CBSE Class 11 Mathematics Sets Assignment Set 02

Sets MCQ Questions with Answers Class 11 Mathematics

Question: In a beauty contest, half the number of experts voted for Mr. A and two thirds voted for Mr. B. 10 voted for both and 6 did not vote for either. How many experts were there in all?

More Question

1. List all the element of the set A = { x : x is an integer x2 ≤ 4} [1]

2. From the sets given below pair the equivalent sets.
A = { 1, 2, 3}, B = {x, y, z, t}, C = {a, b, c} D = {0, a} [1]

3. Write the following as interval
(i) {x : x ϵ R, - 4 < x ≤ 6}
(ii) {x : x ϵ R, 3 ≤ x ≤ 4} [1]

4. If A = {3, 5, 7, 9, 11}, B = {7, 9, 11, 13}, C = {11, 13, 15}
Find (A ∩ B) ∩ ( B ∪ C) [1]

5. Write the set {1/3, 3/5, 5/7. 7/11, 11/13} in set builder form. [1]

6. 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? [4]

7. Let A, B and C be three sets A ∪ B = A ∪ C and A ∩ B = A ∩ C show that B = C [4]

8. If ∪ = {a, e, i. o. u}
A = {a, e, i} And
B = {e, o, u}
C = {a, i, u}
Then verity that A ∩ (B – C) = (A ∩ B) – (A ∩ C) [4]

9. In a town of 10,000 families, it was found that 40% families buy newspaper A, 20% families buy newspaper B and 10% families buy newspaper C. 5% families buy A and B, 3% buy B and C and 4% buy A and C. If 2% families buy all the three papers. Find the no. of families which buy
(i) A only (ii) B only (iii) none of A, B, and C. [6]

10. Two finite sets have m and n elements. The total no. of subsets of the first set is 56 more than the total no. of subsets of second set. Find the value of m and n.

VERY SHORT ANSWER TYPE QUESTIONS (1 MARK)

Which of the following are sets? Justify your answer.

Question 1. The collection of all the months of a year beginning with letter M
Answer: Set.
In simple words: It is a set because we can clearly identify which months begin with the letter M (March, May, June are not, but March and May are).

Exam Tip: A collection is a set only if it is well-defined, meaning any given object can be definitely determined to belong or not belong to it.

Question 2. The collection of difficult topics in Mathematics.
Answer: Not a set.
In simple words: It is not a set because the term 'difficult' is subjective and varies from person to person.

Exam Tip: Subjective terms like 'best', 'dangerous', or 'difficult' make a collection non-well-defined, hence not a set.

Let A = {1,3,5,7,9}. Insert the appropriate symbol \(\in\) or \(\notin\) in blank spaces :- (Question- 3,4)

Question 3. 2 \(\notin\) A
Answer: \(\notin\)
In simple words: 2 is not present in set A.

Exam Tip: Use \(\in\) if an element belongs to the set, and \(\notin\) if it does not.

Question 4. 5 \(\in\) A
Answer: \(\in\)
In simple words: 5 is present in set A.

Exam Tip: Check each element against the given set carefully.

Question 5. Write the set A = { x : x is an integer, -1 \(\le\) x < 4} in roster form
Answer: A = {-1, 0, 1, 2, 3}
In simple words: We list all integers starting from -1 up to, but not including, 4.

Exam Tip: Pay close attention to strict inequalities (<) versus non-strict inequalities (\(\le\)).

Question 6. List all the elements of the set, \(A = \left\{ x : x \in \mathbb{Z}, -\frac{1}{2} < x < \frac{11}{2} \right\}\)
Answer: A = {0, 1, 2, 3, 4, 5}
In simple words: We find all integers between -0.5 and 5.5.

Exam Tip: Convert fractions to decimals to easily identify the valid integer range.

Question 7. Write the set B = {3,9,27,81} in set-builder form.
Answer: B = { x : x = \(3^n\), \(n \in \mathbb{N}\) and 1 \(\le\) n \(\le\) 4 }
In simple words: The elements are powers of 3 from \(3^1\) to \(3^4\).

Exam Tip: Identify the common pattern or geometric progression of the elements.

Which of the following are empty sets? Justify. (Question- 8,9)

Question 8. A = { x : x \(\in\) N and 3 < x < 4}
Answer: Empty set.
In simple words: There is no natural number strictly between 3 and 4.

Exam Tip: An empty set contains no elements and is denoted by \(\phi\).

Question 9. B = { x : x \(\in\) N and \(x^2\) = x}
Answer: Non-empty set.
In simple words: Solving \(x^2 = x\) gives \(x(x - 1) = 0\), so \(x = 0\) or \(x = 1\). Since \(x \in \mathbb{N}\), element 1 belongs to B, making it non-empty (B = {1}).

Exam Tip: Always check the universal condition (like \(x \in \mathbb{N}\)) when solving equations for sets.

Which of the following sets are finite or Infinite? Justify. (Question-10,11)

Question 10. The set of all the points on the circumference of a circle.
Answer: Infinite set.
In simple words: A circle consists of an infinite number of points that cannot be counted.

Exam Tip: Continuous geometric lines and curves contain infinitely many points.

Question 11. B = { x : x \(\in\) N and x is an even prime number}
NAnswer: Finite set.
In simple words: The only even prime number is 2, so B = {2}, which has a countable number of elements.

Question 12. Are sets A = {-2,2}, B = { x : x \(\in\) \(\mathbb{Z}\), \(x^2\) - 4 = 0} equal? Why?
Answer: Yes, they are equal.
In simple words: Solving \(x^2 - 4 = 0\) gives \(x = \pm 2\). Since \(x \in \mathbb{Z}\), B = {-2, 2}, which is identical to set A.

Exam Tip: Two sets are equal if they have exactly the same elements.

Question 13. Write (-5,9] in set-builder form
Answer: { x : x \(\in\) \(\mathbb{R}\), -5 < x \(\le\) 9 }
In simple words: Round bracket means strict inequality (<) and square bracket means inclusive inequality (\(\le\)).

Exam Tip: Match parenthesis to inequalities: '(' or ')' becomes < or >, while '[' or ']' becomes \(\le\) or \(\ge\).

Question 14. Write { x : -3 \(\le\) x < 7} as interval.
Answer: [-3, 7)
In simple words: The inclusive limit -3 uses a square bracket, and the exclusive limit 7 uses a round parenthesis.

Exam Tip: Intervals are written with lower limit first, followed by upper limit.

Question 15. If A = { 1,3,5}, how many elements has P(A)?
Answer: 8 elements.
In simple words: Since \(n(A) = 3\), the number of elements in the power set is \(2^3 = 8\).

Exam Tip: Formula for the number of elements in a power set is \(n[P(A)] = 2^{n(A)}\).

Question 16. Write all the possible subsets of A = {5,6}.
Answer: \(\phi\), {5}, {6}, {5,6}
In simple words: The subsets include the empty set, individual element sets, and the set itself.

Exam Tip: Total subsets for a 2-element set is \(2^2 = 4\).

If A = {2,3,4,5}, B = { 3,5,6,7} find (Question- 17,18)

Question 17. A \(\cup\) B
Answer: {2, 3, 4, 5, 6, 7}
In simple words: Combine all unique elements from both sets.

Exam Tip: Union contains all elements belonging to either A or B or both without repetition.

Question 18. A \(\cap\) B
Answer: {3, 5}
In simple words: Find the common elements present in both sets A and B.

Exam Tip: Intersection contains only the elements common to both sets.

Question 19. If A = {1,2,3,6} , B = {1, 2, 4, 8} find B - A
Answer: {4, 8}
In simple words: Take elements of B and remove any that also appear in A.

Exam Tip: B - A consists of elements that are in B but not in A.

Question 20. If A = {p, q}, B = {p, q, r}, is B a superset of A? Why?
Answer: Yes, because every element of A is also present in B (A \(\subset\) B).
In simple words: Since all elements of A are inside B, A is a subset of B, making B a superset of A.

Exam Tip: If \(A \subset B\), then B is called a superset of A (\(B \supset A\)).

Question 21. Are sets A = {1,2,3,4}, B = { x : x \(\in\) \(\mathbb{N}\) and 5 \(\le\) x \(\le\) 7} disjoint? Why?
Answer: Yes, because \(A \cap B = \phi\).
In simple words: Set B = {5, 6, 7}. There are no common elements between A and B.

Exam Tip: Two sets are disjoint if their intersection is the empty set.

Question 22. If X and Y are two sets such that n(X) = 19, n(Y) = 37 and n(X \(\cap\) Y) = 12, find n(X \(\cup\) Y).
Answer: n(X \(\cup\) Y) = 44
Using the formula:
\(n(X \cup Y) = n(X) + n(Y) - n(X \cap Y)\)
\(n(X \cup Y) = 19 + 37 - 12\)
\(n(X \cup Y) = 56 - 12 = 44\)
In simple words: Add the sizes of both sets and subtract the overlap.

Exam Tip: Memorize the fundamental inclusion-exclusion principle for two sets.

SHORT ANSWER TYPE QUESTIONS (4 MARKS)

Question 23. If U = {1,2,3,4,5,6,7,8,9}, A = {2,3,5,7,9}, B = { 1,2,4,6}, verify
(i) \((A \cup B)' = A' \cap B'\)
(ii) \(B - A = B \cap A' = B - (A \cap B)\)
Answer:
(i) Verification of De Morgan's Law \((A \cup B)' = A' \cap B'\):
\(A \cup B = \{1, 2, 3, 4, 5, 6, 7, 9\}\)
\((A \cup B)' = U - (A \cup B) = \{8\}\)
\(A' = U - A = \{1, 4, 6, 8\}\)
\(B' = U - B = \{3, 5, 7, 8, 9\}\)
\(A' \cap B' = \{8\}\)
Hence, \((A \cup B)' = A' \cap B'\) is verified.

(ii) Verification of set difference properties:
\(B - A = \{1, 4, 6\}\)
\(B \cap A' = \{1, 2, 4, 6\} \cap \{1, 4, 6, 8\} = \{1, 4, 6\}\)
\(A \cap B = \{2\}\)
\(B - (A \cap B) = \{1, 2, 4, 6\} - \{2\} = \{1, 4, 6\}\)
Hence, \(B - A = B \cap A' = B - (A \cap B)\) is verified.
In simple words: We compute each set operation independently and show that the resulting sets are identical.

Exam Tip: Write out each intermediate set clearly to avoid calculation errors.

Question 24. Let A, B be any two sets. Using properties of sets prove that,
(i) \((A - B) \cup B = A \cup B\)
(ii) \((A \cup B) - A = B - A\)
[ Hint : A - B = \(A \cap B'\) and use distributive law.]
Answer:
(i) Prove \((A - B) \cup B = A \cup B\):
LHS = \((A \cap B') \cup B\)
\(= (A \cup B) \cap (B' \cup B)\) (using distributive law)
\(= (A \cup B) \cap U\) (since \(B' \cup B = U\))
\(= A \cup B = RHS\)

(ii) Prove \((A \cup B) - A = B - A\):
LHS = \((A \cup B) \cap A'\)
\(= (A \cap A') \cup (B \cap A')\) (using distributive law)
\(= \phi \cup (B \cap A')\) (since \(A \cap A' = \phi\))
\(= B \cap A' = B - A = RHS\)
In simple words: Apply set identities like complement laws and distributive laws step-by-step.

Exam Tip: Mention the specific set law used at each step for full marks.

Question 25. In a group of 800 people, 500 can speak Hindi and 320 can speak English. Find
(i) How many can speak both Hindi and English?
(ii) How many can speak Hindi only?
Answer:
Let H be the set of people who speak Hindi, and E be the set of people who speak English.
Total people \(n(H \cup E) = 800\)
Hindi speakers \(n(H) = 500\)
English speakers \(n(E) = 320\)

(i) Both Hindi and English \(n(H \cap E)\):
\(n(H \cup E) = n(H) + n(E) - n(H \cap E)\)
\(800 = 500 + 320 - n(H \cap E)\)
\(800 = 820 - n(H \cap E)\)
\(n(H \cap E) = 820 - 800 = 20\)

(ii) Hindi only \(n(H - E)\):
\(n(H - E) = n(H) - n(H \cap E) = 500 - 20 = 480\)
In simple words: We use the intersection formula to find dual speakers, then subtract that from total Hindi speakers to find those who speak Hindi exclusively.

Exam Tip: Define your variables clearly at the start of word problems.

Question 26. A survey shows that 84% of the Indians like grapes, whereas 45% like pineapple. What percentage of Indians like both grapes and pineapple?
Answer:
Let G be the set of people who like grapes, and P be the set of people who like pineapple.
Assuming total percentage is 100%, so \(n(G \cup P) \le 100\%\).
\(n(G) = 84\%\), \(n(P) = 45\%\)
Using formula: \(n(G \cup P) = n(G) + n(P) - n(G \cap P)\)
Assuming everyone likes at least one fruit, \(n(G \cup P) = 100\%\):
\(100 = 84 + 45 - n(G \cap P)\)
\(100 = 129 - n(G \cap P)\)
\(n(G \cap P) = 129 - 100 = 29\%\)
In simple words: The sum of individual percentages exceeds 100%, and the excess represents the overlap of people who like both.

Question 27. 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?
Answer:
Let U be the set of people surveyed, C be cricket players, and T be tennis players.
\(n(U) = 450\)
\(n(C) = 110\)
\(n(T) = 160\)
\(n(C \cap T) = 70\)
First, find number of people who play at least one sport \(n(C \cup T)\):
\(n(C \cup T) = n(C) + n(T) - n(C \cap T)\)
\(n(C \cup T) = 110 + 160 - 70 = 200\)
Number of people who play neither = \(n[(C \cup T)'] = n(U) - n(C \cup T)\)
\(450 - 200 = 250\)
In simple words: Subtract those who play at least one sport from the total survey population to find those who play neither.

Exam Tip: 'Neither A nor B' is given by the complement formula \((A \cup B)'\).

Question 28. In a group of students, 225 students know French, 100 know Spanish and 45 know both. Each student knows either French or Spanish. How many students are there in the group?
Answer:
Let F be French speakers, S be Spanish speakers.
\(n(F) = 225\)
\(n(S) = 100\)
\(n(F \cap S) = 45\)
Since every student knows either French or Spanish, total students = \(n(F \cup S)\):
\(n(F \cup S) = n(F) + n(S) - n(F \cap S)\)
\(n(F \cup S) = 225 + 100 - 45 = 325 - 45 = 280\)
In simple words: Add French and Spanish speakers, then subtract the double-counted students who know both.

Question 29. If A = [-3, 5), B = (0, 6] then find (i) A - B, (ii) A \(\cup\) B
Answer:
(i) \(A - B = [-3, 0]\)
(ii) \(A \cup B = [-3, 6]\)
In simple words: Interval operations are best visualized on a real number line: A - B removes the overlapping positive tail from -3 to 5, leaving [-3, 0], and union spans the entire combined range from -3 to 6.

Exam Tip: Be very careful with interval endpoints (open vs closed brackets) during subtraction.

LONG ANSWER TYPE QUESTIONS (6 MARKS)

Question 30. In a survey it is found that 21 people like product A, 26 people like product B and 29 like product C. If 14 people like product A and B, 15 people like product B and C, 12 people like product C and A, and 8 people like all the three products. Find
(i) How many people are surveyed in all?
(ii) How many like product C only?
Answer:
Let A, B, C be the sets of people liking products A, B, and C respectively.
Given:
\(n(A) = 21\)
\(n(B) = 26\)
\(n(C) = 29\)
\(n(A \cap B) = 14\)
\(n(B \cap C) = 15\)
\(n(C \cap A) = 12\)
\(n(A \cap B \cap C) = 8\)

(i) Total people surveyed \(n(A \cup B \cup C)\):
Using inclusion-exclusion principle:
\(n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(C \cap A) + n(A \cap B \cap C)\)
\(n(A \cup B \cup C) = 21 + 26 + 29 - 14 - 15 - 12 + 8\)
\(n(A \cup B \cup C) = 76 - 41 + 8 = 43\)

(ii) People who like product C only (g region in Venn diagram):
\(n(\text{C only}) = n(C) - n(A \cap C) - n(B \cap C) + n(A \cap B \cap C)\)
\(n(\text{C only}) = 29 - 12 - 15 + 8 = 37 - 27 = 10\)
In simple words: Apply the 3-set union formula for total count, and subtract overlapping product preferences to find single-product enthusiasts.

Exam Tip: Drawing a Venn diagram and labeling regions with variables (a, b, c, d, e, f, g) makes 3-set problems much easier to solve.

Question 31. A college awarded 38 medals in football, 15 in basket ball and 20 in cricket. If these medals went to a total of 50 men and only five men got medals in all the three sports, how many received medals in exactly two of the three sports?
Answer:
Let F, B, C denote the set of men who received medals in Football, Basketball, and Cricket respectively.
Given:
\(n(F) = 38\)
\(n(B) = 15\)
\(n(C) = 20\)
Total men \(n(F \cup B \cup C) = 50\)
Medals in all three \(n(F \cap B \cap C) = 5 = f\)

Using Venn diagram region variables (let b, d, e be the regions representing exactly two sports):
We know total medals awarded = \(n(F) + n(B) + n(C) = 38 + 15 + 20 = 73\)
Also, total medals = \(n(F \cup B \cup C) + [\text{exactly two}] + 2 \times [\text{all three}]\)
More simply, using standard set relations:
\(n(F \cup B \cup C) = n(F) + n(B) + n(C) - [n(F \cap B) + n(B \cap C) + n(C \cap F)] + n(F \cap B \cap C)\)
Let \(S_1 = n(F) + n(B) + n(C) = 73\)
Let \(S_2 = n(F \cap B) + n(B \cap C) + n(C \cap F)\)
\(50 = 73 - S_2 + 5\)
\(50 = 78 - S_2 \implies S_2 = 28\)
The sum of people getting exactly two sports is given by \(S_2 - 3 \times n(F \cap B \cap C)\):
\(= 28 - 3(5) = 28 - 15 = 13\)
In simple words: By comparing total individual sport medals with total distinct athletes, we calculate the overlap of multi-sport medalists.

Exam Tip: Memorize the expanded relations involving \(S_1, S_2, S_3\) for advanced set word problems.