NCERT Exemplar Class 11 Maths Solutions Chapter 1 Sets

NCERT Exemplar Class 11 Maths Solutions Chapter 1

Question:1

Write the following sets in the roster form:

1. $A=\{x:x\in R,2x+11=15\}$

2. $B=\{x:x^2=x,x\in R\}$

3. $C=\{x:x\text{ is a positive factor of a prime number }p\}$

Answer:

(i) Given that: $A=\{x:x\in R,2x+11=15\}$
$\begin{array}{rl}2x+11& =15\\ 2x& =4\\ x& =2\end{array}$
Hence, $A=\{2\}$
(ii) Given that $B=\{x:x^2=x,x\in R\}$
$\begin{array}{rl}{x}^2& =x\\ x^2-x& =0\\ x(x-1)& =0\\ \therefore x=0& x=1\end{array}$
Hence, $B=\{0,1\}$
(iii) Given that $C=\{x:x\text{ is a positive factor of a prime number }p\}$
So the positive factors of prime number $p$ are 1 and $p$. Hence $C=\{1,p\}$

Question:2

Write the following sets in the roster form:

1. $D=\{t:t^3=t,t\in R\}$

2. $E=\{w:\frac{w-2}{w+3}=3,w\in R\}$

3. $F=\{x:x^4-5x^2+6=0,x\in R\}$

Answer:

(i) Given that $D=\{t:t^3=t,t\in R\}$
$\begin{array}{rl}{t}^3& =t\\ t^3-t& =0\\ t(t-1)(t+1)& =0\\ t& =-1,0,1\end{array}$
hence, $D=\{-1,0,1\}$
(ii) Given that $E=\{w:\frac{w-2}{w+3}=3,w\in R\}$
$\begin{array}{rl}\frac{w-2}{w+3}& =3\\ 3w+9& =w-2\\ 2w& =-11\\ w& =-\frac{11}{2}\end{array}$
Hence, $E=\left\{\frac{-11}{2}\right\}$
(iii) Given that $F=\{x:x^4-5x^2+6=0,x\in R\}$
$\begin{array}{rl}{x}^4-5x^2+6& =0\\ (x^2-2)(x^2-3)& =0\end{array}$
$x=\pm \sqrt{2}$ and $x=\pm \sqrt{3}$
Hence, $F=\{-\sqrt{2},+\sqrt{2},-\sqrt{3},+\sqrt{3}\}$

Question:3

If $Y=\{x:x\text{ is a positive factor of the number }{2}^{p-1}(2^p-1)$$\text{, where }{2}^p-1\text{ is a prime number}\}$. Write Y in the roster form.

Answer:

Given that
$Y=\{x:x\text{ is a positive factor of the no. }{2}^{p-1}(2^p-1)\text{, where }{2}^p-1\text{ is a prime number}\}$
The factors of $2^{p-1}$ are $1,2,2^2,2^3,...,2^{p-1}$
The factors of $2^p-1$ are 1 and $2^p-1$
Hence, $Y=\{1,2,2^2,2^3,...,2^{p-1},2^p-1\}$

Question:4

State which of the following statements are true and which are false. Justify your answer.
(i) $35\in \{x:x\text{ has exactly four positive factors}\}$
(ii) $128\in \{y:\text{ the sum of all the positive factors of }y\text{ is }2y\}$
(iii) $3\notin \{x:x^4-5x^3+2x^2-112x+6=0\}$
(iv) $496\notin \{y:\text{ the sum of all the positive factors of }y\text{ is }2y\}$

Answer:

(i) Given that: $35\in \{x:x\text{ has exactly four positive factors}\}$
Factors of 35 are 1, 5, 7, and 35.
Hence, true
(ii) Given that: $128\in \{y:\text{ the sum of all the positive factors of }y\text{ is }2y\}$
Factors of 128 are 1,2,4,8,16,32,64,128
sum of all factors $255\ne 2\times 128$
Hence statement is false.
(iii) Given that: $3\notin \{x:x^4-5x^3+2x^2-112x+6=0\}$
$x^4-5x^3+2x^2-112x+6=0$
Put x = 3
$-366=0$, which is not true, so 3 is not an element of the set
Hence, the statement is true
(iv) Given that: $496\notin \{y:\text{ the sum of all the positive factors of }y\text{ is }2y\}$
The positive factors of 496 are 1,2,4,8,16, 31, 62, 124, 248, and 496
The sum of all positive factors is $992=2\times 496$, which is true, so 496 is an element of the set. Hence, the statement is false.

Question:5

Given $L=\{1,2,3,4\},M=\{3,4,5,6\}$ and $N=\{1,3,5\}$. Verify that $L-(M\cup N)=(L-M)\cap (L-N)$.

Answer:

Given that $L=\{1,2,3,4\},M=\{3,4,5,6\}$ and $N=\{1,3,5\}$.
Now,
$\begin{array}{rl}LHS=L-(M\cup N)& =\{1,2,3,4\}-(\{3,4,5,6\}\cup \{1,3,5\})\\ & =\{1,2,3,4\}-\{1,3,4,5,6\}=\{2\}\end{array}$
$\begin{array}{rl}RHS& =(L-M)\cap (L-N)\\ (L-M)& =\{1,2,3,4\}-\{3,4,5,6\}=\{1,2\}\\ (L-N)& =\{1,2,3,4\}-\{1,3,5\}=\{2,4\}\\ (L-M)& \cap (L-N)=\{1,2\}\cap \{2,4\}=\{2\}\\ LHS& =RHS\end{array}$

Question:6

If A and B are subsets of the universal set U, then show that
(i) A ⊂ A ∪ B
(ii) A ⊂ B ⇔ A ∪ B = B
(iii) (A ∩ B) ⊂ A

Answer:

(i) Given that $A\subset U$ and $B\subset U$
To prove: $A\subset A\cup B$ we have to show that if $x\in A;x\in A\cup B$
Let $x\in A\Rightarrow x\in Aorx\in B\Rightarrow x\in A\cup B$
hence, $A\subset A\cup B$
(ii) $A\subset B$ then let $x\in A\cup B$
$\Rightarrow x\in Aorx\in B$
$\Rightarrow x\in B$
$A\cup B\subset B(i)$
But $B\subset A\cup B(ii)$
From (i) and (ii) we get $A\cup B=B$
Now if $A\cup B=B$, then
let $y\in A\Rightarrow y\in A\cup B\Rightarrow y\in B$
Hence, $A\subset B$
Thus, $A\subset B\Rightarrow A\cup B=B$
iii)
$\begin{array}{l}\text{ Let }x\in A\cap B\\ x\in A\text{ and }x\in B\\ x\in A\\ \text{ Hence, }A\cap B\subset A\end{array}$

Question:7

Given that N = {1, 2, 3,…, 100}. Then write
(i) the subset of N whose elements are even numbers.
(ii) the subset of N whose elements are perfect square numbers.

Answer:

N = {1,2,3,4……100}

1. Required subset whose elements are even = {2,4, 6,8…..,100}

2. Required subset whose elements are perfect squares = {1,4,9 ,16,25,36,49,64,81,100}

Question:8

If X = {1, 2, 3}, if n represents any member of X, write the following sets containing all numbers represented by:
(i) 4n
(ii) n + 6
(iii) $\frac{n}{2}$
(iv) n – 1

Answer:

Given that $X=\{1,2,3\}$
(i) $\{4n|n\in X\}=\{4,8,12\}$
(ii) $\{n+6|n\in X\}=\{7,8,9\}$
(iii) $\{\frac{n}{2}|x\in X\}=\{\frac{1}{2},1,\frac{3}{2}\}$
(iv) $\{(n-1)|x\in X\}=\{0,1,2\}$

Question:9

If Y = {1, 2, 3,…, 10}, and a represents any element of Y, write the following sets, containing all the elements satisfying the given conditions.
(i) $a\in Y$ but $a^2\notin Y$
(ii) a + 1 = 6, $a\in Y$
(iii) a is less than 6 and $a\in Y$

Answer:

Given that: Y = {1, 2, 3,…, 10}
(i) $\{a|a\in Y\text{ but }{a}^2\notin Y\}=\{4,5,6,7,8,9,10\}$
(ii) $\{a|a+1=6,a\in Y\}=\{5\}$
(iii) $\{a|a<6\text{ and }a\in Y\}=\{1,2,3,4,5\}$

Question:10

A, B and C are subsets of Universal Set U. If A = {2, 4, 6, 8, 12, 20}, B = {3, 6, 9, 12, 15}, C = {5, 10, 15, 20} and U is the set of all whole numbers, draw a Venn diagram showing the relation of U, A, B and C.

Answer:

Given that: A, B and C are the subsets of a universal set U.
Where A= {2,4,6,8,12,20}
B = {3, 6,9, 12, 15}
And C = {5, 10, 15, 20}

Question:11

Let U be the set of all boys and girls in a school, G be the set of all girls in the school, B be the set of all boys in the school, and S be the set of all students in the school who take swimming. Some, but not all, students in the school take swimming. Draw a Venn diagram showing one of the possible interrelationships among sets U, G, B and S.

Answer:

Given that U: Set of all boys and girls
G = Set of girls
B = Set of boys
S = Set of all students who take swimming

Question:12

For all sets A, B and C, show that $(A-B)\cap (A-C)=A-(B\cup C)$

Answer:

Let $x\in (A-B)\cap (A-C)$
$$\begin{gathered} \Rightarrow x\in (A-B)\text{ and }x\in (A-C) \\ \Rightarrow x\in A\text{ and }(x\notin B\text{ and }x\notin C) \\ \Rightarrow x\in A\text{ and }(x\notin B\cup C) \end{gathered}$$
So, $(A-B)\cap (A-C)\subset A-(B\cup C)(i)$
Let $y\in A-(B\cup C)$
$$\begin{gathered} y\in A-(B\cup C) \\ \Rightarrow y\in A\text{ and }(y\notin B\cup C) \end{gathered}$$

$\Rightarrow y\in A\text{ and }y\notin B\text{and}y\in A\text{ and }y\notin C$

$$\begin{gathered} \Rightarrow y\in (A-B)\text{ and }y\in (A-C) \\ \Rightarrow y\in (A-B)\cap (A-C) \end{gathered}$$
So, $A-(B\cup C)\subset (A-B)\cap (A-C)(ii)$
From (i) and (ii), we get,
$(A-B)\cap (A-C)=A-(B\cup C)$

Question:13

For all sets A and B, $(A-B)\cup (A\cap B)=A$

Answer:

$\begin{array}{rl}LHS& =(A-B)\cup (A\cap B)\\ & =[(A-B)\cup A]\cap [(A-B)\cup B]\\ & =A\cap (A\cup B)=A=RHS\end{array}$

Question:14

For all sets A, B and C, A – (B – C) = (A – B) – C

Answer:

Let us use the following Venn diagram to solve this question
Step 1: - B - C

Step 2: - A-(B-C)

Step 3: - A-B

Step 4: - (A-B)-C

As steps 2 and 4 are inequal, hence, the statement given is not true.

Question:15

For all sets A, B, and C, if A ⊂ B, then A ∩ C ⊂ B ∩ C

Answer:

Suppose $A\subset B$,
Let $x\in A\cap C$
$$\begin{gathered} x\in A\text{ and }x\in C \\ x\in B\text{ and }x\in C \\ x\in B\cap C \\ \therefore A\cap C\subset B\cap C \end{gathered}$$
Hence, the given statement is true.

Question:16

For all sets A, B, and C, if A ⊂ B, then A ∪ C ⊂ B ∪ C

Answer:

Suppose $A\subset B$,
Let $x\in A\cup C$
$$\begin{gathered} x\in A\text{ or }x\in C \\ x\in B\text{ or }x\in C \\ x\in B\cup C \\ \therefore A\cup C\subset B\cup C \end{gathered}$$
Hence, the given statement is true.

Question:17

For all sets A, B, and C, if A ⊂ C and B ⊂ C, then A ∪ B ⊂ C

Answer:

Suppose $A\subset C,B\subset C$.
Let $x\in A\cup B$
$$\begin{gathered} x\in A\text{ and }x\in B \\ x\in C\text{ and }x\in C[A\subset C\text{ and }B\subset C] \\ \Rightarrow x\in C\Rightarrow A\cup B\subset C \end{gathered}$$
Hence, the given statement is true.

Question:18

For all sets A and B, A ∪ (B – A) = A ∪ B

Answer:

$\begin{array}{rl}A\cup (B-A)& =A\cup (B\cap A^{\prime })\\ & =(A\cup B)\cap (A\cup A^{\prime })& [\text{Distributive law}]\\ & =(A\cup B)\cap U& [A\cup A^{\prime }=U]\\ & =A\cup B=RHS\end{array}$
Hence, the given statement is true.

Question:19

For all sets A and B, A – (A – B) = A ∩ B

Answer:

$\begin{array}{rlrl}LHS& =A-(A-B)=A-(A\cap B^{\prime })& & [A-B=A\cap B^{\prime }]\\ & =A\cap (A\cap B^{\prime }{)}^{\prime }=A\cap [A^{\prime }\cup (B^{\prime }{)}^{\prime }]& & [(A\cap B{)}^{\prime }=A^{\prime }\cup B^{\prime }]\\ & =A\cap (A^{\prime }\cup B)\\ & =(A\cap A^{\prime })\cup (A\cap B)=P\cup (A\cap B)& & \text{Where }P=(A\cap A^{\prime })\\ & =A\cap B=RHS\end{array}$]
Hence, the given statement is proved.

Question:20

For all sets A and B, A – (A ∩ B) = A – B

Answer:

$\begin{array}{rlrl}LHS& =A-(A\cup B)=A\cap (A\cup B{)}^{\prime }& & [A-B=A\cup B^{\prime }]\\ & =A\cap (A^{\prime }\cup B^{\prime })& & [(A\cap B{)}^{\prime }=A^{\prime }\cup B^{\prime }]\\ & =(A\cap A^{\prime })\cup (A\cap B)\\ & =P\cup (A-B)& & [P=(A\cap A^{\prime })\text{ and }A-B=A\cap B^{\prime }]\\ & =(A-B)=R.H.S\end{array}$

Question:21

For all sets A and B, (A ∪ B) – B = A – B

Answer:

$\begin{array}{rlrl}LHS& =(A\cup B)-B=(A\cup B)\cap B^{\prime }& & [A-B=A\cap B^{\prime }]\\ & =(A\cap B^{\prime })\cup (B\cap B^{\prime })\\ & =(A-B)\cup P& & [P=(A\cap A^{\prime })\text{ and }A-B=A\cap B^{\prime }]\\ & =(A-B)=RHS\end{array}$

Question:22

Let
$T=\{x|\frac{x+5}{x-7}-5=\frac{4x-40}{13-x}\}$
Is T an empty set? Justify your answer.

Answer:

Given: $T=\{x|\frac{x+5}{x-7}-5=\frac{4x-40}{13-x}\}$
$\begin{array}{rlrl}\Rightarrow & & \frac{x+5}{x-7}-5& =\frac{4x-40}{13-x}\\ & & \frac{(x+5)-5(x-7)}{x-7}& =\frac{4x-40}{13-x}\\ \Rightarrow & & \frac{x+5-5x+35}{x-7}& =\frac{4x-40}{13-x}\\ \Rightarrow & & -4(x-10)(13-x)& =4(x-10)(x-7)\\ \Rightarrow & & -4(x-10)(13-x+x-7)& =0\\ \Rightarrow & & -4(x-10)6& =0\\ \Rightarrow & & -24(x-10)& =0\\ \Rightarrow & & x-10& =0\\ \Rightarrow & & x=10& \text{ or }T=\{10\}\end{array}$
Hence, T is not an empty set.

Question:23

Let A, B and C be sets. Then show that A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

Answer:

Let $x\in A\cap (B\cup C)$
$\begin{array}{rl}\Rightarrow & x\in A\text{ and }x\in (B\cup C)\\ \Rightarrow & x\in A\text{ and }x\in B\text{ or }x\in C\\ \Rightarrow & x\in A\text{ and }x\in B\text{ or }x\in A\text{ and }x\in C\\ \Rightarrow & x\in (A\cap B)\text{ or }x\in (A\cap C)\\ \Rightarrow & x\in (A\cap B)\cup (A\cap C)\\ \therefore & A\cap (B\cup C)\subset (A\cap B)\cup (A\cap C)& -(i)\end{array}$
Let $y\in (A\cap B)\cup (A\cap C)$
$\begin{array}{rl}\Rightarrow & y\in (A\cap B)\cup (A\cap C)\\ \Rightarrow & y\in (A\cap B)\text{ or }y\in (A\cap C)\\ \Rightarrow & y\in A\text{ and }y\in B\text{ or }y\in A\text{ and }y\in C\\ \Rightarrow & y\in A\text{ and }(y\in B\text{ or }y\in C)\\ \Rightarrow & y\in A\text{ and }(y\in B\cup C)\\ \Rightarrow & y\in A\cap (B\cup C)\\ \therefore & (A\cap B)\cup (A\cap C)\subset A\cap (B\cup C)& -(ii)\end{array}$
From (i) and (ii)
$A\cap (B\cup C)=(A\cap B)\cup (A\cap C)$

Question:24

Out of 100 students, 15 passed in English, 12 passed in Mathematics, 8 in Science, 6 in English and Mathematics, 7 in Mathematics and Science, 4 in English and Science, and 4 in all three. Find how many passed.
(i) in English and Mathematics but not in Science
(ii) in Mathematics and Science but not in English
(iii) in Mathematics only
(iv) in more than one subject only

Answer:

Let the set of students who passed in mathematics be M, the set of students who passed in English be E, and the set of students who passed in science be S.

Then n(U)=100, n(M)=12, n(S)=8,
$n(E\cap M)=6,n(M\cap S)=7,n(E\cap S)=4$ and $n(E\cap M\cap S)=4$
According to the Venn diagram
$n(E\cap M\cap S)=4;e=4$
$n(E\cap M)=6;d+e=6;d=2$
$n(M\cap S)=7;e+f=7;f=3$
$n(E\cap S)=4;b+e=4;b=0$
$n(E)=15;a+b+d+e=15;a=9$
$n(M)=12;d+e+f+g=12;g=3$
$n(S)=8;b+e+f+c=8;c=1$
Thus, the number of students who passed in English and mathematics but not in science = d 2
Number of students who passed in Science and mathematics but not in English = f=3
Number of students who passed in mathematics only = g =3
Number of students who passed in more than one subject = b+ e+ d +f =0+4+2+3 =9

Question:25

In a class of 60 students, 25 students play cricket and 20 students play tennis, and 10 students play both games. Find the number of students who play neither.

Answer:

Let C be the set of students who play cricket and T the set of students who play tennis.
Total number of students = 60 ⇒n(U)=60
Number of students who play cricket = 25 $\Rightarrow$ n(C)=25
Number of students who play tennis = 20 $\Rightarrow$ n(T)=20
Number of students who play both the games = 10 $\Rightarrow n(C\cap T)=10$
Number of students who play any one game$=n(C\cup T)=n(C)+n(T)-nC\cap T=25+20=10=35$
Number of students who play neither $=n(U)-n(C\cup T)=60-35=25$

Question:26

In a town of 10,000 families,, it was found that 40% of families buy newspaper A, 20% of families buy newspaper B, 10% of families buy newspaper C, 5% of families buy A and B, 3% buy B and C, and 4% buy A and C. If 2% of families buy all three newspapers. Find
(a) The number of families which buy newspaper A only.
(b) The number of families which buy none of A, B and C

Answer:

Let A be a set of families which buy newspaper A, B be the set of families which buy newspaper B, and C be the set of families which buy newspaper C
$n(U)=10000,n(A)=40\mathrm{\%},n(B)=20\mathrm{\%},n(c)=10\mathrm{\%},n(A\cap B)=5\mathrm{\%},n(B\cap C)=3\mathrm{\%},$ $n(A\cap C)=4\mathrm{\%}$ and $n(A\cap B\cap C)=2\mathrm{\%}$
Number of families which buy newspaper A only $=n(A)-n(A\cap B)-n(A\cap C)+n(A\cap B\cap C)$
$=10000\times (\frac{40}{100}-\frac{5}{100}-\frac{4}{100}+\frac{2}{100})=10000\times \frac{33}{100}=3300\text{families}$
Number of families which buy none of A, B and C $=n(U)-n(A\cup B\cup C)$
$=n(U)-[n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B\cap C)]$
$=[100-(40+20+10-5-3-4+2)]\times 10000$
$=\frac{40}{100}\times 10000=4000families$

Question:27

In a survey of 200 students of a school, it was found that 120 study Mathematics, 90 study Physics and 70 study Chemistry, 40 study Mathematics and Physics, 30 study Physics and Chemistry, 50 study Chemistry and Mathematics and 20 none of these subjects. Find the number of students who study all three subjects.

Answer:

Let the set of students who study mathematics be M, the set of students who study physics be P, and the set of students who study chemistry be C.
N(U) =200, n(M)=120, n(P)=90, n(C)=70, n(MP) = 40, $n(P\cap C)=30$, $n(C\cap M)=50$ and
$n(M^{\prime }\cap P^{\prime }\cap C^{\prime })=20$
$n(U)-n(M^{\prime }\cap P^{\prime }\cap C^{\prime })=n(M\cup P\cup C)=200-20=180$
$n(M\cup P\cup C)=n(M)+n(P)+n(C)-n(M\cap P)-n(P\cap C)$$-n(C\cap M)+n(M\cap P\cap C)$
$180=120+90+7-40-30-50+n(M\cap P\cap C)$
$180-40=n(M\cap P\cap C)=20$
Hence, the number of students who study all 3 subjects = 20

Question:28

In a group of 50 students, the number of students studying French, English, and Sanskrit was found to be as follows:
French = 17, English = 13, Sanskrit = 15
French and English = 09, English and Sanskrit = 4
French and Sanskrit = 5, English, French and Sanskrit = 3. Find the number of students who study
(i) French only
(ii) English only
(iii) Sanskrit only
(iv) English and Sanskrit but not French
(v) French and Sanskrit but not English
(vi) French and English but not Sanskrit
(vii) at least one of the three languages
(viii) none of the three languages

Answer:

Let F be the set of students who study French, E be the set of students who study English, and S be the set of students who study Sanskrit
$n(U)=5,n(F)=17,n(E)=13,n(S)=15,n(F\cap E)=9,n(E\cap S)=5$ and $n(F\cap E\cap S)=3$
$\begin{array}{rl}n(F\cap E\cap S)=3& \Rightarrow e=3\\ n(F\cap E)=9& \Rightarrow b+e=9& \Rightarrow b=6\\ n(F\cap S)=5& \Rightarrow d+e=5& \Rightarrow d=2\\ n(F)=17& \Rightarrow a+b+d+e=17& \Rightarrow a=6\\ n(E)=13& \Rightarrow b+c+e+f=13& \Rightarrow c=3\\ n(S)=15& \Rightarrow d+e+f+g=15& \Rightarrow g=9\end{array}$
Number of students who study French only =a =6
Number of students who study English only =c =3
Number of students who study Sanskrit only =g =9
Number of students who study English and Sanskrit but not French =f =1
Number of students who study French and Sanskrit but not English =d =2
Number of students who study French and English but not Sanskrit =b = 6
Number of students who study at least one of the three languages = a+b+c+d+e+f+g =6+6+3+2+3+1+9 =30
Number of students who study none of the three languages = 50-30=20

Question:29

Suppose $A_1,A_2,...,A_{30}$ are thirty sets each having 5 elements and $B_1,B_2,...,B_n$ are n sets each with 3 elements, let
$\bigcup _{i=1}^{30}{A}_i=\bigcup _{j=1}^n{B}_j=S$
and each element of S belongs to exactly 10 of the Ai’s and exactly 9 of the B's, S. Then n is equal to
A. 15
B. 3
C. 45
D. 35

Answer:

Number of elements in

$\begin{array}{r}{A}_1\cup A_2\cup A_3\cup ...\cup A_{30}=30\times 5=150\\ (\text{when repetition is allowed})\end{array}$
However, each element is repeated 10 times.
$n(S)=30\times \frac{5}{10}=\frac{150}{10}=15(i)$
Number of elements in $B_1\cup B_2\cup B_3\cup ...\cup B_n=3n(\text{When repetition is not allowed})$
But each element is repeated 09 times
$n(S)=\frac{3n}{9}=\frac{n}{3}$
From (i) and (ii) we get $\frac{n}{3}=15;n=45$. Hence, the correct option is (c)

Question:30

Two finite sets have m and n elements. The number of subsets of the first set is 112 more than that of the second set. The values of m and n are, respectively, (a) 4,7, (b) 7,4, (c) 4,4, (d) 7 7
Answer:

According to the question we have,
${2}^m-2^n=112$

$2^n(2^{m-n}-1)=2^4\times 7$

$2^n=2^4\text{ and }{2}^{m-n}-1=7$

$n=4\text{ and }{2}^{m-n}=1+7=8=2^3$

$$\begin{gathered} n+4\text{ and }m-n=3 \\ m=7 \end{gathered}$$

Question:31

The set (A ∩ B′)′ ∪ (B ∩ C) is equal to
A. A′∪ B ∪ C
B. A′ ∪ B
C. A′ ∪ C′
D. A′ ∩ B

Answer:

The answer is option B.
We know that
$\begin{array}{rlrl}(A\cap B{)}^{\prime }& =A^{\prime }\cup B^{\prime }& & [\text{De Morgan's Law}]\\ (A\cap B{)}^{\prime }\cup (B\cap C)& =[A^{\prime }\cup B^{\prime }]\cup (B\cap C)\\ & =(A^{\prime }\cup B)\cup (B\cap C)& & [(B^{\prime }{)}^{\prime }=B]\\ & =A^{\prime }\cup B\end{array}$

Question:32

Let F1 be the set of parallelograms, F2 the set of rectangles, F3 the set of rhombuses, F4 the set of squares and F5 the set of trapeziums in a plane. Then F1 may be equal to
(a) F2 ∩F3
(b) F3 ∩F4
(c) F2 u Fs
(d) F2 ∪ F3 ∪ F4 ∪ F1

Answer:

The answer is option (d). Rectangles, rhombi, and squares in a plane a parallelograms but trapezia are not
$F_1=F_2\cup F_3\cup F_4\cup F_1$
Hence, (d) is correct.

Question:33

Let S = the set of points inside the square, T = the set of points inside the triangle and C = the set of points inside the circle. If the triangle and circle intersect each other and are contained in a square. Then

(A) S ∩ T ∩ C = φ
(B) S ∪ T ∪ C = C
(C) S ∪ T ∪ C = S
(D) S ∪ T = S ∩ C

Answer:

The answer is option (c)

From the above Venn diagram
$S\cup T\cup C=S$
Hence, the correct option is (C).

Question:34

Let R be the set of points inside a rectangle of sides a and b (a, b > 1) with two sides along the positive direction of the x-axis and y-axis. Then
A. $R=\{(x,y):0\le x\le a,0\le y\le b\}$
B. $R=\{(x,y):0\le x<a,0\le y\le b\}$
C.$R=\{(x,y):0\le x<a,0<y<b\}$
D. $R=\{(x,y):0<x<a,0<y<b\}$

Answer:

The answer is option (d)
R be a set of points inside a rectangle of sides a and b (a, b > 1) with two sides along the positive direction of the x-axis and y-axis
$R=\{(x,y):0<x<a,0<y<b\}$
Hence, the correct option is (d).

Question:35

In a class of 60 students, 25 students play cricket and 20 students play tennis, and 10 students play both games. Then, the number of students who play neither is
(a) 0 (b) 25 (c) 35 (d) 45

Answer:

The answer is option (b).
Let C be the set of students who play cricket and T the set of students who play tennis.
Total number of students = 60 ⇒n(U)=60
Number of students who play cricket = 25 $\Rightarrow$ n(C)=25
Number of students who play tennis = 20 $\Rightarrow$ n(T)=20
Number of students who play both the games = 10 $\Rightarrow n(C\cap T)=10$
Number of students who play any one game$=n(C\cup T)=n(C)+n(T)-nC\cap T=25+20=10=35$
Number of students who play neither $=n(U)-n(C\cup T)=60-35=25$

Question:36

In a town of 840 persons, 450 persons read Hindi, 300 read English and 200 read both. Then the number of persons who read neither is
(a) 210 (b) 290 (c) 180 (d) 260

Answer:

The answer is option (b)
Total number of persons in a town = 840 $\Rightarrow$ n(U) = 840
Number of students who read Hindi = 450 $\Rightarrow$ n(H) = 450
Number of students who play English = 300 $\Rightarrow$ n(E) = 300
Number of students who read both = 200 $\Rightarrow n(H\cap E)=200$
$\begin{array}{rl}n(H\cup E)=n(H)+n(E)-n(H\cap E)& =450+300-200=550\\ n(H^{\prime }\cap E^{\prime })=n(U)-n(H\cup E)& =840-550=290\end{array}$

Question:37

If $X=\{8^n-7n-1:n\in N\}$} and $Y=\{49n-49:n\in N\}$. Then
(A) $X\subset Y$ (B) $Y\subset X$ (C) X = Y (D) $X\cap Y=\varphi$

Answer:

The answer is option (a).
Given that $X=\{8^n-7n-1:n\in N\}=\{0,49,490,...\}$
And $Y=\{49n-49:n\in N\}=\{0,49,98,...\}$
Here, it is clear that every element belonging to the X is also present in Y.
$X\subset Y$. Hence, the correct option is (A).

Question:38

A survey shows that 63% of people watch a News Channel, whereas 76% watch another channel. If x% of the people watch both channels, then
(A) x = 35 (B) x = 63 (C) $39\le x\le 69$ (D) x = 39

Answer:

Let p% of people watch a channel and q% of people watch another channel
$n(p\cap q)=x\mathrm{\%}\text{ and }n(p\cup q)\le 100$
So, $n(p\cup q)\ge n(p)+n(q)-n(p\cap q)$
$100\ge 63+76-x$
$x\ge 39$
Now n(p) = 63.
$n(p\cap q)\le n(p)$
$x\le 63$
So, $39\le x\le 63$
Hence, the correct option is (C).

Question:39

If sets A and B are defined as $A=\{(x,y):y=\frac{1}{x},0\ne x\in R\}$ $B=\{(x,y):y=-x,x\in R\}$, then
(A) A ∩ B = A (B) A ∩ B = B (C) A ∩ B = φ (D) A ∪ B = A

Answer:

The answer is option (c). Given that $A=\{(x,y):y=\frac{1}{x},0\ne x\in R\}$ and $B=\{(x,y):y=-x,x\in R\}$
So, $y=\frac{1}{x}$ and $y=-x$
$\frac{1}{x}\ne -x$
$A\cap B=\varphi$
Thus, option C is correct.

Question:40

If A and B are two sets, then A ∩ (A ∪ B) equals
(A) A (B) B (C) $\varphi$ (D) (A ∪ B)

Answer:

The answer is option (a). Let $x\in A\cap (A\cup B)$
$\begin{array}{rl}\Rightarrow x\in A& \text{ and }x\in (A\cup B)\\ \Rightarrow x\in A& \text{ and }(x\in A\text{ or }x\in B)\\ \Rightarrow (x\in A& \text{ and }x\in A)\text{ or }(x\in A\text{ and }x\in B\\ \Rightarrow x\in A& \text{ or }x\in A\cap B\end{array}$
Hence, A is correct.

Question:41

If A = {1, 3, 5, 7, 9, 11, 13, 15, 17}, B = {2, 4, ..., 18} and N, the set of natural numbers, is the universal set, then A′ ∪ (A ∪ B) ∩ B′) is
(A) φ (B) N (C) A (D) B

Answer:

The answer is option (b).
Given that A ={1,3,5,7,9,11,13,15,17}
B = {2,4,6………..,18}
U=N={1,2,3,4,5,……..}
$\begin{array}{rl}{A}^{\prime }\cup (A\cup B)\cap B^{\prime }& =A^{\prime }\cup [(A\cap B^{\prime })\cup (B\cap B^{\prime })]\\ & =A^{\prime }\cup (A\cap B^{\prime })\cup \varphi & [A\cap A^{\prime }=\varphi ]\\ & =A^{\prime }\cup (A\cap B^{\prime })\\ & =(A^{\prime }\cup A)\cap (A^{\prime }\cup B^{\prime })\\ & =N\cup (A^{\prime }\cup B^{\prime })\\ & =A^{\prime }\cup B^{\prime }\\ & =(A\cap B{)}^{\prime }={\varphi }^{\prime }=N\end{array}$

Question:42

Let S = {x | x is a positive multiple of 3 less than 100}, P = {x | x is a prime number less than 20}. Then n(S) + n(P) is
(A) 34 (B) 41 (C) 33 (D) 30

Answer:

The answer is option (b).
Given that: S={x| x is a positive multiple of 3 < 100}
S = {3, 6, 9, 12, 15, 18 …..99}
n(S) = 33
T = { x| x is a prime number < 20}
T = {2, 3, 5, 7, 11, 13, 17, 19}
n(T) = 8
So, n(S) + n(T) = 33+8 =41
Hence, (b) is correct.

Question:43

If X and Y are two sets and X′ denotes the complement of X, then X ∩ (X ∪ Y)′ is equal to
(A) X (B) Y (C) φ (D) X ∩ Y

Answer:

The answer is option (c)
Let

$\begin{array}{r}x\in X\cap (X\cup Y{)}^{\prime }\\ x\in X\cap (X^{\prime }\cap Y^{\prime })\\ x\in (X\cap X^{\prime })\cap (X\cap Y^{\prime })\\ xx\in \varphi \cap (X\cap Y^{\prime })\\ x\in \varphi \end{array}$

Question:44

The set $\{x\in R:1\le x<2\}$ can be written as ______________.

Answer:

The set $\{x\in R:1\le x<2\}$ can be written as [1,2)

Question:45

When$A=\varphi$, then the number of elements in P(A) is ______________.

Answer:

When$A=\varphi$, then number of elements in P(A) is 1
Here $A=\varphi$, n(A) = 0
$\begin{array}{rl}n[P(A)]& =2^{n(A)}\\ & =2^0\\ & =1\end{array}$

Question:46

If A and B are finite sets such that A ⊂ B, then n (A ∪ B) = ______________.

Answer:

Since,

Question:47

If A and B are any two sets, then A – B is equal to

Answer:

From the Venn diagram
$A-B=A\cap B^{\prime }$

Question:48

Power set of the set A = {1, 2} is ______________.

Answer:

$\text{Power set of }A=P(A)=\{\{1\},\{2\},\{1,2\},\varphi \}$

Question:49

Given the sets A = {1, 3, 5}. B = {2, 4, 6} and C = {0, 2, 4, 6, 8}. Then the universal set of all three sets A, B, and C can be ______________.

Answer:

Given that A = {1,3,5}, B= {2,4,6} and C = {0,2,4,6,8}
Universal set of all given sets is U = {0, 1, 2, 3, 4, 5, 6, 8}

Question:50

If U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, A = {1, 2, 3, 5}, B = {2, 4, 6, 7} and C = {2, 3, 4, 8}. Then
(i) (B ∪ C)′ is ______________. (ii)(C – A)′ is ______________.

Answer:

Given that: U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, A = {1, 2, 3, 5}, B ={2, 4, 6, 7} and C = {2, 3, 4, 8}

1. $(B\cup C{)}^{\prime }=\{2,3,4,6,7,8{\}}^{\prime }=\{1,5,9,10\}$

2. $(C-A{)}^{\prime }=\{1,2,3,5,6,7,9,10\}$

Question:51

For all sets A and B, A – (A ∩ B) is equal to ______________.

Answer:

Since $A-B=A\cap B^{\prime }$
$A-(A\cap B)=A\cap B^{\prime }$

Question:52

If A is any set, then A ⊂ A

Answer:

Every set is a subset of itself. Hence, true.

Question:53

Given that M = {1, 2, 3, 4, 5, 6, 7, 8, 9} and if B = {1, 2, 3, 4, 5, 6, 7, 8, 9}, then $B\not\subset M$

Answer:

M = {1, 2, 3, 4, 5, 6, 7, 8, 9}
And B = {1, 2, 3, 4, 5, 6, 7, 8, 9}
Since B and M have the same elements
M = B, So, $B\not\subset M$ is false

Question:54

The sets {1, 2, 3, 4} and {3, 4, 5, 6} are equal.

Answer:

The two sets do not contain the same elements. Hence, false.

Question:55

$Q\cup Z=Q$, where Q is the set of rational numbers and Z is the set of integers.

Answer:

Since every integer is a rational number
$\therefore Z\subset Q\Rightarrow Q\cup Z=Q$
Hence true

Question:56

Let sets R and T be defined as
$R=\{x\in Z:x\text{ is divisible by 2}\}$
$T=\{x\in Z:x\text{ is divisible by 6}\}$
Then T ⊂ R

Answer:

R = {………….., -8, -6, -4, -2, 0, 2, 4, 6, 8,…..}
And T {……., -18, -12, -6, 0, 6, 12, 18,…….}
Since every element of T is present in R. So, $T\subset R$
Hence, the statement is true.

Question:57

Given$A=\{0,1,2\}$, $B=\{x\in R:0\le x\le 2\}$. Then A = B.

Answer:

Here, A = { 0, 1, 2} B is a set having all real numbers from 0 to 2.
So, $A\ne B$
Hence, the given statement is False.

Question:58

Match the following sets for all sets A, B and C

1. $((A^{\prime }\cup B^{\prime })-A{)}^{\prime }$ (a) $A-B$

2. $[B^{\prime }\cup (B^{\prime }-A^{\prime }){]}^{\prime }$ (b) $A$

3. $(A_B)-(B-C)$ (c) $B$

4. $(A-B)\cap (C-B)$ (d) $(A\times B)\cap (C\times B)$

5. $A\times (B\cap C)$ (e) $(A\times B)\cup (A\times C)$

6. $A\times (B\cup C)$ (f) $(A\cap C)-B$