CAT 2023 — Slot 1
Full paper, free to view. Pick a section below, click any question to see the answer and explanation.
VARC topics covered: Culture & Identity RC, History & Civilization RC, Nature And Environment RC, Odd One Out, Para Completion, Para Summary, Parajumble, Science And Technology RC
For early postcolonial literature, the world of the novel was often the nation. Postcolonial novels were usually [concerned with] national questions. Sometimes the whole story of the novel was taken as an allegory of the nation, whether India or Tanzania. This was important for supporting anti-colonial nationalism, but could also be limiting - land-focused and inward-looking.
My new book "Writing Ocean Worlds" explores another kind of world of the novel: not the village or nation, but the Indian Ocean world. The book describes a set of novels in which the Indian Ocean is at the centre of the story. It focuses on the novelists Amitav Ghosh, Abdulrazak Gurnah, Lindsey Collen and Joseph Conrad [who have] centred the Indian Ocean world in the majority of their novels. . . . Their work reveals a world that is outward-looking - full of movement, border-crossing and south-south interconnection. They are all very different - from colonially inclined (Conrad) to radically anti-capitalist (Collen), but together draw on and shape a wider sense of Indian Ocean space through themes, images, metaphors and language. This has the effect of remapping the world in the reader's mind, as centred in the interconnected global south. . . .
The Indian Ocean world is a term used to describe the very long-lasting connections among the coasts of East Africa, the Arab coasts, and South and East Asia. These connections were made possible by the geography of the Indian Ocean. For much of history, travel by sea was much easier than by land, which meant that port cities very far apart were often more easily connected to each other than to much closer inland cities. Historical and archaeological evidence suggests that what we now call globalisation first appeared in the Indian Ocean. This is the interconnected oceanic world referenced and produced by the novels in my book. . . .
For their part Ghosh, Gurnah, Collen and even Conrad reference a different set of histories and geographies than the ones most commonly found in fiction in English. Those [commonly found ones] are mostly centred in Europe or the US, assume a background of Christianity and whiteness, and mention places like Paris and New York. The novels in [my] book highlight instead a largely Islamic space, feature characters of colour and centralise the ports of Malindi, Mombasa, Aden, Java and Bombay. . . . It is a densely imagined, richly sensory image of a southern cosmopolitan culture which provides for an enlarged sense of place in the world.
This remapping is particularly powerful for the representation of Africa. In the fiction, sailors and travellers are not all European. . . . African, as well as Indian and Arab characters, are traders, nakhodas (dhow ship captains), runaways, villains, missionaries and activists. This does not mean that Indian Ocean Africa is romanticised. Migration is often a matter of force; travel is portrayed as abandonment rather than adventure, freedoms are kept from women and slavery is rife. What it does mean is that the African part of the Indian Ocean world plays an active role in its long, rich history and therefore in that of the wider world.
Which one of the following statements is not true about migration in the Indian Ocean world?
The Indian Ocean world's migration networks connected the global north with the global south.
Geographical location rather than geographical proximity determined the choice of destination for migrants.
The Indian Ocean world's migration networks were shaped by religious and commercial histories of the region.
Migration in the Indian Ocean world was an ambivalent experience.
The Indian Ocean world's migration networks connected the global north with the global south.
The passage describes the Indian Ocean world as connecting the coasts of East Africa, the Arab coasts, and South and East Asia, and describes it as a south-south interconnection. The passage never mentions the Indian Ocean world's migration networks connecting the global north with the global south. The connections described are entirely within and between southern regions.
On the basis of the nature of the relationship between the items in each pair below, choose the odd pair out:
Indian Ocean novels : Outward-looking
Postcolonial novels : Border-crossing
Indian Ocean world : Slavery
Postcolonial novels : Anti-colonial nationalism
Postcolonial novels : Border-crossing
The passage explicitly associates Indian Ocean novels with being outward-looking, border-crossing and south-south interconnection. Postcolonial novels are described as land-focused and inward-looking, supporting anti-colonial nationalism. Border-crossing is specifically a quality of Indian Ocean novels, not postcolonial novels, making the pair postcolonial novels and border-crossing a mismatch.
All of the following statements, if true, would weaken the passage's claim about the relationship between mainstream English-language fiction and Indian Ocean novels EXCEPT:
the depiction of Africa in most Indian Ocean novels is driven by a postcolonial nostalgia for an idyllic past
the depiction of Africa in most Indian Ocean novels is driven by an Orientalist imagination of its cultural crudeness.
very few mainstream English-language novels have historically been set in American and European metropolitan centres.
most mainstream English-language novels have historically privileged the Christian, white, male experience of travel and adventure.
most mainstream English-language novels have historically privileged the Christian, white, male experience of travel and adventure.
The passage's claim about mainstream English-language fiction is that it centres Europe and the US, assumes a background of Christianity and whiteness, and features places like Paris and New York. Option D, stating that most mainstream English-language novels have historically privileged the Christian, white, male experience, actually reinforces the passage's claim rather than weakening it.
All of the following claims contribute to the "remapping" discussed by the passage, EXCEPT:
Indian Ocean novels have gone beyond the specifics of national concerns to explore rich regional pasts.
the world of early international trade and commerce was not the sole domain of white Europeans.
cosmopolitanism originated in the West and travelled to the East through globalisation.
the global south, as opposed to the global north, was the first centre of globalisation.
cosmopolitanism originated in the West and travelled to the East through globalisation.
The passage's remapping involves showing that the Indian Ocean world was the first centre of globalisation, that it is centred in the global south rather than the global north, and that non-European, non-white characters played active roles in world history and trade. Claiming that cosmopolitanism originated in the West and spread east through globalisation directly contradicts the passage's argument that the Indian Ocean world, which is in the global south, was the original centre of globalisation and cosmopolitan culture.
Many human phenomena and characteristics - such as behaviors, beliefs, economies, genes, incomes, life expectancies, and other things - are influenced both by geographic factors and by non-geographic factors. Geographic factors mean physical and biological factors tied to geographic location, including climate, the distributions of wild plant and animal species, soils, and topography. Non-geographic factors include those factors subsumed under the term culture, other factors subsumed under the term history, and decisions by individual people. . . .
[T]he differences between the current economies of North and South Korea . . . cannot be attributed to the modest environmental differences between [them] . . . They are instead due entirely to the different [government] policies . . . At the opposite extreme, the Inuit and other traditional peoples living north of the Arctic Circle developed warm fur clothes but no agriculture, while equatorial lowland peoples around the world never developed warm fur clothes but often did develop agriculture. The explanation is straightforwardly geographic, rather than a cultural or historical quirk unrelated to geography. . . . Aboriginal Australia remained the sole continent occupied only by hunter/gatherers and with no indigenous farming or herding . . . [Here the] explanation is biogeographic: the Australian continent has no domesticable native animal species and few domesticable native plant species. Instead, the crops and domestic animals that now make Australia a food and wool exporter are all non-native (mainly Eurasian) species such as sheep, wheat, and grapes, brought to Australia by overseas colonists.
Today, no scholar would be silly enough to deny that culture, history, and individual choices play a big role in many human phenomena. Scholars don't react to cultural, historical, and individual-agent explanations by denouncing "cultural determinism," "historical determinism," or "individual determinism," and then thinking no further. But many scholars do react to any explanation invoking some geographic role, by denouncing "geographic determinism" . . . Several reasons may underlie this widespread but nonsensical view. One reason is that some geographic explanations advanced a century ago were racist, thereby causing all geographic explanations to become tainted by racist associations in the minds of many scholars other than geographers. But many genetic, historical, psychological, and anthropological explanations advanced a century ago were also racist, yet the validity of newer non-racist genetic etc. explanations is widely accepted today.
Another reason for reflex rejection of geographic explanations is that historians have a tradition, in their discipline, of stressing the role of contingency (a favorite word among historians) based on individual decisions and chance. Often that view is warranted . . . But often, too, that view is unwarranted. The development of warm fur clothes among the Inuit living north of the Arctic Circle was not because one influential Inuit leader persuaded other Inuit in 1783 to adopt warm fur clothes, for no good environmental reason.
A third reason is that geographic explanations usually depend on detailed technical facts of geography and other fields of scholarship . . . Most historians and economists don't acquire that detailed knowledge as part of the professional training.
All of the following are advanced by the author as reasons why non-geographers disregard geographic influences on human phenomena EXCEPT their:
lingering impressions of past geographic analyses that were politically offensive.
belief in the central role of humans, unrelated to physical surroundings, in influencing phenomena.
disciplinary training which typically does not include technical knowledge of geography.
dismissal of explanations that involve geographical causes for human behaviour.
dismissal of explanations that involve geographical causes for human behaviour.
The passage gives three reasons why non-geographers dismiss geographic explanations, the racist associations of past geographic analyses, the tradition of privileging human contingency and individual decisions, and the lack of technical geographic knowledge in most academic training. The author does not list dismissal of geographical explanations as a reason for that same dismissal, since this would be circular and is simply a description of the behaviour being explained rather than a reason for it.
The author criticises scholars who are not geographers for all of the following reasons EXCEPT:
their rejection of the role of biogeographic factors in social and cultural phenomena.
their outdated interpretations of past cultural and historical phenomena.
the importance they place on the role of individual decisions when studying human phenomena
their labelling of geographic explanations as deterministic.
their outdated interpretations of past cultural and historical phenomena.
The passage criticises non-geographer scholars for rejecting geographic explanations as deterministic, for privileging individual decisions and contingency over geographic factors, and for lacking technical geographic knowledge. The passage does not criticise them for making outdated interpretations of past cultural and historical phenomena, it focuses entirely on their methodological bias against geographic explanation.
All of the following can be inferred from the passage EXCEPT:
individual dictat and contingency were not the causal factors for the use of fur clothing in some very cold climates.
agricultural practices changed drastically in the Australian continent after it was colonised.
while most human phenomena result from culture and individual choice, some have bio-geographic origins.
several academic studies of human phenomena in the past involved racist interpretations.
while most human phenomena result from culture and individual choice, some have bio-geographic origins.
The passage does not claim that most human phenomena result from culture and individual choice while only some have biogeographic origins. The passage argues that both geographic and non-geographic factors influence human phenomena, with some cases explained by geography and others by culture or individual decisions, without claiming a hierarchy of frequency between them.
The examples of the Inuit and Aboriginal Australians are offered in the passage to show:
human resourcefulness across cultures in adapting to their surroundings.
how physical circumstances can dictate human behaviour and cultures.
that despite geographical isolation, traditional societies were self-sufficient and adaptive.
how environmental factors lead to comparatively divergent paths in livelihoods and development.
human resourcefulness across cultures in adapting to their surroundings.
The Inuit example shows that warm fur clothing developed in the Arctic because of the cold environment, not because of any cultural or individual decision. The Australian example shows that hunter-gatherer lifestyles persisted because the continent lacked domesticable native species, not because of cultural choice. Both examples are used to demonstrate that physical and environmental circumstances can directly determine human behaviour and cultural practices.
RESIDENTS of Lozère, a hilly department in southern France, recite complaints familiar to many rural corners of Europe. In remote hamlets and villages, with names such as Le Bacon and Le Bacon Vieux, mayors grumble about a lack of local schools, jobs, or phone and internet connections. Farmers of grazing animals add another concern: the return of wolves. Eradicated from France last century, the predators are gradually creeping back to more forests and hillsides. "The wolf must be taken in hand," said an aspiring parliamentarian, Francis Palombi, when pressed by voters in an election campaign early this summer. Tourists enjoy visiting a wolf park in Lozère, but farmers fret over their livestock and their livelihoods. . . .
As early as the ninth century, the royal office of the Luparii—wolf-catchers—was created in France to tackle the predators. Those official hunters (and others) completed their job in the 1930s, when the last wolf disappeared from the mainland. Active hunting and improved technology such as rifles in the 19th century, plus the use of poison such as strychnine later on, caused the population collapse. But in the early 1990s the animals reappeared. They crossed the Alps from Italy, upsetting sheep farmers on the French side of the border. Wolves have since spread to areas such as Lozère, delighting environmentalists, who see the predators' presence as a sign of wider ecological health. Farmers, who say the wolves cause the deaths of thousands of sheep and other grazing animals, are less cheerful. They grumble that green activists and politically correct urban types have allowed the return of an old enemy.
Various factors explain the changes of the past few decades. Rural depopulation is part of the story. In Lozère, for example, farming and a once-flourishing mining industry supported a population of over 140,000 residents in the mid-19th century. Today the department has fewer than 80,000 people, many in its towns. As humans withdraw, forests are expanding. In France, between 1990 and 2015, forest cover increased by an average of 102,000 hectares each year, as more fields were given over to trees. Now, nearly one-third of mainland France is covered by woodland of some sort. The decline of hunting as a sport also means more forests fall quiet. In the mid-to-late 20th century over 2m hunters regularly spent winter weekends tramping in woodland, seeking boars, birds and other prey. Today the Fédération Nationale des Chasseurs, the national body, claims 1.1m people hold hunting licences, though the number of active hunters is probably lower. The mostly protected status of the wolf in Europe—hunting them is now forbidden, other than when occasional culls are sanctioned by the state—plus the efforts of NGOs to track and count the animals, also contribute to the recovery of wolf populations.
As the lupine population of Europe spreads westwards, with occasional reports of wolves seen closer to urban areas, expect to hear of more clashes between farmers and those who celebrate the predators' return. Farmers' losses are real, but are not the only economic story. Tourist venues, such as parks where wolves are kept and the animals' spread is discussed, also generate income and jobs in rural areas.
Which one of the following has NOT contributed to the growing wolf population in Lozère?
A decline in the rural population of Lozère.
An increase in woodlands and forest cover in Lozère.
The shutting down of the royal office of the Luparii.
The granting of a protected status to wolves in Europe.
The shutting down of the royal office of the Luparii.
The Luparii royal office was created to catch wolves and completed its job in the 1930s when the last wolf disappeared. The office was an active hunting institution that contributed to wolf eradication, not to wolf population growth. Its shutting down is not mentioned as a factor in the wolf's return, and even if it were no longer active, the passage attributes the wolf's return to specific other factors.
The inhabitants of Lozère have to grapple with all of the following problems, EXCEPT:
lack of educational facilities.
poor rural communication infrastructure.
livestock losses.
decline in the number of hunting licences.
decline in the number of hunting licences.
The passage explicitly lists concerns about local schools, jobs, phone and internet connections, and livestock losses as problems faced by residents of Lozère. A decline in the number of hunting licences is mentioned as a general French trend, but it is presented as a factor contributing to the wolf's return, not as a problem faced by Lozère's inhabitants themselves.
Which one of the following statements, if true, would weaken the author's claims?
Having migrated out in the last century, wolves are now returning to Lozère.
Unemployment concerns the residents of Lozère.
Wolf attacks on tourists in Lozère are on the rise.
The old mining sites of Lozère are now being used as grazing pastures for sheep.
Wolf attacks on tourists in Lozère are on the rise.
The passage presents wolf tourism, such as wolf parks, as generating income and jobs in rural areas, offering a potential economic benefit that could balance farmers' losses. If wolf attacks on tourists were rising, this would undermine the economic case for wolf tourism and remove the potential solution the author offers to reconcile the competing interests of farmers and environmentalists.
The author presents a possible economic solution to an existing issue facing Lozère that takes into account the divergent and competing interests of:
politicians and farmers.
environmentalists and politicians.
farmers and environmentalists.
tourists and environmentalists.
politicians and farmers.
The passage concludes by noting that farmers' losses are real but are not the only economic story, and that tourist venues discussing wolves generate income and jobs in rural areas. This is a proposed economic solution that directly addresses the competing interests of farmers, who see wolves as a threat, and environmentalists, who celebrate the wolves' return.
[Fifty] years after its publication in English [in 1972], and just a year since [Marshall] Sahlins himself died—we may ask: why did [his essay] "Original Affluent Society" have such an impact, and how has it fared since? . . . Sahlins's principal argument was simple but counterintuitive: before being driven into marginal environments by colonial powers, hunter-gatherers, or foragers, were not engaged in a desperate struggle for meager survival. Quite the contrary, they satisfied their needs with far less work than people in agricultural and industrial societies, leaving them more time to use as they wished. Hunters, he quipped, keep bankers' hours. Refusing to maximize, many were "more concerned with games of chance than with chances of game." . . . The so-called Neolithic Revolution, rather than improving life, imposed a harsher work regime and set in motion the long history of growing inequality . . .
Moreover, foragers had other options. The contemporary Hadza of Tanzania, who had long been surrounded by farmers, knew they had alternatives and rejected them. To Sahlins, this showed that foragers are not simply examples of human diversity or victimhood but something more profound: they demonstrated that societies make real choices. Culture, a way of living oriented around a distinctive set of values, manifests a fundamental principle of collective self-determination. . . .
But the point [of the essay] is not so much the empirical validity of the data—the real interest for most readers, after all, is not in foragers either today or in the Paleolithic—but rather its conceptual challenge to contemporary economic life and bourgeois individualism. The empirical served a philosophical and political project, a thought experiment and stimulus to the imagination of possibilities.
With its title's nod toward The Affluent Society (1958), economist John Kenneth Galbraith's famously skeptical portrait of America's postwar prosperity and inequality, and dripping with New Left contempt for consumerism, "The Original Affluent Society" brought this critical perspective to bear on the contemporary world. It did so through the classic anthropological move of showing that radical alternatives to the readers' lives really exist. If the capitalist world seeks wealth through ever greater material production to meet infinitely expansive desires, foraging societies follow "the Zen road to affluence": not by getting more, but by wanting less. If it seems that foragers have been left behind by "progress," this is due only to the ethnocentric self-congratulation of the West. Rather than accumulate material goods, these societies are guided by other values: leisure, mobility, and above all, freedom. . . .
Viewed in today's context, of course, not every aspect of the essay has aged well. While acknowledging the violence of colonialism, racism, and dispossession, it does not thematize them as heavily as we might today. Rebuking evolutionary anthropologists for treating present-day foragers as "left behind" by progress, it too can succumb to the temptation to use them as proxies for the Paleolithic. Yet these characteristics should not distract us from appreciating Sahlins's effort to show that if we want to conjure new possibilities, we need to learn about actually inhabitable worlds.
We can infer that Sahlins's main goal in writing his essay was to:
put forth the view that, despite egalitarian origins, economic progress brings greater inequality and social hierarchies.
highlight the fact that while we started off as a fairly contented egalitarian people, we have progressively degenerated into materialism.
hold a mirror to an acquisitive society, with examples of other communities that have chosen successfully to be non-materialistic
counter Galbraith's pessimistic view of the inevitability of a capitalist trajectory for economic growth.
hold a mirror to an acquisitive society, with examples of other communities that have chosen successfully to be non-materialistic
The passage states that the real interest of Sahlins's essay is not empirical data about foragers but its conceptual challenge to contemporary economic life and bourgeois individualism. The essay was meant to show that alternatives to the capitalist world's material accumulation genuinely exist, holding a mirror to acquisitive society by pointing to communities that have chosen non-materialistic ways of living.
The author mentions Tanzania's Hadza community to illustrate:
that hunter-gatherer communities' subsistence-level techniques equipped them to survive well into contemporary times.
how pre-agrarian societies did not hamper the emergence of more advanced agrarian practices in contiguous communities.
that forager communities' lifestyles derived not from ignorance about alternatives, but from their own choice.
how two vastly different ways of living and working were able to coexist in proximity for centuries.
that forager communities' lifestyles derived not from ignorance about alternatives, but from their own choice.
The passage states that the Hadza, surrounded by farmers and knowing the alternatives, rejected farming, and that Sahlins used this to show that foragers demonstrate real collective self-determination, making choices about their way of life rather than simply lacking knowledge of alternatives.
The author of the passage mentions Galbraith's "The Affluent Society" to:
show how Galbraith's theories refute Sahlins's thesis on the contentment of pre-hunter-gatherer communities.
contrast the materialist nature of contemporary growth paths with the pacifist content ways of living among the foragers.
document the influence of Galbraith's cynical views on modern consumerism on Sahlins's analysis of pre-historic societies.
show how Sahlins's views complemented Galbraith's criticism of the consumerism and inequality of contemporary society.
show how Sahlins's views complemented Galbraith's criticism of the consumerism and inequality of contemporary society.
The passage describes Sahlins's essay as dripping with New Left contempt for consumerism and explicitly states that the title nods toward Galbraith's book. The passage says the essay brought Galbraith's critical perspective on consumerism and inequality to bear on the contemporary world, and shows how the two works share a critical stance toward capitalist society, meaning Sahlins's views complemented Galbraith's criticism.
The author of the passage criticises Sahlins's essay for its:
cursory treatment of the effects of racism and colonialism on societies.
outdated values regarding present-day foragers versus ancient foraging communities.
critique of anthropologists who disparage the choices of foragers in today's society
failure to supplement its thesis with robust empirical data.
cursory treatment of the effects of racism and colonialism on societies.
The passage explicitly states that while acknowledging the violence of colonialism, racism and dispossession, Sahlins's essay does not thematise them as heavily as we might today, which is a direct criticism of the essay's cursory treatment of racism and colonialism.
There is a sentence that is missing in the paragraph below. Look at the paragraph and decide where (option 1, 2, 3, or 4) the following sentence would best fit.
Sentence: The discovery helps to explain archeological similarities between the Paleolithic peoples of China, Japan, and the Americas.
Paragraph: The researchers also uncovered an unexpected genetic link between Native Americans and Japanese people. (1). During the deglaciation period, another group branched out from northern coastal China and travelled to Japan. (2). "We were surprised to find that this ancestral source also contributed to the Japanese gene pool, especially the indigenous Ainus," says Li. (3). They shared similarities in how they crafted stemmed projectile points for arrowheads and spears. (4). "This suggests that the Pleistocene connection among the Americas, China, and Japan was not confined to culture but also to genetics," says senior author Qing-Peng Kong, an evolutionary geneticist at the Chinese Academy of Sciences.
Option 2
Option 3
Option 1
Option 4
Option 2
The inserted sentence states that the discovery helps explain archaeological similarities between Paleolithic peoples of China, Japan and the Americas. This fits at Option 3, right after the researcher's surprise that the ancestral source contributed to the Japanese gene pool, and right before the closing quote about the Pleistocene connection among the Americas, China and Japan being not just cultural but genetic. The inserted sentence is about archaeological similarities specifically, bridging the genetic finding about Japan with the broader conclusion about connections across China, Japan and the Americas.
There is a sentence that is missing in the paragraph below. Look at the paragraph and decide where (option 1, 2, 3, or 4) the following sentence would best fit.
Sentence: This philosophical cut at one's core beliefs, values, and way of life is difficult enough.
Paragraph: The experience of reading philosophy is often disquieting. When reading philosophy, the values around which one has heretofore organised one's life may come to look provincial, flatly wrong, or even evil. (1). When beliefs previously held as truths are rendered implausible, new beliefs, values, and ways of living may be required. (2). What's worse, philosophers admonish each other to remain unsutured until such time as a defensible new answer is revealed or constructed. Sometimes, philosophical writing is even strictly critical in that it does not even attempt to provide an alternative after tearing down a cultural or conceptual citadel. (3). The reader of philosophy must be prepared for the possibility of this experience. While reading philosophy can help one clarify one's values, and even make one self-conscious for the first time of the fact that there are good reasons for believing what one believes, it can also generate unremediated doubt that is difficult to live with. (4).
Option 1
Option 4
Option 3
Option 2
Option 2
The inserted sentence states that the philosophical challenge to one's core beliefs, values and way of life is difficult enough. This fits at Option 2, right after the statement that when beliefs are rendered implausible, new beliefs and ways of living may be required, and right before the statement that philosophers admonish each other to remain unsutured until a defensible answer is found. The inserted sentence acknowledges how hard the initial challenge is, and the following sentence then adds that things get even worse because philosophers don't even offer an immediate replacement.
Five jumbled up sentences (labelled 1, 2, 3, 4 and 5), related to a topic, are given below. Four of them can be put together to form a coherent paragraph. Identify the odd sentence and key in the number of that sentence as your answer.
1. Having an appreciation for the workings of another person's mind is considered a prerequisite for natural language acquisition, strategic social interaction, reflexive thought, and moral judgment.
2. It is a 'theory of mind' though some scholars prefer to call it 'mentalizing' or 'mindreading', which is important for the development of one's cognitive abilities.
3. Though we must speculate about its evolutionary origin, we do have indications that the capacity evolved sometime in the last few million years.
4. This capacity develops from early beginnings in the first year of life to the adult's fast and often effortless understanding of others' thoughts, feelings, and intentions.
5. One of the most fascinating human capacities is the ability to perceive and interpret other people's behaviour in terms of their mental states.
Sentences 1, 3, 4 and 5 build a coherent argument about the human capacity to understand others' mental states, starting with identifying it as one of the most fascinating human capacities, explaining how it develops from infancy to adulthood, discussing the speculation about its evolutionary origins, and describing its importance for language acquisition, social interaction and moral judgment. Sentence 2 introduces a definition and alternative names for this capacity, describing it as a theory of mind important for cognitive development. This definitional sentence does not fit the developmental and evolutionary narrative the other four sentences build, and the phrase "which is important for the development of one's cognitive abilities" is a weaker restatement of what sentence 1 already captures more specifically.
Five jumbled up sentences (labelled 1, 2, 3, 4 and 5), related to a topic, are given below. Four of them can be put together to form a coherent paragraph. Identify the odd sentence and key in the number of that sentence as your answer.
1. In English, there is no systematic rule for the naming of numbers; after ten, we have "eleven" and "twelve" and then the teens: "thirteen", "fourteen", "fifteen" and so on.
2. Even more confusingly, some English words invert the numbers they refer to: the word "fourteen" puts the four first, even though it appears last.
3. It can take children a while to learn all these words, and understand that "fourteen" is different from "forty".
4. For multiples of 10, English speakers switch to a different pattern: "twenty", "thirty", "forty" and so on.
5. If you didn't know the word for "eleven", you would be unable to just guess it - you might come up with something like "one-teen".
Sentences 1, 2, 4 and 5 build a coherent argument about the irregularity of English number naming, starting with the lack of a systematic rule after ten, describing how the pattern switches for multiples of ten, noting that some English words invert the numbers they name, and illustrating with the example of eleven being unguessable. Sentence 3 talks about children taking time to learn these words and distinguish fourteen from forty, which introduces a child developmental perspective that is separate from the linguistic analysis of the naming system that the other four sentences build.
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. What precisely are the "unusual elements" that make a particular case so attractive to a certain kind of audience?
2. It might be a particularly savage or unfathomable level of depravity, very often it has something to do with the precise amount of mystery involved.
3. Unsolved, and perhaps unsolvable cases offer something that "ordinary" murder doesn't.
4. Why are some crimes destined for perpetual re-examination and others locked into permanent obscurity?
Sentence 4 opens with the broad question of why some crimes get perpetual re-examination and others permanent obscurity. Sentence 1 narrows this by asking specifically what unusual elements make a case attractive to certain audiences. Sentence 2 begins to answer this by suggesting it might be depravity or mystery. Sentence 3 then explains what unsolved cases offer that ordinary murders don't, completing the argument.
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. Algorithms hosted on the internet are accessed by many, so biases in AI models have resulted in much larger impact, adversely affecting far larger groups of people.
2. Though "algorithmic bias" is the popular term, the foundation of such bias is not in algorithms, but in the data; algorithms are not biased, data is, as algorithms merely reflect persistent patterns that are present in the training data.
3. Despite their widespread impact, it is relatively easier to fix AI biases than human-generated biases, as it is simpler to identify the former than to try to make people unlearn behaviors learnt over generations.
4. The impact of biased decisions made by humans is localised and geographically confined, but with the advent of AI, the impact of such decisions is spread over a much wider scale.
Sentence 4 introduces the contrast between localised human bias and the wider-scale impact of AI bias. Sentence 1 follows by explaining why AI bias has a larger impact, since algorithms hosted online are accessed by many. Sentence 2 then clarifies the terminological issue, that algorithmic bias is really a data problem. Sentence 3 closes with the comparative point that AI biases are actually easier to fix than human biases despite their scale.
The passage given below is followed by four alternate summaries. Choose the option that best captures the essence of the passage.
Manipulating information was a feature of history long before modern journalism established rules of integrity. A record dates back to ancient Rome, when Antony met Cleopatra and his political enemy Octavian launched a smear campaign against him with "short, sharp slogans written upon coins." The perpetrator became the first Roman Emperor and "fake news had allowed Octavian to hack the republican system once and for all". But the 21st century has seen the weaponization of information on an unprecedented scale. Powerful new technology makes the fabrication of content simple, and social networks amplify falsehoods peddled by States, populist politicians, and dishonest corporate entities. The platforms have become fertile ground for computational propaganda, 'trolling' and 'troll armies'.
Disinformation, which is mediated by technology today, is not new and has existed since ancient times.
People need to become critical of what they read, since historically, weaponization of information has led to corruption.
Use of misinformation for attaining power, a practice that is as old as the Octavian era, is currently fueled by technology.
Octavian used fake news to manipulate people and attain power and influence, just as people do today
Use of misinformation for attaining power, a practice that is as old as the Octavian era, is currently fueled by technology.
The passage argues that manipulation of information is not new, having existed since ancient Rome with Octavian's smear campaign, but that the 21st century has seen this weaponisation of information on an unprecedented scale due to powerful new technology and social media platforms.
The passage given below is followed by four alternate summaries. Choose the option that best captures the essence of the passage.
Colonialism is not a modern phenomenon. World history is full of examples of one society gradually expanding by incorporating adjacent territory and settling its people on newly conquered territory. In the sixteenth century, colonialism changed decisively because of technological developments in navigation that began to connect more remote parts of the world. The modern European colonial project emerged when it became possible to move large numbers of people across the ocean and to maintain political control in spite of geographical dispersion. The term colonialism is used to describe the process of European settlement, violent dispossession and political domination over the rest of the world, including the Americas, Australia, and parts of Africa and Asia.
As a result of developments in navigation technology, European colonialism led to the displacement of indigenous populations and global political changes in the 16th century.
Colonialism, conceptualized in the 16th century, allowed colonizers to expand their territories, establish settlements, and exercise political power.
Technological advancements in navigation in the 16th century, transformed colonialism, enabling Europeans to establish settlements and exert political dominance over distant regions.
Colonialism surged in the 16th century due to advancements in navigation, enabling British settlements abroad and global dominance.
Technological advancements in navigation in the 16th century, transformed colonialism, enabling Europeans to establish settlements and exert political dominance over distant regions.
The passage argues that colonialism as a general phenomenon is ancient, that it changed decisively in the sixteenth century due to advances in navigation technology, and that modern European colonialism specifically involved using these advances to move large numbers of people across oceans and maintain political control over remote territories.
The schematic diagram below shows 12 rectangular houses in a housing complex. House numbers are mentioned in the rectangles representing the houses. The houses are located in six columns – Column-A through Column-F, and two rows – Row-1 and Row-2. The houses are divided into two blocks - Block XX and Block YY. The diagram also shows two roads, one passing in front of the houses in Row-2 and another between the two blocks.
Some of the houses are occupied. The remaining ones are vacant and are the only ones available for sale.
The road adjacency value of a house is the number of its sides adjacent to a road. For example, the road adjacency values of C2, F2, and B1 are 2, 1, and 0, respectively. The neighbour count of a house is the number of sides of that house adjacent to occupied houses in the same block. For example, E1 and C1 can have the maximum possible neighbour counts of 3 and 2, respectively.
The base price of a vacant house is Rs. 10 lakhs if the house does not have a parking space, and Rs. 12 lakhs if it does. The quoted price (in lakhs of Rs.) of a vacant house is calculated as (base price) + 5 × (road adjacency value) + 3 × (neighbour count).
The following information is also known.
1. The maximum quoted price of a house in Block XX is Rs. 24 lakhs. The minimum quoted price of a house in block YY is Rs. 15 lakhs, and one such house is in Column-E.
2. Row-1 has two occupied houses, one in each block.
3. Both houses in Column-E are vacant. Each of Column-D and Column-F has at least one occupied house.
4. There is only one house with parking space in Block YY.
How many houses are vacant in Block XX?
Step 1:
The quoted price (in lakhs of rupees) is calculated as:
Quoted Price = Base Price + 5 × (Number of adjacent roads) + 3 × (Number of neighbouring occupied houses)
Thus,
QP = 10/12 + 5(0, 1, 2) + 3(0, 1, 2, 3)
Using conditions (1) and (2),
- 24 = 10 + 5 × 1 + 3 × 3
- 15 = 10 + 5 × 1 + 3 × 0 or 15 = 12 + 5 × 0 + 3 × 1
In Block XX, the highest quoted price is Rs. 24 lakh.
Hence, B1, A2, and C2 are occupied, while A1, B2, and C1 are vacant (i.e., available for sale).
Step 2:
From condition (3), E1 and E2 are vacant.
Also, from conditions (2) and (3), exactly one of D1 and F1 is occupied.
For E1,
Quoted Price = 15
= 12 + 5 × 0 + 3 × 1
Therefore, by condition (4), E1 is the only house in Block YY with a parking space.
Further, from condition (3), each of Column D and Column F contains at least one occupied house.
This gives two possibilities:
- If D1 is occupied, then F2 must also be occupied, while D2 may be either occupied or vacant.
- If F1 is occupied, then D2 must be occupied, while F2 may be either occupied or vacant.
Three houses A1, C1 and B2 are vacant in Block XX.
Which of the following houses is definitely occupied?
B1
A1
D2
F2
B1
Step 1:
The quoted price (in lakhs of rupees) is calculated as:
Quoted Price = Base Price + 5 × (Number of adjacent roads) + 3 × (Number of neighbouring occupied houses)
Thus,
QP = 10/12 + 5(0, 1, 2) + 3(0, 1, 2, 3)
Using conditions (1) and (2),
- 24 = 10 + 5 × 1 + 3 × 3
- 15 = 10 + 5 × 1 + 3 × 0 or 15 = 12 + 5 × 0 + 3 × 1
In Block XX, the highest quoted price is Rs. 24 lakh.
Hence, B1, A2, and C2 are occupied, while A1, B2, and C1 are vacant (i.e., available for sale).
Step 2:
From condition (3), E1 and E2 are vacant.
Also, from conditions (2) and (3), exactly one of D1 and F1 is occupied.
For E1,
Quoted Price = 15
= 12 + 5 × 0 + 3 × 1
Therefore, by condition (4), E1 is the only house in Block YY with a parking space.
Further, from condition (3), each of Column D and Column F contains at least one occupied house.
This gives two possibilities:
- If D1 is occupied, then F2 must also be occupied, while D2 may be either occupied or vacant.
- If F1 is occupied, then D2 must be occupied, while F2 may be either occupied or vacant.
House B1 is definitely occupied.
Which of the following options best describes the number of vacant houses in Row-2?
Exactly 3
Either 2 or 3
Either 3 or 4
Exactly 2
Either 2 or 3
Step 1:
The quoted price (in lakhs of rupees) is calculated as:
Quoted Price = Base Price + 5 × (Number of adjacent roads) + 3 × (Number of neighbouring occupied houses)
Thus,
QP = 10/12 + 5(0, 1, 2) + 3(0, 1, 2, 3)
Using conditions (1) and (2),
- 24 = 10 + 5 × 1 + 3 × 3
- 15 = 10 + 5 × 1 + 3 × 0 or 15 = 12 + 5 × 0 + 3 × 1
In Block XX, the highest quoted price is Rs. 24 lakh.
Hence, B1, A2, and C2 are occupied, while A1, B2, and C1 are vacant (i.e., available for sale).
Step 2:
From condition (3), E1 and E2 are vacant.
Also, from conditions (2) and (3), exactly one of D1 and F1 is occupied.
For E1,
Quoted Price = 15
= 12 + 5 × 0 + 3 × 1
Therefore, by condition (4), E1 is the only house in Block YY with a parking space.
Further, from condition (3), each of Column D and Column F contains at least one occupied house.
This gives two possibilities:
- If D1 is occupied, then F2 must also be occupied, while D2 may be either occupied or vacant.
- If F1 is occupied, then D2 must be occupied, while F2 may be either occupied or vacant.
The number of vacant houses in Row-2 is either 2 or 3.
What is the maximum possible quoted price (in lakhs of Rs.) for a vacant house in Column-E?
Step 1:
The quoted price (in lakhs of rupees) is calculated as:
Quoted Price = Base Price + 5 × (Number of adjacent roads) + 3 × (Number of neighbouring occupied houses)
Thus,
QP = 10/12 + 5(0, 1, 2) + 3(0, 1, 2, 3)
Using conditions (1) and (2),
- 24 = 10 + 5 × 1 + 3 × 3
- 15 = 10 + 5 × 1 + 3 × 0 or 15 = 12 + 5 × 0 + 3 × 1
In Block XX, the highest quoted price is Rs. 24 lakh.
Hence, B1, A2, and C2 are occupied, while A1, B2, and C1 are vacant (i.e., available for sale).
Step 2:
From condition (3), E1 and E2 are vacant.
Also, from conditions (2) and (3), exactly one of D1 and F1 is occupied.
For E1,
Quoted Price = 15
= 12 + 5 × 0 + 3 × 1
Therefore, by condition (4), E1 is the only house in Block YY with a parking space.
Further, from condition (3), each of Column D and Column F contains at least one occupied house.
This gives two possibilities:
- If D1 is occupied, then F2 must also be occupied, while D2 may be either occupied or vacant.
- If F1 is occupied, then D2 must be occupied, while F2 may be either occupied or vacant.
The maximum possible quoted price for a vacant house i.e., E2 in Column-E = 10 + 5 × 1 + 3 × 2 = Rs.21 lakhs.
Which house in Block YY has parking space?
E1
F1
F2
E2
E1
Step 1:
The quoted price (in lakhs of rupees) is calculated as:
Quoted Price = Base Price + 5 × (Number of adjacent roads) + 3 × (Number of neighbouring occupied houses)
Thus,
QP = 10/12 + 5(0, 1, 2) + 3(0, 1, 2, 3)
Using conditions (1) and (2),
- 24 = 10 + 5 × 1 + 3 × 3
- 15 = 10 + 5 × 1 + 3 × 0 or 15 = 12 + 5 × 0 + 3 × 1
In Block XX, the highest quoted price is Rs. 24 lakh.
Hence, B1, A2, and C2 are occupied, while A1, B2, and C1 are vacant (i.e., available for sale).
Step 2:
From condition (3), E1 and E2 are vacant.
Also, from conditions (2) and (3), exactly one of D1 and F1 is occupied.
For E1,
Quoted Price = 15
= 12 + 5 × 0 + 3 × 1
Therefore, by condition (4), E1 is the only house in Block YY with a parking space.
Further, from condition (3), each of Column D and Column F contains at least one occupied house.
This gives two possibilities:
- If D1 is occupied, then F2 must also be occupied, while D2 may be either occupied or vacant.
- If F1 is occupied, then D2 must be occupied, while F2 may be either occupied or vacant.
House E1 in Block YY has parking space.
Faculty members in a management school can belong to one of four departments – Finance and Accounting (F&A), Marketing and Strategy (M&S), Operations and Quants (O&Q) and Behaviour and Human Resources (B&H). The numbers of faculty members in F&A, M&S, O&Q and B&H departments are 9, 7, 5 and 3 respectively.
Prof. Pakrasi, Prof. Qureshi, Prof. Ramaswamy and Prof. Samuel are four members of the school's faculty who were candidates for the post of the Dean of the school. Only one of the candidates was from O&Q.
Every faculty member, including the four candidates, voted for the post. In each department, all the faculty members who were not candidates voted for the same candidate. The rules for the election are listed below.
1. There cannot be more than two candidates from a single department.
2. A candidate cannot vote for himself/herself.
3. Faculty members cannot vote for a candidate from their own department.
After the election, it was observed that Prof. Pakrasi received 3 votes, Prof. Qureshi received 14 votes, Prof. Ramaswamy received 6 votes and Prof. Samuel received 1 vote. Prof. Pakrasi voted for Prof. Ramaswamy, Prof. Qureshi for Prof. Samuel, Prof. Ramaswamy for Prof. Qureshi and Prof. Samuel for Prof. Pakrasi.
Which two candidates can belong to the same department?
Prof. Pakrasi and Prof. Samuel
Prof. Pakrasi and Prof. Qureshi
Prof. Qureshi and Prof. Ramaswamy
Prof. Ramaswamy and Prof. Samuel
Prof. Pakrasi and Prof. Qureshi
Step 1:
Using the given information and conditions (1)–(3):
- P received 3 votes.
- Since S voted for P, P received the remaining 2 votes from members of B & H. Therefore, P cannot belong to the B & H department, leaving one candidate from B & H.
- Q received 14 votes.
- Since R voted for Q, Q received 9 votes from F & A and 4 votes from O & Q.
- Hence, Q cannot belong to either F & A or O & Q. Consequently, there is no candidate from F & A, and one candidate belongs to O & Q. The remaining two candidates must come from M & S.
Further,
- Since S voted for P, P and S cannot belong to the same department.
- Since R voted for Q, Q and R cannot belong to the same department.
- Similarly, R and P cannot belong to the same department, and Q and S cannot belong to the same department.
Therefore,
- P and Q belong to the M & S department.
- Q and S belong to B & H and O & Q, in either order.
Step 2:
The vote distribution is as follows:
- P received 3 votes:
- From S
- From 2 members of B & H
- Q received 14 votes:
- From R
- From 9 members of F & A
- From 4 members of O & Q
- R received 6 votes:
- From P
- From 5 members of M & S
- S received 1 vote:
- From Q
The department-wise candidate allocation is:
- F & A: 9 faculty members, 9 members, no candidate
- M & S: 7 faculty members, 5 members, Candidates – P and Q
- O & Q: 5 faculty members, 4 members, Candidate – R or S
- B & H: 3 faculty members, 2 members, Candidate – S or R
Prof. Pakrasi and Prof. Qureshi can belong to the same department M & S.
Which of the following can be the number of votes that Prof. Qureshi received from a single department?
7
9
8
6
9
Step 1:
Using the given information and conditions (1)–(3):
- P received 3 votes.
- Since S voted for P, P received the remaining 2 votes from members of B & H. Therefore, P cannot belong to the B & H department, leaving one candidate from B & H.
- Q received 14 votes.
- Since R voted for Q, Q received 9 votes from F & A and 4 votes from O & Q.
- Hence, Q cannot belong to either F & A or O & Q. Consequently, there is no candidate from F & A, and one candidate belongs to O & Q. The remaining two candidates must come from M & S.
Further,
- Since S voted for P, P and S cannot belong to the same department.
- Since R voted for Q, Q and R cannot belong to the same department.
- Similarly, R and P cannot belong to the same department, and Q and S cannot belong to the same department.
Therefore,
- P and Q belong to the M & S department.
- Q and S belong to B & H and O & Q, in either order.
Step 2:
The vote distribution is as follows:
- P received 3 votes:
- From S
- From 2 members of B & H
- Q received 14 votes:
- From R
- From 9 members of F & A
- From 4 members of O & Q
- R received 6 votes:
- From P
- From 5 members of M & S
- S received 1 vote:
- From Q
The department-wise candidate allocation is:
- F & A: 9 faculty members, 9 members, no candidate
- M & S: 7 faculty members, 5 members, Candidates – P and Q
- O & Q: 5 faculty members, 4 members, Candidate – R or S
- B & H: 3 faculty members, 2 members, Candidate – S or R
Nine can be the number of votes that Prof. Qureshi received from a single department.
If Prof. Samuel belongs to B&H, which of the following statements is/are true?
Statement A: Prof. Pakrasi belongs to M&S.
Statement B: Prof. Ramaswamy belongs to O&Q.
Neither statement A nor statement B
Only statement A
Both statements A and B
Only statement B
Both statements A and B
Step 1:
Using the given information and conditions (1)–(3):
- P received 3 votes.
- Since S voted for P, P received the remaining 2 votes from members of B & H. Therefore, P cannot belong to the B & H department, leaving one candidate from B & H.
- Q received 14 votes.
- Since R voted for Q, Q received 9 votes from F & A and 4 votes from O & Q.
- Hence, Q cannot belong to either F & A or O & Q. Consequently, there is no candidate from F & A, and one candidate belongs to O & Q. The remaining two candidates must come from M & S.
Further,
- Since S voted for P, P and S cannot belong to the same department.
- Since R voted for Q, Q and R cannot belong to the same department.
- Similarly, R and P cannot belong to the same department, and Q and S cannot belong to the same department.
Therefore,
- P and Q belong to the M & S department.
- Q and S belong to B & H and O & Q, in either order.
Step 2:
The vote distribution is as follows:
- P received 3 votes:
- From S
- From 2 members of B & H
- Q received 14 votes:
- From R
- From 9 members of F & A
- From 4 members of O & Q
- R received 6 votes:
- From P
- From 5 members of M & S
- S received 1 vote:
- From Q
The department-wise candidate allocation is:
- F & A: 9 faculty members, 9 members, no candidate
- M & S: 7 faculty members, 5 members, Candidates – P and Q
- O & Q: 5 faculty members, 4 members, Candidate – R or S
- B & H: 3 faculty members, 2 members, Candidate – S or R
Both statements are true.
What best can be concluded about the candidate from O and Q?
It was either Prof. Ramaswamy or Prof. Samuel.
It was either Prof. Pakrasi or Prof. Qureshi.
It was Prof. Samuel.
It was Prof. Ramaswamy.
It was either Prof. Ramaswamy or Prof. Samuel.
Step 1:
Using the given information and conditions (1)–(3):
- P received 3 votes.
- Since S voted for P, P received the remaining 2 votes from members of B & H. Therefore, P cannot belong to the B & H department, leaving one candidate from B & H.
- Q received 14 votes.
- Since R voted for Q, Q received 9 votes from F & A and 4 votes from O & Q.
- Hence, Q cannot belong to either F & A or O & Q. Consequently, there is no candidate from F & A, and one candidate belongs to O & Q. The remaining two candidates must come from M & S.
Further,
- Since S voted for P, P and S cannot belong to the same department.
- Since R voted for Q, Q and R cannot belong to the same department.
- Similarly, R and P cannot belong to the same department, and Q and S cannot belong to the same department.
Therefore,
- P and Q belong to the M & S department.
- Q and S belong to B & H and O & Q, in either order.
Step 2:
The vote distribution is as follows:
- P received 3 votes:
- From S
- From 2 members of B & H
- Q received 14 votes:
- From R
- From 9 members of F & A
- From 4 members of O & Q
- R received 6 votes:
- From P
- From 5 members of M & S
- S received 1 vote:
- From Q
The department-wise candidate allocation is:
- F & A: 9 faculty members, 9 members, no candidate
- M & S: 7 faculty members, 5 members, Candidates – P and Q
- O & Q: 5 faculty members, 4 members, Candidate – R or S
- B & H: 3 faculty members, 2 members, Candidate – S or R
Either Prof. Ramaswamy or Prof. Samuel was from O & Q.
Which of the following statements is/are true?
Statement A: Non-candidates from M&S voted for Prof. Qureshi.
Statement B: Non-candidates from F&A voted for Prof. Qureshi.
Only statement A
Neither statement A nor statement B
Both statements A and B
Only statement B
Only statement B
Step 1:
Using the given information and conditions (1)–(3):
- P received 3 votes.
- Since S voted for P, P received the remaining 2 votes from members of B & H. Therefore, P cannot belong to the B & H department, leaving one candidate from B & H.
- Q received 14 votes.
- Since R voted for Q, Q received 9 votes from F & A and 4 votes from O & Q.
- Hence, Q cannot belong to either F & A or O & Q. Consequently, there is no candidate from F & A, and one candidate belongs to O & Q. The remaining two candidates must come from M & S.
Further,
- Since S voted for P, P and S cannot belong to the same department.
- Since R voted for Q, Q and R cannot belong to the same department.
- Similarly, R and P cannot belong to the same department, and Q and S cannot belong to the same department.
Therefore,
- P and Q belong to the M & S department.
- Q and S belong to B & H and O & Q, in either order.
Step 2:
The vote distribution is as follows:
- P received 3 votes:
- From S
- From 2 members of B & H
- Q received 14 votes:
- From R
- From 9 members of F & A
- From 4 members of O & Q
- R received 6 votes:
- From P
- From 5 members of M & S
- S received 1 vote:
- From Q
The department-wise candidate allocation is:
- F & A: 9 faculty members, 9 members, no candidate
- M & S: 7 faculty members, 5 members, Candidates – P and Q
- O & Q: 5 faculty members, 4 members, Candidate – R or S
- B & H: 3 faculty members, 2 members, Candidate – S or R
Only statement B is true.
A visa processing office (VPO) accepts visa applications in four categories – US, UK, Schengen, and Others. The applications are scheduled for processing in twenty 15-minute slots starting at 9:00 am and ending at 2:00 pm. Ten applications are scheduled in each slot.
There are ten counters in the office, four dedicated to US applications, and two each for UK applications, Schengen applications and Others applications.
Applicants are called in for processing sequentially on a first-come-first-served basis whenever a counter gets freed for their category. The processing time for an application is the same within each category. But it may vary across the categories. Each US and UK application requires 10 minutes of processing time. Depending on the number of applications in a category and time required to process an application for that category, it is possible that an applicant for a slot may be processed later.
On a particular day, Ira, Vijay and Nandini were scheduled for Schengen visa processing in that order. They had a 9:15 am slot but entered the VPO at 9:20 am. When they entered the office, exactly six out of the ten counters were either processing applications, or had finished processing one and ready to start processing the next.
Mahira and Osman were scheduled in the 9:30 am slot on that day for visa processing in the Others category.
The following additional information is known about that day.
1. All slots were full.
2. The number of US applications was the same in all the slots. The same was true for the other three categories.
3. 50% of the applications were US applications.
4. All applicants except Ira, Vijay and Nandini arrived on time.
5. Vijay was called to a counter at 9:25 am.
How many UK applications were scheduled on that day?
Step 1:
The VPO operates for 20 time slots of 15 minutes each, from 9:00 AM to 2:00 PM, and has 10 counters.
From the given information and condition (1), the total number of applications scheduled is:
10 × 20 = 200
According to condition (3), there are 100 US visa applications.
Hence, the remaining 100 applications belong to the other three visa categories.
Using condition (2), the number of US visa applications per slot is:
100 ÷ 20 = 5
Similarly, the combined number of applications for the remaining three categories per slot is also:
100 ÷ 20 = 5
The processing time for each application is fixed within a category, although it may differ across categories.
Therefore, depending on the number of applications in a category and its processing time, an application assigned to a particular slot may actually be processed in a later slot.
The available information is:
- US: 4 counters, Processing time = 10 minutes
- UK: 2 counters, Processing time = 10 minutes
- Schengen: 2 counters, Processing time = Unknown
- Others: 2 counters, Processing time = Unknown
Step 2:
On a particular day, Ira, Vijay, and Nandini were scheduled for Schengen visa processing in that order for the 9:15 AM slot. However, they entered the VPO at 9:20 AM, resulting in a 5-minute delay.
This implies that at least three applicants were scheduled for the Schengen category in each time slot.
Similarly, Mahira and Osman were scheduled for visa processing under the Others category in the 9:30 AM slot.
Therefore, at least two applicants were scheduled for the Others category in every time slot.
The total number of applications scheduled that day was 200.
Out of these, 100 were US visa applications.
Therefore, each 15-minute time slot had:
100 ÷ 20 = 5 US applications
From the previous deductions, each time slot also had:
- 3 Schengen applications
- 2 Others applications
Thus, all 10 application slots in each time slot are accounted for:
- US = 5
- Schengen = 3
- Others = 2
This leaves no applications for the UK category in any time slot.
Hence, the number of UK applications per time slot is 0.
What is the maximum possible value of the total time (in minutes, nearest to its integer value) required to process all applications in the Others category on that day?
Step 1:
The VPO operates for 20 time slots of 15 minutes each, from 9:00 AM to 2:00 PM, and has 10 counters.
From the given information and condition (1), the total number of applications scheduled is:
10 × 20 = 200
According to condition (3), there are 100 US visa applications.
Hence, the remaining 100 applications belong to the other three visa categories.
Using condition (2), the number of US visa applications per slot is:
100 ÷ 20 = 5
Similarly, the combined number of applications for the remaining three categories per slot is also:
100 ÷ 20 = 5
The processing time for each application is fixed within a category, although it may differ across categories.
Therefore, depending on the number of applications in a category and its processing time, an application assigned to a particular slot may actually be processed in a later slot.
The available information is:
- US: 4 counters, Processing time = 10 minutes
- UK: 2 counters, Processing time = 10 minutes
- Schengen: 2 counters, Processing time = Unknown
- Others: 2 counters, Processing time = Unknown
Step 2:
On a particular day, Ira, Vijay, and Nandini were scheduled for Schengen visa processing in that order for the 9:15 AM slot. However, they entered the VPO at 9:20 AM, resulting in a 5-minute delay.
This implies that at least three applicants were scheduled for the Schengen category in each time slot.
Similarly, Mahira and Osman were scheduled for visa processing under the Others category in the 9:30 AM slot.
Therefore, at least two applicants were scheduled for the Others category in every time slot.
The maximum processing time for the Others category is 5 minutes per application.
Since there are 2 Others applications in each time slot and 20 time slots in total, the total processing time is:
= 2 × 5 × 20
= 200 minutes.
Which of the following is the closest to the time when Nandini’s application process got over?
9:35 am
9:37 am
9:45 am
9:50 am
9:45 am
Step 1:
The VPO operates for 20 time slots of 15 minutes each, from 9:00 AM to 2:00 PM, and has 10 counters.
From the given information and condition (1), the total number of applications scheduled is:
10 × 20 = 200
According to condition (3), there are 100 US visa applications.
Hence, the remaining 100 applications belong to the other three visa categories.
Using condition (2), the number of US visa applications per slot is:
100 ÷ 20 = 5
Similarly, the combined number of applications for the remaining three categories per slot is also:
100 ÷ 20 = 5
The processing time for each application is fixed within a category, although it may differ across categories.
Therefore, depending on the number of applications in a category and its processing time, an application assigned to a particular slot may actually be processed in a later slot.
The available information is:
- US: 4 counters, Processing time = 10 minutes
- UK: 2 counters, Processing time = 10 minutes
- Schengen: 2 counters, Processing time = Unknown
- Others: 2 counters, Processing time = Unknown
Step 2:
On a particular day, Ira, Vijay, and Nandini were scheduled for Schengen visa processing in that order for the 9:15 AM slot. However, they entered the VPO at 9:20 AM, resulting in a 5-minute delay.
This implies that at least three applicants were scheduled for the Schengen category in each time slot.
Similarly, Mahira and Osman were scheduled for visa processing under the Others category in the 9:30 AM slot.
Therefore, at least two applicants were scheduled for the Others category in every time slot.
Hence, 9:45 AM is the closest time at which Nandini's visa application processing would have been completed.
Which of the following statements is false?
The application process of Mahira was completed before Nandini’s.
The application process of Osman was completed before Vijay’s.
The application process of Mahira started after Nandini’s.
The application process of Osman was completed before 9:45 am.
The application process of Mahira started after Nandini’s.
Step 1:
The VPO operates for 20 time slots of 15 minutes each, from 9:00 AM to 2:00 PM, and has 10 counters.
From the given information and condition (1), the total number of applications scheduled is:
10 × 20 = 200
According to condition (3), there are 100 US visa applications.
Hence, the remaining 100 applications belong to the other three visa categories.
Using condition (2), the number of US visa applications per slot is:
100 ÷ 20 = 5
Similarly, the combined number of applications for the remaining three categories per slot is also:
100 ÷ 20 = 5
The processing time for each application is fixed within a category, although it may differ across categories.
Therefore, depending on the number of applications in a category and its processing time, an application assigned to a particular slot may actually be processed in a later slot.
The available information is:
- US: 4 counters, Processing time = 10 minutes
- UK: 2 counters, Processing time = 10 minutes
- Schengen: 2 counters, Processing time = Unknown
- Others: 2 counters, Processing time = Unknown
Step 2:
On a particular day, Ira, Vijay, and Nandini were scheduled for Schengen visa processing in that order for the 9:15 AM slot. However, they entered the VPO at 9:20 AM, resulting in a 5-minute delay.
This implies that at least three applicants were scheduled for the Schengen category in each time slot.
Similarly, Mahira and Osman were scheduled for visa processing under the Others category in the 9:30 AM slot.
Therefore, at least two applicants were scheduled for the Others category in every time slot.
When did the application processing for all US applicants get over on that day?
2:00 pm
2:05 pm
2:25 pm
3:40 pm
2:05 pm
Step 1:
The VPO operates for 20 time slots of 15 minutes each, from 9:00 AM to 2:00 PM, and has 10 counters.
From the given information and condition (1), the total number of applications scheduled is:
10 × 20 = 200
According to condition (3), there are 100 US visa applications.
Hence, the remaining 100 applications belong to the other three visa categories.
Using condition (2), the number of US visa applications per slot is:
100 ÷ 20 = 5
Similarly, the combined number of applications for the remaining three categories per slot is also:
100 ÷ 20 = 5
The processing time for each application is fixed within a category, although it may differ across categories.
Therefore, depending on the number of applications in a category and its processing time, an application assigned to a particular slot may actually be processed in a later slot.
The available information is:
- US: 4 counters, Processing time = 10 minutes
- UK: 2 counters, Processing time = 10 minutes
- Schengen: 2 counters, Processing time = Unknown
- Others: 2 counters, Processing time = Unknown
Step 2:
On a particular day, Ira, Vijay, and Nandini were scheduled for Schengen visa processing in that order for the 9:15 AM slot. However, they entered the VPO at 9:20 AM, resulting in a 5-minute delay.
This implies that at least three applicants were scheduled for the Schengen category in each time slot.
Similarly, Mahira and Osman were scheduled for visa processing under the Others category in the 9:30 AM slot.
Therefore, at least two applicants were scheduled for the Others category in every time slot.
The first five US applications require 20 minutes to be processed.
Thereafter, each subsequent batch of five applications is completed in 15 minutes.
Hence, the total time required to process all 100 US applications is:
= 20 + (15 × 19)
= 305 minutes
Therefore, the processing of all US visa applications on that day would be completed at 2:05 PM.
Five restaurants, coded R1, R2, R3, R4 and R5 gave integer ratings to five gig workers – Ullas, Vasu, Waman, Xavier and Yusuf, on a scale of 1 to 5.
The means of the ratings given by R1, R2, R3, R4 and R5 were 3.4, 2.2, 3.8, 2.8 and 3.4 respectively.
The summary statistics of these ratings for the five workers is given below.
* Range of ratings is defined as the difference between the maximum and minimum ratings awarded to a worker. The following is partial information about ratings of 1 and 5 awarded by the restaurants to the workers.
(a) R1 awarded a rating of 5 to Waman, as did R2 to Xavier, R3 to Waman and Xavier, and R5 to Vasu.
(b) R1 awarded a rating of 1 to Ullas, as did R2 to Waman and Yusuf, and R3 to Yusuf.
How many individual ratings cannot be determined from the above information?
Step 1:
Ullas has an average rating of 2.2, so his total score is:
2.2 × 5 = 11
Since his median rating is 2, the middle rating must be 2.
As his mode is also 2, another rating must be 2.
Further, the range is 3, so the remaining two ratings must be 1 and 4 or 2 and 5. The only valid set is:
(1, 2, 2, 2, 4)
Similarly, Vasu has an average rating of 3.8, giving a total score of:
3.8 × 5 = 19
Since the median and mode are both 4, and the range is 3, the only possible ratings are:
(2, 4, 4, 4, 5)
Applying the same logic:
- Waman: (1, 2, 4, 5, 5)
- Xavier: (1, 3, 4, 5, 5)
- Yusuf: (1, 1, 3, 4, 4)
Step 2:
The average ratings awarded by R1, R2, R3, R4, and R5 are 3.4, 2.2, 3.8, 2.8, and 3.4, respectively.
Hence, their total ratings are:
- R1 = 17
- R2 = 11
- R3 = 19
- R4 = 14
- R5 = 17
Using conditions (a) and (b), the partial assignments are:
- Ullas: R1 = 1
- Vasu: R4 = 5
- Waman: R1 = 5, R2 = 1, R3 = 5
- Xavier: R2 = 5, R3 = 5
- Yusuf: R2 = 1, R3 = 1
Since the total rating awarded by R2 is 11, the ratings given by R2 to Ullas and Vasu must be 2 and 2, respectively.
Similarly, the total rating awarded by R3 is 19, so the ratings given by R3 to Ullas and Vasu must be 4 and 4, respectively.
Filling in the remaining values gives:
- Ullas: R1 = 1, R2 = 2, R3 = 4, R4 = 2, R5 = 2
- Vasu: R1 = 4, R2 = 2, R3 = 4, R4 = 5, R5 = 4
- Waman: R1 = 5, R2 = 1, R3 = 5, R4 = 4, R5 = 2
- Xavier: R1 = 3, R2 = 5, R3 = 5, R4 = 1, R5 = 4
- Yusuf: R1 = 4, R2 = 1, R3 = 1, R4 = 3, R5 = 4
From the completed rating assignment, every individual rating can be determined uniquely using the given information.
Therefore, no individual rating remains undetermined.
Hence, the correct answer is 0.
To how many workers did R2 give a rating of 4?
Step 1:
Ullas has an average rating of 2.2, so his total score is:
2.2 × 5 = 11
Since his median rating is 2, the middle rating must be 2.
As his mode is also 2, another rating must be 2.
Further, the range is 3, so the remaining two ratings must be 1 and 4 or 2 and 5. The only valid set is:
(1, 2, 2, 2, 4)
Similarly, Vasu has an average rating of 3.8, giving a total score of:
3.8 × 5 = 19
Since the median and mode are both 4, and the range is 3, the only possible ratings are:
(2, 4, 4, 4, 5)
Applying the same logic:
- Waman: (1, 2, 4, 5, 5)
- Xavier: (1, 3, 4, 5, 5)
- Yusuf: (1, 1, 3, 4, 4)
Step 2:
The average ratings awarded by R1, R2, R3, R4, and R5 are 3.4, 2.2, 3.8, 2.8, and 3.4, respectively.
Hence, their total ratings are:
- R1 = 17
- R2 = 11
- R3 = 19
- R4 = 14
- R5 = 17
Using conditions (a) and (b), the partial assignments are:
- Ullas: R1 = 1
- Vasu: R4 = 5
- Waman: R1 = 5, R2 = 1, R3 = 5
- Xavier: R2 = 5, R3 = 5
- Yusuf: R2 = 1, R3 = 1
Since the total rating awarded by R2 is 11, the ratings given by R2 to Ullas and Vasu must be 2 and 2, respectively.
Similarly, the total rating awarded by R3 is 19, so the ratings given by R3 to Ullas and Vasu must be 4 and 4, respectively.
Filling in the remaining values gives:
- Ullas: R1 = 1, R2 = 2, R3 = 4, R4 = 2, R5 = 2
- Vasu: R1 = 4, R2 = 2, R3 = 4, R4 = 5, R5 = 4
- Waman: R1 = 5, R2 = 1, R3 = 5, R4 = 4, R5 = 2
- Xavier: R1 = 3, R2 = 5, R3 = 5, R4 = 1, R5 = 4
- Yusuf: R1 = 4, R2 = 1, R3 = 1, R4 = 3, R5 = 4
R2 did not give a rating of 4 to any workers.
What rating did R1 give to Xavier?
Step 1:
Ullas has an average rating of 2.2, so his total score is:
2.2 × 5 = 11
Since his median rating is 2, the middle rating must be 2.
As his mode is also 2, another rating must be 2.
Further, the range is 3, so the remaining two ratings must be 1 and 4 or 2 and 5. The only valid set is:
(1, 2, 2, 2, 4)
Similarly, Vasu has an average rating of 3.8, giving a total score of:
3.8 × 5 = 19
Since the median and mode are both 4, and the range is 3, the only possible ratings are:
(2, 4, 4, 4, 5)
Applying the same logic:
- Waman: (1, 2, 4, 5, 5)
- Xavier: (1, 3, 4, 5, 5)
- Yusuf: (1, 1, 3, 4, 4)
Step 2:
The average ratings awarded by R1, R2, R3, R4, and R5 are 3.4, 2.2, 3.8, 2.8, and 3.4, respectively.
Hence, their total ratings are:
- R1 = 17
- R2 = 11
- R3 = 19
- R4 = 14
- R5 = 17
Using conditions (a) and (b), the partial assignments are:
- Ullas: R1 = 1
- Vasu: R4 = 5
- Waman: R1 = 5, R2 = 1, R3 = 5
- Xavier: R2 = 5, R3 = 5
- Yusuf: R2 = 1, R3 = 1
Since the total rating awarded by R2 is 11, the ratings given by R2 to Ullas and Vasu must be 2 and 2, respectively.
Similarly, the total rating awarded by R3 is 19, so the ratings given by R3 to Ullas and Vasu must be 4 and 4, respectively.
Filling in the remaining values gives:
- Ullas: R1 = 1, R2 = 2, R3 = 4, R4 = 2, R5 = 2
- Vasu: R1 = 4, R2 = 2, R3 = 4, R4 = 5, R5 = 4
- Waman: R1 = 5, R2 = 1, R3 = 5, R4 = 4, R5 = 2
- Xavier: R1 = 3, R2 = 5, R3 = 5, R4 = 1, R5 = 4
- Yusuf: R1 = 4, R2 = 1, R3 = 1, R4 = 3, R5 = 4
R1 gave a rating of 3 to Xavier.
What is the median of the ratings given by R3 to the five workers?
Step 1:
Ullas has an average rating of 2.2, so his total score is:
2.2 × 5 = 11
Since his median rating is 2, the middle rating must be 2.
As his mode is also 2, another rating must be 2.
Further, the range is 3, so the remaining two ratings must be 1 and 4 or 2 and 5. The only valid set is:
(1, 2, 2, 2, 4)
Similarly, Vasu has an average rating of 3.8, giving a total score of:
3.8 × 5 = 19
Since the median and mode are both 4, and the range is 3, the only possible ratings are:
(2, 4, 4, 4, 5)
Applying the same logic:
- Waman: (1, 2, 4, 5, 5)
- Xavier: (1, 3, 4, 5, 5)
- Yusuf: (1, 1, 3, 4, 4)
Step 2:
The average ratings awarded by R1, R2, R3, R4, and R5 are 3.4, 2.2, 3.8, 2.8, and 3.4, respectively.
Hence, their total ratings are:
- R1 = 17
- R2 = 11
- R3 = 19
- R4 = 14
- R5 = 17
Using conditions (a) and (b), the partial assignments are:
- Ullas: R1 = 1
- Vasu: R4 = 5
- Waman: R1 = 5, R2 = 1, R3 = 5
- Xavier: R2 = 5, R3 = 5
- Yusuf: R2 = 1, R3 = 1
Since the total rating awarded by R2 is 11, the ratings given by R2 to Ullas and Vasu must be 2 and 2, respectively.
Similarly, the total rating awarded by R3 is 19, so the ratings given by R3 to Ullas and Vasu must be 4 and 4, respectively.
Filling in the remaining values gives:
- Ullas: R1 = 1, R2 = 2, R3 = 4, R4 = 2, R5 = 2
- Vasu: R1 = 4, R2 = 2, R3 = 4, R4 = 5, R5 = 4
- Waman: R1 = 5, R2 = 1, R3 = 5, R4 = 4, R5 = 2
- Xavier: R1 = 3, R2 = 5, R3 = 5, R4 = 1, R5 = 4
- Yusuf: R1 = 4, R2 = 1, R3 = 1, R4 = 3, R5 = 4
The median of the ratings given by R3 to the five workers is 4.
Which among the following restaurants gave its median rating to exactly one of the workers?
R4
R5
R3
R2
R4
Step 1:
Ullas has an average rating of 2.2, so his total score is:
2.2 × 5 = 11
Since his median rating is 2, the middle rating must be 2.
As his mode is also 2, another rating must be 2.
Further, the range is 3, so the remaining two ratings must be 1 and 4 or 2 and 5. The only valid set is:
(1, 2, 2, 2, 4)
Similarly, Vasu has an average rating of 3.8, giving a total score of:
3.8 × 5 = 19
Since the median and mode are both 4, and the range is 3, the only possible ratings are:
(2, 4, 4, 4, 5)
Applying the same logic:
- Waman: (1, 2, 4, 5, 5)
- Xavier: (1, 3, 4, 5, 5)
- Yusuf: (1, 1, 3, 4, 4)
Step 2:
The average ratings awarded by R1, R2, R3, R4, and R5 are 3.4, 2.2, 3.8, 2.8, and 3.4, respectively.
Hence, their total ratings are:
- R1 = 17
- R2 = 11
- R3 = 19
- R4 = 14
- R5 = 17
Using conditions (a) and (b), the partial assignments are:
- Ullas: R1 = 1
- Vasu: R4 = 5
- Waman: R1 = 5, R2 = 1, R3 = 5
- Xavier: R2 = 5, R3 = 5
- Yusuf: R2 = 1, R3 = 1
Since the total rating awarded by R2 is 11, the ratings given by R2 to Ullas and Vasu must be 2 and 2, respectively.
Similarly, the total rating awarded by R3 is 19, so the ratings given by R3 to Ullas and Vasu must be 4 and 4, respectively.
Filling in the remaining values gives:
- Ullas: R1 = 1, R2 = 2, R3 = 4, R4 = 2, R5 = 2
- Vasu: R1 = 4, R2 = 2, R3 = 4, R4 = 5, R5 = 4
- Waman: R1 = 5, R2 = 1, R3 = 5, R4 = 4, R5 = 2
- Xavier: R1 = 3, R2 = 5, R3 = 5, R4 = 1, R5 = 4
- Yusuf: R1 = 4, R2 = 1, R3 = 1, R4 = 3, R5 = 4
Restaurant R4 gave its median rating to exactly one of the workers.
Let n be the least positive integer such that 168 is a factor of 1134ⁿ. If m is the least positive integer such that 1134ⁿ is a factor of 168ᵐ, then m + n equals
9
15
12
24
15
Prime factorize the given numbers.
168 = 2^3 × 3 × 7
1134 = 2 × 3^4 × 7
Therefore,
1134^n = 2^n × 3^(4n) × 7^n
For 168 to be a factor of 1134^n, the powers of each prime in 1134^n must be at least those in 168.
Compare the exponents:
For 2: n ≥ 3
For 3: 4n ≥ 1 → n ≥ 1/4
For 7: n ≥ 1
The largest requirement is n = 3
Now, 1134^3 = 2^3 × 3^12 × 7^3
Next, 168^m = 2^(3m) × 3^m × 7^m
For 1134^3 to be a factor of 168^m, compare the exponents.
For 2: 3m ≥ 3 → m ≥ 1
For 3: m ≥ 12
For 7: m ≥ 3
The largest requirement is m = 12
Therefore, m + n = 12 + 3 = 15
Answer:
B. 15
If x and y are positive real numbers such that logx(x² + 12) = 4 and 3 logy x = 1, then x + y equals
20
68
10
11
10
Given, logx(x² + 12) = 4
Using the definition of logarithm, x4 = x² + 12
Rearrange, x4 − x² − 12 = 0
Let t = x²
Then, t² − t − 12 = 0
Factorizing, (t − 4)(t + 3) = 0
Since x is positive, t = 4
Therefore, x² = 4 → x = 2
Now, 3 logy x = 1 → logy x = 1/3
Using the definition of logarithm, y(1/3) = x
Substitute x = 2, y(1/3) = 2
Cubing both sides, y = 8
Therefore, x + y = 2 + 8 = 10
Answer: C. 10
If √(5x + 9) + √(5x − 9) = 3(2 + √2), then √(10x + 9) is equal to
3√31
4√5
3√7
2√7
3√7
Given, √(5x + 9) + √(5x − 9) = 3(2 + √2)
Square both sides.
LHS = (√(5x + 9) + √(5x − 9))² = (5x + 9) + (5x − 9) + 2√[(5x + 9)(5x − 9)] = 10x + 2√(25x² − 81)
RHS = [3(2 + √2)]² = 9(2 + √2)² = 9(4 + 4√2 + 2) = 54 + 36√2
Therefore, 10x + 2√(25x² − 81) = 54 + 36√2
Divide by 2, 5x + √(25x² − 81) = 27 + 18√2
Since 18√2 is the irrational part, √(25x² − 81) = 18√2 and 5x = 27
Therefore, x = 27/5
Now find, √(10x + 9) = √[10 × (27/5) + 9] = √(54 + 9) = √63 = 3√7
Answer:
C. 3√7
If x and y are real numbers such that x² + (x − 2y − 1)² = −4y(x + y), then the value x − 2y is
0
1
-1
2
1
Given, x² + (x − 2y − 1)² = −4y(x + y)
Expand the square, (x − 2y − 1)² = x² + 4y² + 4y + 1 − 4xy − 2x
Substitute into the given equation, x² + x² + 4y² + 4y + 1 − 4xy − 2x = −4xy − 4y²
The terms −4xy cancel from both sides.
So, 2x² + 4y² − 2x + 4y + 1 = −4y²
2x² + 8y² − 2x + 4y + 1 = 0
Now complete the squares.
2(x² − x) + 8(y² + y/2) + 1 = 0
= 2[(x − 1/2)² − 1/4] + 8[(y + 1/4)² − 1/16] + 1 = 0
= 2(x − 1/2)² − 1/2 + 8(y + 1/4)² − 1/2 + 1 = 0
Therefore, 2(x − 1/2)² + 8(y + 1/4)² = 0
Since both squared terms are non-negative, each must be zero.
Hence, x = 1/2, y = −1/4
Now, x − 2y = 1/2 − 2(−1/4) = 1/2 + 1/2 = 1
Answer:
B. 1
The number of integer solutions of equation 2|x|(x² + 1) = 5x² is
Given, 2|x|(x² + 1) = 5x²
Since the equation contains |x|, consider two cases.
Case 1: x ≥ 0
Then, |x| = x
So, 2x(x² + 1) = 5x² → 2x³ + 2x = 5x² → 2x³ − 5x² + 2x = 0
x(2x² − 5x + 2) = 0
Factorize, x(2x − 1)(x − 2) = 0
Possible solutions are, x = 0, 1/2, 2
Among these, the integer solutions are, x = 0, 2
Case 2: x < 0
Then, |x| = −x
So, −2x(x² + 1) = 5x² → 2x³ + 5x² + 2x = 0
x(2x² + 5x + 2) = 0
Factorize, x(2x + 1)(x + 2) = 0
Possible solutions are, x = 0, −1/2, −2
Since x < 0, discard x = 0.
The only integer solution is, x = −2
Therefore, the integer solutions are, x = −2, 0, 2
Hence, the number of integer solutions is 3
Answer: 3
The equation x³ + (2r + 1)x² + (4r − 1)x + 2 = 0 has -2 as one of the roots. If the other two roots are real, then the minimum possible non-negative integer value of r is
Given, x³ + (2r + 1)x² + (4r − 1)x + 2 = 0
One of the roots is −2.
Substitute x = −2.
(−2)³ + (2r + 1)(−2)² + (4r − 1)(−2) + 2 = 0
−8 + 4(2r + 1) − 2(4r − 1) + 2 = 0
−8 + 8r + 4 − 8r + 2 + 2 = 0
0 = 0
Hence, −2 is a root for every value of r.
Now divide the cubic by (x + 2).
Using synthetic division with −2:
-2 | 1 2r+1 4r−1 2
-2 -4r+2 -2
1 2r−1 1 0
So the remaining quadratic is x² + (2r − 1)x + 1 = 0
For the other two roots to be real, the discriminant must be non-negative.
Discriminant = (2r − 1)² − 4 = 4r² − 4r + 1 − 4 = 4r² − 4r − 3
Therefore, 4r² − 4r − 3 ≥ 0
Now test the non-negative integers.
For r = 0: 4(0)² − 4(0) − 3 = −3 < 0 → Not possible.
For r = 1: 4(1)² − 4(1) − 3 = −3 < 0 → Not possible.
For r = 2: 4(2)² − 4(2) − 3 = 16 − 8 − 3 = 5 > 0 → Possible.
Hence, the minimum non-negative integer value of r is 2
Answer: 2
Let α and β be the two distinct roots of the equation 2x² − 6x + k = 0, such that (α + β) and αβ are the distinct roots of the equation x² + px + p = 0. Then, the value of 8(k - p) is
Given, 2x² − 6x + k = 0
Let its roots be α and β.
Using Vieta's formulas, α + β = 6/2 = 3, αβ = k/2
Now, (α + β) and αβ are the roots of x² + px + p = 0
Therefore, Roots are 3 and k/2.
Again using Vieta's formulas,
Sum of roots = −p → 3 + k/2 = −p ...(1)
Product of roots = p → 3 × k/2 = p ...(2)
From (2), p = 3k/2
Substitute into (1), 3 + k/2 = −3k/2
Multiply throughout by 2, 6 + k = −3k → 4k = −6 → k = −3/2
Now, p = 3k/2 = 3(−3/2)/2 = −9/4
Finally, 8(k − p) = 8[−3/2 − (−9/4)] = 8(−6/4 + 9/4) = 8 × 3/4 = 6
Answer: 6
A mixture P is formed by removing a certain amount of coffee from a coffee jar and replacing the same amount with cocoa powder. The same amount is again removed from mixture P and replaced with same amount of cocoa powder to form a new mixture Q. If the ratio of coffee and cocoa in the mixture Q is 16 : 9, then the ratio of cocoa in mixture P to that in mixture Q is
1 : 3
1 : 2
5 : 9
4 : 9
5 : 9
Let the initial quantity of coffee in the jar be 1 unit.
Suppose a fraction x of the mixture is removed and replaced each time.
After the first replacement (Mixture P)
Coffee remaining = 1 − x
Cocoa added = x
So, Coffee : Cocoa = (1 − x) : x
After the second replacement (Mixture Q)
Initially in mixture P, Coffee = 1 − x, Cocoa = x
A fraction x of the mixture is removed.
Coffee removed = x(1 − x)
Coffee remaining = (1 − x) − x(1 − x) = (1 − x)²
Hence, coffee in mixture Q is (1 − x)²
Given, Coffee : Cocoa = 16 : 9
Therefore, Coffee fraction = 16/(16 + 9) = 16/25
So, (1 − x)² = 16/25 → 1 − x = 4/5 → x = 1/5
Cocoa in Mixture P = x = 1/5
Cocoa in Mixture Q = 1 − Coffee = 1 − 16/25 = 9/25
Required ratio = (1/5) : (9/25) = (5/25) : (9/25) = 5 : 9
Answer:
C. 5 : 9
The minor angle between the hours hand and minutes hand of a clock was observed at 8:48 am. The minimum duration, in minutes, after 8.48 am when this angle increases by 50% is
36/11
2
4
24/11
24/11
At 8:48 am,
Minute hand angle from 12 = 48 × 6° = 288°
Hour hand angle from 12 = 8 × 30 + 48 × 0.5 = 240 + 24 = 264°
Therefore, the minor angle between them is 288° − 264° = 24°
The angle has to increase by 50%.
New angle = 24 + 50% of 24 = 24 + 12 = 36°
Let the required time after 8:48 am be t minutes.
In t minutes, Minute hand moves = 6t°, Hour hand moves = 0.5t°
Hence, the angle increases at 6 − 0.5 = 5.5° per minute
Initially, the angle is 24°.
After t minutes, 24 + 5.5t = 36 → 5.5t = 12 → 11t/2 = 12 → 11t = 24 → t = 24/11 minutes
Therefore, the minimum duration after 8:48 am is 24/11 minutes
Answer:
D. 24/11
Gita sells two objects A and B at the same price such that she makes a profit of 20% on object A and a loss of 10% on object B. If she increases the selling price such that objects A and B are still sold at an equal price and a profit of 10% is made on object B, then the profit made on object A will be nearest to
42%
45%
47%
49%
47%
Let the initial common selling price of both objects be ₹S.
Step 1: Find the cost prices
For object A, Profit = 20% → CP of A = S/1.20 = 5S/6
For object B, Loss = 10% → CP of B = S/0.90 = 10S/9
Step 2: Increase the selling price
Now the new common selling price gives a profit of 10% on object
B.
New SP = 110% of CP of B = 1.10 × 10S/9 = 11S/9
Step 3: Find the profit on object A
CP of A = 5S/6
New SP = 11S/9
Profit % = [(11S/9 − 5S/6) ÷ (5S/6)] × 100
Take LCM of 9 and 6, 11S/9 − 5S/6 = (22S − 15S)/18 = 7S/18
Therefore, Profit % = [(7S/18) ÷ (5S/6)] × 100 = (7/18) × (6/5) × 100 = 7/15 × 100 = 46.67% ≈ 47%
Answer:
C. 47%
Brishti went on an 8-hour trip in a car. Before the trip, the car had travelled a total of x km till then, where x is a whole number and is palindromic, i.e., x remains unchanged when its digits are reversed. At the end of the trip, the car had travelled a total of 26862 km till then, this number again being palindromic. If Brishti never drove at more than 110 km/h, then the greatest possible average speed at which she drove during the trip, in km/h, was
110
90
100
80
100
Let the odometer reading before the trip be x km.
Given,
● Final reading = 26862 km (a palindrome)
● Trip duration = 8 hours
● Maximum speed at any time = 110 km/h
Hence, the maximum possible distance travelled is 110 × 8 = 880 km
Therefore, 26862 − x ≤ 880 → x ≥ 26862 − 880 → x ≥ 25982
To maximize the average speed, we must maximize the distance travelled.
So, we need the smallest palindromic number greater than or equal to 25982.
Check palindromes:
● Mirror 259 → 25952 (less than 25982, so not possible)
● Next palindrome is 26062
Thus, x = 26062
Distance travelled = 26862 − 26062 = 800 km
Average speed = 800 ÷ 8 = 100 km/h
Answer:
C. 100
In an examination, the average marks of 4 girls and 6 boys is 24. Each of the girls has the same marks while each of the boys has the same marks. If the marks of any girl is at most double the marks of any boy, but not less than the marks of any boy, then the number of possible distinct integer values of the total marks of 2 girls and 6 boys is
21
20
22
19
21
Let the marks of each girl be g and the marks of each boy be b.
The average marks of 4 girls and 6 boys is 24.
So, Total marks = 10 × 24 = 240
Therefore, 4g + 6b = 240 → 2g + 3b = 120
Hence, g = (120 − 3b)/2
The conditions are:
● Each girl's marks are at least the marks of a boy.
● Each girl's marks are at most twice the marks of a boy.
So, b ≤ g ≤ 2b
Substitute g = (120 − 3b)/2.
Using g ≥ b: (120 − 3b)/2 ≥ b → 120 − 3b ≥ 2b → 120 ≥ 5b → b ≤ 24
Using g ≤ 2b: (120 − 3b)/2 ≤ 2b → 120 − 3b ≤ 4b → 120 ≤ 7b → b ≥ 120/7
Hence, 120/7 ≤ b ≤ 24
Now we need the total marks of 2 girls and 6 boys.
Required total = 2g + 6b
Using 2g = 120 − 3b, Required total = (120 − 3b) + 6b = 120 + 3b
Since b can take any real value in the interval 120/7 ≤ b ≤ 24, the required total varies from 120 + 3 × (120/7) = 171.43... to 120 + 3 × 24 = 192
Therefore, the possible integer values are 172, 173, ..., 192
Number of integers = 192 − 172 + 1 = 21
Answer:
A. 21
The salaries of three friends Sita, Gita and Mita are initially in the ratio 5 : 6 : 7, respectively. In the first year, they get salary hikes of 20%, 25% and 20%, respectively. In the second year, Sita and Mita get salary hikes of 40% and 25%, respectively, and the salary of Gita becomes equal to the mean salary of the three friends. The salary hike of Gita in the second year is
25%
28%
26%
30%
26%
Let the initial salaries be
● Sita = 5x
● Gita = 6x
● Mita = 7x
After the first year
Sita's salary = 5x × 1.20 = 6x
Gita's salary = 6x × 1.25 = 7.5x
Mita's salary = 7x × 1.20 = 8.4x
After the second year
Sita gets a 40% hike. New salary = 6x × 1.40 = 8.4x
Mita gets a 25% hike. New salary = 8.4x × 1.25 = 10.5x
Let Gita's salary after the second year be G.
Given, Gita's salary becomes equal to the mean salary of all three friends.
So, G = (8.4x + G + 10.5x)/3
Multiply both sides by 3, 3G = 18.9x + G
2G = 18.9x
G = 9.45x
Find Gita's second-year salary hike
Salary before the second-year hike = 7.5x
Increase in salary = 9.45x − 7.5x = 1.95x
Therefore, Salary hike % = (1.95x ÷ 7.5x) × 100 = 26%
Answer:
C. 26%
Arvind travels from town A to town B, and Surbhi from town B to town A, both starting at the same time along the same route. After meeting each other, Arvind takes 6 hours to reach town B while Surbhi takes 24 hours to reach town A. If Arvind travelled at a speed of 54 km/h, then the distance, in km, between town A and town B is
Let Surbhi's speed be v km/h.
Arvind's speed = 54 km/h
After meeting,
● Arvind takes 6 hours to reach
B.
● Surbhi takes 24 hours to reach
A.
Hence, Distance from meeting point to B = 54 × 6 = 324 km
Distance from meeting point to A = 24v
For two people starting at the same time and meeting, (speed ratio) = (distance covered before meeting ratio)
Therefore, 54 : v = 24v : 324
Cross multiply, 54 × 324 = 24v² → v² = (54 × 324)/24 = 729 → v = 27 km/h
Distance from meeting point to A = 24 × 27 = 648 km
Distance from meeting point to B = 324 km
Therefore, total distance = 648 + 324 = 972 km
Answer: 972
Anil invests Rs. 22000 for 6 years in a certain scheme with 4% interest per annum, compounded half-yearly. Sunil invests in the same scheme for 5 years, and then reinvests the entire amount received at the end of 5 years for one year at 10% simple interest. If the amounts received by both at the end of 6 years are same, then the initial investment made by Sunil, in rupees, is
Anil invests ₹22000 for 6 years at 4% p.a., compounded half-yearly.
Rate per half-year = 4% ÷ 2 = 2%
Number of half-years = 6 × 2 = 12
Amount received by Anil = 22000 × (1.02)^12
Let Sunil's initial investment be ₹P.
He invests for 5 years in the same scheme.
Number of half-years = 5 × 2 = 10
Amount after 5 years = P × (1.02)^10
He then reinvests this amount for 1 year at 10% simple interest.
Final amount = P × (1.02)^10 × 1.10
Since both receive the same amount after 6 years,
22000 × (1.02)^12 = P × (1.02)^10 × 1.10
Cancel (1.02)^10 from both sides, 22000 × (1.02)^2 = 1.10P
Now, (1.02)^2 = 1.0404
Therefore, 22000 × 1.0404 = 1.10P → 22888.8 = 1.10P → P = 22888.8 ÷ 1.10 → P = 20808
Answer: 20808
The amount of job that Amal, Sunil and Kamal can individually do in a day, are in harmonic progression. Kamal takes twice as much time as Amal to do the same amount of job. If Amal and Sunil work for 4 days and 9 days, respectively, Kamal needs to work for 16 days to finish the remaining job. Then the number of days Sunil will take to finish the job working alone, is
Let the amount of work done per day by Amal, Sunil and Kamal be A, S and K respectively.
Since the work rates are in Harmonic Progression (HP), their reciprocals are in Arithmetic Progression (AP).
Kamal takes twice as much time as Amal.
Therefore, K = A/2
Taking reciprocals, 1/K = 2/A
Since 1/A, 1/S, 1/K are in AP,
2/S = (1/A + 1/K) = (1/A + 2/
A) = 3/A
Hence, S = 2A/3
Now use the work done.
Amal works for 4 days. Work done = 4A
Sunil works for 9 days. Work done = 9 × (2A/3) = 6A
Kamal works for 16 days. Work done = 16 × (A/2) = 8A
Together they complete the entire job.
So, 4A + 6A + 8A = 1 → 18A = 1 → A = 1/18
Therefore, S = 2A/3 = 2/3 × 1/18 = 1/27
Thus, Sunil alone completes the work in 1 ÷ (1/27) = 27 days
Answer: 27
Let C be the circle x² + y² + 4x − 6y − 3 = 0 and L be the locus of the point of intersection of a pair of tangents to C with the angle between the two tangents equal to 60°. Then, the point at which L touches the line x = 6 is
(6, 6)
(6, 3)
(6, 8)
(6, 4)
(6, 3)
Given, x² + y² + 4x − 6y − 3 = 0
Complete the squares, (x + 2)² + (y − 3)² = 16
Hence,
● Centre of the circle C = (−2, 3)
● Radius = 4
Let P be the point of intersection of the two tangents.
Suppose the distance of P from the centre is d.
In the right triangle formed by the centre, the point of contact, and P,
sin(angle between OP and a tangent) = Radius / OP = 4/d
The angle between the two tangents is 2 × angle between OP and one tangent
Given, 2 × angle = 60°
So, angle = 30°
Therefore, sin 30° = 4/d → 1/2 = 4/d → d = 8
Hence, the locus L is the circle
Centre = (−2, 3)
Radius = 8
Its equation is (x + 2)² + (y − 3)² = 64
The line x = 6 is vertical.
The circle touches this line at its rightmost point.
Rightmost point = (Centre x-coordinate + Radius, Centre y-coordinate) = (−2 + 8, 3) = (6, 3)
Answer:
B. (6, 3)
A quadrilateral ABCD is inscribed in a circle such that AB : CD = 2 : 1 and BC : AD = 5 : 4. If AC and BD intersect at the point E, then AE : CE equals
2 : 1
1 : 2
8 : 5
5 : 8
8 : 5
For a cyclic quadrilateral, if the diagonals intersect at E, then
AE/CE = (AB × AD)/(CB × CD)
(This is a standard result obtained using similar triangles.)
Given, AB : CD = 2 : 1, BC : AD = 5 : 4
Let, AB = 2k, CD = k, BC = 5m, AD = 4m
Now, AE/CE = (AB × AD)/(BC × CD) = (2k × 4m)/(5m × k) = 8/5
Therefore, AE : CE = 8 : 5
Answer:
C. 8 : 5
In a right-angled triangle ΔABC, the altitude AB is 5 cm, and the base BC is 12 cm. P and Q are two points on BC such that the areas of △ABP, △ABQ and △ABC are in arithmetic progression. If the area of ΔABC is 1.5 times the area of △ABP, the length of PQ, in cm, is
Given,
● Altitude AB = 5 cm
● Base BC = 12 cm
Area of △ABC = (1/2) × 12 × 5 = 30 cm²
Given, Area of △ABC = 1.5 × Area of △ABP
So, Area of △ABP = 30 ÷ 1.5 = 20 cm²
Let the area of △ABQ be x.
Since the areas of △ABP, △ABQ and △ABC are in arithmetic progression, 20, x, 30 are in AP.
Therefore, 2x = 20 + 30 → 2x = 50 → x = 25 cm²
Now, Area of △ABP = (1/2) × AB × BP → 20 = (1/2) × 5 × BP → 20 = 2.5 × BP → BP = 8 cm
Similarly, Area of △ABQ = (1/2) × 5 × BQ → 25 = 2.5 × BQ → BQ = 10 cm
Hence, PQ = BQ − BP = 10 − 8 = 2 cm
Answer: 2
For some positive and distinct real numbers x, y and z, if 1/(√y + √z) is the arithmetic mean of 1/(√x + √z) and 1/(√x + √y), then the relationship which will always hold true, is
√x, √z and √y are in arithmetic progression
y, x and z are in arithmetic progression
x, y and z are in arithmetic progression
√x, √y and √z are in arithmetic progression
y, x and z are in arithmetic progression
Let √x = a, √y = b, √z = c, where a, b and c are positive and distinct.
The given condition becomes
1/(b + c) = (1/2) [1/(a + c) + 1/(a + b)]
Multiply both sides by 2, 2/(b + c) = 1/(a + c) + 1/(a + b)
Taking LCM on the RHS, 2/(b + c) = [(a + b) + (a + c)] / [(a + c)(a + b)] = (2a + b + c) / [(a + b)(a + c)]
Cross-multiply, 2(a + b)(a + c) = (b + c)(2a + b + c)
Expand both sides, 2(a² + ab + ac + bc) = 2ab + 2ac + b² + 2bc + c²
Cancel the common terms, 2a² = b² + c²
Since a² = x, b² = y, c² = z, we get 2x = y + z or y + z = 2x
Hence, y, x and z are in arithmetic progression.
Answer:
B. y, x and z are in arithmetic progression
The number of all natural numbers up to 1000 with non-repeating digits is
504
648
738
585
738
We count the natural numbers from 1 to 1000 whose digits do not repeat.
1-digit numbers: These are 1 to 9. Count = 9
2-digit numbers: The first digit can be chosen in 9 ways (1 to 9). The second digit can be chosen in 9 ways (0 to 9 except the first digit). Total = 9 × 9 = 81
3-digit numbers: The first digit can be chosen in 9 ways (1 to 9). The second digit can be chosen in 9 ways (all digits except the first). The third digit can be chosen in 8 ways (excluding the first two digits). Total = 9 × 9 × 8 = 648
Number 1000: 1000 has repeated digit 0. Hence, it is not counted.
Total count = 9 + 81 + 648 = 738
Answer:
C. 738
A lab experiment measures the number of organisms at 8 am every day. Starting with 2 organisms on the first day, the number of organisms on any day is equal to 3 more than twice the number on the previous day. If the number of organisms on the nth day exceeds one million, then the lowest possible value of n is
Let the number of organisms on the nth day be T.
Given, T₁ = 2 and T = 2T₋₁ + 3
Add 3 to both sides. T + 3 = 2(T₋₁ + 3)
Let, U = T + 3
Then, U = 2U₋₁
Also, U₁ = 2 + 3 = 5
Hence, U = 5 × 2ⁿ⁻¹
Therefore, T = 5 × 2ⁿ⁻¹ − 3
We need T > 1,000,000
So, 5 × 2ⁿ⁻¹ − 3 > 1,000,000 → 5 × 2ⁿ⁻¹ > 1,000,003 → 2ⁿ⁻¹ > 200,000.6
Now, 2¹⁷ = 131072, 2¹⁸ = 262144
Since 200,000.6 lies between these two, the smallest possible value is n − 1 = 18
Therefore, n = 19
Answer: 19
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