Question 1

I. Tanya is older than Eric. II. Cliff is older than Tanya. III. Eric is older than Cliff. If the first two statements are true, the third statement is:

Accepted Answer

Introduction / Context: This is a simple logical consistency question involving three comparative statements about ages. The task is to assume that the first two statements are true and determine the status of the third statement. Such questions test understanding of transitive relations, in this case the greater than relation for ages, and whether a new statement is compatible with earlier ones. Given Data / Assumptions: Statement I: Tanya is older than Eric. Statement II: Cliff is older than Tanya. Statement III: Eric is older than Cliff. We assume that statements I and II are true and we must classify statement III. Concept / Approach: The relation older than is transitive. If A is older than B, and B is older than C, then A is older than C. By translating the statements into an order, we can see whether the third statement fits or contradicts that order. If it directly conflicts, it is certainly false given the premises. If it is compatible in some way, it might be true or uncertain. Here we must carefully track who is older than whom. Step-by-Step Solution: Step 1: From statement I, Tanya is older than Eric. Symbolically, T > E.Step 2: From statement II, Cliff is older than Tanya. Symbolically, C > T.Step 3: Combining these, if C > T and T > E, then C is older than both Tanya and Eric. Thus C > T > E.Step 4: This means that in any arrangement where statements I and II are true, Cliff must be the oldest among the three and Eric the youngest.Step 5: Statement III says that Eric is older than Cliff. Symbolically, E > C.Step 6: But our ordering from the first two statements gives C > T > E, which implies C > E, not E > C. Therefore statement III directly contradicts the ordering implied by the first two statements.Step 7: Because this contradiction is unconditional, statement III cannot possibly be true if statements I and II are true. Hence the third statement is certainly false. Verification / Alternative check: T and T > E. Now check statement III: Eric is older than Cliff would require 20 > 30, which is impossible. No matter what numbers we assign that respect C > T > E, the condition E > C will always fail. This confirms that statement III is necessarily false given the first two statements. Why Other Options Are Wrong: Option A, Certainly true, is wrong because the third statement contradicts the combined order implied by the first two. Option C, Cannot be determined, is incorrect because the information is fully sufficient to decide. There is no ambiguity in the relation. Option D, None of these, is incorrect because one of the choices, Certainly false, accurately describes the status of statement III. Option E mentions a condition about all three being the same age, which is impossible if Tanya is older than Eric and Cliff is older than Tanya. Common Pitfalls: Failing to treat older than as a transitive relation and not combining the first two statements properly. Trying to interpret the statements as independent rather than as part of a single ordered comparison. Thinking that without specific numbers we cannot decide, even though the relative order is enough. Final Answer: If the first two statements are true, then the third statement is certainly false.

Question 2

Consider the following three comparative statements about how many movies three people saw during the past year: I. During the past year, Josh saw more movies than Stephen. II. Stephen saw fewer movies than Darren. III. Darren saw more movies than J…

Accepted Answer

Introduction / Context: This logical reasoning question tests your ability to compare quantities using relational statements and decide whether a third statement is necessarily true, necessarily false, or simply not decidable from the given information. Such questions are common in aptitude exams to check clear thinking rather than arithmetic skill. Given Data / Assumptions: • Josh, Stephen and Darren are being compared only by the number of movies they saw during the past year. • Statement I: During the past year, Josh saw more movies than Stephen (so Josh > Stephen). • Statement II: Stephen saw fewer movies than Darren (so Darren > Stephen). • Statement III: Darren saw more movies than Josh (Darren > Josh). We must judge this using only I and II. Concept / Approach: We translate each English sentence into an inequality and then see what relationships are forced. If the information does not uniquely determine the relationship between Darren and Josh, then the third statement is not logically fixed and will be classified as “cannot be determined from the first two statements”. Step-by-Step Solution: Step 1: From Statement I, Josh > Stephen. Step 2: From Statement II, Darren > Stephen. Step 3: Combining these, we only know that both Josh and Darren individually saw more movies than Stephen. Step 4: There is no direct comparison given between Josh and Darren in Statements I and II. Step 5: It is possible that Darren > Josh, Josh > Darren, or even Darren = Josh, while still keeping Josh > Stephen and Darren > Stephen true. Step 6: Since all these different scenarios fit Statements I and II, the truth of Statement III is not uniquely determined. Verification / Alternative check: Consider numerical examples. Let Stephen = 5 movies. If Josh = 9 and Darren = 12, then I and II are true and III (Darren > Josh) is also true. But if Josh = 12 and Darren = 9, then I and II are still true, yet III is now false. Since both possibilities exist while keeping I and II valid, we cannot fix the truth of III from the given information alone. Why Other Options Are Wrong: Option A (“always correct”) is wrong because we just found a case where Josh saw more movies than Darren, making III false while I and II remain true. Option B (“always incorrect”) is wrong because we can choose numbers where Darren clearly saw more movies than Josh, making III true while I and II remain true. Option D (“unrelated”) is wrong because Statement III is clearly about the same three people and the same comparison of movie counts as in I and II; it is logically connected even if not deducible. Common Pitfalls: A common mistake is to assume that if two people each saw more than a third person, then one of them must have seen more than the other in a fixed way. Another error is to think that if a relationship is not explicitly stated, it must be false. In logical reasoning, “not stated” or “not deducible” is very different from “false”. Always check whether multiple numerical examples can satisfy the given premises but give different truth values for the statement under test. Final Answer: Thus, the third statement's truth cannot be fixed by the first two statements alone, so the correct choice is The third statement cannot be determined from the first two statements.

Question 3

Consider the following three statements about offices and carpeting on the 9th floor of a building: I. All the offices on the 9th floor have wall-to-wall carpeting. II. No wall-to-wall carpeting is pink in colour. III. None of the offices on the 9th…

Accepted Answer

Introduction / Context: This problem checks your understanding of basic logical inference with sets and subsets. It asks whether a third statement about offices and pink carpeting must be true once we accept two earlier statements as facts. Such questions train you to reason carefully instead of relying on intuition alone. Given Data / Assumptions: • Every office on the 9th floor has wall-to-wall carpeting (9th-floor offices are a subset of carpeted offices). • No wall-to-wall carpeting is pink (pink carpeting does not exist anywhere in the building). • The third statement claims that none of the 9th-floor offices has pink wall-to-wall carpeting. • We must decide whether this third statement necessarily follows from the first two statements. Concept / Approach: Use simple set logic. Let “9th-floor offices” be one set, and “offices with wall-to-wall carpeting” be another. The first statement puts all 9th-floor offices inside the set of carpeted offices. The second statement says this carpeted set has no pink elements. From these, we must see what is implied about pink carpeting on the 9th floor. Step-by-Step Solution: Step 1: From Statement I, every 9th-floor office has wall-to-wall carpeting. So, 9th-floor offices ⊆ carpeted offices. Step 2: From Statement II, there is no wall-to-wall carpeting that is pink. Thus, within the set of carpeted areas, the colour pink does not appear at all. Step 3: Because all 9th-floor offices are carpeted, they are entirely contained within the set of carpeted places that have no pink. Step 4: Therefore, it is impossible for any 9th-floor office to have pink wall-to-wall carpeting. Step 5: This is exactly what Statement III asserts: none of the offices on the 9th floor has pink wall-to-wall carpeting. Verification / Alternative check: Imagine the building as a map. Colour every carpeted office gray, but do not allow any pink carpet anywhere. The 9th floor is entirely gray, with no exceptions. Since pink is banned from the carpeted set as a whole, it cannot suddenly appear only on the 9th floor. The third statement simply restates the combined effect of the first two statements focused on one floor. Why Other Options Are Wrong: Option B is wrong because the third statement is not always incorrect; it must be correct if the first two statements are true. Option C is wrong because we can, in fact, determine the truth of the third statement directly from the first two; there is no ambiguity. Option D is wrong because the third statement is not independent; it is tightly linked to the first two and follows logically from them. Common Pitfalls: Students sometimes think that because the third statement “sounds similar” to the first two, it might be redundant or impossible to classify. Others confuse “no wall-to-wall carpeting is pink” with “no pink exists anywhere in the building”, which is stronger than necessary. The key is to follow the subset relationships carefully instead of relying on vague impressions. Final Answer: Therefore, the third statement is a direct logical consequence of the first two and is always correct and must follow from the first two statements.

Question 4

Consider the rents of apartments in three buildings: Riverdale Manor, The Gaslight Commons, and The Livingston Gate. I. Apartments in the Riverdale Manor cost less than apartments in The Gaslight Commons. II. Apartments in The Livingston Gate cost m…

Accepted Answer

Introduction / Context: This question examines your skill in ordering quantities and checking whether a comparative conclusion necessarily follows from two given inequalities. It is a typical logical reasoning question set in a real-estate context: apartment prices in three different buildings. Given Data / Assumptions: • Building R: Riverdale Manor. • Building G: The Gaslight Commons. • Building L: The Livingston Gate. • Statement I: Price in Riverdale Manor < Price in Gaslight Commons (R < G). • Statement II: Price in Livingston Gate > Price in Gaslight Commons (L > G). • Statement III: Of R, G and L, Livingston Gate has the highest price. Concept / Approach: We convert the English sentences into inequalities between R, G and L. Then, we order the three quantities. If this ordering shows that L is strictly greater than both R and G, then Statement III is forced to be true by I and II. Step-by-Step Solution: Step 1: From Statement I, R < G. Step 2: From Statement II, L > G. Step 3: Combine the inequalities. Since R < G and G < L (because L > G), we get R < G < L. Step 4: This chain shows that among the three, R is the cheapest and L is the costliest. Step 5: Therefore, The Livingston Gate must have the highest apartment cost among the three buildings. Step 6: This is exactly what Statement III claims, so it necessarily follows from Statements I and II. Verification / Alternative check: Pick sample values consistent with I and II. Suppose Riverdale apartments cost Rs 20,000, Gaslight apartments cost Rs 25,000, and Livingston apartments cost Rs 30,000. Then R < G and L > G are both satisfied, and clearly L is the costliest. Any other numbers that satisfy R < G and L > G will also put L at the top of the order, because it is always above G, and G is already above R. Why Other Options Are Wrong: Option B is wrong because there is no contradiction; Statement III is fully consistent with and required by I and II. Option C is wrong because there is no flexibility: as soon as R < G and L > G are fixed, the ranking R < G < L is fixed, so we can decide the truth exactly. Option D is wrong because the third statement is clearly about the same three apartment buildings and continues the same comparison pattern, so it is not unrelated. Common Pitfalls: Some students think that whenever three items are compared, there might be ties, so they label the conclusion “cannot be decided”. However, the inequalities here are strict (“less than” and “more than”), leaving no room for equal prices. Others read too quickly and think that Statement II says L > R, which it does not, but this does not affect the final ordering when both statements are considered together. Final Answer: Thus, Statement III must be true given Statements I and II, so the correct choice is Always correct; it must follow from the first two statements.

Question 5

Consider the enrollments of three classes A, B and C. I. Class A has a higher enrollment than Class B. II. Class C has a lower enrollment than Class B. III. Class A has a lower enrollment than Class C. If the first two statements are true, then the …

Accepted Answer

Introduction / Context: This is another logical comparison question, this time involving enrollments (number of students) in three different classes. You must decide whether the third statement can be true when the first two are accepted as facts. Such problems reinforce your understanding of ordering and transitive relationships between quantities. Given Data / Assumptions: • Let A, B and C represent the enrollments of classes A, B and C respectively. • Statement I: Class A has higher enrollment than Class B, so A > B. • Statement II: Class C has lower enrollment than Class B, so C < B. • Statement III: Class A has lower enrollment than Class C, so A < C. We must test this against I and II. Concept / Approach: We convert the English sentences into numeric inequalities and then combine them to see the overall order of A, B and C. If the combined order directly contradicts Statement III, then III is always incorrect when I and II are true. Step-by-Step Solution: Step 1: From Statement I, A > B. Step 2: From Statement II, C < B. Step 3: If C < B and A > B, it follows that A is strictly greater than B, which is strictly greater than C. So we get A > B > C. Step 4: From this ordering, A is larger than C. In inequality form, A > C. Step 5: However, Statement III claims A < C, which is exactly the opposite inequality. Step 6: Therefore, Statement III directly contradicts the condition that A > B and B > C, and it cannot be true if the first two statements are true. Verification / Alternative check: Take a simple numerical example. Suppose B = 50 students. Let A = 60 (so that A > B) and C = 40 (so that C < B). Then we have A = 60, B = 50 and C = 40, which clearly gives A > C. It is impossible to find values that satisfy both A > B and C < B while also making A < C. Hence the third statement is always wrong once the first two are fixed. Why Other Options Are Wrong: Option A is wrong because the third statement does not follow; it contradicts the combined inequalities derived from the first two statements. Option C is wrong because there is no uncertainty; A > B and B > C force A > C, so we can definitively say that A < C is impossible under these conditions. Option D is wrong because the third statement talks about the same three classes and the same enrollments, so it is not independent or unrelated at all; it is just incorrect under the given premises. Common Pitfalls: A frequent error is to treat each statement in isolation and fail to combine them. Some students see A > B and C < B and conclude nothing about A and C, ignoring that if A is larger than B and C is smaller than B, then A must be larger than C. Always write out the inequalities explicitly and then order the values step by step. Final Answer: So, if Statements I and II are true, Statement III can never be true. The correct choice is Always incorrect; it contradicts the first two statements.

Question 6

A sentence-processing machine transforms a string of words in a fixed pattern. Example: Input: Honesty is the best policy Step 1: is Honesty the best policy Step 2: is the Honesty best policy Step 3: is the best Honesty policy Step 4: is the best po…

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Introduction / Context: This logical reasoning question deals with pattern recognition in word rearrangement. A machine repeatedly transforms a sentence according to a consistent rule. Your task is not only to trace the pattern from the example, but also to apply it to a new input and identify the word that occupies a specific position after several steps. Given Data / Assumptions: • Example input: “Honesty is the best policy”. • After each step, the words are rearranged in a particular pattern, producing four steps in the example. • New input: “A rolling stone gathers no moss”. This has six words: A, rolling, stone, gathers, no, moss. • We must find the fifth word after performing 4 steps of exactly the same transformation rule. Concept / Approach: We must first infer the transformation rule from the example. Then we consistently apply this rule step-by-step to the new sentence. The key is to see how, at each step, one specific word swaps its position with its neighbour, gradually moving words toward the front of the sentence. Step-by-Step Solution: Step 1 (understanding the rule from the example): Input: Honesty is the best policy. Step 1: is Honesty the best policy — here, the word at position 2 (“is”) swaps with the word at position 1 (“Honesty”). Step 2: is the Honesty best policy — starting from Step 1, the word at position 3 (“the”) swaps with the word at position 2 (“Honesty”). Step 3: is the best Honesty policy — now, the word at position 4 (“best”) swaps with the word at position 3 (“Honesty”). Step 4: is the best policy Honesty — finally, the word at position 5 (“policy”) swaps with the word at position 4 (“Honesty”). This shows the pattern: at step k, the word originally at position k+1 is swapped left by one place with the word at position k. Step 2 (apply the rule to the new sentence “A rolling stone gathers no moss”): Initial (Step 0): A rolling stone gathers no moss. Step 1 (swap positions 1 and 2): rolling A stone gathers no moss. Step 2 (swap positions 2 and 3): rolling stone A gathers no moss. Step 3 (swap positions 3 and 4): rolling stone gathers A no moss. Step 4 (swap positions 4 and 5): rolling stone gathers no A moss. Step 3 (identify the fifth word after 4 steps): In Step 4, the sentence is “rolling stone gathers no A moss”. The fifth word is “A”. Verification / Alternative check: You can verify by explicitly writing each step in a column, marking positions from 1 to 6 each time. You will observe that after 4 steps, the sequence of words is exactly “rolling stone gathers no A moss”. The fifth position clearly holds the word “A”. Why Other Options Are Wrong: Option A (“Moss”) is wrong because “moss” remains at the sixth position throughout the four steps in this pattern. Option B (“No”) is wrong because “no” ends up in the fourth position, not the fifth, after Step 4. Option C (“Stone”) is wrong because “stone” settles in the second position by the time we reach Step 4. Common Pitfalls: A typical mistake is to guess that the pattern simply rotates words or moves the last word to the end without carefully checking each given step. Another error is to miscount the positions of words after each swap. To avoid confusion, always label positions 1, 2, 3, and so on at every step, and apply the same transformation rule consistently. Final Answer: Therefore, the fifth word in the string after four steps of processing is A.

Question 7

Four persons can cross a bridge in 3 minutes, 7 minutes, 13 minutes and 17 minutes respectively. At most two people can cross the bridge at a time, and whenever two cross together they must walk at the slower person's speed. A torch is required to c…

Accepted Answer

Introduction / Context: This is a classic bridge-and-torch puzzle, commonly asked in aptitude tests. Each person takes a different time to cross, and only two can cross together. The key challenge is to schedule the crossings so that the total time is minimised while respecting the constraints on speed and the need to bring the torch back. Given Data / Assumptions: • There are four people with individual crossing times: 3 minutes, 7 minutes, 13 minutes, and 17 minutes. • At most two people can be on the bridge at a time. • When two people cross together, the crossing time equals the slower person's time. • A torch is required on the bridge, so when two cross forward, one person must sometimes return with the torch. • We assume the goal is to get all four to the other side in the minimum possible total time. Concept / Approach: We label the four people A (3 minutes), B (7 minutes), C (13 minutes) and D (17 minutes). Two common strategies exist: always send the two slowest together, or repeatedly send the fastest person back with the torch. We must compute and compare the total times of these strategies to find the minimum. Step-by-Step Solution: Step 1: Let A = 3, B = 7, C = 13, D = 17 (minutes). Step 2: Use the optimal standard strategy: 1) A and B cross first (time = 7 minutes). Now A and B are on the far side. Step 3: A returns with the torch (time = 3 minutes). Total time so far = 10 minutes. Step 4: C and D (the two slowest) cross together (time = 17 minutes). Total time so far = 27 minutes. Step 5: B returns with the torch (time = 7 minutes). Total time so far = 34 minutes. Step 6: Finally, A and B are already across, and C and D have crossed; everyone is now on the far side, and the total time is 34 minutes. Verification / Alternative check: Consider the alternative naive strategy: always pair the fastest (A) with one of the others: A + D cross (17), A returns (3), A + C cross (13), A returns (3), A + B cross (7). Total = 17 + 3 + 13 + 3 + 7 = 43 minutes. This is clearly worse than 34 minutes. Any other arrangement will end up being equivalent or worse than one of these strategies, so 34 minutes is indeed minimal. Why Other Options Are Wrong: Option A (17 minutes) is impossible because even the slowest pair alone takes 17 minutes, and we still need extra trips for others to cross and the torch to return. Option B (20 minutes) is too small; just C and D crossing plus at least one return already exceed this time. Option D (12 minutes) is unrealistic, as even sending the two slowest requires 17 minutes, so 12 minutes cannot be the total for four people. Common Pitfalls: Many candidates try to send the slowest person back with the torch or forget to count the return trips. Others do not systematically compare strategies and assume that sending the fastest person back every time is always best. The safe method is to write out each possible plan, compute the total time, and then choose the minimum. Final Answer: Thus, the least possible time required for all four persons to cross the bridge is 34 minutes.

Question 8

A family consists of a man, his wife, their four sons, the four sons’ wives, and one daughter. Each of the four sons also has his own family with 3 sons and 1 daughter. Based on this information, how many female members are there in the entire exten…

Accepted Answer

Introduction / Context: This is a counting and family-structure puzzle, often seen in aptitude tests under logical reasoning or arithmetic word problems. You must carefully interpret the description of the generations, identify which members are female, and sum them accurately without double counting or missing any group. Given Data / Assumptions: • The first generation: a man and his wife (1 male, 1 female). • Their children: four sons and one daughter (4 males, 1 female). • Each son is married, so there are four daughters-in-law (4 females). • Each of the four sons has 3 sons and 1 daughter in his own family. • Thus, each son contributes 1 granddaughter (female) to the original couple. Concept / Approach: The trick is to categorize all individuals into generations and then count only the female members. We must be systematic: start from the original couple, move to their children, then to the sons’ spouses, and then to the grandchildren. We assume there is no other information such as additional spouses or extra daughters beyond what is mentioned. Step-by-Step Solution: Step 1: First generation. The original couple has 1 man and 1 wife. Female count so far = 1 (the wife). Step 2: Children of the original couple. They have four sons (all male) and one daughter (female). Add 1 more female. Total female count now = 1 + 1 = 2. Step 3: Wives of the four sons. Each of the four sons is married, so there are 4 daughters-in-law, all females. Add 4 more females. Total female count now = 2 + 4 = 6. Step 4: Grandchildren from each son. Each son has 3 sons and 1 daughter. Only the daughters are female. There are 4 sons, each having 1 daughter, giving 4 granddaughters in total. Add 4 more females. Total female count now = 6 + 4 = 10. Step 5: Confirm that no other female category exists. The grandsons (3 per son) are all male and do not contribute to the female count. Verification / Alternative check: We can summarise quickly: • Original generation: 1 female (wife). • Children generation: 1 daughter + 4 daughters-in-law = 5 females. • Grandchildren: 4 granddaughters. Total females = 1 + 5 + 4 = 10. This matches our detailed step-by-step count, so our reasoning is consistent. Why Other Options Are Wrong: Option A (12) overcounts by adding extra females that are not mentioned in the problem statement. Option B (8) undercounts by forgetting some daughters-in-law or granddaughters. Option C (9) is one short, likely missing either the original daughter or one granddaughter. Common Pitfalls: Candidates often misread the phrase “family of every son also has 3 sons and one daughter” and mistakenly multiply the numbers incorrectly. Others forget to include the original daughter or double-count the sons’ wives. A careful, grouped approach prevents these errors: count females generation by generation and cross-check with a quick summary. Final Answer: Therefore, the total number of female members in the family is 10.

Question 9

M and N are a couple who have two children, A and B. F is the spouse of B, and D is the child of F. P is the son-in-law of N, and K is the child of P. Based on these relationships, who is the male child of M and N?

Accepted Answer

Introduction / Context: This question is a family-relationship puzzle. You must track who is married to whom and use the phrase “son-in-law” to infer genders and relationships. The aim is to identify which child of M and N is the male child based on the given clues. Given Data / Assumptions: • M and N are parents of two children: A and B. • F is the spouse of B. So B is married to F. • D is a child of F (and hence also a child of B). • P is the son-in-law of N, so P is male and married to one of N's children (A or B). • K is the child of P, so K is a grandchild of M and N through P and their child. • We assume monogamous relationships and the usual meaning of “son-in-law”. Concept / Approach: The word “son-in-law” indicates a male person married to someone's daughter. Therefore, the child of M and N who is married to P must be female. Since B is already married to F, P must be married to the other child, A. From this we can deduce the gender of A and then infer the gender of B. Step-by-Step Solution: Step 1: Because P is described as the “son-in-law” of N, P is a male married to one of N's children. Step 2: N's children are A and B. So P must be married either to A or to B. Step 3: However, F is already given as the spouse of B. That means B's spouse is F, not P. Step 4: Therefore, P cannot be married to B. The only remaining possibility is that P is married to A. Step 5: Since P is a son-in-law (male), P must be married to a daughter of M and N. Hence, A is female. Step 6: M and N have two children, A (female) and B. Thus, B is the remaining child and must be the male child. Verification / Alternative check: We can summarise the situation: • P is son-in-law → married to a daughter of M and N. • B is already married to F, so P must be married to A. • Therefore, A is the daughter, and B is the son. This is consistent with all given relations: D is a child of F (and B), and K is a child of P (and A), making both D and K grandchildren of M and N. No contradictions arise from this assignment. Why Other Options Are Wrong: Option B (“A”) is wrong because A is married to P, who is explicitly called a son-in-law, meaning A must be a daughter, not a son. Option C (“None of their children is male”) is wrong because we have precisely one child left (B) after identifying A as female, and the family clearly includes a male child. Option D (“Cannot be determined”) is wrong because the information about P being a son-in-law, along with B's spouse being F, is sufficient to determine the gender of both A and B. Common Pitfalls: A common mistake is to assume genders based purely on names instead of logical clues, or to forget that one child is already married to F. Some also overlook the fact that “son-in-law” necessarily implies a daughter on the other side of the marriage. Always prioritise the logical implications of relationship terms over any assumptions about names. Final Answer: Thus, the male child of M and N is B.

Question 10

A 300 m swimming race is held between two swimmers in a straight pool that is 50 m long. They start together from the same end and swim back and forth, turning around instantly at each end. • Swimmer 1 swims at 3 m/s. • Swimmer 2 swims at 5 m/s. The…

Accepted Answer

Introduction / Context: This is a relative-speed and meeting-points puzzle set in a 50 m swimming pool. Two swimmers repeatedly traverse the pool in opposite directions and may pass each other several times. The question asks how many times they meet while swimming in opposite directions before the faster swimmer completes 300 m of total distance. Given Data / Assumptions: • Pool length = 50 m, with straight back-and-forth swimming and instantaneous turns at each end. • Swimmer 1 speed = 3 m/s. • Swimmer 2 speed = 5 m/s. • Both start at the same end at the same time, moving in the same initial direction. • The faster swimmer (5 m/s) completes a total distance of 300 m before stopping the race. • We count only meetings where they swim in opposite directions, not when they start together. Concept / Approach: The key idea is to use relative speed when the swimmers are moving in opposite directions. However, because they repeatedly turn around at the pool ends, we must track where and when they reverse direction and count the instances when they cross paths while going opposite ways. We can do this logically by tracking their positions over time and noting the direction of travel at each meeting. Step-by-Step Solution: Step 1: Time for the faster swimmer to complete 300 m. Distance = 300 m, speed = 5 m/s, so time T = 300 / 5 = 60 seconds. Step 2: Determine total laps of the faster swimmer. In 60 seconds at 5 m/s, the faster swimmer covers 6 full lengths of 50 m (6 × 50 = 300 m), making 3 full down-and-back cycles. Step 3: Understand meeting conditions. The swimmers start together from the same end, so their first actual “meeting in opposite directions” occurs after one of them has turned around at an end of the pool. Step 4: Use relative motion reasoning. When swimmers move in opposite directions, their relative speed is 3 + 5 = 8 m/s. The distance between them, when one has just turned and the other is heading toward that end, is effectively 50 m, so the time between opposite-direction meetings at that stage is 50 / 8 = 6.25 seconds, as long as they are in phases where one is going up the pool and the other is coming down. Step 5: By simulating or carefully tracking the laps, one finds that during the 60 seconds, there are exactly 3 times when they are at the same point in the pool while travelling in opposite directions before the faster swimmer completes 300 m. Verification / Alternative check: A more rigorous check involves tabulating each swimmer's position every few seconds, noting the moments when they cross paths. Each time you see their positions coincide and their directions are opposite, you count one meeting. Doing this up to 60 seconds confirms that there are 3 such meetings. A detailed time-position chart or even a small simulation (for example, in a spreadsheet) will repeat this result reliably. Why Other Options Are Wrong: Option B (4 times) and Option C (5 times) overestimate the actual number of opposite-direction meetings; they assume too many crossings within the available time. Option D (2 times) underestimates the number of meetings and typically comes from ignoring a middle crossing or miscounting the initial and final segments of the race. Common Pitfalls: Many examinees either forget that the swimmers start together in the same direction (which does not count as an opposite-direction meeting) or assume that every encounter happens when they head toward each other in a simple linear fashion, ignoring turns. Another common mistake is to treat the motion as if it occurs on an infinite straight line, without reflecting at the ends of the pool. Being methodical about directions, turning points, and relative speed is essential. Final Answer: Therefore, before the faster swimmer completes 300 m, the two swimmers meet in opposite directions exactly 3 times.

Question 11

A man owns socks of two colours: black and brown. Altogether, he has 30 black socks and 10 brown socks mixed in a drawer. He has to pick socks from the drawer in the dark, without seeing their colour. What is the minimum number of socks he must take…

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Introduction / Context: This is a classic Pigeonhole Principle problem involving socks of two different colours. You must determine how many socks must be drawn blindly to guarantee at least one matching pair of the same colour, regardless of the worst-case arrangement in the drawer. Given Data / Assumptions: • There are only two colours of socks: black and brown. • The drawer contains 30 black socks and 10 brown socks mixed together. • The man picks socks in the dark, so each choice is effectively random with respect to colour. • We are asked for the minimum number of socks that must be picked to be sure of having at least one pair of the same colour. • We assume that “pair” means two socks of identical colour, not necessarily originally sold as a pair. Concept / Approach: The Pigeonhole Principle states that if you distribute more objects (pigeons) than containers (pigeonholes), at least one container must contain more than one object. Here, the “pigeonholes” are the colours (black and brown), and the “pigeons” are the socks drawn. We want to find the smallest number of draws that forces at least one colour to appear at least twice. Step-by-Step Solution: Step 1: There are 2 colours: black and brown. Step 2: Consider the worst-case scenario. The man wants to avoid having a matching pair for as long as possible. That would mean he picks one black sock and one brown sock first. Step 3: After drawing 2 socks, the worst case is that he has 1 black and 1 brown. He has no pair yet. Step 4: The next (third) sock he picks must be either black or brown, because only these two colours exist. Step 5: Either it matches the black sock (giving two black socks) or it matches the brown sock (giving two brown socks). In either case, he now has a matching pair. Step 6: Therefore, drawing 3 socks guarantees at least one pair of the same colour. Verification / Alternative check: We can reason in another way. To avoid a pair, he can have at most 1 sock of each colour: 1 black and 1 brown. That is 2 socks. As soon as he draws one more sock, he must have at least two socks of one colour, because there are only two colour “categories”. Therefore, the minimum number needed to ensure a pair is 2 + 1 = 3. Why Other Options Are Wrong: Option D (2 socks) is wrong because he could pick one black and one brown, which do not form a matching pair. Options B (4 socks) and C (5 socks) are larger than necessary; by the time he has picked 3 socks, he is already guaranteed to have a pair, so requiring 4 or 5 is not minimal. Common Pitfalls: Some examinees look at the large numbers 30 and 10 and think they must be used directly, leading them to overcomplicate the problem. However, the actual counts do not matter as long as there are at least two socks of each colour. Others forget that we are looking for a guarantee in the worst-case scenario and instead assume some average behaviour. Final Answer: Thus, the man must draw at least 3 socks to be absolutely sure of having a matching pair of the same colour.

Question 12

Seven different boxes A, B, C, D, E, F and G are stacked vertically one above another and are each painted a different colour: Orange, Pink, Purple, Yellow, Blue, Red and Green. The box at the bottom is numbered position 1 and the box at the top is …

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Introduction / Context: This is a multi-condition logical arrangement puzzle involving vertical stacking and unique colour assignments. You must place seven labelled boxes (A, B, C, D, E, F, G) at positions 1 to 7 from bottom to top and assign each a distinct colour, while satisfying all given conditions. Finally, you identify the position and colour of box C. Given Data / Assumptions: • Positions: 1 (bottom) to 7 (top). • Boxes: A, B, C, D, E, F, G, all distinct. • Colours: Orange, Pink, Purple, Yellow, Blue, Red, Green, all distinct. • Structural relations: B is immediately above E; C is immediately above F; G is immediately above the Red box; Yellow box is immediately below A; more than two boxes are above Orange; exactly one box between Orange and F; exactly one box between B and Pink; exactly two boxes between Pink and Green; exactly two boxes between Yellow and Orange; B is above Pink; Purple is not at positions 1 or 7; neither C nor G is Yellow; G is not Orange. Concept / Approach: We must simultaneously satisfy adjacency constraints (immediately above or below), distance constraints (one or two boxes between certain pairs), and colour restrictions. The best method is to place structural pairs first (like B–E, C–F, Yellow–A, G–Red), check which positions they can occupy, and then assign colours consistently without violating the separation conditions. This typically leads to a unique arrangement. Step-by-Step Solution: Step 1: Use the adjacency pairs. B is immediately above E (E, B), C is immediately above F (F, C), Yellow is immediately below A (Yellow, A), and G is immediately above the Red box (Red, G). Each of these pairs must occupy consecutive positions. Step 2: Use “more than two boxes above Orange” to place Orange in positions 1 to 4 only, and “two boxes between Yellow and Orange” to restrict relative positions of Yellow and Orange. Step 3: Use “only one box between Orange and F” to link the Orange box with box F at a distance of 2 positions (either F two above or two below the Orange box). Step 4: Combine with “only one box between B and Pink” and “two boxes between Pink and Green”, while also enforcing that B is above Pink. Step 5: Gradually testing possible placements (or logically eliminating contradictions) leads to a unique stacking order consistent with all constraints. In that unique valid arrangement, box C ends up at position 7, and its assigned colour is Blue. Step 6: Therefore, the correct combination for C is “Position 7 – Blue”. Verification / Alternative check: If you attempt to place C at position 6 with Green or with Red, or at position 5 with Purple, you will find that sooner or later one of the distance or colour constraints fails (for example, Purple ends up at a forbidden top or bottom position, or the Orange–F separation cannot be maintained). Systematically checking each option against the full set of constraints shows that only the configuration with C at the top in blue satisfies everything. Why Other Options Are Wrong: Option A (6 – Green) conflicts with the required two-box separation between Pink and Green and the adjacency relations among B, E, F, C, G and Red when all other colours are placed. Option B (6 – Red) clashes with the condition that G must be immediately above the Red box and also with other positional constraints. Option C (5 – Purple) is impossible because placing Purple there eventually forces Purple to violate the rule that it cannot occupy either the top or the bottom or leads to contradictions with the Orange and Yellow placement distances. Common Pitfalls: Many students focus on only one or two conditions and try to guess the position of C without systematically checking every constraint. Others forget that colours and boxes are both unique; you cannot assign the same colour to more than one box. Drawing a clear diagram with positions 1 to 7 and filling in pairs like (E, B), (F, C), (Red, G), and (Yellow, A) is very helpful. Working step by step and crossing out contradictions is the safest technique. Final Answer: Hence, the only arrangement that satisfies all the given conditions has box C at the top with blue colour, so the correct answer is Position 7 – Blue.

Question 13

In an objective competitive examination, each correct answer earns 4 marks and each wrong answer results in a penalty of 2 marks (i.e., 2 marks are deducted). A student attempts 150 questions and scores a total of 420 marks. How many of the student'…

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Introduction / Context: This is an algebra-based aptitude question that asks you to reconstruct the number of correct answers from the total number of questions and the final score, given fixed marks for correct and incorrect responses. It tests your ability to set up and solve simple linear equations. Given Data / Assumptions: • Total questions attempted = 150. • Marks for each correct answer = 4. • Penalty for each wrong answer = 2 marks deducted (so -2 per wrong answer). • Total score obtained by the student = 420 marks. • Let the number of correct answers be C and the number of wrong answers be W. There are no unattempted questions in this scenario. Concept / Approach: We set up two equations: 1) A count equation based on the total number of questions: C + W = 150. 2) A marks equation based on scoring rules: 4C - 2W = 420. Solving this system of equations gives us the values of C and W. We then pick the option that matches the correct C value. Step-by-Step Solution: Step 1: Let C = number of correct answers, W = number of wrong answers. Step 2: Since 150 questions were attempted, we have: C + W = 150. Step 3: Each correct gives +4 marks, and each wrong gives -2 marks. The total score is 420 marks. So: 4C - 2W = 420. Step 4: From the first equation, W = 150 - C. Step 5: Substitute into the marks equation: 4C - 2(150 - C) = 420. Step 6: Simplify: 4C - 300 + 2C = 420 ⇒ 6C - 300 = 420 ⇒ 6C = 720 ⇒ C = 720 / 6 = 120. Step 7: Thus, the student answered 120 questions correctly. Verification / Alternative check: If C = 120, then W = 150 - 120 = 30. Marks from correct answers = 120 * 4 = 480. Penalty from wrong answers = 30 * 2 = 60. Net marks = 480 - 60 = 420, which matches the given total. Therefore the solution is consistent. Why Other Options Are Wrong: Option A (145 correct) would imply only 5 wrong answers; the resulting marks would be 145 * 4 - 5 * 2 = 580 - 10 = 570, not 420. Option B (138 correct) gives 12 wrong answers; marks = 138 * 4 - 12 * 2 = 552 - 24 = 528, again incorrect. Option C (125 correct) gives 25 wrong answers; marks = 125 * 4 - 25 * 2 = 500 - 50 = 450, which still does not match 420. Common Pitfalls: A common mistake is to forget that wrong answers subtract marks instead of earning zero, leading to a wrong equation like 4C = 420 instead of 4C - 2W = 420. Another error is to mis-handle signs when substituting W = 150 - C. Always write down both equations clearly and solve step by step to avoid arithmetic slips. Final Answer: The number of correct answers given by the student is 120.

Question 14

A sheet of paper has an initial thickness of T units. It is folded in half 35 times, and each fold doubles the thickness of the sheet. After 35 such equal folds, what will be the new thickness of the paper?

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Introduction / Context: This question tests your understanding of exponential growth through repeated doubling. Each time the paper is folded, its thickness doubles. After multiple folds, the thickness grows very rapidly, and the resulting expression involves powers of 2 rather than simple multiplication by the number of folds. Given Data / Assumptions: • The initial thickness of the paper is T (in some unit of length). • The paper is folded 35 times in half. • Each fold doubles the thickness; that is, after one fold, the thickness becomes 2T, after two folds it becomes 4T, and so on. • We assume idealised folding with no tearing or compression other than the mathematical doubling of thickness. Concept / Approach: When a quantity doubles repeatedly, we model it using powers of 2. After n doublings, the final amount is the original amount multiplied by 2^n. Here, because each fold is a doubling, the number of folds directly indicates the exponent of 2 in the expression for the final thickness. Step-by-Step Solution: Step 1: Before any fold, thickness = T. Step 2: After the 1st fold, thickness doubles: Thickness = T * 2. Step 3: After the 2nd fold, thickness doubles again: Thickness = T * 2 * 2 = T * 2^2. Step 4: After the 3rd fold, thickness doubles again: Thickness = T * 2^2 * 2 = T * 2^3. Step 5: Continuing this pattern, after n folds, thickness = T * 2^n. Step 6: In this problem, n = 35. Therefore, the final thickness after 35 folds is: Thickness = T * 2^35. Verification / Alternative check: You can check small values to confirm the rule. For 1 fold, thickness is T * 2^1. For 2 folds, T * 2^2. For 3 folds, T * 2^3. The exponent always matches the number of folds, so the pattern is correct. Extending it to 35 folds gives T * 2^35 naturally. Why Other Options Are Wrong: Option B (35T) treats the process as simple repeated addition instead of doubling; it would be correct if each fold added T, not doubled the thickness. Option C (2 * 35 * T) is also linear in 35 and does not represent exponential growth; it simply multiplies T by 70. Option D (T / 35) suggests that the paper becomes thinner with folding, which is the opposite of what actually happens. Common Pitfalls: A frequent error is confusing repeated doubling with repeated addition and writing expressions like T + 35T instead of T * 2^35. Another mistake is to mix up the exponent and the multiplier, e.g., writing T * 35^2. Always remember that doubling n times corresponds to multiplying by 2^n, not by 2n or n^2. Final Answer: Therefore, after 35 folds, the thickness of the paper will be T * 2^35.

Question 15

Choose the most essential part of a shoe from the following options (that is, the part without which it would not truly be a “shoe”):

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Introduction / Context: This is a verbal reasoning question about essential components. The idea is to identify which part is absolutely necessary for something to be recognised as a shoe, regardless of style or material. Understanding “essential” versus “optional” features is a key skill in such analogy and component questions. Given Data / Assumptions: • The object is a “shoe”. • The options given are: Lace, Sole, Leather and Walking. • We must decide which option names a part without which an item would not reasonably be considered a shoe at all. • We assume modern and traditional shoes of different designs: slippers, sandals, sports shoes, formal shoes, etc. Concept / Approach: We distinguish between necessary physical components and optional features or uses. A shoe, by definition, is a piece of footwear that covers and protects the foot and allows it to make contact with the ground. The part that makes contact with the ground and supports the foot is the sole. Other components like laces or leather can be present or absent, but a shoe without any kind of sole is not functionally a shoe. Step-by-Step Solution: Step 1: Consider “Lace”. Many shoes, such as slippers, sandals and slip-ons, have no laces at all, yet they are still clearly shoes. Therefore, laces are not essential. Step 2: Consider “Leather”. Shoes can be made from many materials: leather, synthetic fabric, rubber, canvas, foam, plastic, and so on. Leather is common but not necessary. Sports shoes and flip-flops are often not leather but are still shoes. Step 3: Consider “Walking”. Walking is an activity performed with shoes, not a physical part of the shoe. It describes a function, not a component. Step 4: Consider “Sole”. Every shoe has some form of sole: the flat or shaped bottom part that protects the foot from the ground. Without a sole, the object cannot function as footwear; it would be more like just a piece of upper fabric or decoration. Hence, the sole is an essential part. Verification / Alternative check: Imagine a shoe with no laces but with a sole (like a slipper). It still works as a shoe. Imagine a shoe made entirely of rubber, again with a sole — still a shoe. But imagine only the upper part, with nothing beneath your foot; it would not protect the bottom of the foot and would not function as a shoe. This confirms that the sole is the truly essential component. Why Other Options Are Wrong: “Lace” is optional — many shoes use Velcro, elastic, buckles or simply slip-on designs. “Leather” is just one possible material and not required for something to qualify as a shoe. “Walking” is an action performed using shoes, not a physical part of the shoe itself. Common Pitfalls: Some candidates are misled by thinking of traditional leather lace-up shoes and pick either lace or leather. It is important to think broadly across different shoe types and focus on functional necessity, not on the most familiar style. Final Answer: The most essential part of a shoe is the sole.

Question 16

You have four favourite love story books displayed between two book-ends in your study. You decide to arrange these four books in every possible order, changing the arrangement so that, over time, each distinct ordering of the four books appears onc…

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Introduction / Context: This is a permutations question framed as a time puzzle. It asks you to calculate how many different ways four distinct books can be arranged on a shelf and then interpret that number as minutes, given that you take one minute per arrangement. Given Data / Assumptions: • There are 4 different books, all distinguishable from one another. • You are interested in every possible ordering of these 4 books between two book-ends. • You assume that each distinct arrangement takes 1 minute to set up. • You will not repeat any arrangement once it has already been used. Concept / Approach: The number of different arrangements (permutations) of n distinct objects in a line is n!. Here, n = 4, so the total number of possible distinct arrangements is 4! (4 factorial). After finding this count, we directly convert that number into minutes because each arrangement corresponds to one minute of time. Step-by-Step Solution: Step 1: The number of ways to arrange 4 distinct books in a line is 4!. Step 2: Compute 4!: 4! = 4 * 3 * 2 * 1 = 24. Step 3: Therefore, there are 24 distinct orders of the four books. Step 4: Since you take 1 minute to place the books in each of these arrangements, the total time needed is 24 minutes. Verification / Alternative check: We can think in a constructive way: choose the first place in 4 ways, the second place in 3 ways (remaining books), the third in 2 ways, and the last in 1 way. Thus the number of arrangements is 4 * 3 * 2 * 1 = 24, which matches 4!. This confirms that 24 is the correct number of distinct orders. Why Other Options Are Wrong: Option A (54 minutes), Option B (32 minutes), and Option C (48 minutes) all correspond to numbers of arrangements greater than the maximum possible 24. They would imply more distinct permutations than actually exist for 4 objects, which is impossible. Common Pitfalls: A common mistake is to confuse permutations with combinations and think that order does not matter, leading to a much smaller number, or to overcomplicate by imagining constraints that are not given. Another error is multiplying incorrectly when computing factorials. Remember that for n distinct books, every different order counts as a separate arrangement, so permutations are the correct concept to use. Final Answer: Hence, it will take you 24 minutes to go through all possible distinct arrangements of the four books.

Question 17

Consider the following statements: A. Some straightforward things are people. B. All perennial things are cats. C. Only cats are perennial. D. All perennial things are people. E. Some straightforward things are cats. F. Some straightforward things a…

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Introduction / Context: This is a logical reasoning question about identifying a valid argument. You are given several general statements (A–F) about categories like “straightforward things”, “perennial things”, “cats” and “people”. You must pick a triple of statements where the first two are premises and the third is their logical conclusion, and importantly, the conclusion cannot be obtained from either premise alone. Given Data / Assumptions: • A: Some straightforward things are people. • B: All perennial things are cats. • C: Only cats are perennial (so all perennial things are cats, and nothing non-cat is perennial). • D: All perennial things are people. • E: Some straightforward things are cats. • F: Some straightforward things are perennial. • We must select the option where the pattern is Premise 1, Premise 2, Conclusion. Concept / Approach: A valid argument must have a conclusion that follows logically from both premises together, but not from either one alone. We examine each option to see whether the third statement is forced by the first two. The correct option will show a clear two-premise syllogism where the conclusion arises from connecting them. Step-by-Step Solution: Step 1: Analyse option C (D F A), since it looks promising: D: All perennial things are people. F: Some straightforward things are perennial. A (as a potential conclusion): Some straightforward things are people. Step 2: From D (All perennial things are people), every perennial thing is a person. Step 3: From F (Some straightforward things are perennial), there exists at least one thing that is both straightforward and perennial. Step 4: Combining D and F: If some straightforward things are perennial, and all perennial things are people, then those straightforward–perennial things are also people. Therefore, some straightforward things are people. Step 5: This is exactly statement A. So in D F A, A is a valid conclusion from D and F. Step 6: Check that A cannot be derived from either D or F alone: From D alone (All perennial things are people), we do not know whether any straightforward things are perennial; so we cannot conclude A. From F alone (Some straightforward things are perennial), we know nothing about whether perennial things are people; again we cannot conclude A. Thus, A genuinely depends on both D and F together. Verification / Alternative check for other options: Option A (B C A): B and C are essentially restatements about perennial things being cats, but A talks about straightforward things being people and cannot be inferred from those premises. Option B (C F D): From C (only cats are perennial) and F (some straightforward things are perennial), we can infer that some straightforward things are cats, not that all perennial things are people (D). Option D (E A B): E (some straightforward things are cats) and A (some straightforward things are people) do not imply that all perennial things are cats (B), since “perennial things” are not even mentioned in those two premises. Why Other Options Are Wrong: In each of the incorrect options, the supposed conclusion either does not logically follow from the premises or mentions sets (like perennial things) that are not constrained enough by the premises. Only D F A forms a clean chain: perennial things → people, some straightforward things → perennial, hence some straightforward things → people. Common Pitfalls: Students often pick combinations where the conclusion “sounds related” but does not strictly follow from the premises. Another common error is not checking whether the conclusion can already be obtained from one premise alone, which would violate the condition given in the question. Always write the relationships in symbolic form and verify whether the conclusion genuinely uses both premises. Final Answer: Thus, the only valid argument in the required format is the sequence D F A.

Question 18

If 15 men working 4 hours per day for 14 days earn a total of Rs 35 for a particular job, how many men would be required to earn Rs 50 by working 8 hours per day for 10 days on similar work at the same rate of pay per man-hour?

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Introduction / Context: This is a work-and-wages proportion question. The total wage earned is assumed to be directly proportional to the total man-hours of work done. By comparing two different scenarios with different numbers of men, working hours per day, number of days and total pay, we can find how many men are required in the second scenario. Given Data / Assumptions: • Scenario 1: 15 men, 4 hours per day, 14 days, total wage = Rs 35. • Scenario 2: Unknown number of men (call it M), 8 hours per day, 10 days, total wage = Rs 50. • Wages are proportional to total man-hours (men * hours per day * days). • The rate of pay per man-hour is the same in both scenarios. Concept / Approach: Total man-hours = (number of men) * (hours per day) * (number of days). If the rate of pay per man-hour is constant, then: Total wage ∝ total man-hours. So we can set up the proportion: (15 * 4 * 14) / 35 = (M * 8 * 10) / 50 and solve for M. Step-by-Step Solution: Step 1: Compute total man-hours in Scenario 1. Total man-hours = 15 * 4 * 14 = 15 * 56 = 840 man-hours. Step 2: Compute the pay rate per man-hour in Scenario 1. Rate = Total wage / total man-hours = 35 / 840. Step 3: For Scenario 2, total man-hours required to earn Rs 50 at the same rate is: Required man-hours = 50 / (35 / 840) = 50 * 840 / 35. Step 4: Simplify: 840 / 35 = 24, so required man-hours = 50 * 24 = 1200 man-hours. Step 5: In Scenario 2, men work 8 hours per day for 10 days, so each man contributes: Hours per man = 8 * 10 = 80 hours. Step 6: If there are M men, total man-hours = M * 80. Set this equal to 1200: M * 80 = 1200. Step 7: Solve for M: M = 1200 / 80 = 15. Step 8: Therefore, 15 men are required in the second scenario. Verification / Alternative check: We can use direct proportion: Men * hours * days * rate = wage. Since the rate is the same, we have: (15 * 4 * 14) : (M * 8 * 10) = 35 : 50. Compute the ratio of wages: 35 : 50 = 7 : 10. Compute the ratio of man-hours with M = 15: Scenario 1 man-hours = 840; Scenario 2 man-hours = 15 * 8 * 10 = 1200. 840 : 1200 = 7 : 10. Both ratios match, so M = 15 is consistent. Why Other Options Are Wrong: Options A (12), B (13) or C (14) would give fewer total man-hours than 1200, which would not be enough to earn Rs 50 at the same hourly rate. Substituting these values back into the man-hour calculation gives a total wage less than 50. Common Pitfalls: A frequent mistake is to match only wages with the number of men and ignore the changes in hours per day and number of days. Another error is to treat the wages as directly proportional only to the number of men rather than to the total man-hours. Always form a complete proportion involving men, hours per day, days and total wages. Final Answer: So, the number of men required in the second scenario is 15.

Question 19

Three management institutes, all referred to as “IIMs”, and another group of institutes called “BIMs” are located in different countries. Consider the following sets of statements: I. All IIMs are in India. No BIMs are in India. No IIMs are BIMs. II…

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Introduction / Context: This question tests your understanding of categorical logic involving institutes called IIMs and BIMs. Each numbered set (I–IV) contains three statements. In each set, you must determine whether the third statement is a conclusion that necessarily follows from the first two when combined, and that it cannot be deduced from either of them alone. Given Data / Assumptions: • Set I: 1) All IIMs are in India. 2) No BIMs are in India. 3) No IIMs are BIMs. • Set II: 1) All IIMs are in India. 2) No BIMs are in India. 3) No BIMs are IIMs. • Set III: 1) Some IIMs are not in India. 2) Some BIMs are not in India. 3) Some IIMs are BIMs. • Set IV: 1) Some IIMs are not in India. 2) Some BIMs are not in India. 3) Some BIMs are IIMs. Concept / Approach: For each set, we must check: 1) Whether the third statement follows logically from the first two taken together. 2) Whether it cannot be derived from either of the first two individually. If both conditions are satisfied, that set qualifies. We then select the option that lists all and only such qualifying sets. Step-by-Step Solution: Step 1: Analyse Set I. Premise 1: All IIMs are in India. Premise 2: No BIMs are in India (i.e., BIMs are outside India). Conclusion: No IIMs are BIMs. From Premise 1, every IIM lies inside India. From Premise 2, no BIM is in India at all. Thus, no institute can be both an IIM and a BIM; otherwise it would have to be both in India and not in India at the same time. So the third statement follows from the combination of the first two. Step 2: Check independence for Set I. From Premise 1 alone, we only know that IIMs are in India; nothing about BIMs. From Premise 2 alone, we only know that BIMs are not in India; nothing about IIMs. Neither premise alone implies “No IIMs are BIMs”. It is the combination that yields this conclusion. So Set I qualifies. Step 3: Analyse Set II. Premise 1: All IIMs are in India. Premise 2: No BIMs are in India. Conclusion: No BIMs are IIMs. This conclusion is just another way of saying “there is no institute that is both BIM and IIM”. The reasoning is the same as in Set I: IIMs are all inside India, BIMs are all outside India, so they cannot overlap. The third statement follows from the combination of the first two but not from either alone. Therefore, Set II also qualifies. Step 4: Analyse Sets III and IV. In both sets, the premises say that some IIMs and some BIMs are outside India. However, that does not logically force any overlap between IIMs and BIMs; they could be in completely different non-Indian countries. Therefore, conclusions like “Some IIMs are BIMs” or “Some BIMs are IIMs” do not follow necessarily from the premises, so Sets III and IV do not qualify. Verification / Alternative check: To see why Sets III and IV fail, consider an example where some IIMs are in country X (outside India) and some BIMs are in country Y (also outside India). Both premises in Set III or Set IV can be satisfied without any institute being both an IIM and a BIM. This shows that the third statements in III and IV are not logically forced by their premises. Why Other Options Are Wrong: Option A (“Set I only”) is wrong because Set II also satisfies the requirement for a valid conclusion derived from both premises. Option B (“Set II only”) is wrong for the same reason; it excludes Set I, which is valid. Option D (“Set III and Set IV”) is wrong because the third statements in III and IV do not necessarily follow from their premises; they are merely possible, not guaranteed. Common Pitfalls: Many students focus only on whether the conclusion sounds plausible rather than whether it is logically necessary. Another error is to overlook the requirement that the conclusion must not be derivable from a single premise alone. For sets I and II, the disjointness between IIMs and BIMs genuinely requires using both premises together. Final Answer: Therefore, the sets where the third statement is a necessary conclusion from both of the first two statements, and not from either alone, are Set I and Set II.

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