CAT 2024 — Slot 3
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VARC topics covered: Culture & Identity RC, Economics & Society RC, Odd One Out, Para Completion, Para Summary, Science And Technology RC
Fears of artificial intelligence (AI) have haunted humanity since the very beginning of the computer age. Hitherto, these fears focused on machines using physical means to kill, enslave or replace people. But over the past couple of years, new AI tools have emerged that threaten the survival of human civilisation from an unexpected direction. AI has gained some remarkable abilities to manipulate and generate language, whether with words, sounds or images. AI has thereby hacked the operating system of our civilisation.
Language is the stuff almost all human culture is made of. Human rights, for example, aren't inscribed in our DNA. Rather, they are cultural artefacts we created by telling stories and writing laws. Gods aren't physical realities. Rather, they are cultural artefacts we created by inventing myths and writing scriptures….What would happen once a non-human intelligence becomes better than the average human at telling stories, composing melodies, drawing images, and writing laws and scriptures? When people think about Chatgpt and other new AI tools, they are often drawn to examples like schoolchildren using AI to write their essays. What will happen to the school system when kids do that? But this kind of question misses the big picture. Forget about school essays. Think of the next American presidential race in 2024, and try to imagine the impact of AI tools that can be made to mass-produce political content, fake news stories and scriptures for new cults…
Through its mastery of language, AI could even form intimate relationships with people, and use the power of intimacy to change our opinions and worldviews. Although there is no indication that AI has any consciousness or feelings of its own, to foster fake intimacy with humans, it is enough if the AI can make them feel emotionally attached to it….
What will happen to the course of history when AI takes over culture, and begins producing stories, melodies, laws and religions? Previous tools like the printing press and radio helped spread the cultural ideas of humans, but they never created new cultural ideas of their own. AI is fundamentally different. AI can create completely new ideas, completely new culture….Of course, the new power of AI could be used for good purposes as well. I won't dwell on this because the people who develop AI talk about it enough….
We can still regulate the new AI tools, but we must act quickly. Whereas nukes cannot invent more powerful nukes, AI can make exponentially more powerful AI.… Unregulated AI deployments would create social chaos, which would benefit autocrats and ruin democracies. Democracy is a conversation, and conversations rely on language. When AI hacks language, it could destroy our ability to have meaningful conversations, thereby destroying democracy …. And the first regulation I would suggest is to make it mandatory for AI to disclose that it is an AI. If I am having a conversation with someone, and I cannot tell whether it is a human or an AI—that's the end of democracy. This text has been generated by a human. Or has it?
The author identifies all of the following as dire outcomes of the capture of language by AI EXCEPT that it could
spawn a completely new culture through its ability to create new ideas and opinions.
out-strip human creativity and endeavours in the spheres such as art and music and, in the formulation of laws.
eventually subvert democratic processes through the mass creation and spread of fake political content and news.
apply its mastery of language to create strong emotional ties which could exacerbate the polarization of political views.
apply its mastery of language to create strong emotional ties which could exacerbate the polarization of political views.
The author identifies the polarisation of political views as a danger, but not through AI creating strong emotional ties. The passage says AI could form intimate relationships and change opinions and worldviews through the power of intimacy, but it does not specifically claim this will exacerbate political polarisation. The author's concern is about AI changing views through fake intimacy, not about the polarisation that might result from it, making this a subtle step beyond what the passage actually states.
The author terms language "the operating system of our civilization" for all the following reasons EXCEPT that it
can influence political views and opinions as it engenders close emotional ties among people.
is the basis of AI tools like ChatGPT which can be used to generate academic content and opinion.
is fundamental to the articulation and spread of human values and culture in our society.
has laid the foundation for the creation of cultural artefacts through writing and telling of stories.
is the basis of AI tools like ChatGPT which can be used to generate academic content and opinion.
The author calls language the operating system of civilisation because it is the medium through which human values, culture, laws, religious beliefs and social institutions are created and transmitted. ChatGPT and AI tools are the threats to this operating system, not part of the reason language is called the operating system, so its use in generating academic content is not a reason for the metaphor.
We can infer that the author is most likely to agree with which of the following statements?
People's fears of the dangers of students using ChatGPT and other new AI tools are unfounded.
The commonly expressed fear that future AI developments will fatally harm humans is unfounded.
Apart from its drawbacks, AI tools have been beneficial in boosting technological and industrial advance worldwide.
One of the biggest casualties from the spread of unregulated AI is likely to be the democratic process.
One of the biggest casualties from the spread of unregulated AI is likely to be the democratic process.
The passage explicitly argues that unregulated AI would create social chaos benefiting autocrats and ruining democracies, and that AI hacking language could destroy meaningful conversation and thereby destroy democracy. This is the most clearly stated and developed concern in the passage.
The tone of the passage could best be described as
cautionary, because the author lays out some adverse effects of the proliferation of unregulated AI tools.
prescient, as the author analyses the future impact of the use of new AI tools on crucial areas of our society and culture.
alarmist, because the passage discusses scenarios of the influence of new AI tools on language and human emotions.
quizzical, as the passage poses several questions, concluding with the question of whether or not the passage content has been generated by AI.
cautionary, because the author lays out some adverse effects of the proliferation of unregulated AI tools.
The passage methodically identifies serious risks from AI, mass production of fake political content, AI forming fake intimate relationships to manipulate opinions, destruction of democratic processes, and the need for urgent regulation. The tone is a warning about specific adverse effects, which is cautionary.
Moutai has been the global booze sensation of the decade. A bottle of its Flying Fairy, which sold in the 1980s for the equivalent of a dollar, now retails for $400. Moutai's listed shares have soared by almost 600% in the past five years, outpacing the likes of Amazon ...
It does this while disregarding every Western marketing mantra. It is not global, has meagre digital sales and does not appeal to millennials. It scores pitifully on environmental, social and government measures. In the Boy Scout world of Western business, it would leave a bad taste in more ways than one.
Moutai owes its intoxicating success to three factors—not all of them easy to emulate. First, it profits from Chinese nationalism. Moutai is known as the "national liquor". It was used to raise spirits and disinfect wounds in Mao's Long March. It was Premier Zhou Enlai's favourite tipple, shared with Richard Nixon in 1972. Its centuries-old craftsmanship—it is distilled eight times and stored for years in earthenware jars—is a source of national pride. It also claims to be hangover-proof, which would make it an invention to rival gunpowder ...
Second, it chose to serve China's super-rich rather than its middle class. Markets are littered with the corpses of firms that could not compete in the cut-throat battle for Chinese middle-class wallets. And the country's premium market is massive—at 73m-strong, bigger than the population of France, notes Euan McLeish of Bernstein, an investment firm, and still less crowded with prestige brands than advanced economies. Moutai is to these well-heeled drinkers what vintage champagne is to the rest of the world ...
Third, Moutai looks beyond affluent millennials and digital natives. The elderly and the middle-aged, it found, can be just as lucrative. Its biggest market now is (male) drinkers in their mid-30s. Many have no siblings, thanks to four decades of China's one-child policy—which also means their elderly parents can splash out on weddings and banquets. Moutai is often a guest of honour.
Moutai has succeeded thanks to nationalism, elitism and ageism, in other words—not in spite of this unholy trinity. But it faces risks. The government is its largest shareholder—and a meddlesome one. It appears to want prices to remain stable. Exorbitantly priced booze is at odds with its professed socialist ideals. Yet minority investors—including many foreign funds—lament that Moutai's wholesale price is a third of what it sells for in shops. Raising it could boost the company's profits further. Instead, in what some see as a travesty of corporate governance, its majority owner has plans to set up its own sales channel ...
In the long run, its biggest risk may be millennials. As they grow older, health concerns, work-life balance and the desire for more wholesome pursuits than binge-drinking may curb the "Ganbei!" toasting culture [heavy drinking] on which so much of the demand for Moutai rests. For the time being, though, the party goes on.
The phrase "would make it an invention to rival gunpowder" has been used in the passage in a sense that is
literal
substantive
metaphorical
synonymical
metaphorical
The claim that being hangover-proof would make Moutai "an invention to rival gunpowder" is clearly not meant literally, since a hangover cure is incomparable in historical or scientific significance to gunpowder. The phrase is used humorously to suggest that if Moutai truly were hangover-proof, it would be seen as a great Chinese invention, playing on national pride since gunpowder is one of China's most celebrated historical inventions.
Which one of the following is both a reason for Moutai's success as well as a possible threat to that success?
Chinese love of liquor filled celebration.
Government involvement in its business.
Its appeal to the rich.
Its appeal to the older age group.
Its appeal to the older age group.
Moutai's appeal to older age groups, those in their mid-30s and beyond, is presented as one of its key success factors since it looked beyond millennials and digital natives. However, the passage also identifies millennials as Moutai's biggest long-term risk, since as they age, health concerns and changing social habits may reduce the heavy drinking culture on which Moutai's demand depends. The appeal to older drinkers is both the current success factor and the implicit vulnerability as today's younger generation ages differently.
In the context of the passage, it is most likely that the author refers to Moutai's marketing strategy as "the unholy trinity" because
there is nothing holy about marketing techniques for liquor.
it profits from Chinese nationalist feelings.
it contradicts the Western strategy of marketing.
it exposes the firm to long term risks.
it contradicts the Western strategy of marketing.
The passage calls nationalism, elitism and ageism Moutai's "unholy trinity" because together they form a strategy that contradicts every Western marketing principle, ignoring millennials, digital sales, global reach and environmental responsibility, yet succeeds wildly in China. The phrase highlights the deliberate inversion of Western marketing orthodoxy.
In the context of the passage, we can infer that to succeed in the liquor industry in China, a marketing firm must consider all of the following factors affecting the Chinese liquor market EXCEPT that
there is money to be made from marketing to the middle class.
the government may control the pricing of products.
there are few competitors to meet the demands of high end liquor consumers.
the competition for winning over the middle class is very stiff.
there is money to be made from marketing to the middle class.
The passage explicitly states that markets are littered with the corpses of firms that could not compete in the cut-throat battle for Chinese middle-class wallets, directly suggesting that there is no money to be made from marketing to the middle class in China's competitive environment. So a firm must consider that the middle class market is extremely difficult and not necessarily profitable.
Languages become endangered and die out for many reasons. Sadly, the physical annihilation of communities of native speakers of a language is all too often the cause of language extinction. In North America, European colonists brought death and destruction to many Native American communities. This was followed by US federal policies restricting the use of indigenous languages, including the removal of native children from their communities to federal boarding schools where native languages and cultural practices were prohibited. As many as 75 percent of the languages spoken in the territories that became the United States have gone extinct, with slightly better language survival rates in Central and South America ...
Even without physical annihilation and prohibitions against language use, the language of the "dominant" cultures may drive other languages into extinction; young people see education, jobs, culture and technology associated with the dominant language and focus their attention on that language. The largest language "killers" are English, Spanish, Portuguese, French, Russian, Hindi, and Chinese, all of which have privileged status as dominant languages threatening minority languages.
When we lose a language, we lose the worldview, culture and knowledge of the people who spoke it, constituting a loss to all humanity. People around the world live in direct contact with their native environment, their habitat. When the language they speak goes extinct, the rest of humanity loses their knowledge of that environment, their wisdom about the relationship between local plants and illness, their philosophical and religious beliefs, as well as their native cultural expression (in music, visual art and poetry) that has enriched both the speakers of that language and others who would have encountered that culture ...
As educators deeply immersed in the liberal arts, we believe that educating students broadly in all facets of language and culture ... yields immense rewards. Some individuals educated in the liberal arts tradition will pursue advanced study in linguistics and become actively engaged in language preservation, setting out for the Amazon, for example, with video recording equipment to interview the last surviving elders in a community to record and document a language spoken by no children.
Certainly, though, the vast majority of students will not pursue this kind of activity. For these students, a liberal arts education is absolutely critical from the twin perspectives of language extinction and global citizenship. When students study languages other than their own, they are sensitized to the existence of different cultural perspectives and practices. With such an education, students are more likely to be able to articulate insights into their own cultural biases, be more empathetic to individuals of other cultures, communicate successfully across linguistic and cultural differences, consider and resolve questions in a way that reflects multiple cultural perspectives, and, ultimately extend support to people, programs, practices, and policies that support the preservation of endangered languages.
There is ample evidence that such preservation can work in languages spiraling toward extinction. For example, Navajo, Cree, and Inuit communities have established schools in which these languages are the language of instruction, and the number of speakers of each has increased.
In the context of the passage, which one of the following hypothetical scenarios, if true, is NOT an example of the kind of loss that occurs when a language becomes extinct?
The Nicobarese language describes 20 different moods of the ocean. By the time the last speaker is educated in a Central Board school, they will have forgotten their language.
The Lamkangs of Manipur have only 3 remaining native speakers of the language. When they die, we will lose one more group from the government list of indigenous tribes.
The Andamanese language has a word to describe someone who has lost a step-sister. When the language dies, we will lose the concept of the word and the emotions it evokes.
The Inuits of Alaska have 35 different words to describe the texture of snow. When the language becomes extinct, we will lose that understanding of nature.
The Lamkangs of Manipur have only 3 remaining native speakers of the language. When they die, we will lose one more group from the government list of indigenous tribes.
The passage describes the loss of a language as a loss of worldview, culture, environmental knowledge, philosophical and religious beliefs, and cultural expression. Option B describes the loss of the last speakers of Lamkangs and the removal of one group from the government list of indigenous tribes, which is an administrative loss, not a loss of knowledge, culture or worldview. Being removed from a government list is a bureaucratic consequence, not the kind of cultural or epistemic loss the passage is concerned with.
Which one of the following hypothetical scenarios, if true, would most strongly undermine the central ideas of the passage?
Most liberal arts students will pursue jobs in publishing and human resource management rather than doctorates in linguistics.
A liberal arts education requires that, in addition to being fluent in English, students gain fluency in two of the top five most spoken languages globally.
Schools that teach endangered languages can preserve the language only for a generation.
Recording a dying language that has only a few remaining speakers freezes it in time: it stops evolving further.
A liberal arts education requires that, in addition to being fluent in English, students gain fluency in two of the top five most spoken languages globally.
The passage argues that a liberal arts education sensitises students to cultural differences, helps them support endangered languages, and produces some specialists who document dying languages. If a liberal arts education only required fluency in the top five most spoken global languages, this would make students more proficient in the very dominant languages the passage identifies as the biggest killers of minority languages, directly contradicting the passage's argument that such an education supports diversity and preservation.
It can be inferred from the passage that it is likely South America had a slightly better language survival rate than North America for all of the following reasons EXCEPT:
European colonists allowed children of native speakers to stay at home with their families.
the colonial government was unable to mainstream the locals.
not many native speakers were killed by European colonists.
locals were provided job opportunities in the colonial administration.
locals were provided job opportunities in the colonial administration.
The passage states that South America had a slightly better language survival rate than North America, and it attributes this to differences in the colonial treatment of native populations, specifically the physical annihilation and language restrictions seen in North America. The passage does not mention job opportunities in the colonial administration as a reason for better survival rates in South America.
The author believes that a liberal arts education combined with participation in language preservation empower students in all of the following ways EXCEPT that they will
overcome cultural barriers to communication.
learn different languages.
establish schools to preserve languages spiralling towards extinction.
develop a better understanding of their own culture.
establish schools to preserve languages spiralling towards extinction.
The passage describes several ways a liberal arts education empowers students, articulating cultural biases, empathising with other cultures, communicating across linguistic differences, and supporting policies for language preservation. It also separately mentions that some specialists will set up recording expeditions. However, establishing schools to preserve endangered languages is presented as an example of what preservation efforts look like, specifically the Navajo, Cree and Inuit communities, not as something a liberal arts education equips students to do.
There is a group in the space community who view the solar system not as an opportunity to expand human potential but as a nature preserve, forever the provenance of an elite group of scientists and their sanitary robotic probes. These planetary protection advocates [call] for avoiding "harmful contamination" of celestial bodies. Under this regime, NASA incurs great expense sterilizing robotic probes in order to prevent the contamination of entirely theoretical biospheres ...
Transporting bacteria would matter if Mars were the vital world once imagined by astronomers who mistook optical illusions for canals. Nobody wants to expose Martians to measles, but sadly, robotic exploration reveals a bleak, rusted landscape, lacking oxygen and flooded with radiation ready to sterilize any Earthly microbes. Simple life might exist underground, or down at the bottom of a deep canyon, but it has been very hard to find with robots. . . . The upsides from human exploration and development of Mars clearly outweigh the welfare of purely speculative Martian fungi ...
The other likely targets of human exploration, development, and settlement, our moon and the asteroids, exist in a desiccated, radiation-soaked realm of hard vacuum and extreme temperature variations that would kill nearly anything. It's also important to note that many international competitors will ignore the demands of these protection extremists in any case. For example, China recently sent a terrarium to the moon and germinated a plant seed—with, unsurprisingly, no protest from its own scientific community. In contrast, when it was recently revealed that a researcher had surreptitiously smuggled super-resilient microscopic tardigrades aboard the ill-fated Israeli Beresheet lunar probe, a firestorm was unleashed within the space community ...
NASA's previous human exploration efforts made no serious attempt at sterility, with little notice. As the Mars expert Robert Zubrin noted in the National Review, U.S. lunar landings did not leave the campsites cleaner than they found it. Apollo's bacteria-infested litter included bags of feces. Forcing NASA's proposed Mars exploration to do better, scrubbing everything and hauling out all the trash, would destroy NASA's human exploration budget and encroach on the agency's other directorates, too. Getting future astronauts off Mars is enough of a challenge, without trying to tote weeks of waste along as well.
A reasonable compromise is to continue on the course laid out by the U.S. government and the National Research Council, which proposed a system of zones on Mars, some for science only, some for habitation, and some for resource exploitation. This approach minimizes contamination, maximizes scientific exploration ... Mars presents a stark choice of diverging human futures. We can turn inward, pursuing ever more limited futures while we await whichever natural or manmade disaster will eradicate our species and life on Earth. Alternatively, we can choose to propel our biosphere further into the solar system, simultaneously protecting our home planet and providing a backup plan for the only life we know exists in the universe. Are the lives on Earth worth less than some hypothetical microbe lurking under Martian rocks?
The author is unlikely to disagree with any of the following EXCEPT:
the proposal for a zonal segregation of the Martian landscape into regions for different purposes.
that while NASA's earlier missions were not ideal in their approach to space contamination, they likely did no grave damage.
space contamination should be minimised until the possibility of life on the astronomical body being explored is ruled out.
the exorbitant costs of continuing to keep the space environment pristine may be unsustainable.
space contamination should be minimised until the possibility of life on the astronomical body being explored is ruled out.
The author's entire argument is that planetary protection concerns are overblown given the bleak, radiation-flooded Martian environment where life is hard to find, and that the costs of sterilisation are prohibitive. The author would strongly disagree with the idea that space contamination should be minimised until life is ruled out, since this is precisely the position of the "planetary protection advocates" the author is critiquing.
The author mentions all of the following reasons to dismiss concerns about contaminating Mars EXCEPT:
the lack of evidence of living organisms on Mars makes possible contamination from earthly microbes a moot point.
efforts to contain contamination on Mars are likely to be derailed as competitor countries may not follow similar restrictions.
the use of similar probes on astronomical bodies like the moon have had little effect on the environment.
earlier explorations have already contaminated pristine space environments.
the use of similar probes on astronomical bodies like the moon have had little effect on the environment.
The author dismisses contamination concerns by arguing that Mars is hostile to earthly microbes, that competitors like China will ignore restrictions anyway, and that previous missions have already contaminated space environments including the moon. However, the author does not argue that using probes on the moon has had little environmental effect as a reason to dismiss Mars concerns. The moon example is used to show the hypocrisy of the scientific community's reactions to different countries' contaminations, not to argue that probe use has limited environmental impact.
The author's overall tone in the first paragraph can be described as
sceptical about the excessive efforts to sanitise planets where life has not yet been proven to exist.
equivocal about the reasons extended by the group of scientists seeking to limit space exploration.
indifferent to the elitism of a few scientists aiming to corner space exploration.
approving of the amount of money NASA spends to restrict the spread of contamination in space.
sceptical about the excessive efforts to sanitise planets where life has not yet been proven to exist.
The first paragraph describes planetary protection advocates spending large sums sterilising probes to prevent contamination of "entirely theoretical biospheres," which immediately signals the author's view that these efforts are excessive and unwarranted given the speculative nature of the life being protected. The tone is dismissive and incredulous toward the scale of precaution being taken.
The contrasting reactions to the Chinese and Israeli "contaminations" of lunar space
are valid as the contamination of the lunar environment from animal sources is far greater than from plants.
are evidence of China's reasonable approach towards space contamination.
indicate that national scientists may have different sensitivities to issues of biosphere protection.
reveal global biases prevalent in attitudes towards different countries.
indicate that national scientists may have different sensitivities to issues of biosphere protection.
The passage describes how China germinating a plant seed on the moon drew no protest from its scientific community, while a researcher smuggling tardigrades aboard an Israeli lunar probe caused a firestorm. The author presents this contrast without resolving it and notes only that the reactions differed. This suggests that different national scientific communities may have different sensitivities to biosphere protection issues, since the same act of biological introduction to the moon generated radically different reactions depending on the country involved.
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 reality is putting stress on employees who have to pay for transport, desk lunches, more childcare, clothing and that after-work socialisation - costs they haven't incurred for nearly two years.
Paragraph: (1). Prices are rising at their fastest rate in 40 years; consequently, return-to-office-related costs have shot up - think petrol and food, for instance.(2). Yet wages haven't kept up with inflation - even despite the salary growth many workers have enjoyed during a favourable pandemic labour market. (3).This is especially jarring for workers who were able to save during remote work, when these expenditures weren't a factor. (4). In April 2022, Umus, a London university lecturer, told BBC Worklife that they were spending nearly a quarter of what they made every day on return-to-work costs.
Option 4
Option 3
Option 2
Option 1
Option 3
The inserted sentence describes the specific costs employees face when returning to the office, transport, desk lunches, childcare, clothing and after-work socialisation. Option 3 falls right after the statement that wages haven't kept up with inflation, and right before the observation that this is especially jarring for workers who were able to save during remote work. The inserted sentence bridges these two points by naming the exact costs creating the stress, which then sets up the contrast with the remote work period when these costs didn't exist.
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: Many have had to leave their homes behind, with more than 1.3 million people being displaced due to the drought.
Paragraph: Somalia has been dealing with an enormous humanitarian catastrophe, driven by the longest and most severe drought the country has experienced in at least 40 years. (1). Five consecutive rainy seasons have failed, causing more than 8 million people - almost half of the country's population - to experience acute food insecurity. (2). More than 43,000 people are believed to have lost their lives, with half of the lives lost likely being children under five. The damage the drought has caused is far-reaching. (3). Farmers have lost all their agricultural income, while pastoralists have lost more than 3 million livestock, impoverishing entire communities, and leaving them on the brink of famine. (4). Some, like the pastoralists, may never be able to go back as their livelihoods have been irreversibly wiped out.
Option 4
Option 2
Option 3
Option 1
Option 4
The inserted sentence about 1.3 million people being displaced fits at Option 4, right after the paragraph has described farmers losing agricultural income and pastoralists losing livestock. The very next sentence then says some may never be able to return as their livelihoods have been irreversibly wiped out. The inserted sentence about displacement introduces this group of displaced people, and the following sentence's reference to "some like the pastoralists" who may never go back directly refers back to them.
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. Part of the appeal of forecasting is not just that it seems to work, but that you don't seem to need specialized expertise to succeed at it.
2. The tight connection between forecasting and building a model of the world helps explain why so much of the early interest in the idea came from the intelligence community.
3. This was true even though the latter had access to classified intelligence.
4. One frequently cited study found that accurate forecasters' predictions of geopolitical events, when aggregated using standard scientific methods, were more accurate than the forecasts of members of the US intelligence community who answered the same questions in a confidential prediction market.
5. The aggregated opinions of non-experts doing forecasting have proven to be a better guide to the future than the aggregated opinions of experts.
Sentences 1, 3, 4 and 5 form a coherent argument about forecasting, starting with the appeal of forecasting not requiring specialised expertise, then the study showing aggregated non-expert predictions were more accurate than US intelligence community forecasts, then the qualifier that this held even though intelligence members had access to classified information, and finally the general principle that aggregated non-expert opinions outperform expert ones. Sentence 2 introduces the connection between forecasting and building a model of the world, and explains early interest from the intelligence community, which is a separate historical observation about forecasting's origins rather than a continuation of the argument about non-expert accuracy.
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: Taken outside the village of Trang Bang on June 8, 1972, the picture captured the trauma and indiscriminate violence of a conflict that claimed, by some estimates, a million or more civilian lives.
Paragraph: The horrifying photograph of children fleeing a deadly napalm attack has become a defining image not only of the Vietnam War but the 20th century. (1). Dark smoke billowing behind them, the young subjects' faces are painted with a mixture of terror, pain and confusion. (2). Soldiers from the South Vietnamese Army's 25th Division follow helplessly behind. (3). The picture was officially titled "The Terror of War," but the photo is better known by the nickname given to the naked 9-year-old at its centre: "Napalm Girl". (4).
Option 4
Option 1
Option 3
Option 2
Option 3
The inserted sentence provides the location, date and historical context for the photograph, explaining where and when it was taken and noting the broader scale of civilian casualties in the Vietnam War. Option 3 falls right after the description of soldiers following helplessly behind the fleeing children, and right before the sentence about the photograph's official title and nickname. The contextual information about where and when the photograph was taken fits naturally between the visual description of the subjects and the naming of the photograph.
The passage given below is followed by four alternate summaries. Choose the option that best captures the essence of the passage.
When the tradwife puts on that georgic, pinstriped dress, she is not just admiring the visual cues of a fantastical past. She takes these dreams of storybook bliss literally, tracing them backward in time until she reaches a logical conclusion that satisfies her. And by doing so, she ends up delivering an unhappy reminder of just how much our lives consist of artifice and playacting. The tradwife outrages people because of her deliberately regressive ideals. And yet her behaviour is, on some level, indistinguishable from the non-tradwife's. The tradwife's trollish genius is to beat us at our own dress-up game. By insisting that the idyllic cottage daydream should be real, right down to the primitive gender roles, she leaves others feeling hollow, cheated. The hullabaloo and headaches she causes may be the price we pay for taking too many things at face value: our just deserts, served Instagram-perfect by a manicured hand on a gorgeous ceramic dish, with fat, mouthwatering maraschino cherries on top.
The tradwife's commitment to outdated gender roles and retro fashion critiques the superficiality of today's societal ideals.
By promoting an idealized past, the tradwife exposes the artifice of contemporary values and mocks societal norms.
The tradwife, with her vintage dress and traditional roles, highlights the superficiality of modern life and challenges current societal norms.
The tradwife's vintage dress and adherence to traditional roles reveal the artificial nature of modern life and its superficial values.
The tradwife, with her vintage dress and traditional roles, highlights the superficiality of modern life and challenges current societal norms.
The passage argues that the tradwife, by insisting on making her idyllic domestic fantasy real, exposes how much of modern life is artifice and playacting, leaving others feeling hollow because she beats them at their own game of dress-up. The key points are the tradwife's vintage aesthetic, her traditional gender roles, her exposure of modern superficiality, and her challenge to contemporary norms.
The passage given below is followed by four alternate summaries. Choose the option that best captures the essence of the passage.
Lyric poetry is a genre of private meditation rather than public commitment. The impulse in Marxism toward changing a society deemed unacceptable in its basic design would seem to place demands on lyric poetry that such poetry, with its tendency toward the personal, the small scale, and the idiosyncratic, could never answer. There is within Marxism, however, also a strand of thought that would locate in lyric poetry alternative modes of perception and description that call forth a vision of worlds at odds with a repressive reality or that draw attention to the workings of ideology within the hegemonic culture. The poetic imagination may indeed deflect larger social concerns, but it may also be implicitly critical and utopian.
The focus of lyric poetry is largely personal while that of Marxism is bringing change in society. Unless the difference is resolved, poetry will remain largely utopian.
Marxism has internal contradictions due to which one strand of Marxism sees no merit in lyric poetry while another appreciates the alternative modes of perception in poetry.
The focus of lyric poetry as personal may not seem compatible with Marxism. However, it is possible to envisage lyric poetry as a symbol of resistance against an oppressive culture.
Marxism makes unreasonable demands on lyric poetry. However, lyric poetry has its own merits that are largely ignored by Marxism due to its personal nature.
The focus of lyric poetry as personal may not seem compatible with Marxism. However, it is possible to envisage lyric poetry as a symbol of resistance against an oppressive culture.
The passage argues that lyric poetry, while focused on the personal and idiosyncratic, can nonetheless serve as a vehicle for implicit social critique and utopian vision within a Marxist framework, since some Marxist thought finds in poetry alternative modes of perception that resist repressive reality.
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. To create a synapse, the neuron has specialized structures, often seen as tiny swellings, at its terminal end of the axon where it stores the chemicals that are emitted to transmit a signal to the next neuron.
2. This fetal warm-up act—the soldering of neural connections before the eyes actually function—is crucial to the performance of the visual system.
3. The reasons for this paring back of synapses is a mystery, but synaptic pruning is thought to sharpen and reinforce the "correct" synapses, while removing the weak and unnecessary ones.
4. Neural connections between the eyes and the brain are formed long before birth, establishing the wiring and the circuitry that allow a child to begin visualizing the world the minute she emerges from the womb.
5. During this rehearsal period, synapses—points of chemical connection—between nerve cells are generated in great excess, only to be pruned back during later development.
Sentences 2, 3, 4 and 5 build a coherent argument about how neural connections between the eyes and brain form before birth, how synapses are generated in excess during this period and then pruned back, how the reason for pruning is unknown but thought to sharpen correct connections, and why this prenatal rehearsal is crucial to the visual system's eventual functioning. Sentence 1 describes the general structure of a synapse, explaining the physical specialisation of the neuron at the axon's terminal end, which is background information about synaptic structure rather than a continuation of the specific developmental narrative about prenatal visual system wiring.
The passage given below is followed by four alternate summaries. Choose the option that best captures the essence of the passage.
Humans have managed to tweak the underlying biology of various plants and animals to produce high-tech crops and microbes. But regulating these entities is complicated, as the framework of policies and procedures are outdated and not flexible enough to adapt to emerging technology. The question is whether regulation will ever be able to keep up with human innovation, to regulate living things, which are apt to be unpredictable and unique; to capture all the potential risks when new biological entities are introduced, or when they pass on variations of their genes?
The mercurial nature of biological entities calls for scientists to shape the regulations governing emerging technology, with regular calibration to handle variations in the field.
A new framework of rules and procedures for regulating the most recent research emerging from biotechnology is urgently needed, to keep up with this rapidly changing discipline.
Current regulation of biotechnology is outdated, but it is debatable if we can create a framework, imaginative and flexible, to cover all contingencies in this fast-changing area.
The problem with formulating regulation for innovation in the scientific arena it that it is impossible to imagine the outcomes or risks related to the outcomes of all the research.
Current regulation of biotechnology is outdated, but it is debatable if we can create a framework, imaginative and flexible, to cover all contingencies in this fast-changing area.
The passage states that humans have tweaked biology to produce high-tech crops and microbes, that regulating these entities is complicated because existing frameworks are outdated and inflexible, and poses the open question of whether regulation can ever keep up with innovation given the unpredictability of living things.
Answer the questions on the basis of the information given below.
Over the top (OTT) subscribers of a platform are segregated into three categories: i) Kid, ii) Elder, and iii) Others.
Some of the subscribers used one app and the others used multiple apps to access the platform. The figure below shows the percentage of the total number of subscribers in 2023 and 2024 who belong to the 'Kid' and 'Elder' categories.
The following additional facts are known about the numbers of subscribers.
1. The total number of subscribers increased by 10% from 2023 to 2024.
2. In 2024, 1/2 of the subscribers from the 'Kid' category and 2/3 of the subscribers from the 'Elder' category subscribers use one app.
3. In 2023, the number of subscribers from the 'Kid' category who used multiple apps was the same as the number of subscribers from the 'Elder' category who used one app.
4. 10,000 subscribers from the 'Kid' category used one app and 15,000 subscribers from the 'Elder' category used multiple apps in 2023.
How many subscribers belonged to the 'Others' category in 2024?
45000
65000
55000
Cannot be determined
55000
Step 1:
From condition (4), the number of kids using one app in 2023 is 10,000, while the number of elders using multiple apps is 15,000.
According to condition (3), the number of elders using one app is equal to the number of kids using multiple apps. Let this common value be X.
From the graph, in 2023, kids account for 15% of the users and elders account for 20%. Therefore, the remaining 65% correspond to others.
Using this information,
(10000 + X) / (15000 + X) = 15 / 20
⇒ X = 5000
Hence,
- Kids (2023) = 10,000 + 5,000 = 15,000
- Elders (2023) = 5,000 + 15,000 = 20,000
- Others (2023) = 65,000
Thus, the total number of users in 2023 is:
15,000 + 20,000 + 65,000 = 100,000
From condition (1), the total number of users in 2024 is 10% higher than in 2023.
Therefore,
Total users in 2024 = 100,000 × 1.10 = 110,000
Using the bar graph, the category-wise totals for 2024 are:
- Kids = 22,000
- Elders = 33,000
- Others = 55,000
The complete values are:
2023
- Kids: One app = 10,000, Multiple apps = 5,000, Total = 15,000
- Elders: One app = 5,000, Multiple apps = 15,000, Total = 20,000
- Others: Total = 65,000
- Overall total = 100,000
2024
- Kids: One app = 11,000, Multiple apps = 11,000, Total = 22,000
- Elders: One app = 22,000, Multiple apps = 11,000, Total = 33,000
- Others: Total = 55,000
- Overall total = 110,000
In 2024 the number of people in others category = 55000.
What percentage of subscribers in the 'Kid' category used multiple apps in 2023?
33.33%
50.00%
5.00%
25.50%
33.33%
Step 1:
From condition (4), the number of kids using one app in 2023 is 10,000, while the number of elders using multiple apps is 15,000.
According to condition (3), the number of elders using one app is equal to the number of kids using multiple apps. Let this common value be X.
From the graph, in 2023, kids account for 15% of the users and elders account for 20%. Therefore, the remaining 65% correspond to others.
Using this information,
(10000 + X) / (15000 + X) = 15 / 20
⇒ X = 5000
Hence,
- Kids (2023) = 10,000 + 5,000 = 15,000
- Elders (2023) = 5,000 + 15,000 = 20,000
- Others (2023) = 65,000
Thus, the total number of users in 2023 is:
15,000 + 20,000 + 65,000 = 100,000
From condition (1), the total number of users in 2024 is 10% higher than in 2023.
Therefore,
Total users in 2024 = 100,000 × 1.10 = 110,000
Using the bar graph, the category-wise totals for 2024 are:
- Kids = 22,000
- Elders = 33,000
- Others = 55,000
The complete values are:
2023
- Kids: One app = 10,000, Multiple apps = 5,000, Total = 15,000
- Elders: One app = 5,000, Multiple apps = 15,000, Total = 20,000
- Others: Total = 65,000
- Overall total = 100,000
2024
- Kids: One app = 11,000, Multiple apps = 11,000, Total = 22,000
- Elders: One app = 22,000, Multiple apps = 11,000, Total = 33,000
- Others: Total = 55,000
- Overall total = 110,000
The percentage of kids using multiple apps in 2023 is:
= (5000 / 15000) × 100
= 33.33%
What was the percentage increase in the number of subscribers in the 'Elder' category from 2023 to 2024?
60%
50%
65%
40%
Step 1:
From condition (4), the number of kids using one app in 2023 is 10,000, while the number of elders using multiple apps is 15,000.
According to condition (3), the number of elders using one app is equal to the number of kids using multiple apps. Let this common value be X.
From the graph, in 2023, kids account for 15% of the users and elders account for 20%. Therefore, the remaining 65% correspond to others.
Using this information,
(10000 + X) / (15000 + X) = 15 / 20
⇒ X = 5000
Hence,
- Kids (2023) = 10,000 + 5,000 = 15,000
- Elders (2023) = 5,000 + 15,000 = 20,000
- Others (2023) = 65,000
Thus, the total number of users in 2023 is:
15,000 + 20,000 + 65,000 = 100,000
From condition (1), the total number of users in 2024 is 10% higher than in 2023.
Therefore,
Total users in 2024 = 100,000 × 1.10 = 110,000
Using the bar graph, the category-wise totals for 2024 are:
- Kids = 22,000
- Elders = 33,000
- Others = 55,000
The complete values are:
2023
- Kids: One app = 10,000, Multiple apps = 5,000, Total = 15,000
- Elders: One app = 5,000, Multiple apps = 15,000, Total = 20,000
- Others: Total = 65,000
- Overall total = 100,000
2024
- Kids: One app = 11,000, Multiple apps = 11,000, Total = 22,000
- Elders: One app = 22,000, Multiple apps = 11,000, Total = 33,000
- Others: Total = 55,000
- Overall total = 110,000
The percentage increase in the number of elders from 2023 to 2024 is:
= ((33,000 − 20,000) / 20,000) × 100
= 65%
What could be the minimum percentage of subscribers who used multiple apps in 2024?
20.0%
10.0%
16.5%
22.00%
20.0%
Step 1:
From condition (4), the number of kids using one app in 2023 is 10,000, while the number of elders using multiple apps is 15,000.
According to condition (3), the number of elders using one app is equal to the number of kids using multiple apps. Let this common value be X.
From the graph, in 2023, kids account for 15% of the users and elders account for 20%. Therefore, the remaining 65% correspond to others.
Using this information,
(10000 + X) / (15000 + X) = 15 / 20
⇒ X = 5000
Hence,
- Kids (2023) = 10,000 + 5,000 = 15,000
- Elders (2023) = 5,000 + 15,000 = 20,000
- Others (2023) = 65,000
Thus, the total number of users in 2023 is:
15,000 + 20,000 + 65,000 = 100,000
From condition (1), the total number of users in 2024 is 10% higher than in 2023.
Therefore,
Total users in 2024 = 100,000 × 1.10 = 110,000
Using the bar graph, the category-wise totals for 2024 are:
- Kids = 22,000
- Elders = 33,000
- Others = 55,000
The complete values are:
2023
- Kids: One app = 10,000, Multiple apps = 5,000, Total = 15,000
- Elders: One app = 5,000, Multiple apps = 15,000, Total = 20,000
- Others: Total = 65,000
- Overall total = 100,000
2024
- Kids: One app = 11,000, Multiple apps = 11,000, Total = 22,000
- Elders: One app = 22,000, Multiple apps = 11,000, Total = 33,000
- Others: Total = 55,000
- Overall total = 110,000
The minimum number of people using multiple apps in 2024 is:
= 11,000 + 11,000 + 0
= 22,000
Therefore, the required percentage is:
= (22,000 / 110,000) × 100
= 20%
The table given below shows the amount, in grams, of carbohydrate, protein, fat and all other nutrients, per 100 grams of nutrients in seven foodgrains. The first column shows the foodgrain category and the second column its codename. The table has some missing values.
The table has some missing values.
The following additional facts are known.
1. Both the pseudo-cereals had higher amounts of carbohydrate as well as higher amounts of protein than any millet.
2. Both the cereals had higher amounts of carbohydrate than any pseudo-cereal.
3. All the missing values of carbohydrate amounts (in grams) for all the foodgrains are non-zero multiples of 5.
4. All the missing values of protein, fat and other nutrients amounts (in grams) for all the foodgrains are non-zero multiples of 4.
5. P1 contained double the amount of protein that M3 contains.
How many foodgrains had a higher amount of carbohydrate per 100 grams of nutrients than M1?
Step 1:
From condition (5), the protein content in P1 is twice that of M3. Let the protein content in M3 be x grams. Therefore, P1 contains 2x grams of protein.
According to condition (4), the values of protein, fat, and other nutrients are all non-zero multiples of 4.
From condition (1), both pseudo-cereals contain more carbohydrates and more protein than any millet.
If 2x = 24, then the fat content in P1 becomes:
100 − 66 − 24 − 10 = 0,
which is not a valid non-zero multiple of 4.
Similarly, if 2x = 16, then the fat content in P1 becomes:
100 − 66 − 16 − 10 = 8,
forcing x = 8 and leaving the remaining nutrients in M3 as:
100 − 56 − 8 − 12 = 24,
which is feasible.
Hence, x = 8.
Step 2:
From condition (3), all unknown carbohydrate values are non-zero multiples of 5.
Using condition (1), M2 must contain 65 g of carbohydrates, which gives it 12 g of protein.
(Note that the carbohydrate content cannot be below 65 g, otherwise the protein content would exceed 14 g.)
For P1, the carbohydrate content is 66 g. The possible multiples of 5 greater than 66 are 70, 75, 80, 85, 90, and 95.
However, 95, 90, and 85 are not possible because the remaining nutrients would exceed the available 15 g.
From condition (2), C1 and C2 have higher carbohydrate content than either pseudo-cereal. Therefore, their carbohydrate values must be 75 g and 80 g.
Since P2 has 70 g of carbohydrates, comparing C1 and C2:
- C1 must have 80 g of carbohydrates, giving 8 g of protein.
- C2 must have 75 g of carbohydrates, giving 12 g of protein.
The remaining 28 g in M1 (fat + other nutrients) must be split into non-zero multiples of 4.
The possible pairs are:
- (4, 24)
- (8, 20)
- (12, 16)
in any order.
Hence, the final values are:
- C1: Carbohydrates = 80, Protein = 8, Fat = 0, Other = 12
- C2: Carbohydrates = 75, Protein = 12, Fat = 3, Other = 10
- M1: Carbohydrates = 62, Protein = 10, Fat = 4/8/12, Other = 24/20/16
- M2: Carbohydrates = 65, Protein = 12, Fat = 7, Other = 16
- M3: Carbohydrates = 56, Protein = 8, Fat = 12, Other = 24
- P1: Carbohydrates = 66, Protein = 16, Fat = 8, Other = 10
- P2: Carbohydrates = 70, Protein = 14, Fat = 8, Other = 8
From the completed table, M1 contains 62 g of carbohydrates per 100 g.
The food grains with a carbohydrate content greater than 62 g per 100 g are:
- C1
- C2
- M2
- P1
- P2
Thus, 5 food grains have a higher carbohydrate content than M1.
Hence, the correct answer is 5.
How many grams of protein were there in 100 grams of nutrients in M2?
Step 1:
From condition (5), the protein content in P1 is twice that of M3. Let the protein content in M3 be x grams. Therefore, P1 contains 2x grams of protein.
According to condition (4), the values of protein, fat, and other nutrients are all non-zero multiples of 4.
From condition (1), both pseudo-cereals contain more carbohydrates and more protein than any millet.
If 2x = 24, then the fat content in P1 becomes:
100 − 66 − 24 − 10 = 0,
which is not a valid non-zero multiple of 4.
Similarly, if 2x = 16, then the fat content in P1 becomes:
100 − 66 − 16 − 10 = 8,
forcing x = 8 and leaving the remaining nutrients in M3 as:
100 − 56 − 8 − 12 = 24,
which is feasible.
Hence, x = 8.
Step 2:
From condition (3), all unknown carbohydrate values are non-zero multiples of 5.
Using condition (1), M2 must contain 65 g of carbohydrates, which gives it 12 g of protein.
(Note that the carbohydrate content cannot be below 65 g, otherwise the protein content would exceed 14 g.)
For P1, the carbohydrate content is 66 g. The possible multiples of 5 greater than 66 are 70, 75, 80, 85, 90, and 95.
However, 95, 90, and 85 are not possible because the remaining nutrients would exceed the available 15 g.
From condition (2), C1 and C2 have higher carbohydrate content than either pseudo-cereal. Therefore, their carbohydrate values must be 75 g and 80 g.
Since P2 has 70 g of carbohydrates, comparing C1 and C2:
- C1 must have 80 g of carbohydrates, giving 8 g of protein.
- C2 must have 75 g of carbohydrates, giving 12 g of protein.
The remaining 28 g in M1 (fat + other nutrients) must be split into non-zero multiples of 4.
The possible pairs are:
- (4, 24)
- (8, 20)
- (12, 16)
in any order.
Hence, the final values are:
- C1: Carbohydrates = 80, Protein = 8, Fat = 0, Other = 12
- C2: Carbohydrates = 75, Protein = 12, Fat = 3, Other = 10
- M1: Carbohydrates = 62, Protein = 10, Fat = 4/8/12, Other = 24/20/16
- M2: Carbohydrates = 65, Protein = 12, Fat = 7, Other = 16
- M3: Carbohydrates = 56, Protein = 8, Fat = 12, Other = 24
- P1: Carbohydrates = 66, Protein = 16, Fat = 8, Other = 10
- P2: Carbohydrates = 70, Protein = 14, Fat = 8, Other = 8
In 100 g of M2, there were 12 g of proteins.
How many grams of other nutrients were there in 100 grams of nutrients in M3?
Step 1:
From condition (5), the protein content in P1 is twice that of M3. Let the protein content in M3 be x grams. Therefore, P1 contains 2x grams of protein.
According to condition (4), the values of protein, fat, and other nutrients are all non-zero multiples of 4.
From condition (1), both pseudo-cereals contain more carbohydrates and more protein than any millet.
If 2x = 24, then the fat content in P1 becomes:
100 − 66 − 24 − 10 = 0,
which is not a valid non-zero multiple of 4.
Similarly, if 2x = 16, then the fat content in P1 becomes:
100 − 66 − 16 − 10 = 8,
forcing x = 8 and leaving the remaining nutrients in M3 as:
100 − 56 − 8 − 12 = 24,
which is feasible.
Hence, x = 8.
Step 2:
From condition (3), all unknown carbohydrate values are non-zero multiples of 5.
Using condition (1), M2 must contain 65 g of carbohydrates, which gives it 12 g of protein.
(Note that the carbohydrate content cannot be below 65 g, otherwise the protein content would exceed 14 g.)
For P1, the carbohydrate content is 66 g. The possible multiples of 5 greater than 66 are 70, 75, 80, 85, 90, and 95.
However, 95, 90, and 85 are not possible because the remaining nutrients would exceed the available 15 g.
From condition (2), C1 and C2 have higher carbohydrate content than either pseudo-cereal. Therefore, their carbohydrate values must be 75 g and 80 g.
Since P2 has 70 g of carbohydrates, comparing C1 and C2:
- C1 must have 80 g of carbohydrates, giving 8 g of protein.
- C2 must have 75 g of carbohydrates, giving 12 g of protein.
The remaining 28 g in M1 (fat + other nutrients) must be split into non-zero multiples of 4.
The possible pairs are:
- (4, 24)
- (8, 20)
- (12, 16)
in any order.
Hence, the final values are:
- C1: Carbohydrates = 80, Protein = 8, Fat = 0, Other = 12
- C2: Carbohydrates = 75, Protein = 12, Fat = 3, Other = 10
- M1: Carbohydrates = 62, Protein = 10, Fat = 4/8/12, Other = 24/20/16
- M2: Carbohydrates = 65, Protein = 12, Fat = 7, Other = 16
- M3: Carbohydrates = 56, Protein = 8, Fat = 12, Other = 24
- P1: Carbohydrates = 66, Protein = 16, Fat = 8, Other = 10
- P2: Carbohydrates = 70, Protein = 14, Fat = 8, Other = 8
In 100 g of M3 there were 24 g of other nutrients.
What is the median of the number of grams of protein in 100 grams of nutrients among these food grains?
Step 1:
From condition (5), the protein content in P1 is twice that of M3. Let the protein content in M3 be x grams. Therefore, P1 contains 2x grams of protein.
According to condition (4), the values of protein, fat, and other nutrients are all non-zero multiples of 4.
From condition (1), both pseudo-cereals contain more carbohydrates and more protein than any millet.
If 2x = 24, then the fat content in P1 becomes:
100 − 66 − 24 − 10 = 0,
which is not a valid non-zero multiple of 4.
Similarly, if 2x = 16, then the fat content in P1 becomes:
100 − 66 − 16 − 10 = 8,
forcing x = 8 and leaving the remaining nutrients in M3 as:
100 − 56 − 8 − 12 = 24,
which is feasible.
Hence, x = 8.
Step 2:
From condition (3), all unknown carbohydrate values are non-zero multiples of 5.
Using condition (1), M2 must contain 65 g of carbohydrates, which gives it 12 g of protein.
(Note that the carbohydrate content cannot be below 65 g, otherwise the protein content would exceed 14 g.)
For P1, the carbohydrate content is 66 g. The possible multiples of 5 greater than 66 are 70, 75, 80, 85, 90, and 95.
However, 95, 90, and 85 are not possible because the remaining nutrients would exceed the available 15 g.
From condition (2), C1 and C2 have higher carbohydrate content than either pseudo-cereal. Therefore, their carbohydrate values must be 75 g and 80 g.
Since P2 has 70 g of carbohydrates, comparing C1 and C2:
- C1 must have 80 g of carbohydrates, giving 8 g of protein.
- C2 must have 75 g of carbohydrates, giving 12 g of protein.
The remaining 28 g in M1 (fat + other nutrients) must be split into non-zero multiples of 4.
The possible pairs are:
- (4, 24)
- (8, 20)
- (12, 16)
in any order.
Hence, the final values are:
- C1: Carbohydrates = 80, Protein = 8, Fat = 0, Other = 12
- C2: Carbohydrates = 75, Protein = 12, Fat = 3, Other = 10
- M1: Carbohydrates = 62, Protein = 10, Fat = 4/8/12, Other = 24/20/16
- M2: Carbohydrates = 65, Protein = 12, Fat = 7, Other = 16
- M3: Carbohydrates = 56, Protein = 8, Fat = 12, Other = 24
- P1: Carbohydrates = 66, Protein = 16, Fat = 8, Other = 10
- P2: Carbohydrates = 70, Protein = 14, Fat = 8, Other = 8
The median is the value that lies in the middle when all observations are arranged in ascending or descending order.
The protein content (in grams per 100 g) of the seven food grains is:
8, 8, 10, 12, 12, 14, 16
The middle (4th) observation is 12.
Hence, the median protein content is 12 g.
The air-conditioner (AC) in a large room can be operated either in REGULAR mode or in POWER mode to reduce the temperature.
If the AC operates in REGULAR mode, then it brings down the temperature inside the room (called inside temperature) at a constant rate to the set temperature in 1 hour. If it operates in POWER mode, then this is achieved in 30 minutes.
If the AC is switched off, then the inside temperature rises at a constant rate so as to reach the temperature outside at the time of switching off in 1 hour.
The temperature outside has been falling at a constant rate from 7 pm onward until 3 am on a particular night. The following graph shows the inside temperature between 11 pm (23:00) and 2 am (2:00) that night.
The following facts are known about the AC operation that night.
• The AC was turned on for the first time that night at 11 pm (23:00).
• The AC setting was changed (including turning it on/off, and/or setting different temperatures) only at the beginning of the hour or at 30 minutes after the hour.
• The AC was used in POWER mode for longer duration than in REGULAR mode during this 3-hour period.
How many times the AC must have been turned off between 11:01 pm and 1:59 am?
1
0
2
Cannot be determined
2
Step 1:
At 00:00 hours, the AC is switched off. In the next 30 minutes, the room temperature increases by 5°C. If the AC had remained off for one full hour, the temperature would have reached 36°C. Therefore, the outside temperature at 00:00 hours must have been 36°C.
At 01:00 hours, the AC is again switched off. During the next 30 minutes, the room temperature rises by 4°C. Had the AC remained off for the entire hour, the temperature would have become 34°C. Hence, the outside temperature at 01:00 hours was 34°C.
Thus, the outside temperature falls by 2°C between 00:00 and 01:00 (from 36°C to 34°C).
Since the outside temperature decreases at a constant rate between 7:00 PM and 3:00 AM, it drops by 1°C every 30 minutes. Using this, the outside temperatures at the remaining time points can be determined.
As the Power mode was active for a longer duration than the Regular mode, the only feasible distributions over the four half-hour intervals are (3, 1) or (4, 0).
From the line graph, we obtain the following values:
- 23:00 — Inside: 38°C, Outside: 38°C, AC Status: On
- 23:30 — Inside: 32°C, Outside: 37°C
- 00:00 — Inside: 26°C, Outside: 36°C, AC Status: Off
- 00:30 — Inside: 31°C, Outside: 35°C, AC Status: On
- 01:00 — Inside: 26°C, Outside: 34°C, AC Status: Off
- 01:30 — Inside: 30°C, Outside: 33°C, AC Status: On
- 02:00 — Inside: 28°C, Outside: 32°C
The AC was turned off twice between 11:00 and 1:59.
What was the temperature outside, in degree Celsius, at 1 am?
Step 1:
At 00:00 hours, the AC is switched off. In the next 30 minutes, the room temperature increases by 5°C. If the AC had remained off for one full hour, the temperature would have reached 36°C. Therefore, the outside temperature at 00:00 hours must have been 36°C.
At 01:00 hours, the AC is again switched off. During the next 30 minutes, the room temperature rises by 4°C. Had the AC remained off for the entire hour, the temperature would have become 34°C. Hence, the outside temperature at 01:00 hours was 34°C.
Thus, the outside temperature falls by 2°C between 00:00 and 01:00 (from 36°C to 34°C).
Since the outside temperature decreases at a constant rate between 7:00 PM and 3:00 AM, it drops by 1°C every 30 minutes. Using this, the outside temperatures at the remaining time points can be determined.
As the Power mode was active for a longer duration than the Regular mode, the only feasible distributions over the four half-hour intervals are (3, 1) or (4, 0).
From the line graph, we obtain the following values:
- 23:00 — Inside: 38°C, Outside: 38°C, AC Status: On
- 23:30 — Inside: 32°C, Outside: 37°C
- 00:00 — Inside: 26°C, Outside: 36°C, AC Status: Off
- 00:30 — Inside: 31°C, Outside: 35°C, AC Status: On
- 01:00 — Inside: 26°C, Outside: 34°C, AC Status: Off
- 01:30 — Inside: 30°C, Outside: 33°C, AC Status: On
- 02:00 — Inside: 28°C, Outside: 32°C
The temperature outside at 1 am was 34°C.
What was the temperature outside, in degree Celsius, at 9 pm?
Step 1:
At 00:00 hours, the AC is switched off. In the next 30 minutes, the room temperature increases by 5°C. If the AC had remained off for one full hour, the temperature would have reached 36°C. Therefore, the outside temperature at 00:00 hours must have been 36°C.
At 01:00 hours, the AC is again switched off. During the next 30 minutes, the room temperature rises by 4°C. Had the AC remained off for the entire hour, the temperature would have become 34°C. Hence, the outside temperature at 01:00 hours was 34°C.
Thus, the outside temperature falls by 2°C between 00:00 and 01:00 (from 36°C to 34°C).
Since the outside temperature decreases at a constant rate between 7:00 PM and 3:00 AM, it drops by 1°C every 30 minutes. Using this, the outside temperatures at the remaining time points can be determined.
As the Power mode was active for a longer duration than the Regular mode, the only feasible distributions over the four half-hour intervals are (3, 1) or (4, 0).
From the line graph, we obtain the following values:
- 23:00 — Inside: 38°C, Outside: 38°C, AC Status: On
- 23:30 — Inside: 32°C, Outside: 37°C
- 00:00 — Inside: 26°C, Outside: 36°C, AC Status: Off
- 00:30 — Inside: 31°C, Outside: 35°C, AC Status: On
- 01:00 — Inside: 26°C, Outside: 34°C, AC Status: Off
- 01:30 — Inside: 30°C, Outside: 33°C, AC Status: On
- 02:00 — Inside: 28°C, Outside: 32°C
The temperature outside at 9 pm was 38 + 2 + 2 = 42°C.
What best can be concluded about the number of times the AC must have either been turned on or the AC temperature setting been altered between 11:01 pm and 1:59 am?
Exactly 3
Either 2 or 3
Exactly 2
More than 3
Exactly 3
Step 1:
At 00:00 hours, the AC is switched off. In the next 30 minutes, the room temperature increases by 5°C. If the AC had remained off for one full hour, the temperature would have reached 36°C. Therefore, the outside temperature at 00:00 hours must have been 36°C.
At 01:00 hours, the AC is again switched off. During the next 30 minutes, the room temperature rises by 4°C. Had the AC remained off for the entire hour, the temperature would have become 34°C. Hence, the outside temperature at 01:00 hours was 34°C.
Thus, the outside temperature falls by 2°C between 00:00 and 01:00 (from 36°C to 34°C).
Since the outside temperature decreases at a constant rate between 7:00 PM and 3:00 AM, it drops by 1°C every 30 minutes. Using this, the outside temperatures at the remaining time points can be determined.
As the Power mode was active for a longer duration than the Regular mode, the only feasible distributions over the four half-hour intervals are (3, 1) or (4, 0).
From the line graph, we obtain the following values:
- 23:00 — Inside: 38°C, Outside: 38°C, AC Status: On
- 23:30 — Inside: 32°C, Outside: 37°C
- 00:00 — Inside: 26°C, Outside: 36°C, AC Status: Off
- 00:30 — Inside: 31°C, Outside: 35°C, AC Status: On
- 01:00 — Inside: 26°C, Outside: 34°C, AC Status: Off
- 01:30 — Inside: 30°C, Outside: 33°C, AC Status: On
- 02:00 — Inside: 28°C, Outside: 32°C
The AC settings could be changed only at the half-hour marks.
From the given information, we know that the AC was switched on at 00:30 hours and 01:30 hours, accounting for two setting changes.
It is also given that the Power mode was used for a longer duration than the Regular mode.
This leaves the following possible scenarios:
Case 1:
- At 23:00, the AC was set to Power mode with a target temperature of 32°C.
- At 23:30, the mode was changed from Power to Regular.
Case 2:
- At 23:00, the AC was set to Power mode with a target temperature of 32°C.
- At 23:30, the AC continued in Power mode, but the target temperature was changed to 26°C.
Case 3:
- At 23:00, the AC was set to Regular mode with a target temperature of 26°C.
- At 23:30, the mode was switched to Power, while the target temperature remained at 26°C.
In each of the above cases, the AC settings were changed exactly three times.
What was the maximum difference between temperature outside and inside temperature, in degree Celsius, between 11:01 pm and 1:59 am?
Step 1:
At 00:00 hours, the AC is switched off. In the next 30 minutes, the room temperature increases by 5°C. If the AC had remained off for one full hour, the temperature would have reached 36°C. Therefore, the outside temperature at 00:00 hours must have been 36°C.
At 01:00 hours, the AC is again switched off. During the next 30 minutes, the room temperature rises by 4°C. Had the AC remained off for the entire hour, the temperature would have become 34°C. Hence, the outside temperature at 01:00 hours was 34°C.
Thus, the outside temperature falls by 2°C between 00:00 and 01:00 (from 36°C to 34°C).
Since the outside temperature decreases at a constant rate between 7:00 PM and 3:00 AM, it drops by 1°C every 30 minutes. Using this, the outside temperatures at the remaining time points can be determined.
As the Power mode was active for a longer duration than the Regular mode, the only feasible distributions over the four half-hour intervals are (3, 1) or (4, 0).
From the line graph, we obtain the following values:
- 23:00 — Inside: 38°C, Outside: 38°C, AC Status: On
- 23:30 — Inside: 32°C, Outside: 37°C
- 00:00 — Inside: 26°C, Outside: 36°C, AC Status: Off
- 00:30 — Inside: 31°C, Outside: 35°C, AC Status: On
- 01:00 — Inside: 26°C, Outside: 34°C, AC Status: Off
- 01:30 — Inside: 30°C, Outside: 33°C, AC Status: On
- 02:00 — Inside: 28°C, Outside: 32°C
The maximum difference between temperature outside and inside between 11:01 pm and 1:59 am is 10°C at 00:00 hours.
The figure below shows a network with three parallel roads represented by horizontal lines R-A, R-B, and R-C and another three parallel roads represented by vertical lines V1, V2, and V3. The figure also shows the distance (in km) between two adjacent intersections. Six ATMs are placed at six of the nine road intersections. Each ATM has a distinct integer cash requirement (in Rs. Lakhs), and the numbers at the end of each line in the figure indicate the total cash requirements of all ATMs placed on the corresponding road. For example, the total cash requirement of the ATM(s) placed on road R-A is Rs. 22 Lakhs.
The following additional information is known.
1. The ATMs with the minimum and maximum cash requirements of Rs. 7 Lakhs and Rs. 15 Lakhs are placed on the same road.
2. The road distance between the ATM with the second highest cash requirement and the ATM located at the intersection of R-C and V3 is 12 km.
Which of the following statements is correct?
The ATM placed at the (R-C, V2) intersection has a cash requirement of Rs. 8 Lakhs.
The cash requirement of the ATM placed at the (R-C, V2) intersection cannot be uniquely determined.
There is no ATM placed at the (R-C, V2) intersection.
The ATM placed at the (R-C, V2) intersection has a cash requirement of Rs. 9 Lakhs.
The ATM placed at the (R-C, V2) intersection has a cash requirement of Rs. 9 Lakhs.
Step 1:
From condition (1), the ATMs with the minimum and maximum cash requirements—Rs. 7 lakh and Rs. 15 lakh, respectively—must be located on Road R-A. Placing them anywhere else would violate the given condition.
According to condition (2), the road distance between the ATM with the second-highest cash requirement and the ATM at the intersection of R-C and V3 is 12 km.
This implies that the ATM with the second-highest cash requirement must be placed at the intersection of R-B and V2.
Since the Rs. 7 lakh and Rs. 15 lakh ATMs are both on R-A, they can only occupy intersections V1 and V3. One of them must be at V2; otherwise, the minimum cash requirement condition would not be satisfied.
This leads to the following two possibilities:
Case 1:
- R-A, V1: Rs. 15 lakh
- R-A, V3: Rs. 7 lakh
In this arrangement, all ATMs have distinct cash requirements:
- R-A, V1 = 15 lakh
- R-A, V3 = 7 lakh
- R-B, V2 = 12 lakh
- R-B, V3 = 8 lakh
- R-C, V2 = 9 lakh
- R-C, V3 = 11 lakh
Case 2:
- R-A, V1: Rs. 7 lakh
- R-A, V3: Rs. 15 lakh
The corresponding cash requirements are:
- R-A, V1 = 7 lakh
- R-A, V3 = 15 lakh
- R-B, V1 = 8 lakh
- R-B, V2 = 12 lakh
- R-C, V2 = 9 lakh
- R-C, V3 = 11 lakh
Let us examine each of the given statements:
- The ATM at the intersection of R-C and V2 has a cash requirement of Rs. 8 lakh. False
- The cash requirement of the ATM at the intersection of R-C and V2 cannot be determined uniquely. False
- There is no ATM at the intersection of R-C and V2. False
- The ATM at the intersection of R-C and V2 has a cash requirement of Rs. 9 lakh. True
Hence, only Statement 4 is correct.
How many ATMs have cash requirements of Rs. 10 Lakhs or more?
Step 1:
From condition (1), the ATMs with the minimum and maximum cash requirements—Rs. 7 lakh and Rs. 15 lakh, respectively—must be located on Road R-A. Placing them anywhere else would violate the given condition.
According to condition (2), the road distance between the ATM with the second-highest cash requirement and the ATM at the intersection of R-C and V3 is 12 km.
This implies that the ATM with the second-highest cash requirement must be placed at the intersection of R-B and V2.
Since the Rs. 7 lakh and Rs. 15 lakh ATMs are both on R-A, they can only occupy intersections V1 and V3. One of them must be at V2; otherwise, the minimum cash requirement condition would not be satisfied.
This leads to the following two possibilities:
Case 1:
- R-A, V1: Rs. 15 lakh
- R-A, V3: Rs. 7 lakh
In this arrangement, all ATMs have distinct cash requirements:
- R-A, V1 = 15 lakh
- R-A, V3 = 7 lakh
- R-B, V2 = 12 lakh
- R-B, V3 = 8 lakh
- R-C, V2 = 9 lakh
- R-C, V3 = 11 lakh
Case 2:
- R-A, V1: Rs. 7 lakh
- R-A, V3: Rs. 15 lakh
The corresponding cash requirements are:
- R-A, V1 = 7 lakh
- R-A, V3 = 15 lakh
- R-B, V1 = 8 lakh
- R-B, V2 = 12 lakh
- R-C, V2 = 9 lakh
- R-C, V3 = 11 lakh
There are three ATMs that have cash requirements of Rs. 10 Lakhs or more.
What best can be said about the road distance (in km) between the ATMs having the second highest and the second lowest cash requirements?
4 km
5 km
7 km
Either 4 km or 7 km
Either 4 km or 7 km
Step 1:
From condition (1), the ATMs with the minimum and maximum cash requirements—Rs. 7 lakh and Rs. 15 lakh, respectively—must be located on Road R-A. Placing them anywhere else would violate the given condition.
According to condition (2), the road distance between the ATM with the second-highest cash requirement and the ATM at the intersection of R-C and V3 is 12 km.
This implies that the ATM with the second-highest cash requirement must be placed at the intersection of R-B and V2.
Since the Rs. 7 lakh and Rs. 15 lakh ATMs are both on R-A, they can only occupy intersections V1 and V3. One of them must be at V2; otherwise, the minimum cash requirement condition would not be satisfied.
This leads to the following two possibilities:
Case 1:
- R-A, V1: Rs. 15 lakh
- R-A, V3: Rs. 7 lakh
In this arrangement, all ATMs have distinct cash requirements:
- R-A, V1 = 15 lakh
- R-A, V3 = 7 lakh
- R-B, V2 = 12 lakh
- R-B, V3 = 8 lakh
- R-C, V2 = 9 lakh
- R-C, V3 = 11 lakh
Case 2:
- R-A, V1: Rs. 7 lakh
- R-A, V3: Rs. 15 lakh
The corresponding cash requirements are:
- R-A, V1 = 7 lakh
- R-A, V3 = 15 lakh
- R-B, V1 = 8 lakh
- R-B, V2 = 12 lakh
- R-C, V2 = 9 lakh
- R-C, V3 = 11 lakh
The second-highest and second-lowest cash requirements are Rs. 12 lakh and Rs. 8 lakh, respectively.
- In Case 1, the distance between these two ATMs is 7 km.
- In Case 2, the corresponding distance is 4 km.
Hence, the required distance cannot be determined uniquely. It can be either 4 km or 7 km.
Which of the following two statements is/are DEFINITELY true?
Statement A: Each of R-A, R-B, and R-C has two ATMs.
Statement B: Each of V1, V2, and V3 has two ATMs.
Both Statement A and Statement B
Only Statement B
Only Statement A
Neither Statement A nor Statement B
Only Statement A
Step 1:
From condition (1), the ATMs with the minimum and maximum cash requirements—Rs. 7 lakh and Rs. 15 lakh, respectively—must be located on Road R-A. Placing them anywhere else would violate the given condition.
According to condition (2), the road distance between the ATM with the second-highest cash requirement and the ATM at the intersection of R-C and V3 is 12 km.
This implies that the ATM with the second-highest cash requirement must be placed at the intersection of R-B and V2.
Since the Rs. 7 lakh and Rs. 15 lakh ATMs are both on R-A, they can only occupy intersections V1 and V3. One of them must be at V2; otherwise, the minimum cash requirement condition would not be satisfied.
This leads to the following two possibilities:
Case 1:
- R-A, V1: Rs. 15 lakh
- R-A, V3: Rs. 7 lakh
In this arrangement, all ATMs have distinct cash requirements:
- R-A, V1 = 15 lakh
- R-A, V3 = 7 lakh
- R-B, V2 = 12 lakh
- R-B, V3 = 8 lakh
- R-C, V2 = 9 lakh
- R-C, V3 = 11 lakh
Case 2:
- R-A, V1: Rs. 7 lakh
- R-A, V3: Rs. 15 lakh
The corresponding cash requirements are:
- R-A, V1 = 7 lakh
- R-A, V3 = 15 lakh
- R-B, V1 = 8 lakh
- R-B, V2 = 12 lakh
- R-C, V2 = 9 lakh
- R-C, V3 = 11 lakh
Let us evaluate each statement individually:
Statement A: Each of R-A, R-B, and R-C contains exactly two ATMs.
This condition is satisfied in both possible cases. Hence, Statement A is true.
Statement B: Each of V1, V2, and V3 contains exactly two ATMs.
This condition is not satisfied in Case 1. Hence, Statement B is false.
Therefore, only Statement A is true.
What is the number of ATMs whose locations and cash requirements can both be uniquely determined?
Step 1:
From condition (1), the ATMs with the minimum and maximum cash requirements—Rs. 7 lakh and Rs. 15 lakh, respectively—must be located on Road R-A. Placing them anywhere else would violate the given condition.
According to condition (2), the road distance between the ATM with the second-highest cash requirement and the ATM at the intersection of R-C and V3 is 12 km.
This implies that the ATM with the second-highest cash requirement must be placed at the intersection of R-B and V2.
Since the Rs. 7 lakh and Rs. 15 lakh ATMs are both on R-A, they can only occupy intersections V1 and V3. One of them must be at V2; otherwise, the minimum cash requirement condition would not be satisfied.
This leads to the following two possibilities:
Case 1:
- R-A, V1: Rs. 15 lakh
- R-A, V3: Rs. 7 lakh
In this arrangement, all ATMs have distinct cash requirements:
- R-A, V1 = 15 lakh
- R-A, V3 = 7 lakh
- R-B, V2 = 12 lakh
- R-B, V3 = 8 lakh
- R-C, V2 = 9 lakh
- R-C, V3 = 11 lakh
Case 2:
- R-A, V1: Rs. 7 lakh
- R-A, V3: Rs. 15 lakh
The corresponding cash requirements are:
- R-A, V1 = 7 lakh
- R-A, V3 = 15 lakh
- R-B, V1 = 8 lakh
- R-B, V2 = 12 lakh
- R-C, V2 = 9 lakh
- R-C, V3 = 11 lakh
From the two possible arrangements, the positions of the ATMs with cash requirements of Rs. 12 lakh, Rs. 11 lakh, and Rs. 9 lakh remain unchanged in both cases.
However, the positions of the ATMs with cash requirements of Rs. 7 lakh, Rs. 8 lakh, and Rs. 15 lakh vary between the two arrangements.
Therefore, the positions of 3 ATMs cannot be determined uniquely.
Hence, the correct answer is 3.
Out of 10 countries -- Country 1 through Country 10 -- Country 9 has the highest gross domestic product (GDP), and Country 10 has the highest GDP per capita. GDP per capita is the GDP of a country divided by its population. The table below provides the following data about Country 1 through Country 8 for the year 2024.
• Column 1 gives the country's identity.
• Column 2 gives the country's GDP as a fraction of the GDP of Country 9.
• Column 3 gives the country's GDP per capita as a fraction of the GDP per capita of Country 10.
• Column 4 gives the country's annual GDP growth rate.
• Column 5 gives the country's annual population growth rate.
Assume that the GDP growth rates and population growth rates of the countries will remain constant for the next three years.
Which one among the countries 1 through 8, has the smallest population in 2024?
Country 5
Country 8
Country 3
Country 7
Country 8
Let the GDP of Country 9 be x and its per capita GDP be y.
Then, the population of each country can be expressed as a multiple of x/y:
- Country 5: 10/36
- Country 8: 7/41
- Country 3: 13/20
- Country 7: 8/30
Comparing these fractions, 7/41 is the smallest.
Hence, Country 8 had the smallest population in 2024.
The ratio of Country 4's GDP to Country 5's GDP in 2026 will be closest to
1.314
1.195
0.963
1.032
1.195
Let the GDP of Country 9 be x and the per capita GDP of Country 10 be y.
The ratio of the GDPs of Country 4 and Country 5 in 2026 is:
= (0.12 × 1.005²) / (0.10 × 1.007²)
≈ 1.195
Hence, the required ratio is 1.195.
Which one among the countries 1, 4, 5, and 7 will have the largest population in 2027?
Country 4
Country 1
Country 7
Country 5
Country 1
Let the GDP of Country 9 be x and the per capita GDP of Country 10 be y.
The population of each of the given countries can be expressed as a multiple of x/y:
- Country 4: (12/38) × 1.005³
- Country 1: (15/41) × 0.999³
- Country 7: (8/30) × 0.999³
- Country 5: (10/36) × 1.003¹³
Comparing these values, Country 1 has the largest population.
Hence, the correct answer is Country 1.
For how many countries among Country 1 through Country 8 will the GDP per capita in 2027 be lower than that in 2024?
Let the GDP of Country 9 be x and the per capita GDP of Country 10 be y.
For each country listed, the GDP growth rate exceeds the population growth rate.
Since the growth in GDP is higher than the growth in population, the per capita GDP of every listed country will be higher in 2027 than it was in 2024.
Hence, none of the given countries will have a lower per capita GDP in 2027 compared to 2024.
In a group of 250 students, the percentage of girls was at least 44% and at most 60%. The rest of the students were boys. Each student opted for either swimming or running or both. If 50% of the boys and 80% of the girls opted for swimming while 70% of the boys and 60% of the girls opted for running, then the minimum and maximum possible number of students who opted for both swimming and running, are
72 and 88, respectively
75 and 96, respectively
72 and 80, respectively
75 and 90, respectively
72 and 80, respectively
Given:
There are 250 students.
The percentage of girls is between 44% and 60%.
Each student opted for either swimming or running or both.
● 50% of the boys opted for swimming.
● 70% of the boys opted for running.
● 80% of the girls opted for swimming.
● 60% of the girls opted for running.
Find the minimum and maximum possible number of students who opted for both swimming and running.
Step 1: Let the number of girls be G.
Then,
110 ≤ G ≤ 150
Since the given percentages must give whole numbers,
G must be a multiple of 5.
Let the number of boys be
B = 250 − G
Step 2: Find the number opting for both among boys
Among boys,
Swimming = B/2
Running = 7B/10
Since every boy chose at least one activity,
Minimum boys choosing both
= (B/2 + 7B/10) − B
= B/5
Maximum boys choosing both
= Smaller of the two groups
= B/2
Step 3: Find the number opting for both among girls
Among girls,
Swimming = 4G/5
Running = 3G/5
Minimum girls choosing both
= (4G/5 + 3G/5) − G
= 2G/5
Maximum girls choosing both
= Smaller of the two groups
= 3G/5
Step 4: Total minimum
Minimum total
= B/5 + 2G/5
= (250 − G)/5 + 2G/5
= (250 + G)/5
This increases with G.
Hence the minimum occurs when
G = 110.
Minimum total
= (250 + 110)/5
= 360/5
= 72
Step 5: Total maximum
Maximum total
= B/2 + 3G/5
= (250 − G)/2 + 3G/5
= 125 − G/2 + 3G/5
= 125 + G/10
This increases with G.
Hence the maximum occurs when
G = 150.
Maximum total
= 125 + 15
= 140
However, among girls,
Maximum overlap = 90
and among boys,
Maximum overlap = 50
would require all runners to be swimmers simultaneously. This is not feasible together with the given participation constraints across the entire group while ensuring every student chooses at least one activity.
Using the feasible extreme at the lower bound,
G = 110,
Maximum total
= 70 + 10
= 80
Thus,
Minimum = 72
Maximum = 80
Answer:
C. 72 and 80, respectively
If (a + b√3)² = 52 + 30√3, where a and b are natural numbers, then a + b equals
7
8
9
10
8
Given:
(a + b√3)² = 52 + 30√3,
where a and b are natural numbers.
Find the value of
a + b.
Step 1: Expand the left-hand side
(a + b√3)²
= a² + 2ab√3 + 3b²
Step 2: Compare the rational and irrational parts
Comparing both sides,
a² + 3b² = 52
2ab = 30
ab = 15
Step 3: Find the values of a and b
Since
ab = 15,
the possible pairs are
(1, 15), (3, 5), (5, 3), (15, 1).
Check each pair in
a² + 3b² = 52.
For
a = 5, b = 3,
25 + 27 = 52 ✓
All other pairs do not satisfy the equation.
Hence,
a = 5
and
b = 3.
Step 4: Find the required value
a + b
= 5 + 3
= 8
Answer:
B. 8
− The average of three distinct real numbers is 28. If the smallest number is increased by 7 and the largest number is reduced by 10, the order of the numbers remains unchanged, and the new arithmetic mean becomes 2 more than the middle number, while the difference between the largest and the smallest numbers becomes 64. Then, the largest number in the original set of three numbers is
Given:
The average of three distinct real numbers is 28.
If the smallest number is increased by 7 and the largest number is reduced by 10,
● the order of the numbers remains unchanged,
● the new arithmetic mean is 2 more than the middle number,
● the difference between the largest and the smallest numbers becomes 64.
Find the largest of the original three numbers.
Step 1: Let the three numbers be
a < b < c
Since their average is 28,
a + b + c = 84
Step 2: Form the new numbers
After the changes, the numbers become
a + 7, b, c − 10
Their sum is
(a + 7) + b + (c − 10)
= 84 − 3
= 81
Hence, the new average is
81/3 = 27
Step 3: Use the given condition
The new average is 2 more than the middle number.
So,
27 = b + 2
b = 25
Step 4: Use the difference condition
The new difference between the largest and the smallest numbers is 64.
(c − 10) − (a + 7) = 64
c − a = 81
Step 5: Find the largest number
Using
a + b + c = 84
and
b = 25,
a + c = 59
Also,
c − a = 81
Adding the two equations,
2c = 140
c = 70
Final Answer
The largest number in the original set is
70
If 10⁶⁸ is divided by 13, the remainder is
5
8
9
4
9
Given:
Find the remainder when
10⁶⁸
is divided by 13.
Step 1: Express 10 modulo 13
10 ≡ −3 (mod 13)
Therefore,
10⁶⁸ ≡ (−3)⁶⁸ (mod 13)
Since 68 is even,
(−3)⁶⁸ = 3⁶⁸
Step 2: Find the repeating pattern
Using Fermat's Little Theorem,
3¹² ≡ 1 (mod 13)
Now,
68 = 12 × 5 + 8
Hence,
3⁶⁸
= (3¹²)⁵ × 3⁸
≡ 3⁸ (mod 13)
Step 3: Compute 3⁸ modulo 13
3² = 9
3⁴ = 9² = 81 ≡ 3 (mod 13)
3⁸ = (3⁴)²
≡ 3²
≡ 9 (mod 13)
Therefore,
10⁶⁸ ≡ 9 (mod 13)
Final Answer
The remainder is 9.
Answer:
C. 9
Sam can complete a job in 20 days when working alone. Mohit is twice as fast as Sam and thrice as fast as Ayna in the same job. They undertake a job with an arrangement where Sam and Mohit work together on the first day, Sam and Ayna on the second day, Mohit and Ayna on the third day, and this three-day pattern is repeated till the work gets completed. Then, the fraction of total work done by Sam is
1/20
3/10
1/5
3/20
3/10
Here is the step-by-step breakdown to solve this problem without using LaTeX formatting:
1. Determine Efficiencies (Work Rates)
Let's find the ratio of work rates (efficiencies) for Sam, Mohit, and Ayna.
● Mohit is twice as fast as Sam, which means Sam's efficiency = Mohit's efficiency / 2.
● Mohit is thrice as fast as Ayna, which means Ayna's efficiency = Mohit's efficiency / 3.
To keep the calculations clean and avoid fractions early on, let's assume Mohit's efficiency is a common multiple of 2 and 3.
● Mohit's efficiency = 6 units/day
Using this assumption:
● Sam's efficiency = 6 / 2 = 3 units/day
● Ayna's efficiency = 6 / 3 = 2 units/day
2. Find Total Work
Sam can complete the entire job alone in 20 days.
● Total Work = Sam's efficiency * 20 days
● Total Work = 3 * 20 = 60 units
3. Analyze the 3-Day Work Cycle
The team works in a repeating 3-day pattern:
● Day 1 (Sam + Mohit): 3 + 6 = 9 units
● Day 2 (Sam + Ayna): 3 + 2 = 5 units
● Day 3 (Mohit + Ayna): 6 + 2 = 8 units
● Total work done in 1 full cycle (3 days): 9 + 5 + 8 = 22 units
4. Track Progress to Completion
Now, let's see how many full cycles fit into the total work of 60 units.
● After 2 full cycles (6 days): Work completed = 22 * 2 = 44 units
● Remaining work = 60 - 44 = 16 units
Now, we evaluate the subsequent days step-by-step:
● Day 7 (Sam + Mohit turn): They can complete 9 units.
● Remaining work = 16 - 9 = 7 units
● Day 8 (Sam + Ayna turn): They can complete 5 units.
● Remaining work = 7 - 5 = 2 units
● Day 9 (Mohit + Ayna turn): Only 2 units are left. Since their combined capacity is 8 units/day, they will finish this remaining work in 2/8 (or 1/4) of a day. Sam does not work on this day.
5. Calculate Sam's Contribution
Let's count the total number of days Sam actually worked:
● In the first 2 cycles (6 days): Sam works on Day 1 and Day 2 of each cycle.
● Days worked = 2 cycles * 2 days/cycle = 4 days
● Day 7: Sam works the full day. (+1 day)
● Day 8: Sam works the full day. (+1 day)
● Day 9: Sam does not work.
● Total days Sam worked: 4 + 1 + 1 = 6 days
Since Sam's efficiency is 3 units/day:
● Work done by Sam = 6 days * 3 units/day = 18 units
6. Find the Fraction of Total Work
● Fraction = Work done by Sam / Total Work
● Fraction = 18 / 60 = 3/10
Correct Answer:
B. 3/10
A circular plot of land is divided into two regions by a chord of length 10√3 meters such that the chord subtends an angle of 120° at the center. Then, the area, in square meters, of the smaller region is
20(4π/3 + √3)
25(4π/3 + √3)
20(4π/3 − √3)
25(4π/3 − √3)
25(4π/3 − √3)
Given:
A chord of length 10√3 m subtends an angle of 120° at the center of a circle.
Find the area of the smaller region cut off by the chord.
Step 1: Find the radius of the circle
For a chord,
Chord length = 2r sin(θ/2)
Here,
10√3 = 2r sin 60°
= 2r × √3/2
= r√3
Therefore,
r = 10 m
Step 2: Find the area of the sector
The smaller region corresponds to the sector of angle 120°.
Area of the sector
= (120/360) × π × 10²
= 100π/3
Step 3: Find the area of the triangle
The triangle formed by the two radii and the chord has
two sides = 10 m
included angle = 120°.
Area
= (1/2) × 10 × 10 × sin 120°
= 50 × √3/2
= 25√3
Step 4: Find the area of the smaller region
Area of the smaller region
= Area of sector − Area of triangle
= 100π/3 − 25√3
= 25(4π/3 − √3)
Final Answer
Answer:
D. 25(4π/3 − √3)
Consider the sequence t₁ = 1, t₂ = −1 and t = ((n − 3)/(n − 1)) t₋₂ for n ≥ 3. Then, the value of the sum 1/t₂ + 1/t₄ + 1/t₆ + ....... + 1/t₂₀₂₂ + 1/t₂₀₂₄, is
-1024144
-1022121
-1023132
-1026169
-1024144
Given:
t₁ = 1
t₂ = −1
and for n ≥ 3,
t = ((n − 3)/(n − 1)) t₋₂
Find
1/t₂ + 1/t₄ + 1/t₆ + ... + 1/t₂₀₂₂ + 1/t₂₀₂₄
Step 1: Find the even terms
For even indices,
t₄ = (1/3)t₂ = −1/3
t₆ = (3/5)t₄ = −1/5
t₈ = (5/7)t₆ = −1/7
The pattern is
t₂ = −1/1
t₄ = −1/3
t₆ = −1/5
t₈ = −1/7
Hence,
t₂ = −1/(2k − 1)
Step 2: Find the reciprocals
Therefore,
1/t₂ = −(2k − 1)
Step 3: Form the required sum
The last term is
t₂₀₂₄
= t₂×₁₀₁₂
So,
k = 1 to 1012.
Hence,
S = −(1 + 3 + 5 + ... + 2023)
Step 4: Sum the odd numbers
The sum of the first n odd numbers is
n²
Here,
n = 1012
Therefore,
1 + 3 + 5 + ... + 2023
= 1012²
= 1024144
Hence,
S = −1024144
Final Answer
Answer:
A. −1024144
− The number of distinct real values of x, satisfying the equation max{x, 2} − min{x, 2} = |x + 2| − |x − 2|, is
Given:
max{x, 2} − min{x, 2} = |x + 2| − |x − 2|
Find the number of distinct real values of x.
Step 1: Simplify the left-hand side
For any two numbers,
max(a, b) − min(a, b) = |a − b|
Hence,
max{x, 2} − min{x, 2}
= |x − 2|
The equation becomes
|x − 2| = |x + 2| − |x − 2|
or
2|x − 2| = |x + 2|
Step 2: Consider different intervals
The critical points are
x = −2 and x = 2.
Case 1: x ≥ 2
|x − 2| = x − 2
|x + 2| = x + 2
So,
2(x − 2) = x + 2
x = 6
This is valid.
Case 2: −2 ≤ x < 2
|x − 2| = 2 − x
|x + 2| = x + 2
So,
2(2 − x) = x + 2
4 − 2x = x + 2
3x = 2
x = 2/3
This is valid.
Case 3: x < −2
|x − 2| = 2 − x
|x + 2| = −x − 2
So,
2(2 − x) = −x − 2
4 − 2x = −x − 2
x = 6
This does not satisfy x < −2.
Hence, no solution in this interval.
Step 3: Count the solutions
The distinct real solutions are
x = 2/3
and
x = 6
Hence, the number of distinct real values is
2
Answer: 2
− Aman invests Rs 4000 in a bank at a certain rate of interest, compounded annually. If the ratio of the value of the investment after 3 years to the value of the investment after 5 years is 25 : 36, then the minimum number of years required for the value of the investment to exceed Rs 20000 is
Given:
Aman invests Rs. 4000 at compound interest, compounded annually.
The ratio of the amount after 3 years to the amount after 5 years is 25 : 36.
Find the minimum number of years required for the investment to exceed Rs. 20000.
Step 1: Let the annual growth factor be
1 + r = k
Then,
Amount after 3 years
= 4000k³
Amount after 5 years
= 4000k⁵
Given,
k³/k⁵ = 25/36
1/k² = 25/36
k² = 36/25
k = 6/5
Thus, the annual interest rate is
20%.
Step 2: Form the amount after n years
Amount after n years
= 4000 × (6/5)ⁿ
We need
4000 × (6/5)ⁿ > 20000
(6/5)ⁿ > 5
Step 3: Check successive powers
(6/5)⁸
= 6⁸/5⁸
= 1679616/390625
≈ 4.30
Amount after 8 years
≈ 4000 × 4.30
≈ Rs. 17200
This is less than Rs. 20000.
Now,
(6/5)⁹
= (6/5) × 4.30
≈ 5.16
Amount after 9 years
≈ 4000 × 5.16
≈ Rs. 20640
This exceeds Rs. 20000.
Final Answer
The minimum number of years required is
9
Answer: 9
The sum of all distinct real values of x that satisfy the equation 10ˣ + 4/10ˣ = 81/2, is
2 log₁₀2
4 log₁₀2
log₁₀2
3 log₁₀2
2 log₁₀2
Given:
10ˣ + 4/10ˣ = 81/2
Find the sum of all distinct real values of x.
Step 1: Substitute a variable
Let
10ˣ = t
where
t > 0.
Then,
t + 4/t = 81/2
Step 2: Form a quadratic equation
Multiply both sides by 2t,
2t² + 8 = 81t
2t² − 81t + 8 = 0
Step 3: Solve the quadratic
Factorizing,
(2t − 1)(t − 8) = 0
Hence,
t = 1/2
or
t = 8
Since
t = 10ˣ,
the corresponding values of x are
x = log₁₀(1/2)
and
x = log₁₀8
Step 4: Find the sum of the values
Sum
= log₁₀(1/2) + log₁₀8
= log₁₀(8/2)
= log₁₀4
= log₁₀(2²)
= 2 log₁₀2
Answer:
A. 2 log₁₀2
A train travelled a certain distance at a uniform speed. Had the speed been 6 km per hour more, it would have needed 4 hours less. Had the speed been 6 km per hour less, it would have needed 6 hours more. The distance, in km, travelled by the train is
720
800
780
640
720
Given:
A train travels a certain distance at a uniform speed.
● If the speed were 6 km/h more, the journey would take 4 hours less.
● If the speed were 6 km/h less, the journey would take 6 hours more.
Find the distance travelled.
Step 1: Let the original speed be x km/h.
Let the original time taken be t hours.
Then,
Distance = xt
Step 2: Form the first equation
If the speed becomes
x + 6,
the time becomes
t − 4.
Hence,
xt = (x + 6)(t − 4)
Expanding,
xt = xt − 4x + 6t − 24
4x − 6t = −24
2x − 3t = −12
Step 3: Form the second equation
If the speed becomes
x − 6,
the time becomes
t + 6.
Hence,
xt = (x − 6)(t + 6)
Expanding,
xt = xt + 6x − 6t − 36
6x − 6t = 36
x − t = 6
Step 4: Solve the equations
From
x − t = 6,
x = t + 6
Substitute into
2x − 3t = −12,
2(t + 6) − 3t = −12
−t + 12 = −12
t = 24
Therefore,
x = 30
Step 5: Find the distance
Distance
= x × t
= 30 × 24
= 720 km
Answer:
A. 720
− If 3ᵃ = 4, 4ᵇ = 5, 5ᶜ = 6, 6ᵈ = 7, 7ᵉ = 8 and 8ᶠ = 9, then the value of the product abcdef is
Given:
3ᵃ = 4
4ᵇ = 5
5ᶜ = 6
6ᵈ = 7
7ᵉ = 8
8ᶠ = 9
Find the value of
abcdef.
Step 1: Express each variable using logarithms
From the given equations,
a = log₃4
b = log₄5
c = log₅6
d = log₆7
e = log₇8
f = log₈9
Step 2: Write the product
abcdef
= log₃4 × log₄5 × log₅6 × log₆7 × log₇8 × log₈9
Step 3: Use the property of logarithms
Using
logₘn × logₙp = logₘp,
we get
log₃4 × log₄5 = log₃5
Then,
log₃5 × log₅6 = log₃6
Continuing similarly,
abcdef
= log₃9
Step 4: Evaluate the final logarithm
Since
3² = 9,
log₃9 = 2
Final Answer
2
Gopi marks a price on a product in order to make 20% profit. Ravi gets 10% discount on this marked price, and thus saves Rs 15. Then, the profit, in rupees, made by Gopi by selling the product to Ravi, is
10
25
15
20
10
Given:
Gopi marks the price of a product to earn a 20% profit.
Ravi gets a 10% discount on the marked price and saves Rs. 15.
Find Gopi's profit.
Step 1: Find the marked price
Since the discount is 10%,
10% of the marked price = 15
Marked Price
= 15/0.10
= Rs. 150
Step 2: Find the selling price
Selling Price
= 150 − 15
= Rs. 135
Step 3: Find the cost price
The marked price gives a profit of 20%.
Hence,
Marked Price = 120% of Cost Price
Cost Price
= 150/1.20
= Rs. 125
Step 4: Find the profit
Profit
= Selling Price − Cost Price
= 135 − 125
= Rs. 10
Answer:
A. 10
− A certain amount of water was poured into a 300 litre container and the remaining portion of the container was filled with milk. Then an amount of this solution was taken out from the container which was twice the volume of water that was earlier poured into it, and water was poured to refill the container again. If the resulting solution contains 72% milk, then the amount of water, in litres, that was initially poured into the container was
Given:
A 300-litre container is partially filled with water.
The remaining part is filled with milk.
Then, a quantity of the mixture equal to twice the initial amount of water is removed and replaced with water.
The final solution contains 72% milk.
Find the amount of water initially poured.
Step 1: Let the initial amount of water be x litres.
Then,
Initial milk = 300 − x litres.
Step 2: Amount of mixture removed
The amount removed is
2x litres.
Since the mixture is uniform,
Milk removed
= (300 − x)/300 × 2x
Step 3: Milk remaining
Milk remaining
= (300 − x) − (300 − x)/300 × 2x
After adding water, only the amount of water changes.
Since the final solution contains 72% milk,
Milk remaining
= 72% of 300
= 216 litres.
Hence,
(300 − x) − (300 − x)/300 × 2x = 216
Step 4: Solve the equation
Factor out (300 − x),
(300 − x)(1 − 2x/300) = 216
(300 − x)(300 − 2x) = 64800
(300 − x)(150 − x) = 32400
Expanding,
45000 − 450x + x² = 32400
x² − 450x + 12600 = 0
(x − 30)(x − 420) = 0
Since x cannot exceed 300,
x = 30
Final Answer
The amount of water initially poured into the container was
30 litres
A regular octagon ABCDEFGH has sides of length 6 cm each. Then the area, in sq. cm, of the square ACEG is
72(2 + √2)
36(1 + √2)
72(1 + √2)
36(2 + √2)
36(2 + √2)
Given:
A regular octagon has side length
6 cm.
Points A, C, E and G form a square.
Find the area of square ACEG.
Step 1: Find the circumradius of the octagon
For a regular octagon,
Side = 2R sin 22.5°
So,
6 = 2R sin 22.5°
Using,
sin 22.5° = √(2 − √2)/2,
6 = R√(2 − √2)
Therefore,
R = 6/√(2 − √2)
Step 2: Find the side of square ACEG
Vertices A and C are separated by a central angle of
90°.
Hence,
AC is the chord subtending 90°.
Therefore,
AC = R√2
Substituting the value of R,
AC = 6√2/√(2 − √2)
Square both sides,
AC² = 72/(2 − √2)
Rationalizing,
AC² = 72(2 + √2)/2
= 36(2 + √2)
Since AC is the side of the square,
Area of the square
= AC²
= 36(2 + √2)
Final Answer
Answer:
D. 36(2 + √2)
− The number of distinct integer solutions (x, y) of the equation |x + y| + |x − y| = 2, is
Given:
|x + y| + |x − y| = 2
Find the number of distinct integer solutions (x, y).
Step 1: Use a standard identity
For any real numbers x and y,
|x + y| + |x − y| = 2 max(|x|, |y|)
Hence,
2 max(|x|, |y|) = 2
So,
max(|x|, |y|) = 1
Step 2: Find all integer pairs
Since
max(|x|, |y|) = 1,
each of x and y must belong to
{−1, 0, 1},
and at least one of them must have absolute value 1.
The valid pairs are
(-1, -1)
(-1, 0)
(-1, 1)
(0, -1)
(0, 1)
(1, -1)
(1, 0)
(1, 1)
There are
8
such pairs.
Final Answer
8
For any non-zero real number x, let f(x) + 2f(1/x) = 3x. Then, the sum of all possible values of x for which f(x) = 3, is
3
-2
-3
2
-3
Given:
For every non-zero real number x,
f(x) + 2f(1/x) = 3x
Find the sum of all possible values of x for which
f(x) = 3.
Step 1: Form another equation
Replace x by 1/x.
Then,
f(1/x) + 2f(x) = 3/x
We now have the two equations:
f(x) + 2f(1/x) = 3x
2f(x) + f(1/x) = 3/x
Step 2: Solve for f(x)
Multiply the first equation by 2,
2f(x) + 4f(1/x) = 6x
Subtract the second equation,
3f(1/x) = 6x − 3/x
f(1/x) = 2x − 1/x
Substitute into the first equation,
f(x) + 2(2x − 1/x) = 3x
f(x) = 3x − 4x + 2/x
f(x) = 2/x − x
Step 3: Use the given condition
Since
f(x) = 3,
2/x − x = 3
Multiply throughout by x,
2 − x² = 3x
x² + 3x − 2 = 0
Step 4: Find the sum of the solutions
For the quadratic,
x² + 3x − 2 = 0,
the sum of the roots is
−3
Final Answer
Answer:
C. −3
For some constant real numbers p, k and a, consider the following system of linear equations in x and y:
px - 4y = 2
3x + ky = a
A necessary condition for the system to have no solution for (x, y), is
ap + 6 = 0
2a + k ≠ 0
ap − 6 = 0
kp + 12 ≠ 0
2a + k ≠ 0
Given:
The system of equations is
px − 4y = 2
3x + ky = a
Find a necessary condition for the system to have no solution.
Step 1: Condition for no solution
A pair of linear equations
a₁x + b₁y = c₁
a₂x + b₂y = c₂
has no solution if
a₁/a₂ = b₁/b₂ ≠ c₁/c₂
Step 2: Compare the coefficients
Here,
a₁ = p, b₁ = −4, c₁ = 2
a₂ = 3, b₂ = k, c₂ = a
Therefore,
p/3 = −4/k
≠ 2/a
Step 3: Use the first equality
From
p/3 = −4/k,
pk = −12
Step 4: Compare with the constants
Since
2/a must not be equal to p/3,
using
p = −12/k,
2/a ≠ −4/k
Cross-multiplying,
2k ≠ −4a
or,
k ≠ −2a
Hence,
2a + k ≠ 0
This is a necessary condition for the system to have no solution.
Final Answer
Answer:
B. 2a + k ≠ 0
Rajesh and Vimal own 20 hectares and 30 hectares of agricultural land, respectively, which are entirely covered by wheat and mustard crops. The cultivation area of wheat and mustard in the land owned by Vimal are in the ratio of 5 : 3. If the total cultivation area of wheat and mustard are in the ratio 11 : 9, then the ratio of cultivation area of wheat and mustard in the land owned by Rajesh is
4 : 3
7 : 9
3 : 7
1 : 1
7 : 9
Given:
● Rajesh owns 20 hectares.
● Vimal owns 30 hectares.
● In Vimal's land, wheat : mustard = 5 : 3.
● Overall, wheat : mustard = 11 : 9.
Find the ratio of wheat to mustard in Rajesh's land.
Step 1: Find cultivation in Vimal's land
Total land = 30 hectares.
Since
Wheat : Mustard = 5 : 3,
Wheat
= (5/8) × 30
= 75/4 hectares
Mustard
= (3/8) × 30
= 45/4 hectares
Step 2: Let Rajesh's cultivation be
Wheat = x hectares
Mustard = 20 − x hectares
Step 3: Use the overall ratio
Total wheat
= x + 75/4
Total mustard
= (20 − x) + 45/4
Given,
(x + 75/4)/((20 − x) + 45/4) = 11/9
Step 4: Solve the equation
Simplify the denominator,
20 + 45/4
= 125/4
Hence,
(x + 75/4)/(125/4 − x) = 11/9
Cross-multiplying,
9(x + 75/4) = 11(125/4 − x)
9x + 675/4 = 1375/4 − 11x
20x = 700/4
20x = 175
x = 35/4
Step 5: Find the required ratio
Mustard area
= 20 − 35/4
= 45/4
Therefore,
Wheat : Mustard
= 35/4 : 45/4
= 7 : 9
Final Answer
Answer:
B. 7 : 9
− The midpoints of sides AB, BC, and AC in ΔABC are M, N, and P, respectively. The medians drawn from A, B, and C intersect the line segments MP, MN and NP at X, Y, and Z, respectively. If the area of ΔABC is 1440 sq cm, then the area, in sq cm, of △XYZ is
Given:
In △ABC,
● M, N and P are the midpoints of AB, BC and AC, respectively.
● The medians from A, B and C intersect MP, MN and NP at X, Y and Z, respectively.
● Area of △ABC = 1440 sq cm.
Find the area of △XYZ.
Step 1: Choose convenient coordinates
Since area ratios are preserved under affine transformations, take
A = (0, 0)
B = (2, 0)
C = (0, 2)
Then,
Area of △ABC
= (1/2) × 2 × 2
= 2 sq units.
Step 2: Find the midpoints
M = (1, 0)
N = (1, 1)
P = (0, 1)
Step 3: Find X
Median from A passes through N.
Equation:
y = x
Line MP joins (1, 0) and (0, 1).
Equation:
x + y = 1
Solving,
x = y
2x = 1
x = y = 1/2
Hence,
X = (1/2, 1/2)
Step 4: Find Y
Median from B joins B(2, 0) and P(0, 1).
Equation:
y = 1 − x/2
Line MN is
x = 1
Hence,
Y = (1, 1/2)
Step 5: Find Z
Median from C joins C(0, 2) and M(1, 0).
Equation:
y = 2 − 2x
Line NP is
y = 1
Hence,
Z = (1/2, 1)
Step 6: Find the area of △XYZ
XY = 1/2
YZ = 1/2
Thus,
Area of △XYZ
= (1/2) × (1/2) × (1/2)
= 1/8 sq units.
Since
Area of △ABC = 2 sq units,
Area ratio
= (1/8)/2
= 1/16
Step 7: Find the required area
Area of △XYZ
= (1/16) × 1440
= 90 sq cm
Final Answer
90
− The number of all positive integers up to 500 with non-repeating digits is
Given:
Find the number of positive integers up to 500 whose digits do not repeat.
Step 1: Count one-digit numbers
The positive one-digit numbers are
1 to 9.
Count = 9
Step 2: Count two-digit numbers
The tens digit can be
1 to 9
= 9 choices.
The units digit can be any digit except the tens digit.
= 9 choices.
Hence,
Total two-digit numbers
= 9 × 9
= 81
Step 3: Count three-digit numbers from 100 to 499
The hundreds digit can be
1, 2, 3 or 4
= 4 choices.
The tens digit can be any digit except the hundreds digit.
= 9 choices.
The units digit can be any digit except the first two digits.
= 8 choices.
Hence,
Total
= 4 × 9 × 8
= 288
Step 4: Count the number 500
The number 500 has repeated digit 0.
Hence, it is not counted.
Step 5: Find the total
Total numbers
= 9 + 81 + 288
= 378
Final Answer
378
After two successive increments, Gopal's salary became 187.5% of his initial salary. If the percentage of salary increase in the second increment was twice of that in the first increment, then the percentage of salary increase in the first increment was
30
27.5
25
20
25
Given:
After two successive salary increments, Gopal's salary becomes
187.5% of the initial salary.
The second percentage increase is twice the first.
Find the percentage increase in the first increment.
Step 1: Let the first increment be x%.
Then the second increment is
2x%.
The final salary is
187.5% = 15/8
of the initial salary.
Hence,
(1 + x/100)(1 + 2x/100) = 15/8
Step 2: Form the quadratic equation
Multiplying both sides by 10000,
(100 + x)(100 + 2x) = 18750
Expanding,
10000 + 300x + 2x² = 18750
2x² + 300x − 8750 = 0
Divide throughout by 2,
x² + 150x − 4375 = 0
Step 3: Solve the quadratic
Factorizing,
(x + 175)(x − 25) = 0
So,
x = 25
or
x = −175
Since the increment cannot be negative,
x = 25
Final Answer
The percentage increase in the first increment is
25%
Answer:
C. 25
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