Logical reasoning interviews can show how candidates interpret constraints, separate evidence from assumptions, and explain a conclusion. Recruiters get more useful evidence by asking for a clear chain of reasoning and a check than by rewarding a memorized answer to a familiar riddle. Choose problems relevant to the role, state permitted tools, and provide consistent instructions and appropriate accommodations.

This guide to 100+ Logical Reasoning Interview Questions and Answers contains 120 original exercises across beginner, intermediate, and experienced levels. It mixes deduction, sets, ordering, truth-and-lie statements, quantitative reasoning, probability, data sufficiency, causal interpretation, and workplace decisions. These exercises support structured discussion; they are not a validated intelligence test or a substitute for job-relevant evidence.

How can recruiters address potential AI cheating, deepfakes, impersonation, and unauthorized assistance during logical reasoning interviews?

Ask candidates to explain the assumptions behind a conclusion, construct a counterexample, or solve a small variation with one changed constraint. Clarify which tools and forms of assistance are permitted before the exercise, so legitimate accommodations are not confused with misconduct.

Sherlock AI helps recruiters detect potential AI-assisted cheating, deepfakes, candidate impersonation, and unauthorized real-time assistance during logical reasoning interviews.

Use integrity signals alongside the candidate’s explanation and a consistent review process. A signal alone is not proof of misconduct; give candidates a fair opportunity to clarify unexpected observations.

Logical Reasoning Interview Questions

All facts needed for each exercise are supplied in its question. Treat stated rules as premises for the exercise, not as claims about real people or organizations. When the information does not select a unique answer, identifying that limitation is part of the solution.

Logical Reasoning Interview Questions for Beginners

Start with direct deductions, small sets, simple ordering constraints, explicit numerical rules, and elementary probability.

1. Every urgent ticket is reviewed. Ticket K is urgent. What follows about K?

2. If a backup completes, its indicator is green. The indicator is green. Must the backup have completed?

3. If a record is archived, it is read-only. Record R is not read-only. Can R be archived?

4. A device connects only if it has a valid certificate. Device D connects. What must be true?

5. A file is accepted if and only if it passes both format and checksum checks. It passes format but fails checksum. Is it accepted?

6. At least one of services A and B is running; both may run. A is not running. What follows about B?

7. Exactly one of alerts A and B is active. A is active. Is B active?

8. All analysts are employees. Some employees are mentors. Must some analysts be mentors?

9. No confidential items are public. Some reports are confidential. What follows about those reports?

10. Is “Every submitted form is signed” contradicted by finding an unsigned draft that was never submitted?

11. Of 30 staff, 18 use tool A and 14 use tool B; 8 use both. How many use neither?

12. Every premium account has support. No suspended account has support. Can an account be both premium and suspended?

13. Four jobs A, B, C, D occupy four consecutive slots, one per slot. A precedes B, B precedes C, and C precedes D. What is the order?

14. Four speakers A, B, C, D each speak once. A is first, D last, and B speaks before C. Who speaks second?

15. Three distinct boxes, red, blue, and green, occupy positions 1–3 from left to right. Red is not first, green is not last, and blue is immediately right of red. What is their order?

16. Four tasks A, B, C, D each take one slot. A must precede C, B must precede C, and C must precede D. Which tasks can occupy the first slot?

17. Five people A, B, C, D, E stand in a row. B is immediately right of A, C immediately right of B, D immediately right of C, and E is at an end. What positions can A occupy?

18. Three people Ana, Bo, and Cy receive distinct roles: editor, reviewer, and tester. Ana is not editor or tester; Bo is not tester. Who is tester?

19. A one-hour meeting starts at 9, 10, or 11. Ana is available for starts at 9 or 10; Bo for starts at 10 or 11. Everyone must attend. When can it start?

20. Four distinct parcels P, Q, R, S are delivered one at a time. P is before Q, and R is before S. Must Q be before R?

21. A and B are each either truth-tellers, whose statements are true, or liars, whose statements are false. A says, “B is a liar.” B says, “We are different types.” What are their types?

22. One key is in exactly one of boxes A, B, C. The labels say A: “The key is in B”; B: “The key is not in B”; C: “The key is not in A.” Exactly one label is true. Where is the key?

23. A and B are each truth-tellers or liars, with the usual always-true or always-false statement rule. A says, “We are both liars.” B makes no statement. What are their types?

24. Exactly one of three statements about an integer x is true: x > 5; x is even; x = 4. Given x is one of 2, 4, 6, 7, which values are possible?

25. A sequence starts at 3, and every later term is twice the previous term plus 1. What is the fifth term?

26. A sequence starts at 20 and subtracts 3 at every step. What is its sixth term?

27. A bag has 3 red and 2 blue tokens. Each individual token is equally likely to be drawn. What is the probability of drawing blue?

28. A fair six-sided die is rolled once. Given that the result is even, what is the probability it exceeds 3?

29. Two independent fair coins are tossed. What is the probability of exactly one head?

30. A bag contains 4 red and 2 blue tokens. Two tokens are drawn uniformly without replacement. What is the probability both are blue?

31. A fair six-sided die is rolled once. You receive 12 points for a 6 and 0 otherwise. What is the expected score?

32. A team completes 12 equally sized units in 3 hours at a constant rate. With the same team and rate, how long do 20 units take?

33. A price of $100 rises by 20% and then falls by 20% of the increased price. What is the final price?

34. Three equally weighted scores have an average of 18. Two scores are 15 and 21. What is the third?

35. For a positive integer x, is x even? Statement I: x is divisible by 4. Statement II: x is divisible by 3. Which statements, considered alone, are sufficient?

36. For positive integers x and y, determine x. Statement I: x + y = 10. Statement II: x − y = 2. Is either statement alone sufficient, and are they sufficient together?

37. A company introduces a voluntary workshop, and attendees later outperform non-attendees. Does this alone prove that the workshop caused the difference?

38. A courier says, “Every parcel that leaves before noon arrives today. This parcel arrived today, so it must have left before noon.” Is the conclusion guaranteed?

39. A manager argues, “The revised checklist will reduce errors because it adds a second review.” What additional premise connects the added review to the claimed benefit?

40. In one month, team A records 8 errors in 200 tasks and team B records 6 errors in 100 tasks. Which has the lower observed error rate, and does that establish the better process?

Logical Reasoning Interview Questions for Intermediate

Combine constraints, test counterexamples, count cases, and examine what additional information a conclusion needs.

41. A release is permitted only if both the security review and the rollback test pass. The rollback test failed. What follows about permission, and what remains unknown?

42. A report claims, “Every service has at least one tested recovery procedure.” What exactly would a counterexample have to show?

43. Two reviewers each approved at least one of two proposals. Must there be a proposal approved by both reviewers?

44. An alert is active if and only if sensor B or sensor C is active, where “or” is inclusive. The alert is active and B is inactive. What is C’s state?

45. Exactly one of features A, B, and C will be enabled. The rules are: if A is enabled, B must be enabled; if B is enabled, C must be disabled. Which features can be the sole enabled feature?

46. A policy set says: every archived record is encrypted; record R is archived; record R is not encrypted. If the first two statements are retained, which statement must be withdrawn to restore consistency?

47. Five jobs A, B, C, D, and E run one at a time. A and B must both precede C; C precedes D; D precedes E. How many complete job orders satisfy all constraints?

48. A, B, C, and D occupy four distinct shifts numbered 1 to 4, earliest first. D has shift 4, B is immediately before C, and A is not in shift 1. What is the assignment?

49. Five different boxes A, B, C, D, and E fill shelf positions 1–5 from left to right. D is at 1, E is at 5, C is immediately right of B, and A is left of B. What is the order?

50. Three tasks A, B, and C receive different labels: red, blue, and green, each used once. A is not red, B is not blue, and C is green. Which label does each task receive?

51. Tasks A, B, C, and D take 2, 3, 4, and 1 hours respectively. A and B start at time 0 on separate workers; C starts only after both finish; D starts after C. With unlimited workers and no other delays, what is the earliest completion time?

52. A directed network has only these edges and costs: A→B=3, A→C=2, B→C=1, B→D=4, C→D=6. What is the cheapest route from A to D?

53. Each of A, B, and C is either a truth-teller, whose statement is true, or a liar, whose statement is false. A says “B is a liar”; B says “C is a liar”; C says “A and B are the same type.” What are their types?

54. A, B, and C are each either a truth-teller, whose statement is true, or a liar, whose statement is false. A says “Exactly one of us is a truth-teller”; B says “Exactly two of us are truth-tellers”; C says “A is a liar.” Can their types be determined uniquely?

55. A key is in exactly one of three drawers: red, blue, or green. Exactly two statements are true: A says “It is in red”; B says “It is in blue”; C says “It is not in red.” Which drawer contains it?

56. A, B, and C are each supposed to be either a truth-teller, whose statement is true, or a liar, whose statement is false. A says “B and C are the same type”; B says “A and C are different types”; C says “B is a liar.” Is any assignment consistent with all three statements?

57. Exactly one of A, B, and C caused an outage. A says “B caused it”; B says “C caused it”; C says “B did not cause it.” Exactly one statement is true. Can the culprit be identified uniquely?

58. Among 70 employees, 40 use tool A, 35 use tool B, and 20 use both. How many use neither?

59. In a population of 100, 50 belong to A, 45 to B, and 40 to C. Pairwise intersections, including members of all three, are |A∩B|=20, |A∩C|=15, and |B∩C|=18; all three contain 8. How many belong to none?

60. A team has four engineers and three designers, all distinct. How many unordered three-person committees contain at least one engineer and at least one designer?

61. A bag contains three red and two blue tokens, all equally likely to be drawn. Two tokens are drawn uniformly without replacement. What is the probability of getting exactly one red?

62. Two independent fair six-sided dice are rolled. Given that their sum is 8, what is the probability that at least one die shows 6?

63. For events A and B, P(A)=0.4, P(B)=0.5, and P(A or B)=0.7, with inclusive “or.” Are A and B independent?

64. Ten percent of components are defective. A test flags 80% of defective components and 20% of nondefective components. A randomly selected component is flagged. What is the probability it is defective?

65. Three independent backup checks each pass with probability 1/2. What is the probability that exactly two pass?

66. A sequence is defined by a₁=2 and aₙ₊₁=2aₙ+3 for every positive integer n. What is a₄?

67. Starting with a₁=5, create the sequence by alternately adding 3 and multiplying by 2, beginning with addition. What is a₅?

68. An integer x is between 1 and 10 inclusive. Statement I says x is even; statement II says x is divisible by 3. Which statements are sufficient to determine x: either alone, both together, or neither?

69. A rectangle has positive side lengths and a strictly longer side. Statement I says its perimeter is 20; statement II says its longer side is 6. Which statements are sufficient to determine its area?

70. Machine A completes one batch in 6 hours and machine B completes an identical batch in 12 hours. They work at constant rates on a divisible batch without interfering. How long do they take together?

71. A department of 20 employees has an average score of 70, and a separate department of 30 has an average score of 80. Every employee has one equally weighted score. What is the combined average?

72. A retailer discounts an original price by 15%, then adds a fixed delivery fee of 6. The final total is 108. What was the original price?

73. A support team introduced a chatbot, and its average resolution time fell the next month. No control group or additional data are available. Does this establish that the chatbot caused the improvement?

74. A training experiment randomly assigns employees to two groups. Afterwards, 40% of the training group and 5% of the control group lack outcome measurements. The measured training group scores higher. Is the measured comparison automatically an unbiased estimate of the training effect?

75. A manager argues, “Automating data entry will save ten staff-hours per week, so the team will finish more client projects.” Which missing assumption connects the saved time to the conclusion?

76. Two software plans provide identical required features and service. Plan A costs 100 upfront plus 10 per month; plan B costs 150 upfront plus 5 per month. There are no other costs or discounts. For exactly 12 months, which is cheaper and by how much?

77. A delivery route costs 10 and has a 20% chance of an additional delay penalty of 50. An alternative has a certain total cost of 22. If the sole objective is minimizing expected monetary cost, which route should be chosen?

78. A procurement choice has two possible future states. Supplier A delivers benefits of 8 in state 1 and 6 in state 2; supplier B delivers 7 and 5 respectively. Costs and every other relevant feature are equal, higher benefit is better, and the state probabilities are unknown. Can one supplier still be preferred?

79. Process A succeeds on 81 of 90 easy cases and 2 of 10 hard cases. Process B succeeds on 9 of 10 easy cases and 18 of 90 hard cases. Does A’s higher overall success rate demonstrate better performance within either difficulty group?

80. There is a 40% chance of rain. Doing nothing costs 100 if it rains and 0 otherwise; a precaution costs 30 and eliminates rain damage. A perfect weather report is available for 5 before deciding. With expected cost as the only objective, should the report be purchased?

Logical Reasoning Interview Questions for Experienced

Work through professional decisions involving incomplete information, uncertain outcomes, causal claims, resource constraints, and competing objectives.

81. A release policy says: if deployment proceeds, approval exists; if approval exists, either the audit passed or an exception was signed; no exception was signed; and the audit did not pass. Must deployment be blocked?

82. A reviewer claims: “Every service has some backup operator; therefore one operator backs up every service.” Two services and two operators exist. Does the conclusion follow, and what is the smallest counterexample in this setting?

83. Three feature flags A, B, and C are Boolean. Exactly one of A and B is enabled; B is enabled if and only if C is disabled; and enabling A requires enabling C. List all valid configurations.

84. A specification contains four requirements: R1 says A must be enabled; R2 says A implies B; R3 says B implies C; R4 says C must be disabled. Is the full set inconsistent, and is every requirement necessary for that inconsistency?

85. Policy P permits access exactly when approval A is present and either training T or an exemption E is present. Policy Q permits access exactly when “A and T” or E is present. Are the policies equivalent? Give a separating case.

86. A project must select exactly three distinct modules from A, B, C, D, and E. Selecting A requires B; selecting C requires D; B and D cannot both be selected; and E must be selected. Which selections are valid?

87. Exactly one of three reports is true: R1 says “The incident was caused by A”; R2 says “The incident was caused by A or B”; R3 says “The incident was not caused by C.” Exactly one of A, B, or C caused the incident. Which cause is possible?

88. Three reviewers are each either a truth-teller, whose statement is true, or a liar, whose statement is false. A says “B is a liar.” B says “C is a liar.” C says “A and B are different types.” What are their types?

89. Exactly one of four servers A, B, C, D failed. Reports say R1: “A or B failed”; R2: “B or C failed”; R3: “C or D failed”; R4: “A failed.” Exactly two reports are true. Which failures remain possible?

90. A manufacturing defect affects 2% of units. An alarm detects 90% of defective units and falsely alarms on 5% of nondefective units. A randomly selected unit alarms. What is the probability it is defective?

91. A defect has prior probability 10%. Two tests each have 80% sensitivity and a 10% false-positive rate. Both tests are positive. Can the posterior defect probability be calculated uniquely without knowing their conditional dependence?

92. An investigation starts with prior odds of 1:4 for hypothesis H versus not H. Evidence E is three times as likely under H as under not H. What are the posterior odds and probability of H?

93. A support queue receives 30% of its tickets from system A and 70% from system B. Escalation rates are 20% for A and 5% for B. A randomly chosen escalated ticket is inspected. What is the probability it came from A?

94. Two independent fair coins are tossed once. An automated rule records the trial if at least one coin is heads. You inspect a uniformly chosen recorded trial. What is the probability both coins were heads?

95. A component fails during a shift with probability 0.04. An alarm sounds with probability 0.95 when it fails and 0.02 when it does not. What is the probability the component has not failed given that the alarm stayed silent?

96. A forecast system assigns probability 0.80 to each of 100 binary events. Exactly 80 events occur. Is it perfectly calibrated on this probability group, and what is its mean binary Brier score, defined as the average of (p − y)² with y equal to 1 for occurrence and 0 otherwise?

97. Two services have next-hour outage probabilities 0.10 and 0.20. Their dependence is unknown. What are the tight bounds on the probability at least one service is down?

98. Treatment A succeeds in 81 of 90 low-risk cases and 1 of 10 high-risk cases. Treatment B succeeds in 19 of 20 low-risk cases and 16 of 80 high-risk cases. Which treatment has the higher observed success rate within each group and overall, and does that identify a causal winner?

99. A randomized pilot assigns 100 customers to a new workflow and 100 to control. Only 60 assigned customers use the workflow; 45 of those succeed. Of the 40 nonusers in that arm, 15 succeed. The control arm has 50 successes. What is the intention-to-treat success-rate difference, and should the 75% user rate replace it?

100. In a 100-customer trial arm, outcomes are known for 80 customers: 48 succeeded and 32 failed. The other 20 outcomes are completely unknown. Without assumptions about missingness, what are the sharp bounds on the arm’s success rate?

101. A portfolio has 20 projects in unit A with an average cost of $120,000 and 80 projects in unit B averaging $70,000. What is the portfolio average, and can averaging the two unit averages give the right answer here?

102. In a stipulated causal model, randomized training T increases checklist use M, and M reduces errors Y; there is no other path from T to Y and no confounding. An analyst controls for M and finds no remaining training effect. Does that show training has no total effect?

103. Two teams have known mean completion times of 4 and 10 hours. Statement I says they completed 30 jobs in total. Statement II says the first team completed twice as many jobs as the second. Which statements are sufficient to determine the combined job-level mean?

104. Positive integers x and y count two kinds of item. Statement I says x + y = 11. Statement II says 3x + 5y = 43. Which statements are sufficient to determine x uniquely?

105. A risk-neutral manager chooses one project. Project A pays a net $140,000 with probability 0.40 and loses $20,000 otherwise. Project B pays a certain net $35,000. Which maximizes expected money?

106. A launch yields net profit of $90,000 if successful and a net loss of $30,000 otherwise. Not launching yields zero. For a risk-neutral decision-maker, what success probability makes launching preferable, and what happens at the threshold?

107. Demand will be high with probability 0.30 and low otherwise. Expanding produces net payoffs of $100,000 in high demand and −$20,000 in low demand. Not expanding produces zero. What is the maximum a risk-neutral firm should pay for perfect demand information before deciding?

108. Demand is high with probability 0.30. Expansion pays $100,000 if high and −$20,000 if low; doing nothing pays zero. A forecast is positive with probability 0.80 when demand is high and 0.20 when low. What is its expected decision value before its price?

109. Three plans have losses across states X, Y, Z: A = (2, 9, 4), B = (6, 6, 6), and C = (8, 3, 7). No state probabilities are supplied. Under the minimax-loss rule, which plan is chosen?

110. Three plans have losses across states X, Y, Z: A = (0, 8, 8), B = (4, 5, 5), and C = (9, 0, 6). Under minimax regret, which plan is chosen?

111. Plans A and B yield net payoffs in three possible states: A = (7, 4, 9) and B = (6, 4, 5). The state probabilities are unknown. What comparison is justified for a decision-maker who always prefers more money?

112. A firm can stop now for zero or pay $10,000 for a pilot. The pilot reveals with certainty whether demand is high (probability 0.40) or low. After the pilot, launching pays an additional net $60,000 if high or loses $30,000 if low; stopping pays zero additional. What is the optimal strategy and expected payoff?

113. Tasks A and B take 3 and 2 hours and share one specialist, so they cannot overlap. Task C takes 4 hours, needs another worker, and starts only after A finishes. Task D takes 1 hour and starts only after B and C finish. All tasks are nonpreemptive; workers are otherwise available at time zero. What is the minimum completion time?

114. Five distinct checks A, B, C, D, E run sequentially in any order, but A must precede C, B must precede C, and D must precede E. How many valid orders exist?

115. Among 100 cases, 50 have tag A, 45 tag B, and 40 tag C. Pairwise overlaps, including any triple-tagged cases, are 20 for A and B, 15 for A and C, and 10 for B and C. Exactly 5 have all three tags. How many have exactly one tag and how many have none?

116. A system fails if both redundant units fail. A common-cause event occurs with probability 0.02 and makes both fail. Conditional on no common-cause event, the units fail independently with probability 0.10 each. What is the system failure probability?

117. Three simultaneous jobs A, B, and C each need a different worker. A can be done only by workers 1 or 2; B only by workers 1 or 2; C only by workers 2 or 3. Which worker must take C, and how many feasible assignments exist?

118. Tasks A and B start at time zero and take 4 and 6 days. C takes 3 days after A; D takes 2 days after both A and B; E takes 2 days after both C and D. Unlimited workers are available. What is the earliest finish and how much may A be delayed without delaying the project?

119. A batch contains 8 passing units and 4 defective units. Three distinct units are sampled uniformly without replacement. What is the probability the sample contains exactly one defective unit?

120. A portfolio has a budget of 8 units. Projects A, B, C, D cost 4, 3, 2, 2 units and deliver values 7, 5, 4, 3 respectively. Each can be selected at most once; C requires B; A and D cannot both be selected. Values add with no other effects. Which feasible selection maximizes value?

Logical Reasoning Interview Questions and Answers

Each answer begins with the conclusion, then works through the reasoning and a check or counterexample. Probability and decision answers depend on the explicitly stated sampling, independence, and objective assumptions. These are original exercises; the final references explain general principles rather than supplying the puzzle answers.

Logical Reasoning Interview Questions and Answers for Beginners

Follow the premises carefully and distinguish what must be true from what could be true.

1. Every urgent ticket is reviewed. Ticket K is urgent. What follows about K?

K is reviewed. The rule covers every urgent ticket, and K is explicitly included in that group. Apply the general rule to this particular ticket: urgent(K) implies reviewed(K), and urgent(K) is given, so reviewed(K) follows.

The conclusion is narrower than “K is resolved” or “K is reviewed immediately”; neither timing nor resolution is specified. A useful check is to try to make K urgent but unreviewed. That would directly violate the first premise, so there is no counterexample consistent with the facts.

2. If a backup completes, its indicator is green. The indicator is green. Must the backup have completed?

No. A green indicator is necessary under the stated rule for a completed backup, but the rule does not say it is sufficient. The indicator might also be green in another situation, such as a ready state.

Two possibilities fit the premises: the backup completed and the indicator is green; or it did not complete and the indicator is still green. Since both satisfy “if completed, then green,” completion is underdetermined. To infer completion, I would need the additional reverse rule that the indicator is green only when the backup has completed.

3. If a record is archived, it is read-only. Record R is not read-only. Can R be archived?

No; R is not archived. If it were archived, the rule would require it to be read-only. That conflicts with the explicit fact that it is not read-only.

This applies the contrapositive: archived implies read-only, so not read-only implies not archived. It does not establish whether R is editable by a particular person, because access rights are not specified. The check is direct: assigning both archived and not read-only makes the given conditional false, so that assignment is excluded.

4. A device connects only if it has a valid certificate. Device D connects. What must be true?

D must have a valid certificate. “Connects only if it has a valid certificate” means that a valid certificate is a necessary condition of connecting: connects implies certificate-valid. Because D connects, that necessary condition must hold.

The reverse inference is not available. A valid certificate does not guarantee that a device connects; power, network availability, or other requirements could still matter. I would check the direction by asking which combination the statement forbids. It forbids a connected device without a valid certificate, precisely the alternative to the conclusion here.

5. A file is accepted if and only if it passes both format and checksum checks. It passes format but fails checksum. Is it accepted?

No, the file is not accepted. The biconditional gives an exact acceptance condition: accepted is true precisely when both checks pass. Passing format alone is insufficient because the checksum requirement fails.

Represent the combined condition as format AND checksum. Its value is true AND false, which is false. The “only if” half already rules out acceptance when either required check fails; the “if” half ensures acceptance when both pass. This is stronger than a rule saying that passing both checks is merely one possible way to be accepted.

6. At least one of services A and B is running; both may run. A is not running. What follows about B?

B is running. The premise says that at least one service runs, so the allowed combinations are A alone, B alone, or both. The fact that A is not running removes the first and third possibilities, leaving B alone.

Here “or” is explicitly inclusive, not a claim that exactly one runs. Nevertheless, when one alternative is known to be false, the other must be true under either interpretation. As a check, assigning both services “not running” would contradict the requirement that at least one is running.

7. Exactly one of alerts A and B is active. A is active. Is B active?

B is inactive. “Exactly one” combines two constraints: at least one alert is active, and they cannot both be active. Since A already supplies the required active alert, B must be false to respect the second constraint.

The allowed states before learning A's status are active/inactive and inactive/active. Only the first remains after A is specified as active. This differs from an inclusive “A or B,” which would allow both to be active. Stating the interpretation of “or” is important whenever exclusivity changes the answer.

8. All analysts are employees. Some employees are mentors. Must some analysts be mentors?

No. The employees who are mentors might all be outside the analyst group. For example, let Alex be an analyst and employee but not a mentor, and let Blair be an employee and mentor but not an analyst. Both premises hold, yet no analyst is a mentor.

It is also possible that an analyst is a mentor, so the conclusion is underdetermined rather than necessarily false. The phrase “some employees” guarantees at least one employee-mentor, but does not identify that person's membership in a smaller group. A counterexample is enough to show that the proposed conclusion is not logically required.

9. No confidential items are public. Some reports are confidential. What follows about those reports?

Some reports are not public: specifically, the confidential reports whose existence is given. Choose one such report. Because it is confidential, the rule excluding confidential items from the public group applies to it.

The reasoning does not establish that every report is confidential or that every report is nonpublic. Other reports could be public without violating either premise. The existential statement supplies at least one report to which the universal exclusion applies, avoiding the mistake of assuming that a universal rule by itself guarantees that its subject group contains anything.

10. Is “Every submitted form is signed” contradicted by finding an unsigned draft that was never submitted?

No. The rule applies only to submitted forms. An unsigned draft outside that group is compatible with every submitted form being signed.

To contradict the rule, I need an object that is both a submitted form and unsigned. A signed submitted form supports the rule for that instance but does not prove it universally. An unsigned unsubmitted draft does neither. This distinction keeps the test focused on the actual condition: the relevant counterexample is submitted AND not signed, not merely not signed.

11. Of 30 staff, 18 use tool A and 14 use tool B; 8 use both. How many use neither?

Six staff use neither tool. First count staff using at least one tool: 18 + 14 − 8 = 24. The subtraction removes the second counting of the eight people included in both totals. Then subtract this union from the full staff count: 30 − 24 = 6.

A partition check gives 10 using only A, 6 using only B, 8 using both, and 6 using neither. These groups total 30 and reproduce the reported tool totals: 10 + 8 = 18 and 6 + 8 = 14. No independence assumption is needed.

12. Every premium account has support. No suspended account has support. Can an account be both premium and suspended?

No account can be both. A premium account must have support, while a suspended account cannot have support. Membership in both groups would therefore require support and not support simultaneously.

This conclusion says the groups are disjoint, not that every account belongs to one of them. An account could be neither premium nor suspended. The premises also do not guarantee that either group has members. Checking a proposed overlap is enough: set premium and suspended both true, and the two rules force incompatible support values.

13. Four jobs A, B, C, D occupy four consecutive slots, one per slot. A precedes B, B precedes C, and C precedes D. What is the order?

The order is A, B, C, D. A must precede B, which must precede C, which must precede D. These relations form a complete chain covering all four jobs.

Because each job occupies exactly one of the four slots, there is no unused slot or extra job that could interrupt the chain. A therefore takes slot 1, B slot 2, C slot 3, and D slot 4. Checking the adjacent comparisons confirms all three constraints. The conclusion would be less specific if additional unnamed jobs or empty slots were permitted, but they are explicitly excluded here.

14. Four speakers A, B, C, D each speak once. A is first, D last, and B speaks before C. Who speaks second?

B speaks second. A and D occupy slots 1 and 4, leaving slots 2 and 3 for B and C. Since B must speak before C, B takes slot 2 and C takes slot 3.

The complete schedule is A, B, C, D. Swapping the two middle speakers would violate the only remaining precedence condition. This illustrates a useful method for small scheduling problems: place fixed endpoints first, list the unfilled slots, and then apply relational constraints to the remaining people. No assumptions about speaking duration are required because only order is being asked.

15. Three distinct boxes, red, blue, and green, occupy positions 1–3 from left to right. Red is not first, green is not last, and blue is immediately right of red. What is their order?

The order is green, red, blue. Because blue is immediately right of red, the red-blue pair can occupy only positions 1–2 or 2–3. Red is not first, so positions 1–2 are excluded. Red must be second and blue third, leaving green first.

This arrangement also satisfies the separate condition that green is not last. Checking every condition matters even when one clue already identifies a candidate arrangement. Here all three conditions agree, and the two possible placements of the adjacent pair have been exhausted, establishing uniqueness.

16. Four tasks A, B, C, D each take one slot. A must precede C, B must precede C, and C must precede D. Which tasks can occupy the first slot?

A or B can be first, and no other task can. C needs both A and B before it, while D needs C before it, so neither C nor D can start the schedule.

The orders A, B, C, D and B, A, C, D both satisfy every condition. They demonstrate that each of A and B is genuinely possible rather than merely not ruled out by one clue. There is no rule ordering A relative to B, so selecting only one of them would invent a constraint. The answer is a set of possible first tasks, not one unique complete schedule.

17. Five people A, B, C, D, E stand in a row. B is immediately right of A, C immediately right of B, D immediately right of C, and E is at an end. What positions can A occupy?

A can occupy position 1 or position 2. The four consecutive people must appear as the block A, B, C, D, in that order. A four-person block in a five-person row can start only at position 1 or position 2.

If the block starts at 1, E occupies position 5. If it starts at 2, E occupies position 1. Both satisfy the requirement that E is at an end. Thus the two complete rows are A-B-C-D-E and E-A-B-C-D. The clues determine the internal block order but do not determine which end E takes.

18. Three people Ana, Bo, and Cy receive distinct roles: editor, reviewer, and tester. Ana is not editor or tester; Bo is not tester. Who is tester?

Cy is the tester. Ana cannot be editor or tester, so Ana must be reviewer. That leaves editor and tester for Bo and Cy. Since Bo cannot be tester, Bo must be editor, leaving tester for Cy.

The completed assignment is Ana–reviewer, Bo–editor, Cy–tester. It uses every role exactly once and respects all exclusions. The distinct-role condition is essential: without it, knowing Ana's role would not remove reviewer from the options for Bo and Cy. Writing down the remaining roles after each forced choice prevents that condition from being overlooked.

19. A one-hour meeting starts at 9, 10, or 11. Ana is available for starts at 9 or 10; Bo for starts at 10 or 11. Everyone must attend. When can it start?

The meeting can start at 10 only. Ana's possible starts are {9, 10}; Bo's are {10, 11}. Their common available start is the intersection, {10}.

At 9, Bo is unavailable, and at 11, Ana is unavailable. At 10, both are explicitly available for the one-hour meeting, so the proposed start is feasible. The problem states availability in terms of starts for this meeting, not isolated moments on a calendar, so there is no need to infer whether an available person remains free for the full hour.

20. Four distinct parcels P, Q, R, S are delivered one at a time. P is before Q, and R is before S. Must Q be before R?

No. Q may be before R, but that is not forced. The order P, Q, R, S satisfies both rules and places Q before R. The order R, S, P, Q also satisfies both rules and places Q after R.

The two constraints order P relative to Q and R relative to S, but do not connect the two pairs. These two valid schedules settle the question without listing every possible schedule. When testing whether an ordering is necessary, one valid counterexample disproves necessity; a second contrasting example helps show the remaining uncertainty clearly.

21. A and B are each either truth-tellers, whose statements are true, or liars, whose statements are false. A says, “B is a liar.” B says, “We are different types.” What are their types?

A is a liar and B is a truth-teller. Suppose A were truthful. Then B would be a liar, so they would be different types. But B's statement would then be true, contradicting B's supposed liar status.

Therefore A is a liar. A's claim that B is a liar must be false, so B is truthful. Their types are indeed different, making B's statement true as required. This checks both statements against the resulting assignment. The reasoning depends on the strict puzzle convention that every statement by a liar is false, not merely unreliable.

22. One key is in exactly one of boxes A, B, C. The labels say A: “The key is in B”; B: “The key is not in B”; C: “The key is not in A.” Exactly one label is true. Where is the key?

The key is in A. The first two labels are exact opposites, so exactly one of those two is true regardless of the location. To keep the total number of true labels at one, the third label must be false.

“The key is not in A” is false only when the key is in A. Check that location directly: A's label is false, B's label is true, and C's label is false. There is exactly one true label. If the key were in B or C, the third label would also be true, producing two true labels.

23. A and B are each truth-tellers or liars, with the usual always-true or always-false statement rule. A says, “We are both liars.” B makes no statement. What are their types?

A is a liar and B is a truth-teller. A cannot be truthful: a truthful A asserting that both are liars would assert, among other things, that A is a liar, a contradiction.

So A is a liar, and A's statement must be false. Since A really is a liar, the conjunction “A is a liar and B is a liar” can be false only if B is not a liar. Therefore B is truthful. B's silence supplies no extra statement to evaluate; the result follows entirely from A's compound claim and the specified two-type system.

24. Exactly one of three statements about an integer x is true: x > 5; x is even; x = 4. Given x is one of 2, 4, 6, 7, which values are possible?

The possible values are 2 and 7. For x = 2, the three statements have truth values false, true, false: exactly one is true. For x = 4, they are false, true, true, so there are two. For x = 6, they are true, true, false, also two. For x = 7, they are true, false, false, exactly one.

Thus the information narrows the candidates but does not identify one integer uniquely. Explicitly checking each member of the finite domain is sufficient here; there is no reason to infer an unstated rule selecting 2 over 7 or vice versa.

25. A sequence starts at 3, and every later term is twice the previous term plus 1. What is the fifth term?

The fifth term is 63. Starting with term 1 equal to 3, apply the stated recurrence four times: term 2 is 2 × 3 + 1 = 7; term 3 is 15; term 4 is 31; and term 5 is 63.

The key is to count the initial value as the first term, so only four updates are needed. A check is to add 1 to each term: 4, 8, 16, 32, 64, which doubles each time. Unlike an unexplained number list, this problem supplies the generating rule, making the requested next values determined.

26. A sequence starts at 20 and subtracts 3 at every step. What is its sixth term?

The sixth term is 5. The terms are 20, 17, 14, 11, 8, 5. Moving from the first to the sixth term requires five subtractions, not six, so the calculation is 20 − 5 × 3 = 5.

The explicit constant-step rule makes this an arithmetic sequence with first term 20 and common difference −3. As a check, each adjacent pair differs by −3, and the list contains six terms. Nothing requires the sequence to stop at zero; the requested term happens to remain positive, but later terms could be negative under the same rule.

27. A bag has 3 red and 2 blue tokens. Each individual token is equally likely to be drawn. What is the probability of drawing blue?

The probability is 2/5, or 40%. There are five individual tokens in total, and two of those five are blue. Because each individual token is equally likely, favorable outcomes divided by total outcomes gives 2/5.

The colors are not equally likely merely because there are two color names. The sample space consists of the five tokens, not two equally weighted categories. A complement check gives a red probability of 3/5; red and blue are the only possibilities, and 2/5 + 3/5 = 1. The calculation applies to this single draw.

28. A fair six-sided die is rolled once. Given that the result is even, what is the probability it exceeds 3?

The conditional probability is 2/3. Knowing the roll is even restricts the equally likely possibilities to {2, 4, 6}. Two of these, 4 and 6, exceed 3.

The denominator is therefore three, not six. Using the formula gives the same result: P(even and greater than 3) / P(even) = (2/6) / (3/6) = 2/3. The original die is fair, so the surviving even outcomes remain equally likely after conditioning. The result is not 1/2, which would be the unconditional probability of exceeding 3.

29. Two independent fair coins are tossed. What is the probability of exactly one head?

The probability is 1/2. The four equally likely ordered outcomes are HH, HT, TH, and TT. Exactly one head occurs in HT and TH, giving 2/4 = 1/2.

Independence and fairness make each ordered outcome have probability 1/4. The two mixed outcomes must be counted separately because the first and second coin can produce different results. Treating “two heads,” “one head,” and “no heads” as three equally likely outcomes would be incorrect: the middle category contains two of the four equally likely elementary outcomes.

30. A bag contains 4 red and 2 blue tokens. Two tokens are drawn uniformly without replacement. What is the probability both are blue?

The probability is 1/15. The first draw is blue with probability 2/6. After a blue token is removed, one blue remains among five tokens, so the probability that the second is blue given the first is blue is 1/5. Multiplying gives (2/6) × (1/5) = 1/15.

A combination check reaches the same answer: there is one pair consisting of the two blue tokens, among 15 possible unordered pairs from six tokens. Without replacement is crucial; the blue proportion changes after the first draw, so multiplying 2/6 by itself would not describe this experiment.

31. A fair six-sided die is rolled once. You receive 12 points for a 6 and 0 otherwise. What is the expected score?

The expected score is 2 points per roll. A 6 occurs with probability 1/6 and earns 12 points; the remaining outcomes collectively have probability 5/6 and earn zero. The weighted average is 12 × (1/6) + 0 × (5/6) = 2.

This does not mean that a single roll can produce 2 points: the actual score is either 0 or 12. Expectation describes an average under the stated probability model. As a simple check, summing the six equally likely face-specific scores gives 12; dividing by six also gives 2.

32. A team completes 12 equally sized units in 3 hours at a constant rate. With the same team and rate, how long do 20 units take?

The work takes 5 hours. The team rate is 12 units / 3 hours = 4 units per hour. At the same constant rate, completing 20 units takes 20 / 4 = 5 hours.

The units confirm the calculation: units divided by units per hour yields hours. A proportional check gives 20/12 times the original 3 hours, also 5. This answer relies on the stated equal unit sizes and unchanged constant rate. It does not infer that real teams always scale linearly; setup time, fatigue, or differing task complexity would require a different model if introduced.

33. A price of $100 rises by 20% and then falls by 20% of the increased price. What is the final price?

The final price is $96. The increase takes the price to $100 × 1.20 = $120. The subsequent decrease is 20% of $120, which is $24, leaving $96.

Equivalently, multiply the original price by 1.20 × 0.80 = 0.96. The two percentages do not cancel because they use different bases: the increase adds $20, while the decrease removes $24. The final result is a net 4% decrease from the original price, not a return to $100. Checking the monetary changes makes the base difference explicit.

34. Three equally weighted scores have an average of 18. Two scores are 15 and 21. What is the third?

The third score is 18. An average of 18 across three equally weighted scores means the total is 3 × 18 = 54. The two known scores sum to 15 + 21 = 36, leaving 54 − 36 = 18.

Checking the completed set gives (15 + 21 + 18) / 3 = 18. Equal weighting matters because an ordinary arithmetic average gives each score the same contribution. If the scores had different weights, their weighted total and those weights would be needed. No assumption about their order or a pattern between the scores is necessary.

35. For a positive integer x, is x even? Statement I: x is divisible by 4. Statement II: x is divisible by 3. Which statements, considered alone, are sufficient?

Statement I alone is sufficient; Statement II alone is not. Any integer divisible by 4 can be written as 4k = 2(2k), so it is necessarily even. That determines a “yes” answer without identifying x exactly.

Divisibility by 3 allows both odd and even positive integers. For example, x = 3 gives a “no” answer, while x = 6 gives a “yes,” and both satisfy Statement II. The two statements together are also sufficient, but the important classification is that I already suffices alone. Data sufficiency concerns the requested property, not necessarily the exact value.

36. For positive integers x and y, determine x. Statement I: x + y = 10. Statement II: x − y = 2. Is either statement alone sufficient, and are they sufficient together?

Neither statement alone is sufficient, but together they determine x = 6. Statement I permits, for example, (x, y) = (1, 9) and (2, 8), with different x values. Statement II permits (3, 1) and (4, 2), again with different x values.

Using both equations, add x + y = 10 and x − y = 2 to obtain 2x = 12, so x = 6. Substitution gives y = 4, which is a positive integer and satisfies both statements. The counterexamples establish each individual statement's insufficiency; the solved equations establish joint sufficiency.

37. A company introduces a voluntary workshop, and attendees later outperform non-attendees. Does this alone prove that the workshop caused the difference?

No. The observation establishes an association in this comparison, not the workshop's causal effect. Because attendance is voluntary, attendees may differ beforehand in motivation, experience, workload, or other factors that also influence performance.

A no-effect explanation is consistent with the facts: stronger performers choose to attend, and would have outperformed the others anyway. A beneficial workshop effect is also possible. Comparing pre-workshop performance helps assess one alternative, while a well-designed randomized invitation or assignment could provide stronger causal evidence under its assumptions. The correct conclusion is uncertainty about causation, not that the workshop definitely has no benefit.

38. A courier says, “Every parcel that leaves before noon arrives today. This parcel arrived today, so it must have left before noon.” Is the conclusion guaranteed?

No. The rule says that leaving before noon is sufficient for arrival today; it does not say that it is necessary. A parcel could leave at 1 p.m. and still arrive today without violating the rule.

The courier has reversed the implication: from before-noon implies today, and today, they infer before-noon. That inference is invalid without an additional premise excluding same-day arrival for later departures. The counterexample demonstrates the gap directly. The observed arrival remains useful evidence operationally, but it does not logically determine the departure time from the stated rule alone.

39. A manager argues, “The revised checklist will reduce errors because it adds a second review.” What additional premise connects the added review to the claimed benefit?

A connecting premise is that the second review will catch and correct at least some errors that would otherwise remain, without introducing enough new errors to offset that benefit. Merely adding a review step does not establish that it is effective.

For example, if the second reviewer repeats the same mistaken assumption and changes nothing, the extra review may leave errors unchanged. That counterexample shows why the operational mechanism matters. Even the connecting premise is an assumption needing evidence in practice; useful evidence would compare residual errors under the old and revised processes while accounting for other changes.

40. In one month, team A records 8 errors in 200 tasks and team B records 6 errors in 100 tasks. Which has the lower observed error rate, and does that establish the better process?

Team A has the lower observed error rate: 8/200 = 4%, compared with 6/100 = 6% for team B. Comparing the raw counts alone would reverse the impression because A performed twice as many tasks.

These rates do not by themselves establish that A has the better underlying process. The teams may handle tasks of different difficulty, and one month's observations contain sampling variation. A fair process comparison would clarify comparable task definitions and error recording, then examine uncertainty and relevant differences in workload. The numerical conclusion is definite for this dataset; the broader causal or quality conclusion is not.

Logical Reasoning Interview Questions and Answers for Intermediate

These solutions expose the case split, calculation, or missing assumption that determines the result.

41. A release is permitted only if both the security review and the rollback test pass. The rollback test failed. What follows about permission, and what remains unknown?

The release is not permitted under the stated rule; the security-review result remains unknown. Let P mean permission, S mean security passed, and R mean rollback passed. “Only if” gives P → (S and R), so permission requires R. Because R is false, P must be false.

This is a contrapositive inference, not an assumption that failing any test physically prevents deployment. The rule concerns permission. Both S=true and S=false remain compatible with the premises when permission is false. Checking those two possibilities shows why we cannot infer the security result. Likewise, even if both tests passed, this one-way rule would not itself establish permission: there might be additional requirements.

42. A report claims, “Every service has at least one tested recovery procedure.” What exactly would a counterexample have to show?

A counterexample must identify at least one service with no tested recovery procedure. Finding one untested procedure is not enough: that service might have a different procedure that has been tested.

The claim has the form: for every service s, there exists a procedure p belonging to s such that p is tested. Negating it changes both quantifiers: there exists a service s for which every procedure belonging to s is untested. A service with no recovery procedures also violates the claim, because no tested procedure exists for it. As a check, consider a service with two procedures, one tested and one untested. It satisfies the original claim despite the untested procedure, so that observation cannot refute the report.

43. Two reviewers each approved at least one of two proposals. Must there be a proposal approved by both reviewers?

No. The approvals can be split: reviewer A approves only proposal X, while reviewer B approves only proposal Y. Each reviewer then approved at least one proposal, but neither proposal has both approvals.

The premise is “for every reviewer, there exists an approved proposal.” The proposed conclusion is “there exists a proposal approved by every reviewer.” Exchanging the quantifiers changes the claim. To establish the conclusion, we would need additional information about overlapping approvals. It is also possible that both reviewers approved X, so the premise neither guarantees nor rules out common approval. These two compatible arrangements establish genuine underdetermination rather than merely a failure to find a common proposal in one example.

44. An alert is active if and only if sensor B or sensor C is active, where “or” is inclusive. The alert is active and B is inactive. What is C’s state?

Sensor C must be active. The equivalence says the alert and the expression “B or C” have the same truth value. An active alert therefore requires at least one active sensor. Since B is inactive, C is the only remaining way to satisfy that requirement.

Substitution makes the reasoning explicit: true ↔ (false or C) reduces to true ↔ C, so C=true. This is stronger information than the one-way rule “if a sensor is active, the alert is active,” which would allow other causes of an alert. The inclusive-or convention also permits both sensors to be active in other cases, but the separate premise that B is inactive excludes that possibility here.

45. Exactly one of features A, B, and C will be enabled. The rules are: if A is enabled, B must be enabled; if B is enabled, C must be disabled. Which features can be the sole enabled feature?

B alone and C alone are both allowed; A alone is not. Check the three possible one-feature configurations rather than assuming that the rules select a unique answer.

With A alone, the first rule requires B, contradicting the exactly-one condition. With B alone, the first rule has a false antecedent, and the second rule correctly requires C to be disabled. With C alone, both antecedents are false, so neither implication is violated. Thus the feasible set contains two configurations. The second rule does not say “C is disabled only when B is enabled,” and it does not require B to be enabled. Treating an implication as a command to make its antecedent true would incorrectly eliminate C.

46. A policy set says: every archived record is encrypted; record R is archived; record R is not encrypted. If the first two statements are retained, which statement must be withdrawn to restore consistency?

The statement “R is not encrypted” must be withdrawn. The universal rule applies to R because R is archived, so the first two statements jointly imply that R is encrypted. Keeping the contrary statement would assert both encryption and non-encryption for the same record in the same sense.

Removing that third statement is sufficient: a model in which R is both archived and encrypted satisfies the retained premises. Removing neither is impossible in ordinary classical two-valued logic. The instruction to retain the first two matters; without it, one could instead revise the general policy or the classification of R. This is a consistency exercise, not evidence about which operational record is inaccurate in a real system.

47. Five jobs A, B, C, D, and E run one at a time. A and B must both precede C; C precedes D; D precedes E. How many complete job orders satisfy all constraints?

There are exactly two orders: A, B, C, D, E and B, A, C, D, E. Since C must follow both A and B, and D and E must follow C in that order, A and B must occupy the first two positions.

The only freedom is which of A and B comes first. Their two permutations create two valid schedules, and the remaining three positions are fixed. Checking each listed schedule against all four precedence requirements confirms validity. There is no rule requiring A before B, so silently adding that order would remove a legitimate solution. Conversely, putting either A or B after C would violate a stated dependency, even if the rest of the schedule looked reasonable.

48. A, B, C, and D occupy four distinct shifts numbered 1 to 4, earliest first. D has shift 4, B is immediately before C, and A is not in shift 1. What is the assignment?

The unique assignment is B=1, C=2, A=3, and D=4. With D fixed at 4, the consecutive B–C pair can occupy positions 1–2 or 2–3. It cannot occupy 3–4 because that would place C in D’s shift.

If B–C occupied 2–3, A would be forced into shift 1, contradicting the rule about A. Therefore the pair occupies 1–2 and A takes the remaining shift 3. A final check confirms that all four shifts are distinct, D is last, B is immediately before C, and A is not first. The word “immediately” is essential: merely requiring B somewhere before C would leave additional valid assignments.

49. Five different boxes A, B, C, D, and E fill shelf positions 1–5 from left to right. D is at 1, E is at 5, C is immediately right of B, and A is left of B. What is the order?

The order must be D, A, B, C, E. With the endpoints occupied, A, B, and C fill positions 2–4. The B–C pair can be at 2–3 or 3–4.

If B were at 2 and C at 3, the only remaining place for A would be 4, which is not left of B. Therefore B is at 3 and C at 4, leaving A at 2. This arrangement satisfies every condition, so it is both feasible and unique. Notice that “A is left of B” need not mean immediately left in general; adjacency happens here because only three interior positions remain, not because the wording imposed it.

50. Three tasks A, B, and C receive different labels: red, blue, and green, each used once. A is not red, B is not blue, and C is green. Which label does each task receive?

A is blue, B is red, and C is green. Once C takes green, A and B must divide red and blue between them. A cannot take red, so it takes blue; B then takes red.

Checking the second exclusion confirms that B is not blue. It is useful to verify all constraints even when a subset already forces the answer: a remaining constraint could instead expose an inconsistent puzzle. Here the assignment uses each label exactly once and violates no rule. Without the distinct-label requirement, A and B could both be green, among other possibilities. The uniqueness therefore depends on the full statement, not solely on the two “not” clues.

51. Tasks A, B, C, and D take 2, 3, 4, and 1 hours respectively. A and B start at time 0 on separate workers; C starts only after both finish; D starts after C. With unlimited workers and no other delays, what is the earliest completion time?

The earliest completion time is 8 hours. A finishes at time 2 and B at time 3. Since C needs both predecessors finished, it starts at max(2,3)=3 and finishes at 7. D then runs from 7 to 8.

Adding all durations gives 10 hours, but that incorrectly treats A and B as sequential. The dependency chain through B, C, and D totals 3+4+1=8 hours and is the critical path. No schedule can finish earlier because those three tasks must occur in sequence. The proposed start times attain that lower bound, proving optimality. This result assumes the stated unlimited workers, simultaneous time-zero starts, and absence of setup or transfer time.

52. A directed network has only these edges and costs: A→B=3, A→C=2, B→C=1, B→D=4, C→D=6. What is the cheapest route from A to D?

The cheapest route is A→B→D, with total cost 7. The other possible routes are A→C→D, costing 2+6=8, and A→B→C→D, costing 3+1+6=10.

These exhaust the routes because the listed edges are the only allowed directed connections, and the network has no cycles. Comparing their totals proves that 7 is the unique minimum. Choosing the cheapest first edge, A→C, would be a locally attractive but globally inferior decision. A check on the full path is therefore necessary. The conclusion would need revisiting if edges were bidirectional or additional links existed; neither is permitted by the problem statement, so those alternatives should not be invented.

53. Each of A, B, and C is either a truth-teller, whose statement is true, or a liar, whose statement is false. A says “B is a liar”; B says “C is a liar”; C says “A and B are the same type.” What are their types?

A and C are liars; B is a truth-teller. Let a, b, and c represent truth-teller status. The first statement requires a=not b, and the second requires b=not c. Thus A and C have the same status, while A and B have opposite statuses.

C’s claim that A and B are the same type is consequently false, making C a liar. Then B’s claim that C is a liar is true, and A’s claim that B is a liar is false. This gives a=false, b=true, c=false. Checking each speaker’s status against the truth of that speaker’s own statement verifies consistency and shows that the initial relations leave no second solution.

54. A, B, and C are each either a truth-teller, whose statement is true, or a liar, whose statement is false. A says “Exactly one of us is a truth-teller”; B says “Exactly two of us are truth-tellers”; C says “A is a liar.” Can their types be determined uniquely?

No. There are two consistent assignments: A truthful with B and C lying, or A lying with B and C truthful. The statements do not choose between them.

If A is truthful, exactly one person is truthful. B’s claim of two is then false, and C’s claim that A lies is also false. If A lies, C’s claim is true. There is therefore at least one truthful person, C; but there cannot be exactly one, because that would make A’s statement true. B must consequently also be truthful, giving exactly two and satisfying B’s statement. These two cases exhaust A’s possible types. A count-based clue can therefore produce several internally consistent worlds rather than one answer.

55. A key is in exactly one of three drawers: red, blue, or green. Exactly two statements are true: A says “It is in red”; B says “It is in blue”; C says “It is not in red.” Which drawer contains it?

The blue drawer contains the key. A and C make contradictory statements, so exactly one of them is true regardless of the drawer. To reach the required total of two true statements, B must also be true.

Direct checking confirms this result. Red gives truth values true, false, false: only one truth. Blue gives false, true, true: two truths. Green gives false, false, true: only one truth. The blue case alone satisfies the count. The reasoning uses the guarantee of exactly one key location and exactly two true statements. It does not assume that a person who speaks falsely here would always lie in other conversations.

56. A, B, and C are each supposed to be either a truth-teller, whose statement is true, or a liar, whose statement is false. A says “B and C are the same type”; B says “A and C are different types”; C says “B is a liar.” Is any assignment consistent with all three statements?

No assignment is consistent. C’s statement requires C and B to have opposite types: when B is truthful, C’s statement is false; when B lies, C’s statement is true. Therefore A’s claim that B and C are the same type is false, making A a liar.

With A known to be a liar, B’s statement that A and C differ is true precisely when C is truthful. That requires B and C to have the same type. It contradicts the opposite-type relationship forced by C’s statement. The puzzle is thus unsatisfiable, not merely underdetermined. Assigning a convenient type to one speaker cannot repair it; at least one statement or the truth-teller/liar rules would need to change.

57. Exactly one of A, B, and C caused an outage. A says “B caused it”; B says “C caused it”; C says “B did not cause it.” Exactly one statement is true. Can the culprit be identified uniquely?

No. Either A or B could be the culprit; C cannot. If A caused the outage, the statements are false, false, true, satisfying the one-truth condition. If B caused it, they are true, false, false, also satisfying it.

If C caused it, the statements are false, true, true, violating the required count. This complete case split leaves two possible culprits, so identifying A alone or B alone would overstate the evidence. Another way to see the result is that A’s and C’s statements are opposites and already supply one truth. B’s statement must therefore be false, which rules out C without distinguishing A from B.

58. Among 70 employees, 40 use tool A, 35 use tool B, and 20 use both. How many use neither?

Fifteen employees use neither tool. The number using at least one is 40+35−20=55. The overlap is subtracted once because those 20 employees were counted in both original totals. Subtracting the union from the full group gives 70−55=15.

A disjoint-category check is useful: A only contains 40−20=20 people, B only contains 35−20=15, both contains 20, and neither contains 15. These four categories sum to 70 and reproduce both supplied tool totals. Adding 40 and 35 without adjusting for overlap would produce 75 tool users in a population of 70, a warning that the categories were not mutually exclusive.

59. In a population of 100, 50 belong to A, 45 to B, and 40 to C. Pairwise intersections, including members of all three, are |A∩B|=20, |A∩C|=15, and |B∩C|=18; all three contain 8. How many belong to none?

Ten people belong to none. Inclusion–exclusion gives the union as 50+45+40−20−15−18+8=90. Therefore 100−90=10 are outside all three sets.

The triple intersection is added back because its eight members were counted three times in the individual sets and subtracted three times in the pairwise intersections. They need one net count. For a feasibility check, the pair-only counts are 12, 7, and 10; the single-only counts are 23 for A, 15 for B, and 15 for C. Together with the eight triple members these sum to 90, with no negative category. The explicit intersection convention prevents interpreting the pair counts as excluding the triple group.

60. A team has four engineers and three designers, all distinct. How many unordered three-person committees contain at least one engineer and at least one designer?

There are 30 committees. There are 7 choose 3 = 35 unrestricted committees. Remove the four all-engineer committees and the one all-designer committee, leaving 35−4−1=30.

An independent count separates the valid compositions. Two engineers and one designer give (4 choose 2)×3=18 committees. One engineer and two designers give 4×(3 choose 2)=12. The total is again 30. We use combinations because a committee has no chair or ordered positions; exchanging two members does not create a different committee. The two valid composition categories do not overlap, and the excluded categories exhaust the ways a three-person committee can lack one of the required professions.

61. A bag contains three red and two blue tokens, all equally likely to be drawn. Two tokens are drawn uniformly without replacement. What is the probability of getting exactly one red?

The probability is 3/5. Among the 5 choose 2 = 10 equally likely unordered pairs, there are 3×2=6 pairs with one red and one blue. Hence the probability is 6/10=3/5.

An ordered calculation checks the result: red then blue has probability (3/5)(2/4)=3/10, and blue then red has probability (2/5)(3/4)=3/10. These disjoint outcomes sum to 3/5. The second-draw denominator is four because the first token is not replaced. Using the original red and blue proportions independently on both draws would describe sampling with replacement and would give a different probability. The order of the colors is irrelevant to the target event, but both orders must be included.

62. Two independent fair six-sided dice are rolled. Given that their sum is 8, what is the probability that at least one die shows 6?

The conditional probability is 2/5. The ordered outcomes with sum 8 are (2,6), (3,5), (4,4), (5,3), and (6,2). Each was equally likely before conditioning, so each has conditional probability 1/5 within this five-outcome set.

Two of these outcomes contain a 6, giving 2/5. It would be wrong to count only the unordered pairs {2,6}, {3,5}, and {4,4} as equally likely: the first two have two ordered realizations each, while the double has only one. Conditioning also changes the relevant sample space from 36 outcomes to five. A general probability for seeing a 6 on unrestricted dice therefore cannot be reused without accounting for the sum information.

63. For events A and B, P(A)=0.4, P(B)=0.5, and P(A or B)=0.7, with inclusive “or.” Are A and B independent?

Yes. Inclusion–exclusion gives P(A and B)=0.4+0.5−0.7=0.2. This equals P(A)P(B)=0.4×0.5=0.2, which is the independence condition for these two events.

The implied disjoint probabilities provide a check: A only is 0.2, B only is 0.3, both is 0.2, and neither is 0.3. They are nonnegative and sum to one. Independence is not the same as mutual exclusivity: these events overlap with probability 0.2. Indeed, because both have positive probability, making them mutually exclusive would make them dependent. The supplied probabilities establish independence for A and B as defined, not for every event or measurement in the wider process.

64. Ten percent of components are defective. A test flags 80% of defective components and 20% of nondefective components. A randomly selected component is flagged. What is the probability it is defective?

The probability is 4/13, approximately 30.8%. In an illustrative population of 1,000 components matching the stated rates, 100 are defective and 80 of those are flagged. Of the 900 nondefective components, 180 are flagged. Thus 80 of the 260 flagged components are defective.

Equivalently, Bayes’ rule gives (0.10×0.80)/(0.10×0.80+0.90×0.20)=0.08/0.26=4/13. The answer is not 80%: that is the probability of a flag given a defect, not the reverse conditional. The relatively large nondefective group contributes more false flags than true flags. The result assumes the stated base rate and test rates apply to the randomly selected component’s population.

65. Three independent backup checks each pass with probability 1/2. What is the probability that exactly two pass?

The probability is 3/8. There are three choices for which check fails, and each corresponding outcome has probability (1/2)³=1/8 by independence. The outcomes are disjoint, so their probabilities add to 3/8.

Listing them makes the count transparent: pass–pass–fail, pass–fail–pass, and fail–pass–pass. Including pass–pass–pass would answer “at least two” instead, which has probability 4/8. As a check, the probabilities for zero, one, two, and three passes are 1/8, 3/8, 3/8, and 1/8, summing to one. Independence is a substantive assumption here: identical pass rates alone would not justify multiplying probabilities if all checks were affected by a shared failure.

66. A sequence is defined by a₁=2 and aₙ₊₁=2aₙ+3 for every positive integer n. What is a₄?

The fourth term is 37. Apply the supplied recurrence three times: a₂=2×2+3=7, a₃=2×7+3=17, and a₄=2×17+3=37.

The recurrence, rather than a guess based on a short list, uniquely defines the answer. A useful check is to add 3 to each term: bₙ=aₙ+3 then satisfies bₙ₊₁=2bₙ, starting at b₁=5. Therefore b₄=5×2³=40 and a₄=40−3=37. This also checks the indexing: the fourth term requires three updates from the first term, not four. Without the explicit recurrence, many different rules could agree on a few starting values but disagree about what comes next.

67. Starting with a₁=5, create the sequence by alternately adding 3 and multiplying by 2, beginning with addition. What is a₅?

The fifth term is 38. The updates are a₂=5+3=8, a₃=8×2=16, a₄=16+3=19, and a₅=19×2=38.

The key is to track which operation applies at each transition, not to infer a uniform difference or ratio. From an odd-indexed term to the next even-indexed term, add 3; from an even-indexed term to the next odd-indexed term, multiply by 2. Checking the four transitions against that alternating rule confirms both the value and the indexing. Multiplying first would generate a different sequence, but the problem explicitly rules that out. Because the generation rule is supplied, this is a determinate calculation rather than an ambiguous “spot the pattern” exercise.

68. An integer x is between 1 and 10 inclusive. Statement I says x is even; statement II says x is divisible by 3. Which statements are sufficient to determine x: either alone, both together, or neither?

Both statements together are sufficient, but neither alone is sufficient. Statement I leaves x in {2,4,6,8,10}, while statement II leaves x in {3,6,9}. Their intersection is {6}, so jointly they determine x=6.

A data-sufficiency answer should distinguish the ability to solve from the actual value, so I would state both. The bound from 1 through 10 is essential: without it, every positive multiple of 6 would satisfy both statements. Also, each statement must first be assessed independently with the original range, not with the other statement silently assumed. The explicit candidate sets make that separation clear and verify that the combined premises are consistent rather than yielding an empty intersection.

69. A rectangle has positive side lengths and a strictly longer side. Statement I says its perimeter is 20; statement II says its longer side is 6. Which statements are sufficient to determine its area?

Both statements together determine the area, but neither does alone. From I, the side lengths sum to 10, yet examples 3 and 7 or 4 and 6 give different areas. From II alone, the shorter side could be 2 or 4, again giving different areas.

Combining I and II gives shorter side 10−6=4 and area 6×4=24. The sides are positive and 6 is indeed strictly longer than 4, so the solution satisfies the original conditions. An important check is that perimeter equals twice the sum of the side lengths, not the sum itself. Counterexamples for the individual statements establish insufficiency; solving the combined equations establishes sufficiency.

70. Machine A completes one batch in 6 hours and machine B completes an identical batch in 12 hours. They work at constant rates on a divisible batch without interfering. How long do they take together?

They take 4 hours. A completes 1/6 of a batch per hour and B completes 1/12. Their combined rate is 1/6+1/12=1/4 batch per hour, so one batch requires 1÷(1/4)=4 hours.

A direct contribution check confirms the result: in four hours A completes 4/6=2/3 of the batch and B completes 4/12=1/3, totaling one batch. Averaging the individual completion times would not represent simultaneous work; rates, rather than times, add. The answer depends on the explicit assumptions that the work is divisible, rates are constant, and the machines do not interfere. Setup constraints or an indivisible task could invalidate this simple combined-rate model.

71. A department of 20 employees has an average score of 70, and a separate department of 30 has an average score of 80. Every employee has one equally weighted score. What is the combined average?

The combined average is 76. The first department contributes 20×70=1,400 total score points, and the second contributes 30×80=2,400. Dividing their sum by the total headcount gives 3,800/50=76.

The simple average of the two department averages, 75, would weight the smaller and larger departments equally rather than weighting employees equally. A reasonableness check is that the combined average lies between 70 and 80 and is closer to 80 because the higher-scoring department contains more employees. The calculation assumes the groups are disjoint and that the individual scores share the same scale and weighting, as specified. Overlapping membership would require avoiding double-counting before combining totals.

72. A retailer discounts an original price by 15%, then adds a fixed delivery fee of 6. The final total is 108. What was the original price?

The original price was 120. First remove the fixed delivery fee: the discounted item price is 108−6=102. That amount is 85% of the original price, so the original is 102/0.85=120.

Substituting back checks the order of operations: a 15% discount on 120 removes 18, leaving 102; adding 6 gives the stated total of 108. Dividing the entire total by 0.85 would incorrectly reverse a discount on the delivery fee, which was added afterwards and was not discounted. Likewise, adding 15% to the discounted price would not recover the original because that percentage would use the wrong base. The fixed fee and the percentage change need to be undone separately, in reverse order.

73. A support team introduced a chatbot, and its average resolution time fell the next month. No control group or additional data are available. Does this establish that the chatbot caused the improvement?

No. The timing is compatible with a benefit, but it does not establish causation. The team might have received simpler tickets, added experienced staff, changed its measurement, or experienced a seasonal workload decline during the same period.

Those are alternative explanations, not claims that any particular one actually occurred. A stronger investigation would compare comparable tickets and periods, check staffing and case mix, and ideally use a randomized or otherwise credible comparison design. The current evidence supports the descriptive statement that the measured average fell after introduction. It does not identify what would have happened without the chatbot, which is the missing counterfactual needed for the causal claim. Equally, insufficient causal evidence is not proof that the chatbot had no effect.

74. A training experiment randomly assigns employees to two groups. Afterwards, 40% of the training group and 5% of the control group lack outcome measurements. The measured training group scores higher. Is the measured comparison automatically an unbiased estimate of the training effect?

No. Random assignment does not automatically protect a comparison restricted to people with observed outcomes when outcome data are missing differently across groups. If lower-performing trainees were especially likely to be unmeasured, the remaining training group could look stronger even without the claimed benefit.

The missingness rates alone do not prove bias, but they flag a mechanism that needs investigation. I would examine why outcomes are absent, preserve the original assignment groups in the analysis, and evaluate assumptions or sensitivity bounds for missing outcomes. The crucial distinction is between randomized assignment and the later selection of whose outcomes enter the comparison. A higher observed mean plus random assignment is not sufficient to ignore substantial differential attrition.

75. A manager argues, “Automating data entry will save ten staff-hours per week, so the team will finish more client projects.” Which missing assumption connects the saved time to the conclusion?

The argument needs the saved staff-hours to become usable capacity for project completion, rather than being blocked by another limiting step. For example, if projects are waiting on client approvals or specialist review, removing data-entry work may free time without increasing the number completed.

It also matters that automation does not create offsetting correction work. The stated time saving is an input change, while more completed projects is an output claim; a causal connection between them must be established. A useful check would trace which project tasks staff can actually advance with the freed hours and whether those tasks constrain throughput. This does not show the automation lacks value: reduced workload could be worthwhile even if project counts remain unchanged.

76. Two software plans provide identical required features and service. Plan A costs 100 upfront plus 10 per month; plan B costs 150 upfront plus 5 per month. There are no other costs or discounts. For exactly 12 months, which is cheaper and by how much?

Plan B is cheaper by 10. Plan A costs 100+12×10=220, whereas plan B costs 150+12×5=210. Since the required features and service are identical and no other costs apply, total price is the relevant comparison here.

The break-even point checks the direction: 100+10m=150+5m gives m=10 months. Before ten months A is cheaper, at ten they tie, and beyond ten B is cheaper. Twelve months lies beyond that threshold, consistent with the calculation. Comparing only the monthly fees would miss the upfront difference, while comparing only upfront prices would favor the wrong plan for the stated horizon. The answer should not be generalized to a different subscription length without recomputing.

77. A delivery route costs 10 and has a 20% chance of an additional delay penalty of 50. An alternative has a certain total cost of 22. If the sole objective is minimizing expected monetary cost, which route should be chosen?

Choose the first route, whose expected cost is 20. It incurs the base cost 10 in all cases and the extra 50 with probability 0.20, so its expectation is 10+0.20×50=20, below the alternative’s 22.

An outcome-based check gives the same result: 80% of the time the total is 10, and 20% of the time it is 60; 0.80×10+0.20×60=20. This does not mean the first route always costs less. It has a possible cost of 60, while the alternative is certain. A risk limit or nonlinear concern about delays could change a real decision, but the problem explicitly specifies expected monetary cost as the only objective.

78. A procurement choice has two possible future states. Supplier A delivers benefits of 8 in state 1 and 6 in state 2; supplier B delivers 7 and 5 respectively. Costs and every other relevant feature are equal, higher benefit is better, and the state probabilities are unknown. Can one supplier still be preferred?

Yes. Supplier A strictly dominates B because it provides one more unit of benefit in each possible state. No probability estimate is needed to prefer A under the stated objective and equal-cost assumptions.

If state 1 has probability p and state 2 has probability 1−p, A’s expected benefit is 8p+6(1−p)=6+2p, while B’s is 7p+5(1−p)=5+2p. Their difference is always one for every p between zero and one. The same pointwise comparison holds even without calculating expectations. Dominance is a strong conclusion here because all possible states and relevant differences are supplied. Hidden service risks, different costs, or additional states would require a broader comparison rather than being silently assumed away.

79. Process A succeeds on 81 of 90 easy cases and 2 of 10 hard cases. Process B succeeds on 9 of 10 easy cases and 18 of 90 hard cases. Does A’s higher overall success rate demonstrate better performance within either difficulty group?

No. Both processes succeed on 90% of easy cases and 20% of hard cases. A’s overall success rate is (81+2)/100=83%, while B’s is (9+18)/100=27%. The overall difference comes entirely from their different case mixes.

A handles mostly easy cases and B mostly hard cases. Comparing within each supplied difficulty stratum removes that compositional difference and reveals equal observed rates. This example does not establish that the processes are causally equivalent in every setting; unmeasured differences could still matter. It does show that the aggregate rate alone cannot support the stated claim of better within-group performance. Applying the same easy/hard weights to both processes would give equal weighted success rates.

80. There is a 40% chance of rain. Doing nothing costs 100 if it rains and 0 otherwise; a precaution costs 30 and eliminates rain damage. A perfect weather report is available for 5 before deciding. With expected cost as the only objective, should the report be purchased?

Yes. Without the report, doing nothing has expected cost 0.40×100=40, so the best decision is to take the precaution for 30. With perfect information, take the precaution only when rain is forecast and do nothing otherwise. The expected action cost becomes 0.40×30=12.

Adding the report’s price gives 12+5=17, which is 13 less than the best no-report cost. The gross value of perfect information is therefore 30−12=18, and its net value after the fee is 13. The baseline must be the best decision without information, not automatically doing nothing. This calculation relies on the stated perfection and timing of the report and on the precaution fully eliminating damage.

Logical Reasoning Interview Questions and Answers for Experienced

These solutions make the decision rule explicit and separate model-based conclusions from claims the available evidence cannot establish.

81. A release policy says: if deployment proceeds, approval exists; if approval exists, either the audit passed or an exception was signed; no exception was signed; and the audit did not pass. Must deployment be blocked?

Yes: deployment cannot proceed while all four statements remain true. Let D mean deployment proceeds, A approval exists, P audit passed, and E exception signed. The rules are D implies A and A implies P or E. Since both P and E are false, their disjunction is false. Therefore A must be false, and D must also be false.

This is a backward implication argument, not an assumption that approval alone guarantees deployment. A consistent assignment is D = A = P = E = false, so the premises are not contradictory. The conclusion means the stated policy rules exclude proceeding; it does not establish which operational process should create a missing approval. If the organization changes a premise, the conclusion must be recalculated.

82. A reviewer claims: “Every service has some backup operator; therefore one operator backs up every service.” Two services and two operators exist. Does the conclusion follow, and what is the smallest counterexample in this setting?

No. Assign operator Mira only to service A and operator Noah only to service B. Every service then has a backup, but neither operator backs up both. The premise allows the operator to vary with the service; the conclusion requires a single shared operator.

In notation, “for every service s, there exists an operator o” does not imply “there exists an operator o, for every service s.” Both services and both operators matter in this counterexample: with just one service and a nonempty backup assignment, some operator would cover every service. Checking the two rows of the assignment table verifies the premise, while checking each operator's column disproves the conclusion. Additional evidence of a common operator would be needed to establish the proposed staffing guarantee.

83. Three feature flags A, B, and C are Boolean. Exactly one of A and B is enabled; B is enabled if and only if C is disabled; and enabling A requires enabling C. List all valid configurations.

The valid configurations (A, B, C) are (enabled, disabled, enabled) and (disabled, enabled, disabled). The first rule makes A the negation of B. The second makes C the negation of B as well, so A and C always agree.

If B is disabled, both A and C are enabled, satisfying A's requirement. If B is enabled, both A and C are disabled, and the implication from A to C holds because its antecedent is false. These cases exhaust B's two possible values. The last rule therefore eliminates no additional configurations; it is redundant under the first two rules. “A requires C” should not be silently strengthened to “C requires A,” even though the other rules happen to establish that equivalence here.

84. A specification contains four requirements: R1 says A must be enabled; R2 says A implies B; R3 says B implies C; R4 says C must be disabled. Is the full set inconsistent, and is every requirement necessary for that inconsistency?

Yes. The full set forces A, then B, then C, while also requiring not C. All four requirements are necessary for this particular inconsistency: removing any one leaves a satisfying assignment.

Without R1, set A = B = C = false. Without R2, set A = true and B = C = false. Without R3, set A = B = true and C = false. Without R4, set all three flags true. Each assignment satisfies the remaining three requirements.

This establishes a minimal inconsistent set: no proper subset is inconsistent. It does not identify the requirement the business should discard. That decision needs intent and risk information rather than a logical preference for one rule. A useful review distinguishes diagnosing a conflict from choosing an authorized repair.

85. Policy P permits access exactly when approval A is present and either training T or an exemption E is present. Policy Q permits access exactly when “A and T” or E is present. Are the policies equivalent? Give a separating case.

They are not equivalent. An exemption with no approval separates them: set E = true, A = false, and T = false. Policy P denies access because approval is mandatory, whereas Policy Q permits it through the exemption branch.

Formally, P is A and (T or E), which distributes to (A and T) or (A and E). Q is (A and T) or E. Every case allowed by P is allowed by Q, but Q also accepts unapproved exempt cases. The distinction is therefore directional: Q is more permissive, not merely different in an arbitrary way.

Parentheses and the word “exactly” matter. If the statements only described sufficient conditions, unspecified permission paths could prevent a complete equivalence judgment. Here the complete Boolean definitions make the counterexample decisive.

86. A project must select exactly three distinct modules from A, B, C, D, and E. Selecting A requires B; selecting C requires D; B and D cannot both be selected; and E must be selected. Which selections are valid?

Only {A, B, E} and {C, D, E} are valid. Because E is mandatory, exactly two modules must be selected from A, B, C, and D. Selecting A consumes those two positions with A and B; selecting C similarly requires C and D.

If neither A nor C is selected, the remaining pair must be B and D, which is forbidden. A pair containing A without B, or C without D, violates its dependency. Selecting both A and C would require four non-E modules and also violate the B–D exclusion.

Both reported sets satisfy the exact size, dependency, exclusion, and mandatory-module rules. This explicit feasible-set check is safer than selecting a locally attractive module and discovering an impossible dependency later. No preference between the two valid options was supplied.

87. Exactly one of three reports is true: R1 says “The incident was caused by A”; R2 says “The incident was caused by A or B”; R3 says “The incident was not caused by C.” Exactly one of A, B, or C caused the incident. Which cause is possible?

No cause is possible under all the stated premises. If A caused it, all three reports are true: A is A or B, and A is not C. If B caused it, R1 is false while R2 and R3 are true, giving two true reports. If C caused it, all three reports are false. Thus the stated premises admit no cause.

The correct conclusion is that the reports and the “exactly one true” condition are inconsistent. Listing the truth counts gives A: three, B: two, C: zero; the required count one never occurs. A solver should not force a culprit merely because the question asks for one. The practical next step would be to check which premise was recorded incorrectly, without guessing a replacement.

88. Three reviewers are each either a truth-teller, whose statement is true, or a liar, whose statement is false. A says “B is a liar.” B says “C is a liar.” C says “A and B are different types.” What are their types?

A and C are truth-tellers, and B is a liar. Let a, b, and c indicate truthful types. The statements require a = not b, b = not c, and c = (a differs from b). The first relation already makes A and B different types, so C's statement is true and c is true.

Because C is truthful, B's claim that C is a liar is false; B is therefore a liar. A's claim that B is a liar is true, making A truthful. Substitution checks all three statements against their speakers' types.

The solution is unique: C must be truthful before any discretionary case choice remains. This puzzle assumes each speaker's type governs the single statement shown; it does not require assumptions about motives, knowledge, or how truthful people behave in real interviews.

89. Exactly one of four servers A, B, C, D failed. Reports say R1: “A or B failed”; R2: “B or C failed”; R3: “C or D failed”; R4: “A failed.” Exactly two reports are true. Which failures remain possible?

A, B, and C remain possible; D does not. For A, R1 and R4 are true, while R2 and R3 are false. For B, R1 and R2 are true. For C, R2 and R3 are true. Each of those possibilities gives exactly two true reports.

If D failed, only R3 is true, so D violates the stated report count. The correct result is a set of three candidates, not a single identified server. The rule “exactly two reports are true” constrains the truth pattern but does not tell us which reports are true.

A further observation distinguishing A, B, and C is needed for a unique diagnosis. This is a useful example of satisfying every given constraint while explicitly preserving uncertainty that the evidence does not resolve.

90. A manufacturing defect affects 2% of units. An alarm detects 90% of defective units and falsely alarms on 5% of nondefective units. A randomly selected unit alarms. What is the probability it is defective?

The probability is 18/67, approximately 26.87%. In a hypothetical 10,000-unit population with the stated proportions, 200 units are defective and 180 of them alarm. Of 9,800 nondefective units, 490 alarm. There are therefore 670 alarmed units, of which 180 are defective.

Equivalently, Bayes' rule gives (0.02 × 0.90) divided by [(0.02 × 0.90) + (0.98 × 0.05)]. The numerator is the defective-and-alarmed probability; the denominator includes both ways an alarm can occur.

The 90% detection rate is not the requested posterior probability. The low prior defect rate makes false alarms numerically important. This answer assumes the quoted rates apply to the population from which this unit was randomly selected, rather than a differently filtered group.

91. A defect has prior probability 10%. Two tests each have 80% sensitivity and a 10% false-positive rate. Both tests are positive. Can the posterior defect probability be calculated uniquely without knowing their conditional dependence?

No. The marginal test accuracies do not determine how often both tests are positive within defective and nondefective groups. Two compatible dependence models give different posteriors.

If the tests are conditionally independent given defect status, the joint positive rates are 0.80² = 0.64 for defective units and 0.10² = 0.01 for nondefective units. The posterior is 0.064/(0.064 + 0.009) = 64/73, approximately 87.67%.

If the tests are identical copies of the same result, their joint positive rates remain 0.80 and 0.10. The posterior is 0.08/(0.08 + 0.09) = 8/17, approximately 47.06%. Both models satisfy every stated marginal accuracy. Independence must therefore be supplied or justified; repeating a measurement cannot automatically be treated as independent evidence.

92. An investigation starts with prior odds of 1:4 for hypothesis H versus not H. Evidence E is three times as likely under H as under not H. What are the posterior odds and probability of H?

The posterior odds are 3:4, so the posterior probability of H is 3/7, approximately 42.86%. Updating odds multiplies the prior odds by the likelihood ratio: (1/4) × 3 = 3/4.

To convert odds into probability, divide the H portion by the total: 3/(3 + 4), not 3/4. A concrete compatible model helps check the result: let P(E given H) = 0.60 and P(E given not H) = 0.20. The prior probability of H is 0.20, so the two evidence masses are 0.12 and 0.16. Normalizing gives 0.12/0.28 = 3/7.

The likelihood ratio favors H, raising its probability from 20%, but it does not overcome the prior enough to make H more likely than not.

93. A support queue receives 30% of its tickets from system A and 70% from system B. Escalation rates are 20% for A and 5% for B. A randomly chosen escalated ticket is inspected. What is the probability it came from A?

The probability is 12/19, approximately 63.16%. System A contributes 0.30 × 0.20 = 0.060 of all tickets as escalations. System B contributes 0.70 × 0.05 = 0.035. Therefore the total escalation fraction is 0.095, and A's share within it is 0.060/0.095.

A check using 2,000 tickets gives 600 A tickets with 120 escalations, and 1,400 B tickets with 70 escalations. Of the 190 escalations, 120 came from A. The smaller source can dominate an outcome-selected sample when its outcome rate is much higher.

The answer differs from A's 30% share of the overall queue because the sampling rule selects escalated tickets, not arbitrary tickets. No causal claim about why A escalates more is implied.

94. Two independent fair coins are tossed once. An automated rule records the trial if at least one coin is heads. You inspect a uniformly chosen recorded trial. What is the probability both coins were heads?

The probability is 1/3. Before filtering, the ordered outcomes HH, HT, TH, and TT each have probability 1/4. The recording rule removes only TT, leaving three equally likely recorded outcomes; just HH has two heads.

The explicit recording mechanism matters. This is not the same as observing that a predetermined first coin is heads, which would leave HH and HT and produce probability 1/2. Nor does it involve a person selectively announcing one coin's result using an unspecified strategy.

Checking by conditional probability gives P(HH)/P(at least one head) = (1/4)/(3/4). The uniform selection among recorded trials preserves the conditional distribution created by the rule. Adding an outcome-dependent selection preference after recording could change the answer and would require another model.

95. A component fails during a shift with probability 0.04. An alarm sounds with probability 0.95 when it fails and 0.02 when it does not. What is the probability the component has not failed given that the alarm stayed silent?

The probability is 2352/2357, approximately 99.79%. A nonfailed component with a silent alarm has probability 0.96 × 0.98 = 0.9408. A failed component with a silent alarm has probability 0.04 × 0.05 = 0.002. The total silent-alarm probability is 0.9428.

The requested probability is therefore 0.9408/0.9428. The complementary failure probability is 0.002/0.9428, approximately 0.212%, which provides a second check. Silence sharply lowers the failure estimate but does not prove successful operation.

This calculation concerns the specified shift and component population. It does not imply that alarms on later shifts are independent, or that the same likelihoods apply after maintenance or a change in operating conditions. Those would be new modeling assumptions, not consequences of this update.

96. A forecast system assigns probability 0.80 to each of 100 binary events. Exactly 80 events occur. Is it perfectly calibrated on this probability group, and what is its mean binary Brier score, defined as the average of (p − y)² with y equal to 1 for occurrence and 0 otherwise?

It is exactly calibrated on this observed group, and its mean Brier score is 0.16. Calibration here compares the common predicted probability, 0.80, with the group's observed occurrence fraction, 80/100 = 0.80.

Using the binary Brier convention (p − y)², each of the 80 occurrences contributes (0.80 − 1)² = 0.04, while each of the 20 nonoccurrences contributes (0.80 − 0)² = 0.64. The mean is (80 × 0.04 + 20 × 0.64)/100 = 0.16.

This does not establish perfect forecasting skill, calibration on other probability groups, or future calibration. A constant forecast can match a group's base rate without distinguishing its individual events. The observed calibration statement is deliberately limited to this group and sample; statistical uncertainty about a broader process remains.

97. Two services have next-hour outage probabilities 0.10 and 0.20. Their dependence is unknown. What are the tight bounds on the probability at least one service is down?

The probability lies between 0.20 and 0.30, inclusive. The union formula gives P(A or B) = 0.10 + 0.20 − P(A and B). The overlap cannot be negative and cannot exceed the smaller marginal probability, 0.10. Substituting those extremes gives the bounds.

Both endpoints are achievable. For the lower endpoint, every A outage occurs during a B outage, so the union is just B's 0.20. For the upper endpoint, the outage events are disjoint, so their probabilities add to 0.30.

An independence assumption would yield 0.28, but independence was not stated and is not required to obtain useful bounds. The range is tight because valid joint distributions attain both limits; it is not merely a conservative guess or confidence interval.

98. Treatment A succeeds in 81 of 90 low-risk cases and 1 of 10 high-risk cases. Treatment B succeeds in 19 of 20 low-risk cases and 16 of 80 high-risk cases. Which treatment has the higher observed success rate within each group and overall, and does that identify a causal winner?

B has the higher observed rate in both risk groups, while A has the higher overall rate. In low-risk cases, A achieves 90% versus B's 95%. In high-risk cases, A achieves 10% versus B's 20%. Overall, A achieves 82/100 = 82%, whereas B achieves 35/100 = 35%.

The reversal occurs because A's cases are predominantly low risk and B's predominantly high risk. Aggregating with different group weights can reverse the within-group comparison. The arithmetic is consistent, not evidence that one of the rates is erroneous.

These observational counts alone do not identify a causal winner. Treatment selection, remaining differences within risk groups, and outcome measurement may confound comparisons. The defensible conclusions are the reported descriptive comparisons and the need for an appropriate design or explicit causal assumptions before recommending a treatment.

99. A randomized pilot assigns 100 customers to a new workflow and 100 to control. Only 60 assigned customers use the workflow; 45 of those succeed. Of the 40 nonusers in that arm, 15 succeed. The control arm has 50 successes. What is the intention-to-treat success-rate difference, and should the 75% user rate replace it?

The intention-to-treat difference is 10 percentage points: (45 + 15)/100 − 50/100 = 60% − 50%. All customers are analyzed in their assigned arms, including those who did not use the new workflow.

The 45/60 = 75% user success rate should not replace that comparison as the randomized estimate. Choosing to use the workflow occurs after assignment and may be related to motivation, readiness, or other outcome predictors. The user-only group is therefore no longer protected by the original randomization in the same way.

With valid randomization, comparable outcome measurement, and no relevant interference, the assigned-arm comparison supports an effect of offering the workflow in this pilot setting. It does not by itself estimate the effect of forcing every customer to use it. Sampling uncertainty also remains even though the observed difference is exactly calculable.

100. In a 100-customer trial arm, outcomes are known for 80 customers: 48 succeeded and 32 failed. The other 20 outcomes are completely unknown. Without assumptions about missingness, what are the sharp bounds on the arm’s success rate?

The success rate is between 48% and 68%, inclusive. There are already 48 known successes. If every missing outcome is a failure, the arm has 48 successes out of 100. If every missing outcome is a success, it has 68 out of 100. Both assignments satisfy all the observed data.

The complete-case rate, 48/80 = 60%, falls inside this interval, but treating it as the full-arm rate requires an assumption connecting observed and unobserved customers. The question explicitly supplies none.

These are identification bounds, not a sampling confidence interval. They describe what the missing records permit within this actual group. A model for missingness or recovered outcomes could narrow the interval; a larger sample with the same unknown outcome fraction would not automatically eliminate this missing-data ambiguity.

101. A portfolio has 20 projects in unit A with an average cost of $120,000 and 80 projects in unit B averaging $70,000. What is the portfolio average, and can averaging the two unit averages give the right answer here?

The portfolio average is $80,000 per project. Unit A contributes 20 × $120,000 = $2.4 million, and unit B contributes 80 × $70,000 = $5.6 million. Dividing the combined $8 million by 100 projects gives $80,000.

The unweighted mean of the two unit averages is $95,000, which is not the project-level average because the units contain different numbers of projects. The correct weights are 20/100 and 80/100. The result must also lie closer to $70,000 than to $120,000 because unit B accounts for most projects.

An unweighted average would answer a different question: the mean of the two unit-level averages when each unit receives equal weight. Naming the unit of analysis prevents that valid but different statistic from being substituted for the requested one.

102. In a stipulated causal model, randomized training T increases checklist use M, and M reduces errors Y; there is no other path from T to Y and no confounding. An analyst controls for M and finds no remaining training effect. Does that show training has no total effect?

No. In the stipulated model, checklist use is the mechanism carrying training's effect. Controlling for M holds that mechanism fixed, so the remaining direct effect can be zero even while the total effect is nonzero.

For example, suppose training raises checklist use from 20% to 80%, and error risk is 10% without a checklist versus 2% with one. Without training, error risk is 0.20 × 0.02 + 0.80 × 0.10 = 8.4%. With training it is 0.80 × 0.02 + 0.20 × 0.10 = 3.6%. The total reduction is 4.8 percentage points, despite no direct path.

The calculation illustrates the stated model rather than proving it from observed data. The analyst must distinguish a total-effect question from a direct-effect question before selecting adjustment variables.

103. Two teams have known mean completion times of 4 and 10 hours. Statement I says they completed 30 jobs in total. Statement II says the first team completed twice as many jobs as the second. Which statements are sufficient to determine the combined job-level mean?

Statement II alone is sufficient; statement I alone is not. If the first team completed twice as many jobs, its weight is 2/3 and the second team's is 1/3, so the combined mean is (2 × 4 + 10)/3 = 6 hours. The absolute number of jobs is unnecessary.

Knowing only that there were 30 jobs does not determine the split. A 15–15 split gives 7 hours, while a 20–10 split gives 6 hours; both satisfy statement I and the stated team means. Together, the statements determine counts of 20 and 10, but that adds no information needed for the mean beyond the ratio.

This assumes each team's reported mean is calculated over its own completed jobs and both teams completed at least one job, as required for those means to exist.

104. Positive integers x and y count two kinds of item. Statement I says x + y = 11. Statement II says 3x + 5y = 43. Which statements are sufficient to determine x uniquely?

Both statements together are sufficient, but neither alone is. Subtract three times statement I from statement II: 2y = 10, so y = 5 and x = 6. Substitution gives 6 + 5 = 11 and 18 + 25 = 43.

Statement I alone admits many positive integer pairs, including (1, 10) and (6, 5). Statement II alone also has multiple positive integer solutions: (11, 2), (6, 5), and (1, 8) all produce 43. Thus the positivity and integrality restrictions do not make statement II sufficient by itself.

The unique joint solution follows from two independent linear constraints. Testing integer examples is important here: an equation that looks underdetermined over the reals can sometimes have only one feasible integer solution, but this particular equation does not.

105. A risk-neutral manager chooses one project. Project A pays a net $140,000 with probability 0.40 and loses $20,000 otherwise. Project B pays a certain net $35,000. Which maximizes expected money?

Project A maximizes expected monetary value: 0.40 × $140,000 + 0.60 × (−$20,000) = $44,000, compared with B's certain $35,000. The expected-value advantage is $9,000.

The losses are already net amounts, so no additional project cost should be subtracted. A check is that $44,000 lies between A's two possible outcomes and is much lower than its successful payoff, reflecting the 60% loss probability.

This conclusion depends on the explicit risk-neutral, expected-money objective. It does not say A is certain to outperform B in this single decision. If cash constraints, nonlinear utility, or a binding loss limit were added, they could change the preferred choice. They are not part of the supplied problem, so introducing them silently would replace rather than solve the stated decision.

106. A launch yields net profit of $90,000 if successful and a net loss of $30,000 otherwise. Not launching yields zero. For a risk-neutral decision-maker, what success probability makes launching preferable, and what happens at the threshold?

Launching is preferable when the success probability p exceeds 0.25. Its expected net payoff is 90,000p − 30,000(1 − p) = 120,000p − 30,000. Setting this greater than zero gives p > 1/4.

At p = 0.25, the expected payoff is exactly zero, so launching and not launching tie under the stated objective. Below that threshold, not launching has the higher expected value. Distinguishing strict preference from indifference avoids incorrectly including the break-even point in the preferred region.

The calculation assumes the listed amounts include all consequences valued by the decision-maker and that expected money is the selection criterion. A fixed additional launch fee would shift the threshold; a sunk fee already paid and unrecoverable would not belong in this forward-looking comparison.

107. Demand will be high with probability 0.30 and low otherwise. Expanding produces net payoffs of $100,000 in high demand and −$20,000 in low demand. Not expanding produces zero. What is the maximum a risk-neutral firm should pay for perfect demand information before deciding?

The expected value of perfect information is $14,000. Without information, expansion has expected payoff 0.30 × $100,000 + 0.70 × (−$20,000) = $16,000, so it is better than doing nothing.

With perfect information, the firm expands only when demand is high and does nothing when demand is low. Its expected payoff before buying information becomes 0.30 × $100,000 = $30,000. The improvement over the best uninformed decision is $30,000 − $16,000 = $14,000.

At an information price of exactly $14,000, the firm is indifferent under the expected-money objective; it strictly benefits only below that price. Comparing $30,000 with zero would overstate information value by ignoring the profitable decision already available without the forecast.

108. Demand is high with probability 0.30. Expansion pays $100,000 if high and −$20,000 if low; doing nothing pays zero. A forecast is positive with probability 0.80 when demand is high and 0.20 when low. What is its expected decision value before its price?

The forecast's expected value of sample information is $5,200. A positive forecast has joint high-demand mass 0.30 × 0.80 = 0.24 and low-demand mass 0.70 × 0.20 = 0.14. Expanding after a positive result contributes expected payoff $24,000 − $2,800 = $21,200 across all cases.

For a negative forecast, the corresponding masses are 0.06 and 0.56. Expansion would contribute $6,000 − $11,200, a negative amount, so the optimal action after a negative is not to expand. The informed strategy therefore yields $21,200 before the forecast price.

Without information, expansion yields $16,000. The gain is $21,200 − $16,000 = $5,200, the maximum break-even price. This is below the $14,000 value of perfect information for the same payoffs, providing a useful consistency check.

109. Three plans have losses across states X, Y, Z: A = (2, 9, 4), B = (6, 6, 6), and C = (8, 3, 7). No state probabilities are supplied. Under the minimax-loss rule, which plan is chosen?

Plan B is chosen because its worst possible loss is 6, lower than A's 9 and C's 8. The minimax-loss rule first takes each plan's maximum loss and then selects the plan with the smallest such maximum.

This is not an expected-loss calculation. Without probabilities, averaging the three entries would silently assume equal state weights, an assumption the problem does not supply. The answer is unique under the specified rule because 6 is strictly below both competing worst-case losses.

The decision is deliberately conservative: it protects against the most damaging listed state for each plan. It does not establish that B is best under every attitude toward uncertainty. For example, a decision-maker with credible probabilities could reasonably use a different criterion and obtain a different result.

110. Three plans have losses across states X, Y, Z: A = (0, 8, 8), B = (4, 5, 5), and C = (9, 0, 6). Under minimax regret, which plan is chosen?

Plan B is chosen. The smallest achievable loss in each state is 0 for X, 0 for Y, and 5 for Z. Subtracting these statewise benchmarks gives regrets A = (0, 8, 3), B = (4, 5, 0), and C = (9, 0, 1).

Their maximum regrets are 8, 5, and 9 respectively, so B has the smallest worst-case regret. Regret measures the loss above what the best plan would have achieved if that state had been known in advance. It is not identical to the original loss.

The subtraction must be done within each state, not relative to a single global minimum across the table. No probabilities are needed because minimax regret is the stated criterion. The strict gap between the three maximum regrets makes the selected plan unique.

111. Plans A and B yield net payoffs in three possible states: A = (7, 4, 9) and B = (6, 4, 5). The state probabilities are unknown. What comparison is justified for a decision-maker who always prefers more money?

A weakly dominates B: it pays at least as much in every state and strictly more in the first and third states. Therefore B cannot give a higher payoff than A in any listed state, and A has expected payoff at least as large under every probability distribution over those states.

A strictly higher expected payoff requires positive probability on at least one state where its advantage is strict. If the second state has probability one, both plans pay 4 and their expected payoffs tie. Thus unknown probabilities do not justify claiming a strictly positive expected advantage in every possible distribution.

This comparison assumes the plans differ only in the stated monetary outcomes and that larger payoff is preferred. Hidden costs, timing, or nonmonetary outcomes would require adding relevant consequences to the table before applying dominance.

112. A firm can stop now for zero or pay $10,000 for a pilot. The pilot reveals with certainty whether demand is high (probability 0.40) or low. After the pilot, launching pays an additional net $60,000 if high or loses $30,000 if low; stopping pays zero additional. What is the optimal strategy and expected payoff?

Run the pilot, launch if demand is high, and stop if demand is low. The pilot cost is paid in either outcome. With the optimal contingent action, expected payoff is −$10,000 + 0.40 × $60,000 + 0.60 × $0 = $14,000, exceeding the zero available by stopping now.

The decision is solved backward. Once a high-demand result is known, launching beats stopping; once a low-demand result is known, stopping avoids the additional loss. Then compare the expected payoff of entering that decision tree with the immediate outside option.

The $10,000 becomes sunk after the pilot, so it should not distort the subsequent launch comparison. It must still be included when deciding whether to run the pilot initially. Launching without a pilot was not offered as an available initial action in this problem.

113. Tasks A and B take 3 and 2 hours and share one specialist, so they cannot overlap. Task C takes 4 hours, needs another worker, and starts only after A finishes. Task D takes 1 hour and starts only after B and C finish. All tasks are nonpreemptive; workers are otherwise available at time zero. What is the minimum completion time?

The minimum completion time is 8 hours. Schedule A from time 0 to 3, then B from 3 to 5 on the specialist. Schedule C from 3 to 7 on the other worker, and D from 7 to 8 after both prerequisites finish.

The dependency chain A → C → D already requires 3 + 4 + 1 = 8 hours, so no schedule can finish earlier. The proposed schedule attains that lower bound, proving optimality rather than merely supplying a feasible plan.

If B uses the specialist first, A cannot finish until time 5, C finishes at 9, and D at 10. The order of a shared resource therefore matters even though B itself is not on the longest dependency chain in the optimal schedule. Task durations and availability are treated as deterministic.

114. Five distinct checks A, B, C, D, E run sequentially in any order, but A must precede C, B must precede C, and D must precede E. How many valid orders exist?

There are 20 valid orders. Among the three checks A, B, and C, C must be last, while A and B can appear in either order. Thus two of their six relative orders are valid. Among D and E, one of the two relative orders is valid.

Starting from 5! = 120 complete orders, these independent relative-order constraints retain a fraction (2/6) × (1/2), giving 20. An equivalent constructive count chooses three of the five positions for A, B, C in 10 ways, fills them in either ABC or BAC order, and fills the remaining positions as DE.

The constructive argument checks that no order is counted twice and no valid order is missed. The word “precedes” means earlier somewhere in the sequence, not necessarily immediately before.

115. Among 100 cases, 50 have tag A, 45 tag B, and 40 tag C. Pairwise overlaps, including any triple-tagged cases, are 20 for A and B, 15 for A and C, and 10 for B and C. Exactly 5 have all three tags. How many have exactly one tag and how many have none?

Exactly 60 cases have one tag, and 5 have none. The single-tag counts are A only = 50 − 20 − 15 + 5 = 20; B only = 45 − 20 − 10 + 5 = 20; and C only = 40 − 15 − 10 + 5 = 20.

Inclusion–exclusion gives the union size as 50 + 45 + 40 − 20 − 15 − 10 + 5 = 95, leaving 5 untagged cases. As a separate check, the exactly-two counts are 15, 10, and 5; these sum to 30. Then 60 single-tagged + 30 double-tagged + 5 triple-tagged + 5 untagged = 100.

Subtracting the triple overlap from each reported pair is essential because the question explicitly says pairwise counts include it. Treating those counts as “exactly two” would answer a different problem.

116. A system fails if both redundant units fail. A common-cause event occurs with probability 0.02 and makes both fail. Conditional on no common-cause event, the units fail independently with probability 0.10 each. What is the system failure probability?

The system failure probability is 0.0298, or 2.98%. Partition by the common-cause event. When it occurs, system failure is certain, contributing 0.02. When it does not occur, both units fail with conditional probability 0.10 × 0.10 = 0.01, contributing 0.98 × 0.01 = 0.0098.

Adding the mutually exclusive paths gives 0.02 + 0.0098 = 0.0298. This differs from simply multiplying unconditional unit-failure probabilities. Each unit's marginal failure probability is 0.02 + 0.98 × 0.10 = 0.118, but the shared cause makes their failures dependent.

The independence statement applies only within the no-common-cause condition. Carrying it outside that condition would erase the very dependency that redundancy must account for. The model assumes no additional system-failure mechanism beyond both units failing.

117. Three simultaneous jobs A, B, and C each need a different worker. A can be done only by workers 1 or 2; B only by workers 1 or 2; C only by workers 2 or 3. Which worker must take C, and how many feasible assignments exist?

Worker 3 must take C, and there are two feasible assignments. Jobs A and B together need two distinct workers but can draw only from {1, 2}; they therefore occupy both workers 1 and 2. That leaves worker 3 for C.

The two complete assignments are A–1, B–2, C–3 and A–2, B–1, C–3. Both respect every eligibility and simultaneous-distinct-worker constraint. Assigning worker 2 to C would leave only worker 1 available for both A and B, making completion impossible.

This illustrates why local eligibility is insufficient: worker 2 is individually qualified for C, yet choosing that assignment blocks the full set of jobs. A valid allocation must consider capacity across groups of jobs, not just whether each isolated job has at least one qualified worker.

118. Tasks A and B start at time zero and take 4 and 6 days. C takes 3 days after A; D takes 2 days after both A and B; E takes 2 days after both C and D. Unlimited workers are available. What is the earliest finish and how much may A be delayed without delaying the project?

The earliest project finish is day 10, and A may be delayed by at most 1 day. Initially A finishes at 4 and B at 6. C finishes at 7; D starts at 6 and finishes at 8. E then runs from 8 to 10.

The path B → D → E takes 6 + 2 + 2 = 10 days and is critical. The path A → C → E takes 9 days, leaving 1 day of slack for A. The alternative A → D → E path is shorter.

If A starts at day 1, it finishes at 5, C at 8, and D remains on its original 6-to-8 schedule, so E still finishes at 10. Any greater delay makes C finish after day 8 and pushes E later. This assumes A's duration remains unchanged.

119. A batch contains 8 passing units and 4 defective units. Three distinct units are sampled uniformly without replacement. What is the probability the sample contains exactly one defective unit?

The probability is 28/55, approximately 50.91%. There are C(12, 3) = 220 equally likely unordered samples. A sample with exactly one defective unit selects one of the four defective units and two of the eight passing units, giving C(4, 1) × C(8, 2) = 4 × 28 = 112 favorable samples.

Dividing gives 112/220 = 28/55. As an ordered-draw check, place the defective unit in any of three positions: 3 × (4/12) × (8/11) × (7/10), which gives the same probability.

The changing denominators reflect sampling without replacement. A binomial calculation using an unchanged defect probability of 1/3 on each draw would assume replacement or independent draws and would not be exact for this finite batch.

120. A portfolio has a budget of 8 units. Projects A, B, C, D cost 4, 3, 2, 2 units and deliver values 7, 5, 4, 3 respectively. Each can be selected at most once; C requires B; A and D cannot both be selected. Values add with no other effects. Which feasible selection maximizes value?

There are two optimal selections: {A, B} and {B, C, D}, each with value 12. The cost of {B, C, D} is 3 + 2 + 2 = 7, within budget, and their total value is 5 + 4 + 3 = 12. C's dependency is satisfied and the A–D exclusion is respected.

To prove optimality, split on whether A is selected. With A, D is forbidden. Adding B gives cost 7 and value 12; adding C as well would cost 9 and exceed budget. Without B, C is forbidden, so A alone has value 7. Without A, the full set B, C, D is feasible and maximizes the nonnegative additive values.

Therefore there are two optimal selections, {A, B} and {B, C, D}, both valued at 12. If the question requires a unique choice, the data are insufficient without a tie-breaker; both cost 7, while minimizing the number of selected projects among maximizers would favor {A, B}.

Logical Reasoning Documentation and Further Reading

Problems use stated assumptions and illustrative data. Local enumeration and arithmetic checks validate the specified models, not real-world forecasting accuracy or the validity of a hiring assessment.

These 15 primary educational and scholarly resources cover the principles behind the exercises: formal and informal logic, discrete mathematics, probability, causal models, and decision theory. They are further reading, not sources of the original problem wording or numerical answers.

• Stanford Encyclopedia of Philosophy — Classical Logic

• Stanford Encyclopedia of Philosophy — Informal Logic

• Stanford Encyclopedia of Philosophy — The Logic of Conditionals

• Stanford Encyclopedia of Philosophy — Interpretations of Probability

• Stanford Encyclopedia of Philosophy — Decision Theory

• Stanford Encyclopedia of Philosophy — Causal Models

• OpenStax — Probability Terminology

• OpenStax — Independent and Mutually Exclusive Events

• OpenStax — Two Basic Rules of Probability

• OpenStax — Contingency Tables

• OpenStax — Mean or Expected Value and Standard Deviation

• MIT OpenCourseWare — Mathematics for Computer Science

• MIT OpenCourseWare — Probabilistic Systems Analysis and Applied Probability

• OpenStax — Frequency, Frequency Tables, and Levels of Measurement

• Stanford Encyclopedia of Philosophy — Bayes’ Theorem