[JEE Mains MCQs]
Q1: A survey shows that 73% of the persons working in an office like coffee, whereas 65% like tea. If x denotes the percentage of them, who like both coffee and tea, then x cannot be :
(a) 63
(b) 36
(c) 54
(d) 38
Ans: (b)
C → person like coffee
T → person like Tea
n(C) = 73
n(T) = 65
n(C ∪ T) ≤ 100
n(C) + n(T) – n (C ∩ T) ≤ 100
73 + 65 – x ≤ 100
x ≥ 38
73 – x ≥ 0 ⇒ x ≤ 73
65 – x ≥ 0 ⇒ x ≤ 65
∴ 38 ≤ x ≤ 65
Q2: Let 
where each Xi contains 10 elements and each Yi contains 5 elements. If each element of the set T is an element of exactly 20 of sets Xi’s and exactly 6 of sets Yi’s, then n is equal to:
(a) 30
(b) 50
(c) 15
(d) 45
Ans: (a)
= 50 sets. Given each sets having 10 elements.
So total elements = 50 × 10
sets. Given each sets having 5 elements.
So total elements = 5 × n
Now each element of set T contains exactly 20 of sets Xi.
So number of effective elements in set
Also each element of set T contains exactly 6 of sets Yi.
So number of effective elements in set
Q3: A survey shows that 63% of the people in a city read newspaper A whereas 76% read newspaper B. If x% of the people read both the newspapers, then a possible value of x can be:
(a) 37
(b) 65
(c) 29
(d) 55
Ans: (d)
A ∪ B = 63 - x + x + 76 - x = 139 - x
As 139 - x ≤ 100
⇒ x ≥ 39
From venn diagram, you can see x should be less than 76 and 63 otherwise only A newspaper or only B newspaper reader will be negative number.
Intersection of x ≤ 63 and x ≤ 76 is = x ≤ 63.
∴ 39 ≤ x ≤ 63
From options possible value of x = 55.
Q4: Let R1 and R2 be two relation defined as follows:
R1 = {(a, b) ∈ R2 : a2 + b2 ∈ Q} and
R2 = {(a, b) ∈ R2 : a2 + b2 ∉ Q},
where Q is the set of all rational numbers.
Then:
(a) Neither R1 nor R2 is transitive.
(b) R2 is transitive but R1 is not transitive.
(c) R1 and R2 are both transitive.
(d) R1 is transitive but R2 is not transitive.
Ans: (a)
For R1:
∴ R2 is not transitive.
Q5: Consider the two sets:
A2 = {m ∈ R : both the roots of x2 – (m + 1)x + m + 4 = 0 are real} and B = [–3, 5).
Which of the following is not true?
(a) A ∩ B = {–3}
(b) B – A = (–3, 5)
(c) A ∪ B = R
(d) A - B = ( − ∝ , − 3) ∪ (5, ∝ )
Ans: (d)
Now, let's examine the options.
Option A: A ∩ B = {–3} The intersection of sets A and B would be the set of elements common to both sets. In this case, the only common element is -3. So, option A is true.
Option B: B – A = (–3, 5) The subtraction (or difference) of sets A from B is the set of elements that are in B but not in A. B is [–3, 5), and A is (-∞, -3] U [5, ∞). Subtracting A from B would leave an open interval (-3, 5), not including -3 and 5. So, option B is also true.
Option C: A ∪ B = R The union of sets A and B is the set of elements that are in A, or B, or both. Here, A U B would cover all real numbers. So, option C is true.
Option D: A - B = (-∞, -3) ∪ (5, ∞) The subtraction (or difference) of set B from A is the set of elements that are in A but not in B. B is [–3, 5), and A is (-∞, -3] U [5, ∞). Subtracting B from A would leave (-∞, -3) U [5, ∞), not including -3 and 5. But according to the convention for writing intervals, it should be (-∞, -3) U (5, ∞). So, option D is not true.
Q6: If R = {(x, y) : x, y ∈ Z, x2 + 3y2 ≤ 8} is a relation on the set of integers Z, then the domain of R–1 is:
(a) {0, 1}
(b) {–2, –1, 1, 2}
(c) {–1, 0, 1}
(d) {–2, –1, 0, 1, 2}
Ans: (c)
Given R = {(x, y) : x, y ∈ Z, x2 + 3y2 ≤ 8}
So R = {(0,1), (0,–1), (1,0), (–1,0), (1,1), (1,-1)
(-1,1), (-1,-1), (2,0), (-2,0), (-2,0), (2,1), (2,-1), (-2,1), (-2,-1)}
⇒ R : { -2, -1, 0, 1, 2} → {-1, 0, 1}
∴ R-1 : {-1, 0, 1} → { -2, -1, 0, 1, 2}
∴ Domain of R–1 = {-1, 0, 1}
Q7: If A = {x ∈ R : |x| < 2} and B = {x ∈ R : |x – 2| ≥ 3}; then :
(a) A – B = [–1, 2)
(b) A ∪ B = R – (2, 5)
(c) A ∩ B = (–2, –1)
(d) B – A = R – (–2, 5)
Ans: (d)
A : x ∈ (–2, 2);
B : x ∈ (– ∞ , –1] ∪ [5, ∞ )
⇒ B – A = R – (–2, 5)
[JEE Mains Numericals]
Q8: Set A has m elements and set B has n elements. If the total number of subsets of A is 112 more than the total number of subsets of B, then the value of m.n is ______.
Ans: 28
Number of subsets of A = 2m
Number of subsets of B = 2n
Given = 2m – 2n = 112
∴ m = 7, n = 4 (27 – 24 = 112)
∴ m × n = 7 × 4 = 28
Q9: Let X = {n ∈ N : 1 ≤ n ≤ 50}. If
A = {n ∈ X: n is a multiple of 2} and
B = {n ∈ X: n is a multiple of 7}, then the number of elements in the smallest subset of X containing both A and B is ________.
Ans: 29
X = {1, 2, 3, 4, …, 50}
A = {2, 4, 6, 8, …, 50} = 25 elements
B = {7, 14, 21, 28, 35, 42, 49} = 7 elements
Here n(A∪B) = n(A) + n(B) – n(A∩B)
= 25 + 7 – 3 = 29
where each Xi contains 10 elements and each Yi contains 5 elements. If each element of the set T is an element of exactly 20 of sets Xi’s and exactly 6 of sets Yi’s, then n is equal to:
= 50 sets. Given each sets having 10 elements.
sets. Given each sets having 5 elements.
A ∪ B = 63 - x + x + 76 - x = 139 - x
