CAT 2025 Slot 2 Question Paper with Solutions

VARC — 24 questions

Question 1

The given sentence is missing in the paragraph below. Decide where it best fits among the options 1, 2, 3, or 4 indicated in the paragraph. Sentence: While taste is related to judgment, with thinkers at the time often writing, for example, about “judgments of taste” or using the two terms interchangeably, taste retains a vital link to pleasure, embodiment, and personal specificity that is too often elided in post-Kantian ideas about judgment—a link that Arendt herself was working to restore. Paragraph: ____(1) ____. Denneny focused on taste rather than judgment in order to highlight what he believed was a crucial but neglected historical change. ____(2) ____. Over the course of the seventeenth century and early eighteenth century, across Western Europe, the word taste took on a new extension of meaning, no longer referring specifically to gustatory sensation and the delights of the palate but becoming, for a time, one of the central categories for aesthetic—and ethical—thinking. ____(3) ____. Tracing the history of taste in Spanish, French, and British aesthetic theory, as Denneny did, also provides a means to recover the compelling and relevant writing of a set of thinkers who have been largely neglected by professional philosophy. ____(4) ____.

Answer & explanation

Correct answer: A

The paragraph explains why Denneny focuses on “taste” instead of “judgment” and how this choice helps bring attention to an important but often ignored part of aesthetic and philosophical history. It first shows Denneny pointing out a forgotten historical change, then explains that in the seventeenth and early eighteenth centuries, “taste” grew from a physical sense to a main idea in aesthetics and ethics. Finally, it discusses how tracing this history helps recover overlooked thinkers and ideas. The missing sentence fits best at blank (3) because, by this point in the paragraph, it has already been described how ‘taste’ expanded. The sentence adds to the discussion by explaining why taste is important: even though it was often seen as the same as judgment, it kept its link to pleasure, the body, and personal experience, which later theories often ignored. This directly builds on the previous point about taste’s central role in aesthetics and ethics. Putting the sentence at blank (1) would be too early, since the paragraph has not yet talked about how the concept of ‘taste’ changed over time. Placing it at blank (2) would break up the flow from Denneny’s motivation to the bigger historical story. If it went at blank (4), it would feel awkward because the paragraph there moves on to Denneny’s method and the recovery of overlooked thinkers, not the difference between taste and judgment. So, blank (3) is the most coherent and logical position.

Question 2

The four sentences (labelled 1, 2, 3, and 4) given below, when properly sequenced, would yield a coherent paragraph. Decide on the proper sequencing of the order of the sentences and key in the sequence of the four numbers as your answer. 1. ‘Literature on screen' suggests something more capacious and defining than citation: the possibility that literary adaptations are at once cinema and literature. 2. Even though a growing number of films eligible for Academy Awards for Best Screenplay Based on Material from Another Medium borrow that material from print journalism, franchise characters, television series, comic books, video games and toys, academic studies of adaptation remain stubbornly attached to literature as cinema's natural progenitor. 3. It is as if adaptation studies, by borrowing the cultural cachet of literature, sought to claim its institutional respectability and gravitas even while insuring adaptation's enduring aesthetic and methodological subordination to literature proper. 4. Beneath this contradictory notion of film adaptations as not merely hybrid texts but texts holding dual citizenship in two modes of presentation is an even more pervasive legacy that haunts adaptation studies: the assumption that the primary context within which adaptations are to be studied is literature.

Enter your answer:

Up to 2 decimal places. No negative marking for TITA.

Answer & explanation

Correct answer: 1423

Sentence 1 works best as the opening because it introduces the key idea of what “literature on screen” means and frames the debate by suggesting that adaptations might belong to both cinema and literature. Sentence 4 comes next because it explores this idea further and points out the main assumption in adaptation studies: that literature is the main way to understand adaptations. Sentence 2 then gives clear evidence of this bias by showing the contradiction that even though films now adapt material from many non-literary sources, academic studies still treat literature as cinema’s natural source. Finally, sentence 3 works as the conclusion because it explains why this contradiction continues. It argues that adaptation studies rely on literature’s cultural prestige to gain legitimacy, even if it means putting cinema in a secondary position. The most logical and clear order is therefore 1423.

Question 3

The passage given below is followed by four summaries. Choose the option that best captures the essence of the passage. For millennia, in the process of opening up land for agriculture, gardens, grazing and hunting, humans have created ecological “mosaics”, or “patchworks”: landscapes holding a mixture of habitats, like meadows, gardens and forests. These were not designed as nature reserves, but often catered to hugely diverse animal life. Research indicates that European hay meadows cultivated for animal feed were actually more successful at preserving a vast array of species than meadows explicitly cultivated for biodiversity. Studying the early Holocene, researchers have found that human presence was about as likely to increase biodiversity as reduce it. Of course, not all human-created landscapes have the same value. A paved subdivision with astroturfed lawns is very different to a village with diverse vegetable and flower gardens. But scientists continue to find evidence that the old idea of humans as antithetical to nature is also wrong-headed, and that rosy visions of thriving, human-free environments are more imaginary than real.

Answer & explanation

All of the following statements may be rejected as valid inferences from the passage EXCEPT:

Answer & explanation

Correct answer: C

Which one of the following results for the cross between surface fish (with eyes) and cavefish (without eyes) would invalidate Riddle’s inference from the experiment?

Answer & explanation

Correct answer: C

The experiment tests Riddle’s idea that carotenoid accumulation (yellow fat) is not linked to eye loss but serves another purpose, like reducing inflammation. To disprove this, results must show a direct connection between losing eyes and carotenoid accumulation. Option C would directly invalidate this inference . If only the eyeless offspring had yellow fat, it would provide a strong link between the absence of eyes and carotenoid buildup. This would go against Riddle’s statement that there is “no link between eye presence or size and the accumulation of carotenoids,” and would support the earlier idea that carotenoids build up because cavefish lack eyes. The other options do not challenge her idea. Options A and D show that white fat can appear whether or not the fish have eyes. Option B shows that fish with eyes can also have yellow fat. These results fit with Riddle’s claim that carotenoid buildup does not depend on eye presence. So, only option C would disprove her inference.

Question 19

On the basis of the information in the passage, what is the most likely function of carotenoids in Mexican tetra cavefish?

Answer & explanation

Correct answer: D

The passage explicitly states Riddle’s conclusion: “Instead, Riddle thinks these carotenoids may be another adaptation to suppress inflammation.” This directly points to carotenoids playing a role in controlling inflammation, especially in the context of fat cells becoming large and potentially causing inflammatory damage. Therefore, the correct answer is option D . The other options are either not mentioned in the passage or, if mentioned, are discussed in a different context. Option A is mentioned only as an initial hypothesis that carotenoids might be linked to eye loss, but this idea is explicitly rejected after cross-breeding shows “ no link between eye presence or size and the accumulation of carotenoids. ” Option B is not mentioned at all in connection with carotenoids; nutrient storage is discussed as a general metabolic adaptation, not as a function of carotenoids. Option C refers merely to the observed yellow colour of fat, which the passage treats as a visible effect, not as the biological purpose of carotenoids.

Passage for questions 20–23

The passage below is accompanied by four questions. Based on the passage, choose the best answer for each question. This book takes the position that setting in literature is more than just backdrop, that important insight into literary texts can be made by paying close attention to how authors craft place, as well as to how place functions in a narrative. The authors included in this reference work engage deeply with either real or imagined geographies. They care about how human decisions have shaped landscapes and how landscapes have shaped human practices and values. Some of the best writing is highly vivid, employing the language of the senses because this is the primary means through which humans know physical space. Literature can offer valuable perspectives on physical and cultural geography. Unlike scientific reports, a literary narrative can provide the emotional component missing from the scientific record. In human experience, geographical places have a spiritual or emotional component in addition to and as part of a physical layout and topography. This emotional component, although subjective, is no less “real” than a surveyor’s map. Human consciousness of place is experienced in a multimodal manner. Histories of places live on in many forms, one of which is the human memory or imagination. Both real and imaginary landscapes provide insight into the human experience of place. The pursuit of such a topic speaks to the valuable knowledge produced from bridging disciplines and combining material from both the arts and the sciences to better understand the human condition. The perspectives that most concern cultural geographers are often those regarding movement and migration, cultivation of natural resources, and organization of space. The latter two reflect concerns of the built environment, a topic shared with the field of architectural study. Many of these concerns are also reflected in work sociologists do. Scholars from literary studies can contribute an aesthetic dimension to what might otherwise be a purely ideological approach. Literature can bring together material that spans different branches of science. For example, a literary description of place may involve not only the environment and geography but the noises and quality of light, or how people from different races or classes can experience the same place in different ways linked to those racial or class disparities. Literary texts can also account for the way in which absence—of other people, animals, and so on—affects a human observer or inhabitant. Both literary and scientific approaches to place are necessary, working in unison, to achieve a complete record of an environment. It is important to note that the interdisciplinary nature of this work teaches us that landscapes are not static, that they are not unchanged by human culture. At least part of their identity derives from the people who inhabit them and from the way space can alter and inspire human perspective. The intersection of scientific and literary expression that happens in the study of literary geography is of prime importance due to the complexity of the personal and political ways that humans experience place.

Question 20

Which one of the following is a valid conclusion to draw from the author’s statement that, “The pursuit of such a topic speaks to the valuable knowledge produced from bridging disciplines and combining material from both the arts and the sciences to better understand the human condition.”?

Answer & explanation

Correct answer: C

The sentence "The pursuit of such a topic speaks to the valuable knowledge produced from bridging disciplines and combining material from both the arts and the sciences to better understand the human condition" refers to the importance of "bridging" or intermixing of the material from both arts and sciences in understanding the human condition. The statement that best captures this is provided in option C. Option A is incorrect as "bridging of human condition" is a wrong inference. Option B, although partly correct, misses the message of "bridging" which forms the core of the sentence. Option D is incorrect as it reverses the relationship of the importance of knowledge from arts and science in our understanding of the human condition by writing that human emotions are essential contributors to our understanding of arts and sciences.

Question 21

Which one of the following is not true of the argument in the second paragraph?

Answer & explanation

Correct answer: C

The second paragraph primarily emphasises the importance of literature in providing insights into the emotional dimension of physical and cultural geography, which is all too often missing from scientific records. The paragraph notes that this emotional or spiritual component is primarily subjective and that the history of a place also exists in memory and imagination. Accordingly, options A, B, and D follow. Option C is extreme in that it says that the emotional and spiritual experience of a place can replace a surveyor's map, and is incorrect. Although the emotional and spiritual experience can complement the factual information about geographic and physical world provided by the scientific record, it cannot replace it.

Question 22

The author uses the example of the literary description of place to illustrate that:

Answer & explanation

Correct answer: B

The paragraph states that "For example, a literary description of place may involve not only the environment and geography but the noises and quality of light, or how people from different races or classes can experience the same place in different ways linked to those racial or class disparities." Based on this and the information provided in the passage, it would be wrong to consider the fact that scientific approaches to place are more (or less) accurate than literary ones. The most we can say is that both are useful and complementary. Option A is incorrect. Option B follows from the example of literary description, which examines how people from different backgrounds experience the same place differently, and is therefore the correct answer. Option C mentions architects and the diverse methods they use, which is not the reason why the passage provides the example of literary descriptions. Similarly, option D examines in detail how the absence of people and animals affects the inhabitants and fails to capture the broader message the passage seeks to convey through the example provided. Option B is the correct answer.

Question 23

All of the following statements, if false, would contradict the arguments in the passage, EXCEPT that:

Answer & explanation

Correct answer: D

The passage asserts that "Histories of places live on in many forms, one of which is the human memory or imagination." The passage also implies that these places and their "emotional component, although subjective, is no less 'real' than a surveyor’s map." It follows that, in literature, descriptions of places do not need satellite imagery or other visual aids to give a “real” sense of the place. If this were to be false, the arguments made in the passage would be contradicted. Option A is not the correct answer. From "Both real and imaginary landscapes provide insight into the human experience of place", we can also infer that if option B were to be false, the information would contradict the passage. Option B is not the correct answer. From "Literature can bring together material that spans different branches of science. For example, a literary description of place may involve not only the environment and geography but the noises and quality of light, or how people from different races or classes can experience the same place in different ways linked to those racial or class disparities. Literary texts can also account for the way in which absence—of other people, animals, and so on—affects a human observer or inhabitant", we can infer that literature can be a powerful device that can produce vivid realities through descriptions. If option C were to be false, it would contradict the information provided in the passage. Option C is not the correct answer. From "In human experience, geographical places have a spiritual or emotional component in addition to and as part of a physical layout and topography. This emotional component, although subjective, is no less “real” than a surveyor’s map," we can infer that the physical and emotional experience of a place is subjective. Therefore, if option D were to be false, the arguments provided in the passage will not be contradicted. Option D is the correct answer.

Question 24

The four sentences (labelled 1, 2, 3, and 4) given below, when properly sequenced, would yield a coherent paragraph. Decide on the proper sequencing of the order of the sentences and key in the sequence of the four numbers as your answer. 1. As books age, the cellulose and lignin in the paper begin to break down, releasing a mix of volatile organic compounds into the air. 2. Old books carry a scent that many people instantly recognize—and even love. 3. These compounds are benzaldehyde, which gives off an almond-like scent, vanillin, which smells like vanilla, ethyl hexanol (floral scent), toluene (sweet), and furfural (which has a slightly bready scent). 4. This familiar aroma isn't just dust or mildew; it's actually a result of slow chemical changes happening inside the paper and ink.

Enter your answer:

Up to 2 decimal places. No negative marking for TITA.

Answer & explanation

Correct answer: 2413

Sentence 1 mentions "compounds" while sentence 3 talks about them in detail. Sentence 3 starts with "these compounds" making sentence 1 and 3 an adjacent pair in that order. Similarly, sentence 2 introduces "a scent" that is elaborated upon in sentence 4, implying that sentences 2 and 4 are a pair in that order. The "slow chemical changes" mentioned in sentence 4 are elaborated further in sentences 1 and 3. The most suitable sequence, therefore, would be 2413.

DILR — 22 questions

Passage for questions 25–29

Ananya Raga, Bhaskar Tala, Charu Veena, and Devendra Sur are four musicians. Each of them started and completed their training as students under each of three Gurus — Pandit Meghnath, Ustad Samiran, and Acharya Raghunath between 2013 and 2024, including both the years. Each Guru trains any student for consecutive years only, for a span of 2, 3, or 4 years, with each Guru having a different span. During some of these years, a student may not have trained under these Gurus; however, they never trained under multiple Gurus in the same year. In none of these years, any of these Gurus trained more than two of these students at the same time. When two students train under the same Guru at the same time, they are referred to as Gurubhai, irrespective of their gender. The following additional facts are known. 1. Ustad Samiran never trained more than one of these students in the same year. 2. Acharya Raghunath did not train any of these students during 2015-2018, as well as during 2021-24. 3. Ananya and Devendra were never Gurubhai; neither were Bhaskar and Charu. All other pairs of musicians were Gurubhai for exactly 2 years. 4. In 2013, Ananya and Bhaskar started their trainings under Pandit Meghnath and under Ustad Samiran, respectively.

Question 25

In which of the following years were Ananya and Bhaskar Gurubhai?

Answer & explanation

Correct answer: A

Representing Ananya Raga, Bhaskar Tala, Charu Veena, and Devendra Sur as A, B, C and D, respectively, for easy usage. Also using PM, US, and AR for Pandit Meghnath, Ustad Samiran, and Acharya Raghunath, respectively, for easy reference.

We are given that AR did not train during the periods of 2015-2018 and 2021-2024.

Other information provided is that PM, US, and AR had a span of 2, 3, and 4 consecutive years for each student, in some order. We also know that all the gurus trained each of the students for exactly one span.

In clue 4, we are given that A started the training in 2013 under PM, and B started training in 2013 under US.

Putting all the known information in the table, we get,

Now, if we look at the table, we can see that AR is training for only 4 years, with two years each, separated by a few years. Therefore, the span of AR must be 2 years because we know that he is not teaching continuously for 3 or 4 years.

Since we know that each guru taught all the students, and AR only has 4 years to teach all the students, he must definitely teach 2 students at a time each year, as we are told that no guru taught more than 2 students in a single year. We also know that A and B were taught by PM and US in 2013, respectively. Therefore, we can conclude that C and D were taught by AR during both 2013 and 2014, as AR's span is 2 years.

We can also conclude that A and B were taught by AR during both 2019 and 2020, as AR's span is 2 years, and they were the only students left for AR's class.

We are also given in clue 1 that US did not train more than 1 student in any year, and there are a total of 12 years from 2013 to 2024. If the span of US is 4 years, then he would need a total of 16 years to teach every student, as he is not teaching more than 1 student in any year, but he only taught for 12 years, so the only possibility is for US to have a span of 3 years and PM to have a span of 4 years. We can also conclude that US taught B from 2013 to 2015, as he has a span of 3 years.

We are also given that A, D and B, C were never Gurubhai, and all the other pairs were Gurubhai for exactly 2 years. Therefore, there must be two years each of {A, B}, {A, C}, {B, D}, and {C, D} with classes together. We have already completed the 2 years of {A, B} and {C, D}, so the ones left are two years of {A, C} and {B, D}. We know that the US did not take classes of 2 students together, and we have already filled the schedule of AR, so the only one left is PM, and he must have the other two left.

Since the span of PM is 4, he taught A from 2013 to 2016 and during this span, there must be 2 years of {A, C}, as we know that A will not be taught again after 2016. To have an exact span of 2 years together, C must start in 2015, as it is the only possibility, and he will teach C until 2018, as he has a span of 4 years.

There must be 2 years of {B, D} in the remaining time period of PM. In 2019, A and B were already enrolled with AR, and C had already completed his time with PM. So, the only person possible to have class in 2019 with PM is D, and D will have classes with PM until 2022. Now for 2 years of {B, D}, D must have his classes start from 2021, as it is the only possibility.

There cannot be any gap in the case of US, and in 2016, the only possibility is D as A, C were having classes with PM, and B already had his classes with US. D had classes from 2016 to 2018. In 2019, A, B and D were already having classes, so the only possibility is for C to have classes with US from 2019 to 2021, and finally A will have his classes from 2022 to 2024 with US, as he is the only person left to have classes with US.

The final table looks like,

Ananya and Bhaskar were Gurubhai from 2019 to 2020. Since 2020 is the only option present during this period, it has to be the answer.

Hence, the correct answer is option A.

Question 26

In which year did Charu begin her training under Pandit Meghnath?

Answer & explanation

Correct answer: B

We are given that AR did not train during the periods of 2015-2018 and 2021-2024.

In clue 4, we are given that A started the training in 2013 under PM, and B started training in 2013 under US.

Putting all the known information in the table, we get,

We can also conclude that A and B were taught by AR during both 2019 and 2020, as AR's span is 2 years, and they were the only students left for AR's class.

The final table looks like,

Charu began her training under Pandit Meghnath in 2015.

Hence, the correct answer is option B.

Question 27

In which of the following years were Bhaskar and Devendra Gurubhai?

Answer & explanation

Correct answer: A

We are given that AR did not train during the periods of 2015-2018 and 2021-2024.

In clue 4, we are given that A started the training in 2013 under PM, and B started training in 2013 under US.

Putting all the known information in the table, we get,

We can also conclude that A and B were taught by AR during both 2019 and 2020, as AR's span is 2 years, and they were the only students left for AR's class.

The final table looks like,

Bhaskar and Devendra were Gurubhai from 2021 to 2022. Since 2022 is the only option present during this period, it has to be the answer.

Hence, the correct answer is option A.

Question 28

Which of the following statements is TRUE?

Answer & explanation

Correct answer: D

We are given that AR did not train during the periods of 2015-2018 and 2021-2024.

In clue 4, we are given that A started the training in 2013 under PM, and B started training in 2013 under US.

Putting all the known information in the table, we get,

We can also conclude that A and B were taught by AR during both 2019 and 2020, as AR's span is 2 years, and they were the only students left for AR's class.

The final table looks like,

Statement 1) Charu was training under Ustad Samiran in 2018. This is false as she was training under Pandit Meghnath in 2018.

Statement 2) Ananya was training under Ustad Samiran in 2015. This is false as she was training under Pandit Meghnath in 2015.

Statement 3) Ananya was training under Ustad Samiran in 2018. This is false as she was not training in 2018.

Statement 4) Charu was training under Ustad Samiran in 2019. This is true.

Hence, the correct answer is option D.

Question 29

In how many of the years between 2013-24, were only two of these four musicians training under these three Gurus?

Enter your answer:

Up to 2 decimal places. No negative marking for TITA.

Answer & explanation

Correct answer: 4

We are given that AR did not train during the periods of 2015-2018 and 2021-2024.

In clue 4, we are given that A started the training in 2013 under PM, and B started training in 2013 under US.

Putting all the known information in the table, we get,

We can also conclude that A and B were taught by AR during both 2019 and 2020, as AR's span is 2 years, and they were the only students left for AR's class.

The final table looks like,

Between 2013 and 2024, the years 2017, 2018, 2023, and 2024 were the only years when only two of these four musicians were training under these three Gurus.

Hence, the correct answer is 4.

Passage for questions 30–33

The following charts depict details of research papers written by four authors, Arman, Brajen, Chintan, and Devon. The papers were of four types, single-author, two-author, three-author, and four-author, that is, written by one, two, three, or all four of these authors, respectively. No other authors were involved in writing these papers.

The following additional facts are known. 1. Each of the authors wrote at least one of each of the four types of papers. 2. The four authors wrote different numbers of single-author papers. 3. Both Chintan and Devon wrote more three-author papers than Brajen. 4. The number of single-author and two-author papers written by Brajen were the same.

Question 30

What was the total number of two-author and three-author papers written by Brajen?

Enter your answer:

Up to 2 decimal places. No negative marking for TITA.

Answer & explanation

Correct answer: 4

If all the two-author-type books are counted for both authors separately, then we get the sum to be 4 * 2 = 8.

If all the three-author-type books are counted for all three authors separately, then we get the sum to be 3 * 3 = 9.

If all the four-author-type books are counted for all four authors separately, then we get the sum to be 4 * 2 = 8.

We are given that there are 2 four-author books, so there are exactly two books for each that are of the four-author type.

Putting the values in the table, we get,

We are given that each author has at least one book of each type. In the case of Arman, there are three books left to be assigned, and there are 3 types of books left, so each of them has to be equal to 1 for the above condition to be satisfied.

In the case of Brajen, we are told that he had an equal number of single-author and two-author books.

Case 1: If the equal number is 1, then the three-author type books become 4, and we are given that both Chintan and Devon had more three-author books than Brajen, which is not possible in this case, as both of them combined will be left with 5 books together and whaterver we split 5 books there wont be a case of both of them having more books than Brajen.

Case 2: If the equal number is 2, then the three-author type books become 2, and we are given that both Chintan and Devon had more three-author books than Brajen, which is possible in this case, as both of them combined will be left with 6 books together and the only way to split 6 books such that both of them are greater than 2 is to split them into 3, 3 and it is the only possible way.

Case 3: If the equal number is greater than or equal to 3, then the three-author type books become 0 or negative, which can directly be eliminated, as we are given that all authors wrote at least 1 book of each type.

So, we can eliminate both cases 1 and 3, and we will be left with the case of Brajen having 2 books each of Single-author, Two-author, and Three-author types. Chintan and Devon have three books each of the three-author type.

Filling up the table with these values, we get,

We are given that each author has a different number of single-author type books. The only possibilities for the single-author books of Chintan and Devon are 3 and 4 in some order.

If the single-author books of Chintan are 3, then the double-author books of Chintan become 4, the single-author books of Devon become 4, and the double-author books of Devon become 1.

If the single-author books of Chintan are 4, then the double-author books of Chintan become 3, the single-author books of Devon become 3, and the double-author books of Devon become 2.

Placing both possibilities in the table, we get,

The total number of two-author and three-author papers written by Brajen = 2 + 2 = 4.

Hence, the correct answer is 4.

Question 31

Which of the following statements is/are NECESSARILY true? i. Chintan wrote exactly three two-author papers. ii. Chintan wrote more single-author papers than Devon.

Answer & explanation

Correct answer: A

If all the two-author-type books are counted for both authors separately, then we get the sum to be 4 * 2 = 8.

If all the three-author-type books are counted for all three authors separately, then we get the sum to be 3 * 3 = 9.

If all the four-author-type books are counted for all four authors separately, then we get the sum to be 4 * 2 = 8.

We are given that there are 2 four-author books, so there are exactly two books for each that are of the four-author type.

Putting the values in the table, we get,

In the case of Brajen, we are told that he had an equal number of single-author and two-author books.

Filling up the table with these values, we get,

We are given that each author has a different number of single-author type books. The only possibilities for the single-author books of Chintan and Devon are 3 and 4 in some order.

If the single-author books of Chintan are 3, then the double-author books of Chintan become 4, the single-author books of Devon become 4, and the double-author books of Devon become 1.

If the single-author books of Chintan are 4, then the double-author books of Chintan become 3, the single-author books of Devon become 3, and the double-author books of Devon become 2.

Placing both possibilities in the table, we get,

Statement 1: Chintan wrote exactly three two-author papers.

This is not necessarily true, as he could have written four books as well.

Statement 2: Chintan wrote more single-author papers than Devon.

This is not necessarily true, as the opposite can be possible.

So, neither I nor 2 is necessarily true.

Hence, the correct answer is option A.

Question 32

Which of the following statements is/are NECESSARILY true? i. Arman wrote three-author papers only with Chintan and Devon. ii. Brajen wrote three-author papers only with Chintan and Devon.

Answer & explanation

Correct answer: B

If all the two-author-type books are counted for both authors separately, then we get the sum to be 4 * 2 = 8.

If all the three-author-type books are counted for all three authors separately, then we get the sum to be 3 * 3 = 9.

If all the four-author-type books are counted for all four authors separately, then we get the sum to be 4 * 2 = 8.

We are given that there are 2 four-author books, so there are exactly two books for each that are of the four-author type.

Putting the values in the table, we get,

In the case of Brajen, we are told that he had an equal number of single-author and two-author books.

Filling up the table with these values, we get,

We are given that each author has a different number of single-author type books. The only possibilities for the single-author books of Chintan and Devon are 3 and 4 in some order.

If the single-author books of Chintan are 3, then the double-author books of Chintan become 4, the single-author books of Devon become 4, and the double-author books of Devon become 1.

If the single-author books of Chintan are 4, then the double-author books of Chintan become 3, the single-author books of Devon become 3, and the double-author books of Devon become 2.

Placing both possibilities in the table, we get,

Statement 1: Arman wrote three-author papers only with Chintan and Devon.

There are in total 3 three-author books, and Chintan and Devon are authors of all three of them according to the table. So, Arman wrote the three-author papers only with Chintan and Devon. So, the statement is necessarily true

Statement 2: Brajen wrote three-author papers only with Chintan and Devon.

There are in total 3 three-author books, and Chintan and Devon are authors of all three of them according to the table. So, Brajen wrote the three-author papers only with Chintan and Devon. So, the statement is necessarily true.

So, both I and 2 are necessarily true.

Hence, the correct answer is option B.

Question 33

If Devon wrote more than one two-author papers, then how many two-author papers did Chintan write?

Enter your answer:

Up to 2 decimal places. No negative marking for TITA.

Answer & explanation

Correct answer: 3

If all the two-author-type books are counted for both authors separately, then we get the sum to be 4 * 2 = 8.

If all the three-author-type books are counted for all three authors separately, then we get the sum to be 3 * 3 = 9.

If all the four-author-type books are counted for all four authors separately, then we get the sum to be 4 * 2 = 8.

We are given that there are 2 four-author books, so there are exactly two books for each that are of the four-author type.

Putting the values in the table, we get,

In the case of Brajen, we are told that he had an equal number of single-author and two-author books.

Filling up the table with these values, we get,

We are given that each author has a different number of single-author type books. The only possibilities for the single-author books of Chintan and Devon are 3 and 4 in some order.

If the single-author books of Chintan are 3, then the double-author books of Chintan become 4, the single-author books of Devon become 4, and the double-author books of Devon become 1.

Correct answer: A

Let us try to arrange the hoops and the spheres in increasing order of their diameter according to the information given, In clue 1, we are given that B1 and B6 made a ping on H4, and B5 did not. So, we can definitely say that, (B1, B6) < H4 < B5 --(1) In clue 2, we are given that B4 made a ping on H3, but B1 did not. So, we can definitely say that, B4 < H3 < B1 --(2) In clue 3, we are given that all balls, except B3, made pings on H1. So, we can definitely say that, (B1, B2, B4, B5, B6) < H1 < B3 --(3) In clue 4, we are given that none of the balls, except B2, made a ping on H2. So, we can definitely say that, B2 < H2 < (B1, B3, B4, B5, B6) --(4) Combining (1) and (2), we can definitely say that B4 < H3 < B1 < H4 < B5 --(5) The only ball for which we do not have enough information is about B6, as it can be either less than H3 or greater than H3 and less than H4. Combining (3), (4) and (5), we get, B2 < H2 < B4 < H3 < B1 < H4 < B5 < H1 < B3 The sphere B6 can be placed in 2 positions, giving us two possible inequalities, which are B2 < H2 < (B4, B6) < H3 < B1 < H4 < B5 < H1 < B3 B2 < H2 < B4 < H3 < (B1,B6) < H4 < B5 < H1 < B3 The position of B6 is unclear, and except for that, all the other positions are fixed. We know the correct order of the diameters of the hoops is H2 < H3 < H4 < H1. Hence, the correct answer is option A.

Question 37

What BEST can be said about the total number of pings from all the tests undertaken?

Answer & explanation

Correct answer: A

Let us try to arrange the hoops and the spheres in increasing order of their diameter according to the information given, In clue 1, we are given that B1 and B6 made a ping on H4, and B5 did not. So, we can definitely say that, (B1, B6) < H4 < B5 --(1) In clue 2, we are given that B4 made a ping on H3, but B1 did not. So, we can definitely say that, B4 < H3 < B1 --(2) In clue 3, we are given that all balls, except B3, made pings on H1. So, we can definitely say that, (B1, B2, B4, B5, B6) < H1 < B3 --(3) In clue 4, we are given that none of the balls, except B2, made a ping on H2. So, we can definitely say that, B2 < H2 < (B1, B3, B4, B5, B6) --(4) Combining (1) and (2), we can definitely say that B4 < H3 < B1 < H4 < B5 --(5) The only ball for which we do not have enough information is about B6, as it can be either less than H3 or greater than H3 and less than H4. Combining (3), (4) and (5), we get, B2 < H2 < B4 < H3 < B1 < H4 < B5 < H1 < B3 The sphere B6 can be placed in 2 positions, giving us two possible inequalities, which are B2 < H2 < (B4, B6) < H3 < B1 < H4 < B5 < H1 < B3 B2 < H2 < B4 < H3 < (B1,B6) < H4 < B5 < H1 < B3 The position of B6 is unclear, and except for that, all the other positions are fixed. In the first case, the number of pings possible can be calculated as 4 for B2, 3 for B4, 3 for B6, 2 for B1, 1 for B5 and 0 for B3. In total, the number of pings possible in the 1st case is 4 + 3 + 3 + 2 + 1 = 13. In the second case, the number of pings possible can be calculated as 4 for B2, 3 for B4, 2 for B6, 2 for B1, 1 for B5 and 0 for B3. In total, the number of pings possible in the 2nd case is 4 + 3 + 2 + 2 + 1 = 12. So, the total number of pings can be 12 or 13. Hence, the correct answer is Option A.

Passage for questions 38–41

The Sustainability Index (SI) of a country at a point in time is an integer between 1 and 100. This question is related to SI of six countries - A, B, C, D, E, and F - at three different points in time - 2016, 2020, and 2024. The plot represents the exact changes in their SI, with X-coordinate representing % increase in 2020 from 2016, i.e., (SI in 2020 minus SI in 2016) / (SI in 2016), and Y-coordinate representing % increase in 2024 from 2020. At any point in time, the country with highest SI is ranked 1, while the country with the lowest SI is ranked 6. The following additional facts are known. 1. In 2016, B, C, E, and A had ranks 1, 2, 3, and 4 respectively. 2. F had lower SI than any other country in 2016, 2020, and 2024. 3. In 2024, E was the only country with SI of 90. 4. The range of SI of the six countries was 60 in 2016 as well as in 2024.

Question 38

What was the SI of E in 2016?

Enter your answer:

Up to 2 decimal places. No negative marking for TITA.

Answer & explanation

Correct answer: 60

Let us assume the values of A, B, C, D, E and F to be a, b, c, d, e and f in 2016.

For A:

Value in 2020 = increased by 25% from 2016 = $$\left(1\ +\ \dfrac{25}{100}\right)a$$ = $$\dfrac{5}{4}a$$

Value in 2024 = increased by 50% from 2020 = $$\left(1\ +\ \dfrac{50}{100}\right)\times\ \dfrac{5a}{4}$$ = $$\dfrac{15}{8}a$$

For B:

Value in 2020 = decreased by 25% from 2016 = $$\left(1\ -\ \dfrac{25}{100}\right)b$$ = $$\dfrac{3}{4}b$$

Value in 2024 = decreased by 25% from 2020 = $$\left(1\ -\ \dfrac{25}{100}\right)\times\ \dfrac{3b}{4}$$ = $$\dfrac{9}{16}b$$

For C:

Value in 2020 = decreased by 20% from 2016 = $$\left(1\ -\ \dfrac{20}{100}\right)c$$ = $$\dfrac{4}{5}c$$

Value in 2024 = increased by 40% from 2020 = $$\left(1\ +\ \dfrac{40}{100}\right)\times\ \dfrac{4c}{5}$$ = $$\dfrac{28}{25}c$$

For D:

Value in 2020 = increased by 100% from 2016 = $$\left(1\ +\ \dfrac{100}{100}\right)d$$ = $$2d$$

Value in 2024 = increased by 20% from 2020 = $$\left(1\ +\ \dfrac{20}{100}\right)\times\ 2d$$ = $$\dfrac{12}{5}d$$

For E:

Value in 2020 = increased by 25% from 2016 = $$\left(1\ +\ \dfrac{25}{100}\right)e$$ = $$\dfrac{5}{4}e$$

Value in 2024 = increased by 20% from 2020 = $$\left(1\ +\ \dfrac{20}{100}\right)\times\ \dfrac{5e}{4}$$ = $$\dfrac{3}{2}e$$

For F:

Value in 2020 = increased by 100% from 2016 = $$\left(1\ +\ \dfrac{100}{100}\right)f$$ = $$2f$$

Value in 2024 = decreased by 25% from 2020 = $$\left(1\ -\ \dfrac{25}{100}\right)\times\ 2f$$ = $$\dfrac{3}{2}f$$

We are given that the value of E is 90 in 2024. We can calculate the value of e as,

$$\dfrac{3e}{2}\ =\ 90$$

$$e\ =\ 60$$

The value of E in 2016 is 60, and its value in 2020 can be calculated as,

$$\dfrac{5e}{4}\ =\ \dfrac{5}{4}\times\ 60\ =\ 75$$

We are given that B, C, E, and A had ranks 1, 2, 3, and 4, respectively, in 2016, and F has the lowest rank(6) in all the years.

Putting the values in the table, we get,

SI of E in 2016 is 60.

Hence, the correct answer is 60.

Question 39

What was the SI of F in 2020?

Enter your answer:

Up to 2 decimal places. No negative marking for TITA.

Answer & explanation

Correct answer: 40

Let us assume the values of A, B, C, D, E and F to be a, b, c, d, e and f in 2016.

For A:

Value in 2020 = increased by 25% from 2016 = $$\left(1\ +\ \dfrac{25}{100}\right)a$$ = $$\dfrac{5}{4}a$$

Value in 2024 = increased by 50% from 2020 = $$\left(1\ +\ \dfrac{50}{100}\right)\times\ \dfrac{5a}{4}$$ = $$\dfrac{15}{8}a$$

For B:

Value in 2020 = decreased by 25% from 2016 = $$\left(1\ -\ \dfrac{25}{100}\right)b$$ = $$\dfrac{3}{4}b$$

Value in 2024 = decreased by 25% from 2020 = $$\left(1\ -\ \dfrac{25}{100}\right)\times\ \dfrac{3b}{4}$$ = $$\dfrac{9}{16}b$$

For C:

Value in 2020 = decreased by 20% from 2016 = $$\left(1\ -\ \dfrac{20}{100}\right)c$$ = $$\dfrac{4}{5}c$$

Value in 2024 = increased by 40% from 2020 = $$\left(1\ +\ \dfrac{40}{100}\right)\times\ \dfrac{4c}{5}$$ = $$\dfrac{28}{25}c$$

For D:

Value in 2020 = increased by 100% from 2016 = $$\left(1\ +\ \dfrac{100}{100}\right)d$$ = $$2d$$

Value in 2024 = increased by 20% from 2020 = $$\left(1\ +\ \dfrac{20}{100}\right)\times\ 2d$$ = $$\dfrac{12}{5}d$$

For E:

Value in 2020 = increased by 25% from 2016 = $$\left(1\ +\ \dfrac{25}{100}\right)e$$ = $$\dfrac{5}{4}e$$

Value in 2024 = increased by 20% from 2020 = $$\left(1\ +\ \dfrac{20}{100}\right)\times\ \dfrac{5e}{4}$$ = $$\dfrac{3}{2}e$$

For F:

Value in 2020 = increased by 100% from 2016 = $$\left(1\ +\ \dfrac{100}{100}\right)f$$ = $$2f$$

Value in 2024 = decreased by 25% from 2020 = $$\left(1\ -\ \dfrac{25}{100}\right)\times\ 2f$$ = $$\dfrac{3}{2}f$$

We are given that the value of E is 90 in 2024. We can calculate the value of e as,

$$\dfrac{3e}{2}\ =\ 90$$

$$e\ =\ 60$$

The value of E in 2016 is 60, and its value in 2020 can be calculated as,

$$\dfrac{5e}{4}\ =\ \dfrac{5}{4}\times\ 60\ =\ 75$$

We are given that B, C, E, and A had ranks 1, 2, 3, and 4, respectively, in 2016, and F has the lowest rank(6) in all the years.

Putting the values in the table, we get,

We know that F is ranked 6 and B is ranked 1 in 2016, and we are also given that the range of SI in 2016 and 2024 is 60. This means that the value of b-f = 60, and we can conclude that b is definitely greater than 60, as we know that f is an integer greater than 0.

We know that all the SI values in all the years are integers, and for SI of B to be an integer in 2024, the value of b must be a multiple of 16. The multiples of 16 that are greater than 60 and less than 100 are 64, 80 and 96.

CASE 1: b = 64

If b = 64, then the value of f becomes 4, as we know b - f = 60.

If f = 4, then the value of F in 2024 is 3f/2 = 12/2 = 6.

We already know the value of E in 2024 is 90.

In that case, the range of SI in 2024 is at least 90 - 6 = 84, but we are given that the range is 60 in 2024, which is not possible in this case.

So, we can eliminate this case.

CASE 2: b = 96

If b = 96, then the value of f becomes 36, as we know b - f = 60.

If f = 36, then the value of F in 2024 is 3f/2 = 108/2 = 54.

The value of B in 2024 = 9b/16 = (9 * 96)/16 = 54

We calculated the value of B and F to be 54 in 2024, which is not possible, as we are given that F had a lower SI than any other country in 2024, and we obtained its value to be the same as B.

So, we can eliminate this case.

CASE 3: b = 80

If b = 80, then the value of f becomes 20, as we know b - f = 60.

If f = 20, then the value of F in 2024 is 3f/2 = 60/2 = 30.

The value of B in 2024 = 9b/16 = (9 * 80)/16 = 45

We can see that no condition is violated in this case, as F is less than B in 2024, and the range in 2024 from the obtained values as of now is 90 - 30 = 60.

So, we can conclude that the value of b = 80 and f = 20.

Substituting the values in the table, we get,

So, SI of F in 2020 is 40.

Hence, the correct answer is 40.

Question 40

What was the SI of C in 2024?

Enter your answer:

Up to 2 decimal places. No negative marking for TITA.

Answer & explanation

Correct answer: 84

Let us assume the values of A, B, C, D, E and F to be a, b, c, d, e and f in 2016.

For A:

Value in 2020 = increased by 25% from 2016 = $$\left(1\ +\ \dfrac{25}{100}\right)a$$ = $$\dfrac{5}{4}a$$

Value in 2024 = increased by 50% from 2020 = $$\left(1\ +\ \dfrac{50}{100}\right)\times\ \dfrac{5a}{4}$$ = $$\dfrac{15}{8}a$$

For B:

Value in 2020 = decreased by 25% from 2016 = $$\left(1\ -\ \dfrac{25}{100}\right)b$$ = $$\dfrac{3}{4}b$$

Value in 2024 = decreased by 25% from 2020 = $$\left(1\ -\ \dfrac{25}{100}\right)\times\ \dfrac{3b}{4}$$ = $$\dfrac{9}{16}b$$

For C:

Value in 2020 = decreased by 20% from 2016 = $$\left(1\ -\ \dfrac{20}{100}\right)c$$ = $$\dfrac{4}{5}c$$

Value in 2024 = increased by 40% from 2020 = $$\left(1\ +\ \dfrac{40}{100}\right)\times\ \dfrac{4c}{5}$$ = $$\dfrac{28}{25}c$$

For D:

Value in 2020 = increased by 100% from 2016 = $$\left(1\ +\ \dfrac{100}{100}\right)d$$ = $$2d$$

Value in 2024 = increased by 20% from 2020 = $$\left(1\ +\ \dfrac{20}{100}\right)\times\ 2d$$ = $$\dfrac{12}{5}d$$

For E:

Value in 2020 = increased by 25% from 2016 = $$\left(1\ +\ \dfrac{25}{100}\right)e$$ = $$\dfrac{5}{4}e$$

Value in 2024 = increased by 20% from 2020 = $$\left(1\ +\ \dfrac{20}{100}\right)\times\ \dfrac{5e}{4}$$ = $$\dfrac{3}{2}e$$

For F:

Value in 2020 = increased by 100% from 2016 = $$\left(1\ +\ \dfrac{100}{100}\right)f$$ = $$2f$$

Value in 2024 = decreased by 25% from 2020 = $$\left(1\ -\ \dfrac{25}{100}\right)\times\ 2f$$ = $$\dfrac{3}{2}f$$

We are given that the value of E is 90 in 2024. We can calculate the value of e as,

$$\dfrac{3e}{2}\ =\ 90$$

$$e\ =\ 60$$

The value of E in 2016 is 60, and its value in 2020 can be calculated as,

$$\dfrac{5e}{4}\ =\ \dfrac{5}{4}\times\ 60\ =\ 75$$

We are given that B, C, E, and A had ranks 1, 2, 3, and 4, respectively, in 2016, and F has the lowest rank(6) in all the years.

Putting the values in the table, we get,

CASE 1: b = 64

If b = 64, then the value of f becomes 4, as we know b - f = 60.

If f = 4, then the value of F in 2024 is 3f/2 = 12/2 = 6.

We already know the value of E in 2024 is 90.

In that case, the range of SI in 2024 is at least 90 - 6 = 84, but we are given that the range is 60 in 2024, which is not possible in this case.

So, we can eliminate this case.

CASE 2: b = 96

If b = 96, then the value of f becomes 36, as we know b - f = 60.

If f = 36, then the value of F in 2024 is 3f/2 = 108/2 = 54.

The value of B in 2024 = 9b/16 = (9 * 96)/16 = 54

We calculated the value of B and F to be 54 in 2024, which is not possible, as we are given that F had a lower SI than any other country in 2024, and we obtained its value to be the same as B.

So, we can eliminate this case.

CASE 3: b = 80

If b = 80, then the value of f becomes 20, as we know b - f = 60.

If f = 20, then the value of F in 2024 is 3f/2 = 60/2 = 30.

The value of B in 2024 = 9b/16 = (9 * 80)/16 = 45

We can see that no condition is violated in this case, as F is less than B in 2024, and the range in 2024 from the obtained values as of now is 90 - 30 = 60.

So, we can conclude that the value of b = 80 and f = 20.

Substituting the values in the table, we get,

We know that C is ranked 2nd in 2016, and we know the values of 1st and 3rd to be 80 and 60. We also know that SI of C has to be integers in 2016, 2020 and 2024. So, for the value of SI of C to be an integer in 2024, the value of c has to be a multiple of 25.

So, we know that c lies between 80 and 60, and is also a multiple of 25. The only multiple of 25 that lies between 60 and 80 is 75. So, the value of c has to be 75, and the value of C in 2024 can be calculated as,

C in 2024 = 28c/25 = (28 * 75)/25 = 84.

Hence, the correct answer is 84.

Question 41

What was the SI of B in 2024?

Answer & explanation

Correct answer: D

Let us assume the values of A, B, C, D, E and F to be a, b, c, d, e and f in 2016.

For A:

Value in 2020 = increased by 25% from 2016 = $$\left(1\ +\ \dfrac{25}{100}\right)a$$ = $$\dfrac{5}{4}a$$

Value in 2024 = increased by 50% from 2020 = $$\left(1\ +\ \dfrac{50}{100}\right)\times\ \dfrac{5a}{4}$$ = $$\dfrac{15}{8}a$$

For B:

Value in 2020 = decreased by 25% from 2016 = $$\left(1\ -\ \dfrac{25}{100}\right)b$$ = $$\dfrac{3}{4}b$$

Value in 2024 = decreased by 25% from 2020 = $$\left(1\ -\ \dfrac{25}{100}\right)\times\ \dfrac{3b}{4}$$ = $$\dfrac{9}{16}b$$

For C:

Value in 2020 = decreased by 20% from 2016 = $$\left(1\ -\ \dfrac{20}{100}\right)c$$ = $$\dfrac{4}{5}c$$

Value in 2024 = increased by 40% from 2020 = $$\left(1\ +\ \dfrac{40}{100}\right)\times\ \dfrac{4c}{5}$$ = $$\dfrac{28}{25}c$$

For D:

Value in 2020 = increased by 100% from 2016 = $$\left(1\ +\ \dfrac{100}{100}\right)d$$ = $$2d$$

Value in 2024 = increased by 20% from 2020 = $$\left(1\ +\ \dfrac{20}{100}\right)\times\ 2d$$ = $$\dfrac{12}{5}d$$

For E:

Value in 2020 = increased by 25% from 2016 = $$\left(1\ +\ \dfrac{25}{100}\right)e$$ = $$\dfrac{5}{4}e$$

Value in 2024 = increased by 20% from 2020 = $$\left(1\ +\ \dfrac{20}{100}\right)\times\ \dfrac{5e}{4}$$ = $$\dfrac{3}{2}e$$

For F:

Value in 2020 = increased by 100% from 2016 = $$\left(1\ +\ \dfrac{100}{100}\right)f$$ = $$2f$$

Value in 2024 = decreased by 25% from 2020 = $$\left(1\ -\ \dfrac{25}{100}\right)\times\ 2f$$ = $$\dfrac{3}{2}f$$

We are given that the value of E is 90 in 2024. We can calculate the value of e as,

$$\dfrac{3e}{2}\ =\ 90$$

$$e\ =\ 60$$

The value of E in 2016 is 60, and its value in 2020 can be calculated as,

$$\dfrac{5e}{4}\ =\ \dfrac{5}{4}\times\ 60\ =\ 75$$

We are given that B, C, E, and A had ranks 1, 2, 3, and 4, respectively, in 2016, and F has the lowest rank(6) in all the years.

Putting the values in the table, we get,

CASE 1: b = 64

If b = 64, then the value of f becomes 4, as we know b - f = 60.

If f = 4, then the value of F in 2024 is 3f/2 = 12/2 = 6.

We already know the value of E in 2024 is 90.

In that case, the range of SI in 2024 is at least 90 - 6 = 84, but we are given that the range is 60 in 2024, which is not possible in this case.

So, we can eliminate this case.

CASE 2: b = 96

If b = 96, then the value of f becomes 36, as we know b - f = 60.

If f = 36, then the value of F in 2024 is 3f/2 = 108/2 = 54.

The value of B in 2024 = 9b/16 = (9 * 96)/16 = 54

We calculated the value of B and F to be 54 in 2024, which is not possible, as we are given that F had a lower SI than any other country in 2024, and we obtained its value to be the same as B.

So, we can eliminate this case.

CASE 3: b = 80

If b = 80, then the value of f becomes 20, as we know b - f = 60.

If f = 20, then the value of F in 2024 is 3f/2 = 60/2 = 30.

The value of B in 2024 = 9b/16 = (9 * 80)/16 = 45

We can see that no condition is violated in this case, as F is less than B in 2024, and the range in 2024 from the obtained values as of now is 90 - 30 = 60.

So, we can conclude that the value of b = 80 and f = 20.

Substituting the values in the table, we get,

So, SI of B in 2024 is 45.

Hence, the correct answer is option D.

Passage for questions 42–46

The two most populous cities and the non-urban region (NUR) of each of three states, Whimshire, Fogglia, and Humbleset, are assigned Pollution Measures (PMs). These nine PMs are all distinct multiples of 10, ranging from 10 to 90. The six cities in increasing order of their PMs are: Blusterburg, Noodleton, Splutterville, Quackford, Mumpypore, Zingaloo. The Pollution Index (PI) of a state is a weighted average of the PMs of its NUR and cities, with a weight of 50% for the NUR, and 25% each for its two cities. There is only one pair of an NUR and a city (considering all cities and all NURs) where the PM of the NUR is greater than that of the city. That NUR and the city both belong to Humbleset. The PIs of all three states are distinct integers, with Humbleset and Fogglia having the highest and the lowest PI respectively.

Question 42

What is the PI of Whimshire?

Enter your answer:

Up to 2 decimal places. No negative marking for TITA.

Answer & explanation

Correct answer: 45

We are given that the PMs of the six cities, Blusterburg, Noodleton, Splutterville, Quackford, Mumpypore, and Zingaloo, are in increasing order.

We are also told there exists a single pair of NUR and city such that the PM of NUR is greater than the city and the NUR, city of the pair belongs to Humbleset. So, we can conclude that PMs of NURs of Whimshire and Fogglia are 10 and 20 in some order.

We also know that out of all the cities, Blusterburg has the lowest PM. So, the PM of Blusterberg has to be 30, and the PM of NUR of Humbleset has to be 40 for the above condition to be satisfied, and we can also conclude that Blusterberg is a city of Humbleset.

We can also conclude that the PMs of Noodleton, Splutterville, Quackford, Mumpypore, and Zingaloo are 50, 60, 70, 80, and 90, respectively.

We are given that in the PI calculation, cities have a contribution of 25% and NUR has a contribution of 50%. So, we can calculate the contribution of each city in the calculation of PI by dividing their PI value by 4, and the contribution of NUR can be calculated by dividing the PM value by 2.

We are given that all the PI values are unique integers, and Humbleset has the highest PI value, while Fogglia has the lowest.

For the PI value to be an integer, the cities with decimal contributions must be paired with another city with decimal contributions. Since all the decimals are 0.5, when two cities with 0.5 are added, the sum becomes 1, making the PI an integer.

The cities Splutterville and Mumpypoe must be from the same state; both have integral contributions towards the PI, and if they are added to a decimal, then the result will be a decimal, which should not be the case. These two cities have to belong to either the state of Whimshire or Fogglia, but their combined contribution towards PI is 15 + 20 = 35.

We already know that for the state of Humbleset, the PI is highest, and the two contributions towards the PI are 7.5 and 20, totalling 27.5.

There are three possibilities for the other city of Humbleset, which are Noodleton, Quackford and Zingaloo.

CASE 1: Noodleton

If the other city of Humbleset is Noodleton, then its PI would be 27.5 + 12.5 = 40.

We already obtained the sum of the PIs of one of the other 2 states to be 35 without including the PI of NUR. We know that the NUR contribution for that city would be either 5 or 10. If it is 5, then the PI of that state would be 40, and if it is 10, then the PI of that state would be 45.

However, we are told that the PI values are unique, and Humblest has the highest PI value; in either case, this condition is not satisfied.

So, we can eliminate the case of Noodleton being the second city of Humbleset.

CASE 2: Quackford

If the other city of Humbleset is Quackford, then its PI would be 27.5 + 17.5 = 45.

However, we are told that the PI values are unique, and Humblest has the highest PI value. So, we can eliminate the case of PI of NUR being 10 for the other state.

So, for Humbleset, the PI is 7.5 + 17.5 + 20 = 45.

For one of the other 2 states, the PI is 15 + 20 + 5 = 40

The PI of the other state can be calculated by adding all the leftover PIs, which is 12.5 + 22.5 + 10 = 45.

We can see that PI of Humbleton and one of the other two states is 45, which violates one of the conditions given.

So, we can eliminate the case of Quackford being the second city of Humbleset.

CASE 3: Zingaloo

If the other city of Humbleset is Zingaloo, then its PI would be 27.5 + 22.5 = 50.

If the PI of the other state is 40,

Then, for Humbleset, the PI is 7.5 + 22.5 + 20 = 50.

For one of the other 2 states, the PI is 15 + 20 + 5 = 40

The PI of the other state can be calculated by adding all the leftover PIs, which is 12.5 + 17.5 + 10 = 40.

We can see that PI of the other two states is 40, which violates one of the conditions given.

If the PI of the other state is 45,

Then, for Humbleset, the PI is 7.5 + 22.5 + 20 = 50.

For one of the other 2 states, the PI is 15 + 20 + 10 = 45

The PI of the other state can be calculated by adding all the leftover PIs, which is 12.5 + 17.5 + 5 = 35.

We can see that all the conditions are satisfied in this case, and we can conclude that Zingaloo is the second city of Humbleset, and also we know that PI is the least for Fogglia. So, the PI of Fogglia is 35, and the PI of Whimshire is 45.

We can now assign the cities and the NURs the states and the PM values uniquely based on our calculations. After assigning the table looks like,

We calculated the PI of Whimshire to be 45.

Hence, the correct answer is 45.

Question 43

What is the PI of Fogglia?

Enter your answer:

Up to 2 decimal places. No negative marking for TITA.

Answer & explanation

Correct answer: 35

We are given that the PMs of the six cities, Blusterburg, Noodleton, Splutterville, Quackford, Mumpypore, and Zingaloo, are in increasing order.

We can also conclude that the PMs of Noodleton, Splutterville, Quackford, Mumpypore, and Zingaloo are 50, 60, 70, 80, and 90, respectively.

We are given that all the PI values are unique integers, and Humbleset has the highest PI value, while Fogglia has the lowest.

We already know that for the state of Humbleset, the PI is highest, and the two contributions towards the PI are 7.5 and 20, totalling 27.5.

There are three possibilities for the other city of Humbleset, which are Noodleton, Quackford and Zingaloo.

CASE 1: Noodleton

If the other city of Humbleset is Noodleton, then its PI would be 27.5 + 12.5 = 40.

However, we are told that the PI values are unique, and Humblest has the highest PI value; in either case, this condition is not satisfied.

So, we can eliminate the case of Noodleton being the second city of Humbleset.

CASE 2: Quackford

If the other city of Humbleset is Quackford, then its PI would be 27.5 + 17.5 = 45.

However, we are told that the PI values are unique, and Humblest has the highest PI value. So, we can eliminate the case of PI of NUR being 10 for the other state.

So, for Humbleset, the PI is 7.5 + 17.5 + 20 = 45.

For one of the other 2 states, the PI is 15 + 20 + 5 = 40

The PI of the other state can be calculated by adding all the leftover PIs, which is 12.5 + 22.5 + 10 = 45.

We can see that PI of Humbleton and one of the other two states is 45, which violates one of the conditions given.

So, we can eliminate the case of Quackford being the second city of Humbleset.

CASE 3: Zingaloo

If the other city of Humbleset is Zingaloo, then its PI would be 27.5 + 22.5 = 50.

If the PI of the other state is 40,

Then, for Humbleset, the PI is 7.5 + 22.5 + 20 = 50.

For one of the other 2 states, the PI is 15 + 20 + 5 = 40

The PI of the other state can be calculated by adding all the leftover PIs, which is 12.5 + 17.5 + 10 = 40.

We can see that PI of the other two states is 40, which violates one of the conditions given.

If the PI of the other state is 45,

Then, for Humbleset, the PI is 7.5 + 22.5 + 20 = 50.

For one of the other 2 states, the PI is 15 + 20 + 10 = 45

The PI of the other state can be calculated by adding all the leftover PIs, which is 12.5 + 17.5 + 5 = 35.

We can now assign the cities and the NURs the states and the PM values uniquely based on our calculations. After assigning the table looks like,

We calculated the PI of Fogglia to be 35.

Hence, the correct answer is 35.

Question 44

What is the PI of Humbleset?

Enter your answer:

Up to 2 decimal places. No negative marking for TITA.

Answer & explanation

Correct answer: 50

We are given that the PMs of the six cities, Blusterburg, Noodleton, Splutterville, Quackford, Mumpypore, and Zingaloo, are in increasing order.

We can also conclude that the PMs of Noodleton, Splutterville, Quackford, Mumpypore, and Zingaloo are 50, 60, 70, 80, and 90, respectively.

We are given that all the PI values are unique integers, and Humbleset has the highest PI value, while Fogglia has the lowest.

We already know that for the state of Humbleset, the PI is highest, and the two contributions towards the PI are 7.5 and 20, totalling 27.5.

There are three possibilities for the other city of Humbleset, which are Noodleton, Quackford and Zingaloo.

CASE 1: Noodleton

If the other city of Humbleset is Noodleton, then its PI would be 27.5 + 12.5 = 40.

However, we are told that the PI values are unique, and Humblest has the highest PI value; in either case, this condition is not satisfied.

So, we can eliminate the case of Noodleton being the second city of Humbleset.

CASE 2: Quackford

If the other city of Humbleset is Quackford, then its PI would be 27.5 + 17.5 = 45.

However, we are told that the PI values are unique, and Humblest has the highest PI value. So, we can eliminate the case of PI of NUR being 10 for the other state.

So, for Humbleset, the PI is 7.5 + 17.5 + 20 = 45.

For one of the other 2 states, the PI is 15 + 20 + 5 = 40

The PI of the other state can be calculated by adding all the leftover PIs, which is 12.5 + 22.5 + 10 = 45.

We can see that PI of Humbleton and one of the other two states is 45, which violates one of the conditions given.

So, we can eliminate the case of Quackford being the second city of Humbleset.

CASE 3: Zingaloo

If the other city of Humbleset is Zingaloo, then its PI would be 27.5 + 22.5 = 50.

If the PI of the other state is 40,

Then, for Humbleset, the PI is 7.5 + 22.5 + 20 = 50.

For one of the other 2 states, the PI is 15 + 20 + 5 = 40

The PI of the other state can be calculated by adding all the leftover PIs, which is 12.5 + 17.5 + 10 = 40.

We can see that PI of the other two states is 40, which violates one of the conditions given.

If the PI of the other state is 45,

Then, for Humbleset, the PI is 7.5 + 22.5 + 20 = 50.

For one of the other 2 states, the PI is 15 + 20 + 10 = 45

The PI of the other state can be calculated by adding all the leftover PIs, which is 12.5 + 17.5 + 5 = 35.

We can now assign the cities and the NURs, the states, and the PM values uniquely based on our calculations. After assigning the table looks like,

We calculated the PI of Humbleset to be 50.

Hence, the correct answer is 50.

Question 45

Which pair of cities definitely belong to the same state?

Answer & explanation

Correct answer: D

We are given that the PMs of the six cities, Blusterburg, Noodleton, Splutterville, Quackford, Mumpypore, and Zingaloo, are in increasing order.

We can also conclude that the PMs of Noodleton, Splutterville, Quackford, Mumpypore, and Zingaloo are 50, 60, 70, 80, and 90, respectively.

We are given that all the PI values are unique integers, and Humbleset has the highest PI value, while Fogglia has the lowest.

We already know that for the state of Humbleset, the PI is highest, and the two contributions towards the PI are 7.5 and 20, totalling 27.5.

There are three possibilities for the other city of Humbleset, which are Noodleton, Quackford and Zingaloo.

CASE 1: Noodleton

If the other city of Humbleset is Noodleton, then its PI would be 27.5 + 12.5 = 40.

However, we are told that the PI values are unique, and Humblest has the highest PI value; in either case, this condition is not satisfied.

So, we can eliminate the case of Noodleton being the second city of Humbleset.

CASE 2: Quackford

If the other city of Humbleset is Quackford, then its PI would be 27.5 + 17.5 = 45.

However, we are told that the PI values are unique, and Humblest has the highest PI value. So, we can eliminate the case of PI of NUR being 10 for the other state.

So, for Humbleset, the PI is 7.5 + 17.5 + 20 = 45.

For one of the other 2 states, the PI is 15 + 20 + 5 = 40

The PI of the other state can be calculated by adding all the leftover PIs, which is 12.5 + 22.5 + 10 = 45.

We can see that PI of Humbleton and one of the other two states is 45, which violates one of the conditions given.

So, we can eliminate the case of Quackford being the second city of Humbleset.

CASE 3: Zingaloo

If the other city of Humbleset is Zingaloo, then its PI would be 27.5 + 22.5 = 50.

If the PI of the other state is 40,

Then, for Humbleset, the PI is 7.5 + 22.5 + 20 = 50.

For one of the other 2 states, the PI is 15 + 20 + 5 = 40

The PI of the other state can be calculated by adding all the leftover PIs, which is 12.5 + 17.5 + 10 = 40.

We can see that PI of the other two states is 40, which violates one of the conditions given.

If the PI of the other state is 45,

Then, for Humbleset, the PI is 7.5 + 22.5 + 20 = 50.

For one of the other 2 states, the PI is 15 + 20 + 10 = 45

The PI of the other state can be calculated by adding all the leftover PIs, which is 12.5 + 17.5 + 5 = 35.

We can now assign the cities and the NURs, the states, and the PM values uniquely based on our calculations. After assigning the table looks like,

We can see that Noodleton and Quackford belong to the same state among the options.

Hence, the correct answer is option C.

Question 46

For how many of the cities and NURs is it possible to identify their PM and the state they belong to?

Enter your answer:

Up to 2 decimal places. No negative marking for TITA.

Answer & explanation

Correct answer: 9

We are given that the PMs of the six cities, Blusterburg, Noodleton, Splutterville, Quackford, Mumpypore, and Zingaloo, are in increasing order.

We can also conclude that the PMs of Noodleton, Splutterville, Quackford, Mumpypore, and Zingaloo are 50, 60, 70, 80, and 90, respectively.

We are given that all the PI values are unique integers, and Humbleset has the highest PI value, while Fogglia has the lowest.

We already know that for the state of Humbleset, the PI is highest, and the two contributions towards the PI are 7.5 and 20, totalling 27.5.

There are three possibilities for the other city of Humbleset, which are Noodleton, Quackford and Zingaloo.

CASE 1: Noodleton

If the other city of Humbleset is Noodleton, then its PI would be 27.5 + 12.5 = 40.

However, we are told that the PI values are unique, and Humblest has the highest PI value; in either case, this condition is not satisfied.

So, we can eliminate the case of Noodleton being the second city of Humbleset.

CASE 2: Quackford

If the other city of Humbleset is Quackford, then its PI would be 27.5 + 17.5 = 45.

However, we are told that the PI values are unique, and Humblest has the highest PI value. So, we can eliminate the case of PI of NUR being 10 for the other state.

So, for Humbleset, the PI is 7.5 + 17.5 + 20 = 45.

For one of the other 2 states, the PI is 15 + 20 + 5 = 40

The PI of the other state can be calculated by adding all the leftover PIs, which is 12.5 + 22.5 + 10 = 45.

We can see that PI of Humbleton and one of the other two states is 45, which violates one of the conditions given.

So, we can eliminate the case of Quackford being the second city of Humbleset.

CASE 3: Zingaloo

If the other city of Humbleset is Zingaloo, then its PI would be 27.5 + 22.5 = 50.

If the PI of the other state is 40,

Then, for Humbleset, the PI is 7.5 + 22.5 + 20 = 50.

For one of the other 2 states, the PI is 15 + 20 + 5 = 40

The PI of the other state can be calculated by adding all the leftover PIs, which is 12.5 + 17.5 + 10 = 40.

We can see that PI of the other two states is 40, which violates one of the conditions given.

If the PI of the other state is 45,

Then, for Humbleset, the PI is 7.5 + 22.5 + 20 = 50.

For one of the other 2 states, the PI is 15 + 20 + 10 = 45

The PI of the other state can be calculated by adding all the leftover PIs, which is 12.5 + 17.5 + 5 = 35.

We can now assign the cities and the NURs, the states, and the PM values uniquely based on our calculations. After assigning the table looks like,

We can see that for all 6 cities and 3 NURs, the PMs and the city to which they belong can be identified uniquely. So, in total, we can identify the values uniquely for all 9 of them.

Hence, the correct answer is 9.

Two tangents drawn from a point P and a circle with center O at point Q and R. Point A and B lie on PQ and PR, repectively, such that AB is also a tangent to the same circle. If $$\angle AOB=50^{0}$$, then $$\angle APB$$, in degrees equals

Enter your answer:

Up to 2 decimal places. No negative marking for TITA.

Answer & explanation

Correct answer: 80

We can draw the following diagram as per the given question :

Now, taking $$\triangle AOT$$ and $$\triangle AOQ$$

AO is the common side

AT=AQ (Tangents drawn from an external point are equal in length)

OT=OQ (radius of circle)

So, by S.S.S., $$\triangle\ AOT$$ is congruent to $$\triangle\ AOQ$$

So, by C.P.C.T., $$\angle\ AOT=\angle\ AOQ$$

Similarly, for $$\triangle\ BOT\ $$ and $$\triangle\ BOR\ $$, we can do the same thing,

So, we can say $$\triangle\ BOT$$ and $$\triangle\ BOR$$ will also be congruent

Now, $$\angle\ QOR=\angle\ AOQ+\angle\ AOB+\angle\ ROB=\angle\ AOT+\angle\ AOB+\angle\ BOT=2\angle\ AOB=2\times\ 50^{\circ\ }=100^{\circ\ }$$

Now, in quadrilateral OQPR,

$$\angle\ OQP=\angle\ ORP=90^{\circ\ }$$ (Tangents drawn to a point is perpendicular to the radius drawn at the same point)

Now, sum of angles of a quadrilateral is $$360^{\circ\ }$$

So, $$\angle\ APB+\angle\ QOR=360^{\circ\ }-\left(90^{\circ\ }+90^{\circ\ }\right)=180^{\circ\ }$$

So, $$\angle\ APB=180^{\circ\ }-100^{\circ\ }=80^{\circ\ }$$

Question 60

The number of divisors of $$(2^{6}\times 3^{5}\times 5^{3}\times 7^{2})$$, which are of the form $$(3r+1)$$, where r is a non-negative integer, is

Answer & explanation

Correct answer: D

The divisors of the given number will have the form $$2^a*3^b*5^c*7^d$$ with $$0\le a\le6,\ 0\le b\le5,\ 0\le c\le3,\ 0\le d\le2$$ Because the divisors should be in the form 3r+1, it cannot be divisible by 3, so (b=0). Reduce modulo 3: $$2 \text{mod} 3 = 2, 5 \text{mod} 3 = 2, 7 \text{mod} 3 = 1$$ Hence, $$2^a5^c7^d\equiv 2^{a+c}\cdot1^d \equiv 2^{a+c}\pmod3$$ 2^k will be in the form 3r+1 only when K is even. So, we need a+c to be even $$a\in{0,\dots,6}$$ has 4 even, 3 odd values $$c\in{0,\dots,3}$$ has 2 even, 2 odd Number of (a,c) with (a+c) even is $$4\cdot2 + 3\cdot2 = 8+6=14$$ For each such pair, there are 3 choices for d = 0,1,2. Thus, total divisors in the form (3r+1) equals $$14\times3=42$$.

Question 61

Let ABCDEF be a regular hexagon and P and Q be the midpoints of AB and CD, respectively. Then, the ratio of the areas of trapezium PBCQ and hexagon ABCDEF is

Answer & explanation

Correct answer: B

Let the side of the hexagon be a.

The area of the whole hexagon is going to be $$\frac{3\sqrt{3}a^2}{2}$$

We know that the longest diagonal of a hexagon is 2 times the side of the hexagon.

PQ = $$\frac{\left(BC+AD\right)}{2}= \frac{\left(a+2a\right)}{2} =1.5a$$

The area of the trapezium = Average of parllel sides * Height = $$\frac{\left(BC+PQ\right)}{2}\cdot h=\frac{5}{4}a\cdot h$$

The distance between BC and AD will be $$\sqrt{3}a$$, that is, the length of diagonal BF.

The distance between BC and AD will be $$\frac{\sqrt{3}a}{2}$$

And the height of the trapezium BCQP will be $$\frac{\sqrt{3}a}{4}$$

Area of trapezium = $$\frac{5}{4}a\cdot \frac{\sqrt{3}a}{4}$$ = $$\frac{5\sqrt{3}a^2}{16}$$

Area of hexagon ABCDEF = $$\frac{3\sqrt{3}}{2}a^2$$

We have to find the ratio = $$\frac{5\sqrt{3}a^2}{16}$$ : $$\frac{3\sqrt{3}}{2}a^2$$ = 5:24

Question 62

If a,b,c and d are integers such that their sum is 46, then the minimum possible value of $$(a-b)^{2}+(a-c)^{2}+(a-d)^{2}$$ is

Enter your answer:

Up to 2 decimal places. No negative marking for TITA.

Answer & explanation

Correct answer: 2

Given expression: $$(a-b)^{2}+(a-c)^{2}+(a-d)^{2}$$ The given expression has just the sum of squares of the terms. So, the minimum value is either zero or positive. If we can make all the values equal, we can get zero. But since all the values are integers and the sum 46 is not divisible by 4, we can't make everything equal. So, the nearest four values are 12, 11, 11, 12. With this, the minimum value is $$(12-11)^2+(12-11)^2+(12-12)^2 = 2$$

Question 63

The ratio of expenditures of Lakshmi and Meenakshi is 2 : 3, and the ratio of income of Lakshmi to expenditure of Meenakshi is 6 : 7. If excess of income over expenditure is saved by Lakshmi and Meenakshi, and the ratio of their savings is 4 : 9, then the ratio of their incomes is

Answer & explanation

Correct answer: A

Let Lakshmi's income = A and expenditure = B. Let Meenakshi's income = C and expenditure = D. From B:D = 2:3, take B = 2k and D = 3k. From A:D = 6:7, $$A = \tfrac{6}{7} D = \tfrac{6}{7}\cdot 3k = \tfrac{18}{7}k$$. Lakshmi's saving = $$A - B = \tfrac{18}{7}k - 2k = \tfrac{4}{7}k$$. Let Meenakshi's saving $$= C - D = C - 3k$$. Given the savings ratio $$\frac{\frac{4k}{7}}{C-3k} = \tfrac{4}{9}$$. Solve: $$\dfrac{4}{7}k\cdot\dfrac{9}{4} = C - 3k \Rightarrow \dfrac{9}{7}k = C - 3k \Rightarrow C = \dfrac{30}{7}k$$ So incomes ratio $$A:C = \dfrac{18}{7}k : \dfrac{30}{7}k = 18:30 = 3:5.$$

Question 64

Let $$a_{n}$$ be the $$n^{th}$$ term of a decreasing infinite geometric progression. If $$a_{1}+a_{2}+a_{3}=52$$ and $$a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}=624$$, then the sum of this geometric progression is

Answer & explanation

Correct answer: B

Let the first term be a and the common ratio be r. Given $$a+a r + a r^2 = 52 \quad\text{and}\quad a^2 r + a^2 r^2 + a^2 r^3 = 624.$$ From the first equation $$a(1+r+r^2)=52$$, so $$a=\dfrac{52}{1+r+r^2}$$. Substitute into the second: $$\frac{52^2}{(1+r+r^2)^2}*(r+r^2+r^3)=624$$ $$\frac{r}{(1+r+r^2)}=\frac{3}{13}$$ $$3r^2-10r+3=0$$ $$(3r-1)(r-3)=0$$ Simplify to get an equation in (r); its real solutions are $$r=\tfrac{1}{3}$$ and $$r=3$$. For an infinite geometric progression, we must have (|r|<1), so $$r=\dfrac{1}{3}$$. Now $$a=\frac{52}{1+\tfrac{1}{3}+\tfrac{1}{9}}=\frac{52}{13/9}=36$$ The sum of the infinite progression is $$ S=\frac{a}{1-r}=\frac{36}{1-\tfrac{1}{3}}=\frac{36}{2/3}=54.$$

Question 65

A mixture of coffee and cocoa, 16% of which is coffee, costs Rs 240 per kg. Another mixture of coffee and cocoa, of which 36% is coffee, costs Rs 320 per kg. If a new mixture of coffee and cocoa costs Rs 376 per kg, then the quantity, in kg, of coffee in 10 kg of this new mixture is

Answer & explanation

Correct answer: A

Let coffee price = C Rs/kg and cocoa price = K Rs/kg. From the two given mixtures: $$0.16C + 0.84K = 240$$ $$0.36C + 0.64K = 320$$ Multiply both equations by 100 to remove decimals: $$16C + 84K = 24000$$ $$36C + 64K = 32000$$ Subtract the first from the second: $$(36C+64K)-(16C+84K)=32000-24000$$ $$20C - 20K = 8000 \implies C - K = 400$$ Put $$C=K+400$$ into $$16C+84K=24000$$: $$16(K+400)+84K=24000 $$ $$16K+6400+84K=24000 $$ $$100K = 17600 \implies K = 176$$ So $$C = 176 + 400 = 576$$ (Rs/kg). For the new mixture priced at Rs 376/kg, let the coffee fraction be p. Then $$ p\cdot 576 + (1-p)\cdot 176 = 376$$ Upon solving $$p = \tfrac{1}{2}$$ Thus, coffee is (50%) of the new mixture. In 10 kg of this mixture, coffee = $$10\times 0.5 = 5\text{kg}$$

Question 66

In $$\triangle ABC$$, points D and E are on the sides BC and AC, respectively. BE and AD intersect at point T such that AD:AT=4:3, and BE:BT=5:4. Point F lies on AC such that DF is parallel to BE. Then, BD:CD is

Answer & explanation

Correct answer: B

In the question, it is given $$AD:AT=4:3$$

So,$$TD:AT=1:3$$

Let, $$TD=x,\ AT=3x$$

Also, $$BE:BT=5:4$$

So, $$BT:TE=4:1$$

Let, $$BT=4y,\ TE=y$$

Given, DF is parallel to BE

So, DF is parallel to TE

So, $$\triangle\ ATE~\triangle\ ADF$$ (by A.A., as $$\angle\ ATE=\angle\ ADF$$ and $$\angle\ AET=\angle\ AFD$$)

So, $$\dfrac{AT}{AD}=\dfrac{TE}{DF}$$

So, $$\dfrac{3x}{x+3x}=\dfrac{y}{DF}$$

So, $$DF=\dfrac{4y}{3}$$

Also, since DF is parallel to BE,

So, $$\triangle\ CFD~\triangle\ CBE$$ (by A.A., as $$\angle\ CFD=\angle\ CEB$$ and $$\angle\ CDF=\angle\ CBE$$)

So, $$\dfrac{CD}{BC}=\dfrac{DF}{BE}$$

So, $$\dfrac{CD}{BC}=\dfrac{\dfrac{4y}{3}}{5y}=\dfrac{4y}{15y}=\dfrac{4}{15}$$

So, $$\dfrac{CD}{BD}=\dfrac{4}{11}$$

So, $$BD:CD=11:4$$

Question 67

Ankita is twice as efficient as Bipin, while Bipin is twice as efficient as Chandan. All three of them start together on a job, and Bipin leaves the job after 20 days. If the job got completed in 60 days, the number of days needed by Chandan to complete the job alone, is

Enter your answer:

Up to 2 decimal places. No negative marking for TITA.

Answer & explanation

Correct answer: 340

Let Chandan’s one-day work = x. Then Bipin = 2x and Ankita = 4x. All three work together for the first 20 days. Their combined daily work = $$x + 2x + 4x = 7x$$. Work done in 20 days = $$20\times7x = 140x$$. Bipin leaves after 20 days; Ankita and Chandan continue for the remaining (60-20=40) days. Their combined daily work = $$4x + x = 5x$$. Work done in 40 days = $$40\times5x = 200x$$. Total work = $$140x+200x=340x$$. We know that Chandan does x work in a day. So, the number of days it takes him to finish the work is 340. Answer: 340 days.

Question 68

A certain amount of money was divided among Pinu, Meena, Rinu and Seema. Pinu received 20% of the total amount and Meena received 40% of the remaining amount. If Seema received 20% less than Pinu, the ratio of the amounts received by Pinu and Rinu is

Answer & explanation

Correct answer: C

Let the total amount be 100 units (taking 100 makes percentage calculations easy). Pinu receives 20% of the total, so Pinu gets 20 units. The amount left after giving Pinu his share is 100 - 20 = 80 units. Meena receives 40% of this remaining amount, so Meena gets 40% of 80 = 32 units. Now, the amount left for Rinu and Seema together is 80 - 32 = 48 units. Seema receives 20% less than Pinu. Since Pinu gets 20 units, Seema gets 20 - 20% of 20 = 20 - 4 = 16 units. Out of the 48 units left for Rinu and Seema, Seema’s share is 16 units, so Rinu gets 48 - 16 = 32 units. Therefore, Pinu receives 20 units, and Rinu gets 32 units. The required ratio of the amounts received by Pinu : Rinu = 20 : 32 = 5 : 8.

CAT 2025 — Slot 2

VARC — 24 questions

Question 1

Question 2

Question 3

Passage for questions 4–7

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Passage for questions 10–13

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Passage for questions 16–19

Question 16

Question 17

Question 18

Question 19

Passage for questions 20–23

Question 20

Question 21

Question 22

Question 23

Question 24

DILR — 22 questions

Passage for questions 25–29

Question 25

Question 26

Question 27

Question 28

Question 29

Passage for questions 30–33

Question 30

Question 31

Question 32

Question 33

Passage for questions 34–37

Question 34

Question 35

Question 36

Question 37

Passage for questions 38–41

Question 38

Question 39

Question 40

Question 41

Passage for questions 42–46

Question 42

Question 43

Question 44

Question 45

Question 46

QA — 22 questions

Question 47

Question 48

Question 49

Question 50

Question 51

Question 52

Question 53

Question 54

Question 55

Question 56

Question 57

Question 58

Question 59

Question 60

Question 61

Question 62

Question 63

Question 64

Question 65

Question 66

Question 67