NCERT Solutions for Class 11 Maths Chapter 1

Set builder form of {3, 6, 9, 12} = {x: x = 3n, n∈ N and 1 ≤ n ≤ 4}

A = {x: x is an odd natural number} = {1, 3, 5, 7, 9 …}

The subsets of {1, 2, 3} are Φ, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3} {1, 2, 3}

Here, A = {a, b} and B = {a, b, c} Yes, A ⊂ B. A ∪ B = {a, b, c} = B

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.

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.

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.

Let C denote the set of people who like coffee, and T denote the set of people who like tea n(C ∪ T) = 70, n(C) = 37, n(T) = 52 We know that: n(C ∪ T) = n(C) + n(T) – n(C ∩ T) ∴ 70 = 37 + 52 – n(C ∩ T) ⇒ 70 = 89 – n(C ∩ T) ⇒ n(C ∩ T) = 89 – 70 = 19 Thus, 19 people like both coffee and tea.

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.

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.

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 student 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.

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.