Sets and Functions Online Test Aptitude Questions With Solutions, Online Test On Sets and Functions

Sets and functions are basic mathematical concepts that find applications in different fields like computer science, engineering, economics, and physics. Evaluating your understanding of these concepts and sharpen your problem-solving abilities can be achieved by taking online tests.

Set:

A set is a collection of distinct objects called elements. In an online set test you are asked to identify the properties of sets, such as subsets, intersections, unions.

Function:

A function is a relation between a set of inputs and a set of possible outputs with each input related to exactly one output. In an online function test you may be asked to identify the properties of functions, such as domain, range, inverse, and composition.

The best way to improve your understanding of Sets and Functions and sharpen your skills by utilizing the Free Online Test platform, where you can access various practice questions and their solutions.

Learn these tips and tricks to improve your Set theory calculations.

A – A = Ø

B – A = B∩A

B – A = B – (A∩B)

n(A∪B) = n(A – B) + n(B – A) + n(A∩B)

n(A – B) = n(A∪B) – n(B)

n(A – B) = n(A) – n(A∩B)

(A – B) = A if A∩B = Ø

(A – B) ∩ C = (A∩C) – (B∩C)

A Δ B = (A-B) ∪ (B- A)

n(A‘) = n(∪) – n(A)

Some Important Set Formulas

Laws of Universal Set and Empty set: Ø’ = ∪ and ∪’ = Ø

Law of Double complementation: (A’)’ = A

Complement Law : A∪A’ = U, A⋂A’ = Ø and A’ = U – A

De Morgan’s Law: (A ∪B)’ = A’ ⋂B’ and (A⋂B)’ = A’ ∪ B’

Some Question On Sets and Functions

Question-1)

If p = {a, b, c, d, e, f}, A = {a, b, c}, find (P ∪ A)′

Answer:

P ∪ A = {a, b, c, d, e, f} ∪ {a, b, c}

= {a, b, c, d, e, f} = P

= Hence (P ∪ A)′ = Φ

Question-2)

If N = {a, b, c, d, e, f}, P = {a, b, c}, Q = {c, d, e, f}, and R = {c, d, e} find (P ∪ Q) ∪ R

Answer:

P ∪ Q = { a, b, c } ∪ { c, d, e, f }

= { a, b, c, d, e, f }

∴ (P ∪ Q) ∪ R

= { a, b, c, d, e, f } ∪ { c, d, e }

= { a, b, c, d, e, f } = N

Hence All elements are in set U so Answer is N

Question-3)

If R = { 1, 2, 3 } and S = { 4 , 5 } then what is the value of R × S ?

Answer:

The Cartesian product X × Y of sets X , Y is the set of all ordered pairs ( x , y ).

Two ordered pairs $( x_1 , y_1 ), ( x_2 , y_2 ) $

Given that R = { 1, 2, 3 }

and S = { 4 , 5 }

then, R × S = { 1, 2, 3 } × { 4 , 5 }

⇒ R × S = { 1 } × { 4 , 5 } ,

{ 2 } × { 4 , 5 } , { 3 } × { 4 , 5 }

⇒ R × S = { 1 , 4 } , { 1 , 5 } , { 2 , 4 } , { 2 , 5 } , { 3 , 4 }, { 3 , 5 }

⇒ R × S = { (1 , 4 ) , ( 1 , 5 ) , ( 2 , 4 ) , ( 2 , 5 ) , ( 3 , 4 ), ( 3 , 5) }

Question-4)

Lets keep A and B a two finite sets where n(A) = 20, n(B) = 28 and n(A ∪ B) = 36, so find n(A ∩ B)

Answer:

Using Set Theory Formula: n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

Wherein n(A ∩B) = n(A) + n(B) – n(A ∪B)

= 20 + 28 – 36

= 48 – 36 = 12

Question-5)

If A and B are two sets such that A ∪ B has 18 elements, A has 8 elements, and B has 15 elements, find how many elements does A ∩ B have?

Answer:

From question,

(A ∪ B) = 18,

(A) = 8, (B) = 15

Using the formula

(A ∩ B) = (A) + (B) – (A ∪ B),

So, (A ∩ B) = 8 + 15 – 18 = 5

Question-6)

In a class of 50 students, 35 opted for mathematics and 37 opted for Biology. Find how many students have opted for both Mathematics and Biology:

Answer:

From question,

(M ∪ B) = 50, (here M = math students, B = biology students)

(M) = 35, (B) = 37, (M ∩ B) =?

By formula

(M ∪ B) = (M) + (B) – (M ∩ B)

So, 50 = 35 + 37 – (M ∩ B)

⇒ (M ∩ B) = 35 + 37 – 50

= 72 – 50 = 22