MATHS MCQ

1. The collection of months having names starting with J.

Solution:
As the collection of months having names starting with J is well defined. So, it's a set. Rest are not well defined , hence are not set.

Solution:
Null set is the subset of all given sets as it can lie in all sets.

The only 4 no.s that lie in the given range are -3, 0. -1 and 5.

(A’)’ = A

The set S consists of the square of an integer less than 100
So, S = {0, 1, 4, 9, 16, 25, 36, 49, 64, 81}

Since 80 do not read any
n(S ∪ J) = 450 − 80 = 370....................(S = Science Today; J = Junior Statesman)
By set theory
n(J ∩ S) = n(J) + n(S) − n(J ∪ S)
= 200 + 193 − 370 = 23

A × (B ∪ C) = {a, b, c} × {a, c, d, e, f}.
The above set will consist of 15 ordered pairs and not 3.

A ∪ B must contain all the elements of the bigger set.

3. {a, b, c}

P : there is a rational number x ∈ S such that x > 0
∼P: Every rational number x ∈ S satisfies x ≤ 0

From group theory, integers are closed w.r.t. both addition & multiplication.

Using distributive law of sets Or it is the distributive law itself.

If A and B are disjoint it would mean A is a null set. Otherwise A and B must be equal to A ∩ B AT LEAST.

A is included b is included in the next interval.

1. {2}

3. f

Given, A = {1, 2, 3} and B = {4, 5}.
Since A and B are two disjoint sets i.e. A ∩ B = ϕ then we've,
A − B
= {1, 2, 3}.

Given n(S) = 21, n(T) = 32, n(S ∩ T) = 11
Now n(S) + n(T) = n(S ∩ T) + n(S ∪ T)
⇒ n(S ∪ T) = 21 + 32 - 11 = 42.

Let us find the Relation between the numbers of the set (7, 15, 31).
(7) × 2 + 1 = (15)
(15) × 2 + 1 = (31)
Options A,B,C are not of similar type of above set
This similar type of relation is shown by option D (7, 15, 31).
(5) × 2 + 3 = (13)
(13) × 2 + 3 = (29)

Let A be the left circle and B be the right circle.There are 3 elements in A and 6 elements in B.
The union of A and B contains elements that are in either circle.Thus,the union of A and B will be all of the elements in A along with all of the element B.
However, you have to subtract the elements that are in the overlapping area because you are counting twice.
If A and B don't overlap at all,then the union will ontain 9 elements.If A is completely inside B then the union will only contain 6 elements,which is the minimum no. of elements in the union of A and B.
let A = 1, 2, B = 2, 3
∴ A ∪ B = 1, 2, 3 which is 3 elements.
∴ A has 2 elements, B has 2 elements, and there is 1 element overlapping.
∴ 2 + 2 − 1 = 3! = 3 × 2 × 1 = 6

Since 3 women have neither nose studs nor earrings
n(N∪E) = 15 − 3 = 12
By set theory
n(N∩E) = n(N) + n(E) − n(N∪E)
= 7 + 8 − 12 = 3

3. {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

Let A be the set of students who passed first semester so n(A) = 260
and B be the set of students who passed second semester so n(B) = 210.
Now 170 did not passed any semester.
So, (500 − 170 = 330) students passed atleast one of the semesters.
∴ n(A∪B) = 330
Now n(A ∪ B) = n(A) + n(B) − n(A ∩ B)
330 = 260 + 210 − n(A ∩ B)
n(A ∩ B) = 140

Let H = People who speak Hindi
E = People who speak English
According to the questions,
n(H) = 50, n(E) = 20, n(H ∩ E) = 10
Number of people who speak at least two language = n(H ∪ E)
= n(H) + n(E) − n(H ∩ E)
= 50 + 20 − 10 = 60

Given, A ∩ B = ϕ.
Now,
(A ∪ B′)
= (A′ ∩ B)′ [Using De Morgan's law]
= B′.[ As A′ ∩ B = B − (A ∩ B) = B since A ∩ B = ϕ]

A = {2, 4, 6, 8} and B = {1, 4, 7, 8}
A − B = {2, 6} and B − A = {1, 7}

n(X ∪ Y) = n(X) + n(Y) -n(X ∩ Y)
76 = 45 + n(Y) - 12
n(Y) = 43

To find maximum possible percentage of people drinking either coffee or tea, we can assume everyone drinks at least either of the options.
Hence
a + b = 100
a + 2b = 60 + 70 = 130
b = 30

Given, I is the set of isosceles triangle and E is the equilateral triangles.
We know that every equilateral triangle is an isosceles triangle but the converse is not true.