[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 X_{i} contains 10 elements and each Y_{i} contains 5 elements. If each element of the set T is an element of exactly 20 of sets X_{i}’s and exactly 6 of sets Y_{i}’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 X_{i}.
So number of effective elements in set
Also each element of set T contains exactly 6 of sets Y_{i}.
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 R_{1} and R_{2} be two relation defined as follows:
R_{1} = {(a, b) ∈ R^{2} : a^{2} + b2 ∈ Q} and
R_{2} = {(a, b) ∈ R^{2} : a^{2} + b^{2} ∉ Q},
where Q is the set of all rational numbers.
Then:
(a) Neither R_{1} nor R_{2} is transitive.
(b) R_{2} is transitive but R_{1} is not transitive.
(c) R_{1} and R_{2} are both transitive.
(d) R_{1} is transitive but R_{2} is not transitive.
Ans: (a)
For R_{1}:
∴ R_{2} is not transitive.
Q5: Consider the two sets:
A_{2} = {m ∈ R : both the roots of x^{2} – (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, x^{2} + 3y^{2} ≤ 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, x^{2} + 3y^{2} ≤ 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 = 2^{m}
Number of subsets of B = 2^{n}
Given = 2^{m} – 2^{n} = 112
∴ m = 7, n = 4 (2^{7} – 2^{4} = 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