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CAT 2023 Slot 2: Past Year Question Paper - CAT MCQ


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66 Questions MCQ Test CAT Mock Test Series 2024 - CAT 2023 Slot 2: Past Year Question Paper

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CAT 2023 Slot 2: Past Year Question Paper - Question 1

The central idea of the passage would be undermined if:

Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 1
  • The central idea of the passage is the promotion of sustainable shopping practices, particularly second-hand shopping, as a means to combat the detrimental environmental effects of the fashion industry. But, the passage also discusses the need for consumers to be mindful of the environmental impact of their clothing choices, opting for high-quality items that last longer and shed fewer microfibers.
  • The passage argues that opting for second clothing might not always be beneficial for the environment by highlighting the microfibre pollution that they can potentially cause. Now, if the second-hand clothes being sold were only of higher quality, it would take care of this problem ([They would be well advised to buy] high-quality items that shed less and last longer [as this] combats both microfibre pollution and excess garments ending up in landfills”)
  • So, the correct answer is Option C.
  • Option A is more about the purchasing channel than the nature of the clothes so it does not necessarily undermine the central idea of the passage.
  • Option B supports the central idea by reducing environmental harm.
  • Option D could align with the sustainability goal and support the central idea, so it doesn't necessarily undermine it.
CAT 2023 Slot 2: Past Year Question Paper - Question 2

The act of "thrifting", as described in the passage, can be considered ironic because it:

Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 2

From the context in which the word 'thrifting' is used in the passage, we can conclude that it refers to the purchase of second-hand items at low costs, a practice which is now a trend as consumers get to be 'cool' while also caring for the planet. However, as explained in the passage, the act of thrifting can be considered ironic if, instead of saving the planet, it actually contributes to microfibre pollution of the rivers and oceans. Option A is the correct choice.

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CAT 2023 Slot 2: Past Year Question Paper - Question 3

Based on the passage, we can infer that the opposite of fast fashion, ‘slow fashion’, would most likely refer to clothes that:

Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 3
  • Option A is the correct answer because the passage emphasizes the environmental issues associated with fast fashion, including the wasteful disposal of garments in landfills. The opposite of this disposable and rapid turnover nature of fast fashion would be a more sustainable and durable approach, which aligns with the idea of "slow fashion."
  • The passage suggests that buying high-quality items that last longer is a way to combat the negative environmental impact of the fashion industry. Therefore, 'slow fashion' can be inferred to refer to clothes that are of high quality and long-lasting, promoting a more sustainable and environmentally friendly approach to fashion consumption.
CAT 2023 Slot 2: Past Year Question Paper - Question 4

According to the author, companies like ThredUP have not caught on in the UK for all of the following reasons EXCEPT that:

Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 4

Option D is the correct answer because the passage does not mention or suggest that the British don't buy second-hand clothing. Instead, the passage discusses challenges related to luxury brands and their reluctance to circulate their latest season stock globally at a cheaper price. The reasons mentioned include the financial aspect(Option A), concerns about brand image(Option B), and the desire to avoid devaluing their products(Option D). Therefore, the passage does not attribute the slow adoption of companies like ThredUP in the UK to the British not buying second-hand clothing.

CAT 2023 Slot 2: Past Year Question Paper - Question 5

The author of the passage faults Deneen's conclusions for all of the following reasons, EXCEPT:

Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 5
  • In considering whether Deneen's argument on liberalism is convincing, the author points out Deneen's narrow definition of liberalism is limited to individual freedoms: ' He argues that the essence of liberalism lies in freeing individuals from constraints. In fact, liberalism contains a wide range of intellectual traditions which provide different answers to the question of how to trade off the relative claims of rights and responsibilities, individual expression and social ties..'
  • The author also says Deneen fails to recognise liberalism's ability to reform itself: 'Mr Deneen's fixation on the essence of liberalism leads to the second big problem of his book: his failure to recognise liberalism's ability to reform itself and address its internal problems.'
  • Finally, in the last two lines of the passage, the author states Deneen is wrong in his extreme pessimism about the future of liberalism.
  • Options B, C and D are true.
  • The author does not say that Deneen harks back to premodern notions of liberty. So, option A is the correct answer choice.
CAT 2023 Slot 2: Past Year Question Paper - Question 6

The author of the passage is likely to disagree with all of the following statements, EXCEPT:

Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 6
  • Consider option A. The author is likely to disagree with this. Note the lines, 'He argues that the essence of liberalism lies in freeing individuals from constraints. In fact, liberalism contains a wide range of intellectual traditions which provide different answers to the question of how to trade off the relative claims of rights and responsibilities, individual expression and social ties.'
  • Consider option B. The author starts the passage by saying, 'Over the past four centuries liberalism has been so successful that it has driven all its opponents off the battlefield'. He also argues in the penultimate paragraph that liberalism has the ability to reform itself to remain dominant. So, the author is likely to agree with this option. Option B is the correct choice.
  • Let us also consider options C and D to rule them out.
  • The author is likely to disagree with the statement that claims about liberalism's disintegration are exaggerated and misunderstand its core features. Note the lines, 'Mr Deneen is right to point out that the record of liberalism in recent years has been dismal. He is also right to assert that the world has much to learn from the premodern notions of liberty as self-mastery and self-denial.'
  • The author is also likely to disagree with the idea that if we accept that liberalism is a dying ideal, we must work to find a viable substitute. The author argues against liberation from liberalism and states the liberalism must heed the call to action and reform itself.
  • So, option B is the correct answer choice.
CAT 2023 Slot 2: Past Year Question Paper - Question 7

All of the following statements are evidence of the decline of liberalism today, EXCEPT:

Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 7

All options except B relate to liberalism and the problems caused by its disintegration. Technological advances cannot be considered evidence of the decline of liberalism.

CAT 2023 Slot 2: Past Year Question Paper - Question 8

The author of the passage refers to "the Davos elite" to illustrate his views on:

Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 8

Note the context in which the author talks about the 'Davos elite': 'The biggest enemy of liberalism is not so much atomisation but old-fashioned greed, as members of the Davos elite pile their plates ever higher with perks and share options.' Only option D relates to the greed of the Davos elite. This is the correct answer choice.

CAT 2023 Slot 2: Past Year Question Paper - Question 9

All of the following, if true, can weaken the passage's claim that facts do not speak for themselves, EXCEPT:

Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 9
  • If option B is true, that is, if facts are relative and subject to interpretation, then that strengthens the passage's claim that facts do not speak for themselves. So, option B is the right answer choice.
  • The passage claims that facts do not speak for themselves by arguing that while facts are objective and universal and hold true irrespective of the historian who expresses it, it is the historian who decides to which facts to give the floor, and in what order or context, thereby influencing their interpretation. So, all options except B, if true, weaken the passage's claim.
CAT 2023 Slot 2: Past Year Question Paper - Question 10

If the author of the passage were to write a book on the Battle of Hastings along the lines of his/her own reasoning, the focus of the historical account would be on:

Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 10
  • The main idea of the passage is that facts speak only when the historian calls on them. The author says that it is because historians regard the Battle of Hastings as a major historical event that we are interested in knowing about it. It is the historian's interpretation of facts that we are interested in. So, if the author were to write a book on the Battle of Hastings, the focus of the account would be on subjective interpretations, like exploring the socio-political and economic factors that led to the Battle.
  • Options A and D are easily ruled out as they focus on the importance of facts.
  • Option B is close, as 'nuanced interpretation' is what the author says historians have to focus on. But option B, unlike option C, emphasizes the role of auxillary sciences in helping a historian do his work. The author says relying on facts that can be gathered from auxiliary sciences of history is "a necessary condition" of a historians' work, "but not his essential function". So, option C is better than B.
CAT 2023 Slot 2: Past Year Question Paper - Question 11

According to this passage, which one of the following statements best describes the significance of archaeology for historians?

Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 11
  • Note the context in which the author talks about archaeology and other the "auxiliary sciences" of history: 'But [to] praise a historian for his accuracy is like praising an architect for using well-seasoned timber or properly mixed concrete in his building. It is a necessary condition of his work, but not his essential function. It is precisely for matters of this kind that the historian is entitled to rely on what have been called the "auxiliary sciences" of history-archaeology, epigraphy, numismatics, chronology, and so forth...'
  • The author states auxiliary sciences like archaeology only help historians to ascertain the accuracy of facts. They do not help in the essential function of his work, which is to interpret the facts.
  • Option C is the correct choice.
CAT 2023 Slot 2: Past Year Question Paper - Question 12

All of the following describe the "common-sense view" of history, EXCEPT:

Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 12
  • According to the passage, the "common-sense" view of history is influenced by the positivist view and so it places great weight on facts. In this view, facts are available to the historian in documents, inscriptions, and so on and history can be objective like the sciences if it is derived from historical facts.
  • The author's view is in contrast to the common-sense view. The author believes history is a 'selective' system of cognitive orientations to reality. Facts only speak as the historian interprets them. Option A is the correct answer choice.
CAT 2023 Slot 2: Past Year Question Paper - Question 13

Based on information provided in the passage, all of the following are true, EXCEPT:

Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 13
  • Option A is true, based on the lines, 'The economics of European productions are more appealing, too. American audiences are more willing than before to give dubbed or subtitled viewing a chance. This means shows such as "Lupin", a French crime caper on Netflix, can become global hits.'
  • Option B is true, too: 'In 2015, about 75% of Netflix's original content was American; now the figure is half, according to Ampere, a media-analysis company.'
  • Option C is clearly stated in the passage:'A bigger problem is that national broadcasters still dominate. Streaming services, such as Netflix or Disney+, account for about a third of all viewing hours, even in markets where they are well-established.'
  • Only option D is incorrect: 'Now Netflix has offices across Europe. But ultimately the big decisions rest with American executives.'
CAT 2023 Slot 2: Past Year Question Paper - Question 14

The author sees the rise of Netflix in Europe as:

Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 14

The author clearly sees Netflix as a unifying force in Europe: 'Now Netflix and friends pump the same content into homes across a continent, making culture a cross-border endeavour, too. If Europeans are to share a currency, bail each other out in times of financial need and share vaccines in a pandemic, then they need to have something in common-even if it is just bingeing on the same series.'

CAT 2023 Slot 2: Past Year Question Paper - Question 15

Which one of the following research findings would weaken the author's conclusion in the final paragraph?

Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 15
  • In the last paragraph, the author concludes that Netflix is a unifying force in Europe, making culture "a cross-border endeavour". If there were a wide variance in the popularity and viewing of Netflix shows across different EU countries, then the author's assumption that Netflix is popular across Europe, giving Europeans something to share across borders, is weakened. Option B is the correct answer choice.
  • All other options are unrelated to the author's conclusion in the final paragraph.
CAT 2023 Slot 2: Past Year Question Paper - Question 16

Based only on information provided in the passage, which one of the following hypothetical Netflix shows would be most successful with audiences across the EU?

Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 16
  • Talking about which shows have better appeal, the passage states, 'Not everything works across borders. Comedy sometimes struggles. Whodunits and bloodthirsty maelstroms between arch Romans and uppity tribesmen have a more universal appeal...'. So, a murder mystery drama set in North Africa and France is likely, according to the passage, to be successful with audiences across the EU.
  • Based on the lines above, option D is easily eliminated. The passage declares 'German television is not always built for export', so option B is also ruled out. The passage focuses on translations of European productions and their success. Option A does not relate to this.
CAT 2023 Slot 2: Past Year Question Paper - Question 17

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: And probably much earlier, moving the documentation for kissing back 1,000 years compared to what was acknowledged in the scientific community.

Paragraph: Research has hypothesised that the earliest evidence of human lip kissing originated in a very specific geographical location in South Asia 3,500 years ago.___(1)___. From there it may have spread to other regions, simultaneously accelerating the spread of the herpes simplex virus 1. According to Dr Troels Pank Arbøll and Dr Sophie Lund Rasmussen, who in a new article in the journal Science draw on a range of written sources from the earliest Mesopotamian societies, kissing was already a well-established practice 4,500 years ago in the Middle East.___(2)___. In ancient Mesopotamia, people wrote in cuneiform script on clay tablets.___(3)___. Many thousands of these clay tablets have survived to this day, and they contain clear examples that kissing was considered a part of romantic intimacy in ancient times.___(4)___. "Kissing could also have been part of friendships and family members' relations," says Dr Troels Pank Arbøll, an expert on the history of medicine in Mesopotamia.

Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 17

The given sentence begins with 'and probably much earlier...', so the sentence before must have some reference to time. This narrows down the options to 1 and 2. Of these options 2 makes better sense as the given sentence talks about documentation for kissing and the sentence before 2 talks about written sources that confirm new research findings.

CAT 2023 Slot 2: Past Year Question Paper - Question 18

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: Dualism was long held as the defining feature of developing countries in contrast to developed countries, where frontier technologies and high productivity were assumed to prevail.

Paragraph: ___(1)___. At the core of development economics lies the idea of 'productive dualism': that poor countries' economies are split between a narrow 'modern' sector that uses advanced technologies and a larger 'traditional' sector characterized by very low productivity.___(2)___. While this distinction between developing and advanced economies may have made some sense in the 1950s and 1960s, it no longer appears to be very relevant. A combination of forces have produced a widening gap between the winners and those left behind.___(3)___. Convergence between poor and rich parts of the economy was arrested and regional disparities widened.___(4)___. As a result, policymakers in advanced economies are now grappling with the same questions that have long preoccupied developing economies: mainly how to close the gap with the more advanced parts of the economy.

Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 18

The given sentence is about dualism, long held as a distinction between developing and developed countries. The sentence before option 2 introduces dualism and the sentence after it begins with 'while this distinction between developing and advanced economies..', making option 2 the best choice for fitting in the given sentence.

CAT 2023 Slot 2: Past Year Question Paper - Question 19

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. Self-care particularly links to loneliness, behavioural problems, and negative academic outcomes.
2. "Latchkey children" refers to children who routinely return home from school to empty homes and take care of themselves for extended periods of time.
3. Although self-care generally points to negative outcomes, it is important to consider that the bulk of research has yet to track long-term consequences.
4. In research and practice, the phrase "children in self-care" has come to replace latchkey in an effort to more accurately reflect the nature of their circumstances.
5. Although parents might believe that self-care would be beneficial for development, recent research has found quite the opposite.

Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 19
  • Sentence 3 is the odd one out because it introduces a different perspective. While the other sentences discuss the negative outcomes associated with self-care in children, Sentence 3 suggests a need to consider that the bulk of research has yet to track long-term consequences. This sentence introduces a more neutral or balanced viewpoint that doesn't align with the general theme of the other sentences, which focus on the negative aspects of self-care for children.
  • The remaining sentences (1, 2, 4, and 5) can be put together to form a coherent paragraph discussing the concept of self-care in relation to children, particularly those referred to as "Latchkey children."
CAT 2023 Slot 2: Past Year Question Paper - Question 20

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. The banning of Northern Lights could be considered a precursor to censoring books for “moral”, world view or religious reasons.
  2. Attempts to ban books are attempts to silence authors who have summoned immense courage in telling their stories.
  3. Now the banning and challenging of books in the US has escalated to an unprecedented level.
  4. The widely acclaimed fantasy novel Northern Lights was banned in some parts of the US, and was the second most challenged book in the US.
  5. The American Library Association documented an unparalleled number of reported book challenges in 2022, about 2,500 unique titles.
Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 20

Sentence 2 is the odd one out because it introduces a broader statement about attempts to ban books and silence authors, while the other sentences are specifically focused on the banning and challenging of books in the US. Sentences 1, 3, 4, and 5 collectively form a coherent paragraph discussing the banning of Northern Lights and the escalation of book challenges in the US, while Sentence 2 introduces a different perspective that doesn't directly contribute to the flow of the paragraph regarding the specific incidents and trends mentioned in the other sentences.

CAT 2023 Slot 2: Past Year Question Paper - Question 21

The four sentences (labelled 1, 2, 3 and 4) given below, when properly sequenced, would yield a coherent paragraph. Decide on the proper sequencing of the order of the sentences and key in the sequence of the four numbers as your answer.

  1. Contemporary African writing like ‘The Bottled Leopard’ voices this theme using two children and two backgrounds to juxtapose two varying cultures.
  2. Chukwuemeka Ike explores the conflict, and casts the Western tradition as condescending, enveloping and unaccommodating towards local African practice.
  3. However, their views contradict the reality, for a rich and sustaining local African cultural ethos exists for all who care, to see and experience.
  4. Western Christian concepts tend to deny or feign ignorance about the existence of a genuine and enduring indigenous African tradition.
Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 21
  • Sentence 4 introduces the conflict between Western Christian concepts and indigenous African tradition.This gives context for the discussion to be followed.
  • Now we can see that Sentence 3 starts with “However….” which implies that this sentence is challenging the views presented in Sentence 4 by asserting the existence of a rich indigenous African cultural ethos. Therefore Sentence 3 must be following Sentence 4.
  • Sentence 2 introduces Chuwkuemeka Ike and provides additional information about Chukwuemeka Ike's exploration of the conflict, adding depth to the discussion. Sentence 1 introduces an example of contemporary African writing that explores the conflict. It makes sense that first we give an example and then elaborate on it. Sentence 1 introduces the book; it is a logical inferences that Ike must be the author of this book.
  • Therefore Sentence 2 must be following Sentence 1.

Therefore the correct order is 4-3-1-2.

CAT 2023 Slot 2: Past Year Question Paper - Question 22

The four sentences (labelled 1, 2, 3 and 4) given below, when properly sequenced, would yield a coherent paragraph. Decide on the proper sequencing of the order of the sentences and key in the sequence of the four numbers as your answer.

  1. Like the ants that make up a colony, no single neuron holds complex information like self-awareness, hope or pride.
  2. Although the human brain is not yet understood enough to identify the mechanism by which emergence functions, most neurobiologists agree that complex interconnections among the parts give rise to qualities that belong only to the whole.
  3. Nonetheless, the sum of all neurons in the nervous system generate complex human emotions like fear and joy, none of which can be attributed to a single neuron.
  4. Human consciousness is often called an emergent property of the human brain.
Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 22
  • Sentence 4 introduces the main idea that human consciousness is often referred to as an emergent property of the human brain. This gives context for the discussion to be followed.
  • Now if we consider Sentences 1 and 3 we can see that Sentence 1 ends with “no single neuron holds complex information like self-awareness, hope or pride.” Sentence 3 builds on the idea in Sentence 1 by stating that, nonetheless, the collective activity of all neurons in the nervous system generates complex human emotions like fear and joy “Nonetheless, the sum of all neurons …..”. So we can infer that Sentence 3 must be following Sentence 1. Sentence 3 also supports the notion that emergent properties arise from the interaction of individual components.
  • Finally, Sentence 2 concludes the paragraph by explaining that although the exact mechanism of emergence in the human brain is not fully understood, neurobiologists agree that complex interconnections among the parts give rise to qualities specific to the whole. This sentence wraps up the discussion and reinforces the concept introduced Sentence 4.
  • Therefore the correct order is 4-1-3-2.
CAT 2023 Slot 2: Past Year Question Paper - Question 23

The passage given below is followed by four alternate summaries. Choose the option that best captures the essence of the passage.

Heatwaves are becoming longer, frequent and intense due to climate change. The impacts of extreme heat are unevenly experienced; with older people and young children, those with pre-existing medical conditions and on low incomes significantly more vulnerable. Adaptation to heatwaves is a significant public policy concern. Research conducted among at-risk people in the UK reveals that even vulnerable people do not perceive themselves as at risk of extreme heat; therefore, early warnings of extreme heat events do not perform as intended. This suggests that understanding how extreme heat is narrated is very important. The news media play a central role in this process and can help warn people about the potential danger, as well as about impacts on infrastructure and society.

Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 23
  • The passage discusses the increasing frequency and intensity of heatwaves due to climate change, with vulnerable groups experiencing uneven impacts. It emphasizes that adaptation to heatwaves is a significant public policy concern. The research findings suggest that even vulnerable individuals may not perceive themselves as at risk of extreme heat, highlighting the importance of understanding how extreme heat is narrated. The passage specifically mentions the central role of the news media in warning people about the potential danger of heatwaves and their impacts on infrastructure and society. Option C effectively conveys the primary focus on heatwaves posing a substantial risk and the critical role of the media in alerting the public to this danger.
  • Option A implies a general importance of protection without specifically highlighting the role of the media in alerting people to the risks of heatwaves.
  • Option B acknowledges the vulnerability to heatwaves but it does not emphasize the role of the media in alerting people and suggests a broader critique of measures taken.
  • Option D mentions the need for news stories to become more effective but does not emphasize the central role of the media in alerting people to the risk of heatwaves, as the passage does.
CAT 2023 Slot 2: Past Year Question Paper - Question 24

The passage given below is followed by four alternate summaries. Choose the option that best captures the essence of the passage.

People spontaneously create counterfactual alternatives to reality when they think “if only” or “what if” and imagine how the past could have been different. The mind computes counterfactuals for many reasons. Counterfactuals explain the past and prepare for the future, they implicate various relations including causal ones, and they affect intentions and decisions. They modulate emotions such as regret and relief, and they support moral judgments such as blame. The ability to create counterfactuals develops throughout childhood and contributes to reasoning about other people's beliefs, including their false beliefs.

Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 24
  • The passage discusses the phenomenon of counterfactual thinking, highlighting that people spontaneously create counterfactual alternatives to reality for various reasons. These reasons include explaining the past, preparing for the future, implicating various relations (including causal ones), affecting emotions, and supporting moral judgments. Additionally, the passage mentions that the ability to create counterfactuals develops throughout childhood and contributes to reasoning about other people's beliefs. Option C effectively encompasses these key points, making it the most accurate summary of the passage.
  • Option A focuses primarily on the preparation for the future aspect, neglecting the broader reasons for creating counterfactual alternatives.
  • Option B does not emphasize the developmental aspect and various reasons for creating
  • Option D inaccurately suggests that counterfactual thinking helps reverse past and future actions, which is not the main point of the passage, and it oversimplifies the role of counterfactuals.
CAT 2023 Slot 2: Past Year Question Paper - Question 25

Let a, b, m and n be natural numbers such that a > 1 and b > 1. If ambn = 144145, then the largest possible value of n - m is

Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 25

CAT 2023 Slot 2: Past Year Question Paper - Question 26

The sum of all possible values of x satisfying the equation 24x2 - 22x2 + x + 16 + 22x + 30 = 0, is

Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 26

CAT 2023 Slot 2: Past Year Question Paper - Question 27

For any natural numbers m, n, and k, such that k divides both m + 2n and 3m + 4n, k must be a common divisor of

Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 27

It is given that k divides m+2n and 3m+4n.

Since k divides (m+2n), it implies k will also divide 3(m+2n). Therefore, k divides 3m+6n.

Similarly, we know that k divides 3m+4n.

We know that if two numbers a, and b both are divisible by c, then their difference (a-b) is also divisible by c.

By the same logic, we can say that {(3m+6n)-(3m+4n)} is divisible by k. Hence, 2n is also divisible by k.

Now, (m+2n) is divisible by k, it implies 2(m+2n) =2m+4n is also divisible by k.

Hence, {(3m+4n)-(2m+4n)} = m is also divisible by k.

Therefore, m, and 2n are also divisible by k.

The correct option is C

CAT 2023 Slot 2: Past Year Question Paper - Question 28

Any non-zero real numbers x, y such that y ≠ 3 and will satisfy the condition

Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 28

*Answer can only contain numeric values
CAT 2023 Slot 2: Past Year Question Paper - Question 29

For some positive real number x, if  then the value of 


Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 29

*Answer can only contain numeric values
CAT 2023 Slot 2: Past Year Question Paper - Question 30

The number of positive integers less than 50, having exactly two distinct factors other than 1 and itself, is


Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 30

A positive integer less than 50, having exactly two distinct factors other than 1 and itself, is either a perfect cube below 50 or an integer that is a product of exactly two distinct primes.
Case i)
Perfect cubes below 50 are 23 and 33. So, two numbers here
Case ii)
For the product of two primes to be below 50, the individual primes should be below 25.
(Because, the smallest prime is 2 and multiplying 2 with anything greater than or equal to 25 yields a number greater than or equal to 50.)
2, 3, 5, 7, 11, 13, 17, 19, 23 are prime numbers less than 25.

2, 3, 5, 7 are the primes less than √50, any product of two numbers among them yields a product less than 50.
So, there are 4C2 = 6 pairs here.
11, 13, 17, 19, 23 are the primes greater than √50, any product of two numbers among them yields a product greater than 50.
So, there are 0 pairs here.
Between the two lists 11 and 13 can pair with 2 and 3, while 17, 19, and 23 can only pair with 2.
So, there are 7 pairs here.
So, totally, there are 2 + 6 + 0 + 7 = 15 such numbers.

*Answer can only contain numeric values
CAT 2023 Slot 2: Past Year Question Paper - Question 31

Let k be the largest integer such that the equation (x−1)2 + 2kx + 11 = 0 has no real roots. If y is a positive real number, then the least possible value of k/4y + 9y is


Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 31


Since the equation above has no real roots, the discriminant of the equation should be negative.

The largest integral value that k can take is 4.
Now, we need to minimize k/4y + 9y where k takes the largest integral value and y is positive…

1/y & 9y are both positive real numbers, therefore, their A.M is greater than equal to their G.M.

CAT 2023 Slot 2: Past Year Question Paper - Question 32

In a company, 20% of the employees work in the manufacturing department. If the total salary obtained by all the manufacturing employees is one-sixth of the total salary obtained by all the employees in the company, then the ratio of the average salary obtained by the manufacturing employees to the average salary obtained by the non-manufacturing employees is

Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 32

” In a company, 20% of the employees work in the manufacturing department.”
Ratio of number of manufacturing to non-manufacturing employees = 1 : 4
So, let the number of manufacturing to non-manufacturing employees be x and 4x respectively.
“the total salary obtained by all the manufacturing employees is one-sixth of the total salary obtained by all the employees in the company,”
Ratio of total salaries of manufacturing to non-manufacturing employees = 1 : 5
So, let the total salaries of manufacturing to non-manufacturing employees be y and 5y respectively.
So, the ratio of average salaries of manufacturing to non-manufacturing employees will be 

CAT 2023 Slot 2: Past Year Question Paper - Question 33

Minu purchases a pair of sunglasses at Rs.1000 and sells to Kanu at 20% profit. Then, Kanu sells it back to Minu at 20% loss. Finally, Minu sells the same pair of sunglasses to Tanu. If the total profit made by Minu from all her transactions is Rs.500, then the percentage of profit made by Minu when she sold the pair of sunglasses to Tanu is

Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 33

“Minu purchases a pair of sunglasses at Rs.1000 and sells to Kanu at 20% profit.”
This means, Minu purchased the glasses at Rs.1000 and sold them to Kanu at Rs.1200.
Minu made a profit of Rs. 200 so far.
“Then, Kanu sells it back to Minu at 20% loss.”
This means, Kanu sold the glasses back to Minu at 80% of 1200 = Rs. 960
“Finally, Minu sells the same pair of sunglasses to Tanu. … the total profit made by Minu from all her transactions is Rs.500”
This means, Minu has to make a further profit of Rs. 300. She achieves this by selling the glasses back to Tanu at 960 + 300 = Rs.1260
So the profit % in the transaction is 300/960 x 100 = 31.25%

CAT 2023 Slot 2: Past Year Question Paper - Question 34

Pipes A and C are fill pipes while Pipe B is a drain pipe of a tank. Pipe B empties the full tank in one hour less than the time taken by Pipe A to fill the empty tank. When pipes A, B and C are turned on together, the empty tank is filled in two hours. If pipes B and C are turned on together when the tank is empty and Pipe B is turned off after one hour, then Pipe C takes another one hour and 15 minutes to fill the remaining tank. If Pipe A can fill the empty tank in less than five hours, then the time taken, in minutes, by Pipe C to fill the empty tank is

Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 34

Let A, B and C be the number of hours taken by pipes A, B and C to completely fill (or completely empty) a tank.
So the fraction of the tank filled(or emptied) by them in one hour is 1/A, 1/B, 1/C
“Pipe B empties the full tank in one hour less than the time taken by Pipe A to fill the empty tank”
B = A – 1
“When pipes A, B and C are turned on together, the empty tank is filled in two hours”

If pipes B and C are turned on together when the tank is empty and Pipe B is turned off after one hour, then Pipe C takes another one hour and 15 minutes to fill the remaining tank.”
This means, after pipe C worked for 2 hrs 15 mins (or 9/4 hrs) and the Pipe B draining for 1 hour, the tank got filled.


“Pipe A can fill the empty tank in less than five hours”
A = 3

CAT 2023 Slot 2: Past Year Question Paper - Question 35

Ravi is driving at a speed of 40 km/h on a road. Vijay is 54 meters behind Ravi and driving in the same direction as Ravi. Ashok is driving along the same road from the opposite direction at a speed of 50 km/h and is 225 meters away from Ravi. The speed, in km/h, at which Vijay should drive so that all the three cross each other at the same time, is

CAT 2023 Slot 2: Past Year Question Paper - Question 36

Anil borrows Rs 2 lakhs at an interest rate of 8% per annum, compounded half-yearly. He repays Rs 10320 at the end of the first year and closes the loan by paying the outstanding amount at the end of the third year. Then, the total interest, in rupees, paid over the three years is nearest to

Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 36

“Anil borrows Rs 2 lakhs at an interest rate of 8% per annum, compounded half-yearly. He repays Rs 10,320 at the end of the first year and closes the loan by paying the outstanding amount at the end of the third year.”
This means, the amount paid by Anil at the end of the third year = (2,00,000 × 1.04− 10,320) × 1.044 = 2,40,990.8634 ≅ 2,40,991
So the total amount paid by Anil = 10,320 + 2,40,991 = 2,51,311
So the total interest paid = 2,51,311 − 2,00,000 = 51,311

CAT 2023 Slot 2: Past Year Question Paper - Question 37

The price of a precious stone is directly proportional to the square of its weight. Sita has a precious stone weighing 18 units. If she breaks it into four pieces with each piece having distinct integer weight, then the difference between the highest and lowest possible values of the total price of the four pieces will be 288000. Then, the price of the original precious stone is

Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 37

The price of a precious stone is directly proportional to the square of its weight.”
If W is the weight of the stone and P is the price of that stone, then P = k × W2
For the entire, unbroken stone, the price will be 182 × k = 324 k.
“If she breaks it into four pieces with each piece having distinct integer weight, then the difference between the highest and lowest possible values of the total price of the four pieces will be 288000.”
The minimum profit is achieved when the weights of the broken stones are close to each other, that is, the weights are 3, 4, 5, and 6 units.
In this case the combines worth of the four stones =(32 + 42 + 52 + 62)k = 86k
The maximum profit is achieved when the weights of the broken stones are far from each other, that is, the weights are 1, 2, 3, and 12 units.
In this case the combines worth of the four stones =(1+ 2+ 3+ 122)k = 158k
The difference in the total value = 2,88,000.
158k – 86k = 72k = 2,88,000
k = 4,000
So, the price of the original stone = 324 k = 12,96,000

*Answer can only contain numeric values
CAT 2023 Slot 2: Past Year Question Paper - Question 38

A container has 40 liters of milk. Then, 4 liters are removed from the container and replaced with 4 liters of water. This process of replacing 4 liters of the liquid in the container with an equal volume of water is continued repeatedly. The smallest number of times of doing this process, after which the volume of milk in the container becomes less than that of water, is


Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 38

Every time 10% of the mixture is replace with a pure adulterant (here water), the concentration of the mixture becomes 90% of the initial concentration.
Similarly when a proportion p of a mixture is replaced with pure adulterant, the concentration of the mixture becomes (1 - p) times the previous concentration. (0 ≤ p ≤ 1)

Here p = 4/40 = 0.1 
Initially there was pure milk, so the strength of the mixture is 100% or 1 (in terms of proportion)
After repeating the process n times the concentration of Milk in the mixture will be 1×0.9n
For the volume of milk to be less than the volume of water, p should be less than 0.5
So we are looking for the smallest n, such that 1 × 0.9n < 0.5
This happens when n = 7

*Answer can only contain numeric values
CAT 2023 Slot 2: Past Year Question Paper - Question 39

Jayant bought a certain number of white shirts at the rate of Rs 1000 per piece and a certain number of blue shirts at the rate of Rs 1125 per piece. For each shirt, he then set a fixed market price which was 25% higher than the average cost of all the shirts. He sold all the shirts at a discount of 10% and made a total profit of Rs 51000. If he bought both colors of shirts, then the maximum possible total number of shirts that he could have bought is


Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 39

Let the number of white and black shirts bought by Jayant be w and b respectively.
Then the total Cost Price (CP) =1000 × w + 1125 × b =1000(w + b) + 125 × b
Since the goods are marked up by 25% and then offered at a discount of 10%, the total Selling Price (SP)
= CP × 1.25 × 0.9 = 1.125 CP
This implies that there was a 12.5% of Profit, which is given to be 51, 000
12.5%(CP) = 51,000
CP = 4,08,000
1000(w + b) + 125 × b = 4,08,000
w and b are positive integers (since at least one shirt of each color needs to be purchased.)
To purchase maximum number of shirts, you need to purchase minimum number of the costlier shirts, which are the blue ones…
Observe that the total CP is a multiple of 1000. And for that to happen b should be a multiple of 8 in
1000(w + b) + 125 × b = 4,08,000.
So the minimum value of b = 8, in which case, w = 399.
Hence, the maximum number of shirts that can be purchased = 399 + 8 = 407

*Answer can only contain numeric values
CAT 2023 Slot 2: Past Year Question Paper - Question 40

If a certain amount of money is divided equally among n persons, each one receives Rs 352. However, if two persons receive Rs 506 each and the remaining amount is divided equally among the other persons, each of them receive less than or equal to Rs 330. Then, the maximum possible value of n is


Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 40

Let the total amount be equal to T.
T = n × 352
“However, if two persons receive Rs 506 each and the remaining amount is divided equally among the other persons, each of them receive less than or equal to Rs 330”
 
So, the maximum value that n can take is 16.

CAT 2023 Slot 2: Past Year Question Paper - Question 41

In a rectangle ABCD, AB = 9 cm and BC = 6 cm. P and Q are two points on BC such that the areas of the figures ABP, APQ, and AQCD are in geometric progression. If the area of the figure AQCD is four times the area of triangle ABP, then BP : PQ : QC is

Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 41

Let BP = x, PQ = y and QC = z.

Since Ar(ABP), Ar(APQ) & Ar(AQCD) are in GP and Ar(AQCD) = 4 × Ar(ABP)

x + y + z = 6
x + 2x + 4x – 6 = 6

CAT 2023 Slot 2: Past Year Question Paper - Question 42

A triangle is drawn with its vertices on the circle C such that one of its sides is a diameter of C and the other two sides have their lengths in the ratio a : b. If the radius of the circle is r, then the area of the triangle is

Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 42

Since the angle subtended by the diameter on the circle is a right-angle, such a triangle inscribed in a circle with the diameter as one of its sides will be right angled.

“…and the other two sides have their lengths in the ratio a: b.”
Let the two sides be ax and bx.
The area of the triangle 

*Answer can only contain numeric values
CAT 2023 Slot 2: Past Year Question Paper - Question 43

The area of the quadrilateral bounded by the Y -axis, the line x = 5, and the lines |x − y| − |x − 5| = 2, is


Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 43


 

CAT 2023 Slot 2: Past Year Question Paper - Question 44

Let both the series a1, a2, a3,… and b1, b2, b3… be in arithmetic progression such that the common differences of both the series are prime numbers. If a5 = b9, a19 = b19 and b2 = 0 , then a11 equals

Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 44

Let the common differences of the series a1, a2, a3,… and b1, b2, b3 be p and q respectively.
We are told that p and q are prime numbers.

CAT 2023 Slot 2: Past Year Question Paper - Question 45

If p2 + q2 − 29 = 2pq − 20 = 52 − 2pq, then the difference between the maximum and minimum possible value of (p3 − q3) is

Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 45


*Answer can only contain numeric values
CAT 2023 Slot 2: Past Year Question Paper - Question 46

Let an and bbe two sequences such that an = 13 + 6(n − 1) and bn = 15 + 7(n − 1) for all natural numbers n. Then, the largest three digit integer that is common to both these sequences, is


Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 46


CAT 2023 Slot 2: Past Year Question Paper - Question 47

What is the total number of coins in all the boxes in the 3rd row?

Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 47

We are given that each box contains three sacks. Each sack has a certain number of coins, between 1 and 9, both inclusive.

The average number of coins per sack in the boxes are all distinct integers. The total number of coins in each row is the same. The total number of coins in each column is also the same.

=> The total number of coins in a box range from 3 (1 + 1 + 1) to 27 (9 + 9 + 9)

Since, it is given that the average number of coins per sack in the boxes are all distinct integers => The total number of coins in a box would be 3, 6, 9, 12, 15, 18, 21, 24, 27 => averages of 1, 2, 3, 4,....,9 => Sum = 45.

=> Sum of averages coins in a box in a row or column = 45/3 = 15 [The total number of coins in each row is the same. The total number of coins in each column is also the same.] ==> (1)

Let us represent the final configuration of the sacks in boxes as follows:

Also a bag (x,y) => bag in xth row and yth column.

We are given 2 clues => Table-1 & Table-2

Consider bag (3,1)

=> From Table-1 => Median = 8 & From Table-2 all 3 sacks have more than 5 coins. Also * => There is a 9 in one of the sacks.

=> c, 8, 9 are the coins in bag (3,1), now c > 5 & c + 8 + 9 should be a multiple of 3 => c = 7 is the only possiblility.

=> bag (3,1) has 7, 8, 9 coins with average = 8.

Consider bag (2,1)

Median = 2 and 1 sack has more than 5 coins. Also ** => conditions i & iii should be satisfied.

=> 1, 2, 9 are the coins in bag (2,1) with average = 4

Consider bag (1,2)

Median = 9 and 2 elements are more than 5. Also * => (9 is present & 1 is not present)

=> c, 9, 9 are the coins in bag (1,2) and c is not equal to 1 and less than 5 => c = 3 for c + 18 to be a multiple of 3.

=> 3, 9, 9  are the coins in bag (1,2) with average = 7.

Capturing this info. in the table:

From (1), The average in bag (1,1) is 15 - 4 - 8 = 3.

From (1), The average in bag (1,3) is 15 - 3 - 7 = 5.

Consider bag (1,1)

Avg = 3, 1 sack has more than 5 and ** =>  2 conditions are being satisfied. => (can't be condition-3 => 9 coins as the total sum of coins is it self 3*3 = 9)

=> bag (1,1) has 1, 1, 7 coins with average = 3.

Consider bag (1,3)

Avg. = 5 => Sum = 15.

Median = 6 and 2 sacks have more than 5 and * => (1 condition is satisfied)

Not condition ii as the median is 6 & Not condition iii as the sum of 2 sacks itself will become 6 + 9 = 15

=> 1, 6, c are the coins => For sum = 15 => c = 15 - 1 - 6 = 8

=> bag (1,3) has 1, 6, 8 coins with average = 5.

Consider bag (3,3)

0 sacks have more than 5 coins and ** => conditions i & ii are being satisfied.

=> 1,1,c are the coins. Now c = 1 or 2 or 3 or 4 => c  = 1 or 4 for number of coins to be a multiple of 3.

But c = 1 as no other bag has the possibility to get avg. = 1 as bag (2,2) should have 1, b, c coins and b and c should be more than 1 as only 1*

=> bag (3,3) has 1, 1, 1 coins with average = 1.

Now, we can fill the averages in all the bags.

In bag (2,3) Avg. = 9 => 9, 9, 9  are the coins.

In bag (2,2) => Avg. = 2 => Sum = 6 and only 1* => smallest elemens=t should be 1.

=> 1, b, c are the coins where b + c = 5 and b,c can't be equal to 1 and less than 5 => 2 + 3 = 5 is the only possibility.

=> 1, 2, 3  are the coins with average = 2.

Considering bag (3,2)

Avg. = 6 => Sum = 18.

2 sacks more than 5 coins and ** => 2 sacks have 1 and 9 coins.

=> bag (3,2) has 1, c, 9 coins and c = 18 - 1 - 9 = 8

=> bag (3,2) has 1, 8, 9 coins with average = 6 coins.

==> Final required table, bracket number => average coins per sack in the bag.

Sum of coins in 3rd row = 8*3 + 6*3 + 1*3 = 45.

CAT 2023 Slot 2: Past Year Question Paper - Question 48

How many boxes have at least one sack containing 9 coins?

Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 48

We are given that each box contains three sacks. Each sack has a certain number of coins, between 1 and 9, both inclusive.

The average number of coins per sack in the boxes are all distinct integers. The total number of coins in each row is the same. The total number of coins in each column is also the same.

=> The total number of coins in a box range from 3 (1 + 1 + 1) to 27 (9 + 9 + 9)

Since, it is given that the average number of coins per sack in the boxes are all distinct integers => The total number of coins in a box would be 3, 6, 9, 12, 15, 18, 21, 24, 27 => averages of 1, 2, 3, 4,....,9 => Sum = 45.

=> Sum of averages coins in a box in a row or column = 45/3 = 15 [The total number of coins in each row is the same. The total number of coins in each column is also the same.] ==> (1)

Let us represent the final configuration of the sacks in boxes as follows:

Also a bag (x,y) => bag in xth row and yth column.

We are given 2 clues => Table-1 & Table-2

Consider bag (3,1)

=> From Table-1 => Median = 8 & From Table-2 all 3 sacks have more than 5 coins. Also * => There is a 9 in one of the sacks.

=> c, 8, 9 are the coins in bag (3,1), now c > 5 & c + 8 + 9 should be a multiple of 3 => c = 7 is the only possiblility.

=> bag (3,1) has 7, 8, 9 coins with average = 8.

Consider bag (2,1)

Median = 2 and 1 sack has more than 5 coins. Also ** => conditions i & iii should be satisfied.

=> 1, 2, 9 are the coins in bag (2,1) with average = 4

Consider bag (1,2)

Median = 9 and 2 elements are more than 5. Also * => (9 is present & 1 is not present)

=> c, 9, 9 are the coins in bag (1,2) and c is not equal to 1 and less than 5 => c = 3 for c + 18 to be a multiple of 3.

=> 3, 9, 9 are the coins in bag (1,2) with average = 7.

Capturing this info. in the table:

From (1), The average in bag (1,1) is 15 - 4 - 8 = 3.

From (1), The average in bag (1,3) is 15 - 3 - 7 = 5.

Consider bag (1,1)

Avg = 3, 1 sack has more than 5 and ** => 2 conditions are being satisfied. => (can't be condition-3 => 9 coins as the total sum of coins is it self 3*3 = 9)

=> bag (1,1) has 1, 1, 7 coins with average = 3.

Consider bag (1,3)

Avg. = 5 => Sum = 15.

Median = 6 and 2 sacks have more than 5 and * => (1 condition is satisfied)

Not condition ii as the median is 6 & Not condition iii as the sum of 2 sacks itself will become 6 + 9 = 15

=> 1, 6, c are the coins => For sum = 15 => c = 15 - 1 - 6 = 8

=> bag (1,3) has 1, 6, 8 coins with average = 5.

Consider bag (3,3)

0 sacks have more than 5 coins and ** => conditions i & ii are being satisfied.

=> 1,1,c are the coins. Now c = 1 or 2 or 3 or 4 => c = 1 or 4 for number of coins to be a multiple of 3.

But c = 1 as no other bag has the possibility to get avg. = 1 as bag (2,2) should have 1, b, c coins and b and c should be more than 1 as only 1*

=> bag (3,3) has 1, 1, 1 coins with average = 1.

Now, we can fill the averages in all the bags.

In bag (2,3) Avg. = 9 => 9, 9, 9 are the coins.

In bag (2,2) => Avg. = 2 => Sum = 6 and only 1* => smallest elemens=t should be 1.

=> 1, b, c are the coins where b + c = 5 and b,c can't be equal to 1 and less than 5 => 2 + 3 = 5 is the only possibility.

=> 1, 2, 3 are the coins with average = 2.

Considering bag (3,2)

Avg. = 6 => Sum = 18.

2 sacks more than 5 coins and ** => 2 sacks have 1 and 9 coins.

=> bag (3,2) has 1, c, 9 coins and c = 18 - 1 - 9 = 8

=> bag (3,2) has 1, 8, 9 coins with average = 6 coins.

==> Final required table, bracket number => average coins per sack in the bag.

Bags (2,1), (3,1), (1,2), (3,2), (2,3) have at least 1 sack with 9 coins. => Total of 5 bags.

*Answer can only contain numeric values
CAT 2023 Slot 2: Past Year Question Paper - Question 49

For how many boxes are the average and median of the numbers of coins contained in the three sacks in that box the same?


Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 49

We are given that each box contains three sacks. Each sack has a certain number of coins, between 1 and 9, both inclusive.

The average number of coins per sack in the boxes are all distinct integers. The total number of coins in each row is the same. The total number of coins in each column is also the same.

=> The total number of coins in a box range from 3 (1 + 1 + 1) to 27 (9 + 9 + 9)

Since, it is given that the average number of coins per sack in the boxes are all distinct integers => The total number of coins in a box would be 3, 6, 9, 12, 15, 18, 21, 24, 27 => averages of 1, 2, 3, 4,....,9 => Sum = 45.

=> Sum of averages coins in a box in a row or column = 45/3 = 15 [The total number of coins in each row is the same. The total number of coins in each column is also the same.] ==> (1)

Let us represent the final configuration of the sacks in boxes as follows:

Also a bag (x,y) => bag in xth row and yth column.

We are given 2 clues => Table-1 & Table-2

Consider bag (3,1)

=> From Table-1 => Median = 8 & From Table-2 all 3 sacks have more than 5 coins. Also * => There is a 9 in one of the sacks.

=> c, 8, 9 are the coins in bag (3,1), now c > 5 & c + 8 + 9 should be a multiple of 3 => c = 7 is the only possiblility.

=> bag (3,1) has 7, 8, 9 coins with average = 8.

Consider bag (2,1)

Median = 2 and 1 sack has more than 5 coins. Also ** => conditions i & iii should be satisfied.

=> 1, 2, 9 are the coins in bag (2,1) with average = 4

Consider bag (1,2)

Median = 9 and 2 elements are more than 5. Also * => (9 is present & 1 is not present)

=> c, 9, 9 are the coins in bag (1,2) and c is not equal to 1 and less than 5 => c = 3 for c + 18 to be a multiple of 3.

=> 3, 9, 9 are the coins in bag (1,2) with average = 7.

Capturing this info. in the table:

From (1), The average in bag (1,1) is 15 - 4 - 8 = 3.

From (1), The average in bag (1,3) is 15 - 3 - 7 = 5.

Consider bag (1,1)

Avg = 3, 1 sack has more than 5 and ** => 2 conditions are being satisfied. => (can't be condition-3 => 9 coins as the total sum of coins is it self 3*3 = 9)

=> bag (1,1) has 1, 1, 7 coins with average = 3.

Consider bag (1,3)

Avg. = 5 => Sum = 15.

Median = 6 and 2 sacks have more than 5 and * => (1 condition is satisfied)

Not condition ii as the median is 6 & Not condition iii as the sum of 2 sacks itself will become 6 + 9 = 15

=> 1, 6, c are the coins => For sum = 15 => c = 15 - 1 - 6 = 8

=> bag (1,3) has 1, 6, 8 coins with average = 5.

Consider bag (3,3)

0 sacks have more than 5 coins and ** => conditions i & ii are being satisfied.

=> 1,1,c are the coins. Now c = 1 or 2 or 3 or 4 => c = 1 or 4 for number of coins to be a multiple of 3.

But c = 1 as no other bag has the possibility to get avg. = 1 as bag (2,2) should have 1, b, c coins and b and c should be more than 1 as only 1*

=> bag (3,3) has 1, 1, 1 coins with average = 1.

Now, we can fill the averages in all the bags.

In bag (2,3) Avg. = 9 => 9, 9, 9 are the coins.

In bag (2,2) => Avg. = 2 => Sum = 6 and only 1* => smallest elemens=t should be 1.

=> 1, b, c are the coins where b + c = 5 and b,c can't be equal to 1 and less than 5 => 2 + 3 = 5 is the only possibility.

=> 1, 2, 3 are the coins with average = 2.

Considering bag (3,2)

Avg. = 6 => Sum = 18.

2 sacks more than 5 coins and ** => 2 sacks have 1 and 9 coins.

=> bag (3,2) has 1, c, 9 coins and c = 18 - 1 - 9 = 8

=> bag (3,2) has 1, 8, 9 coins with average = 6 coins.

==> Final required table, bracket number => average coins per sack in the bag.

Average = Median in boxes (3,1), (2,2), (2,3) and (3,3) => 4 boxes.

*Answer can only contain numeric values
CAT 2023 Slot 2: Past Year Question Paper - Question 50

How many sacks have exactly one coin?


Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 50

We are given that each box contains three sacks. Each sack has a certain number of coins, between 1 and 9, both inclusive.

The average number of coins per sack in the boxes are all distinct integers. The total number of coins in each row is the same. The total number of coins in each column is also the same.

=> The total number of coins in a box range from 3 (1 + 1 + 1) to 27 (9 + 9 + 9)

Since, it is given that the average number of coins per sack in the boxes are all distinct integers => The total number of coins in a box would be 3, 6, 9, 12, 15, 18, 21, 24, 27 => averages of 1, 2, 3, 4,....,9 => Sum = 45.

=> Sum of averages coins in a box in a row or column = 45/3 = 15 [The total number of coins in each row is the same. The total number of coins in each column is also the same.] ==> (1)

Let us represent the final configuration of the sacks in boxes as follows:

Also a bag (x,y) => bag in xth row and yth column.

We are given 2 clues => Table-1 & Table-2

Consider bag (3,1)

=> From Table-1 => Median = 8 & From Table-2 all 3 sacks have more than 5 coins. Also * => There is a 9 in one of the sacks.

=> c, 8, 9 are the coins in bag (3,1), now c > 5 & c + 8 + 9 should be a multiple of 3 => c = 7 is the only possiblility.

=> bag (3,1) has 7, 8, 9 coins with average = 8.

Consider bag (2,1)

Median = 2 and 1 sack has more than 5 coins. Also ** => conditions i & iii should be satisfied.

=> 1, 2, 9 are the coins in bag (2,1) with average = 4

Consider bag (1,2)

Median = 9 and 2 elements are more than 5. Also * => (9 is present & 1 is not present)

=> c, 9, 9 are the coins in bag (1,2) and c is not equal to 1 and less than 5 => c = 3 for c + 18 to be a multiple of 3.

=> 3, 9, 9 are the coins in bag (1,2) with average = 7.

Capturing this info. in the table:

From (1), The average in bag (1,1) is 15 - 4 - 8 = 3.

From (1), The average in bag (1,3) is 15 - 3 - 7 = 5.

Consider bag (1,1)

Avg = 3, 1 sack has more than 5 and ** => 2 conditions are being satisfied. => (can't be condition-3 => 9 coins as the total sum of coins is it self 3*3 = 9)

=> bag (1,1) has 1, 1, 7 coins with average = 3.

Consider bag (1,3)

Avg. = 5 => Sum = 15.

Median = 6 and 2 sacks have more than 5 and * => (1 condition is satisfied)

Not condition ii as the median is 6 & Not condition iii as the sum of 2 sacks itself will become 6 + 9 = 15

=> 1, 6, c are the coins => For sum = 15 => c = 15 - 1 - 6 = 8

=> bag (1,3) has 1, 6, 8 coins with average = 5.

Consider bag (3,3)

0 sacks have more than 5 coins and ** => conditions i & ii are being satisfied.

=> 1,1,c are the coins. Now c = 1 or 2 or 3 or 4 => c = 1 or 4 for number of coins to be a multiple of 3.

But c = 1 as no other bag has the possibility to get avg. = 1 as bag (2,2) should have 1, b, c coins and b and c should be more than 1 as only 1*

=> bag (3,3) has 1, 1, 1 coins with average = 1.

Now, we can fill the averages in all the bags.

In bag (2,3) Avg. = 9 => 9, 9, 9 are the coins.

In bag (2,2) => Avg. = 2 => Sum = 6 and only 1* => smallest elemens=t should be 1.

=> 1, b, c are the coins where b + c = 5 and b,c can't be equal to 1 and less than 5 => 2 + 3 = 5 is the only possibility.

=> 1, 2, 3 are the coins with average = 2.

Considering bag (3,2)

Avg. = 6 => Sum = 18.

2 sacks more than 5 coins and ** => 2 sacks have 1 and 9 coins.

=> bag (3,2) has 1, c, 9 coins and c = 18 - 1 - 9 = 8

=> bag (3,2) has 1, 8, 9 coins with average = 6 coins.

==> Final required table, bracket number => average coins per sack in the bag.

Bag (1,1) => 2 sacks with 1 coins, (2,1) => 1 sack, (2,2) => 1 sack, (3,2) => 1 sack, (1,3) => 1 sack, (3,3) => 3 sacks.

=> Total = 2 + 1 + 1 + 1 + 1 + 3 = 9 sacks.

*Answer can only contain numeric values
CAT 2023 Slot 2: Past Year Question Paper - Question 51

In how many boxes do all three sacks contain different numbers of coins?


Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 51

We are given that each box contains three sacks. Each sack has a certain number of coins, between 1 and 9, both inclusive.

The average number of coins per sack in the boxes are all distinct integers. The total number of coins in each row is the same. The total number of coins in each column is also the same.

=> The total number of coins in a box range from 3 (1 + 1 + 1) to 27 (9 + 9 + 9)

Since, it is given that the average number of coins per sack in the boxes are all distinct integers => The total number of coins in a box would be 3, 6, 9, 12, 15, 18, 21, 24, 27 => averages of 1, 2, 3, 4,....,9 => Sum = 45.

=> Sum of averages coins in a box in a row or column = 45/3 = 15 [The total number of coins in each row is the same. The total number of coins in each column is also the same.] ==> (1)

Let us represent the final configuration of the sacks in boxes as follows:

Also a bag (x,y) => bag in xth row and yth column.

We are given 2 clues => Table-1 & Table-2

Consider bag (3,1)

=> From Table-1 => Median = 8 & From Table-2 all 3 sacks have more than 5 coins. Also * => There is a 9 in one of the sacks.

=> c, 8, 9 are the coins in bag (3,1), now c > 5 & c + 8 + 9 should be a multiple of 3 => c = 7 is the only possiblility.

=> bag (3,1) has 7, 8, 9 coins with average = 8.

Consider bag (2,1)

Median = 2 and 1 sack has more than 5 coins. Also ** => conditions i & iii should be satisfied.

=> 1, 2, 9 are the coins in bag (2,1) with average = 4

Consider bag (1,2)

Median = 9 and 2 elements are more than 5. Also * => (9 is present & 1 is not present)

=> c, 9, 9 are the coins in bag (1,2) and c is not equal to 1 and less than 5 => c = 3 for c + 18 to be a multiple of 3.

=> 3, 9, 9 are the coins in bag (1,2) with average = 7.

Capturing this info. in the table:

From (1), The average in bag (1,1) is 15 - 4 - 8 = 3.

From (1), The average in bag (1,3) is 15 - 3 - 7 = 5.

Consider bag (1,1)

Avg = 3, 1 sack has more than 5 and ** => 2 conditions are being satisfied. => (can't be condition-3 => 9 coins as the total sum of coins is it self 3*3 = 9)

=> bag (1,1) has 1, 1, 7 coins with average = 3.

Consider bag (1,3)

Avg. = 5 => Sum = 15.

Median = 6 and 2 sacks have more than 5 and * => (1 condition is satisfied)

Not condition ii as the median is 6 & Not condition iii as the sum of 2 sacks itself will become 6 + 9 = 15

=> 1, 6, c are the coins => For sum = 15 => c = 15 - 1 - 6 = 8

=> bag (1,3) has 1, 6, 8 coins with average = 5.

Consider bag (3,3)

0 sacks have more than 5 coins and ** => conditions i & ii are being satisfied.

=> 1,1,c are the coins. Now c = 1 or 2 or 3 or 4 => c = 1 or 4 for number of coins to be a multiple of 3.

But c = 1 as no other bag has the possibility to get avg. = 1 as bag (2,2) should have 1, b, c coins and b and c should be more than 1 as only 1*

=> bag (3,3) has 1, 1, 1 coins with average = 1.

Now, we can fill the averages in all the bags.

In bag (2,3) Avg. = 9 => 9, 9, 9 are the coins.

In bag (2,2) => Avg. = 2 => Sum = 6 and only 1* => smallest elemens=t should be 1.

=> 1, b, c are the coins where b + c = 5 and b,c can't be equal to 1 and less than 5 => 2 + 3 = 5 is the only possibility.

=> 1, 2, 3 are the coins with average = 2.

Considering bag (3,2)

Avg. = 6 => Sum = 18.

2 sacks more than 5 coins and ** => 2 sacks have 1 and 9 coins.

=> bag (3,2) has 1, c, 9 coins and c = 18 - 1 - 9 = 8

=> bag (3,2) has 1, 8, 9 coins with average = 6 coins.

==> Final required table, bracket number => average coins per sack in the bag.

Bags with different number of coins in all 3 sacks are (2,1), (3,2), (2,2), (3,2), (1,3) => 5 bags.

CAT 2023 Slot 2: Past Year Question Paper - Question 52

What is Akhil's score on Day 1?

Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 52

 

Let the total score of day 1, day 2, day 3, day 4, and day 5 are d1, d2, d3, d4, and d5, respectively. 

The table shows that d1+d2 = 30 ... eq (1), d2+d3 = 31 ... eq (2), d3+d4 = 32 .... eq(3), d4+d5 = 34 ... eq(4)

It is given that participants are ranked each day, with the person having the maximum score being awarded the minimum rank (1) on that day. All participants with a tied score are awarded the best available rank if there is a tie.

It is given that the total score on Day 3 is the same as the total score on Day 4. 

Therefore, d3 = d4 => d3 = d4 = 16, which implies d2 = 15, d5 = 18, and d1 = 15.

The day-wise score is given below:

It is known that Chatur always scores in multiples of 3. His score on Day 2 is the unique highest score in the competition. His minimum score is observed only on Day 1, and it matches Akhil’s score on Day 4.

Hence, only Chatur scored 9 (one time) on Day 2, and no other person scored 9 on any of the given 5 days. Chatur scored 3 only one time, which was on Day 1. Therefore, the scores obtained by Chatur on Day 3, Day 4, and Day 5 are 6, 6, and 6, respectively. It is also known that Akhil's score on Day 4 is the same as the score obtained by Chatur on Day 1. Hence, Akhil's score on Day 4 is 3.

Hence, we get the following table:

From Table 2, we see that the rank of Bimal and Akhil is the same, which is 2. Hence, The score obtained by Akhil and Bimal is the same. Let the score be x. Therefore, 6+2x = 16 => x = 5

The rank of Chatur on Day 5 is 2, and the rank of Bimal is 1, which implies the score obtained by Bimal will be more than Chatur. Hence, Bimal can score either 7 or 8 on Day 5. Therefore, the score obtained by Akhil on Day 5 is either 5 or 4. 

It is given that Bimal’s scores are the same on Day 1 and Day 3. Hence, the score obtained by Bimal on Day 1 is 5, which implies The score obtained by Akhil is 7 on Day 1.

From Table 2, we can see that the rank of Bimal is 3 on Day 2, and the rank of Akhil is 2 on Day 2. Hence, the score of Bimal will be lower than Akhil on Day 2. 

Let the score of Akhil be a, and the score of Bimal be b. Then 9+a+b = 15, and a > b

=> a+b =6, and a> b

Hence, the value of a can be 4/5, and the value of b can be 2/1

Therefore, the final table is given below:

From the table, we can see that the score of Akhil is 7 on day 1.

The correct option is B

CAT 2023 Slot 2: Past Year Question Paper - Question 53

Who attains the maximum total score?

Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 53

 

Let the total score of day 1, day 2, day 3, day 4, and day 5 are d1, d2, d3, d4, and d5, respectively.

The table shows that d1+d2 = 30 ... eq (1), d2+d3 = 31 ... eq (2), d3+d4 = 32 .... eq(3), d4+d5 = 34 ... eq(4)

It is given that participants are ranked each day, with the person having the maximum score being awarded the minimum rank (1) on that day. All participants with a tied score are awarded the best available rank if there is a tie.

It is given that the total score on Day 3 is the same as the total score on Day 4.

Therefore, d3 = d4 => d3 = d4 = 16, which implies d2 = 15, d5 = 18, and d1 = 15.

The day-wise score is given below:

It is known that Chatur always scores in multiples of 3. His score on Day 2 is the unique highest score in the competition. His minimum score is observed only on Day 1, and it matches Akhil’s score on Day 4.

Hence, only Chatur scored 9 (one time) on Day 2, and no other person scored 9 on any of the given 5 days. Chatur scored 3 only one time, which was on Day 1. Therefore, the scores obtained by Chatur on Day 3, Day 4, and Day 5 are 6, 6, and 6, respectively. It is also known that Akhil's score on Day 4 is the same as the score obtained by Chatur on Day 1. Hence, Akhil's score on Day 4 is 3.

Hence, we get the following table:

From Table 2, we see that the rank of Bimal and Akhil is the same, which is 2. Hence, The score obtained by Akhil and Bimal is the same. Let the score be x. Therefore, 6+2x = 16 => x = 5

The rank of Chatur on Day 5 is 2, and the rank of Bimal is 1, which implies the score obtained by Bimal will be more than Chatur. Hence, Bimal can score either 7 or 8 on Day 5. Therefore, the score obtained by Akhil on Day 5 is either 5 or 4.

It is given that Bimal’s scores are the same on Day 1 and Day 3. Hence, the score obtained by Bimal on Day 1 is 5, which implies The score obtained by Akhil is 7 on Day 1.

From Table 2, we can see that the rank of Bimal is 3 on Day 2, and the rank of Akhil is 2 on Day 2. Hence, the score of Bimal will be lower than Akhil's on Day 2.

Let the score of Akhil be a, and the score of Bimal be b. Then 9+a+b = 15, and a > b

=> a+b =6, and a> b

Hence, the value of a can be 4/5, and the value of b can be 2/1

Therefore, the final table is given below:

From the table, we can see that the maximum score is obtained by Chatur. 

*Answer can only contain numeric values
CAT 2023 Slot 2: Past Year Question Paper - Question 54

What is the minimum possible total score of Bimal?


Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 54

Let the total score of day 1, day 2, day 3, day 4, and day 5 are d1, d2, d3, d4, and d5, respectively.

The table shows that d1+d2 = 30 ... eq (1), d2+d3 = 31 ... eq (2), d3+d4 = 32 .... eq(3), d4+d5 = 34 ... eq(4)

It is given that participants are ranked each day, with the person having the maximum score being awarded the minimum rank (1) on that day. All participants with a tied score are awarded the best available rank if there is a tie.

It is given that the total score on Day 3 is the same as the total score on Day 4.

Therefore, d3 = d4 => d3 = d4 = 16, which implies d2 = 15, d5 = 18, and d1 = 15.

The day-wise score is given below:

It is known that Chatur always scores in multiples of 3. His score on Day 2 is the unique highest score in the competition. His minimum score is observed only on Day 1, and it matches Akhil’s score on Day 4.

Hence, only Chatur scored 9 (one time) on Day 2, and no other person scored 9 on any of the given 5 days. Chatur scored 3 only one time, which was on Day 1. Therefore, the scores obtained by Chatur on Day 3, Day 4, and Day 5 are 6, 6, and 6, respectively. It is also known that Akhil's score on Day 4 is the same as the score obtained by Chatur on Day 1. Hence, Akhil's score on Day 4 is 3.

Hence, we get the following table:

From Table 2, we see that the rank of Bimal and Akhil is the same, which is 2. Hence, The score obtained by Akhil and Bimal is the same. Let the score be x. Therefore, 6+2x = 16 => x = 5

The rank of Chatur on Day 5 is 2, and the rank of Bimal is 1, which implies the score obtained by Bimal will be more than Chatur. Hence, Bimal can score either 7 or 8 on Day 5. Therefore, the score obtained by Akhil on Day 5 is either 5 or 4.

It is given that Bimal’s scores are the same on Day 1 and Day 3. Hence, the score obtained by Bimal on Day 1 is 5, which implies The score obtained by Akhil is 7 on Day 1.

From Table 2, we can see that the rank of Bimal is 3 on Day 2, and the rank of Akhil is 2 on Day 2. Hence, the score of Bimal will be lower than Akhil on Day 2.

Let the score of Akhil be a, and the score of Bimal be b. Then 9+a+b = 15, and a > b

=> a+b =6, and a> b

Hence, the value of a can be 4/5, and the value of b can be 2/1

Therefore, the final table is given below:

From the table, we can see that the minimum score obtained by Bimal is 25.

CAT 2023 Slot 2: Past Year Question Paper - Question 55

If the total score of Bimal is a multiple of 3, what is the score of Akhil on Day 2?

Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 55

 

Let the total score of day 1, day 2, day 3, day 4, and day 5 are d1, d2, d3, d4, and d5, respectively.

The table shows that d1+d2 = 30 ... eq (1), d2+d3 = 31 ... eq (2), d3+d4 = 32 .... eq(3), d4+d5 = 34 ... eq(4)

It is given that participants are ranked each day, with the person having the maximum score being awarded the minimum rank (1) on that day. All participants with a tied score are awarded the best available rank if there is a tie.

It is given that the total score on Day 3 is the same as the total score on Day 4.

Therefore, d3 = d4 => d3 = d4 = 16, which implies d2 = 15, d5 = 18, and d1 = 15.

The day-wise score is given below:

It is known that Chatur always scores in multiples of 3. His score on Day 2 is the unique highest score in the competition. His minimum score is observed only on Day 1, and it matches Akhil’s score on Day 4.

Hence, only Chatur scored 9 (one time) on Day 2, and no other person scored 9 on any of the given 5 days. Chatur scored 3 only one time, which was on Day 1. Therefore, the scores obtained by Chatur on Day 3, Day 4, and Day 5 are 6, 6, and 6, respectively. It is also known that Akhil's score on Day 4 is the same as the score obtained by Chatur on Day 1. Hence, Akhil's score on Day 4 is 3.

Hence, we get the following table:

From Table 2, we see that the rank of Bimal and Akhil is the same, which is 2. Hence, The score obtained by Akhil and Bimal is the same. Let the score be x. Therefore, 6+2x = 16 => x = 5

The rank of Chatur on Day 5 is 2, and the rank of Bimal is 1, which implies the score obtained by Bimal will be more than Chatur. Hence, Bimal can score either 7 or 8 on Day 5. Therefore, the score obtained by Akhil on Day 5 is either 5 or 4.

It is given that Bimal’s scores are the same on Day 1 and Day 3. Hence, the score obtained by Bimal on Day 1 is 5, which implies The score obtained by Akhil is 7 on Day 1.

From Table 2, we can see that the rank of Bimal is 3 on Day 2, and the rank of Akhil is 2 on Day 2. Hence, the score of Bimal will be lower than Akhil on Day 2.

Let the score of Akhil be a, and the score of Bimal be b. Then 9+a+b = 15, and a > b

=> a+b =6, and a> b

Hence, the value of a can be 4/5, and the value of b can be 2/1

Therefore, the final table is given below:


In the question, it is given that the total score obtained by Bimal is a multiple of 3, which implies the total score obtained by Bimal is 27, which implies the total score obtained by Akhil is 23.

Akhil will score 23, when his scores on Days 1, 2, 3, 4, and 5 are 7, 4, 5, 3, 4, respectively.

Hence, the score obtained by him on Day 2 is 4.

*Answer can only contain numeric values
CAT 2023 Slot 2: Past Year Question Paper - Question 56

If Akhil attains a total score of 24, then what is the total score of Bimal?


Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 56

Let the total score of day 1, day 2, day 3, day 4, and day 5 are d1, d2, d3, d4, and d5, respectively.

The table shows that d1+d2 = 30 ... eq (1), d2+d3 = 31 ... eq (2), d3+d4 = 32 .... eq(3), d4+d5 = 34 ... eq(4)

It is given that participants are ranked each day, with the person having the maximum score being awarded the minimum rank (1) on that day. All participants with a tied score are awarded the best available rank if there is a tie.

It is given that the total score on Day 3 is the same as the total score on Day 4.

Therefore, d3 = d4 => d3 = d4 = 16, which implies d2 = 15, d5 = 18, and d1 = 15.

The day-wise score is given below:

It is known that Chatur always scores in multiples of 3. His score on Day 2 is the unique highest score in the competition. His minimum score is observed only on Day 1, and it matches Akhil’s score on Day 4.

Hence, only Chatur scored 9 (one time) on Day 2, and no other person scored 9 on any of the given 5 days. Chatur scored 3 only one time, which was on Day 1. Therefore, the scores obtained by Chatur on Day 3, Day 4, and Day 5 are 6, 6, and 6, respectively. It is also known that Akhil's score on Day 4 is the same as the score obtained by Chatur on Day 1. Hence, Akhil's score on Day 4 is 3.

Hence, we get the following table:

From Table 2, we see that the rank of Bimal and Akhil is the same, which is 2. Hence, The score obtained by Akhil and Bimal is the same. Let the score be x. Therefore, 6+2x = 16 => x = 5

The rank of Chatur on Day 5 is 2, and the rank of Bimal is 1, which implies the score obtained by Bimal will be more than Chatur. Hence, Bimal can score either 7 or 8 on Day 5. Therefore, the score obtained by Akhil on Day 5 is either 5 or 4.

It is given that Bimal’s scores are the same on Day 1 and Day 3. Hence, the score obtained by Bimal on Day 1 is 5, which implies The score obtained by Akhil is 7 on Day 1.

From Table 2, we can see that the rank of Bimal is 3 on Day 2, and the rank of Akhil is 2 on Day 2. Hence, the score of Bimal will be lower than Akhil on Day 2.

Let the score of Akhil be a, and the score of Bimal be b. Then 9+a+b = 15, and a > b

=> a+b =6, and a> b

Hence, the value of a can be 4/5, and the value of b can be 2/1

Therefore, the final table is given below:

In the question, it is given that the score obtained by Akhil is 24, which implies the score obtained by Bimal is 26. 

CAT 2023 Slot 2: Past Year Question Paper - Question 57

For which firm(s) can the amounts raised by them be concluded with certainty in each year?

Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 57

In this set, we are told that the amount each firm raised every year increased until it reached a maximum, and then decreased until the firm closed down and no firm raised the same amount of money in two consecutive years.

The increase or decrease can be ± 1± 1 or ± 2± 2. => (1)

We are also told that each firm raised Rs. 1 crore in its first and last year of existence

Consider A:

It raised money for 8 years

=> The raising pattern looks like follows:

1, a, b, c, d, e, f, 1 => where a, b, c,..,, f are the unknown amounts raised.

Also a + b + c + d + e + f = 21 - 2 = 19.

We can observe that 19/6 is slightly greater than 3 => The average amount raised should be around 3.

If a = 3 and f = 3 => b + c + d + e = 13 (not possible) as the minimum case would be (4, 5, 6, 4) => Not possible.

If a = 3 and f = 2 => b + c + d + e = 14 (not possible) as the minimum case would be (4, 5, 4, 3) => Not possible.

=> a = 2 and f = 2 => b + c + d + e = 15 the minimum case is (3, 4, 5, 3) or (3, 5, 4, 3) which gives a sum of 15.

So, the possible cases for A are:

Consider B:

The patterns looks as follows:

1, a, b, 1

If a = 2, b has to be equal to 3 to satisfy (1)

if a = 3, b has to be equal to 2 to satisfy (1)

=> The possible cases for B are:

Consider C:

The pattern looks as follows:

1, ..., 1

Let us assume there are 2 gaps between => a + b = 7 (Not possible) as maximum case would be 1, 3, 2, 1

Let us assume there are 3 gaps between => a + b + c = 7, the minimum case possible is 1, 2, 3, 2, 1 => Satisfies.

Now, if there are 4 gaps => a + b + c + d = 7 => The average value is 7/4 which is less than 2 => Not possible.

=> The possible cases for C are:

Consider D:

The pattern looks as follows:

1, a, b, c, 1

=> a + b + c = 8

When a = 2 and c = 2 => b = 4 => 2, 4, 2 => Satisfies.

When a = 2 and c = 3, b should be 3 (Not satisfying (1))

When a = 3 and c = 3, b should be 2 (Not satisfying (1))

=> The possible cases for D are:

Consider E:

The pattern looks as follows:

1,.....,1

For 1 or 2 gaps, we can't get a sum of 11.

Assume 3 gaps => a + b + c = 11, the maximum case is 3, 5, 3 => Satisfies.

Now, assume 4 gaps

=> a + b + c + d = 11, the minimum case is 2, 3, 4, 2 or 2, 4, 3, 2 which satisfies (1) and 2 + 3 + 4 + 2 = 11.

=> The possible cases for E are:

In summary, the possible cases for all 5 companies is:

We see that only for C and D, we can conclude the amounts raised with certainty.

CAT 2023 Slot 2: Past Year Question Paper - Question 58

What best can be concluded about the total amount of money raised in 2015?

Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 58

In this set, we are told that the amount each firm raised every year increased until it reached a maximum, and then decreased until the firm closed down and no firm raised the same amount of money in two consecutive years.

The increase or decrease can be ± 1 or ± 2. => (1)

We are also told that each firm raised Rs. 1 crore in its first and last year of existence

Consider A:

It raised money for 8 years

=> The raising pattern looks like follows:

1, a, b, c, d, e, f, 1 => where a, b, c,..,, f are the unknown amounts raised.

Also a + b + c + d + e + f = 21 - 2 = 19.

We can observe that 19/6 is slightly greater than 3 => The average amount raised should be around 3.

If a = 3 and f = 3 => b + c + d + e = 13 (not possible) as the minimum case would be (4, 5, 6, 4) => Not possible.

If a = 3 and f = 2 => b + c + d + e = 14 (not possible) as the minimum case would be (4, 5, 4, 3) => Not possible.

=> a = 2 and f = 2 => b + c + d + e = 15 the minimum case is (3, 4, 5, 3) or (3, 5, 4, 3) which gives a sum of 15.

So, the possible cases for A are:

Consider B:

The patterns looks as follows:

1, a, b, 1

If a = 2, b has to be equal to 3 to satisfy (1)

if a = 3, b has to be equal to 2 to satisfy (1)

=> The possible cases for B are:

Consider C:

The pattern looks as follows:

1, ..., 1

Let us assume there are 2 gaps between => a + b = 7 (Not possible) as maximum case would be 1, 3, 2, 1

Let us assume there are 3 gaps between => a + b + c = 7, the minimum case possible is 1, 2, 3, 2, 1 => Satisfies.

Now, if there are 4 gaps => a + b + c + d = 7 => The average value is 7/4 which is less than 2 => Not possible.

=> The possible cases for C are:

Consider D:

The pattern looks as follows:

1, a, b, c, 1

=> a + b + c = 8

When a = 2 and c = 2 => b = 4 => 2, 4, 2 => Satisfies.

When a = 2 and c = 3, b should be 3 (Not satisfying (1))

When a = 3 and c = 3, b should be 2 (Not satisfying (1))

=> The possible cases for D are:

Consider E:

The pattern looks as follows:

1,.....,1

For 1 or 2 gaps, we can't get a sum of 11.

Assume 3 gaps => a + b + c = 11, the maximum case is 3, 5, 3 => Satisfies.

Now, assume 4 gaps

=> a + b + c + d = 11, the minimum case is 2, 3, 4, 2 or 2, 4, 3, 2 which satisfies (1) and 2 + 3 + 4 + 2 = 11.

=> The possible cases for E are:

In summary, the possible cases for all 5 companies is:

Money raised in 2015 is 2 + 1 + 3 + 1 + 0/1 = 7 or 8.

*Answer can only contain numeric values
CAT 2023 Slot 2: Past Year Question Paper - Question 59

What is the largest possible total amount of money (in Rs. crores) that could have been raised in 2013?


Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 59

In this set, we are told that the amount each firm raised every year increased until it reached a maximum, and then decreased until the firm closed down and no firm raised the same amount of money in two consecutive years.

The increase or decrease can be ± 1 or ± 2. => (1)

We are also told that each firm raised Rs. 1 crore in its first and last year of existence

Consider A:

It raised money for 8 years

=> The raising pattern looks like follows:

1, a, b, c, d, e, f, 1 => where a, b, c,..,, f are the unknown amounts raised.

Also a + b + c + d + e + f = 21 - 2 = 19.

We can observe that 19/6 is slightly greater than 3 => The average amount raised should be around 3.

If a = 3 and f = 3 => b + c + d + e = 13 (not possible) as the minimum case would be (4, 5, 6, 4) => Not possible.

If a = 3 and f = 2 => b + c + d + e = 14 (not possible) as the minimum case would be (4, 5, 4, 3) => Not possible.

=> a = 2 and f = 2 => b + c + d + e = 15 the minimum case is (3, 4, 5, 3) or (3, 5, 4, 3) which gives a sum of 15.

So, the possible cases for A are:

Consider B:

The patterns looks as follows:

1, a, b, 1

If a = 2, b has to be equal to 3 to satisfy (1)

if a = 3, b has to be equal to 2 to satisfy (1)

=> The possible cases for B are:

Consider C:

The pattern looks as follows:

1, ..., 1

Let us assume there are 2 gaps between => a + b = 7 (Not possible) as maximum case would be 1, 3, 2, 1

Let us assume there are 3 gaps between => a + b + c = 7, the minimum case possible is 1, 2, 3, 2, 1 => Satisfies.

Now, if there are 4 gaps => a + b + c + d = 7 => The average value is 7/4 which is less than 2 => Not possible.

=> The possible cases for C are:

Consider D:

The pattern looks as follows:

1, a, b, c, 1

=> a + b + c = 8

When a = 2 and c = 2 => b = 4 => 2, 4, 2 => Satisfies.

When a = 2 and c = 3, b should be 3 (Not satisfying (1))

When a = 3 and c = 3, b should be 2 (Not satisfying (1))

=> The possible cases for D are:

Consider E:

The pattern looks as follows:

1,.....,1

For 1 or 2 gaps, we can't get a sum of 11.

Assume 3 gaps => a + b + c = 11, the maximum case is 3, 5, 3 => Satisfies.

Now, assume 4 gaps

=> a + b + c + d = 11, the minimum case is 2, 3, 4, 2 or 2, 4, 3, 2 which satisfies (1) and 2 + 3 + 4 + 2 = 11.

=> The possible cases for E are:

In summary, the possible cases for all 5 companies is:

Maximum money raised in 2013 is 5 + 3 + 1 + 4 + 4 = 17.

CAT 2023 Slot 2: Past Year Question Paper - Question 60

If Elavalaki raised Rs. 3 crores in 2013, then what is the smallest possible total amount of money (in Rs. crores) that could have been raised by all the companies in 2012?

Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 60

In this set, we are told that the amount each firm raised every year increased until it reached a maximum, and then decreased until the firm closed down and no firm raised the same amount of money in two consecutive years.

The increase or decrease can be ± 1 or ± 2. => (1)

We are also told that each firm raised Rs. 1 crore in its first and last year of existence

Consider A:

It raised money for 8 years

=> The raising pattern looks like follows:

1, a, b, c, d, e, f, 1 => where a, b, c,..,, f are the unknown amounts raised.

Also a + b + c + d + e + f = 21 - 2 = 19.

We can observe that 19/6 is slightly greater than 3 => The average amount raised should be around 3.

If a = 3 and f = 3 => b + c + d + e = 13 (not possible) as the minimum case would be (4, 5, 6, 4) => Not possible.

If a = 3 and f = 2 => b + c + d + e = 14 (not possible) as the minimum case would be (4, 5, 4, 3) => Not possible.

=> a = 2 and f = 2 => b + c + d + e = 15 the minimum case is (3, 4, 5, 3) or (3, 5, 4, 3) which gives a sum of 15.

So, the possible cases for A are:

Consider B:

The patterns looks as follows:

1, a, b, 1

If a = 2, b has to be equal to 3 to satisfy (1)

if a = 3, b has to be equal to 2 to satisfy (1)

=> The possible cases for B are:

Consider C:

The pattern looks as follows:

1, ..., 1

Let us assume there are 2 gaps between => a + b = 7 (Not possible) as maximum case would be 1, 3, 2, 1

Let us assume there are 3 gaps between => a + b + c = 7, the minimum case possible is 1, 2, 3, 2, 1 => Satisfies.

Now, if there are 4 gaps => a + b + c + d = 7 => The average value is 7/4 which is less than 2 => Not possible.

=> The possible cases for C are:

Consider D:

The pattern looks as follows:

1, a, b, c, 1

=> a + b + c = 8

When a = 2 and c = 2 => b = 4 => 2, 4, 2 => Satisfies.

When a = 2 and c = 3, b should be 3 (Not satisfying (1))

When a = 3 and c = 3, b should be 2 (Not satisfying (1))

=> The possible cases for D are:

Consider E:

The pattern looks as follows:

1,.....,1

For 1 or 2 gaps, we can't get a sum of 11.

Assume 3 gaps => a + b + c = 11, the maximum case is 3, 5, 3 => Satisfies.

Now, assume 4 gaps

=> a + b + c + d = 11, the minimum case is 2, 3, 4, 2 or 2, 4, 3, 2 which satisfies (1) and 2 + 3 + 4 + 2 = 11.

=> The possible cases for E are:

In summary, the possible cases for all 5 companies is:

Given that E raised 3 in 2013 => in 2012 he could have raised a minimum of 4 crores.

=> Minimum amount is 4 + 1 + 0 + 2 + 4 = 11.

CAT 2023 Slot 2: Past Year Question Paper - Question 61

If the total amount of money raised in 2014 is Rs. 12 crores, then which of the following is not possible?

Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 61

In this set, we are told that the amount each firm raised every year increased until it reached a maximum, and then decreased until the firm closed down and no firm raised the same amount of money in two consecutive years.

The increase or decrease can be ± 1 or ± 2. => (1)

We are also told that each firm raised Rs. 1 crore in its first and last year of existence

Consider A:

It raised money for 8 years

=> The raising pattern looks like follows:

1, a, b, c, d, e, f, 1 => where a, b, c,..,, f are the unknown amounts raised.

Also a + b + c + d + e + f = 21 - 2 = 19.

We can observe that 19/6 is slightly greater than 3 => The average amount raised should be around 3.

If a = 3 and f = 3 => b + c + d + e = 13 (not possible) as the minimum case would be (4, 5, 6, 4) => Not possible.

If a = 3 and f = 2 => b + c + d + e = 14 (not possible) as the minimum case would be (4, 5, 4, 3) => Not possible.

=> a = 2 and f = 2 => b + c + d + e = 15 the minimum case is (3, 4, 5, 3) or (3, 5, 4, 3) which gives a sum of 15.

So, the possible cases for A are:

Consider B:

The patterns looks as follows:

1, a, b, 1

If a = 2, b has to be equal to 3 to satisfy (1)

if a = 3, b has to be equal to 2 to satisfy (1)

=> The possible cases for B are:

Consider C:

The pattern looks as follows:

1, ..., 1

Let us assume there are 2 gaps between => a + b = 7 (Not possible) as maximum case would be 1, 3, 2, 1

Let us assume there are 3 gaps between => a + b + c = 7, the minimum case possible is 1, 2, 3, 2, 1 => Satisfies.

Now, if there are 4 gaps => a + b + c + d = 7 => The average value is 7/4 which is less than 2 => Not possible.

=> The possible cases for C are:

Consider D:

The pattern looks as follows:

1, a, b, c, 1

=> a + b + c = 8

When a = 2 and c = 2 => b = 4 => 2, 4, 2 => Satisfies.

When a = 2 and c = 3, b should be 3 (Not satisfying (1))

When a = 3 and c = 3, b should be 2 (Not satisfying (1))

=> The possible cases for D are:

Consider E:

The pattern looks as follows:

1,.....,1

For 1 or 2 gaps, we can't get a sum of 11.

Assume 3 gaps => a + b + c = 11, the maximum case is 3, 5, 3 => Satisfies.

Now, assume 4 gaps

=> a + b + c + d = 11, the minimum case is 2, 3, 4, 2 or 2, 4, 3, 2 which satisfies (1) and 2 + 3 + 4 + 2 = 11.

=> The possible cases for E are:

In summary, the possible cases for all 5 companies is:

Given that total amount raised in 2014 is 12

=> 3 + 3/2 + 2 + 2 + 1/2 = 12 => 

=> possible case is 3 + 3 + 2 + 2 + 2 = 12.

A) In 2013, B raised 2 crores and E also raised 3/4 crores => Not Possible.

B) In 2013, A could have raised 5/4 and D raised 4 => Possible.

C) In 2014, A raised 3 and B raised 3 => Possible.

D) In 2014, B raised 3 where as E raised 2 => 3 > 2 => Possible.

CAT 2023 Slot 2: Past Year Question Paper - Question 62

What was the total amount spent on tickets (in Rs.) by Bipasha?

Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 62

Consider Statement 2: Anjali took Ride-1 at 11 am after waiting for 30 minutes for Chitra to complete it. It was the only ride where Anjali waited.

This implies that Chitra took Ride 1 at 10 am. Now we also know that she spent Rs 50 and that she left at 11 am. Now, since she did one ride costing Rs 20 at 10, she must have taken Ride-3 at 9 am.

So we get the following table for Chitra.

Now we know that Chitra and Anjali spent Rs 50 before 12:15 pm. It is not possible for Anjali to go on Ride-3 at 10 am as we know that she was waiting for 30 minutes before taking Ride-1 (She was waiting from 10:30 am).

Now, since we know that Ride-1 was the only ride for which she waited, we can say that she took Ride-1 at 11 am and started Ride-3 at 12 am

So we get the following table for Anjali.

Now, we know that Bipasha started her first ride at 11:30 am. We also know that they all spent Rs 50 before 12:15 pm.

Therefore, the first ride Bipasha takes will be Ride-2, costing Rs 50.

So we get the following table for Bipasha.

We know that Ride 3 stops at 1 pm. So the last ride taken by Anjali will either be Ride-2 or Ride-4. Now, considering Statement 4,we know that the last ride taken by Anjali and Bipasha was same and that Bipasha rode it after Anjali. So their last ride can’t be 2.

So the last ride of both Bipasha and Anjali will be 4.

Now if we assume that immediately after ending Ride-3, Anjali goes to Ride-4, then the last ride of Bipasha will be Ride-4 from 2 pm - 3 pm. But we know that Bipasha rode 3 rides. So this case is not possible.

Since Anjali didn’t have any break or waiting time, the only ride she can ride at 1 pm will be Ride 2 and then she will go on Ride-4 from 2 pm to 3 pm.

So we get the following table for Anjali:

Now we know that the last ride that Bipasha took was Ride-4 and that she had a gap of 1.5 hrs before it. This is only possible when she takes one ride between Ride-2 and Ride-4. Since Ride-3 is closed at 1 pm, she can only take Ride 1. So we get the following table for her.

As we can see from the table for Bipasha, she spent a total of 50+20+40= 110

Therefore the required answer is Option C: 110

CAT 2023 Slot 2: Past Year Question Paper - Question 63

Which were all the rides that Anjali completed by 2:00 pm?

Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 63

Consider Statement 2: Anjali took Ride-1 at 11 am after waiting for 30 minutes for Chitra to complete it. It was the only ride where Anjali waited.

This implies that Chitra took Ride 1 at 10 am. Now we also know that she spent Rs 50 and that she left at 11 am. Now, since she did one ride costing Rs 20 at 10, she must have taken Ride-3 at 9 am.

So we get the following table for Chitra.

Now we know that Chitra and Anjali spent Rs 50 before 12:15 pm. It is not possible for Anjali to go on Ride-3 at 10 am as we know that she was waiting for 30 minutes before taking Ride-1 (She was waiting from 10:30 am).

Now, since we know that Ride-1 was the only ride for which she waited, we can say that she took Ride-1 at 11 am and started Ride-3 at 12 am

So we get the following table for Anjali.

Now, we know that Bipasha started her first ride at 11:30 am. We also know that they all spent Rs 50 before 12:15 pm.

Therefore, the first ride Bipasha takes will be Ride-2, costing Rs 50.

So we get the following table for Bipasha.

We know that Ride 3 stops at 1 pm. So the last ride taken by Anjali will either be Ride-2 or Ride-4. Now, considering Statement 4,we know that the last ride taken by Anjali and Bipasha was same and that Bipasha rode it after Anjali. So their last ride can’t be 2.

So the last ride of both Bipasha and Anjali will be 4.

Now if we assume that immediately after ending Ride-3, Anjali goes to Ride-4, then the last ride of Bipasha will be Ride-4 from 2 pm - 3 pm. But we know that Bipasha rode 3 rides. So this case is not possible.

Since Anjali didn’t have any break or waiting time, the only ride she can ride at 1 pm will be Ride 2 and then she will go on Ride-4 from 2 pm to 3 pm.

So we get the following table for Anjali:

Now we know that the last ride that Bipasha took was Ride-4 and that she had a gap of 1.5 hrs before it. This is only possible when she takes one ride between Ride-2 and Ride-4. Since Ride-3 is closed at 1 pm, she can only take Ride 1. So we get the following table for her.

Anjali completed a total of 4 rides, 3 of which were completed at 2. Therefore the answer is Option B: Ride-1, Ride 3, and Ride -2

CAT 2023 Slot 2: Past Year Question Paper - Question 64

Which ride was taken by all three visitors?

Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 64

Consider Statement 2: Anjali took Ride-1 at 11 am after waiting for 30 minutes for Chitra to complete it. It was the only ride where Anjali waited.

This implies that Chitra took Ride 1 at 10 am. Now we also know that she spent Rs 50 and that she left at 11 am. Now, since she did one ride costing Rs 20 at 10, she must have taken Ride-3 at 9 am.

So we get the following table for Chitra.

Now we know that Chitra and Anjali spent Rs 50 before 12:15 pm. It is not possible for Anjali to go on Ride-3 at 10 am as we know that she was waiting for 30 minutes before taking Ride-1 (She was waiting from 10:30 am).

Now, since we know that Ride-1 was the only ride for which she waited, we can say that she took Ride-1 at 11 am and started Ride-3 at 12 am

So we get the following table for Anjali.

Now, we know that Bipasha started her first ride at 11:30 am. We also know that they all spent Rs 50 before 12:15 pm.

Therefore, the first ride Bipasha takes will be Ride-2, costing Rs 50.

So we get the following table for Bipasha.

We know that Ride 3 stops at 1 pm. So the last ride taken by Anjali will either be Ride-2 or Ride-4. Now, considering Statement 4,we know that the last ride taken by Anjali and Bipasha was same and that Bipasha rode it after Anjali. So their last ride can’t be 2.

So the last ride of both Bipasha and Anjali will be 4.

Now if we assume that immediately after ending Ride-3, Anjali goes to Ride-4, then the last ride of Bipasha will be Ride-4 from 2 pm - 3 pm. But we know that Bipasha rode 3 rides. So this case is not possible.

Since Anjali didn’t have any break or waiting time, the only ride she can ride at 1 pm will be Ride 2 and then she will go on Ride-4 from 2 pm to 3 pm.

So we get the following table for Anjali:

Now we know that the last ride that Bipasha took was Ride-4 and that she had a gap of 1.5 hrs before it. This is only possible when she takes one ride between Ride-2 and Ride-4. Since Ride-3 is closed at 1 pm, she can only take Ride 1. So we get the following table for her.

Only Ride-1 was taken by all the visitors. Therefore the correct answer is Option A: Ride-1

*Answer can only contain numeric values
CAT 2023 Slot 2: Past Year Question Paper - Question 65

How many rides did Anjali and Chitra take in total?


Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 65

Consider Statement 2: Anjali took Ride-1 at 11 am after waiting for 30 minutes for Chitra to complete it. It was the only ride where Anjali waited.

This implies that Chitra took Ride 1 at 10 am. Now we also know that she spent Rs 50 and that she left at 11 am. Now, since she did one ride costing Rs 20 at 10, she must have taken Ride-3 at 9 am.

So we get the following table for Chitra.

Now we know that Chitra and Anjali spent Rs 50 before 12:15 pm. It is not possible for Anjali to go on Ride-3 at 10 am as we know that she was waiting for 30 minutes before taking Ride-1 (She was waiting from 10:30 am).

Now, since we know that Ride-1 was the only ride for which she waited, we can say that she took Ride-1 at 11 am and started Ride-3 at 12 am

So we get the following table for Anjali.

Now, we know that Bipasha started her first ride at 11:30 am. We also know that they all spent Rs 50 before 12:15 pm.

Therefore, the first ride Bipasha takes will be Ride-2, costing Rs 50.

So we get the following table for Bipasha.

We know that Ride 3 stops at 1 pm. So the last ride taken by Anjali will either be Ride-2 or Ride-4. Now, considering Statement 4,we know that the last ride taken by Anjali and Bipasha was same and that Bipasha rode it after Anjali. So their last ride can’t be 2.

So the last ride of both Bipasha and Anjali will be 4.

Now if we assume that immediately after ending Ride-3, Anjali goes to Ride-4, then the last ride of Bipasha will be Ride-4 from 2 pm - 3 pm. But we know that Bipasha rode 3 rides. So this case is not possible.

Since Anjali didn’t have any break or waiting time, the only ride she can ride at 1 pm will be Ride 2 and then she will go on Ride-4 from 2 pm to 3 pm.

So we get the following table for Anjali:

Now we know that the last ride that Bipasha took was Ride-4 and that she had a gap of 1.5 hrs before it. This is only possible when she takes one ride between Ride-2 and Ride-4. Since Ride-3 is closed at 1 pm, she can only take Ride 1. So we get the following table for her.

Anjali took 4 rides, and Chitra took 2 rides. Therefore the correct answer is 6

*Answer can only contain numeric values
CAT 2023 Slot 2: Past Year Question Paper - Question 66

What was the total amount spent on tickets (in Rs.) by Anjali?


Detailed Solution for CAT 2023 Slot 2: Past Year Question Paper - Question 66

Consider Statement 2: Anjali took Ride-1 at 11 am after waiting for 30 minutes for Chitra to complete it. It was the only ride where Anjali waited.

This implies that Chitra took Ride 1 at 10 am. Now we also know that she spent Rs 50 and that she left at 11 am. Now, since she did one ride costing Rs 20 at 10, she must have taken Ride-3 at 9 am.

So we get the following table for Chitra.

Now we know that Chitra and Anjali spent Rs 50 before 12:15 pm. It is not possible for Anjali to go on Ride-3 at 10 am as we know that she was waiting for 30 minutes before taking Ride-1 (She was waiting from 10:30 am).

Now, since we know that Ride-1 was the only ride for which she waited, we can say that she took Ride-1 at 11 am and started Ride-3 at 12 am

So we get the following table for Anjali.

Now, we know that Bipasha started her first ride at 11:30 am. We also know that they all spent Rs 50 before 12:15 pm.

Therefore, the first ride Bipasha takes will be Ride-2, costing Rs 50.

So we get the following table for Bipasha.

We know that Ride 3 stops at 1 pm. So the last ride taken by Anjali will either be Ride-2 or Ride-4. Now, considering Statement 4,we know that the last ride taken by Anjali and Bipasha was same and that Bipasha rode it after Anjali. So their last ride can’t be 2.

So the last ride of both Bipasha and Anjali will be 4.

Now if we assume that immediately after ending Ride-3, Anjali goes to Ride-4, then the last ride of Bipasha will be Ride-4 from 2 pm - 3 pm. But we know that Bipasha rode 3 rides. So this case is not possible.

Since Anjali didn’t have any break or waiting time, the only ride she can ride at 1 pm will be Ride 2 and then she will go on Ride-4 from 2 pm to 3 pm.

So we get the following table for Anjali:

Now we know that the last ride that Bipasha took was Ride-4 and that she had a gap of 1.5 hrs before it. This is only possible when she takes one ride between Ride-2 and Ride-4. Since Ride-3 is closed at 1 pm, she can only take Ride 1. So we get the following table for her.

As we can see from the table of Anjali she spent a total of 20 + 30 + 50 + 40 = 140

Therefore the required answer is 140

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