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Page No 1.10:

Question 4:

Are the following sets equal?
A = {x : x is a letter in the word reap}:
B = {x : x is a letter in the word paper};
C = {x : x is a letter in the word rope}.

Answer:

A = {r, e, a, p}
B = {p, a, e, r}
C = {r, o, p, e}
Here, A = B because every element of A is a member of B & every element of B is a member of A.
But every element of C is not a member of A & B.
Also, every element of A and B is not a member of C.
Therefore, we can say that these sets are not equal.

Page No 1.10:

Question 5:

From the sets given below, pair the equivalent sets:
A=1, 2, 3, B = t, p, q, r, s, C = α, β, γ, D = a, e, i, o, u.

Answer:

Two sets A & B are equivalent if their cardinal numbers are equal, i.e., n(A) = n(B).
n(A) = 3
n(B) = 5
n(C) = 3
n(D) =5
Therefore, equivalent sets are (A and C) and (B and D).

Page No 1.10:

Question 6:

Are the following pairs of sets equal? Give reasons.
(i) A = {2, 3}, B = {x : x is a solution of x2 + 5x + 6 = 0};
(ii) A = {x : x is a letter of the word " WOLF"};
    B = {x : x is a letter of the word " FOLLOW"}.

Answer:

(i) A = {2, 3}
    B = {-2, -3}
A is not equal to B because every element of A is not a member of B & every element of B is not a member of A.

(ii) A = {W, O, L, F}
     B = {F, O, L, W}
Here, A = B because every element of A is a member of B & every element of B is a member of A.

Page No 1.10:

Question 7:

From the sets given below, select equal sets and equivalent sets.
A = {0, a}, B = {1, 2, 3, 4} C = {4, 8, 12}, D = {3, 1, 2, 4},
E = {1, 0}, F = {8, 4, 12} G = {1, 5, 7, 11}, H = {a, b}.

Answer:

Equal sets:
(a) B and D, because every element of B is a member of D & every element of D is a member of B.
(b) C and F, because every element of C is a member of F & every element of F is a member of C
                
Equivalent sets:
(a) A, E and H      { n(A) = n(E) =n(H) = 2}
(b) B, D and G     { n(B) = n(D) =n(G) = 4}
(c) C and F           { n(C) = n(F)  = 3}
                        
                  

Page No 1.10:

Question 8:

Which of the following sets are equal?
A = {x : xN, x, < 3},
B = {1, 2}
C = {3, 1}
D = {x : xN, x is odd, x < 5},
E = {1, 2, 1, 1} F = {1, 1, 3}.

Answer:

A = {1, 2}
B = {1, 2}
C = {3, 1}
D = {1, 3}
E = {1, 2, 1, 1} = {1, 2}
F = {1, 1, 3} = {1, 3}
A = B = E and C = D = F

Page No 1.10:

Question 9:

Show that the set of letters needed to spell "CATARACT" and the set of letters needed to spell "TRACT" are equal.

Answer:

Letters required to spell CATARACT are {C, A, T, R}. Let this set be denoted as E.
E = {C, A, T, R}
Letters required to spell TRACT are {T, R, A, C}. Let this set be denoted as F.
F = {T, R, A, C}
The two sets E & F are equal because every element of E is a member of F & every element of F is a member of E.



Page No 1.16:

Question 1:

Which of the following statements are true? Give reason to support your answer.
(i) For any two sets A and B either AB or BA;
(ii) Every subset of an infinite set is infinite;
(iii) Every subset of a finite set is finite;
(iv) Every set has a proper subset;
(v) {a, b, a, b, a, b, ...} is an infinite set;
(vi) {a, b, c} and {1, 2, 3} are equivalent sets;
(vii) A set can have infinitely many subsets.

Answer:

(i) False It is not necessary that for any two sets A & B, either AB or BA.It is not satisfactory always.  Let: A={1,2} & B={α,β,γ}Here, neither AB nor BA.

(ii) False   A={-1,0,1,2,3} is a finite set that is a subset of infinite set Z.     

(iii) True Every subset of a finite set is a finite set.
 
(iv) Falseϕ does not have a proper subset.

(v) False{a,b,a,b,a,b,...} will be equal to {a,b}, which is a finite set.

(vi) True{a,b,c} and {1,2,3} are equivalent sets because the number of elements in both the sets are same.

(vii) FalseIn the set A={1,2}, subsets can be {ϕ}, {1} and {2}, which are finite.

Page No 1.16:

Question 2:

State whether the following statements are true or false:
(i) 1 1, 2, 3
(ii) a{b, c, a}
(iii) a a, b, c
(iv) a, b=a, a, b, b, a 
(v) The set {x ; x + 8 = 8} is the null set.

Answer:

(i) True
(ii) False
It should be written as {a} {b,c,a} or a{b,c,a}.
(iii) False
It should be written as {a} {b,c,a} or a{b,c,a}.
(iv) True
(v) False
The element of the set {x ; x + 8 = 8} is {0}. Therefore, it is not an empty or null set.

Page No 1.16:

Question 3:

Decide among the following sets, which are subsets of which:
A={x : x satisfies x2-8x+12=0},
B = 2, 4, 6, C = 2, 4, 6, 8, ..., D=6.

Answer:

We have:
  A={x : x satisfies x2-8x+12=0.}={2,6}
  B = {2, 4, 6}
  C = {2, 4, 6, 8,...}
  D = {6}
Therefore, we can say that DABC.

Page No 1.16:

Question 4:

Write which of the following statements are true? Justify your answer.
(i) The set of all integers is contained in the set of all set of all rational numbers.
(ii) The set of all crows is contained in the set of all birds.
(iii) The set of all rectangle is contained in the set of all squares.
(iv) The set of all real numbers is contained in the set of all complex numbers.
(v) The sets P = {a} and B = {{a}} are equal.
(vi) The sets A = {x : x is a letter of the word "LITTLE"} and,
B = {x : x is a letter of the word "TITLE"} are equal.

Answer:

(i) True 
A rational number is any mn, where m and n are any integers (n0). Any integer can be put into that form by setting n = 1. Therefore, the set of all integers is contained in the set of all rational numbers.
(ii) True
All crows are birds. Therefore, the set of all crows is contained in the set of all birds.
(iii) False
Every square can be a rectangle, but every rectangle cannot be a square.
(iv) True
Every real number can be written in the (a + bi) form. Thus, we can say that the
set of all real numbers is contained in the set of all complex numbers.
(v) False
P = {a}
B = {{a}} = {P}
P{P}
(vi) True
We have:
A =
{x:x is a letter of the word LITTLE} = {L, I, T, E}
B = {x:x is a letter of the word TITLE} = {T, I, L, E}
Sets A & B are equal because every element of A is a member of B & every element of B is a member of A.

Page No 1.16:

Question 5:

Which of the following statements are correct?
Write a correct form of each of the incorrect statements.
(i) aa, b, c
(ii) aa, b, c
(iii) a{a, b}
(iv) aa, b 
(v) b, ca,b, c
(vi) a, ba, b, c
(vii) ϕa, b
(viii) ϕa, b, c
(ix) x:x+3=3=ϕ

Answer:

Here, (viii) is correct.

The correct forms of each of the incorrect statements are:
(i) aa, b, c
(ii) aa, b, c
(iii) {a}{a, b}
(iv) {a}a, b 
(v) b, ca,b, c
(vi) {a,b}⊄{a,{b,c}}
(vii) ϕa, b
(ix) x:x+3=3ϕ



Page No 1.17:

Question 6:

Let A = {a, b, {c, d}, e}. Which of the following statements are false and why?
(i) c, dA
(ii) c, dA
(iii) c, dA
(iv) aA
(v) aA
(vi) a, b, eA
(vii) a, b, eA
(viii) a, b, cA
(ix) ϕA
(x) ϕA

Answer:

A = {a, b, {c, d}, e}
(i) False
The correct statement would be {c, d}A.
(ii) True
(iii) True
(iv) True
(v) False
The correct statement would be {a}⊂ A or a A.
(vi) True
(vii) False
The correct statement would be a, b, eA.
(viii) False
The correct statement would be {a, b, c} ⊄ A.
(ix) False
A null set is a subset of every set. Therefore, the correct statement would be ϕA.
(x) False
ϕ is an empty set; in other words, this set has no element. It is denoted by ϕ. Therefore, the correct statement would be ϕA.

Page No 1.17:

Question 7:

Let A = {{1, 2, 3}, {4, 5}, {6, 7, 8}}. Determine which of the following is true or false:
(i) 1A
(ii) 1, 2, 3A
(iii) 6, 7, 8A
(iv) 4, 5A
(v) ϕA
(vi) ϕA

Answer:

(i) False
If it could be 1A , then it would be true .

(ii) False
The correct form would be 1, 2, 3A or {1, 2, 3}A.

(iii) True

(iv) True

(v) False
A null set is a subset of every set. Therefore, the correct form would be ϕ  A.

(vi) True

Page No 1.17:

Question 8:

Let A=ϕ, ϕ, 1, 1, ϕ, 2. Which of the following are true?
(i) ϕA
(ii) ϕA
(iii) 1A
(iv) 2, ϕA
(v) 2A
(vi) 2 1A
(vii) 2, 1A
(viii) ϕ, ϕ, 1, ϕA
(ix) ϕA

Answer:

(i) True
(ii) True
(iii) False
The correct form would be 1A.
(iv) True
(v) False
The correct form would be 1A.
(vi) True
(vii) True
(viii) True
(ix) True

Page No 1.17:

Question 9:

Write down all possible subsets of each of the following sets:
(i) {a},
(ii) {0, 1},
(iii) {a, b, c},
(iv) {1, {1}},
(v) ϕ.

Answer:

(i) ϕ ,{a}(ii) ϕ,{0},{1},{0,1}(iii)  ϕ,{a},{b},{c},{a,b},{b,c},{a,c},{a,b,c}(iv) ϕ,{1},{{1}},{1,{1}}(v) ϕ,{ϕ}

Page No 1.17:

Question 10:

Write down all possible proper subsets each of the following sets:
(i) {1, 2},
(ii) {1, 2, 3}
(iii) {1}.

Answer:

(i) {1}, {2}
(ii) {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}
(iii) No proper subsets are there in this set.

Page No 1.17:

Question 11:

What is the total number of proper subsets of a set consisting of n elements?

Answer:

We know that the total number of subsets of a finite set consisting of n elements is 2n.
Therefore, the total number of proper subsets of a set consisting of n elements is 2n-1.

Page No 1.17:

Question 12:

If A is any set, prove that: AϕA=ϕ.

Answer:

To prove: AϕA=ϕ
Proof:
Let:
Aϕ
If A is a subset of an empty set, then A is the empty set.
A=ϕ

Now, let A=ϕ.
This means that A is an empty set.
We know that every set is a subset of itself.
Aϕ
Thus, we have:
AϕA=ϕ

Page No 1.17:

Question 13:

Prove that: AB, B C and C A  A=C.

Answer:

Let xAxB                 ABxC                 BCxA xCAC         ...1It is given tha,CA             ...(2)From 1 and 2, we haveA=C

Page No 1.17:

Question 14:

How many elements has P A, if A=ϕ?

Answer:

Given : A=ϕ This means P(A) ={ϕ}.Hence, P(A) would have one element.

Page No 1.17:

Question 15:

What universal set (s) would you propose for each of the following:
(i) The set of right triangles.
(ii) The set of isosceles triangles.

Answer:

(i) The set of all triangles in a plane
(ii) The set of all triangles in a plane

Page No 1.17:

Question 16:

If X=8n-7n-1:nN and Y=49n-1:nN, then prove that XY.

Answer:

Given:
X=8n-7n-1:nN and Y=49n-1:nN

To prove:
XY
Let: xn=8n-7n-1, nN x1=8-7-1=0For any n2, we have: xn=8n-7n-1=(1+7)n -7n-1xn=C0n+C1n.7+C2n.72+C3n.73+...+Cnn.7n-7n-1xn=1+7n+C2n.72+C3n.73+...+7n-7n-1        [ C0n=1 and C1n=n]xn=72{C2n+C3n.7+C4n72+...+Cnn.7n-2}xn=49{C2n+C3n.7+C4n72+...+Cnn.7n-2}Thus, xn is some positive integral multiple of 49 for all n2. X consists of all those positive integral multiples of 49 that are of the form 49{C2n+C3n.7+C4n72+...+Cnn.7n-2} along with zero.Y={49(n-1):nN} implies that it consists of all integral multiples of 49 along with zero. XY



Page No 1.2:

Question 1:

What is the difference between a collection and a set? Give reasons to support your answer?

Answer:

Well-defined collections are sets.
Example:
The collection of good teachers in a school is not a set, It is a collection.
Thus, we can say that every set is a collection, but every collection is not necessarily a set.
The collection of vowels in English alphabets is a set.

Page No 1.2:

Question 2:

Which of the following collections are sets? Justify your answer:
(i) A collection of all natural numbers less than 50.
(ii) The collection of good hockey players in India.
(iii) The collection of all girls in your class.
(iv) The collection of most talented writers of India.
(v) The collection of difficult topics in mathematics.
(vi) The collection of all months of a year beginning with the letter J.
(Vii) A collection of novels written by Munshi Prem Chand.
(Viii) The collection of all question in this chapter.
(ix) A collection of most dangerous animals of the world.
(x) The collection of prime integers.

Answer:

(i) The collection of all natural numbers less than 50 is a set because it is well defined.
(ii) The collection of good hockey players is not a set because the goodness of a hockey player is not defined here. So, it is not a set.
(iii) The collection of all girls in a class is a set, as it is well defined that all girls of the class are being talked about.
(iv) The collection of the most talented writers of India is a set because it is well defined.
(v) The collection of difficult topics in mathematics is not a set because a topic can be easy for one student while difficult for the other student.
(vi) The collection of all months of a year beginning with the letter J is a set given by {January, June, July}
(vii) A collection of novels written by Munshi Prem Chand is a set because one can determine whether the novel is written by Munshi Prem Chand or not.
(Viii) The collection of all question in this chapter is a set because one can easily check whether it is a question of the chapter or not.
(ix) A collection of most dangerous animals of the world is not a set because we cannot decide whether the animal is dangerous or not.
(x) The collection of prime integers is set given by {2, 3, 5........}

Page No 1.2:

Question 3:

If A = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], then insert the appropriate symbol ∈ or ∉ in each of the following blanks spaces:
(i) 4 ...... A
(ii) −4 ...... A
(iii) 12 ...... A
(iv) 9 ...... A
(v) 0 ...... A
(vi) −2 ...... A

Answer:

(i) 4A
(ii) −4A
(iii) 12A
(iv) 9A
(v) 0A
(vi) −2A



Page No 1.21:

Question 1:

If A and B are two sets such that AB, then find:
(i) AB
(ii) AB

Answer:

From the Venn diagrams given below, we can clearly say that if A and B are two sets such that AB, then

(i) Form the given Venn diagram, we can see that  AB = A


(ii) Form the given Venn diagram, we can see that  AB = B

Page No 1.21:

Question 2:

If A = {1, 2, 3, 4, 5}, B = {4, 5, 6, 7, 8}, C = {7, 8, 9, 10, 11} and D = {10, 11, 12, 13, 14}, find:
(i) AB
(ii) AC
(iii) BC
(iv) BD
(v) ABC
(vi) ABD
(vii) BCD
(viii) ABC
(ix) ABBC
(x) ADBC.

Answer:

Given:
A = {1, 2, 3, 4, 5}, B = {4, 5, 6, 7, 8}, C = {7, 8, 9, 10, 11} and D = {10, 11, 12, 13, 14}
(i) AB = {1, 2, 3, 4, 5, 6, 7, 8}
(ii) AC = {1, 2, 3, 4, 5, 7, 8, 9, 10, 11}
(iii) BC = {4, 5, 6, 7, 8, 9, 10, 11}
(iv) BD = {4, 5, 6, 7, 8, 10, 11, 12, 13, 14}
(v) ABC = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}
(vi) ABD = {1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14}
(vii) BCD = {4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14}
(viii) ABC = {4, 5}
(ix) ABBC = ϕ
(x) ADBC = {4, 5, 10, 11}

Page No 1.21:

Question 3:

Let A=x:xN, B=x:x-2n, nN, C=x:x=2n-1, nN and D = {x : x is a prime natural number}. Find:
(i) AB
(ii) AC
(iii) AD
(iv) BC
(v) BD
(vi) CD

Answer:

A=x:xN={1,2,3,...}B=x:x-2n, nN={2,4,6,8,...} C=x:x=2n-1, nN={1,3,5,7,...}
D = {x:x is a prime natural number.} = {2, 3, 5, 7,...}
(i) AB = B
(ii) AC = C
(iii) AD = D
(iv) BC = ϕ
(v) BD = {2}
(vi) CD = D-{2}

Page No 1.21:

Question 4:

Let A = {3, 6, 12, 15, 18, 21}, B = {4, 8, 12, 16, 20}, C = {2, 4, 6, 8, 10, 12, 14, 16} and D = {5, 10, 15, 20}. Find:
(i) A-B
(ii) A-C
(iii) A-D
(iv) B-A
(v) C-A
(vi) D-A
(vii) B-C
(viii) B-D

Answer:

Given:
A = {3, 6, 12, 15, 18, 21}, B = {4, 8, 12, 16, 20}, C = {2, 4, 6, 8, 10, 12, 14, 16} and D = {5, 10, 15, 20}
(i) A-B  = {3, 6, 15, 18, 21}
(ii) A-C = {3, 15, 18, 21}
(iii) A-D = {3, 6, 12, 18, 21}
(iv) B-A = {4, 8, 16, 20}
(v) C-A = {2, 4, 8, 10, 14, 16}
(vi) D-A = {5, 10, 20}
(vii) B-C = {20}
(viii) B-D = {4, 8, 12, 16}

Page No 1.21:

Question 5:

Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4}, = {2, 4, 6, 8} and C = {3, 4, 5, 6}. Find
(i) A'
(ii) B'
(iii) AC'
(iv) AB'
(v) A''
(vi) B-C'

Answer:

Given:
U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4}, B= {2, 4, 6, 8} and C = {3, 4, 5, 6}
(i) A' = {5, 6, 7, 8, 9}
(ii) B' = {1, 3, 5, 7, 9}
(iii) AC' = {1, 2, 5, 6, 7, 8, 9}
(iv) AB' = {5, 7, 9}
(v) A'' = {1, 2, 3, 4} = A
(vi) B-C' = {1, 3, 4, 5, 6, 7, 9}

Page No 1.21:

Question 6:

Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2, 4, 6, 8} and B = {2, 3, 5, 7}. Verify that
(i) AB'=A'B'
(ii) AB'=A'B'.

Answer:

Given:
U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2, 4, 6, 8} and B = {2, 3, 5, 7}
We have to verify:

(i) AB'=A'B'
LHS
AB ={2,3,4,5,6,7,8}AB '={1,9}

RHS
A'={1,3,5,7,9}B'={1,4,6,8,9}A'B'={1,9}

LHS = RHS
Hence proved.

(ii) AB'=A'B'
LHS
AB={2}AB'={1,3,4,5,6,7,8,9}

RHS
A'={1,3,5,7,9}B'={1,4,6,8,9}A'B'={1,3,4,5,6,7,8,9}

LHS = RHS
Hence proved.



Page No 1.27:

Question 1:

Find the smallest set A such that A1, 2=1, 2, 3, 5, 9.

Answer:

We have to find the smallest set A such that A1, 2=1, 2, 3, 5, 9.

The union of the two sets A & B is the set of all those elements that belong to A or to B or to both A & B.

Thus, A must be {3, 5, 9}.

Page No 1.27:

Question 2:

Let A = {1, 2, 4, 5} B = {2, 3, 5, 6} C = {4, 5, 6, 7}. Verify the following identities:
(i) ABC=ABAC
(ii) ABC=ABAC
(iii) AB-C=AB-AC
(iv) A-BC=AA-BA-C
(v) A-BC=A-BA-C
(vi) ABC=ABAC.

Answer:

Given:
A = {1, 2, 4, 5}, B = {2, 3, 5, 6} and C = {4, 5, 6, 7}
We have to verify the following identities:
(i) ABC=ABAC

LHS
(BC)={5,6}A(BC)={1,2,4,5,6}

RHS
AB={1,2,3,4,5,6}AC={1,2,4,5,6,7}ABAC={1,2,4,5,6}

LHS = RHS
∴ ABC=ABAC

(ii) ABC=ABAC

LHS
(BC)={2,3,4,5,6,7}A(BC)={2,4,5}

RHS
AB={2,5}AC={4,5}ABAC={2,4,5}

LHS = RHS
ABC=ABAC

 (iii) AB-C=AB-AC
 
LHS
(B-C) ={2,3}A(B-C)={2}

RHS
(AB)={2,5}(AC)={4,5}(AB)-(AC)={2}

LHS = RHS
AB-C=AB-AC

 (iv) A-BC=A-BA-C

LHS
(BC)={2,3,4,5,6,7}A-(BC)={1}

RHS
(A-B)={1,4}(A-C)={1,2}(A-B)(A-C)={1}

LHS = RHS
∴ A-BC=A-BA-C

(v) A-BC=A-BA-C

LHS
(BC)={5,6}A-(BC)={1,2,4}

RHS
(A-B)={1,4}(A-C)={1,2}(A-B)(A-C)={1,2,4}

LHS = RHS
A-BC=A-BA-C


(vi) ABC=ABAC

LHS
(BC)=(B-C)(C-B)(B-C)={2,3}(C-B)={4,7}(B-C)(C-B)={2,3,4,7}(BC)={2,3,4,7}A(BC)={2,4}

RHS
(AB)={2,5}(AC)={4,5}(AB)(AC)={(AB)-(AC)}{(AC)-(AB)}(AB)-(AC)={2}(AC)-(AB)={4}{(AB)-(AC)}{(AC)-(AB)}={2,4}(AB)(AC)={2,4} 

LHS = RHS
∴ ABC=ABAC

Page No 1.27:

Question 3:

If U = {2, 3, 5, 7, 9} is the universal set and A = {3, 7}, B = {2, 5, 7, 9}, then prove that:
(i) AB'=A'B'
(ii) AB'=A'B'.

Answer:

Given:
U = {2, 3, 5, 7, 9}
A = {3, 7}
B = {2, 5, 7, 9}

To prove :
(i) AB'=A'B'
(ii) AB'=A'B'

Proof :

(i) LHS:
(AB)={2,3,5,7,9}(AB)'=ϕ

RHS:
A'={2,5,9}B'={3}A'B'=ϕLHS=RHS 

AB'=A'B'

(ii) LHS:
(AB)={7}(AB)'={2,3,5,9}

RHS:
A'={2,5,9}B'={3}A'B'={2,3,5,9}

LHS = RHS

AB'=A'B'

Page No 1.27:

Question 4:

For any two sets A and B, prove that

(i) B ⊂ A ∪ B                          (ii) A ∩ A                          (iii) AA ∩ B = A

Answer:

(i) For all xB

xA or xB

xA ∪ B            (Definition of union of sets)

B ⊂ A ∪ B

(ii) For all A ∩ B

xA and x ∈ B              (Definition of intersection of sets)

xA

⇒ A ∩ A

(iii) Let AB. We need to prove A ∩ B = A.

For all xA

xA and x ∈ B          (AB)

xA ∩ B 

AA ∩ B    
  
Also, AA

Thus, AA ∩ B and AA

A ∩ B = A         [Proved in (ii)]

∴ AA ∩ B = A

Page No 1.27:

Question 5:

For any two sets A and B, show that the following statements are equivalent:
(i) AB
(ii) A-B=ϕ
(iii) AB=B
(iv) AB=A.

Answer:

We have that the following statements are equivalent:
(i) AB
(ii) A-B=ϕ
(iii) AB=B
(iv) AB=A

Proof:
Let ABLet x be an arbitary element of (A-B). Now,x(A-B)xA & xB          (Which is contradictory) Also,ABA-Bϕ                 ...(1) We know that null sets are the subsets of every set.ϕ  A-B     ...(2)From (1) & (2), we get,(A-B)=ϕ(i)=(ii)Now, we have,(A-B)=ϕThat means that there is no element in A that does not belong to B.Now, AB=B(ii)=(iii) We have,AB=BABAB=A(iii)=(iv)We have, AB=AIt should be possible if AB.Now,AB (iv)=(i)We have,(i)=(ii)=(iii)=(iv) Therefore, we can say that all statements are equivalent.

Page No 1.27:

Question 6:

For three sets A, B and C, show that
(i) AB=AC need not imply B = C.
(ii) ABC-BC-A

Answer:

(i) Let A = {2, 4, 5, 6},  B = {6, 7, 8, 9} and C = {6, 10, 11, 12,13}

So, AB=6 and AC=6Hence, AB=AC but BC

(ii)

Let zC-B      ...(1)zC and zBzC and zA       ABzC-A        ...(2)From (1) and (2), we getC-BC-A

Page No 1.27:

Question 7:

For any two sets, prove that:
(i) AAB=A
(ii) AAB=A

Answer:

(i)

LHS = AABLHS=AAAB     LHS=AAB                AABLHS=A = RHS

(ii)

LHS=AABLHS=AAAB     LHS=AAB LHS=A = RHS

Page No 1.27:

Question 8:

Find sets A, B and C such that AB, AC and BC are non-empty sets and ABC=ϕ.

Answer:

Let us consider the following sets,

A = {5, 6, 10 }
B = {6,8,9}
C = {9,10,11}

Clearly, AB=6BC=9, AC=10 and ABC = ϕIt means that, AB,BC and AC are non empty setsand ABC = ϕ

Page No 1.27:

Question 9:

For any two sets A and B, prove that: AB=ϕAB'.

Answer:

Let aAaB          AB=ϕ.

aB'

Thus, aA and aB'  AB'.

Page No 1.27:

Question 10:

If A and B are sets, then prove that A-B, AB and B-A are pair wise disjoint.

Answer:

i  A-B and  ABLet aA-BaA and aBaABHence, A-B and  AB are disjoint sets.ii  B-A and  ABLet aB-AaB and aAaABHence, B-A and  AB are disjoint sets.iii  A-B and B-A A-B=x:xA and xB B-A=x:xB and xAHence,  A-B and  B-A are disjoint sets.

Page No 1.27:

Question 11:

Using properties of sets, show that for any two sets A and B,
ABAB'=A

Answer:

LHS=ABAB'LHS=ABAABB'LHS=ABAABB'LHS=AABB'LHS=AAB'BB'      BB=ϕLHS=AAB'LHS=A=RHS

Page No 1.27:

Question 12:

For any two sets of A and B, prove that:
(i) A'B=UAB
(ii) B'A'AB

Answer:

i Let aA.aUaA'B       U=A'BaB       aA'Hence, AB.

ii Let aA.aA'aB'     B'A'aBHence, AB.

Page No 1.27:

Question 13:

Is it true that for any sets A and B, P AP B=P AB? Justify your answer.

Answer:

False.Let XPAPB XPA or  XPBXA or XBXABXPAB PAPB PAB     ...1Again, let XPABBut XPA or xPB       For example let A=2,5 and B=1,3,4 and take X=1,2,3,4So, XPAPBThus, PAB is not necessarily a subset of PAPB.

Page No 1.27:

Question 14:

Show that for any sets A and B,
(i) A=ABA-B
(ii) AB-A=AB

Answer:

(i)

RHS=ABA-BRHS=ABAB'RHS=ABAABB'RHS=AAB'BB'RHS=AAB'URHS=AAB'RHS=A=LHS

(ii)
LHS=AB-ALHS=ABA'LHS=ABAA'LHS=ABULHS=AB=RHS

Page No 1.27:

Question 15:

Each set X, contains 5 elements and each set Y, contains 2 elements and r=120Xr=S=r=1nYr. If each element of S belong to exactly 10 of the Xr's and to eactly 4 of Yr's, then find the value of n.

Answer:

It is given that each set X contains 5 elements and r=120Xr=S.

nS=20×5=100

But, it is given that each element of S belong to exactly 10 of the Xr's.

∴ Number of distinct elements in S = 10010=10           .....(1)

It is also given that each set Y contains 2 elements and r=1nYr=S.

nS=n×2=2n

Also, each element of S belong to eactly 4 of Yr's.

∴ Number of distinct elements in S = 2n4                  .....(2)

From (1) and (2), we have

2n4=10n=20

Hence, the value of n is 20.



Page No 1.34:

Question 1:

For any two sets A and B, prove that : A'-B'=B-A

Answer:

LHS=A'-B'=A'B''              C-D=CD'=A'B=BA'=B-A                    CD'=C-DRHS=B-A

So, LHS = RHS

Page No 1.34:

Question 2:

For any two sets A and B, prove the following:
(i) AA'B=AB
(ii) A-A-B=AB
(iii) AAB'=ϕ
(iv) A-B=A ΔAB.

Answer:

(i)
LHS=AA'B        =AA'AB        =ϕAB        =AB=RHS
Hence proved.

(ii)
LHS=A-A-B        =A-AB'        =AAB''        =AA'B''        =AA'B        =AA'AB        =AB        =AB=RHS
Hence proved.

(iii)
LHS=AAB'        =AA'B'        =AA'AB'        =ϕAB'           =ϕ=RHS                   ϕA=ϕ
Hence proved.

(iv)
LHS=A ΔAB        =A-ABAB-A        =AAB'ABA'        =AA'B'ABA'        =AA'AB'AA'BA'        =ϕAB'ϕBA'        =AB'ϕ        =AB'        =A-B=RHS     .

Hence proved.

Page No 1.34:

Question 3:

If A, B, C are three sets such that AB, then prove that
C-BC-A.

Answer:

Let aC-BaC and aBaC and aA   AB    aC-A       Hence, C-BC-A 



Page No 1.35:

Question 4:

For any two sets A and B, prove that

(i) AB-B=A-B
(ii) A-AB=A-B
(iii) A-A-B=AB
(iv) AB-A=AB                                  [NCERT EXEMPLAR]
(v) A-BAB=A                                 [NCERT EXEMPLAR]

Answer:

(i)
AB-B=ABB'                                          X-Y=XY'                   =AB'BB'                                Distributive law                   =AB'ϕ                   =AB'                   =A-B

(ii)
A-AB=AAB'                                  X-Y=XY'                   =AA'B'                                 De Morgan law                   =AA'AB'                   =ϕAB'                   =AB'                   =A-B

(iii)
A-A-B=A-AB'                            X-Y=XY'                   =AAB''                   =AA'B''                       De Morgan law                   =AA'B                   =AA'AB                    Distributive law                   =ϕAB                   =AB

(iv)
AB-A=ABA'                         X-Y=XY'                   =ABAA'                 Distributive law                   =AB                     is the universal set                    =AB

(v)
A-BAB=AB'AB           X-Y=XY'                            =AB'B                     Distributive law                            =A                            =A



Page No 1.46:

Question 1:

If A and B are two sets such that n AB=50, n A=28 and n B=32, find n AB.

Answer:

We know:
 n(AB)=n(A)+n(B)-n(AB)50=28+32-n(AB)n(AB)=60-50=10

Page No 1.46:

Question 2:

If P and Q are two sets such that P has 40 elements, PQ has 60 elements and PQ has 10 elements, how many elements does Q have?

Answer:

Given:nP=40nPQ=60nPQ=10To find:nQWe know:nPQ=nP+nQ-nPQ60=40+nQ-10nQ=30

Page No 1.46:

Question 3:

In a school there are 20 teachers who teach mathematics or physics. Of these, 12 teach mathematics and 4 teach physics and mathematics. How many teach physics?

Answer:

Let A be the number of teachers who teach mathematics & B be the number of teachers who teach physics.

Given:nA=12nAB=20nAB=4To find:nBWe know:nAB=nA+nB-nAB20=12+nB-4nB=20-8=12Therefore, 12 teachers teach physics.

Page No 1.46:

Question 4:

In a group of 70 people, 37 like coffee, 52 like tea and each person likes at least one of the two drinks. How many like both coffee and tea?

Answer:

Let A denote the set of the people who like tea & B denote the set of the people who like coffee.

Given:n(AB)=70n(A)=52n(B)=37To find:n(AB)We know: n(AB)=n(A)+n(B)-n(AB)70=52+37-n(AB)n(AB)=19Therefore, 19 people like both tea & coffee.



Page No 1.47:

Question 5:

Let A and B be two sets such that : n A=20, n AB=42 and n AB=4. Find
(i) nB
(ii) n A-B
(iii) n B-A

Answer:

Given:
n A=20, n AB=42 and n AB=4

(i) We know:n(AB)=n(A)+n(B)-n(AB)42=20+n(B)-4n(B)=26(ii) n(A-B)=n(A)-n(AB)n(A-B)=20-4=16(iii) We know that sets follow the commutative property. n(AB)=n(BA)n(B-A)=n(B)-n(BA)n(B-A)=26-4=22

Page No 1.47:

Question 6:

A survey shows that 76% of the Indians like oranges, whereas 62% like bananas. What percentage of the Indians like both oranges and bananas?

Answer:

Let A & B denote the sets of the Indians who like oranges & bananas, respectively.

Given:nA =76%nB=62%nAB=100%nAB=? We know:nAB=nA+nB-nAB100=76+62-nABnAB=38Therefore, 38% of the Indians like both oranges & bananas.

Page No 1.47:

Question 7:

In a group of 950 persons, 750 can speak Hindi and 460 can speak English. Find:
(i) how many can speak both Hindi and English:
(ii) how many can speak Hindi only;
(iii) how many can speak English only.

Answer:

Let A & B denote the sets of the persons who like Hindi & English, respectively.
Given:nA =750nB =460nAB=950(i) We know:nAB=nA+nB-nAB950=750+460-nABnAB=260Thus, 260 persons can speak both Hindi and English.(ii) nA-B=nA-nABnA-B=750-260=490Thus, 490 persons can speak only Hindi.(iii) nB-A=nB-nAB nB-A=460-260                    =200Thus, 200 persons can speak only English.

Page No 1.47:

Question 8:

In a group of 50 persons, 14 drink tea but not coffee and 30 drink tea. Find:
(i) how may drink tea and coffee both;
(ii) how many drink coffee but not tea.

Answer:

Let A & B denote the sets of the persons who drink tea & coffee, respectively .
Given:nAB=50nA=30nA-B=14(i) nA-B=nA-nAB14=30-nABnAB=16Thus, 16 persons drink tea and coffee both.(ii) nAB=nA+nB-nAB50=30+nB-16nB=36We have to find nB-AnB-A=nB-nABnB-A=36-16=20Thus, 20 persons drink coffee but not tea.

Page No 1.47:

Question 9:

In a survey of 60 people, it was found that 25 people read newspaper H, 26 read newspaper T, 26 read newspaper I, 9 read both H and I, 11 read both H and T, 8 read both T and I, 3 read all three newspapers. Find:
(i) the numbers of people who read at least one of the newspapers.
(ii) the number of people who read exactly one newspaper.

Answer:

Given:nH=25nT=26nI=26nHI=9nHT=11nTI=8nHTI=3(i) We know: nHTI=nH+nT+nI-nHT-nTI-nHI+nHTI nHTI=25+26+26-11-8-9+3=52Thus, 52 people can read at least one of the newspapers.(ii) Now, we have to calculate the number of people who read exactly one newspaper.We have:nH+nT+nI-2nHT-2nTI-2nHI+3nHTI=25+26+26-22-16-18+9=30Thus, 30 people can read exactly one newspaper.

Page No 1.47:

Question 10:

Of the members of three athletic teams in a certain school, 21 are in the basketball team, 26 in hockey team and 29 in the football team, 14 play hockey and basket ball 15 play hockey and football, 12 play football and basketball and 8 play all the three games. How many members are there in all?

Answer:

Let A, B & C be the sets of members in basketball team, hockey team & football team, respectively.
Given:nA=21nB=26nC=29nAB=14nBC=15nAC=12n(ABC)=8We know:nABC=nA+nB+nC-nAB-nBC-nAC+nABCnABC=21+26+29-14-15-12+8=43

Therefore, there are 43 members in all teams.

Page No 1.47:

Question 11:

In a group of 1000 people, there are 750 who can speak Hindi and 400 who can speak Bengali. How many can speak Hindi only? How many can speak Bengali? How many can speak both Hindi and Bengali?

Answer:

Let A & B denote the sets of the persons who can speak Hindi & Bengali, respectively.
Given:nAB=1000nA=750nB=400nAB=nA+nB-nAB1000=750+400-nABnAB=150Number of persons who can speak both Hindi and Bengali=nAB=150 Number of persons who can speak only Hindi =nA-B=nA-nAB                                                                                                     =750-150                                                                                                     =600Number of persons who can speak only Bengali=nB-A=nB-nAB                                                                                                        =400-150                                                                                                        =250                                                                                                

Page No 1.47:

Question 12:

A survey of 500 television viewers produced the following information; 285 watch football, 195 watch hockey, 115 watch basketball, 45 watch football and basketball, 70 watch football and hockey, 50 watch hockey and basketball, 50 do not watch any of the three games. How many watch all the three games? How many watch exactly one of the three games?

Answer:

Let F, H B denote the sets of students who watch football, hockey and basketball, respectively.
Also, let U be the universal set.
We have:
n(F) = 285, n(H) = 195, n(B) = 115, n(FB) = 45, n(FH) = 70 and n(HB) = 50
Also, we know:
n(F'H'B') = 50
n(FHB)'= 50
n(U) - n(FHB) = 50
500 - n(FHB) = 50
n(FHB) = 450


Number of students who watch all three games = n(FHB)
n(FHB) - n(F) - n(H) - n(B) + n(FB) + n(FH) + n(HB)
450 - 285 - 195 - 115 + 45 + 70 + 50
20

Number of students who watch exactly one of the three games
= n(F) + n(H) + n(B) - 2{n(FB) + n(FH) + n(HB)} + 3{n(FHB)}
= 285 + 195 + 115 - 2(45 + 70 + 50) + 3(20)
= 325

Page No 1.47:

Question 13:

In a survey of 100 persons it was found that 28 read magazine A, 30 read magazine B, 42 read magazine C, 8 read magazines A and B, 10 read magazines A and C, 5 read magazines B and C and 3 read all the three magazines. Find:
(i) How many read none of three magazines?
(ii) How many read magazine C only?

Answer:

Let A, B C be the sets of the persons who read magazines A, B and C, respectively. Also, let U denote the universal set.
We have: n(U)  = 100
n(A) = 28, n(B) = 30, n(C) = 42, n(AB) = 8, n(AC) = 10, n(BC) = 5 and n(ABC) = 3
Now,
Number of persons who read none of the three magazines = n(A'B'C')
= n(ABC)'
= n(U) - n(ABC)
= n(U) - {n(A) + n(B) + n(C) - n(AB) - n(AC) - n(BC) + n(ABC)}
= 100 - (28 + 30 + 42 -- 10 - 5 + 3)
= 20

Number of students who read magazine C only = n(CA'B')
= n{C(AB)'}
= n(C) - n{C(AB)}
= n(C) - n{(CA) (CB)}
n(C) - n{(CA) + (CB)-(ABC)}
= 42 - (10 + 5 - 3)
= 30

Page No 1.47:

Question 14:

In a survey of 100 students, the number of students studying the various languages were found to be : English only 18, English but not Hindi 23, English and Sanskrit 8, English 26, Sanskrit 48, Sanskrit and Hindi 8, no language 24. Find:
(i) How many students were studying Hindi?
(ii) How many students were studying English and Hindi?

Answer:

Let E, H and S be the sets of students who study English, Hindi and Sanskrit, respectively.
Also, let U be the universal set.
Now, we have:
n(E) = 26, n(S) = 48, n(ES) = 8 and n(SH) = 8
Also,
n(EH') = 23
n(E) - n(EH) = 23
26 - n(EH) = 23
n(EH) = 3
Therefore, the number of students studying English and Hindi is 3

n(EH'S') = 18
n(E) - n{E(HS)'} = 18
26 - n{(EH)(ES)}= 18
26 - {3 + 8 - n(EHS)} = 18
n(EHS) = 3

Also,
n(E'H'S') = 24
n(U) - n(EHS) = 24
n(EHS) = 76

∴ Number of students studying Hindi = n(EHS) - n(E) - n(S) + n(EH) + n(ES) + n(SH) - n(EHS)
= 76 - 24 - 48 + 3 + 8 + 8 - 3
= 18

Page No 1.47:

Question 15:

In a survey it was found that 21 persons liked product P1, 26 liked product P2 and 29 liked product P3. If 14 persons liked products P1 and P2; 12 persons liked product P3 and P1 ; 14 persons liked products P2 and P3 and 8 liked all the three products. Find how many liked product P3 only.

Answer:

Let P1, P2 and P3 denote the sets of persons liking products P1, P2 and P3, respectively.
Also, let U be the universal set.
Thus, we have:
n(P1) = 21, n(P2) = 26 and n(P3) = 29
And,
n(P1P2) = 14, n(P1P3) = 12, n(P2P3) = 14 and n(P1P2P3) = 8

Now,
Number of people who like only product P3:
nP3P1'P2'= nP3P1P2'=n P3 - nP3P1P2=nP3 - nP3P1P3P2=nP3 - nP3P1+nP3P2-nP1P2P3=29 - 12+14-8=11

Therefore, the number of people who like only product P3 is 11



Page No 1.49:

Question 1:

If a set contains n elements, then write the number of elements in its power set.

Answer:

A set having n elements has 2n subsets or elements.

Page No 1.49:

Question 2:

Write the number of elements in the power set of null set.

Answer:

We know that a set of n elements has 2n subsets or elements.
A null set has no element(s) in it.
∴ Number of elements in the power set of null set = 20 = 1

Page No 1.49:

Question 3:

Let A = {x : xN, x is a multiple of 3} and B = {x : xN and x is a multiple of 5}. Write AB.

Answer:

A = {x:xN and x is a multiple of 3.}
   = {3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45,...}
B = {x:xN and x is a multiple of 5.}
   ={5, 10, 15, 20, 25, 30, 35, 40, 45,...}
Thus, we have:
AB = {15, 30, 45,...}
         = {x:xN, where x is a multiple of 15.}

Page No 1.49:

Question 4:

Let A and B be two sets having 3 and 6 elements respectively. Write the minimum number of elements that AB can have.

Answer:

We know that nAB=nA+nB-nAB nAB is minimum when nAB is maximumso, nAB=3Hence, nAB=nA+nB-nAB                              =3+6-3                             = 6

Page No 1.49:

Question 5:

If A = {xC : x2 = 1} and B = {xC : x4 = 1}, then write AB and BA.

Answer:

We have:
A = {xC : x2 = 1}
A = {-1, 1}
And,
B = {xC : x4 = 1}
B = {x4-1 = 0}
B = {x2-1x2+1 =0}
B = {-1, 1, -i, i}
Thus, we get:
AB =
And,
BA = {-i, i}

Page No 1.49:

Question 6:

If A and B are two sets such that AB, then write B' − A' in terms of A and B.

Answer:

B'-A' = Not B - Not A             = Nothing common in them             =

Page No 1.49:

Question 7:

Let A and B be two sets having 4 and 7 elements respectively. Then write the maximum number of elements that AB can have.

Answer:

We know that nAB=nA+nB-nAB nAB is maximum when nAB is minimumso, nAB=0Hence, nAB=nA+nB-nAB                              =4+7-0                             = 11

Page No 1.49:

Question 8:

If A=x, y : y=1x, 0 x  R and B=x, y : y=-x, x  R, then write AB.

Answer:

We have:
A=x, y : y=1x, 0 x  R
   = {1,1, 2,12, 3,13, 4,14, ...}
And,
B=x, y : y=-x, x  R
   = 1, -1, 2, -2, 3, -3, 4, -4, ...
Thus, we get:
AB =

Page No 1.49:

Question 9:

If A=x, y : y=ex, x  R and B=x, y : y=e-x, x  R, then write AB.

Answer:

We have:
A = 0, 1, 1, e, 2, e2, 3, e3, ...
B = 0, 1, 1, e-1, 2, e-2, 3, e-3, ...
Thus, we get:
AB = 0, 1

Page No 1.49:

Question 10:

If A and B are two sets such that n A=20, n B=25 and n AB=40, then write n AB.

Answer:

We have:
nA = 20, nB = 25 and nAB = 40
We know:
nAB = nA + nB - nABnAB = nA + nB -nAB
                   = 20 + 25 - 40
                   = 5

Page No 1.49:

Question 11:

If A and B are two sets such that n A=115, n B=326, n A-B=47, then write n AB.

Answer:

nA = 115, nB = 326 and nA-B = 47Now,nA - nAB = nA-B115 - nAB = 47nAB = 68
Thus, we get:
nAB = nA + nB - nAB
               = 115 + 326 - 68
               = 373

Page No 1.49:

Question 1:

For any set A, (A')' is equal to
(a) A'
(b) A
(c) ϕ
(d) none of these.

Answer:

(b) A

The complement of the complement of a set is the set itself.

Page No 1.49:

Question 2:

Let A and B be two sets in the same universal set. Then, A-B=
(a) AB
(b) A'B
(c) AB'
(d) none of these.

Answer:

(c) AB'
A-B belongs to those elements of A that do not belong to B.
 A-B = AB'

Page No 1.49:

Question 3:

The number of subsets of a set containing n elements is
(a) n
(b) 2n − 1
(c) n2
(d) 2n

Answer:

(d) 2n

The total number of subsets of a finite set consisting of n elements is 2n.



Page No 1.50:

Question 4:

For any two sets A and B, AAB=
(a) A
(b) B
(c) ϕ
(d) none of these.

Answer:

(a) A

A(AB)= (AA)(AB)=A(AB)=A

Page No 1.50:

Question 5:

If A = {1, 3, 5, B} and B = {2, 4}, then
(a) 4  A
(b) 4  A
(c) B  A
(d) none of these.

Answer:

(d) none of these

4A
{4} ⊄ A
B A
Thus, we can say that none of these options satisfy the given relation.

Page No 1.50:

Question 6:

The symmetric difference of A and B is
(a) A-BB-A
(b) A-BB-A
(c) AB-AB
(d) AB-AAB-B

Answer:

(b) A-BB-A
The symmetric difference of A  and B is given by :-
(A-B)(B-A)

Page No 1.50:

Question 7:

The symmetric difference of A = {1, 2, 3} and B = {3, 4, 5} is
(a) {1, 2}
(b) {1, 2, 4, 5}
(c) {4, 3}
(d) {2, 5, 1, 4, 3}

Answer:

(b) {1, 2, 4, 5}
Here,
A = {1, 2, 3} and B = {3, 4, 5}
The symmetric difference of A  and B is given by :-
(A-B)(B-A)
Now, we have:
(A-B)={1,2}(B-A)={4,5}(A-B)(B-A)={1,2,4,5}

Page No 1.50:

Question 8:

For any two sets A and B, A-BB-A=
(a) A-BA
(b) B-AB
(c) AB-AB
(d) ABAB.

Answer:

(c) AB-AB

A-BB-A = AB'BA'=ABA'B'BA'                        Using distribution law=ABAA'B'BB'A'    Using distribution law=ABUUB'A'                    AA'=U=B'B                    =ABB'A'                                            ABU=AB and UB'A'=B'A'       =AB AB'                                          AB'=B'A'=AB AB-AB=AB-AB

Page No 1.50:

Question 9:

Which of the following statements is false:
(a) A-B=AB'
(b) A-B=A-AB
(c) A-B=A-B'
(d) A-B=AB-B.

Answer:

(c) A-B=A-B'
It includes all those elements of A which do not belongs to complement of B which is equal to AB but not equal to
A-B .
Therefore, (c) is false .

Page No 1.50:

Question 10:

For any three sets A, B and C
(a) AB-C=AB-AC
(b) AB-C=AB-C
(c) AB-C=ABAC'
(d) AB-C=AB-AC.

Answer:

(a) AB-C=AB-AC

Let x be any arbitrary element of AB-C.
Thus, we have,
xAB-C x A and x B-C
                     xA and xB and xCxA and xB and xA and xCxAB and xACxAB - ACAB-CAB - ACSimilarly, AB - ACAB-CHence, AB-C=AB - AC

Page No 1.50:

Question 11:

Let A=x : x  R, x  4 and B = x  R : x < 5. Then, n A'B'=
(a) (4, 5]
(b) (4, 5)
(c) [4, 5)
(d) [4, 5]

Answer:

(c) [4, 5)
A=x : x  R, x  4 and B = x  R : x < 5
AB=[4,5)

Page No 1.50:

Question 12:

Let U be the universal set containing 700 elements. If A, B are sub-sets of U such that n A=200, n B=300 and AB=100. Then n A'B'=
(a) 400
(b) 600
(c) 300
(d) none of these.

Answer:

(c) 300
n(A'B') = nAB'
               =nU - n(AB)= 700 - 200+300-100= 300

Page No 1.50:

Question 13:

Let A and B be two sets that n A=16, n B=14, n AB=25. Then, n AB is equal to
(a) 30
(b) 50
(c) 5
(d) none of these

Answer:

We know:
nAB = nA + nB - nAB

Now,
nAB = nA + nB - n(AB)
               = 16 + 14 - 25
               = 5

Page No 1.50:

Question 14:

If A = |1, 2, 3, 4, 5|, then the number of proper subsets of A is
(a) 120
(b) 30
(c) 31
(d) 32

Answer:

(c) 31
The number of proper subsets of any set is given by the formula 2n-1, where n is the number of elements in the set.
Here,
n = 5
∴ Number of proper subsets of A = 25-1 = 31

Page No 1.50:

Question 15:

In set-builder method the null set is represented by
(a) { }
(b) Φ
(c) x : x x
(d) x : x = x

Answer:

(c) x:xx

Page No 1.50:

Question 16:

If A and B are two disjoint sets, then n AB is equal to
(a) n A+nB
(b) n A+nB-nAB
(c) n A+n B+n AB
(d) n A n B
(e) n A-n B

Answer:

(a) n A+nB

Two sets are disjoint if they do not have a common element in them, i.e., AB = .
nAB = nA + nB

Page No 1.50:

Question 17:

For two sets AB=A iff
(a) BA
(b) AB
(c) AB
(d) A=B

Answer:

(a) BA
The union of two sets is a set of all those elements that belong to A or to B or to both A and B.
If AB = A, then BA.

Page No 1.50:

Question 18:

If A and B are two sets such that n A=70, n B=60, n AB=110, then n AB is equal to
(a) 240
(b) 50
(c) 40
(d) 20

Answer:

(d) 20
We have:
n(AB) = nA + nB - n(AB)               =70 + 60 - 110               =20

Page No 1.50:

Question 19:

If A and B are two given sets, then AABc is equal to
(a) A
(b) B
(c) Φ
(d) ABc

Answer:

(d) ABc
A and B are two sets.
AB is the common region in both the sets.
ABc is all the region in the universal set except AB.
Now,
AABc = ABc



Page No 1.51:

Question 20:

If A = {x : x is a multiple of 3} and , B = {x : x is a multiple of 5}, then AB is
(a) AB
(b) AB
(c) AB
(d) AB

Answer:

(b) AB

A = {x:x is a multiple of 3}
A = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, ...

B = {x:x is a multiple of 5.}
B = 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, ...

Now, we have:
A-B = 3, 6, 9, 12, 18, 21, 24, 27, 33, 36, 39, 42, 48, ...
         = AB

Page No 1.51:

Question 21:

In a city 20% of the population travels by car, 50% travels by bus and 10% travels by both car and bus. Then, persons travelling by car or bus is
(a) 80%
(b) 40%
(c) 60%
(d) 70%

Answer:

(c) 60%
Suppose C and B represents the population travel by car and Bus respectively.

nCB = nC + nB - nBC                =0.20 + 0.50 - 0.10                =0.6 or 60%

Page No 1.51:

Question 22:

If AB-B, then
(a) AB
(b) BA
(c) A=Φ
(d) B=Φ

Answer:

(b) BA

Only this case is possible.

Page No 1.51:

Question 23:

An investigator interviewed 100 students to determine the performance of three drinks: milk, coffee and tea. The investigator reported that 10 students take all three drinks milk, coffee and tea; 20 students take milk and coffee; 25 students take milk and tea; 12 students take milk only; 5 students take coffee only and 8 students take tea only. Then the number of students who did not take any of three drinks is
(a) 10
(b) 20
(c) 25
(d) 30

Answer:

Disclaimer: The question in the book has some error, so, none of the options are matching with the solution.
The required information is not available in the question.

Page No 1.51:

Question 24:

Two finite sets have m and n elements. The number of elements in the power set of first set is 48 more than the total number of elements in power set of the second set. Then, the values of m and n are:
(a) 7, 6
(b) 6, 3
(c) 7, 4
(d) 3, 7

Answer:

(c) 6, 4

ATQ :
2m-1=48+2n-12m-2n = 482m-2n = 26-24By comparing we get:m=6 and n=4

Page No 1.51:

Question 25:

In a class of 175 students the following data shows the number of students opting one or more subjects. Mathematics 100; Physics 70; Chemistry 40; Mathematics and Physics 30; Mathematics and Chemistry 28; Physics and Chemistry 23; Mathematics, Physics and Chemistry 18. How many students have offered Mathematics alone?
(a) 35
(b) 48
(c) 60
(d) 22
(e) 30

Answer:

(c) 60
Let M, P and C denote the sets of students who have opted for mathematics, physics, and chemistry, respectively.
Here,
nM = 100, nP = 70 and n(C) = 40
Now,
nMP = 30, nMC = 28, nPC = 23 and nMPC = 18
Number of students who opted for only mathematics:
nMP'C' = nMPC'                         =nM - nMPC                         =nM - nMPMC                         =nM - nMP+nMC-nMPC                         =100 - 30+28-18                         =60
Therefore, the number of students who opted for mathematics alone is 60

Page No 1.51:

Question 26:

Suppose A1,A2,...,A30 are thirty sets each having 5 elements and B1,B2,...,Bn are n sets each with 3 elements. Let i=130Ai=j=1nBj=S and each element of S belong to exactly 10 of the Ai's and exactly 9 of the Bj's, then n is equal to

(a) 15                                (b) 3                                 (c) 45                                (d) 35                               

Answer:

It is given that each set Ai 1i30 contains 5 elements and i=130Ai=S.

nS=30×5=150

But, it is given that each element of S belong to exactly 10 of the Ai's.

∴ Number of distinct elements in S = 15010=15           .....(1)

It is also given that each set Bj 1jn contains 3 elements and j=1nBj=S.

nS=n×3=3n

Also, each element of S belong to eactly 9 of Bj's.

∴ Number of distinct elements in S = 3n9                  .....(2)

From (1) and (2), we have

3n9=15n=45

Thus, the value of n is 45.

Hence, the correct answer is option (c).

Page No 1.51:

Question 27:

Two finite sets have m and n elements. The number of subsets of the first set is 112 more than that of the second. The values of m and n are respectively

(a) 4, 7                                  (b) 7, 4                                  (c) 4, 4                                  (d) 7, 7                                 

Answer:

We know that if a set X contains k elements, then the number of subsets of X are 2k.

It is given that the number of subsets of a set containing m elements is 112 more than the number of subsets of set containing n elements.

2m-2n=1122n2m-n-1=2×2×2×2×72n2m-n-1=2423-1n=4 and m-n=3m-4=3m=7

Thus, the values of m and n are 7 and 4, respectively.

Hence, the correct answer is option (b).

Page No 1.51:

Question 28:

For any two sets A and B, AAB' is equal to

(a) A                              (b) B                              (c) ϕ                              (d) AB                             

Answer:

AAB'=AA'B'                        De Morgan law=AA'B'=ϕB'=ϕ

Hence, the correct answer is option (c).

Page No 1.51:

Question 29:

The set AB''BC is equal to

(a) A'BC                 (b) A'B                    (c) A'C'                 (d) A'B                

Answer:

AB''BC=A'B''BC           De Morgan law=A'BBC=A'CB                          Distributive law

Disclaimer: The question seems to be incorrect or there is some printing mistake in the question. The options given in the question does not match with the answer.

Page No 1.51:

Question 30:

Let F1 be the set of all parallelograms, F2 the set of all rectangles, F3 the set of all rhombuses, F4 the set of all squares and F5 the set of trapeziums in a plane. Then F1 may be equal to

(a) F2F3                        (b) F3F4                        (c) F2F3                       (d) F2F3F4F1                  

Answer:

We know that every rectangle, rhombus and square in a plane is a parallelogram but every trapezium is not a parallelogram.

So, F1 is either of F1 or F2 or F3 or F4.

F1=F1F2F3F4

Hence, the correct answer is option (d).



Page No 1.6:

Question 1:

Describe the following sets in Roster form:
(i) {x : x is a letter before e in the English alphabet};
(ii) {xN : x2 < 25};
(iii) {xN : x is a prime number, 10 < x < 20};
(iv) {xN : x = 2n, nN};
(v) {xR : x > x}.
(vi) {x : x is a prime number which is a divisor of 60}
(vii) {x : x is a two digit number such that the sum of its digits is 8}
(viii) The set of all letters in the word 'Trigonometry'
(ix) The set of all letters in the word 'Better'.

Answer:

Roster form:
In this form, a set is defined by listing elements, separated by commas, within braces {}.
(i) {a, b, c, d}
(ii) {1, 2, 3, 4}
(iii) {11, 13, 17, 19}
(iv) {2, 4, 6, 8, 10,...}
(v) ϕ
(vi) {2, 3, 5}
(vii) {17, 26, 35, 44, 53, 62, 71, 80}
(viii) {T, R, I, G, O, N, M, E, Y}
(ix) {B, E, T, R}

Page No 1.6:

Question 2:

Describe the following sets in set-builder form:
(i) A = {1, 2, 3, 4, 5, 6};
(ii) B=1, 12, 13, 14, 15, ...;
(iii) C = {0, 3, 6, 9, 12, ...};
(iv) D = {10, 11, 12, 13, 14, 15};
(v) E = {0};
(vi) {1, 4, 9, 16, ..., 100}
(vii) {2, 4, 6, 8 .....}
(viii) {5, 25, 125 625}

Answer:

Set-builder form:
To describe a set, a variable x (each element of the set) is written inside braces. Then, after putting a colon, the common property P(x) possessed by each element of the set is written within braces.

(i) {x:xN, x<7}(ii) {x: x=1n, xN}(iii) {x:x=3n, nZ+}(iv) {x:xN, 9<x<16}(v) {x:x=0}(vi) {x2:xN, 1n10}(vii) {x:x=2n, nN}(viii) {5n:nN, 1n4}

Page No 1.6:

Question 3:

List all the elements of the following sets:
(i) A=x:x210,x Z

(ii) B=x:x=12n-1, 1n5

(iii) C=x:x is an integer,-12<x<92

(iv) D = {x : x is a vowel in the word "EQUATION"}

(v) E = {x : x is a month of a year not having 31 days}

(vi) F = {x : x is a letter of the word "MISSISSIPPI"}

Answer:

(i) A={0,±1,±2,±3}(ii) B=1,13,15,17,19(iii) C={0,1,2,3,4}(iv) D={A,E,I,O,U}(v) E={February,April,June,September,November}(vi) F={M,I,S,P}



Page No 1.7:

Question 4:

Match each of the sets on the left in the roster form with the same set on the right described in the set-builder form:

(i) {A, P, L, E} (i) x : x + 5 = 5, x ∈ Z
(ii) {5, −5} (ii) {x : x is a prime natural number and a divisor of 10}
(iii) {0} (iii) {x : x is a letter of the word "RAJASTHAN"}
(iv) {1, 2, 5, 10,} (iv) {x: x is a natural number and divisor of 10}
(v) {A, H, J, R, S, T, N} (v) x : x2 − 25 = 0
(vi) {2, 5} (vi) {x : x is a letter of the word "APPLE"}

Answer:

 (i) {A, P, L, E} is a roster form of {x : x is a letter of the word APPLE}.
(ii) {5, −5} is a roster form of {x : x2 − 25 = 0}.
(iii) {0} is a roster form of  {x : x + 5 = 5, x ∈ Z}.
(iv) {1, 2, 5, 10} is a roster form of {x : x is a natural number and a divisor of 10}.
(v) {A, H, J, R, S, T, N} is a roster form of {x : x is a letter of the word RAJASTHAN}.
(vi) {2, 5} is a roster form of {x : x is a prime natural number and a divisor of 10}.
 

(i) {A, P, L, E} (vi) {x : x is a letter of the word APPLE}
(ii) {5, −5} (v) { x : x2 − 25 = 0}
(iii) {0} (i) {x : x + 5 = 5, x ∈ Z}
(iv) {1, 2, 5, 10} (iv) {x : x is a natural number and a divisor of 10}
(v) {A, H, J, R, S, T, N} (iii) {x : x is a letter of the word RAJASTHAN}
(vi) {2, 5} (ii) {x : x is a prime natural number and a divisor of 10}

Page No 1.7:

Question 5:

Write the set of all vowels in the English alphabet which precede q.

Answer:

The set of vowels in the English alphabet that precede q is {a, e, i, o}.

Page No 1.7:

Question 6:

Write the set of all positive integers whose cube is odd.

Answer:

The set of all positive integers whose cube is odd is {2n + 1 : nZ, n0}.

Page No 1.7:

Question 7:

Write the set 12, 25, 310, 417, 526, 637, 750 in the set-builder form.

Answer:

The set-builder form of the set 12, 25, 310, 417, 526, 637, 750 is nn2+1:nN,n7.



Page No 1.9:

Question 1:

Which of the following are examples of empty set?
(i) Set of all even natural numbers divisible by 5;
(ii) Set of all even prime numbers;
(iii) {x : x2 −2 = 0 and x is rational};
(iv) {x : x is a natural number, x < 8 and simultaneously x > 12};
(v) {x : x is a point common to any two parallel lines}.

Answer:

(i) All natural numbers that end with 0 are even & divisible by 5. Therefore, the given set is not an example of empty set.
(ii) 2 is an even prime number. Therefore, the given set is not an example of empty set.
(iii) There is no rational number whose square is 2 such that x2-2 = 0. Therefore, it is example of empty set.
(iv) It is not possible that x<8 and, at the same time, x>12. Therefore, it is an example of empty set.
(v) There is no common point in two parallel lines. Therefore, it is an example of empty set.

Page No 1.9:

Question 2:

Which of the following sets are finite and which are infinite?
(i) Set of concentric circles in a plane;
(ii) Set of letters of the English Alphabets;
(iii) {xN : x > 5};
(iv) {x = ∈ N : x < 200};
(v) {xZ : x < 5};
(vi) {xR : 0 < x < 1}.

Answer:

(i) There can be infinite concentric circles in a plane. Therefore, it is an infinite set.
(ii) There are 26 letters in the set of English alphabet. Therefore, it is a finite set.
(iii) {xN : x > 5} = {6,7,8,9,...}. There will be infinite numbers. So, it an infinite set.
(iv) There are finite elements in the set {x = ∈ N : x < 200}. Therefore, it is a finite set.
(v) In this set, xZ , so there would be infinite elements in the set {xZ : x < 5}. Therefore, it is an infinite set.
(vi) In this set,  x ∈ R. We know real numbers include all numbers, i.e., decimal numbers, rational numbers and irrational numbers.
So, there would be infinite elements in the set {xR : 0 < x < 1}. Therefore, it is an infinite set.

Page No 1.9:

Question 3:

Which of the following sets are equal?
(i) A=1, 2, 3;
(ii) B=xR : x2-2x+1=0;
(iii) C=1, 2, 2, 3;
(iv) D=xR : x3-6x2+11x-6=0.

Answer:

Two sets A & B are equal if every element of A is a member of B & every element of B is a member of A.
(i) A=1, 2, 3
(ii) B=xR : x2-2x+1=0
Set B would be {1}.
(iii) C=1, 2, 2, 3 
It can be written as {1, 2, 3} because we do not repeat the elements while writing the elements of a set.
C = {1, 2, 3}
(iv) D=xR : x3-6x2+11x-6=0 includes elements {1, 2, 3}.
      ∴ D = {1, 2, 3}
Hence, we can say that A = C = D.



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