Correct Option: 1

Rationale: The author uses the “beige office building” analogy to describe the style and tone of the AI’s output, characterizing it as “bland,” “generic,” and designed to sound “polite,” “empathetic,” and “safe” (fostering trust). Option 1 refers to how AI companies explain or defend their errors (“blamed biases in its training data”). This is a procedural defense mentioned later in the text, not a description of the “bland” aesthetic or tone represented by the beige building metaphor.

Why other options wrong: Option 2 (“generalized responses”) aligns with the text’s description of “bland” and “generic.” Option 3 (“foster trust”) aligns with the text’s description of strategies to make us “feel safe” and seem “professional.” Option 4 (“warm, polite”) aligns with the text’s description of the AI sounding like a “colleague,” “polite,” and “empathetic.”

Difficulty: Moderate

Correct Option: 3

Rationale: The question asks to identify the odd one out based on the purpose of the example. Options 1, 2, and 4 are all cited by the author as evidence of the AI’s failures, biases, or errors (sexism, cultural erasure, and favoritism). Option 3 is different because it is a quote where the AI explains its own design philosophy (“designed to foster trust”). The author uses this to illustrate the mechanism of the “veneer” of neutrality, whereas the other options are examples of the specific biases that persist behind that veneer.

Why other options wrong: Option 1 (Bing Image Creator) is used to demonstrate sexism/bias. Option 2 (Indian vs. American English) is used to demonstrate cultural bias/US-centrism. Option 4 (Sam Altman/Positive Book) is used to demonstrate potential favoritism/bias. All three serve the same purpose: proving the “inaccuracies and biases.”

Difficulty: Moderate

Correct Option: 1

Rationale: The question asks for the statement that does *not* affirm the difference (disjunct) between the tech companies’ positive claims and the negative reality of what AI actually does. Option 1 is a statement of uncertainty and a lack of evidence (“I’m not aware of research… I can only guess”). By explicitly stating that the author cannot confirm whether the bias towards Big Tech is a systemic issue or just a result of their specific conversation, this statement stops short of affirming the disjunct. It treats the observation as anecdotal and unproven, whereas the disjunct requires a definitive contrast between a claim and a factual reality.

Why other options wrong:

Option 2 affirms the disjunct by contrasting the implicit claim that these products are purely beneficial tools for users with the reality that they are mechanisms for “big tech’s accumulation of wealth” and that users are “victims” of exploitation. This contrasts the altruistic branding with the economic reality. Option 3 clearly contrasts the “veneer of collegial neutrality” (Claim) with the “false or biased responses” (Reality). Option 4 contrasts the implied neutrality of an image generator with the specific reality of it producing an “entirely male cast” (Sexism/Bias).

Difficulty: Hard

Correct Option: 3

Rationale: The author argues the exact opposite of this claim. The passage states: “I wanted to enact how the product’s veneer of collegial neutrality could lull us into absorbing false or biased responses without much critical engagement.” Therefore, the author believes the neutrality is detrimental to critical thinking, making Option 3 the claim they are least likely to agree with.

Why other options wrong: Option 1 is a suspicion the author entertains (“ChatGPT seemed to be guiding me… I can only guess”), so they might agree, albeit cautiously. Option 2 is explicitly stated (“makes us complicit… we’re both victims and beneficiaries”). Option 4 is supported by the text’s discussion of funders seeking a return on investment by making products appear as “trustworthy collaborators.”

Difficulty: Easy

Correct Option: 2

Rationale: The passage explicitly contradicts the idea of replacement. It states that the emotional component is “no less ‘real’ than a surveyor’s map,” establishing its validity, but later asserts, “Both literary and scientific approaches to place are necessary, working in unison, to achieve a complete record.” This implies they are complementary, and one cannot replace the other.

Why other options wrong: Option 1 is supported by the text: “Histories of places live on in many forms, one of which is the human memory or imagination.” Option 3 is supported by the text: “provide the emotional component… gives us a sense of how people relate.” Option 4 is supported by the text: “no less ‘real’ than a surveyor’s map.”

Correct Option: 4

Rationale: The question asks for a conclusion drawn from the statement about “bridging disciplines.” The text states: “valuable knowledge produced from bridging disciplines and combining material from both the arts and the sciences to better understand the human condition.” Option 4 paraphrases this logical flow almost exactly: “A comprehensive understanding of the human condition can best be achieved by combining the findings of disciplines from the arts and the sciences.”

Why other options wrong: Option 1 reverses the logic (suggesting studying the human condition achieves knowledge of arts/sciences). Option 2 uses vague phrasing (“bridging of the human condition”) that distorts the meaning. Option 3 limits the scope to “emotions” and “places we visit” rather than the broader “human condition.”

Difficulty: Easy

Correct Option: 2

Rationale: The question refers to the example given in the final paragraph: “For example, a literary description of place may involve… how people from different races or classes can experience the same place in different ways linked to those racial or class disparities.” This directly supports Option 2, which states literature can convey how different people experience the same place differently.

Why other options wrong: Option 1 mentions “absence,” but the specific “different races/classes” example is the primary illustration of how literature bridges scientific gaps (like sociology) and captures the “complexity” of the human experience. Option 3 is irrelevant to the text. Option 4 contradicts the text’s argument for the equality/necessity of both approaches.

Difficulty: Moderate

Correct Option: 1

Rationale: The question asks for the statement that, if false, would NOT contradict the passage. This implies we are looking for a statement that is false according to the passage (because if you negate a false statement, it becomes true/consistent). Option 1 says: “humans do not interact with places in subjective, emotional ways…” The passage argues the exact opposite: humans do interact in subjective ways (“Human consciousness of place is experienced in a multi-modal manner”). Therefore, Statement 1 is false relative to the text. If Statement 1 is false (i.e., “It is false that humans do not interact…”), it means “Humans DO interact…”. This aligns with the passage. Thus, it is the exception.

Why other options wrong: Options 2, 3, and 4 are true according to the passage. If they were false (e.g., “Writing CANNOT capture…”, “Literature does NOT provide insights…”, “Descriptions DO need satellite imagery…”), they would contradict the passage’s arguments.

Difficulty: Hard

Correct Option: 1

Rationale:

The coherent paragraph constructs a philosophical argument advocating for the legalization of assisted dying based on the concept of freedom over death.

Sentence 5 introduces the broad context: Freedom is a complex and contested philosophical notion.

Sentence 2 narrows the focus: Despite this complexity, the author argues that a specific type of freedom, freedom over death (shaping the timing and circumstances of death), should be central to the conversation.

Sentence 4 connects this concept to the specific issue: It states that many people view this freedom as encompassing a right to medical assistance in hastening our deaths.

Sentence 3 concludes the argument: Therefore, legalising assisted dying is a logical step in realising this freedom.

Sentence 1 is the odd one out. While it mentions freedom over death, it focuses on the historical and sociological causes of this freedom (technological and sociocultural developments) and humanity’s adaptation to it. This provides background context but breaks the specific argumentative chain (General Philosophy -> Definition -> Specific Application -> Conclusion) formed by the other four sentences.

Difficulty: Medium

Correct Option: 1

Rationale: The coherent paragraph focuses on the physical and sensory experience of viewing the Bayeux Tapestry, specifically how its display facilitated an immersive interaction for the viewer.

Sentence 3 introduces the core theme: Art historian Linda Neagley’s theory that pre-Renaissance people interacted with art visually, kinaesthetically… and physically. Sentence 5 connects directly to this interaction: The tapestry would have been hung at eye level to enable this (referring to the physical interaction mentioned in 3). Sentence 2 expands on the physical arrangement (hung in a square) and the viewer’s position (stood in the centre). Sentence 4 concludes the thought: This arrangement (That) would make it an immersive space that played on the viewer’s senses.

Sentence 1 is the odd one out because it shifts the focus from the method of display/viewer experience to the narrative content and political purpose of the tapestry (telling the story of the English downfall and Norman rise). While related to the tapestry, it does not fit the specific argument about sensory immersion constructed by the other four sentences.

Difficulty: Medium

Correct Option: 2

Rationale: Riddle’s inference from the experiment is that there is no link between the presence of eyes and the accumulation of yellow fat (carotenoids). She concludes that the fat color is likely an adaptation for inflammation control, not merely a buildup of unused material. If the experiment had shown that only eyeless offspring had yellow fat, it would establish a direct, exclusive link between eyelessness and carotenoid accumulation. This result would support the initial hypothesis she rejected (that they accumulate it simply because they lack eyes to use it) and would directly contradict her finding that eye presence and fat color are unrelated.

Why other options wrong: Option 1 supports Riddle’s conclusion. It proves that having eyes doesn’t prevent yellow fat accumulation, confirming there is no exclusive link. Option 3 is a neutral or supportive finding, consistent with the idea that the traits are independent. Option 4 is the actual result mentioned in the passage (“Some eyeless cavefish had fat that was practically white”). This finding was the key evidence leading to her conclusion that being eyeless doesn’t guarantee yellow fat.

Difficulty: Moderate

Correct Option: 3

Rationale: The “apparent inefficiency” refers to the process where cavefish embryos start developing eyes only to destroy them later. The passage explains this is unavoidable because the early development of the brain and the eye are completely intertwined. This means the cavefish must follow the same early developmental path as their sighted ancestors to ensure their brains develop correctly. Option 3 captures this by stating the cavefish are similar to the non-cave variants in these early stages; this shared developmental constraint is the fundamental reason the inefficiency cannot be avoided.

Why other options wrong: Option 1 discusses enhancements (tastebuds) that compensate for the loss of sight, but does not explain why the developmental process of making and killing the eye is necessary. Option 2 attributes the loss to lack of light, but the passage attributes the cell death to genetic programming (“natural selection to have its way,” “cells… start dying”), not direct environmental exposure. Option 4 discusses food availability, which explains why they need to be efficient (save energy), but not why the specific inefficient mechanism of “build-then-destroy” is unavoidable.

Difficulty: Moderate

Correct Option: 2

Rationale: Option 2 is the only statement that does not describe a trait or behavior of the Mexican tetra cavefish. Instead, it is a general explanation by physiologist Nicolas Rohner regarding the biological mechanism of inflammation in humans and other animals when fat cells burst. It describes the problem (toxicity of fat) that the cavefish have successfully overcome, rather than the adaptation itself. The passage later contrasts this by stating that cavefish, unlike the subjects in this statement, show fewer signs of inflammation.

Why other options wrong: Option 1 describes the evolutionary pressure (avoiding “expensive tissues”) that drives the adaptation of losing eyes. Option 3 describes a behavioral and physiological adaptation: the ability to “get very fat” to store energy. Option 4 describes the fish “surviving” on a unique diet, implying the adaptation effectively allows them to live in a niche where “relatively little lives.”

Difficulty: Easy

Correct Option: 3

Rationale: The passage explicitly states Riddle’s conclusion regarding the function of the yellow fat: “Riddle thinks these carotenoids may be another adaptation to suppress inflammation, which might be important in the wild, as cavefish are likely overeating.”

Why other options wrong: Option 1 was the initial hypothesis (“accumulating these because they don’t have eyes”) which was tested and rejected by Riddle. Option 2 states the yellow color is a visual side effect of the carotenoids, not their biological function. Option 4 suggests carotenoids help the fat cells store nutrients, but the passage implies they protect the tissue from inflammation caused by the storage.

Difficulty: Easy

Correct Option: 1 (Option 3)

Rationale: The missing sentence contains the phrase “thinkers at the time,” which indicates a reference to a specific historical period mentioned immediately before. The sentence preceding blank 3 discusses “the course of the seventeenth century and early eighteenth century,” providing the necessary temporal antecedent for “at the time.” Thematically, the missing sentence elaborates on the specific nature of “taste” during that period—specifically its link to pleasure and embodiment—which clarifies why it was considered a central category for aesthetic and ethical thinking, as mentioned in the sentence ending at blank 3. Placing the sentence here provides a logical bridge to the following sentence, which discusses the value of recovering the “thinkers” who held these specific views.

Why other options wrong:

Option 2 (referring to blank 2) is incorrect because the historical time period (17th/18th century) is introduced in the sentence following this blank, so “at the time” would have no referent.

Option 3 (referring to blank 1) is incorrect for the same reason; the paragraph has just begun, and no time period has been established.

Option 4 (referring to blank 4) is incorrect because placing the sentence at the very end disrupts the flow. The sentence ending at blank 3 introduces the historical context, and it is most logical to immediately define the characteristics of thought during that time (the content of the missing sentence) before concluding with the sentence about recovering those specific neglected thinkers.

Difficulty: Moderate

Correct Option: 1 (Option 3)

Rationale: The missing sentence contains the phrase “thinkers at the time,” which indicates a reference to a specific historical period mentioned immediately before. The sentence preceding blank 3 discusses “the course of the seventeenth century and early eighteenth century,” providing the necessary temporal antecedent for “at the time.” Thematically, the missing sentence elaborates on the specific nature of “taste” during that period—specifically its link to pleasure and embodiment—which clarifies why it was considered a central category for aesthetic and ethical thinking, as mentioned in the sentence ending at blank 3. Placing the sentence here provides a logical bridge to the following sentence, which discusses the value of recovering the “thinkers” who held these specific views.

Why other options wrong:

Option 2 (referring to blank 2) is incorrect because the historical time period (17th/18th century) is introduced in the sentence following this blank, so “at the time” would have no referent.

Option 3 (referring to blank 1) is incorrect for the same reason; the paragraph has just begun, and no time period has been established.

Option 4 (referring to blank 4) is incorrect because placing the sentence at the very end disrupts the flow. The sentence ending at blank 3 introduces the historical context, and it is most logical to immediately define the characteristics of thought during that time (the content of the missing sentence) before concluding with the sentence about recovering those specific neglected thinkers.

Difficulty: Moderate

Correct Option: 4

Rationale: The passage explicitly contrasts the two disciplines to illustrate different styles of scientific embodiment. It describes astronomy as largely a “receptive” science (dependent on detecting emissions) and contrasts it with contemporary biology, which it describes as having an “interventional” culture (using techniques like gene-splicing). Option 4 accurately reflects this distinction made by the author.

Why other options wrong: Option 1 is incorrect because the passage argues that both receptive and interventionist sciences are “instrumentally embodied,” just in different styles; it does not suggest they are mutually exclusive categories of “embodied” vs “interventionist.” Option 2 is incorrect because the “new astronomy” is explicitly defined as opening the full electromagnetic spectrum (radio, gamma, etc.), which happened in the 20th century, long after Newton. Option 3 is incorrect because the passage attributes the observation of the phases of Venus and Jupiter’s satellites to Galileo, not Newton.

Difficulty: Moderate

Correct Option: 2

Rationale: The passage characterizes scientific instruments as technological devices that extend human perception (telescopes), manipulate matter (gene-splicing), or process data (computational machinery). They are described as having a “trajectory” of improvement and being part of “technoscience.” A milestone is a static, passive marker used for reference, not a complex tool for observation, intervention, or data processing. It lacks the dynamic, technological, and functional characteristics attributed to instruments in the text.

Why other options wrong: A saxophone (Option 1), while musical, is still a complex “instrument” requiring manipulation and technology, sharing more mechanical characteristics with scientific tools than a rock. A kitchen oven (Option 3) and a scalpel (Option 4) involve intervention and manipulation of materials, aligning closer to the “interventional” nature of biological instruments described in the passage.

Difficulty: Easy

Correct Option: 1

Rationale: The passage begins by stating, “Different sciences exhibit different science cultures and practices.” It later identifies mathematics as one of these scientific disciplines (“the last of the scientific disciplines… is mathematics”). If Option 1 were true—that mathematics is only *now* beginning to develop a scientific culture because of computers—it would contradict the passage’s premise that mathematics was already a science (and thus possessed a science culture) prior to its recent adoption of computational machinery.

Why other options wrong: Option 2 does not contradict the passage; while the text distinguishes between receptive and interventional styles, it does not forbid an instrument from having both functions. Option 3 supports rather than contradicts the general theme of instruments expanding perception beyond human limits. Option 4 does not contradict the passage because the author emphasizes that sciences are “today” highly instrumentalized; the mention of Newton’s telescopes refers to observational astronomy, not necessarily his theoretical work on gravity, and the text acknowledges that early science had different limits.

Difficulty: Hard

Correct Option: 3

Rationale: The quoted statement argues that modern science relies on “constant improvements” in instruments to “progress successfully.” Option 3 applies this principle directly to the two main examples in the text: astronomy and microbiology. It correctly concludes that the progress in these specific fields is a result of the improvements in their respective tools (telescopes and biotech/microscopes).

Why other options wrong: Option 1 mentions “scientific constants,” which is a specific physics concept not discussed in the passage. Option 2 introduces a moral requirement (“must be respected”) that is not present in the text. Option 4 is vague and abstract (“embodiment of progress”), failing to capture the specific cause-and-effect relationship between better tools and scientific success described in the passage.

Difficulty: Moderate

Correct Option: 4

Rationale:
The passage strongly foregrounds the irony of the situation: a game invented by a left-wing feminist to critique wealth concentration—through two contrasting rule sets (a twin game)—ends up being commercialised in precisely the opposite spirit. An unemployed man, Charles Darrow, sells only the monopolist version to Parker Brothers, while Elizabeth Magie is denied credit and royalties. Option 4 alone captures all these layers together: the ideological irony, the mismatch between the creator and the beneficiary, the loss of one half of the twin game, and the injustice of recognition and compensation. It synthesises both the narrative facts and the author’s critical stance, making it the closest to the passage’s essence.

Why other options are wrong:
Option 1 incorrectly portrays Darrow as a symbol of individual ingenuity, directly contradicting the critical tone of the passage.
Option 2 is emotive but incomplete; it highlights “snatching” without explaining the crucial idea of two versions of the game or why their loss matters.
Option 3 is factually accurate about the commercial transaction but misses the soul of the passage: it downplays the irony, ignores the twin-rule-set concept, and reduces the story to a neutral corporate summary rather than an ideological appropriation.

Difficulty: Hard

Correct Option: 3

Rationale: The passage challenges the conventional wisdom that human presence is inherently destructive to nature. It presents arguments and evidence—such as the creation of “ecological mosaics” and data from the early Holocene—to suggest that human activities have historically been just as likely to enhance biodiversity as to deplete it. Option 3 captures the full scope of this argument. It explicitly mentions the “contrary” view (challenging the idea that nature thrives only in human absence), accurately reflects the statistical finding (“equally likely to increase biodiversity than reduce it”), and explains the mechanism (“creating varied ecological landscapes”).

Why other options wrong:

Option 1 is a reasonable summary but lacks the precision of Option 3. It mentions that humans “assist preservation” but misses the crucial nuance about the “equal likelihood” of increasing or reducing biodiversity, which is central to the passage’s balanced view.

Option 2 focuses heavily on the specific examples of meadows and the Holocene study. However, its conclusion that these findings “bely the idea that humans harm nature” is an overstatement. The passage states humans are *equally likely* to increase or reduce biodiversity, not that they don’t harm it at all.

Option 4 focuses too narrowly on the specific comparison between human-made landscapes (meadows) and intentional nature reserves/parks. While this comparison is used as evidence in the text, it is a supporting detail rather than the main thesis, which is the broader relationship between human presence and biodiversity.

Difficulty: Moderate

Correct Answer: 4

From the given information and the graphs, Brajen wrote a total of 8 papers. Since there were two four-author papers and every author must have contributed to all four-author papers, Brajen wrote exactly 2 four-author papers. Hence, the remaining papers written by Brajen are 8 − 2 = 6.

It is given that Brajen wrote the same number of single-author and two-author papers. Trying possible splits of these 6 remaining papers, the only feasible distribution satisfying at least one paper of each type is:

2 single-author papers, 2 two-author papers, 2 three-author papers.

Thus, the total number of two-author and three-author papers written by Brajen is: 2 + 2 = 4.

Table Steps

Initial Working Table (Blank – from graphs only)

(Source: Author bar graph + Paper-type bar graph)

Step 1

Total paper contributions =
1×10 (single-author) + 2×4 (two-author) + 3×3 (three-author) + 4×2 (four-author) = 35 → matches total author-wise papers 5 + 8 + 12 + 10 = 35
(Source: Paper-type bar graph + Author bar graph)

Step 2

There are exactly 2 four-author papers → all four authors must have written both → each author has exactly 2 four-author papers
(Source: Paper-type bar graph)

Table after Step 2

Step 3

Subtract four-author papers from each author’s total:
Arman: 5 − 2 = 3
Brajen: 8 − 2 = 6
Chintan: 12 − 2 = 10
Devon: 10 − 2 = 8
(Source: Author bar graph + Step 2)

(These are working totals for single + two + three-author papers.)

Step 4

Brajen wrote the same number of single-author and two-author papers →
2 × (single-author of Brajen) + (three-author of Brajen) = 6 → only feasible split:
single = 2, two-author = 2, three-author = 2
(Source: Fact 4 + Step 3)

Table after Step 4

Step 5

Arman must have written at least one of each type and has only 3 papers remaining → forced split:
1 single, 1 two-author, 1 three-author
(Source: Fact 1 + Step 3)

Table after Step 5

Step 6

All authors wrote different numbers of single-author papers →
Arman = 1, Brajen = 2 → remaining single-author papers = 10 − (1 + 2) = 7 → only possible distinct split for Chintan and Devon is {3, 4}
(Source: Fact 2 + Single-author total from graph)

Step 7

Total three-author papers = 3 → total three-author contributions = 9 →
Arman contributes 1, Brajen contributes 2 → remaining contributions = 6 for Chintan and Devon
(Source: Paper-type bar graph + Steps 4 and 5)

Step 8

Chintan and Devon each wrote more three-author papers than Brajen (who wrote 2) → both must have written at least 3 → hence:
three-author of Chintan = 3, three-author of Devon = 3
(Source: Fact 3 + Step 7)

Step 9

Since Chintan and Devon appear in all three three-author papers, every three-author paper must include both → only possible groups are
{Arman, Chintan, Devon} (1 paper) and {Brajen, Chintan, Devon} (2 papers)
(Source: Logical consequence of Step 8)

Step 10

Compute remaining two-author papers using row totals:

Case 1: Chintan single = 3, Devon single = 4 → Chintan two-author = 4, Devon two-author = 1

Case 2: Chintan single = 4, Devon single = 3 → Chintan two-author = 3, Devon two-author = 2

(Source: Step 3 + Step 6 + arithmetic consistency)

Step 11

Since both Case 1 and Case 2 satisfy all graphs and facts, statements assuming a fixed comparison between Chintan and Devon or a fixed two-author count for Chintan are not necessarily true
(Source: Deduction from Step 10)

Step 12

If Devon wrote more than one two-author paper → Case 1 is eliminated → only Case 2 remains → Chintan wrote exactly 3 two-author papers
(Source: Question 4 condition + Step 10)

Final Working Table

Correct Option: 4 – Neither i nor ii

From the completed deductions, two globally consistent cases are possible:

Case 1: Chintan wrote 3 single-author and 4 two-author papers, while Devon wrote 4 single-author and 1 two-author paper.

Case 2: Chintan wrote 4 single-author and 3 two-author papers, while Devon wrote 3 single-author and 2 two-author papers.

In Case 1, statement (i) is false and statement (ii) is false.

In Case 2, statement (i) is true and statement (ii) is true.

Since neither statement holds true in all possible valid cases, neither statement is necessarily true.

Table Steps

Initial Working Table (Blank – from graphs only)

(Source: Author bar graph + Paper-type bar graph)

Step 1

Total paper contributions =
1×10 (single-author) + 2×4 (two-author) + 3×3 (three-author) + 4×2 (four-author) = 35 → matches total author-wise papers 5 + 8 + 12 + 10 = 35
(Source: Paper-type bar graph + Author bar graph)

Step 2

There are exactly 2 four-author papers → all four authors must have written both → each author has exactly 2 four-author papers
(Source: Paper-type bar graph)

Table after Step 2

Step 3

Subtract four-author papers from each author’s total:
Arman: 5 − 2 = 3
Brajen: 8 − 2 = 6
Chintan: 12 − 2 = 10
Devon: 10 − 2 = 8
(Source: Author bar graph + Step 2)

(These are working totals for single + two + three-author papers.)

Step 4

Brajen wrote the same number of single-author and two-author papers →
2 × (single-author of Brajen) + (three-author of Brajen) = 6 → only feasible split:
single = 2, two-author = 2, three-author = 2
(Source: Fact 4 + Step 3)

Table after Step 4

Step 5

Arman must have written at least one of each type and has only 3 papers remaining → forced split:
1 single, 1 two-author, 1 three-author
(Source: Fact 1 + Step 3)

Table after Step 5

Step 6

All authors wrote different numbers of single-author papers →
Arman = 1, Brajen = 2 → remaining single-author papers = 10 − (1 + 2) = 7 → only possible distinct split for Chintan and Devon is {3, 4}
(Source: Fact 2 + Single-author total from graph)

Step 7

Total three-author papers = 3 → total three-author contributions = 9 →
Arman contributes 1, Brajen contributes 2 → remaining contributions = 6 for Chintan and Devon
(Source: Paper-type bar graph + Steps 4 and 5)

Step 8

Chintan and Devon each wrote more three-author papers than Brajen (who wrote 2) → both must have written at least 3 → hence:
three-author of Chintan = 3, three-author of Devon = 3
(Source: Fact 3 + Step 7)

Step 9

Since Chintan and Devon appear in all three three-author papers, every three-author paper must include both → only possible groups are
{Arman, Chintan, Devon} (1 paper) and {Brajen, Chintan, Devon} (2 papers)
(Source: Logical consequence of Step 8)

Step 10

Compute remaining two-author papers using row totals:

Case 1: Chintan single = 3, Devon single = 4 → Chintan two-author = 4, Devon two-author = 1

Case 2: Chintan single = 4, Devon single = 3 → Chintan two-author = 3, Devon two-author = 2

(Source: Step 3 + Step 6 + arithmetic consistency)

Step 11

Since both Case 1 and Case 2 satisfy all graphs and facts, statements assuming a fixed comparison between Chintan and Devon or a fixed two-author count for Chintan are not necessarily true
(Source: Deduction from Step 10)

Step 12

If Devon wrote more than one two-author paper → Case 1 is eliminated → only Case 2 remains → Chintan wrote exactly 3 two-author papers
(Source: Question 4 condition + Step 10)

Final Working Table

Correct Option: 1 – Both i and ii

There are exactly three three-author papers in total. From earlier deductions:

Arman wrote exactly one three-author paper.

Brajen wrote exactly two three-author papers.

Chintan and Devon each wrote three three-author papers.

This implies that Chintan and Devon appear in all three three-author papers. Hence, every three-author paper must include both Chintan and Devon. The only feasible three-author combinations are therefore:

{Arman, Chintan, Devon} – one paper

{Brajen, Chintan, Devon} – two papers

Thus: Arman’s only three-author paper is with Chintan and Devon. Brajen’s three-author papers are also only with Chintan and Devon. Both statements are necessarily true.

Table Steps

Initial Working Table (Blank – from graphs only)

(Source: Author bar graph + Paper-type bar graph)

Step 1

Total paper contributions =
1×10 (single-author) + 2×4 (two-author) + 3×3 (three-author) + 4×2 (four-author) = 35 → matches total author-wise papers 5 + 8 + 12 + 10 = 35
(Source: Paper-type bar graph + Author bar graph)

Step 2

There are exactly 2 four-author papers → all four authors must have written both → each author has exactly 2 four-author papers
(Source: Paper-type bar graph)

Table after Step 2

Step 3

Subtract four-author papers from each author’s total:
Arman: 5 − 2 = 3
Brajen: 8 − 2 = 6
Chintan: 12 − 2 = 10
Devon: 10 − 2 = 8
(Source: Author bar graph + Step 2)

(These are working totals for single + two + three-author papers.)

Step 4

Brajen wrote the same number of single-author and two-author papers →
2 × (single-author of Brajen) + (three-author of Brajen) = 6 → only feasible split:
single = 2, two-author = 2, three-author = 2
(Source: Fact 4 + Step 3)

Table after Step 4

Step 5

Arman must have written at least one of each type and has only 3 papers remaining → forced split:
1 single, 1 two-author, 1 three-author
(Source: Fact 1 + Step 3)

Table after Step 5

Step 6

All authors wrote different numbers of single-author papers →
Arman = 1, Brajen = 2 → remaining single-author papers = 10 − (1 + 2) = 7 → only possible distinct split for Chintan and Devon is {3, 4}
(Source: Fact 2 + Single-author total from graph)

Step 7

Total three-author papers = 3 → total three-author contributions = 9 →
Arman contributes 1, Brajen contributes 2 → remaining contributions = 6 for Chintan and Devon
(Source: Paper-type bar graph + Steps 4 and 5)

Step 8

Chintan and Devon each wrote more three-author papers than Brajen (who wrote 2) → both must have written at least 3 → hence:
three-author of Chintan = 3, three-author of Devon = 3
(Source: Fact 3 + Step 7)

Step 9

Since Chintan and Devon appear in all three three-author papers, every three-author paper must include both → only possible groups are
{Arman, Chintan, Devon} (1 paper) and {Brajen, Chintan, Devon} (2 papers)
(Source: Logical consequence of Step 8)

Step 10

Compute remaining two-author papers using row totals:

Case 1: Chintan single = 3, Devon single = 4 → Chintan two-author = 4, Devon two-author = 1

Case 2: Chintan single = 4, Devon single = 3 → Chintan two-author = 3, Devon two-author = 2

(Source: Step 3 + Step 6 + arithmetic consistency)

Step 11

Since both Case 1 and Case 2 satisfy all graphs and facts, statements assuming a fixed comparison between Chintan and Devon or a fixed two-author count for Chintan are not necessarily true
(Source: Deduction from Step 10)

Step 12

If Devon wrote more than one two-author paper → Case 1 is eliminated → only Case 2 remains → Chintan wrote exactly 3 two-author papers
(Source: Question 4 condition + Step 10)

Final Working Table

Correct Answer: 3

From the two valid cases identified earlier:

In Case 1, Devon wrote only 1 two-author paper.

In Case 2, Devon wrote 2 two-author papers.

Given the condition that Devon wrote more than one two-author paper, Case 1 is eliminated. Only Case 2 remains valid.

In Case 2: Chintan wrote exactly 3 two-author papers.

Thus, under the given condition, the number of two-author papers written by Chintan is fixed and uniquely determined.

Table Steps

Initial Working Table (Blank – from graphs only)

(Source: Author bar graph + Paper-type bar graph)

Step 1

Total paper contributions =
1×10 (single-author) + 2×4 (two-author) + 3×3 (three-author) + 4×2 (four-author) = 35 → matches total author-wise papers 5 + 8 + 12 + 10 = 35
(Source: Paper-type bar graph + Author bar graph)

Step 2

There are exactly 2 four-author papers → all four authors must have written both → each author has exactly 2 four-author papers
(Source: Paper-type bar graph)

Table after Step 2

Step 3

Subtract four-author papers from each author’s total:
Arman: 5 − 2 = 3
Brajen: 8 − 2 = 6
Chintan: 12 − 2 = 10
Devon: 10 − 2 = 8
(Source: Author bar graph + Step 2)

(These are working totals for single + two + three-author papers.)

Step 4

Brajen wrote the same number of single-author and two-author papers →
2 × (single-author of Brajen) + (three-author of Brajen) = 6 → only feasible split:
single = 2, two-author = 2, three-author = 2
(Source: Fact 4 + Step 3)

Table after Step 4

Step 5

Arman must have written at least one of each type and has only 3 papers remaining → forced split:
1 single, 1 two-author, 1 three-author
(Source: Fact 1 + Step 3)

Table after Step 5

Step 6

All authors wrote different numbers of single-author papers →
Arman = 1, Brajen = 2 → remaining single-author papers = 10 − (1 + 2) = 7 → only possible distinct split for Chintan and Devon is {3, 4}
(Source: Fact 2 + Single-author total from graph)

Step 7

Total three-author papers = 3 → total three-author contributions = 9 →
Arman contributes 1, Brajen contributes 2 → remaining contributions = 6 for Chintan and Devon
(Source: Paper-type bar graph + Steps 4 and 5)

Step 8

Chintan and Devon each wrote more three-author papers than Brajen (who wrote 2) → both must have written at least 3 → hence:
three-author of Chintan = 3, three-author of Devon = 3
(Source: Fact 3 + Step 7)

Step 9

Since Chintan and Devon appear in all three three-author papers, every three-author paper must include both → only possible groups are
{Arman, Chintan, Devon} (1 paper) and {Brajen, Chintan, Devon} (2 papers)
(Source: Logical consequence of Step 8)

Step 10

Compute remaining two-author papers using row totals:

Case 1: Chintan single = 3, Devon single = 4 → Chintan two-author = 4, Devon two-author = 1

Case 2: Chintan single = 4, Devon single = 3 → Chintan two-author = 3, Devon two-author = 2

(Source: Step 3 + Step 6 + arithmetic consistency)

Step 11

Since both Case 1 and Case 2 satisfy all graphs and facts, statements assuming a fixed comparison between Chintan and Devon or a fixed two-author count for Chintan are not necessarily true
(Source: Deduction from Step 10)

Step 12

If Devon wrote more than one two-author paper → Case 1 is eliminated → only Case 2 remains → Chintan wrote exactly 3 two-author papers
(Source: Question 4 condition + Step 10)

Final Working Table

Correct Answer: 45
Explanation: Fogglia must have the lowest PI and cannot have its NUR PM greater than either city, so assigning it the smallest PMs (NUR 10, cities 20 and 30) gives PI 35, the minimum possible. Humbleset must contain the only NUR–city inversion, so its NUR must lie between its city PMs. Assigning cities 60 and 80 with NUR 70 produces exactly one inversion and gives PI 50, the highest. The remaining PMs then go to Whimshire, and placing its cities at 40 and 50 with NUR below them yields PI 45.

Table Steps Explanation

City PM order (increasing):
Blusterburg < Noodleton < Splutterville < Quackford < Mumpypore < Zingaloo
Available PMs: 10, 20, 30, 40, 50, 60, 70, 80, 90
(Source: Question statement)

Step 1
Each state has exactly three PMs (2 cities + 1 NUR).
Total PMs are nine distinct multiples of 10 from 10 to 90.
(Source: Question statement)

Step 2
PI formula:
PI = 0.5 × NUR + 0.25 × City1 + 0.25 × City2
PI values are distinct integers.
(Source: Question statement)

Step 3
Exactly one (NUR, City) pair has NUR PM greater than City PM, and both belong to Humbleset.
Therefore:
• In Whimshire and Fogglia: NUR < both cities
• In Humbleset: NUR lies between its two cities
(Source: Given constraint)

Step 4
Fogglia has the lowest PI and Humbleset has the highest PI.
So Fogglia must receive the smallest PM values and Humbleset among the largest.
(Source: PI ordering condition)

Step 5
Assign Fogglia the smallest feasible PMs with NUR below both cities:

Fogglia: NUR = 10; Cities = 20, 40

PI = 0.5×10 + 0.25×20 + 0.25×40 => PI = 5 + 5 + 10 = 20

(Source: Step 2 + Step 4)

Step 6
Remaining PMs after assigning Fogglia: 40, 50, 60, 70, 80, 90
Humbleset must have exactly one city below NUR and one above, so NUR cannot be the maximum or minimum among remaining. (Source: Step 3 logic)

Step 7
Test Humbleset with cities 60 and 80, NUR = 70:
70 > 60 (one valid pair), 70 < 80 (no extra violation)

PI = 0.5×70 + 0.25×60 + 0.25×80 => PI = 35 + 15 + 20 = 70

(Source: Step 3 + Step 4)

Step 8
Remaining PMs for Whimshire: 40, 50, 90

NUR must be less than both cities, to get the middle PI = 50
Whimshire assignment: NUR = 30, Cities = 50 and 90 PI = 0.5×30 + 0.25×50 + 0.25×90 => PI = 15 + 12.5 + 22.5 = 50
(Source: Remaining PMs + PI order)

Step 9
Mapping PMs to city names using the given city order fixes groupings uniquely.
Only Mumpypore (60) and Zingaloo (80) can belong together in Humbleset.
(Source: City PM ranking + Step 7)

Final Working Table

Correct Answer: 35
Explanation: Fogglia is explicitly stated to have the lowest PI and must also have its NUR PM lower than both cities’ PMs. Assigning NUR 10 and the two smallest city PMs, 20 and 30, satisfies all constraints and produces PI = 0.5×10 + 0.25×20 + 0.25×30 = 35. Any higher assignment would violate either the PI ordering or the inversion condition.

Table Steps Explanation

City PM order (increasing):
Blusterburg < Noodleton < Splutterville < Quackford < Mumpypore < Zingaloo
Available PMs: 10, 20, 30, 40, 50, 60, 70, 80, 90
(Source: Question statement)

Step 1
Each state has exactly three PMs (2 cities + 1 NUR).
Total PMs are nine distinct multiples of 10 from 10 to 90.
(Source: Question statement)

Step 2
PI formula:
PI = 0.5 × NUR + 0.25 × City1 + 0.25 × City2
PI values are distinct integers.
(Source: Question statement)

Step 3
Exactly one (NUR, City) pair has NUR PM greater than City PM, and both belong to Humbleset.
Therefore:
• In Whimshire and Fogglia: NUR < both cities
• In Humbleset: NUR lies between its two cities
(Source: Given constraint)

Step 4
Fogglia has the lowest PI and Humbleset has the highest PI.
So Fogglia must receive the smallest PM values and Humbleset among the largest.
(Source: PI ordering condition)

Step 5
Assign Fogglia the smallest feasible PMs with NUR below both cities:

Fogglia: NUR = 10; Cities = 20, 40

PI = 0.5×10 + 0.25×20 + 0.25×40 => PI = 5 + 5 + 10 = 20

(Source: Step 2 + Step 4)

Step 6
Remaining PMs after assigning Fogglia: 40, 50, 60, 70, 80, 90
Humbleset must have exactly one city below NUR and one above, so NUR cannot be the maximum or minimum among remaining. (Source: Step 3 logic)

Step 7
Test Humbleset with cities 60 and 80, NUR = 70:
70 > 60 (one valid pair), 70 < 80 (no extra violation)

PI = 0.5×70 + 0.25×60 + 0.25×80 => PI = 35 + 15 + 20 = 70

(Source: Step 3 + Step 4)

Step 8
Remaining PMs for Whimshire: 40, 50, 90

NUR must be less than both cities, to get the middle PI = 50
Whimshire assignment: NUR = 30, Cities = 50 and 90 PI = 0.5×30 + 0.25×50 + 0.25×90 => PI = 15 + 12.5 + 22.5 = 50
(Source: Remaining PMs + PI order)

Step 9
Mapping PMs to city names using the given city order fixes groupings uniquely.
Only Mumpypore (60) and Zingaloo (80) can belong together in Humbleset.
(Source: City PM ranking + Step 7)

Final Working Table

Correct Answer: 50
Explanation: Humbleset has the highest PI and is the only state where a NUR PM exceeds a city PM. This requires placing the NUR between its two cities. Using cities with PMs 60 and 80 and assigning the NUR PM as 70 creates exactly one inversion and gives PI = 0.5×70 + 0.25×60 + 0.25×80 = 50, higher than the other states’ PIs and consistent with all constraints.

Table Steps Explanation

City PM order (increasing):
Blusterburg < Noodleton < Splutterville < Quackford < Mumpypore < Zingaloo
Available PMs: 10, 20, 30, 40, 50, 60, 70, 80, 90
(Source: Question statement)

Step 1
Each state has exactly three PMs (2 cities + 1 NUR).
Total PMs are nine distinct multiples of 10 from 10 to 90.
(Source: Question statement)

Step 2
PI formula:
PI = 0.5 × NUR + 0.25 × City1 + 0.25 × City2
PI values are distinct integers.
(Source: Question statement)

Step 3
Exactly one (NUR, City) pair has NUR PM greater than City PM, and both belong to Humbleset.
Therefore:
• In Whimshire and Fogglia: NUR < both cities
• In Humbleset: NUR lies between its two cities
(Source: Given constraint)

Step 4
Fogglia has the lowest PI and Humbleset has the highest PI.
So Fogglia must receive the smallest PM values and Humbleset among the largest.
(Source: PI ordering condition)

Step 5
Assign Fogglia the smallest feasible PMs with NUR below both cities:

Fogglia: NUR = 10; Cities = 20, 40

PI = 0.5×10 + 0.25×20 + 0.25×40 => PI = 5 + 5 + 10 = 20

(Source: Step 2 + Step 4)

Step 6
Remaining PMs after assigning Fogglia: 40, 50, 60, 70, 80, 90
Humbleset must have exactly one city below NUR and one above, so NUR cannot be the maximum or minimum among remaining. (Source: Step 3 logic)

Step 7
Test Humbleset with cities 60 and 80, NUR = 70:
70 > 60 (one valid pair), 70 < 80 (no extra violation)

PI = 0.5×70 + 0.25×60 + 0.25×80 => PI = 35 + 15 + 20 = 70

(Source: Step 3 + Step 4)

Step 8
Remaining PMs for Whimshire: 40, 50, 90

NUR must be less than both cities, to get the middle PI = 50
Whimshire assignment: NUR = 30, Cities = 50 and 90 PI = 0.5×30 + 0.25×50 + 0.25×90 => PI = 15 + 12.5 + 22.5 = 50
(Source: Remaining PMs + PI order)

Step 9
Mapping PMs to city names using the given city order fixes groupings uniquely.
Only Mumpypore (60) and Zingaloo (80) can belong together in Humbleset.
(Source: City PM ranking + Step 7)

Final Working Table

Correct Answer: (2) Noodleton, Quackford

Explanation: If we force all three state PIs to be integers and respect the ordering Fogglia < Whimshire < Humbleset, the only consistent set of PIs is 20, 50, and 70. Humbleset must have the highest PI, 70. This requires NUR 70 and cities 60 and 80, which are Mumpypore and Zingaloo.

Fogglia must have the lowest PI, 20. This requires NUR 10 and cities 20 and 40, which are Blusterburg and Splutterville.

This leaves the remaining city PMs, 30 and 50, for Whimshire to achieve the middle PI of 50. The remaining cities, Noodleton (30) and Quackford (50), must therefore belong to Whimshire. Since Noodleton and Quackford are uniquely grouped together, they definitely belong to the same state.

Table Steps Explanation

City PM order (increasing):
Blusterburg < Noodleton < Splutterville < Quackford < Mumpypore < Zingaloo
Available PMs: 10, 20, 30, 40, 50, 60, 70, 80, 90
(Source: Question statement)

Step 1
Each state has exactly three PMs (2 cities + 1 NUR).
Total PMs are nine distinct multiples of 10 from 10 to 90.
(Source: Question statement)

Step 2
PI formula:
PI = 0.5 × NUR + 0.25 × City1 + 0.25 × City2
PI values are distinct integers.
(Source: Question statement)

Step 3
Exactly one (NUR, City) pair has NUR PM greater than City PM, and both belong to Humbleset.
Therefore:
• In Whimshire and Fogglia: NUR < both cities
• In Humbleset: NUR lies between its two cities
(Source: Given constraint)

Step 4
Fogglia has the lowest PI and Humbleset has the highest PI.
So Fogglia must receive the smallest PM values and Humbleset among the largest.
(Source: PI ordering condition)

Step 5
Assign Fogglia the smallest feasible PMs with NUR below both cities:

Fogglia: NUR = 10; Cities = 20, 40

PI = 0.5×10 + 0.25×20 + 0.25×40 => PI = 5 + 5 + 10 = 20

(Source: Step 2 + Step 4)

Step 6
Remaining PMs after assigning Fogglia: 40, 50, 60, 70, 80, 90
Humbleset must have exactly one city below NUR and one above, so NUR cannot be the maximum or minimum among remaining. (Source: Step 3 logic)

Step 7
Test Humbleset with cities 60 and 80, NUR = 70:
70 > 60 (one valid pair), 70 < 80 (no extra violation)

PI = 0.5×70 + 0.25×60 + 0.25×80 => PI = 35 + 15 + 20 = 70

(Source: Step 3 + Step 4)

Step 8
Remaining PMs for Whimshire: 40, 50, 90

NUR must be less than both cities, to get the middle PI = 50
Whimshire assignment: NUR = 30, Cities = 50 and 90 PI = 0.5×30 + 0.25×50 + 0.25×90 => PI = 15 + 12.5 + 22.5 = 50
(Source: Remaining PMs + PI order)

Step 9
Mapping PMs to city names using the given city order fixes groupings uniquely.
Only Mumpypore (60) and Zingaloo (80) can belong together in Humbleset.
(Source: City PM ranking + Step 7)

Final Working Table

Correct Answer: 9
Explanation: Applying all constraints fixes Fogglia’s PMs as 10, 20, and 30, Humbleset’s as 60, 70, and 80, and Whimshire receives the remaining PMs. The city ordering then assigns each PM to a specific city, and each NUR also becomes uniquely located. As a result, the PM and state for all six cities and all three NURs are completely determined, giving a total of 9 identifiable entities.

Table Steps Explanation

City PM order (increasing):
Blusterburg < Noodleton < Splutterville < Quackford < Mumpypore < Zingaloo
Available PMs: 10, 20, 30, 40, 50, 60, 70, 80, 90
(Source: Question statement)

Step 1
Each state has exactly three PMs (2 cities + 1 NUR).
Total PMs are nine distinct multiples of 10 from 10 to 90.
(Source: Question statement)

Step 2
PI formula:
PI = 0.5 × NUR + 0.25 × City1 + 0.25 × City2
PI values are distinct integers.
(Source: Question statement)

Step 3
Exactly one (NUR, City) pair has NUR PM greater than City PM, and both belong to Humbleset.
Therefore:
• In Whimshire and Fogglia: NUR < both cities
• In Humbleset: NUR lies between its two cities
(Source: Given constraint)

Step 4
Fogglia has the lowest PI and Humbleset has the highest PI.
So Fogglia must receive the smallest PM values and Humbleset among the largest.
(Source: PI ordering condition)

Step 5
Assign Fogglia the smallest feasible PMs with NUR below both cities:

Fogglia: NUR = 10; Cities = 20, 40

PI = 0.5×10 + 0.25×20 + 0.25×40 => PI = 5 + 5 + 10 = 20

(Source: Step 2 + Step 4)

Step 6
Remaining PMs after assigning Fogglia: 40, 50, 60, 70, 80, 90
Humbleset must have exactly one city below NUR and one above, so NUR cannot be the maximum or minimum among remaining. (Source: Step 3 logic)

Step 7
Test Humbleset with cities 60 and 80, NUR = 70:
70 > 60 (one valid pair), 70 < 80 (no extra violation)

PI = 0.5×70 + 0.25×60 + 0.25×80 => PI = 35 + 15 + 20 = 70

(Source: Step 3 + Step 4)

Step 8
Remaining PMs for Whimshire: 40, 50, 90

NUR must be less than both cities, to get the middle PI = 50
Whimshire assignment: NUR = 30, Cities = 50 and 90 PI = 0.5×30 + 0.25×50 + 0.25×90 => PI = 15 + 12.5 + 22.5 = 50
(Source: Remaining PMs + PI order)

Step 9
Mapping PMs to city names using the given city order fixes groupings uniquely.
Only Mumpypore (60) and Zingaloo (80) can belong together in Humbleset.
(Source: City PM ranking + Step 7)

Final Working Table

Correct Answer: 6

Explanation: From the conditions, B2 is the only ball that pings H2, so B2’s diameter is the smallest. All balls except B3 ping H1, so B3 is the largest ball. From H4, B1 and B6 ping but B5 does not, implying B5 is larger than both B1 and B6, and H4 lies between them in size. From H3, B4 pings but B1 does not, so B4 is smaller than B1, and H3 lies between them. Using all constraints, B1 pings exactly on H1 and H4 (2 pings), B2 pings on all four hoops (4 pings), and B3 pings on none (0 pings). Hence total pings by B1, B2, and B3 = 2 + 4 + 0 = 6.

Steps to fill the table

(Ping means ball diameter ≤ hoop diameter)
(Source: Question statement)

Step 1
All balls except B3 made a ping on H1 → B1, B2, B4, B5, B6 ≤ H1 and B3 > H1
(Source: Clue 3)

Step 2
None of the balls except B2 made a ping on H2 → only B2 ≤ H2, all others > H2
(Source: Clue 4)

From Steps 1 and 2:
H2 < B1, B3, B4, B5, B6 < H1
(Source: Direct comparison)

Step 3
B4 made a ping on H3, but B1 did not → B4 ≤ H3 < B1
(Source: Clue 2)

Step 4
B1 and B6 made pings on H4, but B5 did not → B1 ≤ H4, B6 ≤ H4 < B5
(Source: Clue 1)

This gives B1 < B5 and B6 < B5
(Source: Diameter implication)

Step 5
From earlier steps:
B2 is the smallest ball (only one passing H2)
B3 is the largest ball (fails H1)
Also B4 < B1 < B5

So partial order becomes:
B2 < B4 < B1 < B5 < B3, with B6 between B1 and B5 or between B4 and B1
(Source: Combined deductions)

Step 6
Since B4 passes H3 and B1 fails H3 → H3 < B1
Since B1 passes H4 → B1 ≤ H4
Thus hoops satisfy:
H2 < H3 < H4 < H1
(Source: Steps 1, 3, and 4)

Step 7
Filling all forced pings based on final orderings:

(Source: All constraints applied)

Step 8
Pings required for Q10:
B1 = 2, B2 = 4, B3 = 0 → total = 6
(Source: Step 7)

Step 9
Total guaranteed pings = B1(2) + B2(4) + B3(0) + B5(1) = 7
B4 can have 2 or 3, B6 can have 2 or 3
So total pings = 12 or 13
(Source: Step 7 optional passes)

Final Deductions
Ball order (small → large):
B2 < B4 < B1 < B6 < B5 < B3

Hoop order (small → large):
H2 < H3 < H4 < H1

Correct Answer: 2

Explanation: From the deductions, B2 is the smallest and B3 is the largest ball. We also have B4 < B1 and B1 < B5, giving B4 < B1 < B5 < B3 as a necessary chain. The relative order of B1 and B6, however, is not fixed: both B1 and B6 ping H4 and fail H3, so both lie on the same side of those hoops without a strict comparison between them. Therefore, the statement “B1 < B6 < B3” is not necessarily true, while the other options are consistent with all valid configurations.

Steps to fill the table

(Ping means ball diameter ≤ hoop diameter)
(Source: Question statement)

Step 1
All balls except B3 made a ping on H1 → B1, B2, B4, B5, B6 ≤ H1 and B3 > H1
(Source: Clue 3)

Step 2
None of the balls except B2 made a ping on H2 → only B2 ≤ H2, all others > H2
(Source: Clue 4)

From Steps 1 and 2:
H2 < B1, B3, B4, B5, B6 < H1
(Source: Direct comparison)

Step 3
B4 made a ping on H3, but B1 did not → B4 ≤ H3 < B1
(Source: Clue 2)

Step 4
B1 and B6 made pings on H4, but B5 did not → B1 ≤ H4, B6 ≤ H4 < B5
(Source: Clue 1)

This gives B1 < B5 and B6 < B5
(Source: Diameter implication)

Step 5
From earlier steps:
B2 is the smallest ball (only one passing H2)
B3 is the largest ball (fails H1)
Also B4 < B1 < B5

So partial order becomes:
B2 < B4 < B1 < B5 < B3, with B6 between B1 and B5 or between B4 and B1
(Source: Combined deductions)

Step 6
Since B4 passes H3 and B1 fails H3 → H3 < B1
Since B1 passes H4 → B1 ≤ H4
Thus hoops satisfy:
H2 < H3 < H4 < H1
(Source: Steps 1, 3, and 4)

Step 7
Filling all forced pings based on final orderings:

(Source: All constraints applied)

Step 8
Pings required for Q10:
B1 = 2, B2 = 4, B3 = 0 → total = 6
(Source: Step 7)

Step 9
Total guaranteed pings = B1(2) + B2(4) + B3(0) + B5(1) = 7
B4 can have 2 or 3, B6 can have 2 or 3
So total pings = 12 or 13
(Source: Step 7 optional passes)

Final Deductions
Ball order (small → large):
B2 < B4 < B1 < B6 < B5 < B3

Correct Answer: 1

Explanation: From H2, only B2 pings, so H2 is the smallest hoop. From H1, all except B3 ping, and since B3 is the largest ball, H1 must be the largest hoop. From H3 and H4, we know H3 allows B4 but not B1, while H4 allows B1 and B6 but not B5. This places the hoops in increasing order as H2 < H3 < H4 < H1. Hence option 1 is correct.

Steps to fill the table

(Ping means ball diameter ≤ hoop diameter)
(Source: Question statement)

Step 1
All balls except B3 made a ping on H1 → B1, B2, B4, B5, B6 ≤ H1 and B3 > H1
(Source: Clue 3)

Step 2
None of the balls except B2 made a ping on H2 → only B2 ≤ H2, all others > H2
(Source: Clue 4)

From Steps 1 and 2:
H2 < B1, B3, B4, B5, B6 < H1
(Source: Direct comparison)

Step 3
B4 made a ping on H3, but B1 did not → B4 ≤ H3 < B1
(Source: Clue 2)

Step 4
B1 and B6 made pings on H4, but B5 did not → B1 ≤ H4, B6 ≤ H4 < B5
(Source: Clue 1)

This gives B1 < B5 and B6 < B5
(Source: Diameter implication)

Step 5
From earlier steps:
B2 is the smallest ball (only one passing H2)
B3 is the largest ball (fails H1)
Also B4 < B1 < B5

So partial order becomes:
B2 < B4 < B1 < B5 < B3, with B6 between B1 and B5 or between B4 and B1
(Source: Combined deductions)

Step 6
Since B4 passes H3 and B1 fails H3 → H3 < B1
Since B1 passes H4 → B1 ≤ H4
Thus hoops satisfy:
H2 < H3 < H4 < H1
(Source: Steps 1, 3, and 4)

Step 7
Filling all forced pings based on final orderings:

(Source: All constraints applied)

Step 8
Pings required for Q10:
B1 = 2, B2 = 4, B3 = 0 → total = 6
(Source: Step 7)

Step 9
Total guaranteed pings = B1(2) + B2(4) + B3(0) + B5(1) = 7
B4 can have 2 or 3, B6 can have 2 or 3
So total pings = 12 or 13
(Source: Step 7 optional passes)

Final Deductions
Ball order (small → large):
B2 < B4 < B1 < B6 < B5 < B3

Correct Answer: 3

Explanation: From the full ordering, pings can be counted ball by ball. B1 pings on H1 and H4 (2). B2 pings on all four hoops (4). B3 pings on none (0). B4 pings on H1, H3, and possibly H4 (2 or 3). B5 pings on H1 only (1). B6 pings on H1 and H4, and may or may not ping H3 (2 or 3). Adding these, the total number of pings is either 12 or 13 depending on whether B6 (or B4) fits through H3. No configuration allows fewer than 12 or more than 13. Therefore, the best statement is “12 or 13”.

Steps to fill the table

(Ping means ball diameter ≤ hoop diameter)
(Source: Question statement)

Step 1
All balls except B3 made a ping on H1 → B1, B2, B4, B5, B6 ≤ H1 and B3 > H1
(Source: Clue 3)

Step 2
None of the balls except B2 made a ping on H2 → only B2 ≤ H2, all others > H2
(Source: Clue 4)

From Steps 1 and 2:
H2 < B1, B3, B4, B5, B6 < H1
(Source: Direct comparison)

Step 3
B4 made a ping on H3, but B1 did not → B4 ≤ H3 < B1
(Source: Clue 2)

Step 4
B1 and B6 made pings on H4, but B5 did not → B1 ≤ H4, B6 ≤ H4 < B5
(Source: Clue 1)

This gives B1 < B5 and B6 < B5
(Source: Diameter implication)

Step 5
From earlier steps:
B2 is the smallest ball (only one passing H2)
B3 is the largest ball (fails H1)
Also B4 < B1 < B5

So partial order becomes:
B2 < B4 < B1 < B5 < B3, with B6 between B1 and B5 or between B4 and B1
(Source: Combined deductions)

Step 6
Since B4 passes H3 and B1 fails H3 → H3 < B1
Since B1 passes H4 → B1 ≤ H4
Thus hoops satisfy:
H2 < H3 < H4 < H1
(Source: Steps 1, 3, and 4)

Step 7
Filling all forced pings based on final orderings:

(Source: All constraints applied)

Step 8
Pings required for Q10:
B1 = 2, B2 = 4, B3 = 0 → total = 6
(Source: Step 7)

Step 9
Total guaranteed pings = B1(2) + B2(4) + B3(0) + B5(1) = 7
B4 can have 2 or 3, B6 can have 2 or 3
So total pings = 12 or 13
(Source: Step 7 optional passes)

Final Deductions
Ball order (small → large):
B2 < B4 < B1 < B6 < B5 < B3

Correct Answer: 2020 

From the conditions, Ananya and Bhaskar must be Gurubhai for exactly two years. Gurubhai status can arise only when two students train under the same Guru in the same year. Ustad Samiran never trains more than one student in a year, so no Gurubhai relationships can occur under him. Acharya Raghunath trains only in two isolated two‑year blocks (2013–14 and 2019–20), and each block can host at most one Gurubhai pair. Since Ananya cannot be paired with Devendra and Bhaskar cannot be paired with Charu, the available Raghunath slots are already forced onto other admissible pairs. Hence, Ananya and Bhaskar’s Gurubhai years must occur under Pandit Meghnath. Using the final consistent year‑wise schedule, Ananya and Bhaskar are together under Pandit Meghnath during 2020 and one other adjacent year. Among the given answer options, 2020 is the only year that necessarily belongs to their Gurubhai overlap irrespective of how the remaining flexible years are arranged. Therefore, 2020 is the correct answer.

Steps to make the table:

Phase 1: Fix training spans and Raghunath slots

1. Each guru trains for a different duration: 2, 3, and 4 continuous years.

2. Acharya Raghunath is available only in two isolated windows: 2013–14 and 2019–20. Hence, his training span must be exactly 2 years.

3. This leaves Pandit Meghnath and Ustad Samiran with spans of 3 years and 4 years respectively.

4. To satisfy the given gurubhai overlap conditions, assign Meghnath a 3-year span and Samiran a 4-year span.

5. From Fact 4, Ananya starts with Meghnath in 2013 and Bhaskar starts with Samiran in 2013.

6. Therefore, neither Ananya nor Bhaskar can take Raghunath’s 2013–14 slot, as that would overlap with their starting guru.

7. Hence, Ananya and Bhaskar must take Raghunath in 2019–20, leaving Charu and Devendra to take Raghunath in 2013–14.

Phase 2: Fix schedules for Ananya and Bhaskar

8. Ananya must complete her 3-year training under Meghnath starting in 2013.

   So, Ananya studies under Meghnath from 2013 to 2015.

9. Bhaskar must complete his 4-year training under Samiran starting in 2013.

   So, Bhaskar studies under Samiran from 2013 to 2016.

10. Bhaskar’s Meghnath training must occur after his Raghunath period of 2019–20.

    The only viable placement is 2022–2024.

11. Ananya’s Samiran training must begin after her Raghunath period (2019–20) and must not overlap Bhaskar’s Meghnath slot excessively.

    The consistent placement is 2021–2024.

Phase 3: Fix schedules for Charu and Devendra

12. Devendra must be gurubhai with Bhaskar for exactly two years under Samiran.

13. Bhaskar’s Samiran span is 2013–16, so Devendra’s 4-year Samiran span must overlap this window for exactly two years.

14. The only placement satisfying this is 2015–2018, overlapping in 2015 and 2016.

15. Devendra’s Meghnath training must follow his Samiran period.

16. To ensure Bhaskar and Devendra are gurubhai under Meghnath in the year 2022, Devendra’s Meghnath span is placed from 2020 to 2022.

17. Charu must be gurubhai with Ananya for exactly two years, partially under Meghnath and partially under Samiran.

18. Ananya’s Meghnath span is 2013–15, so Charu’s 3-year Meghnath span must overlap this for one year.

    The only placement is 2015–2017.

19. To complete the gurubhai requirement with Ananya, Charu’s 4-year Samiran span must overlap Ananya’s Samiran span in exactly one year.

20. Placing Charu’s Samiran span from 2018 to 2021 achieves this overlap in 2021.

Final Working Table (2013–2024)

Spans: M=3 yrs, S=4 yrs, R=2 yrs.
R Groups:
Group 1 (R=2013–2014): C, D
Group 2 (R=2019–2020): A, B

Correct Answer: 2015 

Charu must be Gurubhai with Ananya for exactly two years. This overlap cannot occur under Ustad Samiran, and Acharya Raghunath has only limited two‑year windows that are already constrained by other mandatory Gurubhai pairs. Hence, the Ananya–Charu Gurubhai overlap must occur under Pandit Meghnath. Ananya begins training under Pandit Meghnath in 2013, and Pandit Meghnath’s span must be four consecutive years to accommodate all required overlaps. For Charu to overlap with Ananya for exactly two years under Pandit Meghnath, Charu’s start year under Pandit Meghnath must be precisely two years after Ananya’s start, so that their overlap is neither more nor less than two years. If Charu started earlier than 2015, the overlap would exceed two years; if she started later, the overlap would be reduced to one or zero years. Hence, Charu must begin training under Pandit Meghnath in 2015, making this the only viable and forced answer.

Steps to make the table:

Phase 1: Fix training spans and Raghunath slots

1. Each guru trains for a different duration: 2, 3, and 4 continuous years.

2. Acharya Raghunath is available only in two isolated windows: 2013–14 and 2019–20. Hence, his training span must be exactly 2 years.

3. This leaves Pandit Meghnath and Ustad Samiran with spans of 3 years and 4 years respectively.

4. To satisfy the given gurubhai overlap conditions, assign Meghnath a 3-year span and Samiran a 4-year span.

5. From Fact 4, Ananya starts with Meghnath in 2013 and Bhaskar starts with Samiran in 2013.

6. Therefore, neither Ananya nor Bhaskar can take Raghunath’s 2013–14 slot, as that would overlap with their starting guru.

7. Hence, Ananya and Bhaskar must take Raghunath in 2019–20, leaving Charu and Devendra to take Raghunath in 2013–14.

Phase 2: Fix schedules for Ananya and Bhaskar

8. Ananya must complete her 3-year training under Meghnath starting in 2013.

   So, Ananya studies under Meghnath from 2013 to 2015.

9. Bhaskar must complete his 4-year training under Samiran starting in 2013.

   So, Bhaskar studies under Samiran from 2013 to 2016.

10. Bhaskar’s Meghnath training must occur after his Raghunath period of 2019–20.

    The only viable placement is 2022–2024.

11. Ananya’s Samiran training must begin after her Raghunath period (2019–20) and must not overlap Bhaskar’s Meghnath slot excessively.

    The consistent placement is 2021–2024.

Phase 3: Fix schedules for Charu and Devendra

12. Devendra must be gurubhai with Bhaskar for exactly two years under Samiran.

13. Bhaskar’s Samiran span is 2013–16, so Devendra’s 4-year Samiran span must overlap this window for exactly two years.

14. The only placement satisfying this is 2015–2018, overlapping in 2015 and 2016.

15. Devendra’s Meghnath training must follow his Samiran period.

16. To ensure Bhaskar and Devendra are gurubhai under Meghnath in the year 2022, Devendra’s Meghnath span is placed from 2020 to 2022.

17. Charu must be gurubhai with Ananya for exactly two years, partially under Meghnath and partially under Samiran.

18. Ananya’s Meghnath span is 2013–15, so Charu’s 3-year Meghnath span must overlap this for one year.

    The only placement is 2015–2017.

19. To complete the gurubhai requirement with Ananya, Charu’s 4-year Samiran span must overlap Ananya’s Samiran span in exactly one year.

20. Placing Charu’s Samiran span from 2018 to 2021 achieves this overlap in 2021.

Final Working Table (2013–2024)

Spans: M=3 yrs, S=4 yrs, R=2 yrs.
R Groups:
Group 1 (R=2013–2014): C, D
Group 2 (R=2019–2020): A, B

Correct Answer: 2022 

Bhaskar and Devendra are required to be Gurubhai for exactly two years. As with other pairs, this cannot happen under Ustad Samiran. Acharya Raghunath’s two‑year blocks can accommodate only one Gurubhai pair per block and are already compelled to host other valid pairs after applying the “never Gurubhai” restrictions. Therefore, Bhaskar and Devendra must overlap under Pandit Meghnath. From the final year‑wise allocation, Bhaskar’s and Devendra’s Pandit Meghnath spans overlap exactly for two years. Among the listed options, 2022 is one of those overlapping years and is the only option that is guaranteed to be part of the Bhaskar–Devendra Gurubhai overlap across all valid configurations. The other years either fall outside their common Pandit Meghnath window or violate other fixed constraints. Hence, 2022 is the correct answer.

Steps to make the table:

Phase 1: Fix training spans and Raghunath slots

1. Each guru trains for a different duration: 2, 3, and 4 continuous years.

2. Acharya Raghunath is available only in two isolated windows: 2013–14 and 2019–20. Hence, his training span must be exactly 2 years.

3. This leaves Pandit Meghnath and Ustad Samiran with spans of 3 years and 4 years respectively.

4. To satisfy the given gurubhai overlap conditions, assign Meghnath a 3-year span and Samiran a 4-year span.

5. From Fact 4, Ananya starts with Meghnath in 2013 and Bhaskar starts with Samiran in 2013.

6. Therefore, neither Ananya nor Bhaskar can take Raghunath’s 2013–14 slot, as that would overlap with their starting guru.

7. Hence, Ananya and Bhaskar must take Raghunath in 2019–20, leaving Charu and Devendra to take Raghunath in 2013–14.

Phase 2: Fix schedules for Ananya and Bhaskar

8. Ananya must complete her 3-year training under Meghnath starting in 2013.

   So, Ananya studies under Meghnath from 2013 to 2015.

9. Bhaskar must complete his 4-year training under Samiran starting in 2013.

   So, Bhaskar studies under Samiran from 2013 to 2016.

10. Bhaskar’s Meghnath training must occur after his Raghunath period of 2019–20.

    The only viable placement is 2022–2024.

11. Ananya’s Samiran training must begin after her Raghunath period (2019–20) and must not overlap Bhaskar’s Meghnath slot excessively.

    The consistent placement is 2021–2024.

Phase 3: Fix schedules for Charu and Devendra

12. Devendra must be gurubhai with Bhaskar for exactly two years under Samiran.

13. Bhaskar’s Samiran span is 2013–16, so Devendra’s 4-year Samiran span must overlap this window for exactly two years.

14. The only placement satisfying this is 2015–2018, overlapping in 2015 and 2016.

15. Devendra’s Meghnath training must follow his Samiran period.

16. To ensure Bhaskar and Devendra are gurubhai under Meghnath in the year 2022, Devendra’s Meghnath span is placed from 2020 to 2022.

17. Charu must be gurubhai with Ananya for exactly two years, partially under Meghnath and partially under Samiran.

18. Ananya’s Meghnath span is 2013–15, so Charu’s 3-year Meghnath span must overlap this for one year.

    The only placement is 2015–2017.

19. To complete the gurubhai requirement with Ananya, Charu’s 4-year Samiran span must overlap Ananya’s Samiran span in exactly one year.

20. Placing Charu’s Samiran span from 2018 to 2021 achieves this overlap in 2021.

Final Working Table (2013–2024)

Spans: M=3 yrs, S=4 yrs, R=2 yrs.
R Groups:
Group 1 (R=2013–2014): C, D
Group 2 (R=2019–2020): A, B

Correct Answer: Charu was training under Ustad Samiran in 2019 

Ustad Samiran trains exactly one student per year, so assigning students to him is tightly constrained. Bhaskar must start under Ustad Samiran in 2013, fixing the early part of Samiran’s schedule. Once the required Gurubhai overlaps under Pandit Meghnath and Acharya Raghunath are fixed, the remaining unassigned years for Ustad Samiran get forced. Tracing the only arrangement that satisfies all constraints shows that Charu’s three‑year consecutive span under Ustad Samiran necessarily includes 2019. This placement avoids forbidden Gurubhai overlaps, ensures Charu completes training under all three Gurus, and keeps Samiran restricted to a single student per year. None of the other statements can be guaranteed without violating at least one given condition. Therefore, the statement “Charu was training under Ustad Samiran in 2019” is the only one that is necessarily true.

Steps to make the table:

Phase 1: Fix training spans and Raghunath slots

1. Each guru trains for a different duration: 2, 3, and 4 continuous years.

2. Acharya Raghunath is available only in two isolated windows: 2013–14 and 2019–20. Hence, his training span must be exactly 2 years.

3. This leaves Pandit Meghnath and Ustad Samiran with spans of 3 years and 4 years respectively.

4. To satisfy the given gurubhai overlap conditions, assign Meghnath a 3-year span and Samiran a 4-year span.

5. From Fact 4, Ananya starts with Meghnath in 2013 and Bhaskar starts with Samiran in 2013.

6. Therefore, neither Ananya nor Bhaskar can take Raghunath’s 2013–14 slot, as that would overlap with their starting guru.

7. Hence, Ananya and Bhaskar must take Raghunath in 2019–20, leaving Charu and Devendra to take Raghunath in 2013–14.

Phase 2: Fix schedules for Ananya and Bhaskar

8. Ananya must complete her 3-year training under Meghnath starting in 2013.

   So, Ananya studies under Meghnath from 2013 to 2015.

9. Bhaskar must complete his 4-year training under Samiran starting in 2013.

   So, Bhaskar studies under Samiran from 2013 to 2016.

10. Bhaskar’s Meghnath training must occur after his Raghunath period of 2019–20.

    The only viable placement is 2022–2024.

11. Ananya’s Samiran training must begin after her Raghunath period (2019–20) and must not overlap Bhaskar’s Meghnath slot excessively.

    The consistent placement is 2021–2024.

Phase 3: Fix schedules for Charu and Devendra

12. Devendra must be gurubhai with Bhaskar for exactly two years under Samiran.

13. Bhaskar’s Samiran span is 2013–16, so Devendra’s 4-year Samiran span must overlap this window for exactly two years.

14. The only placement satisfying this is 2015–2018, overlapping in 2015 and 2016.

15. Devendra’s Meghnath training must follow his Samiran period.

16. To ensure Bhaskar and Devendra are gurubhai under Meghnath in the year 2022, Devendra’s Meghnath span is placed from 2020 to 2022.

17. Charu must be gurubhai with Ananya for exactly two years, partially under Meghnath and partially under Samiran.

18. Ananya’s Meghnath span is 2013–15, so Charu’s 3-year Meghnath span must overlap this for one year.

    The only placement is 2015–2017.

19. To complete the gurubhai requirement with Ananya, Charu’s 4-year Samiran span must overlap Ananya’s Samiran span in exactly one year.

20. Placing Charu’s Samiran span from 2018 to 2021 achieves this overlap in 2021.

Final Working Table (2013–2024)

Spans: M=3 yrs, S=4 yrs, R=2 yrs.
R Groups:
Group 1 (R=2013–2014): C, D
Group 2 (R=2019–2020): A, B

Correct Answer: 4 

Each musician trains for a total of 9 years (2 + 3 + 4), so total training‑years across all musicians equals 36. These 36 training‑years are distributed over 12 calendar years, giving an average of 3 musicians training per year. However, due to Ustad Samiran’s restriction of training at most one student per year and Acharya Raghunath’s long inactive stretches, the distribution cannot be uniform. Some years will necessarily have fewer than three musicians training. When the final consistent schedule is examined, exactly four years emerge in which only two musicians are training, with the remaining musicians either between training spells or constrained by inactive Gurus. Any reduction below four would force compensating years with more than four trainees, which is impossible. Hence, the number of years with exactly two musicians training is fixed at four.

Steps to make the table:

Phase 1: Fix training spans and Raghunath slots

1. Each guru trains for a different duration: 2, 3, and 4 continuous years.

2. Acharya Raghunath is available only in two isolated windows: 2013–14 and 2019–20. Hence, his training span must be exactly 2 years.

3. This leaves Pandit Meghnath and Ustad Samiran with spans of 3 years and 4 years respectively.

4. To satisfy the given gurubhai overlap conditions, assign Meghnath a 3-year span and Samiran a 4-year span.

5. From Fact 4, Ananya starts with Meghnath in 2013 and Bhaskar starts with Samiran in 2013.

6. Therefore, neither Ananya nor Bhaskar can take Raghunath’s 2013–14 slot, as that would overlap with their starting guru.

7. Hence, Ananya and Bhaskar must take Raghunath in 2019–20, leaving Charu and Devendra to take Raghunath in 2013–14.

Phase 2: Fix schedules for Ananya and Bhaskar

8. Ananya must complete her 3-year training under Meghnath starting in 2013.

   So, Ananya studies under Meghnath from 2013 to 2015.

9. Bhaskar must complete his 4-year training under Samiran starting in 2013.

   So, Bhaskar studies under Samiran from 2013 to 2016.

10. Bhaskar’s Meghnath training must occur after his Raghunath period of 2019–20.

    The only viable placement is 2022–2024.

11. Ananya’s Samiran training must begin after her Raghunath period (2019–20) and must not overlap Bhaskar’s Meghnath slot excessively.

    The consistent placement is 2021–2024.

Phase 3: Fix schedules for Charu and Devendra

12. Devendra must be gurubhai with Bhaskar for exactly two years under Samiran.

13. Bhaskar’s Samiran span is 2013–16, so Devendra’s 4-year Samiran span must overlap this window for exactly two years.

14. The only placement satisfying this is 2015–2018, overlapping in 2015 and 2016.

15. Devendra’s Meghnath training must follow his Samiran period.

16. To ensure Bhaskar and Devendra are gurubhai under Meghnath in the year 2022, Devendra’s Meghnath span is placed from 2020 to 2022.

17. Charu must be gurubhai with Ananya for exactly two years, partially under Meghnath and partially under Samiran.

18. Ananya’s Meghnath span is 2013–15, so Charu’s 3-year Meghnath span must overlap this for one year.

    The only placement is 2015–2017.

19. To complete the gurubhai requirement with Ananya, Charu’s 4-year Samiran span must overlap Ananya’s Samiran span in exactly one year.

20. Placing Charu’s Samiran span from 2018 to 2021 achieves this overlap in 2021.

Final Working Table (2013–2024)

Spans: M=3 yrs, S=4 yrs, R=2 yrs.
R Groups:
Group 1 (R=2013–2014): C, D
Group 2 (R=2019–2020): A, B

Correct Answer: 60

Let E₍₂₀₁₆₎ = x

From graph: E increases 25% then 20%

SI 2024 = 1.25x × 1.20 = 1.5x

Given SI 2024 = 90

1.5x = 90 → x = 60

Correct Answer: 40

Let F₍₂₀₁₆₎ = y

From graph: F increases 100% in 2020, then drops 25%

F₍₂₀₂₀₎ = 2y

F₍₂₀₂₄₎ = 0.75 × 2y = 1.5y

Since F is lowest in 2024, and range in 2024 = 60

Highest in 2024 = 90 → lowest = 30

So 1.5y = 30 → y = 20

Thus F₍₂₀₂₀₎ = 2y = 40

Correct Answer: 84

Let C₍₂₀₁₆₎ = z

From graph: −25% then +40%

C₍₂₀₂₀₎ = 0.75z

C₍₂₀₂₄₎ = 1.4 × 0.75z = 1.05z

Using 2016 ranks: B > C > E > A and E₍₂₀₁₆₎ = 60

Trying integer values consistent with range = 60 and F lowest = 20

C₍₂₀₁₆₎ = 80 works

Then C₍₂₀₂₄₎ = 1.05 × 80 = 84

Sustainability Index Scatter Graph Explained | CAT 2025 Slot 2 DILR | Graph Modern | Easy

Correct Answer: Option 3 (45)

Let B₍₂₀₁₆₎ = b

From graph: −25% then −25% (From the graph, B shows a 25% decrease from 2016 to 2020 and a further 25% decrease from 2020 to 2024)

B₍₂₀₂₀₎ = 0.75b

B₍₂₀₂₄₎ = 0.75 × 0.75b = 0.5625b

From 2016 ranks and range = 60

Highest = B = 80, lowest (F) = 20

So B₍₂₀₁₆₎ = 80

B₍₂₀₂₄₎ = 0.5625 × 80 = 45

Correct answer A) 5 : 8

Suppose total = 100.

So Pinu got 20, Rinu got 32 thus ratio = 20 : 32 = 5 : 8

Correct answer A) 3 : 5

Let Lakshmi’s expenditure = 2x, Meenakshi’s expenditure = 3x (since their expenditures are in ratio 2:3).

Lakshmi’s income is given to be in ratio 6:7 with Meenakshi’s expenditure  so Lakshmi’s income = (6/7) × (3x) = (18/7) x.

Then Lakshmi’s saving = income − expenditure = (18/7 x) − 2x = (4/7) x.

Let Meenakshi’s income = y, so her saving = y − 3x.

Given their savings ratio is 4:9 , (4/7 x) : (y − 3x) = 4 : 9.

So (4/7 x) / (y − 3x) = 4/9 , solving gives y = (30/7) x.

Hence incomes are (18/7 x) : (30/7 x) = 18 : 30 = 3 : 5

Correct Answer D) 14

Given: CP = 1650, MP = 2200

Condition: profit % = discount %

Correct Answer is c) 8%

Let the annual rate of interest be r%. The present value of the two instalments must equal the loan amount.

1000 = 530 / (1 + r/100) + 594 / (1 + r/100)²

Concept of upstream and Downstream:

Sneha’s speed = 6 km/h, let river speed = x km/h

Time difference: 14/(6-x) – 14/(6+x) = 0.8 , x = 1 km/h

Rita’s effective speed

Upstream = 5 – 1 = 4 km/h

Downstream = 5 + 1 = 6 km/h

Time to go and return = 100 min = 5/3 h

d/4 + d/6 = 5/3 , d = 4 km

Total distance:

Total distance = 4 + 4 = 8 km

Answer: 8 km

Correct Answer (c) 5 kg

Let the price of coffee be C and that of cocoa be K (per kg).

From the first mixture (16% coffee, price 240):
0.16C + 0.84K = 240

From the second mixture (36% coffee, price 320):
0.36C + 0.64K = 320

Multiply both equations by 100:
16C + 84K = 24000
36C + 64K = 32000

Solve these two:
From the pair, you get K = 176 and then C = 576.

Now for the new mixture with price 376 and coffee fraction x:
576x + 176(1 − x) = 376→576x + 176 − 176x = 376 → 400x = 200 → x = 0.5

So the new mixture has 50% coffee.
Therefore, in 10 kg of the new mixture, coffee = 0.5 × 10 = 5 kg.

Shortcut method:

Coffee % goes from 16% → 36%, Price goes from 240 → 320

So an increase of 20% coffee increases price by 80 ⇒ Each 1% coffee increases price by 4.

Now compare old price to new price  → Price of new mixture = 376   →  Difference from cheaper mixture = 376 − 240 = 136

So increase in coffee %  = 136 ÷ 4 = 34%

Hence coffee % in new mixture = 16% + 34% = 50%

So in 10 kg, coffee = 50% of 10 = 5 kg

Correct Answer: 49 copies

First 7 days avg = 60 ⇒ total = 7 × 60 = 420

First 8 days avg = 63 ⇒ total = 8 × 63 = 504

8th day sales = 504 − 420 = 84

9th day sales = 84 − 11 = 73

Avg from 2nd to 9th day = 66 ⇒ total (2nd–9th) = 8 × 66 = 528

Total (1st–9th) = 504 + 73 = 577

1st day sales = 577 − 528 = 49

Correct Answer 340

Let Chandan’s efficiency be 1 unit/day.

Then Bipin’s efficiency = 2 units/day and Ankita’s efficiency = 4 units/day.

For the first 20 days, all three work together, so work done = (4 + 2 + 1) × 20 = 140 units.

For the remaining 40 days, only Ankita and Chandan work, so work done = (4 + 1) × 40 = 200 units.

Total work = 140 + 200 = 340 units.

Since Chandan alone does 1 unit of work per day, he would complete the job in 340 days.

Correct Answer D) 14

Given: CP = 1650, MP = 2200

Condition: profit % = discount %

Correct Answer: Option C

Given equations:

3x² − 5x + p = 0

2x² − 2x + q = 0

For a quadratic ax² + bx + c = 0, sum of roots = −b/a.

Step 1: Use the fact that r is a common root

Since r satisfies both equations:

From (1): 3r² − 5r + p = 0  →  p = 5r − 3r²

From (2): 2r² − 2r + q = 0  →  q = 2r − 2r²

Step 2: Express sum of the “other” roots

Equation (1): Sum of roots = 5/3. So, other root of (1) = 5/3 − r

Equation (2): Sum of roots = 1. So, other root of (2) = 1 − r

Required sum = (5/3 − r) + (1 − r) = 8/3 − 2r

Step 3: Eliminate r using p and q

From p = 5r − 3r²; From q = 2r − 2r²

Rewrite both: p = r(5 − 3r); q = r(2 − 2r)

Solve for r in linear combination form:

Multiply second by 3/2: (3/2)q = 3r − 3r²

Now subtract from p: p − (3/2)q = (5r − 3r²) − (3r − 3r²) = 2r

So: r = (p − 3q/2)/2 = p/2 − 3q/4

Step 4: Substitute r into required sum Required sum = 8/3 − 2r = 8/3 − 2(p/2 − 3q/4) = 8/3 − p + 3q/2.

Correct Answer: A) 20

Given equation:
9^(x² + 2x − 3) − 4·3^(x² + 2x − 2) + 27 = 0

Step 1: Write everything with base 3

9 = 3², so ⇒ 9^(x² + 2x − 3) = 3^(2x² + 4x − 6)

Also, 3^(x² + 2x − 2) = 3·3^(x² + 2x − 3)

Let t = 3^(x² + 2x − 3), where t > 0.

Then the equation becomes: t² − 4(3t) + 27 = 0 ⇒ t² − 12t + 27 = 0

Step 2: Solve the quadratic in t

t² − 12t + 27 = 0 ⇒ (t − 3)(t − 9) = 0 ⇒  So t = 3 or t = 9

Step 3: Convert back to x

Case 1: 3^(x² + 2x − 3) = 3 ⇒ x² + 2x − 3 = 1 ⇒ x² + 2x − 4 = 0

Roots: x = −1 ± √5 ⇒ Product of roots = −4

Case 2: 3^(x² + 2x − 3) = 9 = 3² ⇒ x² + 2x − 3 = 2 ⇒ x² + 2x − 5 = 0

Roots: x = −1 ± √6 ⇒  Product of roots = −5

Step 4: Product of all possible values of x

Total product = (product from case 1) × (product from case 2)
= (−4) × (−5) = 20

Given: (m + 2n)(2m + n) = 27

Since m and n are integers, (m + 2n) and (2m + n) must be integer factors of 27.

Possible factor pairs (m + 2n, 2m + n) are:
(1, 27), (3, 9), (9, 3), (27, 1),
(−1, −27), (−3, −9), (−9, −3), (−27, −1)

We now solve each pair as a system: m + 2n = A; 2m + n = B

From these:
Multiply first equation by 2: 2m + 4n = 2A
Subtract from second: (2m + n) − (2m + 4n) = B − 2A
⇒ −3n = B − 2A ⇒ n = (2A − B) / 3
Then m = A − 2n

Maximum possible valid value of 2m − 3n = 17

Use X maro G Strategy put x = -5 and x = -10

cc

Minimum possible value is 2

Since a, b, c, d must be integers, they must be the integers closest to 11.5, which are 11 and 12.

To satisfy the sum constraint a + b + c + d = 46, we can take values 11, 11, 12, 12.

Expression becomes: (11 − 12)² + (11 − 12)² + (11 − 11)² = 1 + 1 + 0 = 2

Correct Answer: 42

N = 2⁶ × 3⁵ × 5³ × 7²

We want the number of divisors of N that are of the form (3r + 1), i.e.

divisors ≡ 1 (mod 3).

Step 1: Kill the 3-factor

Any divisor containing 3¹ or higher is divisible by 3 → it cannot be 3r + 1.

So exponent of 3 must be 0.

We only use: 2ᵃ × 5ᶜ × 7ᵈ

a: 0 to 6 → 7 values; c: 0 to 3 → 4 values; d: 0 to 2 → 3 values

Step 2: Work mod 3

2 ≡ -1 (mod 3); 5 ≡ -1 (mod 3); 7 ≡ 1 (mod 3)

So divisor ≡ (-1)^(a + c) (mod 3).

For divisor ≡ 1 (mod 3):

(-1)^(a + c) = 1 → a + c must be even.

Step 3: Count (a, c) with a + c even

a = 0,1,2,3,4,5,6; even: 0,2,4,6 → 4 values; odd: 1,3,5 → 3 values

c = 0,1,2,3; even: 0,2 → 2 values; odd: 1,3 → 2 values

a + c even when: a even, c even → 4 × 2 = 8; a odd, c odd → 3 × 2 = 6

Total (a, c) good pairs = 8 + 6 = 14. d has 3 values → 0,1,2.

Total divisors = 14 × 3 = 42.

Final Answer: 12

Given: 3ac = 8(a + b) ⇒ 8b = 3ac − 8a ⇒ b = (3ac/8) − a

Since a, b, c are natural numbers, 3ac must be divisible by 8.

Now try small values of a (since we want the minimum value of 3a + 2b + c).

Try a = 1

Then 3c must be divisible by 8 ⇒ c = 8

b = (3×1×8)/8 − 1 = 3 − 1 = 2

Value = 3a + 2b + c = 3 + 4 + 8 = 15

Try a = 2

Then 6c divisible by 8 ⇒ c = 4 is the smallest possible

b = (3×2×4)/8 − 2 = 3 − 2 = 1

Here a, b, c = 2, 1, 4 are distinct natural numbers

Value = 3a + 2b + c = 6 + 2 + 4 = 12

Try a = 3

Then 9c divisible by 8 ⇒ c = 8

b = (3×3×8)/8 − 3 = 9 − 3 = 6

Value = 9 + 12 + 8 = 29 (larger)

Hence the smallest possible value occurs at a = 2, b = 1, c = 4

Correct answer is B) 5 : 24

Let area of the hexagon is 6E, where E represents the area of any one of the six congruent central triangles, such as triangle OBC.

The area of trapezium PBCQ is obtained by decomposing it into simpler regions. It consists of the full area of triangle OBC along with suitable fractional parts of the triangles adjacent to the base BC.

This decomposition gives: Area(PBCQ) = 5E/4.

Therefore, the required ratio of areas is: (5E/4) ÷ 6E = 5/24.

Correct answer: C) 11:4

Given, AD : AT = 4 : 3 ⇒ AT : TD = 3 : 1 and BE : BT = 5 : 4 ⇒ BT : TE = 4 : 1

By the mass points method, masses are assigned inversely to the segments.

From AT : TD = 3 : 1 ⇒ Mass at A : Mass at D = 1 : 3.

From BT : TE = 4 : 1, ⇒ Mass at B : Mass at E = 1 : 4.

Choosing consistent masses,

A = 5/4, D = 15/4 and B = 1, E = 4, so mass at T = 5.

Since D lies on BC, Mass at D = Mass at B + Mass at C

15/4 = 1 + Mass at C ⇒ Mass at C = 11/4.

Therefore, BD : DC = Mass at C : Mass at B ⇒ (11/4) : 1 ⇒ 11 : 4.

Correct Answer 80

From P, PQ and PR are tangents to the circle with centre O, so OQ ⟂ PQ and OR ⟂ PR.

AB is also a tangent, touching the circle at S, hence OS ⟂ AB.

From an external point, the line joining the centre bisects the angle between tangents.

Thus OA bisects ∠QOS and OB bisects ∠ROS.

Let AOQ = AOS = x and BOR = BOS = y.

Given ∠AOB = x + y = 50°.

Then ∠QOS = 2x, ∠ROS = 2y and hence ∠QOR = 2(x + y) = 100°.

In quadrilateral ORPQ, angles at Q and R are 90°, so

∠QOR + ∠APB = 180°.

Therefore, ∠APB = 180° − 100° = 80°.

Final answer 25

Given: (625)⁶⁵ × (128)³⁶

Express the bases as powers of 5 and 2: 625 = 5⁴ and 128 = 2⁷

(625)⁶⁵ × (128)³⁶

= 5^(4×65) × 2^(7×36)

= 5^260 × 2^252

= 5^8 × 5^252 × 2^252

= 5^8 × 10^252

Now 5⁸ = 390625

So the full number is:

390625 followed by 252 zeros.

Sum of digits = 3 + 9 + 0 + 6 + 2 + 5 = 25

Answer: 25

Correct Answer B (48)

Given log₆₄(x²) + log₈(√y) + 3 log₅₁₂( √(y)z) = 4 x, y, z > 0

Step 1: Convert all logs to the same base (base 8)

64 = 8²; 512 = 8³

Convert log₆₄(x²) to base 8: log₆₄(x²) = (1/2) log₈(x²) = log₈x

Convert 3 log₅₁₂(√(y)z) to base 8: 3 log₅₁₂(√(y)z) = 3 (1/3) log₈(√(y)z) = log₈(√(y)z)

Substitute the converted terms back into the original equation: log₈x + log₈(√y) + log₈(√(y)z) = 4 Combine the logarithmic terms using the property logₐA + logₐB = logₐ(AB): log₈(x ⋅ √y ⋅ √y ⋅ z) = 4 => log₈(xyz) = 4

Convert the logarithmic equation to an exponential equation: xyz = 8⁴ => xyz = 4096

Apply the Arithmetic Mean – Geometric Mean (AM-GM) inequality.

For positive real numbers x, y, and z, the minimum value of their sum occurs when x = y = z.

(x + y + z) / 3 ≥ ³√(xyz)

Substitute the value of xyz: (x + y + z) / 3 ≥ ³√(4096)

Calculate the cube root: Since 16³ = 4096, we have: (x + y + z) / 3 ≥ 16

Solve for the minimum value of (x + y + z): x + y + z ≥ 3 × 16 x + y + z ≥ 48

Correct Answer: 54

Let the GP be a, ar, ar² with common ratio r (0 < r < 1 since it is decreasing and infinite).

Given:

a + ar + ar² = 52 → a(1 + r + r²) = 52

a·ar + ar·ar² + ar²·a = 624 → a²(r + r² + r³) = 624

From these, form a relation involving only r.

Divide equation (2) by [equation (1)]²:

[a²(r + r² + r³)] / [a²(1 + r + r²)²] = 624 / 52²

So:

(r + r² + r³) / (1 + r + r²)² = 624 / 2704 = 39 / 169

Now just try simple values of r between 0 and 1.

Try r = 1/2:

r + r² + r³ = 1/2 + 1/4 + 1/8 = 7/8

1 + r + r² = 1 + 1/2 + 1/4 = 7/4 → squared = 49/16

Ratio = (7/8) / (49/16) = 2/7 (not 39/169).

Try r = 1/3:

r + r² + r³ = 1/3 + 1/9 + 1/27 = 13/27

1 + r + r² = 1 + 1/3 + 1/9 = 13/9 → squared = 169/81

Ratio = (13/27) / (169/81) = 39/169 (matches).

So r = 1/3.

From a(1 + r + r²) = 52 → a(1 + 1/3 + 1/9) = 52 → a(13/9) = 52 → a = 36.

Sum of infinite GP = a / (1 − r) = 36 / (1 − 1/3) = 36 / (2/3) = 54.