} \times 300 = 100$ . Remaining players $= 300 - 100 = 200$ ; football only $= 60%$ of $200 = 120$ . Not playing both (neither) $= 60$ . So both $= n (U) - [n_{0} (V) + n_{0} (F) + n (V \cup F)] = 300 - (100 + 120 + 60) = 20$ . Then $n (V) = n_{0} (V) + both = 100 + 20 = 120$ and $n (F) = n_{0} (F) + both = 120 + 20 = 140$ . Ratio $n (V) : n (F) = 120 : 140 = 6 : 7$ .

33Textbookex1.1-q10

Among 65 players participated in a survey, 11 play volleyball only and 33 play cricket only. If the number of players who play cricket is the double of the number of players who play volleyball, find the number of players who play both and the number of players who does not play both by using Venn diagram.

Let $V$ = volleyball, $C r$ = cricket, and both $= b$ . Let $n (V) = v$ , so $n (C r) = 2 v$ . $n_{0} (V) = v - b = 11$ and $n_{0} (C r) = 2 v - b = 33$ . Subtracting: $(2 v - b) - (v - b) = 33 - 11$ , so $v = 22$ . Then $b = v - 11 = 22 - 11 = 11$ . So $n (V) = 22$ , $n (C r) = 44$ . Total playing at least one $= n_{0} (V) + b + n_{0} (C r) = 11 + 11 + 33 = 55$ . Players who do not play both $= n (U) - n (V \cup C r) = 65 - 55 = 10$ .

34Textbookex1.1-q11

In a survey of 80 people, 60 like orange only and 10 like both orange and apple. The number of people who like orange is 5 times the number of people who like apple. By using the Venn diagram, find the number of people who like apples only and those who do not like both the fruits.

Note (assumption used): read literally, "60 like orange" together with "orange total = 5 x apple total" and "neither = ?" does not reproduce the textbook's published answer; taking the 60 as orange-ONLY ( $n_{0} (O) = 60$ ) makes the problem fully consistent and matches the answer key, so that reading is used below. Let $O$ = orange, $A$ = apple. $n (U) = 80$ , $n_{0} (O) = 60$ , $n (O \cap A) = 10$ , and $n (O) = 5 \times n (A)$ . Then $n (O) = n_{0} (O) + n (O \cap A) = 60 + 10 = 70$ . So $n (A) = \frac{n ( O )}{5} = \frac{70}{5} = 14$ . Apple only $n_{0} (A) = n (A) - n (O \cap A) = 14 - 10 = 4$ . People who like neither fruit $= n (U) - [n_{0} (O) + n (O \cap A) + n_{0} (A)] = 80 - (60 + 10 + 4) = 6$ .

Exercise 1.2

1Textbookex1.2-q1-a

In the given Venn diagram, the elements of sets $P$ , $Q$ and $R$ are illustrated. Based on this, find the values of the following sets. $n (P)$

Counting all elements inside circle $P$ (including overlaps): $n (P) = 7$ . Note: this question is answered by directly counting the labelled elements inside each region of the given diagram. The individual element labels in the scanned figure were not all fully legible, so the region counts below are taken to match the textbook's own published answer key rather than re-listed element-by-element; the method (count elements in each region, or in the union) is otherwise identical to Example 1 of section 1.2.

2Textbookex1.2-q1-b

$n (Q)$

Counting all elements inside circle $Q$ (including overlaps): $n (Q) = 6$ .

3Textbookex1.2-q1-c

$n (P \cup Q \cup R)$

Counting every element inside at least one of the three circles: $n (P \cup Q \cup R) = 14$ .

4Textbookex1.2-q1-d

$n_{0} (P)$

Counting elements only in $P$ (not in $Q$ or $R$ ): $n_{0} (P) = 4$ .

5Textbookex1.2-q1-e

$n_{0} (R)$

Counting elements only in $R$ : $n_{0} (R) = 3$ .

6Textbookex1.2-q1-f

$n (P \cap R)$

Counting elements common to $P$ and $R$ (including the triple overlap): $n (P \cap R) = 2$ .

7Textbookex1.2-q1-g

$n (P \cup Q \cup R)$

Counting elements outside all three circles: $n (P \cup Q \cup R) = 4$ .

8Textbookex1.2-q1-h

$n_{0} (P \cap Q)$

Counting elements in $P \cap Q$ only, excluding the triple overlap: $n_{0} (P \cap Q) = 1$ .

9Textbookex1.2-q1-i

$n (P \cap Q \cap R)$

Counting the element common to all three circles: $n (P \cap Q \cap R) = 1$ .

10Textbookex1.2-q2-a

If $U = {$ positive integers less than 30 $}$ , $P = {$ multiples of 2 less than 30 $}$ , $Q = {$ multiples of 3 less than 30 $}$ and $R = {$ multiples of 5 less than 30 $}$ , show the relation between the sets $P$ , $Q$ and $R$ in a Venn diagram and verify the following relations: $n (P \cup Q) = n (P) + n (Q) - n (P \cap Q)$

RHS $= n (P) + n (Q) - n (P \cap Q) = 14 + 9 - 4 = 19$ . Directly, $P \cup Q$ has $14 + 9 - 4 = 19$ elements (no double-counting of the $4$ common multiples of 6). So LHS $=$ RHS $= 19$ ; verified. First list the sets. $P = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28}$ , $n (P) = 14$ . $Q = {3, 6, 9, 12, 15, 18, 21, 24, 27}$ , $n (Q) = 9$ . $R = {5, 10, 15, 20, 25}$ , $n (R) = 5$ . $P \cap Q$ = multiples of 6 $= {6, 12, 18, 24}$ , $n (P \cap Q) = 4$ . $Q \cap R$ = multiples of 15 $= {15}$ , $n (Q \cap R) = 1$ . $P \cap R$ = multiples of 10 $= {10, 20}$ , $n (P \cap R) = 2$ . $P \cap Q \cap R$ = multiples of 30 (less than 30) $= {}$ , $n (P \cap Q \cap R) = 0$ .

11Textbookex1.2-q2-b

$n (P \cup Q \cup R) = n (P) + n (Q) + n (R) - n (P \cap Q) - n (Q \cap R) - n (R \cap P) + n (P \cap Q \cap R)$

RHS $= n (P) + n (Q) + n (R) - n (P \cap Q) - n (Q \cap R) - n (R \cap P) + n (P \cap Q \cap R) = 14 + 9 + 5 - 4 - 1 - 2 + 0 = 21$ . Directly listing $P \cup Q \cup R$ (all multiples of 2, 3 or 5 below 30) also gives $21$ elements. So LHS $=$ RHS $= 21$ ; verified.

12Textbookex1.2-q2-c

$n (P \cup Q \cup R) = n (P - Q) + n (Q - R) + n (R - P) + n (P \cap Q \cap R)$

$n (P - Q)$ = elements of $P$ not in $Q$ $= 14 - 4 = 10$ . $n (Q - R)$ = elements of $Q$ not in $R$ $= 9 - 1 = 8$ . $n (R - P)$ = elements of $R$ not in $P$ $= 5 - 2 = 3$ . RHS $= 10 + 8 + 3 + 0 = 21 = n (P \cup Q \cup R)$ ; verified.

13Textbookex1.2-q3-a

Question 3 If $n (U) = 100$ , $n (M) = 45$ , $n (E) = 50$ , $n (S) = 35$ , $n (M \cap E) = 20$ , $n (E \cap S) = 20$ , $n (S \cap M) = 15$ and $n (M \cap E \cap S) = 5$ , then find $n (M \cup E \cup S)$ .

$n (M \cup E \cup S) = n (M) + n (E) + n (S) - n (M \cap E) - n (E \cap S) - n (S \cap M) + n (M \cap E \cap S) = 45 + 50 + 35 - 20 - 20 - 15 + 5 = 80$ . So $n (M \cup E \cup S) = n (U) - n (M \cup E \cup S) = 100 - 80 = 20$ .

14Textbookex1.2-q3-b

If $n (U) = 105$ , $n (A) = 40$ , $n (B) = 35$ , $n (C) = 30$ , $n (A \cap B) = 15$ , $n (B \cap C) = 12$ , $n (A \cap B \cap C) = 6$ and $n (A \cup B \cup C) = 30$ , then find $n (A \cap C)$ .

$n (A \cup B \cup C) = n (U) - n (A \cup B \cup C) = 105 - 30 = 75$ . Also $n (A \cup B \cup C) = n (A) + n (B) + n (C) - n (A \cap B) - n (B \cap C) - n (A \cap C) + n (A \cap B \cap C)$ : $75 = 40 + 35 + 30 - 15 - 12 - n (A \cap C) + 6 = 84 - n (A \cap C)$ . So $n (A \cap C) = 84 - 75 = 9$ .

15Textbookex1.2-q3-c

If $n (U) = 120$ , $n (M) = 50$ , $n (E) = 40$ , $n (S) = 45$ , $n (M \cap E) = 15$ , $n (E \cap S) = 15$ , $n (S \cap M) = 15$ and $n (M \cup E \cup S) = 15$ , then find $n (M \cap E \cap S)$ .

$n (M \cup E \cup S) = n (U) - n (M \cup E \cup S) = 120 - 15 = 105$ . Also $105 = 50 + 40 + 45 - 15 - 15 - 15 + n (M \cap E \cap S) = 90 + n (M \cap E \cap S)$ . So $n (M \cap E \cap S) = 105 - 90 = 15$ .

16Textbookex1.2-q3-d

If $n (A \cup B \cup C) = 105$ , $n_{0} (A) = 25$ , $n_{0} (B) = 25$ , $n_{0} (C) = 15$ , $n_{0} (A \cap B) = 15$ , $n_{0} (A \cap C) = 10$ and $n (A \cap B \cap C) = 10$ , then find $n_{0} (B \cap C)$ .

Using $n (A \cup B \cup C) = n_{0} (A) + n_{0} (B) + n_{0} (C) + n_{0} (A \cap B) + n_{0} (B \cap C) + n_{0} (A \cap C) + n (A \cap B \cap C)$ : $105 = 25 + 25 + 15 + 15 + n_{0} (B \cap C) + 10 + 10 = 100 + n_{0} (B \cap C)$ . So $n_{0} (B \cap C) = 5$ .

17Textbookex1.2-q4-a

Question 4 Out of 90 students who participated in an examination, 43 passed in Science, 40 in Mathematics and 38 in Nepali. Among them, 13 passed in Science and Mathematics, 18 in Mathematics and Nepali as well as 16 passed in Science and Nepali. Using the information, answer: i) Show the information in the Venn diagram. ii) Find the number of students who did not pass in any subject.

Flag: the problem as given does not state how many students passed all three subjects, so the triple intersection cannot be derived uniquely from the stated data; the natural default (no student passed all three, i.e. $n (S \cap M \cap N) = 0$ ) is used below. i) Let $S$ =Science, $M$ =Mathematics, $N$ =Nepali. $n (U) = 90$ , $n (S) = 43$ , $n (M) = 40$ , $n (N) = 38$ , $n (S \cap M) = 13$ , $n (M \cap N) = 18$ , $n (S \cap N) = 16$ , $n (S \cap M \cap N) = 0$ (assumed). ii) $n (S \cup M \cup N) = n (S) + n (M) + n (N) - n (S \cap M) - n (M \cap N) - n (S \cap N) + n (S \cap M \cap N) = 43 + 40 + 38 - 13 - 18 - 16 + 0 = 74$ . Students who did not pass any subject $= n (U) - n (S \cup M \cup N) = 90 - 74 = 16$ .

18Textbookex1.2-q4-b

In a survey of a group, 60 like tea, 45 like coffee, 30 like milk, 25 like coffee and tea, 20 like milk and tea, 15 like coffee and milk and 10 like all three drinks. Based on the information, answer: i) Show the information in a Venn diagram. ii) Find how many people were surveyed.

i) Let $T$ =tea, $C$ =coffee, $M k$ =milk. $n (T) = 60$ , $n (C) = 45$ , $n (M k) = 30$ , $n (C \cap T) = 25$ , $n (M k \cap T) = 20$ , $n (C \cap M k) = 15$ , $n (T \cap C \cap M k) = 10$ . ii) $n (T \cup C \cup M k) = n (T) + n (C) + n (M k) - n (C \cap T) - n (M k \cap T) - n (C \cap M k) + n (T \cap C \cap M k) = 60 + 45 + 30 - 25 - 20 - 15 + 10 = 85$ . So $85$ people were surveyed (assuming everyone surveyed liked at least one drink).

19Textbookex1.2-q4-c

In a survey among 60 students, 23 played volleyball, 15 played basketball and 20 played cricket. If 7 played volleyball and basketball, 5 played basketball and cricket, 4 played volleyball and cricket but 15 played neither of the games. Based on this information, answer: i) Show the information in a Venn diagram. ii) Find how many students played all the three games. iii) How many played only volleyball and cricket?

i) Let $V$ =volleyball, $B$ =basketball, $C r$ =cricket. $n (V) = 23$ , $n (B) = 15$ , $n (C r) = 20$ , $n (V \cap B) = 7$ , $n (B \cap C r) = 5$ , $n (V \cap C r) = 4$ , $n (V \cup B \cup C r) = 15$ , $n (U) = 60$ . ii) $n (V \cup B \cup C r) = n (U) - 15 = 45$ . Also $n (V \cup B \cup C r) = n (V) + n (B) + n (C r) - n (V \cap B) - n (B \cap C r) - n (V \cap C r) + n (V \cap B \cap C r)$ : $45 = 23 + 15 + 20 - 7 - 5 - 4 + x = 42 + x$ , so $x = n (V \cap B \cap C r) = 3$ . iii) Only volleyball and cricket $= n_{0} (V \cap C r) = n (V \cap C r) - x = 4 - 3 = 1$ .

20Textbookex1.2-q5-i

Out of total students who participated in an examination, 40% passed in Science, 45% in Mathematics and 50% in Nepali. Similarly, 10% passed in Science and Mathematics, 20% in Mathematics and Nepali as well as 15% in Science and Nepali but 5% failed in all the three subjects. Based on the information, answer the following questions: Find the percentage of students who passed in all the three subjects.

Let $S$ =Science, $M$ =Maths, $N$ =Nepali (all in \%). $n (S) = 40$ , $n (M) = 45$ , $n (N) = 50$ , $n (S \cap M) = 10$ , $n (M \cap N) = 20$ , $n (S \cap N) = 15$ , failed-all (neither) $= 5$ , so $n (S \cup M \cup N) = 100 - 5 = 95$ . Using $n (S \cup M \cup N) = n (S) + n (M) + n (N) - n (S \cap M) - n (M \cap N) - n (S \cap N) + x$ : $95 = 40 + 45 + 50 - 10 - 20 - 15 + x = 90 + x$ , so $x = n (S \cap M \cap N) = 5%$ .

21Textbookex1.2-q5-ii

Find the percentage of students who passed in only one subject.

Only Science: $n_{0} (S) = n (S) - n (S \cap M) - n (S \cap N) + x = 40 - 10 - 15 + 5 = 20%$ . Only Maths: $n_{0} (M) = 45 - 10 - 20 + 5 = 20%$ . Only Nepali: $n_{0} (N) = 50 - 20 - 15 + 5 = 20%$ . Only one subject $= 20 + 20 + 20 = 60%$ .

22Textbookex1.2-q5-iii

Find the percentage of students who passed in only two subjects.

Only two subjects $=$ at least one $-$ only one $-$ all three $= 95 - 60 - 5 = 30%$ .

23Textbookex1.2-q5-iv

Find the percentage of students who passed in at least one subject.

At least one subject $= n (S \cup M \cup N) = 95%$ .

24Textbookex1.2-q5-v

Show the information in a Venn diagram.

The Venn diagram has: only $S = 20%$ , $S \cap M$ only $= 5%$ , only $M = 20%$ , all three $= 5%$ , $S \cap N$ only $= 10%$ , $M \cap N$ only $= 15%$ , only $N = 20%$ , outside $= 5%$ .

25Textbookex1.2-q6-a

The following information was obtained from a survey on a questionnaire whether they read Yubamanch or Madhupark or Muna conducted among some people in a community: 30 read Yubamanch, 25 read Madhupark, 15 read both Yubamanch and Muna, 9 read Madhupark only, 11 read Muna only, 5 read Yubamanch and Madhupark only but 10 read neither of the newspapers. Based on this information, answer the following questions: Show the information in a Venn diagram.

Let $Y$ =Yubamanch, $M$ =Madhupark, $N$ =Muna. $n (Y) = 30$ , $n (M) = 25$ , $n (Y \cap N) = 15$ , $n_{0} (M) = 9$ , $n_{0} (N) = 11$ , $n_{0} (Y \cap M) = 5$ , $n (Y \cup M \cup N) = 10$ . Let $n (Y \cap M \cap N) = t$ . Then only- $Y$ $= n (Y) - n_{0} (Y \cap M) - n (Y \cap N) = 30 - 5 - 15 + t$ ... solved algebraically in part b) below; the Venn diagram is filled once $t$ is found. Flag: the scanned source also mentions a figure of "12" for Yubamanch-and-Madhupark that is inconsistent with the rest of the data (likely an OCR duplication of the "5 read Yubamanch and Madhupark only" figure); it is not used below, and the values used give a fully consistent system matching the textbook's own answer key.

26Textbookex1.2-q6-b

Find the total number of people who participated in the survey.

Let $t = n (Y \cap M \cap N)$ . Only- $Y$ $= n (Y) - n_{0} (Y \cap M) - n (Y \cap N) = 30 - 5 - 15 = 10$ (this is already exclusive of $M$ and $N$ overlaps other than what's subtracted, since $n (Y \cap N)$ already includes $t$ ). Only $Y \cap N$ (excluding triple) $= n (Y \cap N) - t = 15 - t$ . Only $M \cap N$ (excluding triple) $= n_{0} (M)$ is Madhupark-only $= 9$ , so total for $M$ : $n (M) = n_{0} (M) + n_{0} (Y \cap M) + [only M \cap N] + t$ , i.e. $25 = 9 + 5 + [only M \cap N] + t$ , giving only- $M \cap N = 11 - t$ . Total surveyed $= [only Y = 10] + [n_{0} (M) = 9] + [n_{0} (N) = 11] + [n_{0} (Y \cap M) = 5] + [15 - t] + [11 - t] + t + 10 = 71 - t$ . From the textbook's answer, total $= 64$ , so $t = 7$ .

27Textbookex1.2-q6-c

Find the number of people who read exactly two newspapers.

With $t = 7$ : only $Y \cap N = 15 - 7 = 8$ , only $M \cap N = 11 - 7 = 4$ . Exactly two newspapers $= n_{0} (Y \cap M) + [only Y \cap N] + [only M \cap N] = 5 + 8 + 4 = 17$ .

28Textbookex1.2-q6-d

Find the number of people who read Muna.

$n (N) = n_{0} (N) + [only Y \cap N] + [only M \cap N] + t = 11 + 8 + 4 + 7 = 30$ .

29Textbookex1.2-q7-a

In a survey among 90 people, who were asked which language film they like, 48 like Nepali, 40 like English, 31 like Hindi, 24 like Nepali and English, 19 like Hindi and English, 6 like all the three languages and 21 did not like any. Then, How many people did like both Nepali and Hindi films?

Let $N p$ =Nepali, $E$ =English, $H$ =Hindi. $n (U) = 90$ , $n (N p) = 48$ , $n (E) = 40$ , $n (H) = 31$ , $n (N p \cap E) = 24$ , $n (H \cap E) = 19$ , $n (N p \cap H \cap E) = 6$ , $n (N p \cup H \cup E) = 21$ . First $n (N p \cup H \cup E) = 90 - 21 = 69$ . Using $n (N p \cup H \cup E) = n (N p) + n (E) + n (H) - n (N p \cap E) - n (H \cap E) - n (N p \cap H) + n (N p \cap H \cap E)$ : $69 = 48 + 40 + 31 - 24 - 19 - n (N p \cap H) + 6 = 82 - n (N p \cap H)$ . So $n (N p \cap H) = 82 - 69 = 13$ .

30Textbookex1.2-q7-b

How many people did not like Hindi films?

People who did not like Hindi $= n (U) - n (H) = 90 - 31 = 59$ .

31Textbookex1.2-q7-c

How many people did not like both Nepali and Hindi films?

People who did not like both Nepali and Hindi $= n (U) - n (N p \cap H) = 90 - 13 = 77$ .

Mixed Exercise

1Textbookmixed-q1-a

There are two overlapping sets $A$ and $B$ shown alongside in a Venn diagram where $n_{0} (A) = 16 + x$ , $n_{0} (B) = 5 x$ , $n (A \cap B) = y$ and $n (A \cup B) = x$ . Then, answer the following questions: Insert the above information by drawing a Venn diagram.

The Venn diagram shows two overlapping circles $A$ and $B$ inside $U$ : the $A$ -only region is $16 + x$ , the overlap $A \cap B$ is $y$ , the $B$ -only region is $5 x$ , and the region outside both circles is $x$ .

2Textbookmixed-q1-b

If $n (A) = n (B)$ , find the value of $n (A \cup B)$ .

$n (A) = n_{0} (A) + y = 16 + x + y$ and $n (B) = n_{0} (B) + y = 5 x + y$ . Setting $n (A) = n (B)$ : $16 + x + y = 5 x + y$ , so $16 = 4 x$ , giving $x = 4$ . Since $n (A \cup B) = x$ , $n (A \cup B) = 4$ .

3Textbookmixed-q1-c

If $n (U) = 50$ , find the ratio of $n (A \cap B)$ and $n (A \cup B)$ .

Continuing with the condition $n (A) = n (B)$ from part (b), so $x = 4$ . Since $n (U) = 50$ : $n_{0} (A) + n (A \cap B) + n_{0} (B) + n (A \cup B) = 50$ , i.e. $(16 + x) + y + 5 x + x = 50$ , so $16 + 7 x + y = 50$ , giving $7 x + y = 34$ . With $x = 4$ : $28 + y = 34$ , so $y = 6$ . Ratio $n (A \cap B) : n (A \cup B) = y : x = 6 : 4 = 3 : 2$ . (Note: this part is only solvable by continuing to use the part (b) condition $n (A) = n (B)$ , since the diagram alone does not fix both $x$ and $y$ ; this is flagged as an assumption needed to match a determinate answer.)

4Textbookmixed-q2-a

$A$ and $B$ are the subsets of a universal set $U$ such that $n (U) = 100$ , $n (A - B) = 32 + x$ , $n (B - A) = 5 x$ , $n (A \cap B) = x$ and $n (A \cup B) = y$ . Show the above information in a Venn diagram.

The Venn diagram shows: $A$ -only region $= 32 + x$ , overlap $A \cap B = x$ , $B$ -only region $= 5 x$ , and outside both $= y$ , all inside $U$ with $n (U) = 100$ .

5Textbookmixed-q2-b

If $n (A) = n (B)$ , find the value of $n (A \cap B)$ .

$n (A) = n (A - B) + n (A \cap B) = (32 + x) + x = 32 + 2 x$ . $n (B) = n (B - A) + n (A \cap B) = 5 x + x = 6 x$ . Setting $n (A) = n (B)$ : $32 + 2 x = 6 x$ , so $32 = 4 x$ , giving $x = 8$ . Since $n (A \cap B) = x$ , $n (A \cap B) = 8$ .

6Textbookmixed-q2-c

Find the value of $n (A \cup B)$ . By what percent is $n (A \cap B)$ more or less than $n (A \cup B)$ ? Find it.

Continuing with $x = 8$ from part (b). Since $n (U) = 100$ : $(32 + x) + x + 5 x + y = 100$ , i.e. $32 + 7 x + y = 100$ . With $x = 8$ : $32 + 56 + y = 100$ , so $y = 12$ , i.e. $n (A \cup B) = 12$ . Comparing $n (A \cap B) = 8$ with $n (A \cup B) = 12$ : $n (A \cap B)$ is less than $n (A \cup B)$ by $\frac{12 - 8}{12} \times 100% \approx 33.3%$ ; equivalently, $n (A \cup B)$ is $50%$ more than $n (A \cap B)$ (since $\frac{12 - 8}{8} \times 100% = 50%$ ).

7Textbookmixed-q3-a

According to a survey of 93 women of a community, the number of women engaged in agriculture is 80 and that in sewing is 71 but the number of women engaged in other jobs is 10. Present the information in a Venn diagram by finding the cardinality of sets.

Let $A g$ = agriculture, $S w$ = sewing. $n (U) = 93$ , $n (A g) = 80$ , $n (S w) = 71$ , $n (A g \cup S w) = 10$ (engaged in other jobs only, i.e. neither agriculture nor sewing). So $n (A g \cup S w) = 93 - 10 = 83$ .

8Textbookmixed-q3-b

Find how many women were engaged in both agriculture and sewing.

$n (A g \cup S w) = n (A g) + n (S w) - n (A g \cap S w)$ : $83 = 80 + 71 - n (A g \cap S w)$ , so $n (A g \cap S w) = 151 - 83 = 68$ .

9Textbookmixed-q3-c

By how many times the number of women engaged in agriculture only is more than the number of women engaged in sewing only? Calculate it.

Agriculture only $= n (A g) - n (A g \cap S w) = 80 - 68 = 12$ . Sewing only $= n (S w) - n (A g \cap S w) = 71 - 68 = 3$ . Ratio $= \frac{12}{3} = 4$ , so agriculture only is $4$ times sewing only.

10Textbookmixed-q4-a

According to a survey of 1000 farmers in a community, the number of farmers cultivating potatoes was 800 and the number of farmers cultivating tomatoes was 500 but 50 farmers cultivated crops other than these. Show the information in a Venn diagram by finding the cardinality of sets.

Let $P$ =potato, $T$ =tomato. $n (U) = 1000$ , $n (P) = 800$ , $n (T) = 500$ , $n (P \cup T) = 50$ (cultivate other crops only). So $n (P \cup T) = 1000 - 50 = 950$ .

11Textbookmixed-q4-b

Find the number of farmers who cultivate both.

$n (P \cup T) = n (P) + n (T) - n (P \cap T)$ : $950 = 800 + 500 - n (P \cap T)$ , so $n (P \cap T) = 1300 - 950 = 350$ .

12Textbookmixed-q4-c

Write the number of farmers who cultivate potato only and that of tomato only in ratio.

Potato only $= n (P) - n (P \cap T) = 800 - 350 = 450$ . Tomato only $= n (T) - n (P \cap T) = 500 - 350 = 150$ . Ratio $= 450 : 150 = 3 : 1$ .

13Textbookmixed-q5-a

In a survey of 400 people of a community, it was found that the ratio of the people having motorcycle license only and car license only was 5:3. Among them, one-fourth of the people had license of both vehicles but 60 did not have any license. Show the above information in a Venn diagram.

Let $M t$ =motorcycle, $C$ =car. $n (U) = 400$ , $n_{0} (M t) : n_{0} (C) = 5 : 3$ so let $n_{0} (M t) = 5 k$ , $n_{0} (C) = 3 k$ ; $n (M t \cap C) = 100$ (see note); $n (M t \cup C) = 60$ . Assumption: "one-fourth of the people had license of vehicles" is read as "one-fourth of the total had license of both vehicles", i.e. $n (M t \cap C) = \frac{1}{4} \times 400 = 100$ ; this reading reproduces the textbook's published answers exactly.

14Textbookmixed-q5-b

From the above information, how many people had license of each vehicle?

From $n (U) = n_{0} (M t) + n (M t \cap C) + n_{0} (C) + n (M t \cup C)$ : $400 = 5 k + 100 + 3 k + 60$ , so $8 k = 240$ , giving $k = 30$ . So $n_{0} (M t) = 150$ , $n_{0} (C) = 90$ . Motorcycle license $n (M t) = n_{0} (M t) + n (M t \cap C) = 150 + 100 = 250$ . Car license $n (C) = n_{0} (C) + n (M t \cap C) = 90 + 100 = 190$ . So $250$ people had motorcycle licenses and $190$ had car licenses.

15Textbookmixed-q5-c

Find the number of people who had license of motorcycle.

From part (b), $n (M t) = 250$ .

16Textbookmixed-q6-a

The information of the students of a school whether they like volleyball, football or cricket is as follows: 30 like volleyball and football, 20 like volleyball and cricket as well as 35 like football and cricket. 10 like all the three games; football, volleyball and cricket but 5 like neither of the games. Represent the given information in cardinality of sets.

Let $V$ =volleyball, $F$ =football, $C r$ =cricket. Given: $n (V \cap F) = 30$ , $n (V \cap C r) = 20$ , $n (F \cap C r) = 35$ , $n (V \cap F \cap C r) = 10$ , $n (V \cup F \cup C r) = 5$ . So only-two overlaps are $n_{0} (V \cap F) = 30 - 10 = 20$ , $n_{0} (V \cap C r) = 20 - 10 = 10$ , $n_{0} (F \cap C r) = 35 - 10 = 25$ . Flag: as transcribed, this problem gives only the three pairwise-overlap totals, the triple overlap, and the "neither" count — it does not state $n (V)$ , $n (F)$ , $n (C r)$ individually (the count in each sport on its own), which the standard formula needs to pin down a unique total. The scanned source is therefore missing information here (most likely each sport's own total was cut off in the OCR). The figures below reproduce the textbook's own published answers (230 and 19.57%) as the final results, with the part of the method that is derivable from the given data shown explicitly; the individual only-one-sport counts implied by the diagram could not be independently re-derived from the data as transcribed.

17Textbookmixed-q6-b

Show the information in a Venn diagram.

The Venn diagram places $10$ at the centre (all three), $20$ , $10$ , $25$ in the three pairwise-only regions, and $5$ outside all three circles; the three single-sport-only regions are filled once the individual sport totals are known (see the flag above).

18Textbookmixed-q6-c

Find the total number of students in the school.

Per the textbook's published answer key, the total number of students in the school is $230$ (this could not be independently re-derived from the data as transcribed — see the flag above).

19Textbookmixed-q6-d

What percentage of students like football only?

Per the textbook's published answer key, football only is approximately $19.57%$ of the total students (consistent with a total of $230$ and football-only $\approx 45$ students).

20Textbookmixed-q7-a

The following information from a survey of 45 people of different lingual groups of a community is obtained: 25 speak Nepal Bhasa, 23 speak Tamang and 15 speak Maithili. 12 speak Nepal Bhasa and Tamang, 5 speak Nepal Bhasa and Maithili as well as 10 speak Tamang and Maithili. 4 speak all the three languages. Based on the information, answer the following questions: Show the above information in a Venn diagram.

Let $N b$ =Nepal Bhasa, $T a$ =Tamang, $M a$ =Maithili. $n (U) = 45$ , $n (N b) = 25$ , $n (T a) = 23$ , $n (M a) = 15$ , $n (N b \cap T a) = 12$ , $n (N b \cap M a) = 5$ , $n (T a \cap M a) = 10$ , $n (N b \cap T a \cap M a) = 4$ .

21Textbookmixed-q7-b

Find how many people speak the language other than these languages; Nepal Bhasa, Tamang and Maithili.

$n (N b \cup T a \cup M a) = n (N b) + n (T a) + n (M a) - n (N b \cap T a) - n (T a \cap M a) - n (N b \cap M a) + n (N b \cap T a \cap M a) = 25 + 23 + 15 - 12 - 10 - 5 + 4 = 40$ . People speaking some other language $= n (U) - n (N b \cup T a \cup M a) = 45 - 40 = 5$ .

22Textbookmixed-q7-c

How many people speak only one language? Find.

Only Nepal Bhasa: $n_{0} (N b) = n (N b) - n (N b \cap T a) - n (N b \cap M a) + n (N b \cap T a \cap M a) = 25 - 12 - 5 + 4 = 12$ . Only Tamang: $n_{0} (T a) = 23 - 12 - 10 + 4 = 5$ . Only Maithili: $n_{0} (M a) = 15 - 5 - 10 + 4 = 4$ . Only one language $= 12 + 5 + 4 = 21$ .

23Textbookmixed-q7-d

How many people speak both Nepal Bhasa and Tamang but do not speak the Maithili language?

Nepal Bhasa and Tamang but not Maithili $= n_{0} (N b \cap T a) = n (N b \cap T a) - n (N b \cap T a \cap M a) = 12 - 4 = 8$ .

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