CLEVERB EDUTECH PVT LTD

Survey no:97, Belahalli main road, near srinivasapura,

Yelahanka
Bangalore Urban, Karnataka 560064
MATHEMATICS TEST

Topics covered
Mathematics
Sets - Venn diagram and operation on sets, Types of sets, Sets & methods of representation,

(3) 6 (4) Cannot be determined

Subject : Mathematics No. of Questions : 20
8. If U = {1, 2, 3, 4, 5, 6, 7, 8, 9, }, A = {2, 4, 6, 8} and
1. In a certain town, 25% of the families own a phone and 15% B = {2, 3, 5, 7, } then (A ∪ B) , (A
′ ′ ′
∩ B ), (AΔB) is equal to
own a car;65% families own neither a phone nor a car and
2,000 families own both a car and a phone. Consider the (1) {1, 9}, {2, 8}, {3, 4, 5, 6, 7, 8, }
following three statements: (2) {1, 9}, {1, 9}, {3, 4, 5, 6, 7, 8}
(i) 5% families own both a car and a phone (3) {1, 9}, {1, 9}, {5.6, 7, 8}
(ii) 35% families own either a car or a phone (iii) 40,000
(4) None of the above
families live in the town
Then 9. If A = {ϕ, {ϕ}} then the power set of A is
(1) Only (i) and (iii) are correct (1) A (2) {ϕ, {ϕ}, A}
(2) Only (ii) and (iii) are correct (3) {ϕ, {ϕ}, {{ϕ}}, A} (4) None
(3) All (i), (ii) and (iii) are correct
(4) Only (i) and (ii) are correct 10. If A 1 ⊂ A2 ⊂ A3 . . . ⊂ A79 and n (A ) = i + 2 then ,
i

79

2. Which of the following is an empty set n ( ⋂ Ai ) =

i=1

(1) {ϕ}
(1) 80 (2) 3 (3) 90 (4) 91
(2) {0}
(3) The set of all natural numbers less than 1 11. If set A has |A| = 3 elements and satisfies:
(4) The set of all even prime numbers (i) 0 ∈ A and (ii) if x ∈ A, y ∈ A and x ≠ y, then x + y ∈ A
and 10 is an element of A, what is the other element of A
3. Which of the folling is the empty set? excluding 0 and 10?
(1) {x/x is real, and x
2
+ 1 = 0} (1) −20 (2) 20 (3) 10 (4) 1
(2) {x : x is real, and x
2
= x + 2}
12. Let set A be a 90-element subset of , {1, 2, 3, . . . , 100}, and
(3) {x : x is real, and x
2
− 1 = 0} let S be the sum of the elements of A. Find the number of
(4) {x : x is real, and x
2
− 9 = 0} possible values of S.
(1) 901 (2) 798
4. Let P = {θ : sin θ − cos θ = √2 cos θ}and
(3) 900 (4) none of these
Q = {θ : sin θ + cos θ = √ 2 sin θ} be two sets.

Then 13. Two sets A and B are as under

A = {(a, b) ∈ R × R : |a − 5| < 1 and |b − 5| < 1} ;
(1) P ⊂ Q and Q − P ≠ ϕ (2) Q ⊄ P
2 2
B = {(a, b) ∈ R × R : 4(a − 6) + 9(b − 5) ≤ 36} ,
(3) P = Q (4) P ⊄ Q

Then:
5. If A and B are subsets of U = {1, 2, 3, 4, 5, 6, 7, 8, 9} such that
C C C (1) Neither A ⊂ B nor B ⊂ A
A ∩ B = {1, 9}, A ∩ B = {2}, A ∩ B = {4, 6, 8},

what is the set A ? (2) B ⊂ A

(3) A ⊂ B
(1) {2, 3, 5, 7} (2) {3, 4, 9}
(4) A ∩ B = ϕ(an empty set)
(3) {1, 2, 6, 7} (4) {5, 8}

14. The set {x : x is a positive integer less than 6 and 3 x

− 1 s
6. If A = {3, 5, 7, 9} , B = {1, 3, 5} , an even number} in roster form is
μ = {1, 2, 3, 4, 5, 7, 9} then (A ∪ B)
′
=
(1) {1, 2, 3, 4, 5} (2) {1, 2, 3, 4, 5, 6, }

(1) {2, 4} (2) {2, 4, 6, 8}

(3) {2, 4, 6, } (4) {1, 3, 5}
(3) {1, 2, 4, 7, 9} (4) ∅
15. Let F be the set of parallelograms, F the set of
1 2

7. A and B are any two non-empty sets and A is a proper rectangles, F the set of rhombuses, F the set of squares
3 4

subset of B. If n (A) = 5, then find the minimum possible and F the set of trapeziums in the plane. Then F may be
5 1

value of n (A Δ B) equal to
(1) Is 1 (2) Is 5 (1) F2 ∩ F3 (2) F3 ∩ F4

https://cleverb.in/branch/exam/generate-paper/56 MATHEMATICS TEST

 (3) F2 ∪ F5 (4) F1 ∪ F2 ∪ F3 ∪ F4 students is selected at random, then the probability that the
student selected has opted neither for NCC nor for NSS is
16. If A and B are two sets, then A ∩ (A ∪ B) equals to 1 5 1 2
(1) (2) (3) (4)
(1) A (2) B (3) ϕ (4) A ∩ B 6 6 3 3

17. A relation on the set A = {x : |x| < 3, x ∈ Z} , where is the 19. If X and Y are two sets and X denotes the complement of
′

X, then X ∩ (X ∪ Y ) is equal to
′
set of integers is defined by R = {(x, y) : y = |x| , x ≠ −1} .
Then the number of elements in the power set of is (1) X (2) Y (3) ϕ (4) A ∩ B

(1) 32 (2) 16 (3) 8 (4) 64

20. The number of elements in the set
18. In a class of 60 students, 40 opted for NCC, 30 opted for {x ∈ R : (|x| − 3) |x + 4| = 6} is equal to

NSS and 20 opted for both NCC and NSS. If one of these
(1) 2 (2) 1 (3) 3 (4) 4

https://cleverb.in/branch/exam/generate-paper/56 MATHEMATICS TEST

 CLEVERB EDUTECH PVT LTD
Survey no:97, Belahalli main road, near srinivasapura,
Yelahanka
Bangalore Urban, Karnataka 560064
MATHEMATICS TEST

Answer Key
Mathematics

1) 3 2) 3 3) 1 4) 3 5) 1 6) 1 7) 1 8) 2 9) 3 10) 2

11) 3 12) 1 13) 3 14) 1 15) 4 16) 1 17) 2 18) 1 19) 3 20) 1

Hint And Solutions

10. (2) The solution to this problem is similar to the first
1. (3) n (P ) = 25%
problem
n (C) = 15%

n (P
′ ′
∪ C ) = 65% 11. (3) Let A be the set Since (10 + a) ∈ A from (ii), we have
′ 10 + a = 0 or 10 + a = 10
⇒ n(P ∪ C) = 65%
which implies a = −10 or a = 0. Since |A| = 3this implies
n (P ∪ C) = 35%
|A| = {0, 10, −10}
n (P ∩ C) = n (P ) + n (C) − n (P ∪ C)
and the other element of A excluding 0 and 10 is -10.
25 + 15 − 35 = 5%

x × 5% = 2000 12. (1) The smallest S is 1 + 2+. . . +90 = 91 ⋅ 45 = 4095. The

x = 40, 000 largest S is 11 + 12+. . . +100 = 111 ⋅ 45 = 4995 . All
numbers between 4095 and 4995 are possible values of
2. (3) The set of all natural numbers less than 1.
S, so the number of possible values of S is
3. (1) Clearly, x + 1 = 0 ⇒ x 2 2
= −1 4995 − 4095 + 1 = 901

⇒ x is not real Alternatively, for ease of calculation, let set B be a 10-

element subset of {1, 2, 3, . . . , 100}, and let T be the sum
4. (3) sin θ − cos θ = √2 cos θ
of the elements of B. Note that the number of possible S is
⇒ sin θ = cos θ + √2 cos θ the number of possible . The smallest possible T is
2 − 1 1 + 2+. . . +10 = 55 and the largest is
= (√2 + 1) cos θ = (
91 + 92+. . . +100 = 955 , so the number of possible values
) cos θ
√2 − 1
of T, and therefore S, is 955 − 55 + 1 = 901
⇒ (√2 − 1) sin θ = cos θ

⇒ sin θ + cos θ = √2 sin θ

13. (3) Given |a − 5| < 1 and |b − 5| < 1
2 2

∴ P = Q (a − 6) (b − 5)
i.e., 4 < a, b < 6 and + ≤ 1
9 4
5. (1) We have Taking axes as a-axis and b-axis
(A
C
∩ B) ∪ (A
C
∩ B
C
) = A
C
∩ (B ∪ B
C
) Hence, the square that represent the set A, like wise the
= A
C
∪ U
ellipse represents the set B.
C Clearly from the above graph, we see the set A is inside
⇒ A = {1, 9} ∪ {4, 6, 8} ⇒ A = {2, 3, 5, 7}.
the set B. i.e., A ⊂ B
6. (1) A ∪ B = {1, 3, 5, 7, 9} ⇒ (A ∪ B)
′
= {2, 4}

7. (1) Given tat

A ∪ A ⊂ B ⇒ A − B = ϕ ⇒ n (A − B) = 0

A ⊂ B ⇒ n(A − B) = 0, also given that n (A) = 5

Thus the minimum number of elements in B is 6.
we have, n (A Δ B) = n (A) + n (B) − 2n (A ∩ B)
⇒ n (A Δ B) = n (A) + n (B) − 2n (A)< br>

(A ⊂ B ⇒ A ∩ B = A)

= n (B) − n (A)

= 6 − 5 = 1

8. (2) Since A ∪ B = {2, 3, 4, 5, 6, 7, 8} Now (A ∪ B) ′

= {1, 9 } .
Likewise A ∩ B = {1, 9) and
′ ′

AΔB = (A − B) ∪ (B − A) = {3, 4, 5, 6, 7, 8}
14. (1) Given that x is a positive integer less than 6. That is
9. (3) (A) = {ϕ, {ϕ}} x ∈ z
+
and x < 6 . Hence the set is {1, 2, 3, 4, 5}. And also
P (A) = {ϕ, {ϕ}, {{ϕ}}, A} given the other set 3 − 1 is an even number. The required
x

set in roster form is {1, 2, 3, 4, 5}.

https://cleverb.in/branch/exam/generate-paper/56 MATHEMATICS TEST
15. (4) Given, F = The set of parallelograms
1 nor for NSS are 60 − 50 = 10
F = The set of rectangles, F = The set of rhombuses 10 1
2 3
So, P (Neither C nor S) = =
F = The set of squares
4 60 6

F = The set of trapeziums

5

By definition of parallelogram, opposite sides are equal 19. (3) X ∩ (X ∪ Y )

′
= X ∩ (X
′
∩ Y
′
)
′
and parallel. [(A ∪ B)
′
= A ∩ B ]
′

In rectangles, rhombuses and squares all have opposite = (X ∩ X ) ∩ (X ∩ Y

′ ′
)
sides equal and parallel, therefore = ϕ ∩ (X ∩ Y
′
) = ϕ
F2 ⊂ F1 , F3 ⊂ F1 , F4 ⊂ F1
[ϕ ∩ A = ϕ]
∴ F1 = F1 ∪ F2 ∪ F3 ∪ F4

16. (1) A ∩ (A ∪ B) = A
20. (1) Case-1 x ≤ −4
(−x − 3) (−x − 4) = 6

⇒ (x + 3) (x + 4) = 6
2
⇒ x + 7x + 6 = 0

⇒ x = −1 or − 6

but x ≤ −4

x = −6, x = −1 is rejected

case - 2 x ∈ (−4, 0)
(−x − 3) (x + 4) = 6
2
⇒ −x − 7x − 12 − 6 = 0
2
⇒ x + 7x + 18 = 0

D < 0 No solution
17. (2) A = {x : |x < 3| , x ∈ Z} case -3 x ≥ 0
A = {−2, −1, 0, 1, 2} (x − 3) (x + 4) = 6
2
R = {(x, y) : y = |x| , x ≠ −1} ⇒ x + x − 12 − 6 = 0
2
R = {(−2, 2) , (0, 0) , (1, 1) , (2, 2)} ⇒ x + x − 18 = 0

Number of elements in the power set of R −1 ± √1 + 72

x =

18. (1) Let C and S represents the set of students who opted 2

√73 − 1
for NCC and NSS respectively. ∴ x = only
∴ n (C) = 40, n (S) = 30, n (C ∩ S) = 20 2

Now, n (C ∪ S) = 40 + 30 − 20 = 50 Here x =
−1 − √73
is rejected
So, number of students who has opted neither for NCC 2

https://cleverb.in/branch/exam/generate-paper/56 MATHEMATICS TEST