Question 1

Logical reasoning — Pictures, storybooks, and words (choose the statement that must be true) Facts:  Pictures can tell a story.  All storybooks have pictures.  Some storybooks have words. Which statement must also be a fact?

Accepted Answer

Introduction / Context: The problem mixes existence statements (“some”) with universals (“all”). We must identify a conclusion that holds in every scenario consistent with the premises, without adding assumptions like “better than” or “very simple,” which are not given. Given Data / Assumptions: Pictures can tell a story. Every storybook has pictures. Some storybooks also contain words. Concept / Approach: Because all storybooks have pictures, any storybook that has words must have both words and pictures. Statements comparing pictures and words by quality or implying simplicity are not grounded in the premises. Step-by-Step Solution: I: “Pictures can tell a story better than words can.” No comparative claim is provided. Not a must.II: “The stories in storybooks are very simple.” No statement about complexity is given. Not a must.III: “Some storybooks have both words and pictures.” Since all storybooks have pictures and some have words, those “some” necessarily have both. III is a must. Verification / Alternative check: Let S be the set of storybooks, W ⊆ S the subset with words. Because every element of S has pictures, every element of W has pictures too. If W is nonempty (given), then “some storybooks have both” is guaranteed. Why Other Options Are Wrong: I only / II only: Introduce unsupported comparisons or qualities. None of the statements is a known fact.: Incorrect because III is compelled. I and III only: Includes I, which is not given. Common Pitfalls: Adding value judgments or comparative claims not present in the facts; forgetting that “all S have P” combined with “some S have Q” implies “some S have both P and Q.” Final Answer: III only

Question 2

Logical reasoning — Pens, colors, and writing (choose the statement that must be true) Facts:  Some pens don't write.  All blue pens write.  Some writing utensils are pens. Which statement must also be a fact?

Accepted Answer

Introduction / Context: This question requires strict adherence to the given facts without importing everyday knowledge (e.g., “pens are writing utensils”). Although that is usually true, statement 3 only says “Some writing utensils are pens,” which is weaker than “All pens are writing utensils.” We must avoid assuming more than is stated. Given Data / Assumptions: Exist at least some nonfunctioning pens. All blue pens are functioning (they write). There exists at least one pen that is a writing utensil; however, it is not stated that all pens are writing utensils. Concept / Approach: We test each candidate for necessity across all models consistent with the premises. To affirm I or II, we would need guaranteed overlap between the nonwriting pens or blue pens and the subset “writing utensils.” The facts do not force that overlap. Step-by-Step Solution: I: “Some writing utensils don't write.” We do not know whether any of the nonwriting pens are among the pens that are writing utensils. It is possible (consistent with the facts) that all nonwriting pens fall outside the “writing utensil” subset mentioned. Not a must.II: “Some writing utensils are blue.” Existence of blue pens is not stated; “All blue pens write” is conditional, not existential. Even if blue pens exist, we do not know whether any blue pen lies in the “writing utensil” subset. Not a must.III: “Some blue writing utensils don't write.” This is incompatible with “All blue pens write,” and we have no other blue writing utensils in evidence. Not a must. Verification / Alternative check: Create a model where only black pens exist, some don't write, and the subset of pens that are “writing utensils” are all functioning. All premises hold, but I and II fail; hence none is necessary. Why Other Options Are Wrong: I only / I and II only / II and III only / I and III only: Each includes at least one statement not forced by the premises. Common Pitfalls: Assuming “all pens are writing utensils” or assuming the existence of blue pens not stated by the facts. Final Answer: None of the statements is a known fact.

Question 3

Logical reasoning — Islands and volcanoes (choose the statement that must be true) Facts:  Islands are surrounded by water.  Maui is an island.  Maui was formed by a volcano. Which statement must also be a fact?

Accepted Answer

Introduction / Context: This is a straightforward application of categorical logic. From a universal statement about islands and a specific statement identifying Maui as an island, we can infer a specific property about Maui. We must resist adding universal claims about volcanoes that are not provided. Given Data / Assumptions: All islands are surrounded by water. Maui belongs to the set of islands. Maui's geological origin involves a volcano, but nothing is said about other islands or all volcanoes. Concept / Approach: Apply universal instantiation: if all elements of a set have property P, and X is a member of that set, then X has property P. Do not generalize from a single example (Maui) to all islands or all volcanoes. Step-by-Step Solution: From 1) and 2): Since Maui is an island and every island is surrounded by water, conclude that Maui is surrounded by water.Statement II (“All islands are formed by volcanoes”) is not implied by Maui's formation.Statement III (“All volcanoes are on islands”) is also not implied; many volcanoes exist on continents. Verification / Alternative check: Generic logic pattern: All A are B; X is A; therefore X is B. This matches 1) and 2). Statements II and III attempt unjustified universal generalizations. Why Other Options Are Wrong: II only / II and III only: Overgeneralize beyond the data. None of the statements is a known fact.: Incorrect because statement I is certain. I and II only: Includes an unjustified generalization (II). Common Pitfalls: Confusing an example (“Maui was formed by a volcano”) with a universal law applying to all islands or all volcanoes. Final Answer: I only

Question 4

Logical deduction – evaluate which statement must be true Facts: Robert owns four vehicles. Two of the vehicles are red. One of the vehicles is a minivan. Question: Which of the following statements is guaranteed by the facts?

Accepted Answer

Introduction / Context: This critical reasoning problem asks you to determine what must be true, given three short facts about Robert’s vehicles. The test of necessity is strict: a statement must follow in every possible arrangement consistent with the facts, not just in some convenient scenario. Given Data / Assumptions: Robert owns exactly four vehicles. Exactly two of the four vehicles are red. At least one vehicle type is identified: one vehicle is a minivan. We do not know the types of the other vehicles or the colors beyond “two are red.” Concept / Approach: Test each proposed statement against multiple consistent cases. If you can create even one valid case that makes a statement false, then that statement is not a necessary fact. Only a statement true in all valid cases can be marked as “must be true.” Step-by-Step Solution: Evaluate I: “Robert has a red minivan.” The facts allow the minivan to be either red or not red. For example, two red sedans + one blue minivan + one blue truck fits all facts without a red minivan. So I is not necessary.Evaluate II: “Robert has three cars.” The facts never mention cars. He could have one minivan, one motorcycle, one SUV, one pickup; or many other combinations. So II is not necessary.Evaluate III: “Robert’s favorite color is red.” Preference is never discussed. Not necessary.Since none of I, II, or III must be true, the correct choice is that none is a known fact. Verification / Alternative check: Construct two contrasting valid scenarios (with and without a red minivan) to confirm that I is not forced. Similar constructions show II and III are unfounded. Why Other Options Are Wrong: “I only”, “II only”, “II and III only”, or “I and II only” all assert something not guaranteed by the original facts. Common Pitfalls: Assuming typical real-world distributions (e.g., most vehicles are cars) or inferring preferences (favorite color) from counts. Final Answer: None of the statements is a known fact.

Question 5

Chase-and-order logic – who makes the tackle? Four defenders chase a receiver: Calvin is directly behind the ball carrier; Jenkins and Burton are side-by-side behind Calvin; Zeller is behind both Jenkins and Burton. Calvin misses and falls; Burton t…

Accepted Answer

Introduction / Context: This is a spatial-order reasoning question set in a sports context. The specific running order of players determines who can reach the ball carrier when some defenders fall or trip. Given Data / Assumptions: Order from front to back before mishaps: Receiver > Calvin > (Jenkins and Burton side-by-side) > Zeller. Events: Calvin attempts a tackle, misses, and falls. Burton trips (and thus cannot tackle). No other movement constraints are provided; assume players behind can overtake if unobstructed. Concept / Approach: Eliminate anyone who is incapacitated or too far to arrive first. With Calvin down and Burton tripping, identify the next closest active defender: Jenkins. Zeller is behind Jenkins and thus farther away. Step-by-Step Solution: Initial stack: Receiver → Calvin → Jenkins/Burton (same level) → Zeller.Calvin falls; he cannot tackle now.Burton trips; he also cannot tackle now.Jenkins remains the front-most standing defender.Therefore, Jenkins is the one positioned to make the tackle. Verification / Alternative check: If both Calvin and Burton are out, only Jenkins (closest) and Zeller remain. Distance advantage means Jenkins reaches first barring new information. Why Other Options Are Wrong: Calvin: already fell.Burton: tripped and is out of the play.Zeller: further back than Jenkins.“No one” contradicts that Jenkins is still active and closest. Common Pitfalls: Overlooking that Jenkins and Burton were at the same level; with Burton tripping, Jenkins becomes the next viable tackler. Final Answer: Jenkins

Question 6

Work-crew swaps – determine final positions A crew paints Mr. Field’s house: Michael (front), Ross (back alley), Jed (north side), Shawn (south side). Michael swaps with Jed; then Jed swaps with Shawn. Where is Shawn after both swaps?

Accepted Answer

Introduction / Context: This sequencing puzzle tracks personnel swaps around a house. You must follow two consecutive position exchanges to determine one worker’s final location. Given Data / Assumptions: Initial positions: Michael = front; Ross = back; Jed = north; Shawn = south. Swap 1: Michael ↔ Jed. Swap 2: Jed ↔ Shawn (using Jed’s new position from Swap 1). No other changes occur. Concept / Approach: Apply the swaps strictly in order, updating positions after each. Keep a simple table or running list so that you do not revert Jed to his original position before the second swap. Step-by-Step Solution: Start: M(front), R(back), J(north), S(south).After Swap 1 (M ↔ J): M(north), J(front), R(back), S(south).After Swap 2 (J ↔ S): J(south), S(front), M(north), R(back).Therefore, Shawn ends up at the front of the house. Verification / Alternative check: Reverse the process mentally: If Shawn is at front at the end, then just before Swap 2 he must have been at south, and Jed at front—consistent with Swap 1’s aftermath. Why Other Options Are Wrong: North/Back/South do not match the final tracked position for Shawn after both swaps.“Roof” is an irrelevant distractor not supported by the scenario. Common Pitfalls: Accidentally applying the second swap to Jed’s original location instead of his post–Swap 1 location. Final Answer: in front of the house

Question 7

Workday trades – who worked on Tuesday? Four temps each work a different day (Mon–Thu). Ms. Johnson was scheduled Monday but traded with Mr. Carter (originally Wednesday). Ms. Falk traded with Mr. Kirk (originally Thursday). After all switches, who …

Accepted Answer

Introduction / Context: This scheduling logic question involves two swaps across a Monday–Thursday window. The challenge is to track original assignments and then apply trades to deduce the person who ends up on Tuesday. Given Data / Assumptions: Each of four workers is assigned one unique day among Monday, Tuesday, Wednesday, Thursday. Ms. Johnson was originally Monday; Mr. Carter was originally Wednesday; Mr. Kirk was originally Thursday; Ms. Falk is the remaining person (thus originally Tuesday by elimination). Trades: Johnson ↔ Carter; Falk ↔ Kirk. Concept / Approach: First recover every original day by elimination. Then apply each trade to swap days between the two named people. The worker not involved in a particular trade retains their current day after that trade. Step-by-Step Solution: Original schedule: Johnson = Mon; Carter = Wed; Kirk = Thu; Falk = Tue (by elimination).Trade 1 (Johnson ↔ Carter): Johnson → Wed; Carter → Mon; Falk remains Tue; Kirk remains Thu.Trade 2 (Falk ↔ Kirk): Falk → Thu; Kirk → Tue; Johnson remains Wed; Carter remains Mon.Final schedule: Mon = Carter; Tue = Kirk; Wed = Johnson; Thu = Falk. Verification / Alternative check: Check uniqueness (each day has exactly one worker) and that each trade simply swapped the two names’ days—both conditions hold. Why Other Options Are Wrong: Carter moved to Monday; Johnson to Wednesday; Falk to Thursday. None of these is Tuesday.“No one worked Tuesday” contradicts the setup that each day has one worker. Common Pitfalls: Forgetting to assign Falk’s original day by elimination before executing the second trade. Final Answer: Mr. Kirk

Question 8

Eyewitness reliability – choose the most probable combined description of the mugger Four witnesses each give a different description; which description is most likely correct?

Accepted Answer

Introduction / Context: This problem tests consensus reasoning. When multiple eyewitnesses give differing descriptions, the best estimate of truth usually combines the most frequently repeated attributes across reports. Given Data / Assumptions: We have four conflicting descriptions, but options synthesize common traits. Typical approach: identify features (height, build, age) that appear most often among accounts. No single original report is guaranteed correct; we infer the most probable composite. Concept / Approach: Use the “majority overlap” heuristic. If “tall” occurs in most accounts and “thin” occurs in most accounts, and “middle-aged” is more common than “young,” select the option combining those prevailing traits. Step-by-Step Solution: Scan options for common traits: “tall” appears in multiple choices; “thin” appears in several; “middle-aged” appears more often than “young.”Identify the choice combining these frequent descriptors: “tall, thin, and middle-aged.”Thus, option combining the modal traits is most plausible as a consensus reconstruction. Verification / Alternative check: If two attributes tie, prefer the one that best maximizes overlap across all three categories simultaneously; here, tall + thin + middle-aged yields the highest joint frequency. Why Other Options Are Wrong: Average height or average weight reduce overlap relative to “tall” and “thin.”“Young” appears less often than “middle-aged,” lowering likelihood.An entirely different build/height (e.g., short, heavy) is not supported by majority features. Common Pitfalls: Focusing on one attribute in isolation instead of maximizing agreement across all three. Final Answer: He was tall, thin, and middle-aged.

Question 9

Partner assignment rule – who should work with Baxter today? Rule: No pair may work together more than seven consecutive class periods. Current streaks: Adam–Baxter = 7 in a row; Carter–Dennis = 3 in a row; Carter refuses to work with Adam.

Accepted Answer

Introduction / Context: This classroom pairing puzzle applies a constraint on consecutive pairings. Once two students have studied together seven periods in a row, they cannot be paired again immediately. Additional preferences also apply (Carter refuses to work with Adam). Given Data / Assumptions: Adam and Baxter have reached the maximum 7-period streak and cannot be paired this period. Carter and Dennis have a 3-period streak but have no rule preventing a change. Carter does not want to work with Adam. We must choose Baxter’s partner among available classmates. Concept / Approach: Exclude prohibited pairs first. Baxter cannot be with Adam due to the seven-in-a-row cap. Carter refuses Adam, but that does not affect Carter pairing with Baxter. Dennis can partner with Baxter, but we should also preserve flexibility and avoid creating new conflicts if unnecessary. Step-by-Step Solution: Remove Adam as a partner for Baxter (violates the 7-in-a-row rule).Consider Carter: allowed, and does not violate any preference (Carter’s objection is to Adam).Consider Dennis: also allowed; however, there is no stated need to maintain Carter–Dennis specifically.Any valid partner distinct from Adam is acceptable; Carter is a straightforward compliant choice. Verification / Alternative check: Pairing Carter with Baxter respects all constraints: the seven-in-a-row cap and Carter’s refusal (which applies only to Adam). Why Other Options Are Wrong: Adam: explicitly prohibited by the seven-in-a-row rule.Dennis: feasible, but the question seeks the best assignment under given notes; Carter is directly indicated as acceptable and resolves his refusal without conflict.Forest: presumably the teacher; not a student partner.“No one” contradicts the implied requirement to form pairs. Common Pitfalls: Misreading Carter’s refusal as a restriction on Carter’s general pairing rather than only with Adam. Final Answer: Carter

Question 10

Seat numbering logic – determine Inez’s seat Henry is in seat 253; Marla sits to Henry’s right in seat 254; George sits to Henry’s left; Inez sits to George’s left. What seat is Inez in?

Accepted Answer

Introduction / Context: This is a directional/number-line reasoning puzzle. Stadium seats typically increase by 1 to the right, decrease by 1 to the left. Using these conventions, we can trace positions step by step. Given Data / Assumptions: Henry sits in seat 253. Marla is to Henry’s right in seat 254, confirming that right = +1. George is to Henry’s left, which will be 253 − 1 = 252. Inez is to the left of George, so one more seat to the left from 252. Concept / Approach: Apply consistent left/right numbering increments. Once the convention is confirmed by Marla’s seat, proceed leftwards for George and then for Inez. Step-by-Step Solution: Start with Henry: 253.Right of Henry: 254 (Marla) → confirms numbering direction.Left of Henry: 252 (George).Left of George: 251 (Inez). Verification / Alternative check: Count directly: 253 (Henry) → 252 (left, George) → 251 (left, Inez). All consistent. Why Other Options Are Wrong: 254 is Marla; 255 and 256 are to the right of Marla; 252 is George’s seat. Common Pitfalls: Inverting left/right numbering or skipping a seat when counting. Final Answer: 251

Question 11

Team selection – which gymnast moves up to Level 1? Rule: Move the gymnast with the biggest prize and the most experience. Data: Roberta (bronze; 7 competitions). Jamie (silver; fewer comps than Roberta). Beth (higher medal than Jamie; more comps th…

Accepted Answer

Introduction / Context: This ranking puzzle asks you to identify the gymnast with the highest achievement and the greatest experience. “Biggest prize” refers to medal hierarchy (gold > silver > bronze). “Most experience” refers to the largest number of prior competitions. Given Data / Assumptions: Roberta: bronze; 7 competitions. Jamie: silver; fewer competitions than Roberta (< 7). Beth: higher medal than Jamie (thus gold) and more competitions than Roberta (> 7). Michele: bronze; 3 competitions. Concept / Approach: Apply a two-criteria filter: first check medal ranking, then break ties by experience or confirm the leader on both counts. The problem states the coaches will move the gymnast who satisfies both: biggest prize and most experience simultaneously. Step-by-Step Solution: Medal comparison: Beth = gold; Jamie = silver; Roberta = bronze; Michele = bronze. Beth leads on medal.Experience comparison: Beth has competed more times than Roberta’s 7, so Beth > 7; Jamie has fewer than 7; Michele = 3.Beth dominates both criteria: top medal and highest experience.Therefore, Beth moves to Level 1. Verification / Alternative check: No other gymnast beats Beth on either metric simultaneously—she is uniquely best on both measures. Why Other Options Are Wrong: Roberta: respectable experience but lower medal (bronze).Jamie: better than bronze but fewer competitions than Roberta and less than Beth on medal.Michele: lowest experience and bronze medal.“None” contradicts the given data that Beth outranks all others on both axes. Common Pitfalls: Treating the criteria as either/or when the instruction requires the best on both simultaneously. Final Answer: Beth

Question 12

Relative ages – who is oldest and gets the extra slice? Randy is 2 months older than Greg. Greg is 3 months younger than Ned. Kent is 1 month older than Greg. The oldest gets the extra piece. Who is it?

Accepted Answer

Introduction / Context: This is a simple relative-age comparison. Converting all ages to differences from a single baseline makes it easy to rank the four friends by age and identify the oldest. Given Data / Assumptions: Randy = Greg + 2 months. Greg = Ned − 3 months (i.e., Ned = Greg + 3 months). Kent = Greg + 1 month. Oldest friend receives the extra pizza slice. Concept / Approach: Use a reference variable for Greg’s age and express all others in terms of Greg. Then compare the offsets to determine who has the largest value (oldest). Step-by-Step Solution: Let Greg’s age be G (as a baseline).Randy = G + 2.Kent = G + 1.Ned = G + 3.Compare: Ned (G + 3) > Randy (G + 2) > Kent (G + 1) > Greg (G). Ned is oldest. Verification / Alternative check: Order check: Each inequality is consistent with the given pairwise differences. Why Other Options Are Wrong: Randy and Kent are older than Greg but still younger than Ned.Greg is explicitly younger than both Randy and Ned.“Tie” conflicts with distinct offsets (+3, +2, +1, 0). Common Pitfalls: Accidentally reversing the “younger than” relation when converting to offsets. Final Answer: Ned

Question 13

Seniority ordering – who becomes math department chair? Seniority (highest to lowest): Mr. Temple > Mr. Rhodes > Ms. West > Ms. Brody (given: Temple has more than West; Rhodes has more than West but less than Temple; West has more than Brody). Mr. T…

Accepted Answer

Introduction / Context: This is an ordering problem based on seniority. The chairperson is chosen by seniority, but the most senior candidate (Mr. Temple) declines the role, so we must select the next most senior person. Given Data / Assumptions: Ms. West has less seniority than Mr. Temple, but more than Ms. Brody. Mr. Rhodes has more seniority than Ms. West, but less than Mr. Temple. Mr. Temple does not want the job. Chair goes to the most senior available candidate. Concept / Approach: Construct a strict ordering from the comparisons, then remove Mr. Temple from consideration and choose the top remaining person. Step-by-Step Solution: From “West < Temple” and “West > Brody,” place: Temple > West > Brody.From “Rhodes > West” and “Rhodes < Temple,” place Rhodes between Temple and West.Complete order: Temple > Rhodes > West > Brody.Temple declines; therefore, Rhodes is next in line and becomes chair. Verification / Alternative check: Double-check that each comparative statement is satisfied by the final ranking; all constraints hold. Why Other Options Are Wrong: Temple declined. West and Brody have less seniority than Rhodes. “Joint chair” is not part of the selection rule given. Common Pitfalls: Misplacing Rhodes relative to West or forgetting to remove Temple after his refusal. Final Answer: Mr. Rhodes

Question 14

Distance reasoning — estimating Alexis’s distance from the airport  Danielle has been visiting friends in Ridge-wood for two weeks and flies out early tomorrow. Most of her friends live fairly close to the airport. Madison lives 10 miles away. Franc…

Accepted Answer

Introduction / Context: This critical-reasoning question tests your ability to interpret relational distance constraints and choose a value that satisfies all conditions simultaneously. The prompt names multiple friends and fixes their distances from the airport. You must locate Alexis on the number line relative to those anchor points. Given Data / Assumptions: Madison is 10 miles from the airport. Frances is 5 miles from the airport. Samantha is 7 miles from the airport. Alexis is farther than Frances (so > 5 miles) but closer than Samantha (so < 7 miles). We are selecting an approximate distance from the given options. Concept / Approach: The strategy is to translate the text into inequalities and then check which answer options are within the valid interval. Because the statements are strict (farther than, closer than), Alexis must lie strictly between 5 and 7 miles from the airport. Any option equal to 5 or 7 or outside this range must be eliminated. Step-by-Step Solution: Represent Frances's distance as 5 and Samantha's as 7 on a number line.Alexis is said to be farther than Frances → Alexis > 5.Alexis is also closer than Samantha → Alexis < 7.Combine to get the strict inequality: 5 < Alexis < 7.Scan options: nine (9) ✗, seven (7) ✗, eight (8) ✗, six (6) ✓, five (5) ✗. Verification / Alternative check: Why Other Options Are Wrong: nine miles: 9 is not less than 7, so Alexis would not be closer than Samantha. seven miles: Alexis must be closer than 7, not equal to 7. eight miles: 8 is greater than 7, violating the ‘‘closer than Samantha’’ condition. five miles: Alexis must be farther than Frances; equality fails. Common Pitfalls: Final Answer: six miles

Question 15

Work-schedule logic — deducing Nurse Kemp’s consecutive night shifts  Nurse Rogers has worked 5 night shifts in a row. Nurse Kemp has worked more consecutive night shifts than Rogers. Nurse Calvin has worked 8 in a row, which is fewer than Kemp. Nur…

Accepted Answer

Introduction / Context: This puzzle involves inequality chaining and a sum constraint. You are given runs of consecutive night shifts for four nurses. The goal is to compute Nurse Kemp’s count using “greater than”, “less than”, and “more than combined” conditions. It is a typical test of bounded reasoning and integer feasibility. Given Data / Assumptions: Nurse Rogers: 5 consecutive night shifts. Nurse Kemp: more than Rogers → Kemp > 5. Nurse Calvin: 8 consecutive night shifts and less than Kemp → Kemp > 8. Nurse Miller: 15 consecutive night shifts, which is more than the sum of Kemp and Rogers → 15 > Kemp + 5. All counts are whole numbers. Concept / Approach: Convert each statement to numeric inequalities and intersect their solution sets. Because these are small integers, quick arithmetic will reveal the only value that satisfies every condition simultaneously. Step-by-Step Solution: From “Kemp more than Rogers”: Kemp > 5.From “Calvin less than Kemp” with Calvin = 8: Kemp > 8.From “Miller has more than Kemp and Rogers combined”: 15 > Kemp + 5 → Kemp < 10.Combine: Kemp is an integer with Kemp > 8 and Kemp < 10 → Kemp = 9. Verification / Alternative check: Why Other Options Are Wrong: eight: contradicts “Kemp > 8”. ten: violates 15 > Kemp + 5 because 10 + 5 = 15 is not “more than”. eleven: fails 15 > Kemp + 5 since 11 + 5 = 16. seven: violates Kemp > 8 and Kemp > 5. Common Pitfalls: Overlooking that “more than combined” uses strict greater than, not greater than or equal to. Also, some solvers forget that all quantities are integers, which is essential for narrowing the value to a single number. Final Answer: nine

Question 16

Chasing positions — who is directly behind the dog after the moves?  Children chase a dog whose leash has broken. Initial order behind the dog: James, then Ruby, then Rachel. Max is initially ahead of the dog walking the other way; after they pass, …

Accepted Answer

Introduction / Context: This problem tests dynamic ordering. Positions change as participants speed up or change lanes. You must carefully track who ends up directly behind the dog after two children move alongside it and another child inserts into the line mid-chase. Given Data / Assumptions: Initial order from front: Dog → James → Ruby → Rachel. Max is ahead of the dog but walking the opposite direction; after they pass, he turns and joins behind Ruby. James then moves up to run alongside the dog on the left (so he is no longer behind). Ruby then moves up to run alongside the dog on the right (so she is also no longer behind). Concept / Approach: Track the queue step by step. “Alongside” removes a runner from the single-file line behind the dog. The next person in the queue becomes the new “directly behind”. Insertions must respect the described location (“behind Ruby”) before Ruby moves up—then adjust after Ruby advances. Step-by-Step Solution: Initial single file: Dog, James, Ruby, Rachel.Max turns and joins behind Ruby → Dog, James, Ruby, Max, Rachel.James accelerates to run alongside the dog (left) → line behind dog becomes: Ruby, Max, Rachel.Ruby accelerates to run alongside the dog (right) → line behind dog becomes: Max, Rachel.Therefore, the child directly behind the dog is Max. Verification / Alternative check: Why Other Options Are Wrong: James: he is explicitly alongside on the left, not behind. Ruby: she is explicitly alongside on the right, not behind. Rachel: she is behind Max; not directly behind the dog. None of the above: incorrect because Max fits perfectly. Common Pitfalls: Forgetting that “alongside” means leaving the single-file order behind the leader. Another error is inserting Max in the wrong place or not reconsidering the order after Ruby moves forward. Final Answer: Max

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