# Set Theory _ Ele Number System_AIEEE_9-5-2010 by nehalwan

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```									                                       SET THEORY & ELE. NUMBER SYSTEM( AIEEE level)
9-5-2010                                                                           1

Section - I (Single Answer Type)

1.      If A = {a, b, {c}, φ} ; then which one is true.                                               4 × 2220
⇒       y =            = 60
148
(a)      c∈A                     (b)     {c} ⊂ A
6.      If a natural number X is divided by 7, then the set
(c*)     {c} ∈ A                 (d)     {φ} ∈ A
of all possible remainders is.
Ans. c                                                                                 (a)      {2, 3, 5}              (b)     {1, 3, 5, 7}
Sol.    Q {c} is an element of A,                                                      (c)      {1, 2, 3, 5, 7}        (*d)    {0, 1, 2, 3, 4, 5, 6}
Q {c} ∈ A,                                                             Ans. d

2.      Which one of the given sets is empty.                                  Sol.    On dividing any natural no. by 7, the possible remain-

(a)      {x : x is an even prime number}                                       der can be 0, 1,2,3,4,5,6

(b*)     {x : x2 + 1 = 0 and x ∈ N}                                            Q        Set of possible remainders is {0,1,2,3,4,5,6}

(c)      {x : x2 = x and x ∈ N}                                        7.      If A = {3x : x = N}, B = {5x : x = 5N} then de-

(d)      {x : G.C.D. (x, 1) = 1 and x ∈ N}                                     scribe set A ∩ B.

Ans. b                                                                                 (a)      {3x : x ∈ N}           (b)     {5x : x ∈ N}

Sol.    x2 + 1 =0 ⇒ x2 = -1               ⇒ x = ± −1 ∉ N                               (*c)     {15x : x ∈ N}          (d)     none of these

⇒        Set in opt(b) is empty                                        Ans     c

Sol.    A = {3, 6 9, 12, 15 ......}
3.
.in
If N = set of natural numbers, W = set of whole

numbers, Z = set of integers, then which set can

serve as the universal set for this problem.
B = {5, 10, 15, 20, .....}

(A ∩ B) = {15, 30, 45 ........}
.co
(a)      N                       (b)     W                                     = {15 x : x ∈ N}

(c)      Z = set of positive integer
+
8.      The product of two +ve integer is 24 & their
(d*)     Z                                                                     greatest common divisor is 2. Their least com-
Ans. d                                                                                 mon divisor is
du

Sol :                                                                                  (a)      12                     (b)     24

Q       N, W, Z are the subsets of set Z (set of integers)                             (*c) 1                          (d)     2

Q       Z can serve as a universal set for this problem.                       Ans. c
ite

4.      If X and Y are two sets such that n(X) = 17, n(Y) =                    Sol.    Least Common divisor of any two natural no. is                  One

28 and n(X ∪ Y) = 40, then n(X ∩ Y) is equal to.                               (1) Note : L.C.Divisor means Least +ve common Di-

(a)      15                      (b*)    5                                     visor.
mi

(c)      25                      (d)     10                            9.      The values of x, y, z from the venn-
A x        4       y       B
Ans. b                                                                                 diagram shown, if is known that                     6
5
8

Sol.    n(x ∩ y) = n(x) + n(y) - n(x ∪ y)                                              x, y, z are three consecutive                           z       C

=        17 + 28 - 40 = 5                                                      prime numbers and n(A ∪ B ∪ C)

5.      If the H.C.F. and L.C.M. of two numbers are re-                                = 33 are.

spectively 4 and 2220 respectively. If one of the                              (a)      2, 3, 7                (b*)    2, 3, 5

numbers is 148, then what is the other number.                                 (c)      5, 7, 11               (d)     No solution

(a*)     60                      (b)     90                            Ans. b

(c)      30                      (d)     40                            Sol.    n(A ∪ B ∪ C) = 33

Ans. a                                                                                  ⇒       x + y + z + 4 + 5 + 6 + 8 = 33

Sol.    x . y = H.C.F. x L.C.M. ⇒ 148 y = 4 x 2220                                      ⇒       x + y + z = 10

S.C.O 73, Sec - 20 C, Chandigarh, 2775646, 9417317019, S.C.F 73, Sector 10, & S.C.O. 214, Sec. 14, Panchkula, 2583315, 9463087706
SET THEORY & ELE. NUMBER SYSTEM( AIEEE level)
9-5-2010                                                                  2
Q        x, y, z are numbers having sum equal to 10.                   Ans. a

⇒        x, y , z should be 2, 3, 5 respectively.                      Sol.    Total no of divsors = 15

10.    If U = { 2, 3, 5, 7, 11, 13, 17 } ; A = { 2, 3, 5, 13};                         ⇒      (a + 1)(b + 1) = 15

B = {3, 5, 7, 17}. The set representing A ∆ B will                              ⇒      a + 1 = 3, b + 1 = 5 or a + 1 = 5, b + 1 = 3 or
be :                                                                                   a + 1 = 1 , b + 1 = 15 or a + 1 = 15, b + 1 = 1

(a)      {2, 7, 13, 17}          (b)      {2, 7, 13}                            ⇒      a = 2 , b = 4 or a = 4, b = 2, or a = 0, b = 14 or

(c)      {2, 13, 17}             (d)      none of these                                a = 14, b = 0

Ans. a                                                                                 ⇒      (a, b) can be (2, 4); (4, 2); (0, 4); (4, 0)

Sol.   A = U − A = {7,11,17} and B = U - B = {2, 11, 13}                              But in last two pairs N = paqb becomes q14 and p14

A∆B = (A − B) ∪ (B − A) = {7, 17} ∪ {2, 13}                                    respectively      i.e. they have only one prime and not

=        {2, 7, 13, 17}                                                        two distinct prime.

11.    If A & B are two disjoint sets then which one true                             ∴        Last two pairs are rejected

(a)      A ⊂ BC and B ⊄ AC                                                     ∴       Only two pairs of a and b are possible.

(b)      A ⊄ BC and B ⊂ AC                                             14.     The number of the proper common divisors of x =

(c)      A ⊂ BC and B ⊂ AC                                                     (2)4 (3)5 and y = (2)3 . (3)6 is :

(d)      None of these                                                         (a)     23                      (b)     24

Ans. c

Sol.   A and B are disjoint sets, ⇒
.in   A ∩ B = {}                      Ans. a
(c)     25                      (d)     none

x ∈ A ⇒ x ∉ B [ Q A and B have no element in common]                   Sol.    Sol.    x = (2)4 (3)5 and y = (2)3 . (3)6 H.C.F. of divisors
.co
of x and y = (2)3 (3)5
⇒ x ∈ Bc

⇒ A ⊂ Bc                                                                       Common divisors of x & y are the divisors of ((2)3 (3)5)

similarly     B ⊂ Ac                                                           = 4 . 6 = 24
du

12.    The number of proper even factors of 684 is :                          ∴       Number of proper common divisors = 24 – 1 = 23.

(a)      12                      (b)      16                           15.     Number of non-empty subsets of set containing

(c)      4                       (d*)     11                                   10 elements in all are

(a)     1023                    (b)     1024
ite

Ans. d

Sol.   684 = (2)2 × (3)2 × (19)1
(c)     210 – 2                 (d)     29

No. of even factors of 684                                    Ans     a
∴
Sol.    Number of subsets of set containing 10 elements
mi

=         (3)(3)(2) = 12
= 210 . 1024.
∴        No. of proper even factors = 12 - 1 = 11
∴       Number of subsets containing 1 empty subset φ.
13.    A natural number N is primerly factorised as paqb,

where p and q are two distinct primes. Such that                               Number non empty containing 1 empty subset

total number of divisors of N be 15 then numbers                               = 1240 – 1 = 1023.

of possible values of (a, b) will be :                                 ∴       option (a)

(a)      2                       (b)      4

(c)      1                       (d)      None of these

S.C.O 73, Sec - 20 C, Chandigarh, 2775646, 9417317019, S.C.F 73, Sector 10, & S.C.O. 214, Sec. 14, Panchkula, 2583315, 9463087706
SET THEORY & ELE. NUMBER SYSTEM( AIEEE level)
9-5-2010                                                                  3
(Assertion & Reason Type)

17.     A:      A quadrilateral having equal sides can be
16.    A:       LCM of √2 & 3 is 3√2
rhombus or square.
R:       If x & y are two coprime numbers then their
R:      A compound statement with an exclusive OR
LCM is x. y.
is true when only one component statement
Ans. D
is true not both.
Sol.   Multiples of √2 = √2, 2√2, 3√2 .......
Ans. a
Multiples of 3 = 3, 6, 6, 9 .....
Sol.    A is true and R is true
∴      LCM does not exist
18.     A:      x is divisible by 15, this implies x is divisible
∴      A is false but R is true.                                                              by 3 and x is divisible by 5

∴      (d) option                                                                     R:      x is not divisible by 3 or 5, implies x is not

divisible by 15.

Ans. A

Comprehensive Type)

.in
Paragraph for question nos. 19 - 22

If A and B are two sets, then the set of all elements
21.     Let C = set of players of cricket

B = set of players of basketball

which belong to A but not B i.e. n (A – B) = { x : x ∈ A                              H = set of players of Hockey, and n (C) = 40, n (B)
.co
but x ∉ B}, & the set of all elements which belong to                                 = 39, n(H) = 36, n( C ∩ B) = 9, n(C ∩ H) = 15, n(B ∩

only one of the sets A & B is known as Symmetric                                      H) = 11 n(C ∩ B ∩ H) = 5, then the number of players

Difference of sets A and B, denoted by A ∆ B i.e. A ∆ B =                             who play only cricket is :
du

(A - B) ∪ (B - A) = {x :x ∈ A or x ∈ B but x ∉ ( A ∩ B) }.                            (*a)    21                      (b)     22

Now answer the following questions.                                                   (c)     23                      (d)     24

Ans. a
19.    If A = {a, e, i, o, u} and B = {e, i, u, m, n} then (A
ite

~ B) is :                                                              Sol.     n(only cricket)

(a)      {a, e}                  (*b)    {a, o}                                =       n(c) - n(C ∩ B)-n(C ∩ H)+n(C ∩ B ∩ H)

(c)      {o, u}                  (d)     {u}                                   =       40 - 9 - 15 + 5         =       21
mi

Ans. b                                                                        22.     If AC = {1, 3, 5, 6, 9, 10} & BC = {1, 2, 3, 5, 6, 8}

Sol.   A ~ B = {x : x ∈ A and x ∉ B}                    =       {a, 0}                then (A ∪ B)C is :

20.    If A = {a, e, i, o, u} and B = {e, i, u, m, n}, then A                         (a)     {1, 3, 5, 9}

∆ B is :                                                                       (b)     {1, 3, 5, 6, 7, 8, 9, 10}

(a)      {a, e, m, n}            (b)     {o, m, n}                             (c)     {3, 5, 6, 9}            (*d)    {1, 3, 5, 6}

(*c)     {a, o, m, n}            (d)     {e, o, m, n}                  Ans. d

Ans. c                                                                        Sol.    By Demorgan’s theorem

Sol.   A ∆ B = (A - B) ∪ (B - A)                                                       (A ∪ B)c = Ac ∩ Bc

=        {a, o} ∪ {m, n}                                                       ={1,3,5,6,9,10} ∩ {1,2,3,5,6,8}

=        {a, o, m, n}                                                          ={1,3,5,6}

S.C.O 73, Sec - 20 C, Chandigarh, 2775646, 9417317019, S.C.F 73, Sector 10, & S.C.O. 214, Sec. 14, Panchkula, 2583315, 9463087706
SET THEORY & ELE. NUMBER SYSTEM( AIEEE level)
9-5-2010                                                                      4

Paragraph for Question nos. 23 - 26

Given A        =       {x : x = 3m, m ∈ N &       x ≤ 17} ,       B     =              n(A ∪ (B ∩ C)) = 6

{y    : y = 2k , k ∈ W & k ≤ 6} & C = {x : x is prime number                   25.     Number of elements in ( A − C ) ∪ (C − B) is equal

& x ≤ 7}. Where W is the set of whole numbers and N is                                 to

the set of natural numbers.             Based on this information                      (a)     6               (b*)    7               (c)        8

now answer the following questions                                                     (d)     10

Sol.    A = {3, 6, 9, 12, 15}                                                  Sol.    (A – C) ∪ (C – B) = {6, 9, 12, 15} ∪ {3, 5, 7}

B = {0, 2, 4, 6, 8, 10, 12}                                                                            = {3, 5, 6, 7, 9, 12, 15}

C = {2, 3, 5, 7}                                                                                       number of element = 7.

23.     Number of elements in ( A ∪ B) will be                                 26.     Number of elements in A ∆ B (The symmetric differ-

(a)      9               (b)     6               (c)     8                     ence of A & B) is

(d*)     None of these                                                         (a)     12              (b)     10              (c*)       8

Sol.    (A ∪ B) = {0, 2, 3, 4, 6, 8, 9, 10, 12, 15}                                    (d)     None of these

n(A ∪ B) = 10
Sol.     (A ∆ B) = (A – B) ∪ (B – A)
24.

(a)      12              (b)
.in
Number of elements in A ∪ (B ∩ C) will be

10              (c)     8
= {3, 9, 15} ∪ {0, 2, 4, 8, 10}

n(A ∆ B) = {0, 2, 3, 4, 8, 9, 10, 15}
(*d)     None of these
.co
number of elements = 8.
Sol.    (A ∪ (B ∩ C)) = {2, 3, 6, 9, 12, 15}

Section II (More than one Type)
du

27.     Which of following sets is/are empty sets                              Sol.
(a)      A = {x : x ∈ R and x – 4 = 0}
2
(a)     A = {5}
(b)      B = {x : x ∈ R and x + 4 = 0}
4
(b)     B = {6, –6}
(c)      C = {x : x ∈ N and x = 1}   3
(c)     C = {6}
ite

(d)      D = {x : x ∈ N and x + 2x + 1 = 0}
2
(d)     x2 = – 1                ∴       x is not real
Ans. b, d                                                                      ∴       D = {}                  ∴       a, c are singleton sets
Sol.    A = {2, –2}              ∴       non empty                             29.     If U = {1, 2, 3, 4, 5, 6, 7, 8, 9} A = {x : x ∈ N, 30 <
mi

(a)      x =–4
4
∴       not reals                             x2 < 70} B = {x : x is a prime number less than
∴       B=φ                                   10} then which of following is false ?
(c)      C = {1}                 ∴       non empty                             (a)     A' ∩ B' = {1, 4, 9}             (b)     A–B={2,3,5}
(d)      D = {–1}                ∴       empty                                 (c)     B – A = {6, 8}          (d)     A∆B = {2,3,5,6,8}.
but x ∈ N               ∴       (b), (d)                      Ans. b, c
28.     Which of the following is a singleton set ?                            Sol.    A = {6, 7, 8}           B = {2, 3, 5, 7}
(a)      A = {x : |x| = 5, x ∈ N}                                              (a)     A' ∩ B'
(b)      A = {x : |x| = 6, x ∈ Z}                                              {1, 2, 3, 4, 5, 9} ∩ {1, 4, 6, 8, 9} ∩ {1, 4, 9} is true
(c)      A = {x : 5 < x ≤ 6, x ∈ Z}                                    (b)     A – B = {6, 8} is true
(d)      A = {x : x ∈ R and x + 1 = 0 }
2
(c)     B – A = {2, 3, 5} is false
Ans. a, c                                                                      (d)     A ∆ B = (A – B) ∪ (B – A) = {2, 3, 5, 6, 8} is true

S.C.O 73, Sec - 20 C, Chandigarh, 2775646, 9417317019, S.C.F 73, Sector 10, & S.C.O. 214, Sec. 14, Panchkula, 2583315, 9463087706
SET THEORY & ELE. NUMBER SYSTEM( AIEEE level)
9-5-2010                                                                  5

∴         (b), (c) are false.

30.    Consider the following relations :
A = (A ∩ B) ∪ (A – B) is correct (3) is false
(1)       A-B = A – (A ∩ B) (2)               A =(A ∩ B) ∪ (A – B)
∴         1 and 2 are true that is (a) and (b) is the correct
(3)       A – (B ∪ C) = (A-B) ∪ (A – C)
which of these is correct
31.       If A = {x : x is a factor of 800} then which of the
*(a)      1                (*b)        2
following is true ?
(c)       3                (d)         None of these
(a)       n(A) = 18           (b)     n[p(A)] = 2 18
Ans. a, b
(c)       n(A) = 10           (d)     n[(p(A)] = 2 10
Sol.   A – B = A – (A ∩ B) is correct.
Ans. a, b
A          B
Sol.      800 = 25 . 52

∴         n(A) = no. of factor of A & n(P (A)) = 2 18 = 6×3 = 18.

∴         So (a), (b) is correct..
A-B
A-(A ∩ B)

Section III (Column Matching Type)

32.    Match the shaded portions in Venn diagram in column A with their rotations in column B.

Column A                                                          Column B

A
.in
B
.co
(i)                                                                         (a)         A – (B ∪ C)

C

A
du

(ii)                                                                        (b)         AC ∩ (B ∪ C)
C                        B
ite

A
B
mi

(iii)                                                                       (c)         A ∪ (B ∩ C) ~ (A ∩ B)

C

B
A

(iv)                                                                        (d)         CC ∩ (A – B)

C

(e)         {(A – B) ∪ (B – C) ∪ (C – A)}

Ans. (i)         c                (ii)        a, d         (iii) b                 (iv)        e

S.C.O 73, Sec - 20 C, Chandigarh, 2775646, 9417317019, S.C.F 73, Sector 10, & S.C.O. 214, Sec. 14, Panchkula, 2583315, 9463087706
SET THEORY & ELE. NUMBER SYSTEM( AIEEE level)
9-5-2010                                                                  6

33.              Column A                                                          Column B

(i)       {x : x2 = 25 and x is positive even integer}                      (a)           Singleton set

(ii)      {x : x ∈ R and 1 < x < 1.01}                                      (b)           Infinite set

(iii) Set of all concentric circle with centre (0, 0)                       (c)           Uncountable set

(iv)      A = {1, 2, 3}, B = {3, 4}, C = {3, 4, 5}                          (d)           null set

then A ∩ (B ∩ C)

(e)           countable set

Ans. (i)         d, e            (ii)    c, b            (iii) b, c                (iv)          a, e

Solution.                                                                                ber and are uncountable.

(i)    {+5, –5} there is no positive even integer                              (iv)      (B ∩ C) = {3, 4}

∴      It is null set                                                                    A ∩ (B ∩ C) = {1, 2, 3} ∩ {3, 4} = {3}

(ii)   There are infinite real numbers between two num-                        ∴         Singeton set.

Section IV (Subjective Type)

34.    In a city, three daily newspapers A, B, C are                                     and       1546 = xq2 + r

published 42% of the people in that city read A,                                  on Subtraction. 248 = x (q2 − q1 )

⇒         x is a divisor of 248

.in
28% read B and C 36% read A and C 8% do not

read any of the three newspapers. The percentage

of persons who read all the three papers is
Now, two digit factors of 248 are 31 and 62

∴         If x = 31, then 1298 = (31) (41) +27

⇒ λ = 27
.co
Ans. 25%
If x = 62, then, 1298 = (62) (20) + 58
Sol.   Let the no. of persons in the city by 100. Then we have
⇒ λ = 58
n(A) = 42, n(B) = 51, b(C) = 68;
λ → 27, 58
du

n(A ∩ B) = 30, n(B ∩ C) = 28, n(A ∩ C) = 36
∴         sum = 27 + 58 = 85
n(A ∪ B ∪ C) = 100 – 8 = 92.
36.       If x = 560, and
Using
ite

(i)       L = Number of proper divisors of x
n(A ∪ B ∪ C) = n(A) + n(B) + n(C)– n(A ∩ B) – n(B ∩ C)
(ii)      M = Number of proper odd divisors of x
– n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C)
(iii)     N = Number of proper even divisors of x
Substituting the above values, we have
mi

Find L – (M + N)
92 = 42 + 51 + 68 – 30 – 28 – 36 + n(A ∩ B ∩ C)
Ans. 0
⇒      n(A ∩ B ∩ C) = 92 – 161 + 94
Sol.      560 = (2)4 . (5)1 . (7)1
⇒      n(A ∩ B ∩ C) = 92 – 67 = 25.
(i)       Total No. of divisors of
Hence, 25% of the people read all the three papers.
x = (5) × (2) × (2) = 20
35.    When 1298 and 1546 are divided by a certain two

digit number, the remainders are equal. Find sum                                  L = Number of proper divisors of x = 20 - 2 = 18

of all possible remainders.                                             (ii)      No. of odd divisors of x = (b + 1) (c + 1)

Ans. 85                                                                                  = ( 2) (2) = 4

Sol.   Let x be the two digit numbers                                          ∴         M = no. of proper odd divisors = 4 – 1 = 3

∴         1298 = xq1 + r                                                (iii) No. of even divisors of x = (a) ( b +1)( c + 1)

S.C.O 73, Sec - 20 C, Chandigarh, 2775646, 9417317019, S.C.F 73, Sector 10, & S.C.O. 214, Sec. 14, Panchkula, 2583315, 9463087706
SET THEORY & ELE. NUMBER SYSTEM( AIEEE level)
9-5-2010                                                                  7

= ( 4) (2) (2) = 16                                                            passed in chemistry. Maximum 19 students passed

∴      N = No of proper even divisors of x = 16 - 1 = 15                              in Maths and Physics, Maximum 29 students

passed in Maths and Chemistry and Maximum 20
∴      L – (M + N) = 0 Ans.
students in Physics and Chemistry. Find the number
37.    A = {x : x = n2 and n = 4, 5, 6} B = {y : y = n ∈ N
of maximum students passed in three subjects.
and 15 ≤ y ≤ 30, n ∈ N} and C = {p ∈ A and p ∉ B}
Ans. 14
Find number of proper odd factors of p. C = {p ∈

A}.                                                                    Sol.    Q       n(M ∪ P ∪ C) = 50, n(M) = 37, n(P) = 24

Ans. 2                                                                                n(C)=43, n(M ∩ P) ≤ 19, n(M ∩ C) ≤ 29,

Sol.   A = {16, 25, 36} B = {16, 25}                                                  n(P ∩ C) ≤ 20

(A – B) = {36} = p                                                             Now, n(M ∪ P ∪ C) = n (M)+n(P) + n(C)

∴        p = 36 = 22 . 32.                                                     – n (M ∩ P) -n (M ∩ C)- n(P ∩ C)+n (M ∩ P ∩C)

number of odd factors = 1.(3) = 3                                       ⇒      50 = 37 +24 +43 – n (M ∩ P) – n (M ∩ C) – n(P ∩ C)

+ n(M ∩ P ∩ C)
number of odd proper factors = 3 – 1 = 2
⇒      n(M ∩ P ∩ C) =n (M ∩ P) +n(M ∩ C) + N(P ∩ C) – 54
38.    Out of 50 students, in examinations of Maths,

Physics and Chemistry, 37 students passed in                            ⇒      n (M ∩ P ∩ C) ≤ 19 + 29 +20 – 54

.in
Maths, 24 students in Physics and 43 students                           ⇒      n (M ∩ P ∩ C) ≤ 14
.co
du
ite
mi

S.C.O 73, Sec - 20 C, Chandigarh, 2775646, 9417317019, S.C.F 73, Sector 10, & S.C.O. 214, Sec. 14, Panchkula, 2583315, 9463087706

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