Sets Class 9 Extra Questions Advanced Maths Chapter 1

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Class 9 Sets Extra Questions

Sets Class 9 Very Short Question Answer

Question 1.
What is the difference between a collection and a set? Give reason to support your answer.
Answer:
Every set is a collection, but a collection is not necessarily a set. Only well-defined collection are sets.
e.g. Collection of most talented writers of India is a collection, but it is not a set.

Question 2.
Find the number of subsets of the set A = {1, 4, 5}.
Answer:
Given, set A = {1, 4, 5}
Number of elements = 3
Then, the number of subsets of set A = 2ⁿ = 2³ = 8 subsets

Question 3.
List all the proper subsets of the set A = {a, b}.
Answer:
The proper subsets of the set A = {a, b] are Φ, {a}, {b}.

Question 4.
Write the following set in roster form.
{x : x is a prime number and a divisor of 6}
Answer:
Given, {x : x is a prime number and a divisor of 6} So, 2 and 3 are the only prime numbers, which are divisor of 6, so in roster form {2, 3}.

Question 5.
Write {2, 4, 8, 16} in set-builder form.
Answer:
The set-builder form is {x : x = 2ⁿ, n ∈ N, 1 ≤ n ≤ 4}.

Question 6.
Find number of subsets, if
A = {x : x² + 3 < 2, x ∈ R}.
Answer:
Given, A = {x : x² + 3 < 2, x ∈ R}
Set A is a null set, so there will be only one subset Φ itself.
So, the number of subset is 1.

Question 7.
State true or false 7747 ∈ {x : x is a multiple of 37}.
Answer:
Here, 7747 is not the multiple of 37.
So, given statement is false.

Question 8.
Let A be a set and n(A) = 5 then find the value of n[P(A)]
Answer:
Hint n [P(A)] = 2⁵
= 32

Question 9.
If set X = Set of rational numbers and set Y = Set of irrational numbers then find X ∪ Y and X ∩ Y.
Answer:
Given, X = Set of rational numbers
and Y = Set of irrational numbers
X ∪ Y = Set of rational numbers ∪ Set of irrational numbers
= Set of real numbers (rational + irrational) and X ∪ Y = Φ

Question 10.
If A = {3, 5} and B = {4, 6} then find A ∩ B.
Answer:
Given, A = {3, 5} and B = {4, 6}
∴ A ∩ B = Φ

Question 11.
If U = {1, 2, 3 …. 10} and A = {2, 4, 6} then find A’.
Answer:
Given, U = {1, 2, 3 …… 10} and A = {2, 4, 6}.
Now, A’ = U – A
= {1, 2, 3, …. 10} – {2, 4, 6} = {1, 3, 5, 7, 8, 9, 10}

Sets Class 9 Short Question Answer

Question 1.
Which of the following sets are empty set, equal set, finite set and infinite set.
A = {x : x is a solution of x² – 5x + 6 = 0}
B = {x : x² – 16x + 55 = 0 and x² = 25}
C = {x : \(\frac{-1}{2}\) ≤ x ≤ \(\frac{1}{2}\)} D = {x : 0 ≤ 4x² ≤ 1}
E = {x : x is a natural number and x ≥ 100}
F = {x : x is a whole number and 2 < x < 3}
Answer:
We have, A = {x : x is a solution of x² – 5x + 6 = 0}
Now, x² – 5x + 6 = 0
⇒ x² – 3x – 2x + 6 = 0
⇒ x(x – 3) – 2(x – 3) = 0
⇒ (x – 3) (x – 2) = 0 ⇒ x = 2,3
A = {3, 2}
So, it is a finite set.
B = {x : x² – 16x + 55 = 0 and x² = 25}
Now, x² – 16x + 55 = 0 and x² = 25
⇒ x² – 11x – 5x + 55 = 0 and x = ±\(\sqrt{25}\)
⇒ x(x – 11) – 5(x – 11) = 0 and x = ±5
⇒ (x – 11) (x – 5) = 0 and x = ±5
∴ x = 5
So, it is also a finite and singleton set.
C = {x : –\(\frac{1}{2}\) ≤ x ≤ \(\frac{1}{2}\)
and D = {x: 0 ≤ 4x² ≤ 1}
= {x : 0 ≤ < x² ≤ \(\frac{1}{4}\)
= {x : –\(\frac{1}{2}\) ≤ x ≤ \(\frac{1}{2}\)}
⇒ C = D
They are equal sets.
E = {x : x ∈ N and x ≥ 100}
∴ It is an infinite set.
F = {x : x is a whole number and 2 < x < 3}
There is no whole number between 2 and 3.
∴ It is an empty set.

Question 2.
If X = {1, 2, 3) and n represents any member of X, write the following sets containing all numbers represented by
(i) 4n
(ii) n + 6
(iii) \(\frac{n}{2}\)
(iv) n – 1
Answer:
(i) {4, 8,12}
(ii) {7, 8, 9}
(iii) {\(\frac{1}{2}\), 1, \(\frac{3}{2}\)}
(iv) {0, 1, 2}

Question 3.
Given set A = {1, 2, 3, 4, …, 100), write the subset
(i) X of A whose elements are multiple of 7.
(ii) Y of A, whose elements are represented by x + 3, where x ∈ A
Answer:
(i) {7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98}
(ii) {4, 5, 6, … , 100}

Question 4.
Let A and B be two sets. Then, prove that
A = B ⇔ A ⊆ B and B ⊆ A.
Answer:
To prove that two sets A and B are equal if and only if A ⊆ B and B ⊆ A we need to show two implications.
(i) If A = B then A ⊆ B and B ⊆ A
(ii) If A ⊆ B and B ⊆ A then A = B
(i) If A = B
By definition of set equality, A and B have same elements.
Then, A ⊆ B (every element of A is in B)
and B ⊆ A (every element of B is in A).
Thus, if A = B then A ⊆ B and B ⊆ A.

(ii) If A ⊆ B and B ⊆ A
By definition of subset, A ⊆ B means every element of A is also an element of B and B ⊆ A means every element of B is in A.
So, A = B [by definition of equality]
Thus, A = B if and only if A ⊆ B and B ⊆ A.

Question 5.
Two finite sets have m and n elements. The total number of subsets of the first set is 56 more than the total number of subsets of the second set. Find the values of m and n.
Answer:
Let A and B be such sets i.e. n(A) = m and n(B) = n.
So, total number of subsets of set A = 2^m
and total number of subsets of set B = 2ⁿ.
Given, 2^m – 2ⁿ = 56
⇒ 2ⁿ (2^{m – n} – 1) = 2³.7
⇒ 2ⁿ (2^{m – n} – 1) = 2³ (8 – 1) = 2³ (2³ – 1)
⇒ n = 3 and m – n = 3
⇒ n = 3 and m = 6

Question 6.
Write down the power set of the following sets.
(i) {1, 2, {1}, {1, 2, 3}}
(ii) {x, y, z, {x, y}}
Answer:
(i) Let A = {1, 2, {1},{1, 2, 3}}
∴ P(A) = {Φ, {1}, {2}, {{1}}, {{1, 2, 3}}, {1, 2},
{1, {1}}, {1,{1, 2, 3}}, {2, {1}}, {2, {1,2, 3}},
{{1}, {1, 2, 3}}, {1, 2, {1}},{1,2, {1, 2, ,3}}{1, {1},
{1, 2, 3}}, {2, {1}, {1, 2, 3}}, {1, 2, {1},{1, 2, 3}}}

(ii) Let B = {x, y, z,{x, y}}
∴ P(B) = {Φ, {x}, {y}, {z}, {{x, y}}, {x, y}, {x, z},
{x,{x, y}}, {y, z}, {y, {x, y}}, {z,{x, y}}, {x, y, z},
{x, y, {x, y}}, {x, z,{x, y}}, {y, z,{x, y}}, {x, y, z,{x, y}}}

Question 7.
If A = {1, 2, 3, 4, 5}, B = { 1, 3, 5, 8} and C = {2, 5, 7, 8} then verify that A – (B ∪ C) = (A – B) ∩ (A – C).
Answer:
Given, A = {1, 2, 3, 4, 5}, B = {1, 3, 5, 8} and C = {2, 5, 7, 8}
Now, B ∪ C = {1, 2, 3, 5, 7, 8}
∴ A – (B ∪ C = {4}
and (A – B) ∩ (A – C) = {2, 4} ∩ {1, 3, 4} = {4}

Question 8.
If U = {a, b, c, d, e, f}, A = {a, b, c}, B = {c, d, e, f}, C = {c, d, e) and D = {d, e, f} then tabulate the following sets.
(i) A ∩ D
(ii) A ∩ C
(iii) U ∩ D
(iv) A ∪ Φ
(v) (U ∩ Φ)’
(vi) (U ∪ A)’
Answer:
Given, U = {a, b, c, d, e, f}, A = {a, b, c},
B = {c, d, e, f}, C = {c, d, e} and D = {d, e, f}

(i) A ∩ D = {a, b, c} ∩ {d, e, f} = Φ

(ii) A ∩ C = {a, b, c} ∩ {c, d, e} = {c}

(iii) U ∩ D = {a, b, c, d, e, f} ∩ {d, e, f} = {d, e, f}

(iv) A ∪ Φ = {a, b, c} ∪ Φ) = {a, b, c}

(v) (U ∩ Φ)’ = U – (U ∩ Φ)
Now, (U ∩ Φ) = {a, b, c, d, e, f} ∩ Φ = Φ
∴ U – (U ∩ Φ) = {a, b, c, d, e, f} – Φ
= {a, b, c, d, e, f} = U

(vi) (U ∪ A)’ = [U – (U ∪ A)
Now, (U ∪ A) = {a, b, c, d, e, f} ∪ {a, b, c}
= {a, b, c, d, e, f}
∴ U – (U ∪ A) = {a, b, c, d, e, f} – {a, b, c, d, e, f}
= Φ

Question 9.
If n(A ∪ B) = 25, n(A) = 12 and n(A – B) = 8 then write the number of elements in A ∩ B and B – A.
Answer:
Hint n(A – B) = n(A) – n(A ∩ B)
⇒ n(A ∩ B) = 12 – 8 = 4
∵ n(B) = n(A ∪ B) + n(A ∩ B) – n(A)
= 25 + 4 – 12 = 17
∴ n(B – A) = n(B) – n(B ∩ A)
n(A ∩ B) = 4 and n(B – A) = 13

Sets Class 9 Long Question Answer

Question 1.
Let T = {x : \(\frac{x+5}{x-7}\) – 5 = \(\frac{4 x-40}{13-x}\)}.
Is T an empty set? Justify your answer.
Answer:

⇒ (4x – 40) (x – 7) + (4x – 40) (13 – x) = 0
⇒ 24(x – 10) = 0 ⇒ x = 10 ⇒ T = {10}
Hence, T is not an empty set.

Question 2.
In a group of 80 people, 55 drink tea and 45 drink coffee. If every person atleast one of the two beverages, find the number of people who drink only tea.
Answer:
Hint n(T ∪ C) = 80, n(T) = 55 and n(C) = 45
∴ n(T ∪ C) = n(T) + n(C) – n(T ∩ C)
⇒ 80 = 55 + 45 – n(T ∩ C)
⇒ 80 = 100 – n(T ∩ C)
⇒ n(T ∩ C) = 100 – 80 = 20
Now, only tea = n(T) – n(T ∩ C) = 55 – 20 = 35

Question 3.
Let U = {1, 2, 3, 4, 5, 6, 8}, A = {2, 3, 4} and B = {3, 4, 5}.
Show that
(A ∪ B)’ = A’ ∩ B’ and (A ∩ B)’ = A’ ∪ B’.
Answer:
We have, U = {1, 2, 3, 4, 5, 6, 8}, A = {2, 3, 4}
and B = {3, 4, 5}
(i) A ∪ B = {2, 3, 4} ∪ {3, 4, 5} = {2, 3, 4, 5}
∴ (A ∪ B)’ = U – (A ∪ B)
= {1, 2, 3, 4, 5, 6, 8} – {2, 3, 4, 5}
= {1, 6, 8} ……….. (i)
A’ – U – A
= {1, 2, 3, 4, 5, 6, 8} – {2, 3, 4}
= {1, 5, 6, 8}
B’ = [U – B = {1, 2, 3, 4, 5, 6, 8} – {3, 4, 5}
= {1, 2, 6, 8}
A’ ∩ B’ = {1, 5, 6, 8} ∩ {1, 2, 6, 8}
= {1, 6, 8} ………… (ii)
From Eqs. (i) and (ii), we get
(A ∪ B)’ = A’ ∩ B’

(ii) A ∩ B = {2, 3, 4} ∩ {3, 4, 5} = {3, 4}
∴ (A ∩ B)’ = U – (A ∩ B)
= {1, 2, 3, 4, 5, 6, 8} – {3, 4}
= {1, 2, 5, 6, 8} …….(ii)
Now, A’ ∪ B’ = {1, 5, 6, 8} ∪ {1, 2, 6, 8}
= {1, 2, 5, 6, 8} ……….. (iv)
From Eqs. (iii) and (iv), we get
(A ∩ B)’ =A’ ∪ B’

Question 4.
A survey of 500 television viewers produced the following information, 285 watch Football, 195 watch Hockey, 115 watch Basketball, 45 watch Football and Basketball, 70 watch Football and Hockey, 50 watch hockey and Bassketball, 50 do not watch any of the three games. How many watch all the three games?
Answer:
Hint n (F) = 285, n(H) = 195, n(B) = 115,
n(F ∩ B) = 45, n(F ∩ H) = 70, n(H ∩ B) = 50
\(n(\bar{F} \cap \bar{B} \cap \bar{F})\) = 50, n(F ∪ H ∪ B) = 500
∵ n(F ∪ H ∪ B) = n(F) + n(H) + n(B)
– n(F ∩ B) – n(F ∩ H) – n(H ∩ B) + n(F ∩ H ∩ B)
= 20

Question 5.
An investigator interviewed 100 students to determine their preferences for the three drinks. Milk (M), Coffee (C) and Tea (T). He reported that 10 students had all the three drinks M, C, T; 20 had M and C, 30 had C and T, 25 had M and T, 12 had M only, 5 had C only, 8 had T only. Using venn diagram, find how many did not take any of the three drinks?
Answer:
Hint n(U) = 100, n(M ∩ C ∩ T) = 10,
n(M ∩ C) = 20, n(C ∩ T) = 30, n(M ∩ T) = 25,
n(M only) = 12, n(C only) = 5 and n(T only) = 8

Now, n(M ∪ C ∪ T) = 12 + 15 + 10 + 10 + 5 + 20 + 8
= 80
∴ \(n(\bar{M} \cap \bar{C} \cap \bar{T})\) = n(U) – n(M ∪ C ∪ T)
= 100 – 80 = 20

Sets Class 9 Case Based Questions

Question 1.
Aarti explained operations on sets to her younger sister Pooja then wrote three sets as A = {2, 3, 6,7}, B = {4, 5, 8} and C = {x : x is a prime number less than 9}. She asked her sister Pooja that the following question will judge how much you have understood. She asked her younger sister to solve and write down answer.
(i) A ∪ B
(ii) A ∩ B
(iii) (a) (A ∪ B) ∩ C Or (b)(A ∩ C) – B
Answer:
We have, A = {2, 3, 6, 7}, B = {4, 5, 8} and C = {2, 3, 5, 7}.
(i) A ∪ B = {2, 3, 6, 7} ∪ {4, 5, 8} = {2, 3, 4, 5, 6, 7, 8}

(ii) A ∩ B = {2, 3, 6, 7} ∩ {4, 5, 8} = Φ

(iii) (a) (A ∪ B) ∩ C = {2, 3, 4, 5, 6, 7, 8} ∩ {2, 3, 5, 7} = {2, 3, 5, 7}
Or
(b) (A ∩ C) – B
Now, A ∩ C = {2, 3, 6, 7} ∩ {2, 3, 5, 7} = {2, 3, 7}
(A ∩ C) – B = {2, 3, 7} – {4, 5, 8} = {2, 3, 7}

Question 2.
Consider the universal set U = {x : x ∈ N, x < 20),
where N is the set of natural numbers. Let the sets A, B and C be defined as follows.
A = {x : x ∈ N, x is an even prime number)
B = {x : x ∈ N, x is an odd number and 5 < x ≤ 11}
and C = {x² : x ∈ N, x < 4} then
(i) write the sets A, B and C in roster form.
(ii) is set A subset of set B? Justify your answer.
(iii) (a) Calculate (A ∪ B)’ C.
Or
(b) Calculate (A ∩ C) ∪ (B ∩ C).
Answer:
Given, U = {x : x ∈ N, x < 20},
A = {x : x ∈ N, x is an even prime number},
B = {x : x ∈ N, x is an odd number and 5 < x ≤ 11}
and C = {x^{2 : x ∈ N, x < 4}.}

(i) Set A in roster form, A = {2}
Set B in roster form, B = {7, 9, 11}
Set C is roster form, C = {1², 2², 2, 3³} = {1, 4, 9}

(ii) No, A is not a subset of B because the element 2, which belongs to set A, but does not belong to set B.

(iii) (a) From part (i),
A = {2}, B = {7, 9, 11} and C = {1, 4, 9}
Also, U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19}
∴ A ∪ B = {2, 7, 9, 11}
Now, (A ∪ B)’ = U – (A ∪ B)
= {1, 3, 4, 5, 6, 8, 10, 12, 13, 14, 15, 16, 17, 18, 19}
Hence, (A ∪ B)’ ∩ C = {1, 4}
Or
(b) A ∩ C = {2} ∩ {1, 4, 9} = Φ
and B ∩ C = {7, 9, 11} ∩ {1, 4, 9} = {9}
So, (A ∩ C) ∪ (B ∩ C) = Φ ∪ {9} = {9}

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