This NCERT solutions for class 11 Maths Chapter 1 is prepared by our expert faculty at eSaral according to the latest update on the term-wise CBSE Syllabus for 2021-22.
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The questions and answers are as follows:
Question 1.
Which of the following are sets. Explain your answer.
(a). The collection of all days in a week which have the first letter S.
(b). The collection of ten most famous singers of India.
(c). A group of the best football strikers in the world.
(d). The collection of all girls in your school.
(e). The collection of all odd numbers below 50.
(f). A collection of poems written by the poet Shakespeare.
(g). The collection of all prime numbers.
(h). The collection of questions in a science book.
(i). A collection of the most dangerous reptiles in India.
Solution:
(i) The collection of all months of a year beginning with the letter J is a well-defined collection of objects because one can identify a month that belongs to this collection. Hence, this collection is a set.
(ii) The collection of the ten most talented writers of India is not a well-defined collection because the criteria for determining a writer’s talent vary from person to person. Hence, this collection is not a set.
(iii) A team of eleven best cricket batsmen of the world is not a well-defined collection because the criteria for determining a batsman’s talent may vary from person to person. Hence, this collection is not a set.
(iv) The collection of all boys in your class is a well-defined collection because you can identify a boy who belongs to this collection. Hence, this collection is a set.
(v) The collection of all-natural numbers less than 100 is a well-defined collection because one can identify a number that belongs to this collection. Hence, this collection is a set.
(vi) A collection of novels written by the writer Munshi Prem Chand is a well-defined collection because one can identify a book that belongs to this collection. Hence, this collection is a set.
(vii) The collection of all even integers is a well-defined collection because one can identify an even integer that belongs to this collection.
Hence, this collection is a set.
(viii) The collection of questions in this chapter is a well-defined collection because one can identify a question that belongs to this chapter. Hence, this collection is a set.
(ix) The collection of most dangerous animals in the world is not a well-defined collection because the criteria for determining the dangerousness of an animal can vary from person to person. Hence, this collection is not a set.
Question 2.
Let P = {2, 3, 4, 5, 6, 7}. Insert the correct symbol inside the given blank spaces below:
(a). 2 . . . . . . . . . . P
(b). 9 . . . . . . . . . P
(c). 11 . . . . . . . . P
(d). 4 . . . . . . . . P
(e). 0 . . . . . . . . P
(f). 7 . . . . . . . . P
Solution:
(i) 5 ∈ A
(ii) 8 ∉ A
(iii) 0 ∉ A
(iv) 4 ∈ A
(v) 2 ∈ A
(vi) 10 ∉ A
Question 3.
Write the given sets in roster form:
(a). P = {y: y is an integer and -4 < y < 6}.
(b). Q = {y: y is a natural number which is <8}
(c). R = {y: y is a 2 digit natural number in which the sum of its digits is 9}
(d). S = {y: y is a prime number which is a divisor of 70}
(e). T = The set of all letters in the word ELEPHANT
(f). U = The set of all letters in the word DIVISION
Solution:
(i) A = {x: x is an integer and –3 < x < 7}
The elements of this set are –2, –1, 0, 1, 2, 3, 4, 5, and 6 only.
Therefore, the given set can be written in roster form as
A = {–2, –1, 0, 1, 2, 3, 4, 5, 6}
(ii) B = {x: x is a natural number less than 6}
The elements of this set are 1, 2, 3, 4, and 5 only.
Therefore, the given set can be written in roster form as
B = {1, 2, 3, 4, 5}
(iii) C = {x: x is a two-digit natural number such that the sum of its digits is 8}
The elements of this set are 17, 26, 35, 44, 53, 62, 71, and 80 only.
Therefore, this set can be written in roster form as
C = {17, 26, 35, 44, 53, 62, 71, 80}
(iv) D = {x: x is a prime number which is a divisor of 60}
📷
60 = 2 × 2 × 3 × 5
The elements of this set are 2, 3, and 5 only.
Therefore, this set can be written in roster form as D = {2, 3, 5}.
Questions.4.
Write the given sets in set-builder form:
(a). {4, 8, 12, 16, 20}
(b). {3, 9, 27, 81}
(c). {4, 16, 64, 256, 1024}
(d). {1, 3, 5, 7…}
(e). {1, 8, 27, 64….1000}
Solution:
📷
Question 5.
In a survey of 450 people, it was found that 110 play cricket, 160 play tennis, and 70 play both cricket as well as tennis. How many play neither cricket nor tennis?
Solution:
23
Question 6.
In a town of 10,000 families, it was found that 40% of families buy newspaper A, 20% families buy newspaper B and 10% of families buy newspaper C. 5% of families buy A and B, 3%, buy B and C and 4% buy A and C.
If 2% of families buy all the three newspapers, find the no of families which buy(1) A only (2) B only (3) none of A, B and C (4) exactly two newspapers (5) exactly one newspaper (6) A and C but not B (7) at least one of A, B, C.
Solution:
{2, 3, 5}]
Question 7.
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 value of m and n.
Solution:
[m = 6 and n = 3]
Question 8.
Write the following sets in the set-builder form: (3, 6, 9, 12).
Solution:
Set builder form of {3, 6, 9, 12} = {x: x = 3n, n∈ N and 1 ≤ n ≤ 4}
Question 9.
List all the elements of the following set: A = {x: x is an odd natural number}.
Solution:
A = {x: x is an odd natural number} = {1, 3, 5, 7, 9 …}
Question 2.
Question 10.
Write down all the subsets of the following set: {1, 2, 3}.
Solution:
The subsets of {1, 2, 3} are Φ, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3} {1, 2, 3}
Question 11.
Let A = {a, b}, B = {a, b, c}. Is A ⊂ B? What is A ∪ B?
Solution:
Here, A = {a, b} and B = {a, b, c} Yes, A ⊂ B. A ∪ B = {a, b, c} = B
Question 12.
In a group of 400 people, 250 can speak Hindi, and 200 can speak English. How many people can speak both Hindi and English?
Solution:
Let H be the set of people who speak Hindi, and E be the set of people who speak English ∴ n(H ∪ E) = 400, n(H) = 250, n(E) = 200 n(H ∩ E) =? We know that: n(H ∪ E) = n(H) + n(E) – n(H ∩ E) ∴ 400 = 250 + 200 – n(H ∩ E) ⇒ 400 = 450 – n(H ∩ E) ⇒ n(H ∩ E) = 450 – 400 ∴ n(H ∩ E) = 50 Thus, 50 people can speak both Hindi and English.
Question 13.
If S and T are two sets such that S has 21 elements, T has 32 elements and S ∩ T has 11 elements, how many elements does S ∪ T have?
Solution:
It is given that: n(S) = 21, n(T) = 32, n(S ∩ T) = 11 We know that: n (S ∪ T) = n (S) + n (T) – n (S ∩ T) ∴ n (S ∪ T) = 21 + 32 – 11 = 42 Thus, the set (S ∪ T) has 42 elements.
Question 14.
● If X and Y are two sets such that X has 40 elements, X ∪Y has 60 elements and X ∩Y has 10 elements, how many elements does Y have?
Solution:
It is given that: n(X) = 40, n(X ∪ Y) = 60, n(X ∩ Y) = 10 We know that: n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y) ∴ 60 = 40 + n(Y) – 10 ∴ n(Y) = 60 – (40 – 10) = 30 Thus, the set Y has 30 elements.
Question 15.
In a group of 70 people, 37 like coffee, 52 like tea, and each person likes at least one of the two drinks. How many people like both coffee and tea?
Solution:
It is given that: n(X) = 40, n(X ∪ Y) = 60, n(X ∩ Y) = 10 We know that: n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y) ∴ 60 = 40 + n(Y) – 10 ∴ n(Y) = 60 – (40 – 10) = 30 Thus, the set Y has 30 elements.
Question 16.
In a group of 65 people, 40 like cricket, 10 like both cricket and tennis. How many like tennis only and not cricket? How many like tennis?
Solution:
Let C denote the set of people who like cricket, and T denote the set of people who like tennis ∴ n(C ∪ T) = 65, n(C) = 40, n(C ∩ T) = 10 We know that: n(C ∪ T) = n(C) + n(T) – n(C ∩ T) ∴ 65 = 40 + n(T) – 10 ⇒ 65 = 30 + n(T) ⇒ n(T) = 65 – 30 = 35 Therefore, 35 people like tennis. Now, (T – C) ∪ (T ∩ C) = T Also, (T – C) ∩ (T ∩ C) = Φ ∴ n (T) = n (T – C) + n (T ∩ C) ⇒ 35 = n (T – C) + 10 ⇒ n (T – C) = 35 – 10 = 25 Thus, 25 people like only tennis.
Question 17.
In a committee, 50 people speak French, 20 speak Spanish, and 10 speak both Spanish and French. How many speak at least one of these two languages?
Solution:
Let F be the set of people in the committee who speak French and S be the set of people in the committee who speak Spanish ∴ n(F) = 50, n(S) = 20, n(S ∩ F) = 10 We know that: n(S ∪ F) = n(S) + n(F) – n(S ∩ F) = 20 + 50 – 10 = 70 – 10 = 60 Thus, 60 people in the committee speak at least one of the two languages.
Question 18.
In a survey of 600 students in a school, 150 students were found to be taking tea and 225 taking coffee, 100 were taking both tea and coffee. Find how many students were taking neither tea nor coffee?
Solution:
Let U be the set of all students who took part in the survey. Let T be the set of students taking tea. Let C be the set of students taking coffee. Accordingly, n(U) = 600, n(T) = 150, n(C) = 225, n(T ∩ C) = 100 To find: Number of students taking neither tea nor coffee i.e., we have to find n(T’ ∩ C’). n(T’ ∩ C’) = n(T ∪ C)’ = n(U) – n(T ∪ C) = n(U) – [n(T) + n(C) – n(T ∩ C)] = 600 – [150 + 225 – 100] = 600 – 275 = 325 Hence, 325 students were taking neither tea nor coffee.
Question 19.
In a group of students 100 students know Hindi, 50 know English, and 25 know both. Each of the students knows either Hindi or English. How many students are there in the group?
Solution:
Let U be the set of all students in the group. Let E be the set of all students who know English. Let H be the set of all students who know Hindi. ∴ H ∪ E = U Accordingly, n(H) = 100 and n(E) = 50 n( H U E ) = n(H) + n(E) – n(H ∩ E) = 100 + 50 – 25 = 125 Hence, there are 125 students in the group.
Question 20.
List all the elements from the given sets:
(a). P = {y: y is even natural number}
(b). Q = {y: y is an integer
Solution:
(a) P = {y: y is an even natural number} = {2, 4, 6, 8, 10, …….}
(b) Q = {y: y is an integer 0, 1, 2, 3, 4, 5 ……..}
CONCLUSION
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