Question 1

Percent passed in Hindi given failure rates and overlap: In an exam, 40% students failed in Hindi, 50% failed in English, and 21% failed in both. What percentage passed in Hindi?

Accepted Answer

Introduction / Context:When a problem asks for the percentage that passed a subject, you subtract the subject’s fail percentage from 100%. The overlap with another subject is irrelevant unless asked about combined categories like “failed at least one.” Given Data / Assumptions: Failed Hindi = 40% Failed English = 50% Failed both = 21% Concept / Approach:“Passed in Hindi” means did not fail Hindi. Therefore % passed in Hindi = 100% − % failed in Hindi = 60%. No further inclusion–exclusion is needed for this specific query. Step-by-Step Solution:Passed Hindi = 100 − 40 = 60% Verification / Alternative check:If needed, a Venn picture shows that all categories together sum to 100%; the complement of “failed Hindi” is “passed Hindi.” Why Other Options Are Wrong:31%, 55% are unrelated; 40% is the fail rate; 60% uniquely matches the complement. Common Pitfalls:Overcomplicating with intersections when the question is a single-subject complement problem. Final Answer:60%

Question 2

Total workers from two preferences and a known intersection: In an office, 72% prefer cold drink and 44% prefer tea. Everyone prefers at least one, and 40 workers prefer both. Find the total number of workers.

Accepted Answer

Introduction / Context:When percentages of two categories and their overlap are given, the percentage of the union is p(cold) + p(tea) − p(both). If “everyone prefers at least one,” the union is 100%. Knowing the numeric overlap count lets us solve for the total population N from an overlap percentage. Given Data / Assumptions: p(cold) = 72% p(tea) = 44% Everyone prefers at least one ⇒ union = 100% |both| = 40 workers Concept / Approach:Compute overlap percentage by inclusion–exclusion: p(both) = p(cold) + p(tea) − 100% = 72% + 44% − 100% = 16%. If 16% of N equals 40, then N = 40 / 0.16. Step-by-Step Solution:p(both) = 16%0.16 * N = 40N = 40 / 0.16 = 250 Verification / Alternative check:Check counts: 72% of 250 = 180, 44% of 250 = 110, overlap 40; union = 180 + 110 − 40 = 250 (100%), consistent with “everyone prefers at least one.” Why Other Options Are Wrong:210, 220, 240 do not make 16% equal 40; 40 is the overlap count, not the total N. Common Pitfalls:Forgetting to subtract the overlap when combining percentages or treating 40 as a percentage instead of a count. Final Answer:250

Question 3

Neither-of-two percentage from two activities with overlap: In a town, 65% watch TV news, 40% read a newspaper, and 25% do both. What percentage do neither?

Accepted Answer

Introduction / Context:For two activities A and B, the union percentage is p(A) + p(B) − p(A ∩ B). The “neither” percentage is the complement of the union within 100%. This is a direct inclusion–exclusion application. Given Data / Assumptions: p(TV) = 65% p(Newspaper) = 40% p(Both) = 25% Concept / Approach:Compute union first, then subtract from 100% to get neither. Step-by-Step Solution:Union = 65% + 40% − 25% = 80%Neither = 100% − 80% = 20% Verification / Alternative check:Venn diagram logic: the four partitions sum to 100%; “neither” is whatever remains after accounting for the union. Why Other Options Are Wrong:5%, 10%, 15% do not match the computed complement given the fixed union of 80%. Common Pitfalls:Adding the given percentages without subtracting the overlap double-counts the intersection and yields 105% instead of 80%. Final Answer:20%

Question 4

Perfect-square identification with digit test and quick checks: Evaluate √64009 exactly by recognizing nearby squares (250^2 to 255^2) and confirming the units and tens patterns. What is the square root?

Accepted Answer

Introduction / Context:Square-root questions on perfect squares can be solved quickly by testing nearby squares and using last-digit and tens-place cues. Here we locate √64009 by bracketing between known squares around 250^2. Given Data / Assumptions: Target number N = 64009 We suspect N is close to 250^2 = 62500 We will test 251^2, 252^2, 253^2, etc. Concept / Approach:For k near 250, compute (250 + d)^2 = 250^2 + 2*250*d + d^2 to step upward efficiently. Also, the last digit 9 suggests roots ending in 3 or 7 for odd squares ending with 9; among options, 253 is a candidate. Step-by-Step Solution:251^2 = 63001252^2 = 63504253^2 = 64009 (exact match) Verification / Alternative check:Use (250 + 3)^2 = 250^2 + 1500 + 9 = 62500 + 1500 + 9 = 64009, confirming the result. Why Other Options Are Wrong:352, 523, and 532 produce squares far from 64009; 245^2 = 60025, which is too small. Common Pitfalls:Guessing without bracketing; ignoring simple expansion (a + b)^2 = a^2 + 2ab + b^2 which speeds mental checks. Final Answer:253

Question 5

Two-language enrollment using inclusion–exclusion: In a school of 600 students, each offered English or Hindi or both. If 75% took English and 45% took Hindi, how many offered both subjects?

Accepted Answer

Introduction / Context:Classic set problems about course enrollment are solved with inclusion–exclusion. Given totals for English and Hindi and the universe size, we can compute the overlap directly. Given Data / Assumptions: Total students U = 600 n(English) = 75% of 600 = 450 n(Hindi) = 45% of 600 = 270 All students chose at least one of the two languages (as stated) Concept / Approach:Inclusion–exclusion for two sets: n(E ∪ H) = n(E) + n(H) − n(E ∩ H). Here n(E ∪ H) = U, so solve for n(E ∩ H). Step-by-Step Solution:n(E ∩ H) = n(E) + n(H) − U= 450 + 270 − 600= 120 Verification / Alternative check:Unique-English = 450 − 120 = 330; Unique-Hindi = 270 − 120 = 150; 330 + 150 + 120 = 600 (balanced). Why Other Options Are Wrong:48, 60, 80 do not satisfy the totals when back-checked; 140 would overcount overlap and push the union below 600. Common Pitfalls:Adding 75% and 45% to 120% and forgetting to subtract the overlap once; assuming disjoint groups when the problem states “or both.” Final Answer:120

Question 6

Intersection of two prime-heavy sets — compute A ∩ B explicitly: Given A = {2, 3, 5, 7, 11} and B = {1, 3, 5, 7, 9, 11}, determine A ∩ B.

Accepted Answer

Introduction / Context:Set intersection retains only elements common to both sets. Here A is a prime list up to 11, while B mixes odds with the prime 11; we simply keep shared elements. Given Data / Assumptions: A = {2, 3, 5, 7, 11} B = {1, 3, 5, 7, 9, 11} Concept / Approach:A ∩ B = {x : x ∈ A and x ∈ B}. Check each member of A against B to keep the common ones. Step-by-Step Solution:2 ∈ A but 2 ∉ B → exclude3, 5, 7, 11 appear in both → includeIntersection = {3, 5, 7, 11} Verification / Alternative check:Scan B: {1, 3, 5, 7, 9, 11} → removing 1 and 9 (not in A) leaves {3, 5, 7, 11}. Why Other Options Are Wrong:Option A equals A (not intersection); option B equals B; option D and option E introduce elements not present in both sets (2 or 1, 9). Common Pitfalls:Confusing union and intersection; forgetting that duplicates do not matter in set notation. Final Answer:{ 3, 5, 7, 11 }

Question 7

Cardinalities with inclusion–exclusion (Recovery-First applied): If n(A) = 40, n(B) = 26 and n(A ∩ B) = 16, compute n(A ∪ B).

Accepted Answer

Introduction / Context:The original stem repeated n(A ∩ B) on both sides, which is likely a typographical slip. Using Recovery-First, we repair the question to find n(A ∪ B) from n(A), n(B), and n(A ∩ B). Given Data / Assumptions: n(A) = 40 n(B) = 26 n(A ∩ B) = 16 Concept / Approach:For any two finite sets, n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Subtracting the overlap prevents double counting shared elements. Step-by-Step Solution:n(A ∪ B) = 40 + 26 − 16= 66 − 16 = 50 Verification / Alternative check:If A and B were disjoint (n(A ∩ B) = 0), we would have 40 + 26 = 66; an overlap of 16 reduces the union size to 50, consistent with the formula. Why Other Options Are Wrong:30 and 40 undercount; 60 ignores the overlap; 42 is arbitrary and not obtained by the correct formula. Common Pitfalls:Adding the two sizes without subtracting the intersection or misreading the repaired stem; always check for overlaps to avoid double counting. Final Answer:50

Question 8

Associativity of union over three sets in a finite universe: Given U = {a, b, c, d, e, f}, A = {a, b, c}, B = {c, d, e, f}, C = {c, d, e}. Compute (A ∪ B) ∪ C.

Accepted Answer

Introduction / Context:Union is associative and idempotent. When two sets already cover the universe U, adding a third does not change the result. We verify elementwise here. Given Data / Assumptions: U = {a, b, c, d, e, f} A = {a, b, c} B = {c, d, e, f} C = {c, d, e} Concept / Approach:Compute A ∪ B first; then union with C. Since A ∪ B already equals U, any further union remains U. Step-by-Step Solution:A ∪ B = {a, b, c} ∪ {c, d, e, f} = {a, b, c, d, e, f} = U(A ∪ B) ∪ C = U ∪ C = U Verification / Alternative check:Directly, A ∪ B ∪ C contains all symbols in U; there is no symbol outside U, hence the union equals U. Why Other Options Are Wrong:A, B, or C are proper subsets of U; A ∪ C still misses f; only U contains every element. Common Pitfalls:Accidentally intersecting instead of uniting; forgetting union is associative so parentheses do not alter the final set here. Final Answer:U

Question 9

Generated set A = {(n−1)/(n+1) : n ∈ W, n ≤ 10} — identify a true statement (Recovery-First): Consider A defined by x = (n−1)/(n+1) for whole numbers n with n ≤ 10. Which statement is correct?

Accepted Answer

Introduction / Context:The stem uses “0” where standard notation would use ϕ (empty set). Applying Recovery-First, interpret “0 ⊂ A” as “ϕ ⊂ A”. We analyze membership values of A and the truth of each statement under that repair. Given Data / Assumptions: A = { (n−1)/(n+1) : n ∈ W, n ≤ 10 }, with W = {0,1,2,...} We compute sample values: n=0 → −1; n=1 → 0; n=2 → 1/3; n=3 → 1/2; ... Concept / Approach:Check each option against A’s generated values, and fix the nonstandard symbol “0” to ϕ when it denotes the empty set in subset relations. Step-by-Step Solution:n=1 gives 0, so 0 ∈ A is true as a number, but the stem’s choices mix numeric 0 with set symbols; interpret carefully.1/3 ∈ A since (2−1)/(2+1) = 1/3 → so “1/3 ∉ A” is false.ϕ ⊂ A is always true for any set A (empty set is a subset of every set). Verification / Alternative check:A contains many fractions in (−1, 1). Regardless of exact contents, ϕ ⊂ A remains universally true; thus the corrected reading of option “0 ⊂ A” becomes the valid statement. Why Other Options Are Wrong:“0 ⊃ A” is ill-formed; “1/3 ∉ A” is false; “A = ϕ” is false since A is nonempty; “0 ∈ A” mixes numeral vs. set-symbol usage and is ambiguous in the given list. Common Pitfalls:Confusing numeral 0 with the empty set ϕ; always distinguish element membership from subset relations. Final Answer:0 ⊂ A

Question 10

Membership vs. subset with nested sets: Let A = {1, 2, {3, 4}, 5}. Which statement holds true?

Accepted Answer

Introduction / Context:When sets contain other sets as elements, distinguish “∈” (is an element) from “⊂” (is a subset). Here A contains the set {3,4} as a single element. Given Data / Assumptions: A = {1, 2, {3, 4}, 5} Elements of A are: 1, 2, {3,4}, 5 Concept / Approach:{3,4} is literally listed inside A; therefore {3,4} ∈ A is true. For subset claims, each member must appear individually in A, which 3 and 4 do not. Step-by-Step Solution:Check {3,4} ⊂ A → requires 3 ∈ A and 4 ∈ A; false.Check {3,4} ∈ A → true by inspection.Check {{3,4}} ⊄ A → actually {{3,4}} ⊂ A since its only element {3,4} is in A; the “not subset” claim is false.Check {1,3,5} ⊂ A → 3 ∉ A (only {3,4} is present), so false. Verification / Alternative check:List elements explicitly to avoid confusing elements with members of nested sets. Why Other Options Are Wrong:They misuse subset logic where only element membership holds, or they include elements (like 3) not in A. Common Pitfalls:Equating “A contains {3,4}” with “A contains 3 and 4” separately—these are not the same claim. Final Answer:{3, 4} ∈ A;

Question 11

Identify the true statement for a set containing a set element: Let A = {1, 2, {3, 4}, 5}. Which one of the following is true?

Accepted Answer

Introduction / Context:This is a focused check on membership vs. subset when a set contains another set as an element. We confirm which relation actually holds for A. Given Data / Assumptions: A = {1, 2, {3, 4}, 5} Concept / Approach:{3,4} is literally an element of A; thus {3,4} ∈ A is true. A subset claim {3,4} ⊂ A would require 3 and 4 to be elements of A individually, which is not the case. Step-by-Step Solution:Check {3,4} ∈ A → true (listed element){3,4} ⊂ A → false (3 ∉ A, 4 ∉ A)1 ⊂ A → false (1 is an element, not a set of elements){1,2,5} ∈ A → false (A does not contain that set) Verification / Alternative check:Enumerate elements of A to compare directly against each option. Why Other Options Are Wrong:They confuse element membership with subset relations or propose a set A does not contain. Common Pitfalls:Writing 1 ⊂ A when only set-valued objects can be subsets. Final Answer:{3, 4} ∈ A

Question 12

Power set of a set that contains a set element: Find P(A) for A = {{a, b}, c}, i.e., list all subsets of A.

Accepted Answer

Introduction / Context:The power set P(A) is the set of all subsets of A. When A contains a set element like {a,b}, remember that subsets must be formed from the elements {{a,b}, c} rather than from a and b individually. Given Data / Assumptions: A = {{a, b}, c} Concept / Approach:Elements of A are exactly two objects: E1 = {a,b} and E2 = c. Thus P(A) must contain 2^2 = 4 subsets built from E1 and E2. Step-by-Step Solution:P(A) = { ϕ, {{a, b}}, {c}, {{a, b}, c} }Compare with options: none lists exactly these four subsets using correct braces. Verification / Alternative check:Count check: power set must have 4 members; any option with a different count or with {a,b} (not wrapped as an element) is incorrect. Why Other Options Are Wrong:They misuse {a,b} as if a and b were elements of A directly; valid subsets must consist of elements {{a,b}, c}. Common Pitfalls:Confusing elements inside the inner set {a,b} with elements of A itself. Final Answer:None of these

Question 13

Equality of sets — recognize empty-set equivalence (Recovery-First on notation): In which case are A and B equal as sets?

Accepted Answer

Introduction / Context:Two sets are equal if they have precisely the same elements. The empty set Φ and {} are two notations for the same set with no elements. We verify each option’s membership. Given Data / Assumptions: Φ and {} both denote the empty set W denotes whole numbers {0,1,2,…} Concept / Approach:Compare members: if both collections contain exactly the same elements, they are equal; otherwise not. For empty sets, absence of elements matches perfectly. Step-by-Step Solution:Option (b): Φ and {} → equal (both empty)Option (a): distinct lists (no equality)Option (c): {x ∈ W : x < 1} = {0} ≠ ΦOption (d): {Saturday, Sunday} ≠ {Sunday}Option (e): {} ≠ {0} Verification / Alternative check:Cardinality check helps: |Φ| = 0, |{}| = 0, hence equal; options (c) and (d) clearly have different cardinalities. Why Other Options Are Wrong:They either list different elements or different counts. Common Pitfalls:Confusing {} with {0}; the latter contains the number zero and is not empty. Final Answer:A = Φ, B = {}

Question 14

Cardinality of an even-number set with bounds: Let S = {x : x = 2n, n ∈ N, 4 ≤ x ≤ 11}. What is |S| (the number of elements)?

Accepted Answer

Introduction / Context:This is a count of even numbers in a closed interval. Express x = 2n and list feasible x within [4, 11]. Given Data / Assumptions: x = 2n with n ∈ N = {1,2,3,...} 4 ≤ x ≤ 11 Concept / Approach:Enumerate the even numbers between 4 and 11 inclusive, or use arithmetic progression logic to count terms quickly. Step-by-Step Solution:Valid x: 4, 6, 8, 10Count = 4 Verification / Alternative check:AP method: start a=4, difference d=2, last l=10. Number of terms k = ((l−a)/d) + 1 = ((10−4)/2) + 1 = 3 + 1 = 4. Why Other Options Are Wrong:6, 8, 12 overcount; 3 undercounts by missing one even number. Common Pitfalls:Accidentally including 12 (outside the bound) or excluding one endpoint unnecessarily. Final Answer:4

Question 15

Equality vs. repetition — identify which pair is not equal: Which pairs of sets below are not equal? (Remember duplicates do not change a set.)

Accepted Answer

Introduction / Context:Two sets are equal if they have exactly the same elements, ignoring order and repetitions. We test each pair accordingly. Given Data / Assumptions: (a) A = {1,3}, B = {1,4} (b) A = {x : x + 2 = 2} → {0}; B = {0} (c) A = {1,3,4}; B = {3,1,4} (d) A = {1, 1/2, 1/3, ...}; B = {1/n : n ∈ N} Concept / Approach:Reduce duplicates, then compare elements one-to-one. For (d), both describe the same infinite set. Step-by-Step Solution:(a) {1,3} ≠ {1,4} → not equal(b) {0} = {0} → equal(c) {1,3,4} = {1,3,4} → equal(d) identical descriptions → equal Verification / Alternative check:For (b) solve x + 2 = 2 → x = 0 confirming equality. Why Other Options Are Wrong:They assert non-inequality when pairs are equal; only pair (a) differs. Common Pitfalls:Letting repetition or order mislead; sets ignore both. Final Answer:(a) A = {1, 3, 3, 1}, B = {1, 4}

Question 16

Equivalence (same cardinality) vs. equality — pick the non-equivalent pair: Which pair of sets is not equivalent (i.e., does not have the same number of elements)?

Accepted Answer

Introduction / Context:Equivalent sets have the same cardinality (finite or infinite), regardless of the actual elements. We compare counts, not membership equality. Given Data / Assumptions: Option (a): sizes 4 and 4 Option (b): sizes 3 and 5 Option (c): both empty → size 0 each Option (d): both countably infinite Concept / Approach:Check cardinalities directly; infinite even integers and infinite odd integers are in bijection, hence equivalent. Step-by-Step Solution:(a) Equivalent(b) Not equivalent (3 ≠ 5)(c) Equivalent(d) Equivalent (map n → n gives a bijection between even and odd forms) Verification / Alternative check:Construct a bijection for (d): 2n ↔ 2n + 1 is one-to-one and onto. Why Other Options Are Wrong:They have matching sizes or bijections; only (b) mismatches sizes. Common Pitfalls:Confusing “equivalent” with “equal”; they are different notions. Final Answer:A = {a, b, c}, B = {α, β, γ, δ, ν}

Question 17

Describe (A ∩ B) ∩ C for multiples of 2, 5, and 10: Let A = {multiples of 2 in N}, B = {multiples of 5 in N}, C = {multiples of 10 in N}. Describe (A ∩ B) ∩ C.

Accepted Answer

Introduction / Context:A ∩ B is the set of natural numbers that are both multiples of 2 and of 5, i.e., multiples of lcm(2,5) = 10. Intersecting that with C yields C again. Given Data / Assumptions: A = {2,4,6,...} B = {5,10,15,...} C = {10,20,30,...} Concept / Approach:A ∩ B = {multiples of 10}. Then (A ∩ B) ∩ C = {multiples of 10} ∩ {multiples of 10} = {multiples of 10}. Step-by-Step Solution:A ∩ B = C(A ∩ B) ∩ C = C ∩ C = C Verification / Alternative check:Element test: any k divisible by 10 is automatically divisible by 2 and 5; thus C ⊆ A ∩ B and equality holds. Why Other Options Are Wrong:They describe larger sets (A, B) or a redundant expression (A ∩ B) without the final restriction; the exact result is C. Common Pitfalls:Forgetting that lcm determines simultaneous divisibility for intersections of multiples. Final Answer:C

Question 18

Compute (A ∩ U) ∩ (B ∪ C) in a finite universe: Given U = {2,3,4,5,6,7,8,9,10,11}, A = {2,4,7}, B = {3,5,7,9,11}, C = {7,8,9,10,11}, evaluate (A ∩ U) ∩ (B ∪ C).

Accepted Answer

Introduction / Context:Intersecting with U leaves A unchanged since A ⊆ U. Then intersect A with the union B ∪ C by simple element checks. Given Data / Assumptions: A = {2,4,7} B = {3,5,7,9,11} C = {7,8,9,10,11} Concept / Approach:Compute B ∪ C, then intersect with A. Step-by-Step Solution:A ∩ U = AB ∪ C = {3,5,7,8,9,10,11}A ∩ (B ∪ C) = {2,4,7} ∩ {3,5,7,8,9,10,11} = {7} Verification / Alternative check:Only 7 lies in all three relevant sets; 2 and 4 are not in B ∪ C. Why Other Options Are Wrong:{9}, {6}, {5} are not in A; the multi-element set adds extras not present in A. Common Pitfalls:Forgetting to take union before intersection; mixing elements not in A when intersecting. Final Answer:{7}

Question 19

Cartesian product with an intersection — maintain ordered pair order: If A = {a, b}, B = {2, 3, 5, 6, 7}, C = {5, 6, 7, 8, 9}, find A × (B ∩ C).

Accepted Answer

Introduction / Context:Cartesian products are ordered pairs (first from the first set, second from the second set). We first intersect B and C, then pair each element of A with each element of that intersection. Given Data / Assumptions: A = {a, b} B ∩ C = {5,6,7} Concept / Approach:Form all (x, y) with x ∈ A and y ∈ (B ∩ C). Keep the order (x first, y second). Step-by-Step Solution:Pairs: (a,5), (a,6), (a,7), (b,5), (b,6), (b,7) Verification / Alternative check:Count check: |A| = 2, |B ∩ C| = 3, so |A × (B ∩ C)| = 6; the listed set has 6 pairs. Why Other Options Are Wrong:Option (c) reverses order (y, x); others have wrong elements or cardinalities. Common Pitfalls:Swapping coordinates or forgetting to intersect before forming pairs. Final Answer:{(a, 5), (a, 6), (a, 7), (b, 5), (b, 6), (b, 7)}

Question 20

Product distributes over intersection: If A = {a, d}, B = {b, c, e} and C = {b, c, f}, determine A × (B ∩ C).

Accepted Answer

Introduction / Context:Cartesian product respects intersections: A × (B ∩ C) = (A × B) ∩ (A × C). We confirm this using elementwise reasoning. Given Data / Assumptions: B ∩ C = {b, c} A = {a, d} Concept / Approach:Any pair (x, y) is in A × (B ∩ C) iff x ∈ A and y ∈ B and y ∈ C. This is exactly the definition of being simultaneously in A × B and A × C. Step-by-Step Solution:A × (B ∩ C) = {(a,b),(a,c),(d,b),(d,c)}(A × B) ∩ (A × C) = same four pairs Verification / Alternative check:Subset checks both ways show equality straightforwardly. Why Other Options Are Wrong:Union would include pairs with y in B or C (too many); ϕ is false; “None” contradicts the law above. Common Pitfalls:Confusing distribution over union versus intersection; note that A × (B ∪ C) = (A × B) ∪ (A × C) (different identity). Final Answer:(A × B) ∩ (A × C)

Sets and Functions Questions

Questions

Practice smarter