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CAT Mock Test- 7 - CAT MCQ


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66 Questions MCQ Test CAT Mock Test Series 2024 - CAT Mock Test- 7

CAT Mock Test- 7 for CAT 2024 is part of CAT Mock Test Series 2024 preparation. The CAT Mock Test- 7 questions and answers have been prepared according to the CAT exam syllabus.The CAT Mock Test- 7 MCQs are made for CAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for CAT Mock Test- 7 below.
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CAT Mock Test- 7 - Question 1

The central idea of the passage is that

Detailed Solution for CAT Mock Test- 7 - Question 1

The central idea of the passage is that political action is necessary to solve large-scale problems like global poverty. This is evident in the passage when it states that effective altruism focuses on the internal moral economy of the giver rather than addressing real-world problems, and that it does not provide a plan for the elimination of poverty itself. The passage also criticizes the effective altruism movement for its political pessimism and consumer-hero hubris, arguing that it overlooks the potential for achieving far more good through institutional reform and political action. Therefore, option 1 is the correct answer.

CAT Mock Test- 7 - Question 2

All of the following options represent causes reflective of effective altruism EXCEPT:

Detailed Solution for CAT Mock Test- 7 - Question 2

An integral component of effective altruism is the presence of "organisations most efficient at translating those donations into gains in human well-being." The perceived impact of these institutions is positive and uplifting, especially concerning the poor or needy. Causes that highlight this positive impact on human well-being are reflective of effective altruism. Moreover, from the lines 
"The underlying problem is that effective altruism's distinctive combination of political pessimism and consumer-hero hubris forecloses the consideration of promising possibilities for achieving far more good. Singer and other effective altruist philosophers believe that their most likely customers find institutional reform too complicated and political action too impersonal and hit and miss to be attractive."
We can infer that effective altruism does not typically involve political action and institutional reform.
Option A: The NGO here is utilising the collected funds to build schools in rural regions: a gesture with a positive impact. Hence, this can be eliminated as the potential answer. 
Option B: The charity organisation ensures the well being of the poverty-stricken individuals of the slums through their healthcare drives offered for free; this again indicates positive gains for people. We can, thereby, eliminate Option B as the answer.
Option C: Again, there is some perceptible positive impact that stems from the gesture of offering free legal aid to the poor and marginalised people. Thus, we can discard option C as a possible answer.
Option D: Funding of protest with some political agenda does not reflect any observable impact on human well being (unlike the other cases). Also, this is more in line with what the author wants (political action, changing institutions, holding government accountable) which is in contrast with effective altruism. Hence, Option D is the correct choice.

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CAT Mock Test- 7 - Question 3

The passage makes all of the following claims about effective altruism, EXCEPT

Detailed Solution for CAT Mock Test- 7 - Question 3

{This ‘effective altruism’ movement has two components: i) encouraging individuals in the rich world to donate more; and ii) encouraging us to donate more rationally to the organisations most efficient at translating those donations into gains in human well-being.}
From these lines, we can see that 'effective altruism' supports rational donation, which means donating to efficient organisations which will make the best out of it. Hence, Option A can be inferred.
From the same lines, we can infer that affluent people are encouraged to donate more. Hence B can also be inferred.
In the passage, it has been given that an assumption of limited resources is made. Then a goal is decided upon considering the maximisation of these fixed resources. This often results in focus upon superficial problems, which seem to generate more value, instead of the root problem of poverty. Hence C can also be inferred.
Option D is a distortion. In the last paragraph, the author states that "Singer and others have been making this argument for nearly 50 years, yet the level of private donations remain orders of magnitude below what would be required to eliminate global poverty, however efficiently allocated." Here, the author is talking about cumulative donations, not individual donations. And secondly, the author is concerned about private donations' inability to tackle global poverty. Societal problems is a broad term and could involve other extraneous issues as well. Hence, option D cannot be inferred.

CAT Mock Test- 7 - Question 4

According to the author, the efficiency view promoted by effective altruism

Detailed Solution for CAT Mock Test- 7 - Question 4

In the third paragraph, the author states that "...If one takes the efficiency view promoted by the effective altruism movement then one assumes a fixed set of resources and the choice of which goal to aim for follows from a calculation of how to maximize the expected value those resources can generate; i.e. the means justifies the end..." So, based on the value maximization of the resources available, the goal is chosen. Option C captures this point correctly and hence, is the correct answer. 
In this context, the author does not talk about varying the funding to achieve a goal. Hence option A is irrelevant.
The author does not talk about abandoning difficult goals just because they require prolonged commitment. Thus, B can be eliminated too.
Option D has not been discussed in the passage. It talks about people choosing goals because of personal satisfaction rather than the constraint of resources they have. No such claim has been made.
Hence, Option C is the correct choice.

CAT Mock Test- 7 - Question 5

In the third paragraph, why does the author remark that "ironically, the roles have reversed?"?

Detailed Solution for CAT Mock Test- 7 - Question 5

The author makes the following observation in the third paragraph- {The robots are now humanity’s saviors, suppressing bad human mass behaviour online with increasingly sophisticated filtering algorithms. We once obsessed about how to restrain machines we could not predict or control — now we worry about how to use machines to restrain humans we cannot predict or control.}
So, humans feared that self-improving machines would go out of control and that it would be difficult for humans to restrain them. However, in a role reversal, the machines today help curb bad human behaviour online. Hence, ironically, the roles have reversed. 
Only option B conveys this inference and hence is the right answer.
Option A is outside the scope of discussion. Machines evolving to differentiate between good and bad has not been mentioned anywhere. Hence they can be eliminated.
Though an irony, Option C has not been mentioned in the passage as such. Hence can be eliminated.
Option D is close but contains the distortion that we cannot do without machines. The passage does not say that machines are indispensable. Hence, D can be eliminated.

CAT Mock Test- 7 - Question 6

According to the author, what was the reason for the shift toward the fear of chaotic human behavior?

Detailed Solution for CAT Mock Test- 7 - Question 6

In the fourth paragraph, the author mentions that humans feared the consequences of attaining their desires through complicated machines. And hence, they attempted to rectify their desires. However, when they lost hope in this endeavour, they shifted away from the fear of unpredictable robots and toward the fear of chaotic human behaviour, flipping the problem on its head.
Option A conveys the idea elucidated above. Option A is the answer.
Option B is partially correct but misses out on the remedying theme.
Options C and D have not been implied in the passage.

CAT Mock Test- 7 - Question 7

The risk in Wiener's distinction between what we desire and what actually happens, in the end, is that:

Detailed Solution for CAT Mock Test- 7 - Question 7

In the last paragraph, the author posits the following in regard to Wiener's distinction- "Instead of examining carefully whether our desires are in fact good, we may simply assume they are, and so blame bad behavior on the messy cooperation between ourselves and the “system.” That is, we assume, without proper examination, that our desires are good, and instead choose to blame the system and other factors for the bad consequences. Option C conveys this idea and is the answer. 
Options  B and D have not been implied in the passage.
Option A is close, but the latter part is extraneous to the discussion.

CAT Mock Test- 7 - Question 8

Which of the following could be an example of Weiner's desire-outcome disparity argument?
I. A weapons system, which cannot be stopped easily, starts bombing after receiving an erroneous command.
II. An AI program developed to mitigate global warming starts eliminating a fraction of the human population to complete its objective.
III. A Social media platform allows groups of militants to communicate their plans and coordinate their attacks.

Detailed Solution for CAT Mock Test- 7 - Question 8

Weiner's desire-outcome disparity is about advanced systems which achieve undesired outcomes because of their inherent nature and them acting on misleading substitutes of the objectives fed.
Since the command itself was erroneous, hence the system acted on incorrect information instead of acting on a misleading substitute of the objective. Hence I does not fit the argument.
The AI program was designed to mitigate global warming. But eliminating humans to achieve the same is an undesired outcome, which arises due to the program acting on the misleading substitute of the objective assigned. The substitute being 'achieving the goal without a care for the humans', which the scientists would not have desired. Hence II fits the argument.
Though the author mentions that Weiner did not talk about social media, he also mentions that an analogy can be drawn between the two, hence the argument can be extended to include the misuse of social media, as mentioned in Statement III.
Hence Statements II and III fit Weiner's argument.

CAT Mock Test- 7 - Question 9

Which of the following statements CANNOT be inferred from the passage concerning the Nordic regions?
I. Progressive policies have not addressed the presence of gender stereotypes in the workplace.
II. Competition among women has exacerbated the income gap between men and women.
III. Occupational gender segregation has led to segregation in education on major subjects.
IV. Social norms discourage women from taking up certain occupations.

Detailed Solution for CAT Mock Test- 7 - Question 9

In the last line of the third paragraph, the author notes that " Progressive legislation and policy have not prevented the continuation of gender stereotypes." Hence, though they have failed in arresting the continuation of gender stereotypes, they have tried to address them. Statement I is wrong.
Statement II has not been implied anywhere in the passage.
Statement III is a vague statement and attempts to generalise a singular observation. In the second paragraph, the author links occupation segregation in STEM industries with segregation in education on these subjects. This cannot be generalised. Furthermore, we cannot say if occupational segregation causes educational segregation or the other way around. Hence, II is not inferrable.
According to the intergovernmental study,  “social norms continue to restrict occupational choices”. Statement IV can be inferred.
Statements I, II and III cannot be inferred. Option B is the answer.

CAT Mock Test- 7 - Question 10

All of the following have been discussed about gender equality in Sweden, EXCEPT:

Detailed Solution for CAT Mock Test- 7 - Question 10

"However, behind women participation statistics and progressive policies, gender stereotypes prevail, particularly in the workplace, and women in the region say that there is still a lot of work to be done."
Though referred to as one of the most gender-equal countries, the above lines show that Sweden is still far from being gender-equal. Hence Option A can be inferred.
In the fourth and fifth paragraphs, the author states the following regarding the gender equality situation in Sweden- "Men were also encouraged to take care of the family. “There’s this dual-earner/dual-carer ideal that Swedish gender equality is based on,” says Björklund. Policies in Sweden have since then focused on facilitating that work-family balance."
Option B can be clearly inferred from these lines. 
"The modern concept of gender equality has its foundations in the postwar welfare state. In Sweden, it was motivated by the need for more women in the workforce after the Second World War". Option C has been discussed as well.
Option D is a distortion. In the last paragraph, the author says that the national trait of "Swedishness" has been hijacked by far-right parties. However, we cannot say if Swedishness was introduced by them or that it was introduced to push anti-immigrant policies. As this has not been discussed, we can say that it is the right answer.

CAT Mock Test- 7 - Question 11

The central idea in the fifth paragraph is that

Detailed Solution for CAT Mock Test- 7 - Question 11

{"...However, the expectations on women to be full-time workers, self-sacrificing mothers and still have leisure time have put unrealistic pressure on this ideal. Expectations on men are not as high, and Björklund says that fathers can get away with being less caring than mothers - an idea underpinned by traditional stereotypes and middle-class values..."} In the fifth paragraph, the author talks about how the existing stereotypes lead to a disparity in the expectations from men and women to maintain a work-life balance. The difference in expectation is then highlighted. The author mentions how, in reality, gender stereotypes lead people to expect a lot more from women regarding caregiving responsibilities. Option C conveys the idea correctly.
Option A asserts that men are expected to fare better in the earner role, which has not been implied in the passage. 
Option B states that traditional stereotypes allow men to ignore their caregiving duties. The passage states that men can get away with doing much lesser than women - and not completely ignoring their responsibilities. Moreover, the impact on women due to the imbalanced expectations is missed out in this option. Hence, we can eliminate Option B. 
Option D can be safely eliminated. The author does not discuss the need for a new Swedish dual-earner/dual-giver ideal but talks about the threat to the same.
Hence, Option C is the correct answer.

CAT Mock Test- 7 - Question 12

Which of the following is likely to be the next course of discussion?

Detailed Solution for CAT Mock Test- 7 - Question 12

The author concludes the passage on a foreboding note, highlighting how certain discussions now assume gender equality has gone too far. In the same paragraph, the author discusses how the disparity among different women groups, especially white, middle-class and immigrant women has been used as a tool against immigration. Hence, any discussion about how gender equality has not gone too far and how it is needed for all women would be the right way to continue the line of thought. Option C gives a rebuttal of sorts to the xenophobic idea presented in the last paragraph. Hence, C is the aptest answer.
The author does not explicitly discuss diversity in workplaces. It does not connect with any previous discussion. Hence, option A can be eliminated.
Option B is narrow. The author does not discuss anything related to intersectional groups in the passage and is focused on the larger group of women. 
Option D is extreme and has not been implied in the passage. Moreover, the focus would shift from gender equality to minorities which is inconsistent with the passage.

CAT Mock Test- 7 - Question 13

The purpose of the last three paragraphs is to:

Detailed Solution for CAT Mock Test- 7 - Question 13

The last three paragraphs talk about how genetic research aided by SPARK has helped in the treatment and mitigation of various kinds of autism. The author further says that the addition of rich data to this will help provide better analysis and support. So the author tries to show the benefits of SPARK and how the addition of rich data will help provide better support. Option C captures the points correctly.
Option A contains a distortion in the second half. The passage does not imply that Spark can reach its true potential only through greater participation. Hence, A can be eliminated. 
The author does not talk about various developments but the achievements due to a particular method. Additionally, the need for more academic participation has not been emphasised. Hence Option B is a distortion.
The author does not focus on the plight of children suffering from autism. Instead, he highlights how SPARK helps these children and how promoting participation in genetic research could render valuable data. Hence Option D is incorrect.

CAT Mock Test- 7 - Question 14

Which of the following is the author most likely to agree with?

Detailed Solution for CAT Mock Test- 7 - Question 14

{"Essentially, we will take a page from the playbook that oncologists use to treat certain types of cancer-based upon their genetic signatures and apply targeted therapeutic strategies to help people with autism."} The above line means that just like oncologists treat different types of cancers based on their genetic signatures differently, we need to use targeted treatment strategies to treat people with autism. But option A says that we need oncologists to treat autism, which distorts the author's message. Hence A is incorrect.
{"... learn if and how different genetic variants contribute to autistic behaviours, we need not only the largest but also the most diverse group of participants." }The author says that to learn about the contribution of genetic variants to autistic behaviours, we would need a large and diverse data set. Hence, we can clearly infer option B.
It is not mentioned in the passage that the treatment strategies used to treat cancer can be extended to autism. Option C is incorrect.
{"We now know that genes play a central role in the causes of these “autisms,” which are the result ..." } Option D is incorrect because it is quite the opposite of what has been mentioned in the line above. Moreover, the author says 'missed' during newborn screening, means something that was there but was not detected because of an error.
Hence, the author would most likely agree with option B.

CAT Mock Test- 7 - Question 15

Which of the following cannot be inferred?
I. The effectiveness of 'big data' is determined not by its size but by its diversity.
II. Genes are a major factor influencing autism.
III. Consumption of a diet containing low levels of phenylalanine helps decrease the level of amino acid in the blood.

Detailed Solution for CAT Mock Test- 7 - Question 15

"..we need not only the largest but also the most diverse group of participants."
From this line, we can infer that along with a large dataset, we also need diversity to make big data effective. But is has not been mentioned that the size of the data is not important. Hence Statement I is a distortion and cannot be inferred.
"We now know that genes play a central role in the causes of these autisms,...”
From this line, we can infer that the role of genes in causing autism is prominent. Hence Statement II can be inferred.
In paragraph 7, it is implied that a diet containing low levels of phenylalanine was given to stabilize the level of amino acid in the blood. It has not been implied that the level of amino acid is decreased because of it. Hence Statement III cannot be inferred.

CAT Mock Test- 7 - Question 16

All of the following have been discussed in the passage as benefits of having richer and bigger datasets EXCEPT:

Detailed Solution for CAT Mock Test- 7 - Question 16

Option A: While the first half of A's statement is true, the latter half appears distorted. There is no mention of "untreatable diseases", and 'phenylketonuria' has been presented in a different context. Hence, Option A is the correct choice.
Option B: {The more people who participate, the deeper and richer these data sets become, catalyzing research that is expanding our knowledge of both biology and behaviour to develop more precise approaches to medical and behavioural issues.} Option B has been presented as an advantage of having larger and richer datasets. Hence, we can eliminate Option B.
Option C: {Access to rich data sets draws talented scientists to the field of autism science to develop new methods of finding patterns in the data, better predicting associated behavioural and medical issues, and, perhaps, identifying more effective supports and treatments.} The author highlights the utility of richer datasets in drawing more talent to the field of autism science. Thus, Option C can be eliminated. 
Option D: {To ensure an accurate inventory of all the major genetic contributors, and learn if and how different genetic variants contribute to autistic behaviours, we need not only the largest but also the most diverse group of participants.}. The use of larger datasets in developing an accurate repository has been presented above. Therefore, Option D can be rejected as the potential answer. 
Hence, Option A is the correct choice.

*Answer can only contain numeric values
CAT Mock Test- 7 - Question 17

The four sentences (labelled 1, 2, 3, 4) 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. On the contrary, the industry is happy reducing the wage bills, doing mechanisation and raising its profits.
2. During the pandemic, nearly 31 million families have moved down from the middle class and nearly 100 million people have lost jobs.
3. The industries that are most likely to create employment, i.e. the medium and small industries, are going down under and the large ones which do not create employment are the poster boys.
4. They are the ones that will get the 6 per cent productivity-linked incentive from the tax paid by the average taxpayers, with unknown consequences.


Detailed Solution for CAT Mock Test- 7 - Question 17

Sentence 2 is the opening sentence since it highlights the damaging effect of the pandemic, especially with regard to the loss of jobs. The author draws a parallel between this group and the industries in general by continuing the idea in 1, which mentions the measures taken by certain industries. 
34 is a bloc. 3 describes the industrial scenario at large - how large-scale industries are doing well, despite laying off employees and not doing enough to arrest the pandemic's effects. Sentence 4 talks about the tax incentives these large-scale industries get, further critiquing the current situation. Hence, 2134 form a coherent paragraph.

*Answer can only contain numeric values
CAT Mock Test- 7 - Question 18

The four sentences (labelled 1, 2, 3, 4) 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. For others, Trump’s loss makes him into a loser—especially damaging given how much Trump hates losers.
2. In the more immediate future, however, no one will remain as personally angry about it as Trump.
3. Polls show that many Republicans believe the 2020 election was tainted, and the damage that will do to faith in democracy in the long term is dangerous.
4. Some followers who saw him as a man who could challenge the establishment will view his defeat as proof that politics is irredeemable, and will slide into apathy and disengagement.


Detailed Solution for CAT Mock Test- 7 - Question 18

Statement 3 is independent and has to be the opening sentence; it highlights the damage to the faith in democracy and the deleterious effect this could have.  41 is a bloc, describing why the republican loss in the elections could be damaging in the long term. The author covers how one faction of followers might become apathetic or disengaged from politics. The perception with regard to Trump is further elaborated in 1. Hence, 4 and 1 together extend on the damage that is mentioned in 3. On the other hand, 2 talks about ramifications in the immediate future. Sentence 2 has to precede 4 because 4 has the pronoun 'him'. Hence, the reference to Trump must have been made before sentence 4. Therefore, 3241 is the correct answer. 

*Answer can only contain numeric values
CAT Mock Test- 7 - Question 19

Five sentences related to a topic are given below. Four of them can be put together to form a meaningful and coherent short paragraph. Identify the odd one out.
1. Without robust national privacy safeguards, entire databases of citizen information are ready for purchase, whether to predatory loan companies, law enforcement agencies, or even malicious foreign actors.
2. Federal privacy bills that don’t give sufficient attention to data brokerage will therefore fail to tackle an enormous portion of the data surveillance economy.
3. Data brokerage is a threat to democracy.
4. This is why the largest data brokers are lobbying more aggressively in Washington.
5. This will leave civil rights, national security, and public-private boundaries vulnerable in the process.


Detailed Solution for CAT Mock Test- 7 - Question 19

After reading all the sentences, it can be inferred that the passage discusses data brokerage and the threats it poses to the proper functioning of a democracy.
Sentence 3 is independent and is the opening sentence. 12 is a bloc as it discusses the situation in the absence of strong data safeguards. 5 has to follow 2 as it describes the larger consequences- the word 'this' is the link. Hence, 3125 form a coherent paragraph.
Sentence 4 does not fit the context as it talks about data brokers lobbying more aggressively. Though it is not an independent sentence, none of the other sentences logically lead to the idea discussed in sentence 4. Hence 4 is the odd one out here.

*Answer can only contain numeric values
CAT Mock Test- 7 - Question 20

Five sentences related to a topic are given below. Four of them can be put together to form a meaningful and coherent short paragraph. Identify the odd one out.
1. Researchers see signs of this in sperm whales in the Galápagos and the Caribbean, in humpbacks across the South Pacific, in Arctic belugas, and in the Pacific Northwest’s killer whales.
2. Today many scientists believe some whales and dolphins, like humans, have distinct cultures.
3. Whale culture, it seems, is rattling timeworn conceptions of ourselves.
4. The possibility is prompting new thinking about how some marine species evolve.
5. Cultural traditions may help drive genetic shifts, altering what it means to be a whale.


Detailed Solution for CAT Mock Test- 7 - Question 20

After reading all the sentences, it can be inferred that the passage talks about the existence of distinct cultures among whales and dolphins and how this may have shaped their evolutionary paths. 
Sentence 2 introduces the topic and is the opening sentence. 1 follows 2 as it provides further evidence to the scientists' claim. 45 is a bloc. 4 talks about the new findings changing our understanding of how marine species evolve. 5 builds on that thought, highlighting how cultural shifts may have caused genetic shifts. 
Sentence 3, on the other hand, asserts that the new findings could alter the way we understand ourselves. None of the other sentences alludes to this line of thought. Hence, 3 is out of context.

*Answer can only contain numeric values
CAT Mock Test- 7 - Question 21

The four sentences (labelled 1, 2, 3, and 4) 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. However, the very ubiquity of such technology poses critical questions about data privacy and individual autonomy.

2. The integration of advanced algorithms and everyday technology has streamlined countless processes, improving efficiency exponentially.

3. As we entrust more of our lives to digital gatekeepers, the parameters of the debates shift from technical concerns to ethical dilemmas.

4. Such an environment, teeming with both subtle surveillance and convenience, highlights the double-edged nature of progress.


Detailed Solution for CAT Mock Test- 7 - Question 21

The proper sequencing of the sentences is:
2. The integration of advanced algorithms and everyday technology has streamlined countless processes, improving efficiency exponentially.
4. Such an environment, teeming with both subtle surveillance and convenience, highlights the double-edged nature of progress.
1. However, the very ubiquity of such technology poses critical questions about data privacy and individual autonomy.
3. As we entrust more of our lives to digital gatekeepers, the parameters of the debates shift from technical concerns to ethical dilemmas.

This sequence creates a coherent paragraph by first introducing the integration of advanced algorithms and everyday technology and how it has improved efficiency. Then, it raises the issue of the double-edged nature of progress, emphasizing the presence of both surveillance and convenience. This leads to a discussion about the critical questions regarding data privacy and individual autonomy that arise due to the ubiquity of technology. Finally, it concludes by stating that as we entrust more of our lives to digital gatekeepers, the focus of the debates shifts from technical concerns to ethical dilemmas.

CAT Mock Test- 7 - Question 22

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

Urban biodiversity plays a crucial role in enhancing city residents' quality of life. It aids in pollution reduction, provides psychological benefits, and helps regulate urban temperatures. Despite its importance, urban biodiversity is often overlooked in city planning, with priority given to industrial and residential development. This neglect can lead to the degradation of local ecosystems and loss of species, potentially triggering broader environmental consequences. Integrating green spaces and biodiversity considerations into urban development plans is imperative for sustainable city living.

Detailed Solution for CAT Mock Test- 7 - Question 22

The passage discusses the critical importance of urban biodiversity and how its neglect in urban planning can lead to various environmental issues. Option B is correct as it encapsulates the core argument for the integration of biodiversity for a sustainable urban future, whereas the other options either contradict or miss the main point of the passage.

CAT Mock Test- 7 - Question 23

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

The surge in freelance and gig-based employment is reshaping the traditional work environment. While offering flexibility and autonomy, the gig economy also ushers in challenges such as job insecurity, lack of benefits, and an inconsistent income stream. Governments struggle to catch up with this evolving employment model, often leaving workers unprotected under labor laws. Moving forward requires reimagining labor policies to provide gig workers with adequate security and rights, ensuring a fair work environment.

Detailed Solution for CAT Mock Test- 7 - Question 23

The passage discusses the transformative effect of the gig economy on traditional employment, highlighting both opportunities and challenges, particularly the need for updated labor policies. Option B encompasses these nuances, focusing on the need for policy evolution in response to this new work style, while the other options fail to capture the full picture or present false statements.

CAT Mock Test- 7 - Question 24

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

Modern dietary patterns, driven by a quest for convenience and speed, have led to an increase in processed food consumption and a decline in nutritional quality. This shift contributes to various health issues, including obesity, diabetes, and heart disease, straining healthcare systems globally. Public health initiatives emphasize returning to whole foods and balanced diets, but this change is hindered by advertising, fast-paced lifestyles, and food deserts. Addressing these challenges requires comprehensive strategies that include public education, improved food policies, and collaboration with food producers for healthier options.

Detailed Solution for CAT Mock Test- 7 - Question 24

The passage connects modern dietary habits with health challenges and underscores the need for a holistic approach to promote healthier diets. Option A captures this connection and the necessity for a multi-pronged strategy involving education, policy, and industry collaboration. The other options either provide contradictory information or oversimplify the issue.

CAT Mock Test- 7 - Question 25

Directions: Read the question carefully and answer the question that follows:

Mr. Wilson works in a zoo, where he has to count the number of animals born in a month. He collates the data for the entire month at the end of the particular month. However, during a certain period of twenty months, from month 1 to month 20, Mr. Wilson counted the number of animals born on only ten months. He then plotted the ten points on a graph sheet and connected consecutive points on the graph using straight lines, forming a line graph.


Further, it is also known that
(1) the difference in number of animals born between any two consecutive counting that Mr. Wilson took was not more than 40
(2) the difference in the number of animals born counted by Mr. Wilson for the third time and for the fourth time was 40
(3) the second count that Mr. Wilson took was the same as the eighth count that he took.

Q. What is the average number of animals born between second count and third count that Mr. Wilson counted?

Detailed Solution for CAT Mock Test- 7 - Question 25

Since Mr. Wilson plotted only ten points and connected consecutive points using straight lines, wherever there is a change in the slope of the line, Mr. Wilson must have made a count.
Hence, Mr. Wilson must have counted the number of animals born on months 1, 4, 7, 10, 14, 18 and 20 and the number of animals that were born were 300, 330, 270, 240, 280, 360 and 340 respectively.
The differences in number of animals born between these counting are 30, 60, 30, 40, 80 and 20 respectively.
From (1), he must have made at least one count between month 4 and month 7 and at least one count between month 14 and month 18. These account for nine counts in total.
From (3), Mr. Wilson's second count is the same as the eighth count. The second count can be on month 2 or month 3 or month 4.
In any of the three cases, for the eighth count to be the same as the second count, the eighth count must have been made from month 14 to month 18.

Case 1: If second count was on month 2:

If second count was on month 2, (i.e., of 310 animals), the eighth count cannot be on any month because no month from month 14 to month 18 have the number of animals born as 310.

Case 2: If second count was on month 4:

Similarly, If second count was on month 4, (i.e., of 330 animals), the eighth count cannot be on any month because no month from month 14 to month 18 have the number of animals born as 310.

Case 3: If second count was on month 3:

The second count on month 3 (of 320 animals) is possible because month 16. has the number of animals born as 320.
So, eighth count must be month 16.

So, the third count is on month 4 (of 330 animals).

From (2), the number of animals born on fourth count must be either 290 or 370 (inferable from the graph).
So, the fourth count must be on month 6.

The following table presents the counting that Mr. Wilson made:

Average number of animals born between second count and third count = 

CAT Mock Test- 7 - Question 26

Directions: Read the question carefully and answer the question that follows:

Mr. Wilson works in a zoo, where he has to count the number of animals born in a month. He collates the data for the entire month at the end of the particular month. However, during a certain period of twenty months, from month 1 to month 20, Mr. Wilson counted the number of animals born on only ten months. He then plotted the ten points on a graph sheet and connected consecutive points on the graph using straight lines, forming a line graph.


Further, it is also known that
(1) the difference in number of animals born between any two consecutive counting that Mr. Wilson took was not more than 40
(2) the difference in the number of animals born counted by Mr. Wilson for the third time and for the fourth time was 40
(3) the second count that Mr. Wilson took was the same as the eighth count that he took.

Q. What is the difference between the fifth count and the ninth count that Mr. Wilson took?

Detailed Solution for CAT Mock Test- 7 - Question 26

Since Mr. Wilson plotted only ten points and connected consecutive points using straight lines, wherever there is a change in the slope of the line, Mr. Wilson must have made a count.
Hence, Mr. Wilson must have counted the number of animals born on months 1, 4, 7, 10, 14, 18 and 20 and the number of animals that were born were 300, 330, 270, 240, 280, 360 and 340 respectively.
The differences in number of animals born between these counting are 30, 60, 30, 40, 80 and 20 respectively.
From (1), he must have made at least one count between month 4 and month 7 and at least one count between month 14 and month 18. These account for nine counts in total.
From (3), Mr. Wilson's second count is the same as the eighth count. The second count can be on month 2 or month 3 or month 4.
In any of the three cases, for the eighth count to be the same as the second count, the eighth count must have been made from month 14 to month 18.

Case 1: If second count was on month 2:

If second count was on month 2, (i.e., of 310 animals), the eighth count cannot be on any month because no month from month 14 to month 18 have the number of animals born as 310.

Case 2: If second count was on month 4:

Similarly, If second count was on month 4, (i.e., of 330 animals), the eighth count cannot be on any month because no month from month 14 to month 18 have the number of animals born as 310.

Case 3: If second count was on month 3:

The second count on month 3 (of 320 animals) is possible because month 16. has the number of animals born as 320.
So, eighth count must be month 16.

So, the third count is on month 4 (of 330 animals).

From (2), the number of animals born on fourth count must be either 290 or 370 (inferable from the graph).
So, the fourth count must be on month 6.

The following table presents the counting that Mr. Wilson made:

The difference between the fifth count and the ninth count that Mr. Wilson took = 360 - 270 = 90

CAT Mock Test- 7 - Question 27

Directions: Read the question carefully and answer the question that follows:

Mr. Wilson works in a zoo, where he has to count the number of animals born in a month. He collates the data for the entire month at the end of the particular month. However, during a certain period of twenty months, from month 1 to month 20, Mr. Wilson counted the number of animals born on only ten months. He then plotted the ten points on a graph sheet and connected consecutive points on the graph using straight lines, forming a line graph.


Further, it is also known that
(1) the difference in number of animals born between any two consecutive counting that Mr. Wilson took was not more than 40
(2) the difference in the number of animals born counted by Mr. Wilson for the third time and for the fourth time was 40
(3) the second count that Mr. Wilson took was the same as the eighth count that he took.

Q. On which of the following months did Mr. Wilson count the number of animals born?

Detailed Solution for CAT Mock Test- 7 - Question 27

Since Mr. Wilson plotted only ten points and connected consecutive points using straight lines, wherever there is a change in the slope of the line, Mr. Wilson must have made a count.
Hence, Mr. Wilson must have counted the number of animals born on months 1, 4, 7, 10, 14, 18 and 20 and the number of animals that were born were 300, 330, 270, 240, 280, 360 and 340 respectively.
The differences in number of animals born between these counting are 30, 60, 30, 40, 80 and 20 respectively.
From (1), he must have made at least one count between month 4 and month 7 and at least one count between month 14 and month 18. These account for nine counts in total.
From (3), Mr. Wilson's second count is the same as the eighth count. The second count can be on month 2 or month 3 or month 4.
In any of the three cases, for the eighth count to be the same as the second count, the eighth count must have been made from month 14 to month 18.

Case 1: If second count was on month 2:

If second count was on month 2, (i.e., of 310 animals), the eighth count cannot be on any month because no month from month 14 to month 18 have the number of animals born as 310.

Case 2: If second count was on month 4:

Similarly, If second count was on month 4, (i.e., of 330 animals), the eighth count cannot be on any month because no month from month 14 to month 18 have the number of animals born as 310.

Case 3: If second count was on month 3:

The second count on month 3 (of 320 animals) is possible because month 16. has the number of animals born as 320.
So, eighth count must be month 16.

So, the third count is on month 4 (of 330 animals).

From (2), the number of animals born on fourth count must be either 290 or 370 (inferable from the graph).
So, the fourth count must be on month 6.

The following table presents the counting that Mr. Wilson made:

Mr. Wilson counted the number of animals born on the 6th month.

CAT Mock Test- 7 - Question 28

Directions: Read the question carefully and answer the question that follows:

Mr. Wilson works in a zoo, where he has to count the number of animals born in a month. He collates the data for the entire month at the end of the particular month. However, during a certain period of twenty months, from month 1 to month 20, Mr. Wilson counted the number of animals born on only ten months. He then plotted the ten points on a graph sheet and connected consecutive points on the graph using straight lines, forming a line graph.


Further, it is also known that
(1) the difference in number of animals born between any two consecutive counting that Mr. Wilson took was not more than 40
(2) the difference in the number of animals born counted by Mr. Wilson for the third time and for the fourth time was 40
(3) the second count that Mr. Wilson took was the same as the eighth count that he took.

Q. What is the average of the number of animals born during the ten months that Mr. Wilson counted?

Detailed Solution for CAT Mock Test- 7 - Question 28

Since Mr. Wilson plotted only ten points and connected consecutive points using straight lines, wherever there is a change in the slope of the line, Mr. Wilson must have made a count.
Hence, Mr. Wilson must have counted the number of animals born on months 1, 4, 7, 10, 14, 18 and 20 and the number of animals that were born were 300, 330, 270, 240, 280, 360 and 340 respectively.
The differences in number of animals born between these counting are 30, 60, 30, 40, 80 and 20 respectively.
From (1), he must have made at least one count between month 4 and month 7 and at least one count between month 14 and month 18. These account for nine counts in total.
From (3), Mr. Wilson's second count is the same as the eighth count. The second count can be on month 2 or month 3 or month 4.
In any of the three cases, for the eighth count to be the same as the second count, the eighth count must have been made from month 14 to month 18.

Case 1: If second count was on month 2:

If second count was on month 2, (i.e., of 310 animals), the eighth count cannot be on any month because no month from month 14 to month 18 have the number of animals born as 310.

Case 2: If second count was on month 4:

Similarly, If second count was on month 4, (i.e., of 330 animals), the eighth count cannot be on any month because no month from month 14 to month 18 have the number of animals born as 310.

Case 3: If second count was on month 3:

The second count on month 3 (of 320 animals) is possible because month 16. has the number of animals born as 320.
So, eighth count must be month 16.

So, the third count is on month 4 (of 330 animals).

From (2), the number of animals born on fourth count must be either 290 or 370 (inferable from the graph).
So, the fourth count must be on month 6.

The following table presents the counting that Mr. Wilson made:

The average number of animals born counted by Mr. Wilson =

*Answer can only contain numeric values
CAT Mock Test- 7 - Question 29

Directions: Read the question carefully and answer the question that follows:

Mr. Wilson works in a zoo, where he has to count the number of animals born in a month. He collates the data for the entire month at the end of the particular month. However, during a certain period of twenty months, from month 1 to month 20, Mr. Wilson counted the number of animals born on only ten months. He then plotted the ten points on a graph sheet and connected consecutive points on the graph using straight lines, forming a line graph.


Further, it is also known that
(1) the difference in number of animals born between any two consecutive counting that Mr. Wilson took was not more than 40
(2) the difference in the number of animals born counted by Mr. Wilson for the third time and for the fourth time was 40
(3) the second count that Mr. Wilson took was the same as the eighth count that he took.

Q. How many months had elapsed before Mr. Wilson began his fourth count? Key in the value.


Detailed Solution for CAT Mock Test- 7 - Question 29

Since Mr. Wilson plotted only ten points and connected consecutive points using straight lines, wherever there is a change in the slope of the line, Mr. Wilson must have made a count.
Hence, Mr. Wilson must have counted the number of animals born on months 1, 4, 7, 10, 14, 18 and 20 and the number of animals that were born were 300, 330, 270, 240, 280, 360 and 340 respectively.
The differences in number of animals born between these counting are 30, 60, 30, 40, 80 and 20 respectively.
From (1), he must have made at least one count between month 4 and month 7 and at least one count between month 14 and month 18. These account for nine counts in total.
From (3), Mr. Wilson's second count is the same as the eighth count. The second count can be on month 2 or month 3 or month 4.
In any of the three cases, for the eighth count to be the same as the second count, the eighth count must have been made from month 14 to month 18.

Case 1: If second count was on month 2:

If second count was on month 2, (i.e., of 310 animals), the eighth count cannot be on any month because no month from month 14 to month 18 have the number of animals born as 310.

Case 2: If second count was on month 4:

Similarly, If second count was on month 4, (i.e., of 330 animals), the eighth count cannot be on any month because no month from month 14 to month 18 have the number of animals born as 310.

Case 3: If second count was on month 3:

The second count on month 3 (of 320 animals) is possible because month 16. has the number of animals born as 320.
So, eighth count must be month 16.

So, the third count is on month 4 (of 330 animals).

From (2), the number of animals born on fourth count must be either 290 or 370 (inferable from the graph).
So, the fourth count must be on month 6.

The following table presents the counting that Mr. Wilson made:

Since Mr. Wilson's fourth count was for month number 6, this means that 6 months had already elapsed.

*Answer can only contain numeric values
CAT Mock Test- 7 - Question 30

Directions: Refer to the illustrations below and answer the question that follows.


Total units sold in 2014 = 2,52,000
Total units sold in 2015 = 3,24,000
Percent defective units of total units sold for each product:

Q. How many defective units of product A were sold in 2014 and 2015?


Detailed Solution for CAT Mock Test- 7 - Question 30

Number of defective units of product A sold in 2014 and 2015 = 2,52,000 x 0.15 x 0.25 + 3,24,000 x 0.20 x 0.45 = 9450 + 29,160 = 38,610

CAT Mock Test- 7 - Question 31

Directions: Refer to the illustrations below and answer the question that follows.


Total units sold in 2014 = 2,52,000
Total units sold in 2015 = 3,24,000
Percent defective units of total units sold for each product:

Q. Out of the total units of product B that were sold in 2014 and 2015, what percent was defective?

Detailed Solution for CAT Mock Test- 7 - Question 31

Total units of product B that were sold in 2014 and 2015 = 2,52,000 x 0.18 + 3,24,000 x 0.29 = 1,39,320
Total defective units of product B that were sold in 2014 and 2015 = 2,52,000 x 0.18 x 0.3 + 3,24,000 x 0.29 x 0.4 = 51,192
Required percentage = (51,192 x 100)/1,39,320 = 36.74%

CAT Mock Test- 7 - Question 32

Directions: Refer to the illustrations below and answer the question that follows.


Total units sold in 2014 = 2,52,000
Total units sold in 2015 = 3,24,000
Percent defective units of total units sold for each product:

Q. What number (approx) of total products sold in 2014 was defective?

Detailed Solution for CAT Mock Test- 7 - Question 32

Total products sold in 2014 = 2,52,000
Total defective products sold in 2014
= (2,52,000 × 0.15 × 0.25) + (2,52,000 × 0.18 × 0.3) + (2,52,000 × 0.29 × 0.28) + (2,52,000 × 0.38 × 0.39)
= 9,450 + 13,608 + 20,462 + 37,346 = 80,867 (approx)

CAT Mock Test- 7 - Question 33

Directions: Refer to the illustrations below and answer the question that follows.


Total units sold in 2014 = 2,52,000
Total units sold in 2015 = 3,24,000
Percent defective units of total units sold for each product:

Q. What is the percent increase in the number of defective units of product C sold from 2014 to 2015?

Detailed Solution for CAT Mock Test- 7 - Question 33

Required percent increase = (3,24,000 x 0.31 x 0.35 - 2,52,000 x 0.29 x 0.3 ) x 100/(2,52,000 x 0.29 x 0.3) = (35,154 - 21,924) x 100/21,924 = 60.34%

CAT Mock Test- 7 - Question 34

Directions: Refer to the illustrations below and answer the question that follows.


Total units sold in 2014 = 2,52,000
Total units sold in 2015 = 3,24,000
Percent defective units of total units sold for each product:

Q. Of the total number of units sold in 2015, how many were defective?

Detailed Solution for CAT Mock Test- 7 - Question 34

Total number of defective units sold in 2015 = (3,24,000 × 0.20 × 0.45) + (3,24,000 × 0.30 × 0.4) + (3,24,000 × 0.30 × 0.35) + (3,24,000 × 0.20 × 0.25)
= 29,160 + 38,880 + 34,020 + 16,200
= 1,18,260

*Answer can only contain numeric values
CAT Mock Test- 7 - Question 35

Directions: Study the following information to answer the given question.

A certain candy manufacturing company manufactures candies and pack them in exactly five types of packets, which contain 20, 30, 70, 170 and 190 candies each. Each of five persons Earl, Amy, Spencer, Tyler and Nicole bought similar packing of candies and no two persons among the five bought same packing of candies. The numbers of candies that Earl, Amy, Spencer, Tyler and Nicole bought from the shopkeeper were 1020, 3570, 3990, 2380 and 1470, in that order. Further, the number of packets that each person bought from the shopkeeper was distinct. However, all of them realised that they had bought some extra candies, so they returned the excess candies in such a way that they made packets similar to those made by the company and each of Earl, Amy, Spencer, Tyler and Nicole returned in the form of 2 packets, 4 packets, 1 packet, 2 packets and 1 packet, respectively, but in packet(s) different from the type the person had bought from the shopkeeper.

Q. What is the total number of packets that the five persons bought from the shopkeeper?


Detailed Solution for CAT Mock Test- 7 - Question 35

It is given that Earl bought 1020 candies. He could have bought the packets of type 20/30/170. Amy bought 3570 candies. He could have bought the packets of type 30/70/170. Spencer bought 3990 candies. He could have bought the packets of type 30/70/190. Tyler bought 2380 candies. He could have bought the packets of type 20/70/170. Nicole bought 1470 candies. He could have bought the packets of type 30/70. Since only Spencer could have bought the packets of type 190, he must have bought the packets of type 190 candies. The number of packets that he would have bought will be 21. If Nicole bought packs of 70 candies, he must have bought 21 packets. This is not possible. Hence, Nicole bought the packets containing 30 candies each. Amy could not have bought the packets of type 170, as she would have bought 21 packets. Hence, Amy must have bought packets containing 70 candies each. He must have bought 51 packets. Earl could not have bought packets containing 20 candies each as he would have then bought 51 packets. Hence, Earl must bought packets containing 170 candies. Tyler must have bought packets containing 20 candies each.
The following table provides the type of packet and the number of packets bought by each of the five persons:

Total number of packets that the five friends bought from the shopkeeper = 246

CAT Mock Test- 7 - Question 36

Directions: Study the following information to answer the given question.

A certain candy manufacturing company manufactures candies and pack them in exactly five types of packets, which contain 20, 30, 70, 170 and 190 candies each. Each of five persons Earl, Amy, Spencer, Tyler and Nicole bought similar packing of candies and no two persons among the five bought same packing of candies. The numbers of candies that Earl, Amy, Spencer, Tyler and Nicole bought from the shopkeeper were 1020, 3570, 3990, 2380 and 1470, in that order. Further, the number of packets that each person bought from the shopkeeper was distinct. However, all of them realised that they had bought some extra candies, so they returned the excess candies in such a way that they made packets similar to those made by the company and each of Earl, Amy, Spencer, Tyler and Nicole returned in the form of 2 packets, 4 packets, 1 packet, 2 packets and 1 packet, respectively, but in packet(s) different from the type the person had bought from the shopkeeper.

Q. Which of the following cannot be the number of candies Earl purchased?

Detailed Solution for CAT Mock Test- 7 - Question 36

It is given that Earl bought 1020 candies. He could have bought the packets of type 20/30/170. Amy bought 3570 candies. He could have bought the packets of type 30/70/170. Spencer bought 3990 candies. He could have bought the packets of type 30/70/190. Tyler bought 2380 candies. He could have bought the packets of type 20/70/170. Nicole bought 1470 candies. He could have bought the packets of type 30/70. Since only Spencer could have bought the packets of type 190, he must have bought the packets of type 190 candies. The number of packets that he would have bought will be 21. If Nicole bought packs of 70 candies, he must have bought 21 packets. This is not possible. Hence, Nicole bought the packets containing 30 candies each. Amy could not have bought the packets of type 170, as she would have bought 21 packets. Hence, Amy must have bought packets containing 70 candies each. He must have bought 51 packets. Earl could not have bought packets containing 20 candies each as he would have then bought 51 packets. Hence, Earl must bought packets containing 170 candies. Tyler must have bought packets containing 20 candies each.
The following table provides the type of packet and the number of packets bought by each of the five persons:

Earl returned 2 packets to the shopkeeper. None of these two packets can be of type 170. If the candies were 760, Earl must have returned 260 candies. This must be done by returning one packet of 190 candies and one packet of 70 candies. Hence, this is possible. If the number of candies were 920, Earl must have returned one packet of 30 candies and one packet of 70 candies. If the number of candies were 800, Earl must have returned one packet of 30 candies and one packet of 190 candies. If the number of candies were 820, Earl must have returned 200 candies. This can be done only if he returned the packets of 30 and 170 candies. Since this is not possible, the number of candies that Earl purchased cannot be 820.

CAT Mock Test- 7 - Question 37

Directions: Study the following information to answer the given question.

A certain candy manufacturing company manufactures candies and pack them in exactly five types of packets, which contain 20, 30, 70, 170 and 190 candies each. Each of five persons Earl, Amy, Spencer, Tyler and Nicole bought similar packing of candies and no two persons among the five bought same packing of candies. The numbers of candies that Earl, Amy, Spencer, Tyler and Nicole bought from the shopkeeper were 1020, 3570, 3990, 2380 and 1470, in that order. Further, the number of packets that each person bought from the shopkeeper was distinct. However, all of them realised that they had bought some extra candies, so they returned the excess candies in such a way that they made packets similar to those made by the company and each of Earl, Amy, Spencer, Tyler and Nicole returned in the form of 2 packets, 4 packets, 1 packet, 2 packets and 1 packet, respectively, but in packet(s) different from the type the person had bought from the shopkeeper.

Q. How many of the following statements are definitely true?
I. The number of packets that Earl purchased is an even number.
II. The number of packets that Amy purchased is an even number.
III. The number of packets that Spencer purchased is an odd number.
IV. The number of packets that Tyler purchased is an even number.

Detailed Solution for CAT Mock Test- 7 - Question 37

It is given that Earl bought 1020 candies. He could have bought the packets of type 20/30/170. Amy bought 3570 candies. He could have bought the packets of type 30/70/170. Spencer bought 3990 candies. He could have bought the packets of type 30/70/190. Tyler bought 2380 candies. He could have bought the packets of type 20/70/170. Nicole bought 1470 candies. He could have bought the packets of type 30/70. Since only Spencer could have bought the packets of type 190, he must have bought the packets of type 190 candies. The number of packets that he would have bought will be 21. If Nicole bought packs of 70 candies, he must have bought 21 packets. This is not possible. Hence, Nicole bought the packets containing 30 candies each. Amy could not have bought the packets of type 170, as she would have bought 21 packets. Hence, Amy must have bought packets containing 70 candies each. He must have bought 51 packets. Earl could not have bought packets containing 20 candies each as he would have then bought 51 packets. Hence, Earl must bought packets containing 170 candies. Tyler must have bought packets containing 20 candies each.
The following table provides the type of packet and the number of packets bought by each of the five persons:

Statement I and III are correct.

CAT Mock Test- 7 - Question 38

Directions: Study the following information to answer the given question.

A certain candy manufacturing company manufactures candies and pack them in exactly five types of packets, which contain 20, 30, 70, 170 and 190 candies each. Each of five persons Earl, Amy, Spencer, Tyler and Nicole bought similar packing of candies and no two persons among the five bought same packing of candies. The numbers of candies that Earl, Amy, Spencer, Tyler and Nicole bought from the shopkeeper were 1020, 3570, 3990, 2380 and 1470, in that order. Further, the number of packets that each person bought from the shopkeeper was distinct. However, all of them realised that they had bought some extra candies, so they returned the excess candies in such a way that they made packets similar to those made by the company and each of Earl, Amy, Spencer, Tyler and Nicole returned in the form of 2 packets, 4 packets, 1 packet, 2 packets and 1 packet, respectively, but in packet(s) different from the type the person had bought from the shopkeeper.

Q. Who among the following bought the highest number of packets?

Detailed Solution for CAT Mock Test- 7 - Question 38

It is given that Earl bought 1020 candies. He could have bought the packets of type 20/30/170. Amy bought 3570 candies. He could have bought the packets of type 30/70/170. Spencer bought 3990 candies. He could have bought the packets of type 30/70/190. Tyler bought 2380 candies. He could have bought the packets of type 20/70/170. Nicole bought 1470 candies. He could have bought the packets of type 30/70. Since only Spencer could have bought the packets of type 190, he must have bought the packets of type 190 candies. The number of packets that he would have bought will be 21. If Nicole bought packs of 70 candies, he must have bought 21 packets. This is not possible. Hence, Nicole bought the packets containing 30 candies each. Amy could not have bought the packets of type 170, as she would have bought 21 packets. Hence, Amy must have bought packets containing 70 candies each. He must have bought 51 packets. Earl could not have bought packets containing 20 candies each as he would have then bought 51 packets. Hence, Earl must bought packets containing 170 candies. Tyler must have bought packets containing 20 candies each.
The following table provides the type of packet and the number of packets bought by each of the five persons:

Tyler bought the highest number of packets (119).

CAT Mock Test- 7 - Question 39

Directions: Study the following information to answer the given question.

A certain candy manufacturing company manufactures candies and pack them in exactly five types of packets, which contain 20, 30, 70, 170 and 190 candies each. Each of five persons Earl, Amy, Spencer, Tyler and Nicole bought similar packing of candies and no two persons among the five bought same packing of candies. The numbers of candies that Earl, Amy, Spencer, Tyler and Nicole bought from the shopkeeper were 1020, 3570, 3990, 2380 and 1470, in that order. Further, the number of packets that each person bought from the shopkeeper was distinct. However, all of them realised that they had bought some extra candies, so they returned the excess candies in such a way that they made packets similar to those made by the company and each of Earl, Amy, Spencer, Tyler and Nicole returned in the form of 2 packets, 4 packets, 1 packet, 2 packets and 1 packet, respectively, but in packet(s) different from the type the person had bought from the shopkeeper.

Q. Which of the following statements is correct?

Detailed Solution for CAT Mock Test- 7 - Question 39

It is given that Earl bought 1020 candies. He could have bought the packets of type 20/30/170. Amy bought 3570 candies. He could have bought the packets of type 30/70/170. Spencer bought 3990 candies. He could have bought the packets of type 30/70/190. Tyler bought 2380 candies. He could have bought the packets of type 20/70/170. Nicole bought 1470 candies. He could have bought the packets of type 30/70. Since only Spencer could have bought the packets of type 190, he must have bought the packets of type 190 candies. The number of packets that he would have bought will be 21. If Nicole bought packs of 70 candies, he must have bought 21 packets. This is not possible. Hence, Nicole bought the packets containing 30 candies each. Amy could not have bought the packets of type 170, as she would have bought 21 packets. Hence, Amy must have bought packets containing 70 candies each. He must have bought 51 packets. Earl could not have bought packets containing 20 candies each as he would have then bought 51 packets. Hence, Earl must bought packets containing 170 candies. Tyler must have bought packets containing 20 candies each.
The following table provides the type of packet and the number of packets bought by each of the five persons:

Total number of packets bought by Amy and Spencer = 51 + 21 = 72
Number of packets bought by Tyler = 119
So, 'the total number of packets bought by Amy and Spencer is less than that bought by Tyler' is true.

CAT Mock Test- 7 - Question 40

Directions: Read the information given below and answer the question that follows.

Eight chess players – Carlson, Caruana, Nakamura, Anand, Taplov, Vidit, Harikrishna and Boris – played three different format tournaments – Classic, Rapid and Biltz to increase their FIDE Ratings– such that each player played exactly one format. Two of the eight players played Classic, three played Rapid and three played Biltz. The increase in FIDE Ratings of each player is distinct. Further, the lowest increase in the rating in Classic format is higher than the highest increase in the rating in Rapid Format. Also the lowest increase in the rating in Rapid format is higher than the highest increase in the rating in Biltz Format.

It is also known that:
1. At least two players increased their ratings more than Anand's increase, and Anand did not increase his rating less than what Harikrishna did.
2. Vidit did not play Biltz and his increase of rating is just less than the increase of Boris.
3. Carlson got the highest increase in rating in Biltz.
4. Nakamura was not one with the lowest increase in his rating.
5. Caruana, who did not play Biltz, increased his rating less than what Harikrishna did.

Q. Who got the highest increase in rating in Rapid format?

Detailed Solution for CAT Mock Test- 7 - Question 40

The two players who played Classic got highest increase in their FIDE Ratings, the three players who played Rapid got increase in their FIDE Ratings less than Classic players and the three players who played Biltz got increase in their FIDE Ratings less than Rapid players.
Hence, we can order the eight players from 1 to 8 such that 1st and 2nd are Classic, and 3rd, 4th and 5th are Rapid, and 6th, 7th and 8th are Biltz.
Carlson got the highest increase in rating in Biltz. So, his order is 6.

From conditions 1, 2 and 5, it is clear that Anand, Harikrishna, Caruana, Vidit, Boris not played Biltz. So, Carlson, Nakamura and Taplov played Biltz.
From condition 4, Nakamura did not get the lowest increase in his rating. So, Nakamura's order is 7 and Taplov's order is 8.

From conditions 1 and 5, at least two players increase their ratings more than Anand did, and Anand did not increase his ratings less than Harikrishna did, and Caruana increased his rating less than Harikrishna did. So, the order of Anand, Harikrishna and Caruana is 3, 4 and 5, respectively.
From condition 2, Vidit, who did not played Biltz, has his increase of rating at just less than the increase of Boris. So, the order of Boris and Vidit is 1 and 2, respectively.

Anand got the highest increase in rating in Rapid format.

CAT Mock Test- 7 - Question 41

Directions: Read the information given below and answer the question that follows.

Eight chess players – Carlson, Caruana, Nakamura, Anand, Taplov, Vidit, Harikrishna and Boris – played three different format tournaments – Classic, Rapid and Biltz to increase their FIDE Ratings– such that each player played exactly one format. Two of the eight players played Classic, three played Rapid and three played Biltz. The increase in FIDE Ratings of each player is distinct. Further, the lowest increase in the rating in Classic format is higher than the highest increase in the rating in Rapid Format. Also the lowest increase in the rating in Rapid format is higher than the highest increase in the rating in Biltz Format.

It is also known that:
1. At least two players increased their ratings more than Anand's increase, and Anand did not increase his rating less than what Harikrishna did.
2. Vidit did not play Biltz and his increase of rating is just less than the increase of Boris.
3. Carlson got the highest increase in rating in Biltz.
4. Nakamura was not one with the lowest increase in his rating.
5. Caruana, who did not play Biltz, increased his rating less than what Harikrishna did.

Q. How many players got increase in ratings higher than the Nakamura?

Detailed Solution for CAT Mock Test- 7 - Question 41

The two players who played Classic got highest increase in their FIDE Ratings, the three players who played Rapid got increase in their FIDE Ratings less than Classic players and the three players who played Biltz got increase in their FIDE Ratings less than Rapid players.
Hence, we can order the eight players from 1 to 8 such that 1st and 2nd are Classic, and 3rd, 4th and 5th are Rapid, and 6th, 7th and 8th are Biltz.
Carlson got the highest increase in rating in Biltz. So, his order is 6.

From conditions 1, 2 and 5, it is clear that Anand, Harikrishna, Caruana, Vidit, Boris not played Biltz. So, Carlson, Nakamura and Taplov played Biltz.
From condition 4, Nakamura did not get the lowest increase in his rating. So, Nakamura's order is 7 and Taplov's order is 8.

From conditions 1 and 5, at least two players increase their ratings more than Anand did, and Anand did not increase his ratings less than Harikrishna did, and Caruana increased his rating less than Harikrishna did. So, the order of Anand, Harikrishna and Caruana is 3, 4 and 5, respectively.
From condition 2, Vidit, who did not played Biltz, has his increase of rating at just less than the increase of Boris. So, the order of Boris and Vidit is 1 and 2, respectively.

CAT Mock Test- 7 - Question 42

Directions: Read the information given below and answer the question that follows.

Eight chess players – Carlson, Caruana, Nakamura, Anand, Taplov, Vidit, Harikrishna and Boris – played three different format tournaments – Classic, Rapid and Biltz to increase their FIDE Ratings– such that each player played exactly one format. Two of the eight players played Classic, three played Rapid and three played Biltz. The increase in FIDE Ratings of each player is distinct. Further, the lowest increase in the rating in Classic format is higher than the highest increase in the rating in Rapid Format. Also the lowest increase in the rating in Rapid format is higher than the highest increase in the rating in Biltz Format.

It is also known that:
1. At least two players increased their ratings more than Anand's increase, and Anand did not increase his rating less than what Harikrishna did.
2. Vidit did not play Biltz and his increase of rating is just less than the increase of Boris.
3. Carlson got the highest increase in rating in Biltz.
4. Nakamura was not one with the lowest increase in his rating.
5. Caruana, who did not play Biltz, increased his rating less than what Harikrishna did.

Q. Which of the following pairs of players played the same format?

Detailed Solution for CAT Mock Test- 7 - Question 42

The two players who played Classic got highest increase in their FIDE Ratings, the three players who played Rapid got increase in their FIDE Ratings less than Classic players and the three players who played Biltz got increase in their FIDE Ratings less than Rapid players.
Hence, we can order the eight players from 1 to 8 such that 1st and 2nd are Classic, and 3rd, 4th and 5th are Rapid, and 6th, 7th and 8th are Biltz.
Carlson got the highest increase in rating in Biltz. So, his order is 6.

From conditions 1, 2 and 5, it is clear that Anand, Harikrishna, Caruana, Vidit, Boris not played Biltz. So, Carlson, Nakamura and Taplov played Biltz.
From condition 4, Nakamura did not get the lowest increase in his rating. So, Nakamura's order is 7 and Taplov's order is 8.

From conditions 1 and 5, at least two players increase their ratings more than Anand did, and Anand did not increase his ratings less than Harikrishna did, and Caruana increased his rating less than Harikrishna did. So, the order of Anand, Harikrishna and Caruana is 3, 4 and 5, respectively.
From condition 2, Vidit, who did not played Biltz, has his increase of rating at just less than the increase of Boris. So, the order of Boris and Vidit is 1 and 2, respectively.

CAT Mock Test- 7 - Question 43

Directions: Read the information given below and answer the question that follows.

Eight chess players – Carlson, Caruana, Nakamura, Anand, Taplov, Vidit, Harikrishna and Boris – played three different format tournaments – Classic, Rapid and Biltz to increase their FIDE Ratings– such that each player played exactly one format. Two of the eight players played Classic, three played Rapid and three played Biltz. The increase in FIDE Ratings of each player is distinct. Further, the lowest increase in the rating in Classic format is higher than the highest increase in the rating in Rapid Format. Also the lowest increase in the rating in Rapid format is higher than the highest increase in the rating in Biltz Format.

It is also known that:
1. At least two players increased their ratings more than Anand's increase, and Anand did not increase his rating less than what Harikrishna did.
2. Vidit did not play Biltz and his increase of rating is just less than the increase of Boris.
3. Carlson got the highest increase in rating in Biltz.
4. Nakamura was not one with the lowest increase in his rating.
5. Caruana, who did not play Biltz, increased his rating less than what Harikrishna did.

Q. Who among the following was the player who got minimum increase in his rating?

Detailed Solution for CAT Mock Test- 7 - Question 43

The two players who played Classic got highest increase in their FIDE Ratings, the three players who played Rapid got increase in their FIDE Ratings less than Classic players and the three players who played Biltz got increase in their FIDE Ratings less than Rapid players.
Hence, we can order the eight players from 1 to 8 such that 1st and 2nd are Classic, and 3rd, 4th and 5th are Rapid, and 6th, 7th and 8th are Biltz.
Carlson got the highest increase in rating in Biltz. So, his order is 6.

From conditions 1, 2 and 5, it is clear that Anand, Harikrishna, Caruana, Vidit, Boris not played Biltz. So, Carlson, Nakamura and Taplov played Biltz.
From condition 4, Nakamura did not get the lowest increase in his rating. So, Nakamura's order is 7 and Taplov's order is 8.

From conditions 1 and 5, at least two players increase their ratings more than Anand did, and Anand did not increase his ratings less than Harikrishna did, and Caruana increased his rating less than Harikrishna did. So, the order of Anand, Harikrishna and Caruana is 3, 4 and 5, respectively.
From condition 2, Vidit, who did not played Biltz, has his increase of rating at just less than the increase of Boris. So, the order of Boris and Vidit is 1 and 2, respectively.

CAT Mock Test- 7 - Question 44

Directions: Read the information given below and answer the question that follows.

Eight chess players – Carlson, Caruana, Nakamura, Anand, Taplov, Vidit, Harikrishna and Boris – played three different format tournaments – Classic, Rapid and Biltz to increase their FIDE Ratings– such that each player played exactly one format. Two of the eight players played Classic, three played Rapid and three played Biltz. The increase in FIDE Ratings of each player is distinct. Further, the lowest increase in the rating in Classic format is higher than the highest increase in the rating in Rapid Format. Also the lowest increase in the rating in Rapid format is higher than the highest increase in the rating in Biltz Format.

It is also known that:
1. At least two players increased their ratings more than Anand's increase, and Anand did not increase his rating less than what Harikrishna did.
2. Vidit did not play Biltz and his increase of rating is just less than the increase of Boris.
3. Carlson got the highest increase in rating in Biltz.
4. Nakamura was not one with the lowest increase in his rating.
5. Caruana, who did not play Biltz, increased his rating less than what Harikrishna did.

Q. The increase in ratings of how many players is between Harikrishna and Nakamura?

Detailed Solution for CAT Mock Test- 7 - Question 44

The two players who played Classic got highest increase in their FIDE Ratings, the three players who played Rapid got increase in their FIDE Ratings less than Classic players and the three players who played Biltz got increase in their FIDE Ratings less than Rapid players.
Hence, we can order the eight players from 1 to 8 such that 1st and 2nd are Classic, and 3rd, 4th and 5th are Rapid, and 6th, 7th and 8th are Biltz.
Carlson got the highest increase in rating in Biltz. So, his order is 6.

From conditions 1, 2 and 5, it is clear that Anand, Harikrishna, Caruana, Vidit, Boris not played Biltz. So, Carlson, Nakamura and Taplov played Biltz.
From condition 4, Nakamura did not get the lowest increase in his rating. So, Nakamura's order is 7 and Taplov's order is 8.

From conditions 1 and 5, at least two players increase their ratings more than Anand did, and Anand did not increase his ratings less than Harikrishna did, and Caruana increased his rating less than Harikrishna did. So, the order of Anand, Harikrishna and Caruana is 3, 4 and 5, respectively.
From condition 2, Vidit, who did not played Biltz, has his increase of rating at just less than the increase of Boris. So, the order of Boris and Vidit is 1 and 2, respectively.

CAT Mock Test- 7 - Question 45

AB and CD are 2 parallel chords of a circle which has a radius of 10 units. Parallel chords are 14 units apart such that one of the chords has a length of 12 units. What is the length of non-parallel side of trapezium ABCD

Detailed Solution for CAT Mock Test- 7 - Question 45

Since the radius of the circle is 10 and the chords are 14 units apart, both of them have to be on the opposite side of the centre, let CD = 12. It can be shown as

E and F are the midpoint of respective chords.
From the properties of the circle, we know that OF is the perpendicular bisector of DC. Thus DF = 6 and OD =10(radius). It gives OF = 8 (from Pythagoras theorem)
Since it is given that EF = 14, it implies that EO = 14-OF =14-8 = 6
EOB is also a right-triangle such that EO = 6 and OB = 10. This gives EB = 8 or AB = 16.
Since they are symmetrical the trapezium will look like this

From pythogoras we find that AD2 = 142 + 22 = 196 + 4 = 200
AD = 10√2

CAT Mock Test- 7 - Question 46

Among all the possible permutation of the word " VENKATESHAN" What is the probability that a word is chosen such that both the "A"s are before both the "E"s?

Detailed Solution for CAT Mock Test- 7 - Question 46

There are a total of 11 alphabets in VENKATESHAN which when arranged in ascending order are A, A, E, E, H K, N, N, S, T and V.
Total possible permutation are =  
Now we need to find the cases where all the "A"s are before "E"s. There are total of 2 A and 2 E. If we pick places for A and E, they can be placed in one way. 4 places can be picked in 11C4 ways.
Remaining 7 alphabets can be arranged in 7!/2! ways.
The number of ways possible  = 
Probability = 
Probability = 

CAT Mock Test- 7 - Question 47

A rectangle ABCD exists such that AB = 6 units and BC = 12 units. E and F trisect the diagonal AC in 3 equal parts. G is centroid of the triangle formed by DEF. What is the area of DGBC?

Detailed Solution for CAT Mock Test- 7 - Question 47

Let us draw a rectangle on the coordinate axis. With (0,0) be one of the edge A. B can be (6,0), C(6,12) and D(0,12).  E and F divide the AC in 3 equal part. E has to be (2,4) and F (4,8)
G is  centroid of DEF, Coordinates of G will be 

Upon finding G the required diagram looks like this

Area DGBC = Area GBC + Area DCG
Area DGBC = 
Area DGBC = 24+12 = 36 sq units

CAT Mock Test- 7 - Question 48

How many 4-digit numbers exist which is divisible by 3 or 8?

Detailed Solution for CAT Mock Test- 7 - Question 48

The divisibility rule of 3 states that the sum of digits has to be divisible by 3 in order to be divisible by 3.
Such numbers are 1002,1005,1008.....9999 : 3000 numbers.
The divisibility rules of 8 imply that the last 3 digits have to be divisible by 8: Thus "dcba" has to be divisible by 8. Such number are : 1000,1008,..9992 : 1125.
We have included the number which are divisible by 24 twice. Thus we have to remove those extra numbers.
Such number are : 1008,1032,...9984: 375 numbers.
The numbers of 4-digit number  which are divisible by 3 or 8: 3000+1125-375 = 3750.

CAT Mock Test- 7 - Question 49

What is the ratio of the shaded region to non-shaded region in the following diagram

ABC is an equilateral triangle and D, E and F are the midpoint of the sides.

Detailed Solution for CAT Mock Test- 7 - Question 49

Since ABC is an equilateral triangle and DEF is the midpoint. Thus they will divide the ABC in 4 equilateral triangle of side half of that of ABC
Let side of ABC = 2a units. Then the side of the smaller triangle is a unit.
Shaded region = Area of 3 small circles + Area DEF - Area of 1 small circle
Shaded region = Area of 2 small circles + Area DEF 
Circles are an incircle of equilateral triangle of side "a"
Radius of this incircle 
Area of 1 small circle = 
Area of 2 small circle 
Area of triangle DEF = 
Shaded area = 
Unshaded area = Total area - Shaded area
Unshaded area =  - Shaded area
Unshaded area = 
Unshaded area = 
Ratio = 

*Answer can only contain numeric values
CAT Mock Test- 7 - Question 50

A cunning rice merchant uses a faulty weighing machine to cheat. While buying the rice from the whole seller he uses a faulty machine which shows the quantity 20% less than actually on the scale. While selling the rice, he uses another faulty scale which shows 10% extra weight than that on the scale. He marks up the price of rice by 20% on the rate he bought from the wholesaler. Sherlock found out about this scam and asked the rice seller to give approximately a d% discount on the current marked price so that the rice seller doesn't make a profit or a loss. What is the value of [d]?
[X] represents the largest integer less than or equal to X.


Detailed Solution for CAT Mock Test- 7 - Question 50

Let the rate at which the whole seller sells the rice be Rs x per kg.
But the cunning merchant uses the faulty scale to trick the wholesaler. When the wholeseller puts in 1kg of rice, the machine will show 0.8kg of rice. Thus merchant will get 1 kg of rice for the price of 0.8x0.8x and thus Actual selling price for the wholeseller = Actual cost-price for the merchant = Rs0.8x = 4x/5 per kg.
The rate at which he sells the rice = Rs 1.2x per kg.
When the merchant puts 1 kg of rice on the weighing machine, the machine will show 1.1 kg.
As per the rate the customer has to pay 1.2x x 1.1 = 1.32x  for 1 kg of rice. 
Actual selling price for the merchant= Rs 1.32x per kg
Let Sherlock by y kgs of rice.
Cost price to the merchant = 0.8xy
Actual selling price for the merchant = 1.32 xy
Discount = d
In order to be no-profit no loss

*Answer can only contain numeric values
CAT Mock Test- 7 - Question 51

An absent-minded shopkeeper interchanged the markup of m% and discount d% of a product. So, instead of making a profit of 8%, he made a loss of 12% on the product. What is the value of d2 + m2?


Detailed Solution for CAT Mock Test- 7 - Question 51

Let the cost price be x
When markup percent is m% and discount given is d% then (1+m%)(1−d%)x=1.08x
Upon expanding we get 

When markup percent is d% and discount given is m% then
(1+d%)(1−m%)x = 0.88x
Upon expanding we get

Adding (I) and (II) we get,

Substracting  (I) and (II) we get,

or m-d =10
Squaring both sides, m2 + d2 - 2md = 100
m2 + d2 - 400 = 100
m2 + d2 = 500

CAT Mock Test- 7 - Question 52

Siddhart is a pen-seller. He sells a pen such that the profit earned by selling 10 pens is equal to the cost price of 1 pen. The discount he gave on the marked price of the pen is 3 times the profit he earned by selling the pen. What approximate discount % should he offer the customer if he wishes to have a profit of 20%

Detailed Solution for CAT Mock Test- 7 - Question 52

Let the profit be P and cost price be C.P
Then 10 P = C.P ....(I)
Selling Price = Cost Price + Profit = 10 P +P = 11P
Discount given  = 3P.
Marked price = Selling Price + Discount = 11P+3P = 14P.
Let the discount given to ensure profit of 20% be d%.

*Answer can only contain numeric values
CAT Mock Test- 7 - Question 53

A number N is given by = 25 x 38 x 42 x 53 x 6 x 72
If a certain factor of N is not divisible by 9, the probability that the factor is an odd number is m/n where 'n' is a natural number less than 20. Find the value of m+n. 


Detailed Solution for CAT Mock Test- 7 - Question 53

Let us start by writing the N in empirical form which is 25 x 38 x 24 x 53 x 2 x 3 x 72
Which equals 210 x 39 x 53 x 72
Let A = number of factors not divisible by 9
B = Number of Factors which are odd
We have to find 
(A∩B) = Number of factors not divisible by 9 and are odd. It implies it does not have 2 as a factor and the highest power of 3 can be  1. Such type of factors are = (1+1)(3+1)(2+1) = 24
A = Number of factors that are not divisible by 9. Thus the maximum power of 3 they can have is 1. Such number are = (10+1)(1+1)(3+1)(2+1) = 264
Probability = 24/264 = 1/11
m+n = 12.

*Answer can only contain numeric values
CAT Mock Test- 7 - Question 54

Ram invests a total of Rs 1,50,000 into two schemes in the ratio 2:1. Both of the schemes have interest rates which are integers. After 2 years, Ram received a total of 1,93,000 in return. If one of the interest rates is 20%, and the interest rate of the other is R%, what is the value of R? Assume the interest to be compounded annually


Detailed Solution for CAT Mock Test- 7 - Question 54

Let the scheme be Scheme I and Scheme II. Money spent on scheme I = 
Money on scheme II = 50,000
We don't know which of the scheme has an interest rate of 20%. Thus there will be 2 cases
Case I: Scheme I has a 20% return rate, In this case, let r be return rate for scheme II

Here we can see that the RHS is less than 1, the value of rr has to be negative. Which is not possible. Thus this case is rejected
Case II: Scheme II has a 20% return rate, In this case, let r be return rate for scheme I

Therefore r = 10. Which is an integer

CAT Mock Test- 7 - Question 55

Let N be a natural number < 1500 exist such that when N divided by 7 has the same remainder when N3  is divided by 7. How many such values of N are there?

Detailed Solution for CAT Mock Test- 7 - Question 55

We have to find values of N such that N and Nboth have the same remainder when divided by 7
Thus (N mod 7) =(N3 mod 7)
It can be written as 
(N mod 7) = (N x N x N) mod7 = (Nmod7) x (Nmod7) x (Nmod7).
Thus it can be implied that
(N mod 7) = (N mod 7)3
When N = 1 then (Nmod7) = (Nmod7)3 = 1 so N =1 is a solution
When N = 2 then (Nmod7) = 2 and (Nmod7)3 = 1 so N ≠ 2

When N = 7 then the remainder is 0. Thus N has to be of form 7k, 7k+1 and 7k+6 to meet the required criteria
N< 1500 of from 7k = 7,14, .... 1498 which is 214 terms
N<1500 of from 7k+1 = 1, 8,15, .... 1499 which is again 215 terms
N<1500 of from 7k+6 = 6,15, .... 1497 which is again 214 terms
Total terms = 214+215+214 = 643.

CAT Mock Test- 7 - Question 56

A man was driving from city X to city Y. After travelling for  3 hours, his car develops a fault and his speed reduced to 1/3rd of his original speed. He was originally supposed to complete the journey in 17 hours. What was % increase in his overall travelling time?

Detailed Solution for CAT Mock Test- 7 - Question 56

t/3 = 14 Let the original speed of the car be  xx km/hr
Since he was originally supposed to cover the distance in 17 hours, The total distance of the journey is 17x km.
After the malfunction, his speed will be x/3. Let him take time "t" hours to complete after the vehicle malfunction. Equating the distances we get
3x + (x/3)t = 17x
Cancelling x from both sides
We get t/3 = 14
t = 42 hours.
Initial travelling time = 17 hours. 
Final travelling time = 3+42 = 45 hours.
% increase in time = 

*Answer can only contain numeric values
CAT Mock Test- 7 - Question 57

In a race. A beats B by 120m or 20 seconds. B beats C by 50m or 10 seconds. What is the sum of the speeds of A, B and C? (in m/sec)


Detailed Solution for CAT Mock Test- 7 - Question 57

300m race A=10m/s. B = 6m/s C = 5m/s
Let the distance of the race be x meters. Let the speed of A, B and C be Va, Vand Vc (in m/sec)
A beats B by 120m or 20 seconds 
Thus 
Putting the value of x/Va from (I) to (II)

It gives Vb = 6 m/s
Now coming to B beats C by 50m or 10 seconds

From (III) and  (IV) we get

or V= 5m/sec
From (III)

Cross multiplying 
5x=6(x−50) or x= 300 meters.

Sum = 10+6+5 = 21

CAT Mock Test- 7 - Question 58

A, B and C are 3 friends who have to go to another city Z. A has a bike that has an average speed of 40 km/hr. B and C both can walk with an average speed of 5km/hr. To minimise the time taken for them to reach the other city it was decided that  C will start walking towards the city Z while A will gives a lift to B for a certain distance after which A will drop-off B and B will start walking towards Z. A will go back to pick up C and then will start moving towards city Z. They all reach the destination at the same time. What percentage of the overall journey time, did B spend walking?

Detailed Solution for CAT Mock Test- 7 - Question 58


A and B start on bike and A drops B at R. The time taken is T1. Hence , PR 40T1.
At the same time, C walks from P to Q. Distance covered = 5T1.
Distance between A and C = 40T1 - 5T1 = 35T1.
Relative speed between A and C = 40 + 5 = 45.
Time taken to meet = 35/45T1 = 7/9T1
Now, in the meantime, B also walks from R to T. His speed is same as C, so distance travelled is same, i.e, 5 x 7/9T1 = 35/9T1.
Hence, the distnce between the bike and B when the bike starts from S is ST, i.e, 35T1.
Now, let they meet at Z after time T2.
Relative speed = 40 - 5 = 35.
Hence, 35 T2 = 35T1
T2 = T1
Hence, time taken by the overall journey = T1 + 7/9T1 + T1 = 25/9T1
Time spent by B walking = Toatl time - T1 = 25/9T1-T1 = 16/9T1.
Hence, percentage time = 16/25 x 100 = 64%.
 

Alternate solution:
Let us divide all the activities into certain time periods for simplification. Let all of them start at time t = 0.
(1) From time t = 0 to t1. A and B will move towards Z on bike and C will walk towards Z. At exactly t1, A will drop B and start moving back to pick up C. Distance covered by A = 40t1, Distance covered by B = 40t1 and Distance covered by C = 5(t2 - t1) and distance by A in this time period = - 40(t2 - t1). We have taken"-" as A is moving in opposite direction 3) From time t = t2 to t3. A has picked up C and now both are on bike moving towards city Z. 0. B is walking towards city Z. They all reach at city Z at same time t3 Distance travelled by A = 40(t3 - t2). Distance covered by B = 5(t3 - t2). Distance travelled by C = 40 (t3 - t2)
Let the overall distance be d
Distance travelled by A = 40t1 - 40 (t2 - t1) + 40(t3 - t2) = d
Thus 40(2t1 + t3 - 2t2) = d ...(I)
Distance travelled by B = 40t1 + 5(t2 - t1)+5(t3-t2)
35t1 + 5t3 = d ...(II)
Distance travelled by C = 5t1 + 5(t2 - t1) + 40(t3 - t2) = d
40t3 - 35t2 = d ...(III)
Equating (II) and (III) 
35t1 + 5t3 = 40t3 - 35t2
t1 = t3 - t2 ...(IV)
Equating (I) and (III)
80t1 + 40t3 - 80t2 = 40t3 - 35t2
⇒ 80t1  = 45t2
Or t2 = (16/9)t1 ...(V)
Putting (V) in(IV)

From time t1 to t3 B was walking and total journey time was t3
% time on walking = 
% time on walking = 
% time walking = 

*Answer can only contain numeric values
CAT Mock Test- 7 - Question 59

What is the sum of the first 20 numbers in the series 8,24,62,122, 204


Detailed Solution for CAT Mock Test- 7 - Question 59

Let us look at the series 8, 24, 62, 122
Their first level difference = 24-8 = 16, 62-24= 38 and 122-62= 60
Their second level difference = 38-16 = 22, 60-38 = 22 which are same. So this series varies with power of 2
Let Tn  represent the term of series. Then Tn can be assumed as an2 + bn + c

From (III)-(II) we get 5a+b = 38
From (II)-(I) we get 3a+b =16
Eliminating b we get 2a = 22 or a=11. Thus b = -17 and c = 14

Sum = 31570-3570+280 = 28280

CAT Mock Test- 7 - Question 60

2 men and a woman can complete a task in a certain number of days. If 13 men and 12 women work on the same task they can complete it in 1/8th of days required earlier. The number of days taken by 13 men and 12 women is equal to the number of days taken by 4 men and 27 children to finish the same task.
If 2 children and a man can complete another task in 12 days. What is the minimum number of days required by 3 women and 1 child to do the same task?
Even if they have to work for a fraction of a day, it will be counted as a whole day. 

Detailed Solution for CAT Mock Test- 7 - Question 60

Let the amount of work done by a man in 1 day be "m" units and the amount of work done by a woman in 1 day be "w" units. Let's assume they take N days to complete
As per the first line we can say that 
(2m + w) x N = (N/8)(13m + 12w)
On cancelling the common factors and rearranging we get
(2m + w) x 8  = (13 m + 12w)
Which implies 16m + 8w = (13m + 12w)
or 3m = 4w ...(I)
It is also given that it is equal to 4 men and 27 children. Let a child to "c" units of work in a day
Then
(N/8) (13m + 12w) = (N/8)(4m +27c)
Or, 13m + 12w = 4m + 27c
Putting value of w from (I)
13m+9m=4m+27c
or, 18m=27c or 2m = 3c...(II)
Given that another task takes 12 days to be completed by 2 children and a man. Let us assume that it takes x days to be completed by 3 women and 1 child. Equating the amount of work done

Cancelling m and cross multiplying

9.6 days is needed. 10 is the next biggest number .

CAT Mock Test- 7 - Question 61

There are 3 types of sugar, type A, type B and type C. The cost of each of the is Rs 35/kg, Rs, 40/kg and Rs 42/kg. The shopkeeper mixes them in a fixed ratio. He sold 40 kg of the mixture for Rs 1900 and had a profit of 25%. If the ratio of type A, type B and type C in the ratio x:y:z where x,y,z are integers. What is the minimum value of x+y+z?

Detailed Solution for CAT Mock Test- 7 - Question 61

Shopkeeper sold 40 kg of mixed sugar for Rs 1900. Selling price per kg = 1900/40 = Rs 1900/40 = Rs 47.5/kg
Cost price per kg = Rs 38/kg
SInce 35 < 38 < 40.
We have to find the ratio of sugar of mixture I containing type A and B and mixture 2 containing A and C
For mixture 1 let the quantity of type A and type B be a:b
Then 35a + 40b = 38(a+b) or 3a = 2b. Thus  a:b = 2:3....(I)
For mixture 2 let the quantity of type A and type C be a:c
 35a + 42c = 38(a+c) or 3a = 4c or  a:c = 4:3 ...(II)
(I)+(II) gives a:b:c = 6:3:3. Which can be simplified to 2:1:1

*Answer can only contain numeric values
CAT Mock Test- 7 - Question 62

f(x) and g(x) are 2 polynomial functions such that f(g(x)) = x2 + 12x + 33 and g(f(x)) = x2 + 4x + 5. it is given that f(26) = 46. What is maximum possible value of f(3)?


Detailed Solution for CAT Mock Test- 7 - Question 62

We know that f(g(x)) =x2 + 12x +33
Thus f(g(f(x))) = f(x2) + 12f(x) + 33
Putting g(f(x)) = x2 + 4x + 5
Thus f(x2 + 4x + 5) = f(x)2 + 12f(x) + 33
Given f(26) = 46
putting x2 + 4x + 5 = 26
Or x2 + 4x - 21 = 0
Or (x+7)(x-3) =0.
Putting x = 3 in f(x2 + 4x + 5) = f(x)2 + 12f(x) + 33 
f(26) = 46 = f(3)2+12f(3)+33
Or f(3)2 + 12f(3) - 13 = 0
(f(3)+13)(f(3)-1) = 0 Thus f(3) = 1 or -13
1 > -13 so it is the answer

CAT Mock Test- 7 - Question 63

Fresh grapes have 75% water and the remaining 25% as fruit pulp. A shopkeeper bought a certain quantity of fresh grapes and let it dry so that the fraction of pulp in the grapes increase by 1/3rd of the original quantity. If the ratio of the evaporated water to the pulp is m:n where m and n are the lowest possible integers. Which of the following is the has maximum absolute value?

Detailed Solution for CAT Mock Test- 7 - Question 63

Let 100kg of fresh grapes be there. Then 25 kgs is the pulp and 75 kg is the weight of water,
Initial fraction of pulp = 25% = 1/4
FInal fraction of pulp = 
SInce the weight of pulp will not change 
Putting pulp = 25kgs

This gives updated weight of water as 50 kgs.
Evaporated water =  75-50 = 25
Ratio = 25:25 = 1:1 = m:n
a) |n-5m| =|1-5| = 4
b) |n+m| = 1+1 =2
c) |6m-3n| = 6-3 = 3
d)  |m-3n| = |1-3| = 2
A has the greatest value

*Answer can only contain numeric values
CAT Mock Test- 7 - Question 64

Let aa be the value of the common area bounded by the following 2 equations:

What is the value of 4a?


Detailed Solution for CAT Mock Test- 7 - Question 64

Since both the terms of x and  y have modulus, it will be symmetric in all 4 quadrants. We can find the value in the first quadrant and multiply t by 4 to get the desired result
Thus, y - x ≥ 5  and y + x ≤ 10

Area shaded in red shows the feasible region in quadrant 1. Its area = (1/2)5x2.5=6.25
Overall area = a = 4 x 6.25 = 25
4a = 4 x 25 = 100

CAT Mock Test- 7 - Question 65

Rajesh is 10 years younger to Baskar. 10 years back, Rajesh's age was two-thirds that of Baskar's. How old is Baskar now?

Detailed Solution for CAT Mock Test- 7 - Question 65

Let the present age of Baskar be 'b' and that of Rajesh be "r". So, r=b−10 ...(1)
10 years back Rajesh was ( r−10) years old. 10 years back Baskar was (b−10) years old.
The question states that 10 years back Rajesh was two thirds as old as Baskar was. i.e., (r−10)=(2/3)×(b−10) ...(2)
Cross multiplying, we get 3(r−10)=2(b−10) or 3r−30=2b−20 ..(2)
From eqn (1) we can substitute r as (b−10) in eqn ( 2) So, 3(b−10)−30=2b−20
or 3b−30−30=2b−20
or b=40
The present age of Baskar is 40 years.

CAT Mock Test- 7 - Question 66

At how many points do the following curves C1 and C2 intersect?

Detailed Solution for CAT Mock Test- 7 - Question 66

From curve C1 we can see that x ≥ 0 and y can be either positive or negative. Thus either intersection happens in the first quadrant or fourth quadrant.
x1/2 = |y| or x = y2 ...(I)) from C1
Putting this in the C2 equation
y2(y) = -4y2 + y
On rearranging we get y3 + 4y2 - y  = 0
y(y2 + 4y - 1) = 0
Discriminate of (y2 + 4y - 1) is positive so it must have 2 distinct roots ≠ 0
y(y2 + 4y - 1) = 0 has 3 distinct roots. 
For each value of y there will be unique x such that x1/2 = |Y|.  

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