# ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 5 Sets Check Your Progress

## ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 5 Sets Check Your Progress

Question 1.
Write the following sets in tabular form and also in set builder form:
(i) The set of even integers which lie between -6 and 10.
(ii) The set of two digit numbers which are perfect square.
(iii) {factors of 42}
Solution:
(i) Given set = {-4, -2, 0, 2, 4, 6, 8)
(tabular form)
or {x : x = 2n, n ∈ I and -3 < n < 5}
(set builder form)
(ii) The set can be written as { 16, 25, 36, 49, 64, 81 } (tabular form)
or {x : x = n2, n ∈ N and 4 ≤ n ≤ 9)
(set builder form)
(iii) The set can be written as { 1, 2, 3, 6, 7, 14, 21, 42) (tabular form)
or {x : x ¡s a factor of 42)
(set builder form)

Question 2.
Write the following sets in roster form:
(i) {x : x = 5n, n ∈ I and -3 < n ≤ 13}
(ii) {x : x = n2, n ∈ W and n < 5}
(iii) {x : x = n2 – 2, n ∈ W and n < 4}
Solution:
The set can be written as
(i) Integers lie between -2 and 3 are -2, -1, 0, 1, 2, 3.
Given x = 5n, putting n = -2, -1,0, 1, 2, 3, we get
x = 5 × -2, 5 × -1, 5 × 0, 5 × 1, 5 × 2, 5 × 3,
= -10, -5, 0, 5, 10, 15
Set = {-10, -5, 0, 5, 10, 15)
(tabular form)
(ii) Whole numbers less than 5 are 0, 1, 2, 3, 4.
Given x = n2, putting n 0, 1, 2, 3, 4, we get
x = 02, 12, 22, 32, 42 = 0, 1, 4, 9, 16
Given set = {0, 1, 4, 9, 16) (roster form)
Whole numbers less than 4 are 0, 1, 2, 3
Given x = n2 – 2, putting n = 0, 1, 2, 3, We get
x = 02 – 2, 12 – 2, 22 – 2, 32 – 2
= -2, -1, 2, 7
Given set = {-2, -1, 2, 7}(roster form)

Question 3.
Write the following sets in set builder form:
(i) {-14, -7, 0, 7, 14, 21, 28}
(ii) {1, 2, 3, 6, 9, 18}
Solution:
(i) {x | x = 7n, n ∈ I and -2 ≤ n ≤ 4} (set builder form)
(ii) Given set = {x | x ∈ N, x is a factor of 18} (set builder form)

Question 4.
Classify the following sets into the finite set, infinite set the empty set. In the case of a (non-empty) finite set, mention the cardinal number.
(i) The set of even prime numbers.
(ii) {multiples of 9}
(iii) {x : x is a prime factor of 84}
(iv) {x : 2x + 5 = 1, x ∈ N}
(v) {x : x is a month of a year having less than 30 days}
(vi) {x | x is a month of a leap year having 28 days}
Solution:
(i) It is a finite set having 1 element. So, cardinal number = 1
(ii) It is an infinite set as it has the unlimited number of different elements.
Because, if we write it in roster form, the given set = {9, 18, 27, 36, ……….}
(iii) Prime factors of 84 = 2, 3, 7.
The set can be written as = {2, 3, 7}
It is a finite set having 3 elements.
(iv) 2x + 5 = 1
⇒ 2x = 1 – 5
⇒ 2x = -4
⇒ x = -2
But x ∈ N and Natural numbers are {1, 2, 3, …….}
It is an empty set.
(v) {x : x is a month of a year having less than 30 days}
= February
It is a finite set as it is one element.
(vi) {x | x is a month of a leap year having 28 days}
= Φ
It is an empty set as there is no month in the leap year which has 28 days.

Question 5.
In the following, determine whether A and B are equivalent sets and if so, whether A = B.
(i) A = {1, 3, 5}, B = {Red, Blue, Green}
(ii) A = {prime factors of 70}, B = {prime factors of 60}
(iii) A = {even natural numbers less than 10}, B = {odd natural numbers less than 10}
Solution:
(i) A ↔ B as n (A) = 3 = n (B)
But A ≠ B because, they have different elements.
(ii) Prime factors of 70 = 2, 5, 7
A = {2, 5, 7}
Prime factors of 60 = 2, 3, 5
B = {2, 3, 5}
A ↔ B as n (A) = n (B)
But A ≠ B
they have not the same elements.
(iii) If we write A and B in tabular form, we get
A = {2, 4, 6, 8}
B = {1, 3, 5, 7, 9}
So, n (A) ≠ n (B)
A is not equivalent to B.

Question 6.
Let P = {letters of SCHOOL} and Q = {letters of FALSE}, then State whether each of the following statement is true or false for the above sets:
(i) P ⊂ Q
(ii) Q ⊂ P
(iii) P ↔ Q
Solution:
If P = {letters of SCHOOL}
Q = {letters of FALSE}
P = {S, C, H, O, L} and Q = {F, A, L, S, E}
(i) P ⊂ Q – False.
(ii) Q ⊂ P – False.
(iii) P ↔ Q – True.
{Both have equal number of elements}

Question 7.
State whether each of the following statement is true or false for the sets A, B and C where
A = {x | x ∈ N, x < 40 and x is a multiple of 6}
B = {x | x ∈ W, x ≤ 40 and x is a multiple of 8}
C = {x | x is a factor of 28}.
(i) A ↔ B
(ii) B ↔ C
(iii) A ↔ C
Solution:
If we write A, B and C in tabular form, we get 32,
A = {6, 12, 18, 24, 30, 36},
B = {0, 8, 16, 24, 40}
and C = {1, 2, 4, 7, 14, 28}
(i) A ↔ B True, because n (A) = 6 = n (B)
(ii) B ↔ C True, because n (B) = 6 = n (C)
(iii) A ↔ C True, because n (A) = 6 = n (C)