RowQ
The Vault
RowQ
The Vault
CBSE Class 11 Maths · 11 questions · 26 marks
Almost every idea you meet later this year — a domain, a solution set, a sample space — is quietly a set, so this chapter is the vocabulary lesson for the whole syllabus. You will learn to describe a collection in roster and set-builder form, count its subsets, and combine collections using union, intersection, difference and complement. Venn diagrams then turn wordy survey problems into arithmetic you can do in two lines.
A set A contains 6 elements. The number of subsets of A is:
Answer
The number of subsets is 64. For a set with n elements each element can either be included in a subset or left out, giving 2 independent choices per element. So the total number of subsets is 2ⁿ. Here n = 6, so the count is 2⁶ = 64. The answer 12 comes from wrongly using 2n, 36 from n², and 720 from 6! — none of which count subsets.
If n(A) = 17, n(B) = 23 and n(A ∪ B) = 31, then n(A ∩ B) equals:
Answer
n(A ∩ B) = 9. Use n(A ∪ B) = n(A) + n(B) - n(A ∩ B). Substituting, 31 = 17 + 23 - n(A ∩ B) = 40 - n(A ∩ B). So n(A ∩ B) = 40 - 31 = 9. As a check, the union splits as 8 (only A) + 14 (only B) + 9 (both) = 31.
Which of the following is the empty set?
Answer
{x : x ∈ R, x² + 9 = 0} is the empty set. Solving x² + 9 = 0 gives x² = -9, and the square of a real number can never be negative, so no real x qualifies and the set is ∅. The others are all non-empty: x² = 49 gives {-7, 7}; the natural numbers strictly between 3 and 8 are {4, 5, 6, 7}; and the integers from -2 to 2 are {-2, -1, 0, 1, 2}.
For any two subsets A and B of a universal set U, (A ∪ B)' is equal to:
Answer
(A ∪ B)' = A' ∩ B'. This is De Morgan's first law. An element lies outside A ∪ B exactly when it is outside A and also outside B, which is precisely the description of A' ∩ B'. The option A' ∪ B' is the complement of A ∩ B, not of A ∪ B, so it is the second De Morgan law misapplied here.
Assertion (A): The sets P = {3, 6, 9} and Q = {x : x is a multiple of 3, x ∈ N, x ≤ 9} are equal. Reason (R): Two sets are equal when every element of the first belongs to the second and every element of the second belongs to the first.
Answer
Both A and R are true and R is the correct explanation of A. Writing Q in roster form, the natural multiples of 3 that are at most 9 are 3, 6 and 9, so Q = {3, 6, 9}. Every element of P lies in Q and every element of Q lies in P, so P = Q and Assertion A is true. Reason R states exactly the definition of equality of sets, and it is the criterion actually used to conclude P = Q, so R correctly explains A.
Write the set {x : x ∈ Z, x² ≤ 10} in roster form.
Answer
We need all integers whose square does not exceed 10. Testing outward from 0: 0² = 0, (±1)² = 1, (±2)² = 4, (±3)² = 9 — all at most 10. But (±4)² = 16, which is greater than 10, so 4 and -4 and anything beyond are excluded. Hence the set in roster form is {-3, -2, -1, 0, 1, 2, 3}, which has 7 elements.
In a class of 60 students, 34 play chess and 28 play carrom, while 11 play both games. Find how many students play at least one of the two games, and how many play neither.
Answer
Let C be the set of chess players and K the set of carrom players. Given n(C) = 34, n(K) = 28 and n(C ∩ K) = 11. Students playing at least one game: n(C ∪ K) = n(C) + n(K) - n(C ∩ K) = 34 + 28 - 11 = 51. Students playing neither game = total - n(C ∪ K) = 60 - 51 = 9. So 51 students play at least one game and 9 play neither.
If U = {1, 2, 3, 4, 5, 6, 7, 8}, A = {2, 4, 6, 8} and B = {1, 2, 3, 4}, find A' and A - B.
Answer
A' = U - A consists of the elements of U that are not in A. Removing 2, 4, 6, 8 from U leaves A' = {1, 3, 5, 7}. A - B consists of the elements of A that do not belong to B. From A = {2, 4, 6, 8}, the elements 2 and 4 also lie in B, so they are removed. Hence A - B = {6, 8}.
In a colony of 120 households, 68 subscribe to newspaper A, 55 to newspaper B and 47 to newspaper C. Also, 28 subscribe to both A and B, 22 to both B and C, 25 to both A and C, and 12 subscribe to all three. Find (i) the number of households that subscribe to at least one newspaper, (ii) the number that subscribe to none, and (iii) the number that subscribe to exactly one newspaper.
Answer
Step 1 — List the data: n(A) = 68, n(B) = 55, n(C) = 47, n(A ∩ B) = 28, n(B ∩ C) = 22, n(A ∩ C) = 25, n(A ∩ B ∩ C) = 12, total = 120. Step 2 — (i) Apply the three-set formula: n(A ∪ B ∪ C) = 68 + 55 + 47 - 28 - 22 - 25 + 12. First 68 + 55 + 47 = 170, then 28 + 22 + 25 = 75, so 170 - 75 = 95, and 95 + 12 = 107. So 107 households subscribe to at least one newspaper. Step 3 — (ii) Households subscribing to none = 120 - 107 = 13. Step 4 — (iii) For exactly one newspaper, subtract the overlaps from each set: Only A = 68 - 28 - 25 + 12 = 27. Only B = 55 - 28 - 22 + 12 = 17. Only C = 47 - 22 - 25 + 12 = 12. Exactly one = 27 + 17 + 12 = 56. Step 5 — Check: exactly two = (28 - 12) + (22 - 12) + (25 - 12) = 16 + 10 + 13 = 39, and 56 + 39 + 12 = 107, which matches part (i).
(a) Prove that (A ∪ B)' = A' ∩ B' for any two subsets A and B of a universal set U. (b) Verify the result for U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, A = {2, 3, 5, 7} and B = {3, 6, 7, 9}.
Answer
(a) We show that each side is a subset of the other. Let x ∈ (A ∪ B)'. Then x ∈ U and x ∉ A ∪ B, so x is neither in A nor in B. Hence x ∈ A' and x ∈ B', which means x ∈ A' ∩ B'. Therefore (A ∪ B)' ⊆ A' ∩ B'. Conversely, let x ∈ A' ∩ B'. Then x ∉ A and x ∉ B, so x cannot belong to A ∪ B, giving x ∈ (A ∪ B)'. Therefore A' ∩ B' ⊆ (A ∪ B)'. Since each set contains the other, (A ∪ B)' = A' ∩ B'. (b) A ∪ B = {2, 3, 5, 6, 7, 9}, so (A ∪ B)' = U - (A ∪ B) = {1, 4, 8, 10}. A' = {1, 4, 6, 8, 9, 10} and B' = {1, 2, 4, 5, 8, 10}. Their common elements are A' ∩ B' = {1, 4, 8, 10}. Both sides equal {1, 4, 8, 10}, so the law is verified.
Read the following and answer the questions that follow: A campus cafe recorded the orders of 200 customers on a rainy Monday. Exactly 118 customers ordered tea, 96 ordered coffee, and 40 ordered both tea and coffee. Every remaining customer ordered only a snack. Let T be the set of tea drinkers and C the set of coffee drinkers. (a) Find the number of customers who ordered at least one hot drink. (b) How many customers ordered neither tea nor coffee? (c) How many ordered tea but not coffee? (d) How many ordered exactly one of the two drinks?
Answer
(a) n(T ∪ C) = n(T) + n(C) - n(T ∩ C) = 118 + 96 - 40 = 174. So 174 customers ordered at least one hot drink. (b) Customers ordering neither = 200 - n(T ∪ C) = 200 - 174 = 26. (c) Tea but not coffee means n(T - C) = n(T) - n(T ∩ C) = 118 - 40 = 78. (d) Coffee but not tea = 96 - 40 = 56. So exactly one drink = 78 + 56 = 134. Equivalently, n(T ∪ C) - n(T ∩ C) = 174 - 40 = 134, which confirms the answer.
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