Question 1

Product and set difference identity: If A = {a, d}, B = {b, c, e} and C = {b, c, f}, evaluate A × (B − C).

Accepted Answer

Introduction / Context:Cartesian products interact with difference: A × (B − C) = (A × B) − (A × C). We compute B − C and check the identity. Given Data / Assumptions: B − C = {e} A = {a, d} Concept / Approach:Pairs in A × (B − C) must have second coordinate in B but not in C. That equals all pairs from A × B with those from A × C removed. Step-by-Step Solution:A × (B − C) = {(a,e), (d,e)}(A × B) = {(a,b),(a,c),(a,e),(d,b),(d,c),(d,e)}(A × C) = {(a,b),(a,c),(d,b),(d,c)}Difference = {(a,e),(d,e)} Verification / Alternative check:Elementwise condition matches exactly: keep y ∈ B and y ∉ C. Why Other Options Are Wrong:They either include too many pairs or the wrong second components. Common Pitfalls:Forgetting to remove all pairs whose second coordinate lies in C. Final Answer:(A × B) − (A × C)

Question 2

Intersection of two Cartesian products: For A = {1,2,3}, B = {2,3,4}, C = {1,3,4}, D = {2,4,5}, compute (A × B) ∩ (C × D).

Accepted Answer

Introduction / Context:A well-known identity: (A × B) ∩ (C × D) = (A ∩ C) × (B ∩ D). We justify it by unpacking ordered-pair membership conditions. Given Data / Assumptions: A = {1,2,3}, B = {2,3,4} C = {1,3,4}, D = {2,4,5} Concept / Approach:(x,y) lies in both products iff x ∈ A and x ∈ C, and y ∈ B and y ∈ D. Thus x ∈ A ∩ C and y ∈ B ∩ D. Step-by-Step Solution:A ∩ C = {1,3}B ∩ D = {2,4}(A ∩ C) × (B ∩ D) = {(1,2),(1,4),(3,2),(3,4)} Verification / Alternative check:Direct enumeration from both sides matches—pairs not satisfying both coordinate conditions drop out. Why Other Options Are Wrong:(A ∩ D) × (B ∩ C) swaps intersections incorrectly; ϕ is false since intersections are nonempty; “None” contradicts the identity. Common Pitfalls:Interchanging first and second coordinate intersections; always match first-with-first, second-with-second. Final Answer:(A ∩ C) × (B ∩ D)

Question 3

Sets and Venn Logic – English-only speakers from group totals: In a group of 50 people, 35 speak Hindi, 25 speak both Hindi and English, and every person speaks at least one of Hindi or English. How many people speak only English (i.e., English but …

Accepted Answer

Introduction / Context:This question tests basic inclusion-exclusion on two overlapping sets (Hindi speakers and English speakers). We are asked for the count of people who speak only English given totals and overlap, with the assurance that everyone speaks at least one language. Given Data / Assumptions: Total group = 50 Hindi speakers H = 35 Both languages (H ∩ E) = 25 Every person speaks at least one of Hindi or English (no outsiders) Concept / Approach:For two sets, |H ∪ E| = |H| + |E| − |H ∩ E|. Here |H ∪ E| = 50. Solve for |E|, then subtract the overlap to get English-only. Step-by-Step Solution:50 = 35 + |E| − 25|E| = 50 − 35 + 25 = 40English-only = |E| − |H ∩ E| = 40 − 25 = 15 Verification / Alternative check:Compute Hindi-only: 35 − 25 = 10. Then 10 (Hindi-only) + 15 (English-only) + 25 (both) = 50 (consistent). Why Other Options Are Wrong:10 is Hindi-only; 20 and 40 conflict with inclusion-exclusion constraints. Common Pitfalls:Subtracting the overlap from the total directly or mixing up Hindi-only with English-only. Final Answer:15

Question 4

Vehicle ownership distribution – cars first, then motorcycles: In a locality of 80 families, 20% of the families own a car (one each). Of the remaining families, 50% own a motorcycle (one each). How many families do not own any vehicle?

Accepted Answer

Introduction / Context:This is a sequential-partition problem: first select car owners, then among the remainder select motorcycle owners. The groups are disjoint by construction (motorcycle selection happens after removing car-owning families). Given Data / Assumptions: Total families = 80 Car owners = 20% of 80 = 16 Remaining families after cars = 80 − 16 = 64 Motorcycle owners = 50% of remaining = 32 Concept / Approach:Add the two disjoint sets of vehicle owners and subtract from the total to get the count with no vehicle. Step-by-Step Solution:Total with a vehicle = 16 + 32 = 48No-vehicle families = 80 − 48 = 32 Verification / Alternative check:Proportions check: 32/80 = 40% have no vehicles, consistent with the sequential selection. Why Other Options Are Wrong:30 and 36 misapply percentages; 23 is unrelated to 80. Common Pitfalls:Taking 50% of the original 80 instead of the remainder (64). Final Answer:32

Question 5

Pass in both subjects – complement of union of failures: In an examination, 30% of students failed in Hindi, 45% failed in English, and 20% failed in both. What percentage passed in both subjects?

Accepted Answer

Introduction / Context:Passing both subjects is the complement of failing at least one subject. The failure sets overlap; use inclusion-exclusion, then subtract from 100%. Given Data / Assumptions: % fail Hindi = 30% % fail English = 45% % fail both = 20% Concept / Approach:|Fail H ∪ Fail E| = 30 + 45 − 20 = 55. Therefore, % passing both = 100 − 55. Step-by-Step Solution:Fail at least one = 55%Pass both = 100% − 55% = 45% Verification / Alternative check:Venn logic confirms the complement relationship between passing both and failing at least one. Why Other Options Are Wrong:35%, 35.7%, 40% do not match the inclusion-exclusion result. Common Pitfalls:Adding 30 and 45 without subtracting the overlap 20. Final Answer:45%

Question 6

Set difference A − B with small numeric sets: Let A = {1, 2, 3, 4, 5} and B = {2, 4, 6}. Find A − B (elements in A that are not in B).

Accepted Answer

Introduction / Context:Set difference A − B collects items that belong to A but not to B. It is a fundamental operation in set theory and appears frequently in data filtering and logic puzzles. Given Data / Assumptions: A = {1, 2, 3, 4, 5} B = {2, 4, 6} Concept / Approach:Remove from A every element that appears in B. Elements 2 and 4 are common and are excluded from A − B. Step-by-Step Solution:Start with A: {1, 2, 3, 4, 5}Remove 2 and 4 (present in B)Result: {1, 3, 5} Verification / Alternative check:Symmetric difference would include 2 and 4 differently, but set difference strictly excludes A’s elements present in B. Why Other Options Are Wrong:{2, 3, 5} and {1, 4, 5} keep elements that should be removed; {2, 3, 6} includes 6, which is not in A. Common Pitfalls:Confusing A − B with B − A or with symmetric difference. Final Answer:{1, 3, 5}

Question 7

Power set construction for a 2-element set: Write the power set of A = {8, 9} (i.e., the set of all subsets of A, including the empty set and A itself).

Accepted Answer

Introduction / Context:The power set of a set A is the set of all its subsets. For a set with n elements, the power set has 2^n subsets, including the empty set and A itself. Given Data / Assumptions: A = {8, 9} n = 2 elements Concept / Approach:List all subsets of a 2-element set: the empty subset, both singleton subsets, and the full set. Step-by-Step Solution:Subsets: ϕ, {8}, {9}, {8, 9}Power set P(A) = {ϕ, {8}, {9}, {8, 9}} Verification / Alternative check:Count check: 2^2 = 4 subsets appear, as required. Why Other Options Are Wrong:{ϕ, {8}, {9}} omits {8, 9}; {{8, 9}, {8}} is incomplete; 'All of these' cannot be correct because the listed options are not all valid power sets. Common Pitfalls:Forgetting to include the empty set, or thinking order matters (it does not in sets). Final Answer:{ϕ, {8}, {9}, {8, 9}}

Question 8

Comparable sets (subset or superset relation exists): Which pair of sets is comparable (i.e., one is a subset of the other)?

Accepted Answer

Introduction / Context:Two sets are comparable if one is contained in the other. This checks basic understanding of subset relations and differences between elements and sets-as-elements. Given Data / Assumptions: (a) A = {1, 3, 5}, B = {2, 3, 5, 6} (b) A = {x ∈ N : x ≤ 10} = {1,2,…,10}, B = {1,2,…,10,11} (c) A includes the element {4,5}, not two separate elements 4 and 5 Concept / Approach:Check whether every element of one set is in the other (subset test) for each pair. Step-by-Step Solution:(a) Neither contains the other (1 ∉ B and 2,6 ∉ A)(b) {1..10} ⊂ {1..11} → comparable(c) {4,5} as a single element prevents A ⊆ B or B ⊆ A Verification / Alternative check:Count and membership checks confirm only (b) satisfies comparability. Why Other Options Are Wrong:(a) lacks mutual containment; (c) confuses element vs subset; 'Both (a) and (b)' is invalid because (a) fails. Common Pitfalls:Misreading {4,5} as two elements instead of one element which is itself a set. Final Answer:A = {x: x ∈ N and x ≤ 10}, B = {1, 2, 3, …, 10, 11}

Question 9

At least one game – count using inclusion-exclusion: In a class, 50 students play cricket, 20 play football, and 10 play both. How many students play at least one of these two games?

Accepted Answer

Introduction / Context:Counting those who play at least one of two sports requires inclusion-exclusion to avoid double-counting those who play both. Given Data / Assumptions: |C| = 50 (cricket) |F| = 20 (football) |C ∩ F| = 10 Concept / Approach:|C ∪ F| = |C| + |F| − |C ∩ F|. Step-by-Step Solution:|C ∪ F| = 50 + 20 − 10 = 60 Verification / Alternative check:At least one = only cricket (40) + only football (10) + both (10) = 60. Why Other Options Are Wrong:45, 55, 65 do not respect the overlap structure. Common Pitfalls:Adding 50 and 20 without subtracting the overlap. Final Answer:60

Question 10

Identify the finite set among descriptions: Which of the following sets is finite?

Accepted Answer

Introduction / Context:Finite vs infinite sets is a foundational classification. Sets defined by bounded inequalities over natural numbers are finite; sets described by open-ended rules typically are infinite. Given Data / Assumptions: N denotes positive integers Quadrilaterals on a plane can vary continuously Primes and multiples extend without bound Concept / Approach:Translate each description and determine whether it yields finitely many elements. Step-by-Step Solution:(a) Primes: infinite set(b) Quadrilaterals in the plane: uncountably many (varying side lengths and angles)(c) x^2 ≤ 25 with x ∈ N → x ∈ {1,2,3,4,5} (finite)(d) Multiples of 3 in N: infinite Verification / Alternative check:Count in (c) is 5 elements, confirming finiteness. Why Other Options Are Wrong:They describe sets with endlessly many elements by rule or continuous variation. Common Pitfalls:Including 0 in N; even then, (c) remains finite. Final Answer:{x: x ∈ N and x^2 − 25 ≤ 0}

Question 11

Identify the infinite set among descriptions: Which of the following sets is infinite?

Accepted Answer

Introduction / Context:An infinite set has no finite bound on its elements. Geometric constructions often yield infinite families (e.g., circles with the same center and varying radii). Given Data / Assumptions: Even prime numbers: only {2} (finite) Rivers in one country: finite list Concentric circles about a fixed center: radius can vary over infinitely many positive values Concept / Approach:Check whether there is a natural limiting count. Concentric circles admit unboundedly many radii. Step-by-Step Solution:(a) = {2} → finite(b) Finite by geography(c) Infinite family by any distinct positive radius Verification / Alternative check:Between any two radii, more radii exist (continuum), reinforcing infinitude. Why Other Options Are Wrong:They are finite sets by definition or context. Common Pitfalls:Misinterpreting 'even prime' as many primes; there is only one even prime. Final Answer:Set of all concentric circles

Question 12

Cardinality of distinct letters in a word: Find the number of distinct letters in the set {x : x is a letter of the word ASSASSINATION}.

Accepted Answer

Introduction / Context:Cardinality here means the count of unique elements. For a word, it is the number of distinct letters ignoring repetition. Given Data / Assumptions: Word: ASSASSINATION Concept / Approach:List unique letters present in the word, each counted once in the set. Step-by-Step Solution:Distinct letters: A, S, I, N, T, OCount = 6 Verification / Alternative check:Scan the word left-to-right ensuring no new letters beyond {A,S,I,N,T,O} appear. Why Other Options Are Wrong:4 and 2 underestimate; 8 overestimates the unique letters. Common Pitfalls:Counting repeated letters multiple times. Final Answer:6

Question 13

Truth among set relations for BOWL / ELBOW / BELLOW: Let A = letters of 'BOWL', B = letters of 'ELBOW', C = letters of 'BELLOW'. Which statement is false?

Accepted Answer

Introduction / Context:We compare sets made from letters of words (duplicates ignored). The task is to identify the false statement among four relations. Given Data / Assumptions: A = {B, O, W, L} B = {E, L, B, O, W} C = {B, E, L, O, W} Concept / Approach:Compute each relation by direct comparison of membership. Step-by-Step Solution:A ⊂ B: true (A's four letters are all in B)B = C: true (same five letters)B ⊃ C: true if interpreted as superset allowing equality'B is a proper subset of C': false because B and C are equal Verification / Alternative check:Enumerations confirm equality of B and C, invalidating the 'proper subset' claim. Why Other Options Are Wrong:They are true under standard set interpretations (A ⊂ B, B ⊃ C, B = C). Common Pitfalls:Confusing superset '⊃' with proper superset; here, equality makes 'proper subset' false. Final Answer:B is a proper subset of C.

Question 14

Identify disjoint pairs among given sets: Which of the following pairs are disjoint? (i) {1,2,3,4} and {x ∈ N : 4 ≤ x ≤ 6} (ii) {a, e, i, o, u} and {c, d, e, f} (iii) {x : x is even integer} and {x : x is odd integer}

Accepted Answer

Introduction / Context:Disjoint sets have empty intersection. We check each pair for any shared elements. Given Data / Assumptions: (i) {1,2,3,4} and {4,5,6} (ii) vowels and {c,d,e,f} (iii) even integers and odd integers Concept / Approach:Look for any common member between the two sets in each pair. Step-by-Step Solution:(i) Common element 4 → not disjoint(ii) Common element e → not disjoint(iii) No integer is both even and odd → disjoint Verification / Alternative check:Intersections: (i) {4}, (ii) {e}, (iii) ϕ. Why Other Options Are Wrong:They identify pairs that are not disjoint or select multiple when only one applies. Common Pitfalls:Overlooking boundary elements like 4 in (i). Final Answer:(iii)

Question 15

Find |X ∩ Y| using union and individual sizes: If |X ∪ Y| = 18, |X| = 8, and |Y| = 15, how many elements are in X ∩ Y?

Accepted Answer

Introduction / Context:The inclusion-exclusion principle for two sets relates the sizes of each set, their union, and their intersection. Given Data / Assumptions: |X ∪ Y| = 18 |X| = 8 |Y| = 15 Concept / Approach:|X ∪ Y| = |X| + |Y| − |X ∩ Y|. Solve for |X ∩ Y|. Step-by-Step Solution:18 = 8 + 15 − |X ∩ Y||X ∩ Y| = 8 + 15 − 18 = 5 Verification / Alternative check:Substitute back to verify union size remains 18. Why Other Options Are Wrong:7, 9, 11 contradict inclusion-exclusion with given numbers. Common Pitfalls:Forgetting to subtract the intersection when adding set sizes. Final Answer:5

Question 16

Recovering |B| from |A|, |A ∪ B|, and |A ∩ B|: If set A has 40 elements, |A ∪ B| = 60 elements, and |A ∩ B| = 10 elements, how many elements does set B have?

Accepted Answer

Introduction / Context:This uses inclusion-exclusion solved for |B|. The original stem had a symbol typo; we apply the standard interpretation |A ∩ B| = 10 via Recovery-First to make it solvable. Given Data / Assumptions: |A| = 40 |A ∪ B| = 60 |A ∩ B| = 10 (repaired) Concept / Approach:|A ∪ B| = |A| + |B| − |A ∩ B| → solve for |B|. Step-by-Step Solution:60 = 40 + |B| − 10|B| = 60 − 30 = 30 Verification / Alternative check:Check: 40 + 30 − 10 = 60 (consistent). Why Other Options Are Wrong:40, 45, 50 contradict the union size when back-substituted. Common Pitfalls:Reading the misprinted symbol literally without applying standard set identity. Final Answer:30

Question 17

Minimum number who can speak Hindi only (guaranteed count): In a group of 1000 people, 750 can speak Hindi and 400 can speak English. What is the minimum number who must be Hindi-only speakers?

Accepted Answer

Introduction / Context:Without explicit overlap, exact counts of language-only groups cannot be uniquely determined. However, a minimum guaranteed number for Hindi-only can be found using inclusion-exclusion and the constraint that at most 100% of people exist. Given Data / Assumptions: Total people = 1000 Hindi speakers H = 750 English speakers E = 400 No statement that everyone speaks at least one language Concept / Approach:The size of the overlap satisfies |H ∩ E| ≥ |H| + |E| − 1000 = 150 (pigeonhole bound). Minimizing Hindi-only means maximizing the overlap. Step-by-Step Solution:Minimum overlap = 150Hindi-only minimum = |H| − |H ∩ E| = 750 − 150 = 600 Verification / Alternative check:An arrangement meeting the bound exists: 150 bilingual, 600 Hindi-only, 250 English-only or outsiders adjusted accordingly. Why Other Options Are Wrong:650, 750, 800 exceed the guaranteed minimum; the question asks for the minimum that must be Hindi-only. Common Pitfalls:Subtracting 400 from 750 directly (ignores total cap) or assuming everyone speaks at least one language when not stated. Final Answer:600

Question 18

Neither egg nor meat – count pure vegetarians via union: In a town of 5000 people, 3200 eat eggs, 2500 eat meat, and 1500 eat both egg and meat. How many are pure vegetarians (eat neither egg nor meat)?

Accepted Answer

Introduction / Context:To find those who eat neither of two categories, compute the size of the union (eat at least one) and subtract from the total population. Given Data / Assumptions: Total = 5000 Egg eaters = 3200 Meat eaters = 2500 Both = 1500 Concept / Approach:|Egg ∪ Meat| = 3200 + 2500 − 1500 = 4200. Pure vegetarians are the complement in the population. Step-by-Step Solution:Neither = 5000 − 4200 = 800 Verification / Alternative check:Sum of disjoint parts: Only-egg (1700) + Only-meat (1000) + Both (1500) + Neither (800) = 5000. Why Other Options Are Wrong:600, 850, 900 do not align with the union computation. Common Pitfalls:Forgetting to subtract the overlap when computing the union. Final Answer:800

Question 19

Both coffee and tea – given at least one of the two: In a group of 70 people, 37 like coffee and 52 like tea. Each person likes at least one of coffee or tea. How many like both?

Accepted Answer

Introduction / Context:When everyone likes at least one of two options, the size of the union equals the total. We can recover the intersection from inclusion-exclusion. Given Data / Assumptions: Total = 70 |Coffee| = 37 |Tea| = 52 Everyone likes at least one → |Coffee ∪ Tea| = 70 Concept / Approach:|C ∩ T| = |C| + |T| − |C ∪ T|. Step-by-Step Solution:|C ∩ T| = 37 + 52 − 70 = 19 Verification / Alternative check:Only-coffee = 37 − 19 = 18; Only-tea = 52 − 19 = 33; 18 + 19 + 33 = 70. Why Other Options Are Wrong:17, 21, 23 contradict the computed overlap. Common Pitfalls:Using 37 + 52 as if no overlap existed. Final Answer:19

Question 20

Cartesian product distributes over union: Given A = {a, d}, B = {b, c, e}, and C = {b, c, f}, compute A × (B ∪ C).

Accepted Answer

Introduction / Context:For sets, the Cartesian product distributes over union: X × (Y ∪ Z) = (X × Y) ∪ (X × Z). This parallels algebraic distributivity. Given Data / Assumptions: A = {a, d} B = {b, c, e} C = {b, c, f} Concept / Approach:Use the identity above to avoid listing all ordered pairs explicitly. Step-by-Step Solution:B ∪ C = {b, c, e, f}A × (B ∪ C) = {(a,x), (d,x) : x ∈ {b,c,e,f}}This equals (A × B) ∪ (A × C) by distributivity Verification / Alternative check:Explicit enumeration yields 8 ordered pairs, the same as the union of A × B (6 pairs) and A × C (6 pairs) minus their overlap A × {b,c} (4 pairs). Why Other Options Are Wrong:Intersection gives too few pairs; empty set is impossible. Common Pitfalls:Confusing set-theoretic union/intersection with arithmetic operations. Final Answer:(A × B) ∪ (A × C)

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