Set Theory Model Questions Set 1 Practice Questions Answers Test with Solutions & More Shortcuts
Set Theory PRACTICE TEST [2 - EXERCISES]
Set Theory Model Questions Set 1
Set Theory Model Questions Set 2
Question : 21
If X= {a, {b}, c}, Y = {{a}, b, c} and Z = {a, b, {c}}, then (X ∩ Y) ∩ Z equals to
a) {Φ}
b) {{a}, {b}, {c}}
c) {a, b, c}
d) Φ
Answer »Answer: (d)
(X ∩ Y) = {a,{b},c} ∩ {{a},b,c} = c
Now, (X ∩ Y) ∩ Z
= c ∩ {a, b,{c}} = Φ
Question : 22
In the quadratic equation $x^2$ + ax + b = 0, a and b can take any value from the set {1, 2, 3, 4}. How many pairs of values of a and b are possible in order that the quadratic equation has real roots?
a) 8
b) 7
c) 6
d) 16
Answer »Answer: (b)
$x^2$ + ax + b = 0 | value 0f (a, b) |
$a^2$ - ab ≥ 0 | [1, 2, 3, 4] |
for real roots | |
$1^2$ - 4 × 1 ≥ 0 | (1, 1) |
not possible
(2, 1) → $2^2$ – 4 × 1 ≥ 0 possible
(3, 1) → possible
(3, 2), (4, 1), (4, 2) (4, 3) (4, 4) → possible
So → possible values can be possible
Question : 23
In a competitive examination, 250 students have registered. Out of these, 50 students have registered for Physics, 75 students for Mathematics and 35 students for both Mathematics and Physics. What is the number of students who have registered neither for Physics nor for Mathematics?
a) 150
b) 100
c) 90
d) 160
Answer »Answer: (d)
Number of students registered for
Physics n(P) = 50
Mathematics n(M) = 75
Number of students registered for both subjects n(P ∩ M) = 35.
Number of students, registered for either physics or mathematics
n(P ∪ M) = n(P) + n(M) – n(P ∩ M).
= 50 + 75 – 35 = 90
∴ Number of students registered neither for physics nor for mathematics
n($\ov{P ∪ M}$) = 250 – 90 = 160.
Question : 24
Let :
P = Set of all integral multiples of 3
Q = Set of all integral multiples of 4
R = Set of all integral multiples of 6
Consider the following relations:
I. P ∪ Q = R
II. P ⊂ R
III. R ⊂ (P ∪ Q)
Which of the relations given above is/are correct?
a) Only III
b) Only II
c) Only I
d) II and III
Answer »Answer: (a)
Here, P = {..., –6, –3, 0, 3, 6, ...}
Q = {..., –8, – 4, 0, 4, 8, ...}
and R = {..., –36, –6, 0, 6, 36, ...}
I. P ∪ Q = {..., –8, –6, – 4, –3, 0 3, 4, 6, 8, ...} ≠ R
II. Here, P ⊄ R
III. Here, R ⊂ (P ∪ Q) is true.
Question : 25
If A = {x : x is an odd integer} and B = {x : $x^2$ – 8x + 15 = 0}. Then, which one of the following is correct?
a) B ⊆ A
b) A ⊆ B
c) A = B
d) A ⊆ $B^C$
Answer »Answer: (a)
Given that,
A = {x : x is an odd integer}
and B = {x : $x^2$ – 8x + 15 = 0}
= (x : $x^2$ – 5x – 3x + 15 = 0)
= {x : x (x – 5) – 3(x – 5) = 0}
= {x : (x – 5) (x – 3) = 0} = {3, 5}
Since, B has the odd elements,
∴ B ⊆ A
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