Subject: Compulsory Mathematics

The number of distinct element in a given set A is called the cardinal number of A. It is denoted by n(A). If A = { 1, 2, 3, 4, 5 }, then the cardinality of set A is denoted by n (A) = 5.

The cardinality of set A is defined as the number of elements in the set A and is denoted by n(A).

For example, if A = {a,b,c,d,e} then cardinality of set A i.e.n(A) = 5

Let A and B are two subsets of a universal set U. Their relation can be shown in Venn-diagram as:

$$ n(A) = n_o( A) + n(A \cap B)$$

$$\text{or,}\: n(A) - n (A \cap B)= n_o(A)$$

$$ n(B) = n_o(B) + n(A \cap B)$$

$$\text {or,}\: n(B) - n(A \cap B) = n_o(B)$$

Also,

\begin{align*} n(A∪B) &= n_o(A) + n(A∩B) + n_o(B)\\ n(A∪B) &= n(A) - n(A∩B) + n(A∩B) + n(B) - n(A∩B)\\ n(A∪B) &= n(A) + n(B)- n(A∩B)\\ \therefore n(A∪B) &= n(A) + n(B) - n(A∩B)\\ \end{align*}

If A and B are disjoint sets then:

\(n(A \cap B) = 0, n(A \cup B) =n(A) + n(B)\)

Again,

\(n(U) = n(A \cup B) + n(\overline {A\cup B)}\)

If \(n(\overline {A \cup B)}\)=0, then \( n(U) = n(A \cup B)\)

Let A, B and C are three non-empty and intersecting sets, then:

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

In Venn-diagram

\(n(A)\) = Number of elements in set A.

\(n(B)\) = Number of elements in set B.

\(n(C)\)=Number of element in set C.

\(n_o(A)\) = Number of elements in set A only.

\(n_o(B)\) = Number of elements in set B only.

\(n_o(C)\) = Number of elements in set C only.

\(n_o(A \cap B)\) = Number of elements in set A and B only.

\(n_o(B \cap C)\) = Number of elements in set B and C only.

\(n_o(C \cap A)\) = Number of elements in set A and C only.

\(n(A \cap B \cap C)\) = Number of elements in set A, B and C.

\begin{align*} n(A \cup B \cup C) &= n_o(A) +n_o(B) +n_o(C) +n_o(A \cap B) +n_o(B \cap C) +n_o(C \cap A) + n(A \cap B \cap C)\\ &= n(A) - n_o(A \cap B) - n_o(C \cap A) - n(A \cap B \cap C) + n(B) - n_o(B \cap C) - n_o(C\cap B) - n(A \cap B \cap C)

+ n(C) - n_o(A \cap C) - n_o(B \cap C) - n(A \cap B \cap C) + n_o(A \cap B) +n_o(B \cap C) +n_o(C \cap A) + n(A \cap B \cap C)\\ &= n(A) + n(B) + n(C) - [n_o(A \cap B) +n(A \cap B \cap C)] - [n_o(A \cap B) +n(A \cap B \cap C)] - [n_o(B \cap C) +n(A \cap B \cap C)] - [n_o(C \cap A) +n(A \cap B \cap C)]+n(A \cap B \cap 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)\\ \end{align*}

$$\boxed{\therefore (A \cup B \cup C)= n(A) + n(B) + n(C) - n(A \cap B) -n(B \cap C) -n(C \cap A) +n(A \cap B \cap C)} $$

If A, B and C are disjoint sets,

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

- The cardinality of a set is a positive integer but it is not decimal. So, n(A) is not equal to 50% because 50% = 0.5.
- If A, B and C are disjoint sets, \(n(A \cup B \cup C) = n(A) + n(B) + n(C).\)
- \((A \cup B \cup C)= n(A) + n(B) + n(C) - n(A \cap B) -n(B \cap C) -n(C \cap A) +n(A \cap B \cap C)\)

- It includes every relationship which established among the people.
- There can be more than one community in a society. Community smaller than society.
- It is a network of social relationships which cannot see or touched.
- common interests and common objectives are not necessary for society.

A and B are two subsets of a universal set 'U' in which n(U) = 43, n(A) = 25, n(B) = 18 and n(A ∩ B)= 7 then,

(i) Show it in Venn-diagram

(ii) Find the value of \(\overline {A \cup B} \)

Solution:

n(U) = 43

n(A) = 25

n(B) = 18

n(A ∩ B)= 7

The Venn-diagram of above information is as follow:

\begin{align*}\overline {A \cup B} &= ? \\ n(A \cup B) &= n(A) + n(B) - n(A \cap B) \\ &= 25 + 18 -7 \\ &= 43 - 7 \\ &= 36\\ Again, n(\overline {A \cup B}) &= n(U) - n(A \cup B) \\ &= 43 - 36\\ &= 7 Ans. \end{align*}

In the survey of 80 people, it was found that 47 people liked to see Nepali movies, 31 liked to see Hindi movies and 21 does not like to see both movies. (i) By how many people liked to see only Nepali movies?

(ii) By how many people liked to see only Hindi movies?

(iii) Represent the above information in Venn-diagram.

**Solution:**

Let N and H be the set of people who like to see Nepali movies and Hindi movies respectively

From question, n(U) = 80, n(N) = 47, n(H) = 31 , \((\overline {N \cup H})\) = 21

(i)

\begin{align*} n(N \cup H) &= n(U) - (\overline {N \cup H}) \\ &= 80 - 21 \\ &= 59 \\ Again, n(N \cup H) &= n(N) + n(H) - n(N \cap H)\\ or \: 59 &= 47 + 31 -n(N \cap H) \\ or, \: 59 &= 78 -n(N \cap H) \\ or, \:n(N \cap H) &= 78 - 59 \\ &= 19 \\ n_o(N) &= n(N) -n(N \cap H) \\ &= 47 -19 \\ &= 28 Ans. \end{align*}

(ii)

The no. of people who like to see Hindi movies only

\begin{align*} n_o (H) &= n(H) -n(N \cap H) \\ &= 31 - 19 \\ &= 12 \end{align*}

(iii)

The above information is shown in the Venn-diagram.

A group of 65 people, 40 like football, 10 like both football and volleyball.

(i) How many people like Volleyball?

(ii) How many people like football only?

(iii) How many people like volleyball only?

**solution:**

Let F and V represent the sets of people who like football and volleyball respectively and U be the universal sets.

From question,

\(n(F) = 40, n(F \cap V) = 10, \: n(U) = n(F \cup V) = 65 , \: n(V) = ? \)

(i)

We know that, \begin{align*} n(F \cup V) &= n(F) + n(V) - n(F \cap V) \\ or, \: 65 &= 40 + n(V) - 10 \\ or, \: 65 &= 30 + n(V)\\ or, \: n(V) &= 65 - 30 \\ \therefore n(V) &= 35 Ans. \end{align*}

(ii)

\( n_o (F) = n(F) - n(F \cap V) \: = 40 -10 \: = 30 Ans. \)

(iii)

\( n_o (V) = n(V) -n(F \cap V) \: = 35 - 10 \: = 25 Ans. \)

In a survey of 100 people. It was found that 65 liked folk songs, 55 liked modern songs and 35 liked folk song as well as modern songs.

(i) Draw the Venn diagram to illustrate the fact.

(ii) How many people did not like both songs?

**Solution:**

Let, F & M denote the set of people who like folk and modern song respectively. U be the universal set. From question,

n(U) = 100

n(F) = 65

n(M) = 55

n(F∩M) = 35

(i)

The above information represents in Venn diagram as follow:

From Venn-diagram

\begin{align*} n(F \cup M) &= n(F) + n(M) - n(F \cap M)\\ &= 65 + 55 - 35 \\&= 85 \: Ans. \end{align*}

(ii)

\begin{align*} n(\overline {F \cup M} ) &= 100 - 85 \\ &= 15 Ans. \end{align*}

In a class of 30 students, 20 students like to play cricket and 15 like to play volleyball. Also, each student like to play at least one of the two games. How many students like to play both games? Illustrate the above information by venn-diagram.

**Solution:**

Let, C and V represent the sets of student who like to play cricket and volleyball. Let, U be the universal set.

From question, \( n(C) = 20, \: n(V) = 15, \: n(U) = n(C \cup V) = 30, \: n(C \cap V) = ? \)

We know that,

\begin{align*} n(C \cup V) &= n(C) + n(V) - n(C \cap V) \\ or, \: 30 &= 20 + 15 -n(C \cap V) \\or, \:n(C \cap V) &= 35 - 30\\ \therefore \: n(C \cap V) &= 5 \: Ans. \end{align*}

The above information is shown in the following venn-diagram.

In a survey of 120 students, It is found that 17 drink nor coffee, 88 drink tea and 26 drink coffee. By drawing Venn-diagram find out the number of students who drink both tea and coffee.

**Solution:**

Let, T and C represent the set of students who like to drink tea & coffee respectively.

Let, U be the universal set.

From question, \(n(U) = 120, \: n(T) = 88, \: n(C) = 26, \: n(\overline{T \cup C})= 17, \: Let \: n(T \cap C)= x\)

The above information is shown in Venn-diagram.

We know that,

\begin{align*} n(T \cup C) &= n(U) - n(\overline{T \cup C})\\ n(T \cup C) &= 120 - 17 \\ &= 103 \\ Again, \: n(T \cup C) &= n(T) + n(C) - n(T \cap C)\\ or, \: 103 &= 88 + 26 -n(T \cap C) \\ or, \: 103 &= 114 - n(T \cap C)\\ \therefore \:n(T \cap C) &= 114 - 103 \\ &= 11 \: Ans. \end{align*}

In a class of 130 students, 70 students like tea, 40 students like coffee, 30 students don't like both. Find the number of students who like both and illustrate this information by means of Venn-diagram.

**Solution:**

Let, T and C denote sets of students who like tea and coffee respectively. Let, U be the universal sets.

From question, \( n(U)= 130,\: n(T)=70, \: n(C)=40, n(\overline {T \cup C})= 30 \)

We know that,

\begin{align*} n(T \cup C ) &= n(U) - n(\overline{T \cup C})\\ n(T \cup C ) &= 130 - 30 \\ &= 100 \: Ans. \\ Again, \\n(T \cup C ) &= n(T) + n(C) - n(T \cap C)\\ or, \: 100 &= 40 + 70 -n(T \cap C)\\ or, \: 100 &= 110 -n(T \cap C) \\ \therefore n(T \cap C) &= 10 \: Ans. \\ \end{align*}

The Venn-diagram showing the given information as follows.

131 students were asked how they travelled daily 56 said they used the autorickshaw, 103 said they used the bus, 65 said they used the bus but never an autorickshaw.

(i) How many students used the autorickshaw but not the bus?

(ii) How many students used neither the autorickshaw nor the bus?

**Solution:**

Let, A and B denote the set of students used the autorickshaw and bus. Let, U be the universal set,

From question,

\( n(U) = 131, \: n(A) = 56, \: n(B) = 103, \: n_o (B) = 65 \\ Let, n(A \cap B )= x, \: n(\overline{A \cup B}) = y \)

The above information is shown in Venn-diagram as follows.

From Venn-diagram,

\begin{align*} x + 65 &= 103 \\ or, \: x &= 103 - 65 \\ \therefore x= 38 \\ \end{align*}

(i) \begin{align*} \text {The no. of students used autorickshaw only} \: n_o (A) &= 56 - x \\ &= 56 - 38 \\ &= 18 \: Ans. \end{align*}

(ii)\begin{align*} n(\overline {A \cup B}) &= n(U) - n(A \cup B)\\ &= 131 - (65 + 38 + 18) \\ &= 131 - 121\\ &= 10 \: Ans. \end{align*}

In a survey of 2400 tourist who visited Nepal, it was found that 1650 liked to visit Bhaktapur, 850 liked to visit Lalitpur and 150 did not like to visit both place.

(i) Represent the above information in a Venn-diagram.

(ii) How many were there who liked to visit Bhaktapur only?

**Solution:**

Let, B and L denote the set of tourist who like to visit Bhaktapur and Lalitpur respectively. Let, U be the universal set.

From question,

\( n(U) = 2400, \: n(B) = 1650, \: n(L) = 850, \: n(\overline{B \cup L}) = 150 \)

(i)

The above information is shown in Venn-diagram as follows.

(ii)

\begin{align*} n(B \cup L) &= n(U) - n(\overline{B \cup L})\\ or, \:n(B \cup L) &= 2400 - 150 \\ &= 2250 \\ Again, \\n(B \cup L) &= n(B) + n(L) - n(B \cap L) \\ or, \: 2250 &= 1650 + 850 -n(B \cap L)\\ or, \:n(B \cap L) &= 2500 - 2250 \\ \therefore \: n(B \cap L) &= 250 \: Ans. \\ \end{align*}

(iii)

\begin{align*} n_o (B) &= n(B) - n(B \cap L)\\ &= 850 -250 \\ &= 600 \: Ans. \end{align*}

In a school, 60 students are passed in mathematics 45 passed in science and 30 passed in both subjects.

(i) How many students are passed in mathematics only?

(ii) How many students are passed in science only?

(iii) How many students are in the school?

(iv) Represent the above information in Venn-diagram.

**Solution:**

Let, M and S denote the set of students who passed in mathematics and Science respectively. Let, U be the universal set,

From question,

\( n(M) = 60 \: n(S) = 45, \: n(M \cap S )= 30\)

(i) \begin{align*} n_o (M) &= n(M) - n(M \cap S)\\ &= 60 -30 \\ &= 30 \: Ans. \end{align*}

(ii)\begin{align*} n_o (S) &= n(S) - n(M \cap S)\\ &= 45 -30 \\ &= 15 \: Ans. \end{align*}

(iii) \begin{align*}n(M \cup S) &= n(M) + n(S) -n(M \cap S)\\ &= 60 + 45 - 30 \\ &= 75 \: Ans. \end{align*}

(iv)The above information is shown in Venn-diagram as follows.

In a class of 55 students, 15 students liked Maths but not English and 18 liked English but not Maths. If 5 students did not like both, how many students liked both subjects? Represent the above information in a Venn-diagram.

**Solution:**

Let, M and E denote the set of students who like Maths and English respectively. Let, U be the universal set.

From question,

\( n(U)=55, \: n_o(M) = 15, \: n_o(E) = 18, \: n(\overline{M \cup E} )= 5 \\ Let \: n(M \cap E)= x \)

Now,

\begin{align*} n(M \cup E) &= n(U) - n(\overline{M \cup E})\\ &= 55 -5 \\ &= 50 \end{align*}

The above information is shown in Venn-diagram as follows.

From Venn-diagram,

\begin{align*} 15 + x + 18 &= 50\\ or, \: x + 33 &= 50\\ or, \: x &= 50 - 33 \\ &= 17 \end{align*}

\( \therefore \text {The no. of students who like both subject is 17.} \)

50 students in a classroom like Mathematics or Science or both. Out of them 20 like both subjects. The ratio number of students who like Mathematics to those who like Science is 3 : 2

(i) Find the number of students who like Science only.

(ii) Show the above information in the Venn-diagram.

**Solution:**

Let, M and S denote the set of students who like Mathematics and Science respectively. Let, U be the universal set,

From question,

\( n(U)=50 \: = n(M \cup S), \: n(M \cap S) = 20, \: Let, \: n(M)=3x \: and \: n(S)=2x \)

We know that,

\begin{align*} n(M \cup S) &= n(M) + n(S) - n(M \cap S)\\ 50 &= 3x + 2x - 20 \\ or, \: 5x &= 50 + 20\\ or, \: x &= \frac{70}{5}\\ &= 14 \end{align*}

(i)\begin{align*} \text {The no. of student who like Mathematics, n(M)} &= 3x \\ &= 3 \times 14 \\ &= 42 \: Ans. \end{align*}

(ii)\begin{align*} \text {The no. of student who like Science, n(S)} &= 2x \\ &= 2 \times 14 \\ &= 28 \end{align*}

So, \( n_o(S) = n(S) - n(M \cap S) \: \:= 28 - 20 \: \: = 8\: Ans. \)

(iii) The above information is shown in Venn-diagram as follows:

In a class of 25 students 12 have taken Mathematics, 8 have taken Mathematics but not Biology. Find the number of students who have taken both Mathematics and Biology and those who have taken Biology but not Mathematics. Solve by making Venn-diagram.

**Solution:**

Let, M and B denote the set of students who have taken Mathematics and Biology respectively. Let, U be the universal set.

From question,

\( n(M \cup B)=25 = n(U), \: n(M) = 12, \: n_o(M) = 8 \)

Now,

\begin{align*} n(M \cap B)&= n(M) - n_o(M) \\ &= 12 - 8 \\ &= 4 \end{align*}

\begin{align*} n_o(B) &= n(M \cup B) - n(M) \\ &= 25 - 12 \\ &= 13 \end{align*}

The no. of students who have taken Maths & Biology = 4 Ans.

The no. of students who have taken Biology but not Math = 13 Ans.

The above information is shown in Venn-diagram as follows:

In an examination 40% of students passed in mathematics only and 30% passed in science only. If 10% of students were failed in both subject.

(i) What is the percentage of student passed in both subject?

(ii) What percentage of student passed in mathematics?

(iii) Represent all in Venn-diagram.

**Solution:**

Let, M and S denote the set of students who passed in mathematics and Science respectively. Let, U be the universal set,

From question,

\( n_o(M) = 40\%, \: n_o(S) = 30\%, \: n(\overline{M \cup S })= 10\%, \: n(U) = 100\% \)

\begin{align*}n(M \cup S)&= n(U) -n(\overline{M \cup S }) \\ &= 100\% - 10\% \\ &= 90\% \end{align*}

(i)

\begin{align*} n(M \cup S) &=n_o(M) +n_o(S) + n(M \cap S)\\ 90\% &= 40\% + 30\% + n(M \cap S)\\ or, \: 90\% - 70\% &= n(M \cap S)\\ \therefore n(M \cap S) &= 20\% \end{align*}

(ii)

\begin{align*} n(M) &= n_o(M) + n(M \cap S)\\ &= 40\% + 20\% \\ &= 60\% \end{align*}

(iii)

The above information is shown in Venn-diagram as follows.

In a survey of a group of people, It was found 70% of the people liked folk songs, 60% liked modern song, 4000 people like both of them and 10% liked none of them then

(i) Draw a Venn-diagram to illustrate the above information.

(ii) Find the total number of people in the survey.

**Solution:**

Let F & M denote the set of people who liked folk & modern song respectively.

Let U be the universal set. From question,

\( n(F)= 70\% , \: n(M)= 60\%, \: n(\overline{F \cup M})= 10\%, \: n(F \cap M)= 4000, \: n(U)= 100\% \)

The above information represents in Venn-diagram as follows:

\begin{align*} n(F \cup M) &= n(U) - n(\overline{F \cup M}) \\ &= 100\% -10\% \\ &= 90\% \end{align*}

From Venn-diagram

\begin{align*} 70\% - x\% + x\% + 60\% - x\% &=90\% \\ or, \: x &= (130 - 90)\% \\ \therefore x &= 40\% \end{align*}

Let total number of people be 'y'

\begin{align*} 40\% \; of\; y &= 4000 \\ or, \: y \times \frac{40}{100} &= 4000\\ or, \: y &= \frac{4000 \times 100}{40} \\ \therefore y &= 10000 \: Ans. \end{align*}

From the Venn-diagram along side find

\( (i) n(A \cap B), (ii) n(A \cup B), (iii) n(\overline {A \cup B}) \)

Solution:

From question, \( n(U)=30, \: n(A)=20,\: n(B)= 10 \)

\( 3x + y = n(A) \\ 3x + y = 20 \: \: \: \: \: .............(1)\)

\( x + y = n(B) \\ x + y = 10 \: \: \: \: \: \: ..............(2)\)

Subtracting eqn (2) from eqn (1)

\begin{array}{rrrr} 3x&+ &y&=&20 \\ x&+&y&=&10 \\ -&&-&&-\\ \hline\\ &&2x&=&10\\\end{array}

\begin{align*} x &= \frac{10}{2} \\ \therefore x &= 5 \end{align*}

Putting value of x in eq^{n} (1)

\begin{align*} 3x + y &= 20 \\ or, \: 3 \times 5 + y &= 20 \\ or, \: y &= 20 - 15 \\ y &= 5 \end{align*}

(i) \begin{align*} n(A \cap B) &= y\\ &= 5 \: Ans.\end{align*}

(ii) \begin{align*} n(A \cup B) &= 3x + y + x \\ &= 3 \times 5 + 5 + 5 \\ &= 15 + 10\\ &= 25 \: Ans. \end{align*}

(iii) \begin{align*}n(\overline {A \cup B}) &= n(U) - n(A \cup B) \\ &= 30 - 25 \\ &= 5 \end{align*}

In a survey of a group of people, It was found 77% of the people liked folk songs, 63% liked modern songs and 5% liked none of them. If 135 people liked both of them .

(i) Find the total number of people in the survey.

(ii) How many liked folk song only?

(iii) Represent the above information in Venn-diagram.

**Solution:**

Let, F and M be the sets of people who liked folk and modern songs.

Let U be the universal set.

From Question,

\( n(U) = 100\%, \: n(F) = 77\%, \: n(M) = 63\%, \: n(\overline {F \cup M} )= 5\% \), n(M∩N) = 135

\( Let, \: n(M \cap N) = x\% \)

We know that,

\begin{align*} n(F \cup M) &= n(U) - n(\overline {F \cup M})\\ &= 100\% - 5\% \\ &= 95\% \\ Again,\\ n(F \cup M) &= n(F) + n(M) - n(F \cap M)\\ 95\% &= 77\% + 63\% -x\%\\ or, \: x\% &= 140\% - 95\%\\ \therefore x &= 45\%\end{align*}

(i) From question,

\begin{align*} 45\% \: of \: n(U) &= 135\\ or, \: n(U) \times \frac{45}{100} &= 135\\ or, \: n(U) &= \frac {135 \times 100}{45}\\ \therefore n(U) &= 300 \: Ans.\end{align*}

(ii)

\begin{align*}n(F) &= 300 \:of \: 77\%\\ &= 300 \times \frac {77}{100}\\ &= 231\\ Now, \\ n_o (F) &= n(F) - n(F \cap M)\\ &= 231 - 135 \\ &=96 \: Ans. \end{align*}

(iii)

The above information represent in Venn-diagram as follow:

In a group of 95 students the ratio of student who like mathematics and science is 4 : 5. If 10 of them like both subject and 15 of them like none of the subjects then by drawing Venn-diagram find how many of them

(i) Like only mathematics.

(ii) Like only science.

**Solution:**

Let M and S be the student who liked math and science respectively.

Let U be the universal sets then,

From question,

\( n(U)= 95, \: Let\: n(M)= 4x, \: n(S)= 5x, \: n(M \cap S)= 10, \: n(\overline{M \cup S})= 15 \)

The above information is shown in the venn-diagram.

\begin{align*} n(M \cup S) &=n(U) - n(\overline{M \cup S})\\ &= 95 -15 \\ &= 80 \end{align*}

\begin{align*} n(M \cup S) &=n_o(M)+n_o(S) + n(M \cap S)\\ 80 &= 4x - 10 + 10 + 5x -10\\ or, \: 80 &= 9x - 10\\ \therefore x &= \frac{90}{9} &= 10 \end{align*}

(a) The number of students who liked mathematics only

\begin{align*} 4x - 10 &= 4 \times 10 - 10 \\ &= 40 - 10 \\ &= 30 \: Ans. \end{align*}

(b) The number of students who liked science only

\begin{align*} 5x - 10 &= 5 \times 10 - 10 \\ &= 50 - 10 \\ &= 40 \: Ans. \end{align*}

In an examination, out of 200 students 70 passed in English, 80 in Mathematics, 60 in Nepali, 35 in English as well as in Mathematics, 25 in English as well as in Nepali, 35 in math as well as nepali and 10 passed in all three subjects.

(a) Draw the Venn-diagram showing above information.

(b) Find the no. of students who fail in all subjects?

**Solution:**

Let E, M and N represents the set of students who passed in English, Mathematics and Nepali respectively. Let U be the universal set.

\( n(U) = 200, \: \: \: \: \: n(E) = 70, \: \: \:\: \: n(M) = 80\: \:\: \: \: n(N) = 60, \\ n(E \cap M)=35, \: \:\: \: \: n(E \cap N)=25 \: \:\: \: n(M \cap N)=35 \: \:\: \: n(E \cap M \cap N) = 10 \)

(a) The above information is shown in venn-diagram as follows.

(b) \begin{align*} n(E \cap M \cap N) &= n(E) + n(M) + n(N) - n(E \cap M) - n(M \cap N) - n(N \cap E) + n(E \cap M \cap N)\\ &= 70 + 80 + 60 - 35 -25 -35 + 10 \\ &= 125 \end{align*}

\begin{align*} n(\overline{E \cap M \cap N}) &= n(U) -n(E \cap M \cap N) \\ &=200 - 125 \\ &= 75 \: Ans. \end{align*}

In a certain examination, 58% failed in English, 39% in Account and 25% in Statistics,32% in English and Account, 19% in English and Statics, 17% in Account and Statics and 13% in all three subjects.

(a) Draw a venn-diagram

(b) What percentage did not fail in any subjects.

Let E, A and S denote the set of students who failedin English, Account and statistics respectively. Let U be the universal set.

From question,

\( n(U) = 100\% , \: n(E)=58\%, \: n(A) = 39\%, \: n(S)= 25\%, \: n(E \cap A)=32\%, \\ \: n(A \cap S)= 17\%, \: n(E \cap S)=19\% \: n(E \cap A \cap S)= 13\% \)

(a)The above information is shown in the venn-diagram as follow:

\begin{align*}n(E \cup A \cup S) &= n(E) + n(A) + n(S) - n(E \cap A) - n(A \cap S) - n(S \cap E) + n(E \cap A \cap S)\\ &= (58 + 39 + 25 - 32 - 17 - 19 - 13)\% \\ &= 67\% \end{align*}

(b) \begin{align*} n(\overline {E \cup A \cup S}) &= n(U) - n(E \cup A \cup S )\\ &= 100\% -67\% \\ &= 33\% \: Ans. \end{align*}

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