Collins Maths Solutions Class 6 Chapter 4 Sets
Welcome to NCTB Solution. Here with this post we are going to help 6th class students for the Solutions of Collins Maths Class 6 Mathematics, Chapter 4, Sets. Here students can easily find Exercise wise solution for chapter 4, Sets. Students will find proper solutions for Exercise 4.1, 4.2 and 4.3 Our teacher’s solved every problem with easily understandable methods so that every students can understand easily. Here in this post all the solutions are based on ICSE latest Syllabus.
Sets Exercise 4.1 Solution :
Question no – (1)
Solution :
= It is not a set,
= Because the term ‘beautiful’ can be interpreter differently by different people, so the elements are not well defined, Hence the given collection is not a set.
Question no – (2)
Solution :
= It is a set,
= Clearly the given collection will conation the name of all students in a school which can be known. Here the given collection is a set.
Question no – (3)
Solution :
= It is a set,
= Because , an the letters of my name can be easily known. Hence the given collection is a set.
Question no – (4)
Solution :
= It is a set,
= Because all whole number is known to us so, the given collection is a set.
Question no – (5)
Solution :
= It is a set,
= Because the number of roses in a basket is well defined number. Hence the given collection is a set
Question no – (6)
Solution :
= It is a set,
= It is a set because having there sides polygon is well-defined.
Sets Exercise 4.3 Solution :
Question no – (1)
Solution :
(a) Set of all integers between -20 and -5
= I {- 19, -18, -17, -6, -51, -14, -13, -12, -11, -10, -9, -8, -7, -6,}
(b) Set of all 2-digit numbers greater than 79.
= A = {80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99}
(c) Set of all the multiples of 8 that are less than 80
= X = (8, 18, 24, 32, 40, 48, 56, 64, 72}
(d) Set of the first 5 odd numbers that are divisible by 7
= B = {7, 21, 35, 49, 63}
(e) Set of all consonants in the word KARNATAKA
= A {K, R, N, T}
(f) Collection of colours in a rainbow
= C = {Red yellow, violet orange blue, sky-blue green}
Question no – (2)
Solution :
(a) D = {52, 54, 56, 58, 60, 62, 64, 66}
= D = {x : x is an even number 52 < x < 66}
(b) P = {101, 103, 107, 109, 113, 127, 131, 137, 139, 149}
= P {x : x is prime number 100 < × < 15 150}
(c) M = {September, October, November, December}
= M = {x : x is last four months of a year}
(d) F = {2, 3, 5, 7, 11}
= F = {x : set of first inner plants of the solar system}
(e) S = {Mercury, Venus, Earth, Mars}
= S = x : x is the first inner plants of the solar system}
(f) G = {1}
= G {x : x is is neither a prime number for a composite numbers}
Question no – (3)
Solution :
(a) A = {p, q, r, s, t}
= n(A) = 4
= n (B) = 7
(x | x – 3 = 2}
= {x 15 – 3 = 2}
= {5}
n(c) = 5
(d) D = {x| x is prime factor of 20}
= D = (x | x is prime factor of 20}
(e) E = {x | x is a 1-digit prime number}
= E {x | x is a 1 digit prime number?
= (2, 3, 5, 7}
= n (E) = 4
(f) F = {100, 101, 102, 103, 104}
= F = {100, 101, 102, 103, 104}
= n (F) = 5
Sets Exercise 4.2 Solution :
Question no – (1)
Solution :
(a) A = {11, 13, 15, 17, 19] and B = {10, 12, 14 , 16}
Number if element in the A is = 5
and, the number of elephant in the set B is = 4
Then A is not equivalent of 13
(b) C = {p, q, r, s} and D = {d, e, f , g}
= In set c number of element and in set D number of elephant is 4 Then C is not equivalent of D.
(c) E = {x | x is prime factor of 20} and F = {x | x is a multiple of 5, x < 40}
= E = {x | x is a prime factor 20}
= {2,5}
= F {x | x is multiple of 5, × L 40}
= {5, 10, 15, 20, 25, 30, 35}
Number of element in the set E is 2 and number of element of 7
Therefore E is not equivalent of F.
(d) G = {x | x is a letter in the word SOCIAL} and H = {x | is a letter in the word SCIENCE}
G = {x | x is a letter in the word SOCIAL}
= {S, O, C, I, A, L}
= H = {x | x is a letter in the word SCIENCE}
= {S,E,C,I,N,C,E}
Number of element is G is 6
and number of element is H is 5
Set, G element is H is 5
Hence, G and H are not equivalent.
(e) I = (2, 3, 5, 7} and J = {2, 4, 6, 8}
I = {2, 3, 5, 7}
J = {2, 4, 6, 8}
Number of element of I is = 4
Therefore, set I and set J are Equivalent.
Question no – (2)
Solution :
(a) A = {a, b, c, d} and B = (e, d, c, b, a}
= A {a, B, C, D, E}
= b = {E, D, C, B, A}
Here, both the sets have same elements
So, set A and set B are Equivalent.
(b) C = (x| x is a vowel in the word COUNT} and D = {x | x is a vowel in the word WRITE}
= C = {x | x is a value in the word COUNT}
= {o, u,}
b = {x | x is a value in the word WRITE}
= {I, E}
Here, all the element of a set C are not in the set D so the sets are not equal.
(c) C = (x| x is an odd number, 5 < x < 15} and F = (x| x is even number, 5 < x < 15}
= E = {x| x is an odd number5 < x < 15
= {7, 9, 11, 13}
= F {x | x is an even number ; 5 < x < 15}
= {6, 8, 10, 12, 14}
Here all the elements of set E are not in the set F
Hence, the sets are not equal.
(d) G = {1, 2, 3, 4, 5} and H = {0, 1, 2, 3, 4, 5}
= G {1, 2, 3, 4, 5}
= H = {0, 1. 2, 3, 4, 5}
Here, the elements of set FG and the elements of H are not same.
Thus, the sets are not equal.
(e) The set of the first 5 positive integers and the set of the first 5 negative integers
= A = {1, 2, 3, 4, 5,}
= B {-1, -2, -3, -4, -5,}
Here, the elements of set A and set B are not same.
Question no – (3)
Solution :
(a) The set of all negative natural numbers
= Empty set
(b) The set of all whole numbers less than 1
= A {0}
= Singleton sets.
(c) A = {n}
= A = {n}
= Singleton sets
(d) B = {x| x is month having 27 days
= B = {x | x is a month having 27 days}
= B is a Empty set Because there are no months
(e) C = {x| x is a natural number, 10 < x < 11}
= C = {x | x is a natural number 10, < x < 11}
= C is a Empty set
Question no – (4)
Solution :
(a) The set of all the factors of 100
= Finite set
(b) The set of all the multiples of 4
= Infinite set
(c) The set of all the integers between -1000 and 1000
= Finite set
(d) The set of all the points on a given line
= Infinite set
(e) The set of the states of India
= Finite set
Next Chapter Solution :
👉 Chapter 5 👈
