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CAT Mock Test- 10 (November 12) - CAT MCQ


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CAT Mock Test- 10 (November 12) - Question 1

Read the passage and answer the following questions:
To better understand how the brain underlies selfhood, we need to understand its complex form; its intricate structure at the level of connections between neurons. After all, understanding biological structure has revealed the nature of many diverse life forms. Plants thrive because their typically broad leaves are perfect for transducing light energy into vital chemical energy. Similarly, eyes, whether human or insect, enable the transduction of light from one’s surroundings into electrical signals within the nervous system. These impulses carry information that represents features of the surrounding environment. But when it comes to the relationship between structure and function, brains have remained an enigma. There’s a lot more to them than to other organs that have specific functions, such as eyes, hearts or even hands. These organs can now be surgically replaced. Yet, even if a brain transplant were possible, you couldn’t just switch your brain with another person’s and maintain the same mind. Upon birth, a person’s brain structure is largely prescribed by experience in the womb and their unique genetic code. As we age, experience continues to imprint unique changes on the brain’s neural connectivity, increasing connections in some areas while decreasing them in others, accumulating reroutes upon reroutes as a person ages and learns, gaining knowledge and experience. Additionally, there are alterations in the strength of existing connections. These processes are especially evident in twins, whose brains are strikingly similar when born. However, as they grow, learn and experience the world, their brains diverge, and their essential selves become increasingly unique.
Although there are indeed anatomical regions that appear to serve relatively specific functions, one’s memory is not formed, stored or recalled within the activity of any single brain region. Certain structures, such as the amygdala and the hippocampus, play key roles but trying to find memory in one specific area is simply impossible. It would be like trying to listen to Beethoven’s Fifth but hearing only the strings. Instead, memory, in its broadest sense, lies in the uniqueness of a brain’s entire connective structure, known as the connectome. The connectome consists of its complete network of neurons and all the connections between them, called synapses. It is argued that, fundamentally, ‘you are your connectome’.
Mapping a connectome at the level of single neurons, however, is currently impossible in a living animal. Instead, animal brains must be extracted, perfused with a fixative such as formaldehyde and sliced up as many times as possible before being analysed structurally in order to painstakingly find individual neurons and trace their paths. To achieve this, the properties of each new slice are recorded using various microscopy techniques. Once that’s been done, patterns of electrical flow can be estimated from different neuron types and from connections that excite or inhibit other neurons. What’s crucial is that the extracted brain is preserved accurately enough to maintain its intricate, complex connectome before it’s sliced up.
Currently, it’s unlikely that any human brain has been preserved with its entire connectome perfectly intact as our brains degrade too quickly after death.

The author mentions the example of plants and eyes at the beginning to... 

Detailed Solution for CAT Mock Test- 10 (November 12) - Question 1

Let us pay heed to the introduction: "To better understand how the brain underlies selfhood, we need to understand its complex form; its intricate structure at the level of connections between neurons. After all, understanding biological structure has revealed the nature of many diverse life forms. Plants thrive because their typically broad leaves are perfect for transducing light energy into vital chemical energy. Similarly, eyes, whether human or insect, enable the transduction of light from one’s surroundings into electrical signals within the nervous system. These impulses carry information that represents features of the surrounding environment. But when it comes to the relationship between structure and function, brains have remained an enigma..."
The author begins with the claim that a better grasp of the underlying structure or the complex form of the brain might aid in understanding its role in selfhood. He furthers his claim by mentioning that this has been the case with many diverse life forms - gaining a deeper awareness of their biological structure enabled a better understanding of their function. The example of plants and the eyes have been mentioned in this regard: to supplement this assertion. However, at the same time, the author hints at how the brain does not completely adhere to this relation (of structure and function), and there are unknown elements present associated with the brain functioning (associated cognitive processes). Option C captures this correctly.
Option A: The author's focus is not on making the readers understand the structure of plants or the eyes; instead, the biological structure is tied up to a corresponding function Option A does not capture the author's intention and hence, can be eliminated. 
Option B: The statement here appears to be far fetched. The author doesn't assert that understanding the biological structure "always" reveals the nature of life forms; he specifies that this is the case in some situations and with certain entities. Thus, we can reject Option B.
Option D: The author clearly states that extending the same idea (as is the case with plants and the eyes) to the brain would be inappropriate due to the inherent complexity). Option D deviates from this idea and is, therefore, incorrect.
Hence, Option C is the correct answer. 

CAT Mock Test- 10 (November 12) - Question 2

Read the passage and answer the following questions:
To better understand how the brain underlies selfhood, we need to understand its complex form; its intricate structure at the level of connections between neurons. After all, understanding biological structure has revealed the nature of many diverse life forms. Plants thrive because their typically broad leaves are perfect for transducing light energy into vital chemical energy. Similarly, eyes, whether human or insect, enable the transduction of light from one’s surroundings into electrical signals within the nervous system. These impulses carry information that represents features of the surrounding environment. But when it comes to the relationship between structure and function, brains have remained an enigma. There’s a lot more to them than to other organs that have specific functions, such as eyes, hearts or even hands. These organs can now be surgically replaced. Yet, even if a brain transplant were possible, you couldn’t just switch your brain with another person’s and maintain the same mind. Upon birth, a person’s brain structure is largely prescribed by experience in the womb and their unique genetic code. As we age, experience continues to imprint unique changes on the brain’s neural connectivity, increasing connections in some areas while decreasing them in others, accumulating reroutes upon reroutes as a person ages and learns, gaining knowledge and experience. Additionally, there are alterations in the strength of existing connections. These processes are especially evident in twins, whose brains are strikingly similar when born. However, as they grow, learn and experience the world, their brains diverge, and their essential selves become increasingly unique.
Although there are indeed anatomical regions that appear to serve relatively specific functions, one’s memory is not formed, stored or recalled within the activity of any single brain region. Certain structures, such as the amygdala and the hippocampus, play key roles but trying to find memory in one specific area is simply impossible. It would be like trying to listen to Beethoven’s Fifth but hearing only the strings. Instead, memory, in its broadest sense, lies in the uniqueness of a brain’s entire connective structure, known as the connectome. The connectome consists of its complete network of neurons and all the connections between them, called synapses. It is argued that, fundamentally, ‘you are your connectome’.
Mapping a connectome at the level of single neurons, however, is currently impossible in a living animal. Instead, animal brains must be extracted, perfused with a fixative such as formaldehyde and sliced up as many times as possible before being analysed structurally in order to painstakingly find individual neurons and trace their paths. To achieve this, the properties of each new slice are recorded using various microscopy techniques. Once that’s been done, patterns of electrical flow can be estimated from different neuron types and from connections that excite or inhibit other neurons. What’s crucial is that the extracted brain is preserved accurately enough to maintain its intricate, complex connectome before it’s sliced up.
Currently, it’s unlikely that any human brain has been preserved with its entire connectome perfectly intact as our brains degrade too quickly after death.

Which of the following sentences is/are true as per the passage?

Detailed Solution for CAT Mock Test- 10 (November 12) - Question 2

"...There’s a lot more to them than to other organs that have specific functions, such as eyes, hearts or even hands. These organs can now be surgically replaced..." Although the author does render us with examples highlighting that this might be the case, there is more to the picture that needs to be considered. The author introduces an additional constraint: "...Yet, even if a brain transplant were possible, you couldn’t just switch your brain with another person’s and maintain the same mind...". Hence, we cannot definitively state that Option A is true (inadequate information). 
We can consider Option B to be true from the following lines- "These processes are especially evident in twins, whose brains are strikingly similar when born. However, as they grow, learn and experience the world, their brains diverge, and their essential selves become increasingly unique."
"...Currently, it’s unlikely that any human brain has been preserved with its entire connectome perfectly intact as our brains degrade too quickly after death..." Although the author makes this assertion, the future of the preservation of the human brain is not a subject that is touched upon. Hence, Option C is out of scope.
"...Mapping a connectome at the level of single neurons, however, is currently impossible in a living animal..." Although this comment is made, we can eliminate Option D  on the same grounds as Option C (out of scope).
Hence, Option B is the correct answer.

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CAT Mock Test- 10 (November 12) - Question 3

Read the passage carefully and answer the following questions:
We must dispense with the idea that democracy is like a torch that gets passed from one leading society to another. The core feature of democracy - that those who rule can do so only with the consent of the people - wasn’t invented in one place at one time: it evolved independently in a great many human societies. Over several millennia and across multiple continents, early democracy was an institution in which rulers governed jointly with councils and assemblies of the people. Classical Greece provided particularly important instances of this democratic practice, and it’s true that the Greeks gave us a language for thinking about democracy, including the word demokratia itself. But they didn’t invent the practice. The core feature of early democracy was that the people had power, even if multiparty elections didn’t happen. The people, or at least some significant fraction of them, exercised this power in many different ways. In some cases, a ruler was chosen by a council or assembly, and was limited to being first among equals. In other instances, a ruler inherited their position, but faced constraints to seek consent from the people before taking actions both large and small.
The first difference between early democracy and our democracies today is that this earlier form of rule was a small-scale phenomenon. In Classical Athens those who had the right to participate in politics tended to do so in a very direct and intensive way, particularly in local assemblies. In modern democracy, participation is very broad, but it’s also not deep; for most of us, it’s limited to voting in elections every few years, and in between these moments others make the decisions. The potential risk of this arrangement, as has been noted by astute observers since the birth of modern republics, is that citizens might grow distrustful of the people who are actually running government on a daily basis and of the special influences to which they might be subject. One way to address the problem of scale is to delegate much more power to states, provinces and localities. But on crucial issues of foreign trade, diplomacy or pressing constitutional questions, it’s impractical for individual states, regions or provinces to set their own policy. If large scale has the potential to lead to distrust and disengagement in a democracy, then a closely related problem is that of polarisation, which can take many forms, such as that involving tensions between different classes of people in the same location, or a difference of opinions between people living in different locations.
The absence of a state bureaucracy was a chief reason why early democracy proved to be such a stable form of rule for so many societies. With little autonomous power - apart from the ability to persuade - those who would have liked to rule as autocrats found themselves without the means to do so. The flipside of this was that, in many early democracies, those who were unhappy with a central decision could simply refuse to participate or even decamp to a new locality. Modern democracy lacks the same protections from central power that early democracies enjoyed. At the same time, having a powerful central state can allow a society to achieve goals such as universal education and prosperity, to name but a few. The question then is how to live with a state while preserving democracy.

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

Detailed Solution for CAT Mock Test- 10 (November 12) - Question 3

" In modern democracy, participation is very broad, but it’s also not deep; for most of us, it’s limited to voting in elections every few years, and in between these moments others make the decisions. The potential risk of this arrangement, as has been noted by astute observers since the birth of modern republics, is that citizens might grow distrustful of the people who are actually running government on a daily basis..."
From the above lines, it is clear that when people don't participate or engage in decision making, they begin to lose trust in the government. So, if people participate sporadically, as mentioned in option A, there will be little trust between the government and the public. Hence, the author will not agree with sentence A.
On the other hand, option D proposes transparency and openness. The author is pro-public participation as can be seen from the passage. Option D, if implemented, could foster trust and improve participation and thus would be something the author agrees with.
" At the same time, having a powerful central state can allow a society to achieve goals such as universal education and prosperity, to name but a few."
Even though the passage says that the presence of state bureaucracy allows the centre to be more powerful, it also mentions some of the important goals that can be achieved only with a powerful central state. So, option B is incorrect.
The author never implies something that is mentioned in option C as it is not feasible since modern democracies are large-scale and would disrupt the working of the government.

CAT Mock Test- 10 (November 12) - Question 4

Read the passage carefully and answer the following questions:
We must dispense with the idea that democracy is like a torch that gets passed from one leading society to another. The core feature of democracy - that those who rule can do so only with the consent of the people - wasn’t invented in one place at one time: it evolved independently in a great many human societies. Over several millennia and across multiple continents, early democracy was an institution in which rulers governed jointly with councils and assemblies of the people. Classical Greece provided particularly important instances of this democratic practice, and it’s true that the Greeks gave us a language for thinking about democracy, including the word demokratia itself. But they didn’t invent the practice. The core feature of early democracy was that the people had power, even if multiparty elections didn’t happen. The people, or at least some significant fraction of them, exercised this power in many different ways. In some cases, a ruler was chosen by a council or assembly, and was limited to being first among equals. In other instances, a ruler inherited their position, but faced constraints to seek consent from the people before taking actions both large and small.
The first difference between early democracy and our democracies today is that this earlier form of rule was a small-scale phenomenon. In Classical Athens those who had the right to participate in politics tended to do so in a very direct and intensive way, particularly in local assemblies. In modern democracy, participation is very broad, but it’s also not deep; for most of us, it’s limited to voting in elections every few years, and in between these moments others make the decisions. The potential risk of this arrangement, as has been noted by astute observers since the birth of modern republics, is that citizens might grow distrustful of the people who are actually running government on a daily basis and of the special influences to which they might be subject. One way to address the problem of scale is to delegate much more power to states, provinces and localities. But on crucial issues of foreign trade, diplomacy or pressing constitutional questions, it’s impractical for individual states, regions or provinces to set their own policy. If large scale has the potential to lead to distrust and disengagement in a democracy, then a closely related problem is that of polarisation, which can take many forms, such as that involving tensions between different classes of people in the same location, or a difference of opinions between people living in different locations.
The absence of a state bureaucracy was a chief reason why early democracy proved to be such a stable form of rule for so many societies. With little autonomous power - apart from the ability to persuade - those who would have liked to rule as autocrats found themselves without the means to do so. The flipside of this was that, in many early democracies, those who were unhappy with a central decision could simply refuse to participate or even decamp to a new locality. Modern democracy lacks the same protections from central power that early democracies enjoyed. At the same time, having a powerful central state can allow a society to achieve goals such as universal education and prosperity, to name but a few. The question then is how to live with a state while preserving democracy.

Which of the following cannot be inferred from the passage?
I. Classical Greece did not just invent the practice of democracy, but also gave us the word 'demokratia'.
II. The advantage of early democracies over modern ones is the very direct and intensive participation of the people in local assemblies.
III. The presence or absence of state bureaucracy is a vital factor affecting the stability of a democratic rule.

Detailed Solution for CAT Mock Test- 10 (November 12) - Question 4

Statement I: In the introductory paragraph of the passage, it is mentioned that "it’s true that the Greeks gave us a language for thinking about democracy, including the word demokratia itself. But they didn’t invent the practice." So, statement I cannot be inferred.
Statement II: The passage does differentiate between early and modern democracies on the basis of depth and extent of participation of the people in the process of decision making but does not tell us that this makes one advantageous over the other. Statement II cannot be inferred.
Statement III: "The absence of a state bureaucracy was a chief reason why early democracy proved to be such a stable form of rule for so many societies. With little autonomous power - apart from the ability to persuade - those who would have liked to rule as autocrats found themselves without the means to do so." makes statement III inferrable as it states that it is a 'chief' reason and there was no scope for an autocratic rule.
Hence, Statements I and II
cannot be inferred from the passage. Option A is the correct answer. 

CAT Mock Test- 10 (November 12) - Question 5

Read the passage carefully and answer the following questions:
Green energy investment is hot again in the U.S. To some, the new boom will raise the specter of the clean-tech bust that followed a streak of exuberance a decade ago. But there are reasons to believe that this time the trend is no bubble or mirage.
The most basic reason is that the fundamental underlying technology has matured in a way it simply hadn’t a decade ago. In 2009, the levelized cost of solar photovoltaic electricity was $359 per megawatt-hour — more than four times as expensive as electricity from a natural gas plant. By 2019, solar PV had fallen in price to $40 per megawatt-hour, 28% cheaper than gas. That’s an 89% decline in 10 years, with more cost drops yet to come. Meanwhile, lithium-ion batteries have experienced a similar drop in prices.
That order-of-magnitude drop in costs makes all the difference. First of all, it means that solar and wind aren’t risky new technologies. Solyndra, a solar manufacturer, failed in 2011 because it was trying to market an innovative new kind of solar cell, which ended up being too expensive when the tried-and-true design came down in cost. Future investments in solar won’t have to bet on any difficult technological breakthroughs. Batteries might be a different story — lots of money is being thrown at startups trying to create solid-state batteries, which would be a true breakthrough. But Tesla Inc. is doing just fine with the old kind, so that sector is probably going to do OK as well. Venture investing does well when it doesn’t have to bet on “hard tech”, and much of clean tech is no longer hard.
Second, cost drops in clean energy mean that success doesn’t depend on government intervention. In the earlier boom, fickle government subsidies were often necessary for capital-intensive energy companies to succeed. Now, even though President Joe Biden is planning a big push into clean-energy investment, the market is investing quite a lot in renewables all on its own.
Finally, investors have probably learned their lesson. Clean energy itself was never a good fit for venture. It’s capital intensive, since buying solar panels and wind turbines entails a lot of money up front; venture capital tends to focus on cheap, small investments that scale. And instead of companies creating highly differentiated products and new markets, as in software, clean electricity companies are basically all trying to provide the same commodified product.
This time around, venture capitalists are letting bigger investors handle the build out of solar and wind, and finding other niches where low-cost, differentiated startups can add value — such as solar services and financing, lab-grown meat and electric vehicles. Some of those bets are certainly going to fail, but that’s always the case in private equity. The success of Tesla — now with a market cap of almost $700 billion, or 28 times the amount that was lost in the clean-tech bust — demonstrates the time-honored principle that a few big hits can compensate for a lot of little failures.
In other words, clean tech is entering the final stage of the famous Gartner Hype Cycle — a pattern that describes the progression of emerging technologies and business models, starting with an innovation that sees expectations climb and then crash, before they finally rise again to sustained productivity. The clean-tech bust, like the dot-com bust in 2000, was a case of investor enthusiasm for a new technology outstripping the technology itself. But just as few today would question the value of companies like Google and Facebook that came into their own during a trough in investor enthusiasm,  eventually the value of clean technology won’t be in question.

The central idea of the passage is that

Detailed Solution for CAT Mock Test- 10 (November 12) - Question 5

In the passage, the author opines that the clean-tech industry is witnessing a surge in investments, and unlike the last time, which ended in a clean-tech bust, this time, it may well be the start of a sustainable phase for the industry. He also provides several reasons to justify his stance. Additionally, he believes that the investors have learned their lessons from the clean-tech bust and are ready to make mature investment decisions.
Option B conveys the above inference and is the answer. The other options are either extreme or tangential to the discussion.
Option A is true, but it is not the central idea of the passage.
Option C fails to capture all the relevant points. In the fifth paragraph, the author states "Clean energy itself was never a good fit for venture. It’s capital intensive, since buying solar panels and wind turbines entails a lot of money up front; venture capital tends to focus on cheap, small investments that scale." Hence, clean energy was never a good fit for venture capitalists, according to the author. The dearth of differentiated products cannot be quoted as the sole reason for this hypothesis. The capital-intensive nature of the industry is equally important, if not more. Hence, option C can be eliminated.
Option D can be eliminated as well. It is tangential to the discussion. Furthermore, in the second paragraph, the author mentions that the clean-tech industry has evolved, which is a major reason why there might not be another clean-tech bust. So, it is not possible to say that investor overenthusiasm alone leads to bubbles and economic crises.

CAT Mock Test- 10 (November 12) - Question 6

The passage given below is followed by four summaries. Choose the option that best captures the author’s position.
Language bias in academia is something that researchers have been facing for a long time. Not only are the vast majority of scientific papers published in English, the ‘correctness’ of English used in them is a factor that determines their acceptance into top journals—which, incidentally, also publish exclusively in English. It discourages non-English speaking researchers as it prioritizes the purity of the language over the content of their research. This also puts them at a disadvantage professionally, as they’re robbed of the most common, popular platforms that can further their research and careers. In addition, the overwhelming reliance on, and preference of, English favours only research that looks at the world in specific predisposed ways, brought on by the use of English, such as the tendency to prescribe indigenous knowledge as ‘folklore’ and not something that could have factual validity. It discounts anything that digresses from the norm, even when the information might be highly relevant and important.

Detailed Solution for CAT Mock Test- 10 (November 12) - Question 6

In the passage, the author discusses how language bias in academia could deny non-English speaking researchers the opportunity to get published in top journals and subsequently earn recognition and progress in their careers. In addition to this professional disadvantage, established predispositions could also lead to the repudiation of certain research information that is factually correct.  Comparing the options, option B comes closest to conveying this inference.
Option A is extreme. The passage talks about professional disadvantage and disillusionment. But, discrimination is too strong a word in this context. Furthermore, it has not been implied that the bias leads to the researchers abandoning their works.
Option C comes close but talks about a drop in research quality, which has not been implied.
Option D, too, is correct for the most part but talks about the publication of sub-standard articles, which the author does not mention anywhere in the passage.
Hence, option B is the answer .

*Answer can only contain numeric values
CAT Mock Test- 10 (November 12) - Question 7

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. Gaura Devi led 27 of the village women to the site and confronted the loggers.
2. On 26 March 2018, a Chipko movement conservation initiative was marked by Google Doodle on its 45th anniversary.
3. On 25 March 1974, the day the lumbermen were to cut the trees, the men of Reni village and DGSS workers were in Chamoli, diverted by the state government and contractors to a fictional compensation payment site, while back home labourers arrived by the truckload to start logging operations.
4. When all talking failed, and the loggers started to shout and abuse the women, threatening them with guns, the women resorted to hugging the trees to stop them from being felled.
5. A local girl, on seeing them, rushed to inform Gaura Devi, the head of the village Mahila Mangal Dal, at Reni village.


Detailed Solution for CAT Mock Test- 10 (November 12) - Question 7

Statement 2 is the odd one because the incident narrated by the other sentences need not necessarily refer to the Chipko Movement.
Further, when the other sentences are arranged in sequence 3514, we get a coherent para.
Clearly, 3 will be the starting sentence as it starts by narrating the incident and will be followed by 5 because 'them' in 5 refers to the 'labourers' in 3. Sentence 1 will follow 5 as 5 tells us about a girl informing Gaura Devi, and 1 tells us how she reacted to the information. Finally, 4 follows 1 as it elaborates on the confrontation. Hence, the sequence 3514.

*Answer can only contain numeric values
CAT Mock Test- 10 (November 12) - Question 8

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, such unbridled advancements have also escalated global issues like climate change, resource depletion, and significant socio-economic disparities.

2. The 21st century has been heralded as an era of technological marvels, transforming every facet of human life with unprecedented innovations.

3. These challenges necessitate a reconfiguration of global priorities towards sustainability and equitable progress, demanding collaborative, not isolated, innovation.

4. Consequently, the narrative of development has increasingly become a balancing act between embracing technological growth and acknowledging its consequential perils.


Detailed Solution for CAT Mock Test- 10 (November 12) - Question 8

The sequence starts with the marvels of the current era (2), introduces the downside of these advancements (1), reflects on the resulting global narrative (4), and concludes with a call for a new direction (3).

*Answer can only contain numeric values
CAT Mock Test- 10 (November 12) - Question 9

A lock is guarded by a security code that serves as the password. The code is in the following form:


Where a, b, c, d and g are numbers and e, f are alphabets. The following information is also known about the respective characters:
1. 'a' is a prime number from 1 to 10.
2. The 2-digit number 'bc' is also a prime number such that b = a + 1.
3. 'd' = 'bc' % 'a', that is, d is the remainder when the 2-digit number 'bc' is divided by 'a'.
4. 'e' is the alphabet at the index of n, alphabet[n], where n is the highest number of occurrences of any digit(0 to 9) among a,b,c,d. Here,alphabet[1] = A, alphabet[2] = B,..., alphabet[26] = Z.
5. 'f' is the alphabet either preceding or succeeding the alphabet 'e' when all English alphabets are arranged alphabetically.
6. 'g' is the remainder when the sum of digits a, b, c and d is divided by 10. g = (a+b+c+d)%10.
Based on the information given above, answer the questions that follow.

If we consider all possible codes, which digit(0 to 9) does not occur even once in any of the codes? Enter -1 if all 10 digits appear at least once in any of the codes.


Detailed Solution for CAT Mock Test- 10 (November 12) - Question 9

a can be 2, 3, 5, 7.
Hence, when a = 2, b = 3, bc = 31, 37
When a = 3, b = 4, bc = 41, 43, 47
When a = 5, b = 6, bc = 61, 67
When a = 7, b = 8, bc = 83, 89.
Hence, the different possibilities are,
2311 _ _ _
2371 _ _ _
3412 _ _ _
3431 _ _ _
3472 _ _ _
5611 _ _ _
5672 _ _ _
7836 _ _ _
7895 _ _ _
Now, when it is 2311 _ _ _, 1 occurs twice, which is the highest, hence e = alphabet[2] = B.
Similarly, when it is 2371 _ _ _, the highest occurrence of any digit is once, hence e = alphabet[1] = A.
Similarly, we get the remaining as well,
2311B _ _
2371A _ _
3412A _ _
3431B _ _
3472A _ _
5611B _ _
5672A _ _
7836A _ _
7895A _ _
Hence, e is always either B or A. If e is B. f can be either A or C, if e is A, f can only be B. Hence, we get the following codes.
2311BA _
2311BC _
2371AB _
3412AB _
3431BA _
3431BC _
3472AB _
5611BA _
5611BC _
5672AB _
7836AB _
7895AB _
Now, G has to be calculated. For, 2311BA _, g = (2 + 3 + 1 + 1)%10 = 7
Similarly, we calculate for others as well,
2311BA7
2311BC7
2371AB3
3412AB0
3431BA1
3431BC1
3472AB6
5611BA3
5611BC3
5672AB0
7836AB4
7895AB9
As we can see every digit appears at least once if all codes are considered.
Hence, the answer is -1 as asked in the question.

*Answer can only contain numeric values
CAT Mock Test- 10 (November 12) - Question 10

A lock is guarded by a security code that serves as the password. The code is in the following form:


Where a, b, c, d and g are numbers and e, f are alphabets. The following information is also known about the respective characters:
1. 'a' is a prime number from 1 to 10.
2. The 2-digit number 'bc' is also a prime number such that b = a + 1.
3. 'd' = 'bc' % 'a', that is, d is the remainder when the 2-digit number 'bc' is divided by 'a'.
4. 'e' is the alphabet at the index of n, alphabet[n], where n is the highest number of occurrences of any digit(0 to 9) among a,b,c,d. Here,alphabet[1] = A, alphabet[2] = B,..., alphabet[26] = Z.
5. 'f' is the alphabet either preceding or succeeding the alphabet 'e' when all English alphabets are arranged alphabetically.
6. 'g' is the remainder when the sum of digits a, b, c and d is divided by 10. g = (a+b+c+d)%10.
Based on the information given above, answer the questions that follow.

If we consider all possible codes, how many distinct alphabets appear at least once in any of the codes?


Detailed Solution for CAT Mock Test- 10 (November 12) - Question 10

a can be 2, 3, 5, 7.
Hence, when a = 2, b = 3, bc = 31, 37
When a = 3, b = 4, bc = 41, 43, 47
When a = 5, b = 6, bc = 61, 67
When a = 7, b = 8, bc = 83, 89.
Hence, the different possibilities are,
2311 _ _ _
2371 _ _ _
3412 _ _ _
3431 _ _ _
3472 _ _ _
5611 _ _ _
5672 _ _ _
7836 _ _ _
7895 _ _ _
Now, when it is 2311 _ _ _, 1 occurs twice, which is the highest, hence e = alphabet[2] = B.
Similarly, when it is 2371 _ _ _, the highest occurrence of any digit is once, hence e = alphabet[1] = A.
Similarly, we get the remaining as well,
2311B _ _
2371A _ _
3412A _ _
3431B _ _
3472A _ _
5611B _ _
5672A _ _
7836A _ _
7895A _ _
Hence, e is always either B or A. If e is B. f can be either A or C, if e is A, f can only be B. Hence, we get the following codes.
2311BA _
2311BC _
2371AB _
3412AB _
3431BA _
3431BC _
3472AB _
5611BA _
5611BC _
5672AB _
7836AB _
7895AB _
Now, G has to be calculated. For, 2311BA _, g = (2 + 3 + 1 + 1)%10 = 7
Similarly, we calculate for others as well,
2311BA7
2311BC7
2371AB3
3412AB0
3431BA1
3431BC1
3472AB6
5611BA3
5611BC3
5672AB0
7836AB4
7895AB9
As we can see only A, B and C appear. So, there are 3 alphabets.

CAT Mock Test- 10 (November 12) - Question 11

A lock is guarded by a security code that serves as the password. The code is in the following form:


Where a, b, c, d and g are numbers and e, f are alphabets. The following information is also known about the respective characters:
1. 'a' is a prime number from 1 to 10.
2. The 2-digit number 'bc' is also a prime number such that b = a + 1.
3. 'd' = 'bc' % 'a', that is, d is the remainder when the 2-digit number 'bc' is divided by 'a'.
4. 'e' is the alphabet at the index of n, alphabet[n], where n is the highest number of occurrences of any digit(0 to 9) among a,b,c,d. Here,alphabet[1] = A, alphabet[2] = B,..., alphabet[26] = Z.
5. 'f' is the alphabet either preceding or succeeding the alphabet 'e' when all English alphabets are arranged alphabetically.
6. 'g' is the remainder when the sum of digits a, b, c and d is divided by 10. g = (a+b+c+d)%10.
Based on the information given above, answer the questions that follow.

How many codes are possible?

Detailed Solution for CAT Mock Test- 10 (November 12) - Question 11

a can be 2, 3, 5, 7.
Hence, when a = 2, b = 3, bc = 31, 37
When a = 3, b = 4, bc = 41, 43, 47
When a = 5, b = 6, bc = 61, 67
When a = 7, b = 8, bc = 83, 89.
Hence, the different possibilities are,
2311 _ _ _
2371 _ _ _
3412 _ _ _
3431 _ _ _
3472 _ _ _
5611 _ _ _
5672 _ _ _
7836 _ _ _
7895 _ _ _
Now, when it is 2311 _ _ _, 1 occurs twice, which is the highest, hence e = alphabet[2] = B.
Similarly, when it is 2371 _ _ _, the highest occurrence of any digit is once, hence e = alphabet[1] = A.
Similarly, we get the remaining as well,
2311B _ _
2371A _ _
3412A _ _
3431B _ _
3472A _ _
5611B _ _
5672A _ _
7836A _ _
7895A _ _
Hence, e is always either B or A. If e is B. f can be either A or C, if e is A, f can only be B. Hence, we get the following codes.
2311BA _
2311BC _
2371AB _
3412AB _
3431BA _
3431BC _
3472AB _
5611BA _
5611BC _
5672AB _
7836AB _
7895AB _
Now, G has to be calculated. For, 2311BA _, g = (2 + 3 + 1 + 1)%10 = 7
Similarly, we calculate for others as well,
2311BA7
2311BC7
2371AB3
3412AB0
3431BA1
3431BC1
3472AB6
5611BA3
5611BC3
5672AB0
7836AB4
7895AB9
As we can see there are 12 different valid codes.

CAT Mock Test- 10 (November 12) - Question 12

60 students were shortlisted for their Personal Interview by the Indian School of Management. But the interview process was to be conducted online and hence students were sent the KOOM meeting link for the same. All the 60 students were to join the KOOM common room at 9:00 AM sharp and then the admission co-ordinators would send in a certain number of students to different waiting rooms- X, Y and Z prior to their one on one interviews with the panellists. There were 3 panellists- A, B and C who were conducting interviews for the candidates waiting in the rooms X, Y and Z respectively. The capacity for the waiting rooms X, Y and Z were 5, 10 and 15 respectively. 
The 60 students were given a rank from 1 to 60 based on their composite score and it is known that no two students got the same score. To start off the interviews, the ones with top 5 ranks are sent to panellist A. The 10 below them are sent to B and students ranked 16 to 30 were sent to C. When the last person from the waiting room X, Y and Z gives an interview, the next set of students are immediately sent to the waiting room as soon as the interview gets over and the priority is maintained as students with higher ranks are sent to X, Y and Z respectively if the interviews of the previous batch end simultaneously. 
It is known that:
1. Panellist A takes an interview for 15 minutes and takes a break of 5 minutes only after interviewing 3 candidates 
2. Panellist B takes an interview for 10 minutes and takes a break of 5 minutes after interviewing 5 students.
3. Panellist C takes an interview of 5 minutes and takes a break of 5 minutes after interviewing 4 students.
4. Lunch break is scheduled between 12:00 noon to 12:30 PM and students are not cut abruptly if their interview began before 12:00 noon, i.e. the interviewers take the break only after finishing the interview if it started before 12:00 noon.
5. Batches of 5, 10 and 15 students enter the waiting room together such that students who enter the waiting room at a time have continuous ranks. For example, if the interviewer X is ending up the interview of the last student from the previous batch, 5 students with continuous ranks (say 21-25) enter the waiting room X together.
6. Students allotted to a particular waiting room is only interviewed by the designated interviewer and is not interviewed by anyone else even if the other interviewer is free.
7. If panellist A goes into the lunch break after having interviewed 1 student, he will again take his 5 minute break after interviewing 2 students post the lunch break so as to satisfy the condition 1 always. The same applies for panellist B and C who follow conditions 2 and 3 respectively.

When does interviewer B finish his last interview?

Detailed Solution for CAT Mock Test- 10 (November 12) - Question 12

Students ranked from 1 to 5 will be interviewed by A and thus are moved to the waiting room X.
Similarly, students ranked 6 to 15 (10 students) will be interviewed by B and are moved to waiting room Y.
And students ranked 16 to 30 (15 students) will be interviewed by C and are moved to waiting room Z.
A takes a break after every 3 interviews. B and C take breaks after every 5 and 4 interviews respectively.
So, a map of the first few students will look like:

Here, M, N, X and Y are the ranks of the students. M to M+4 are the set of 5 students who will be interviewed once the interview of the student ranked 5th gets over and similarly for others.
We see that interviewer A finishes his first batch of interviews first (at 10:20) and hence M must be 31. Students ranked 31 to 35 move to waiting room X.
Next to finish interviewing his/her first batch of interviewees is B (at 10:30) and hence Y=36. Students ranked from 36 to 50 will move into waiting room Z.
Finally, interviewer Y will get the next 10 students ranked from 51 to 60 and hence X=51.
We do not require N.
Interviewer A interviews students ranked from 1 to 5, 31 to 35.
Interviewer B interviews students ranked from 6 to 15 and 51 to 60.
Interviewer C interviews students ranked from 16 to 30 and 36 to 50.

An updated table will look like:

We can break down the interviews of students ranked 41 to 50 for interviewer C as:
Each set of 4 consecutively ranked students will be interviewed in 20 minutes. The fifth student will be interviewed after 20+5(break) minutes= 25 minutes.
So, the interview of student ranked 41 starts at 11:05 am and that of student ranked 45 starts at 11:05+25= 11:30 am. Similarly the interview of student ranked 49 starts at 11:55 am and so on.
We can see from the table that interviewer B ends his interviews at 1:00 pm.

CAT Mock Test- 10 (November 12) - Question 13

60 students were shortlisted for their Personal Interview by the Indian School of Management. But the interview process was to be conducted online and hence students were sent the KOOM meeting link for the same. All the 60 students were to join the KOOM common room at 9:00 AM sharp and then the admission co-ordinators would send in a certain number of students to different waiting rooms- X, Y and Z prior to their one on one interviews with the panellists. There were 3 panellists- A, B and C who were conducting interviews for the candidates waiting in the rooms X, Y and Z respectively. The capacity for the waiting rooms X, Y and Z were 5, 10 and 15 respectively.
The 60 students were given a rank from 1 to 60 based on their composite score and it is known that no two students got the same score. To start off the interviews, the ones with top 5 ranks are sent to panellist A. The 10 below them are sent to B and students ranked 16 to 30 were sent to C. When the last person from the waiting room X, Y and Z gives an interview, the next set of students are immediately sent to the waiting room as soon as the interview gets over and the priority is maintained as students with higher ranks are sent to X, Y and Z respectively if the interviews of the previous batch end simultaneously. 
It is known that:
1. Panellist A takes an interview for 15 minutes and takes a break of 5 minutes only after interviewing 3 candidates 
2. Panellist B takes an interview for 10 minutes and takes a break of 5 minutes after interviewing 5 students.
3. Panellist C takes an interview of 5 minutes and takes a break of 5 minutes after interviewing 4 students.
4. Lunch break is scheduled between 12:00 noon to 12:30 PM and students are not cut abruptly if their interview began before 12:00 noon, i.e. the interviewers take the break only after finishing the interview if it started before 12:00 noon.
5. Batches of 5, 10 and 15 students enter the waiting room together such that students who enter the waiting room at a time have continuous ranks. For example, if the interviewer X is ending up the interview of the last student from the previous batch, 5 students with continuous ranks (say 21-25) enter the waiting room X together.
6. Students allotted to a particular waiting room is only interviewed by the designated interviewer and is not interviewed by anyone else even if the other interviewer is free.
7. If panellist A goes into the lunch break after having interviewed 1 student, he will again take his 5 minute break after interviewing 2 students post the lunch break so as to satisfy the condition 1 always. The same applies for panellist B and C who follow conditions 2 and 3 respectively.

When was the interview of the student ranked 53 finished?

Detailed Solution for CAT Mock Test- 10 (November 12) - Question 13

Students ranked from 1 to 5 will be interviewed by A and thus are moved to the waiting room X.
Similarly, students ranked 6 to 15 (10 students) will be interviewed by B and are moved to waiting room Y.
And students ranked 16 to 30 (15 students) will be interviewed by C and are moved to waiting room Z.
A takes a break after every 3 interviews. B and C take breaks after every 5 and 4 interviews respectively.
So, a map of the first few students will look like:

Here, M, N, X and Y are the ranks of the students. M to M+4 are the set of 5 students who will be interviewed once the interview of the student ranked 5th gets over and similarly for others.
We see that interviewer A finishes his first batch of interviews first (at 10:20) and hence M must be 31. Students ranked 31 to 35 move to waiting room X.
Next to finish interviewing his/her first batch of interviewees is B (at 10:30) and hence Y=36. Students ranked from 36 to 50 will move into waiting room Z.
Finally, interviewer Y will get the next 10 students ranked from 51 to 60 and hence X=51.
We do not require N.
Interviewer A interviews students ranked from 1 to 5, 31 to 35.
Interviewer B interviews students ranked from 6 to 15 and 51 to 60.
Interviewer C interviews students ranked from 16 to 30 and 36 to 50.
An updated table will look like:

We can break down the interviews of students ranked 41 to 50 for interviewer C as:
Each set of 4 consecutively ranked students will be interviewed in 20 minutes. The fifth student will be interviewed after 20+5(break) minutes= 25 minutes.
So, the interview of student ranked 41 starts at 11:05 am and that of student ranked 45 starts at 11:05+25= 11:30 am. Similarly the interview of student ranked 49 starts at 11:55 am and so on.
Student ranked 53 was interviewed by B. We can see from the table that a set of 5 interviews get their interviews done in 50 minutes (including the 5 min break).
Students with rank 51 finish his interview at 11:00 AM.
Student with rank 52 finish at 11:10 AM and the one ranked 53 finish at 11:20 AM

CAT Mock Test- 10 (November 12) - Question 14

60 students were shortlisted for their Personal Interview by the Indian School of Management. But the interview process was to be conducted online and hence students were sent the KOOM meeting link for the same. All the 60 students were to join the KOOM common room at 9:00 AM sharp and then the admission co-ordinators would send in a certain number of students to different waiting rooms- X, Y and Z prior to their one on one interviews with the panellists. There were 3 panellists- A, B and C who were conducting interviews for the candidates waiting in the rooms X, Y and Z respectively. The capacity for the waiting rooms X, Y and Z were 5, 10 and 15 respectively.
The 60 students were given a rank from 1 to 60 based on their composite score and it is known that no two students got the same score. To start off the interviews, the ones with top 5 ranks are sent to panellist A. The 10 below them are sent to B and students ranked 16 to 30 were sent to C. When the last person from the waiting room X, Y and Z gives an interview, the next set of students are immediately sent to the waiting room as soon as the interview gets over and the priority is maintained as students with higher ranks are sent to X, Y and Z respectively if the interviews of the previous batch end simultaneously.
It is known that:
1. Panellist A takes an interview for 15 minutes and takes a break of 5 minutes only after interviewing 3 candidates 
2. Panellist B takes an interview for 10 minutes and takes a break of 5 minutes after interviewing 5 students.
3. Panellist C takes an interview of 5 minutes and takes a break of 5 minutes after interviewing 4 students.
4. Lunch break is scheduled between 12:00 noon to 12:30 PM and students are not cut abruptly if their interview began before 12:00 noon, i.e. the interviewers take the break only after finishing the interview if it started before 12:00 noon.
5. Batches of 5, 10 and 15 students enter the waiting room together such that students who enter the waiting room at a time have continuous ranks. For example, if the interviewer X is ending up the interview of the last student from the previous batch, 5 students with continuous ranks (say 21-25) enter the waiting room X together.
6. Students allotted to a particular waiting room is only interviewed by the designated interviewer and is not interviewed by anyone else even if the other interviewer is free.
7. If panellist A goes into the lunch break after having interviewed 1 student, he will again take his 5 minute break after interviewing 2 students post the lunch break so as to satisfy the condition 1 always. The same applies for panellist B and C who follow conditions 2 and 3 respectively.

When did the 50th ranked student finish his/her interview?

Detailed Solution for CAT Mock Test- 10 (November 12) - Question 14

Students ranked from 1 to 5 will be interviewed by A and thus are moved to the waiting room X.
Similarly, students ranked 6 to 15 (10 students) will be interviewed by B and are moved to waiting room Y.
And students ranked 16 to 30 (15 students) will be interviewed by C and are moved to waiting room Z.
A takes a break after every 3 interviews. B and C take breaks after every 5 and 4 interviews respectively.
So, a map of the first few students will look like:

Here, M, N, X and Y are the ranks of the students. M to M+4 are the set of 5 students who will be interviewed once the interview of the student ranked 5th gets over and similarly for others.
We see that interviewer A finishes his first batch of interviews first (at 10:20) and hence M must be 31. Students ranked 31 to 35 move to waiting room X.
Next to finish interviewing his/her first batch of interviewees is B (at 10:30) and hence Y=36. Students ranked from 36 to 50 will move into waiting room Z.
Finally, interviewer Y will get the next 10 students ranked from 51 to 60 and hence X=51.
We do not require N.
Interviewer A interviews students ranked from 1 to 5, 31 to 35.
Interviewer B interviews students ranked from 6 to 15 and 51 to 60.
Interviewer C interviews students ranked from 16 to 30 and 36 to 50.
An updated table will look like:

We can break down the interviews of students ranked 41 to 50 for interviewer C as:
Each set of 4 consecutively ranked students will be interviewed in 20 minutes. The fifth student will be interviewed after 20+5(break) minutes= 25 minutes.
So, the interview of student ranked 41 starts at 11:05 am and that of student ranked 45 starts at 11:05+25= 11:30 am. Similarly the interview of student ranked 49 starts at 11:55 am and so on.
Student ranked 46 was interviewed by C. We can see from the table that a set of 4 interviews get their interviews done in 25 minutes (including the 5 min break).
Students with rank 41 to 44 get their interviews done between 11:05 and 11:25.
Students with rank 45 to 48 get their interviews done between 11:30 and 11:50.
The student ranked 49 is the last one to finish his interview before lunch at exactly 12:00 noon.
So, the student with a rank of 50 starts his interview at 12:30, finishes at 12:35.

CAT Mock Test- 10 (November 12) - Question 15

60 students were shortlisted for their Personal Interview by the Indian School of Management. But the interview process was to be conducted online and hence students were sent the KOOM meeting link for the same. All the 60 students were to join the KOOM common room at 9:00 AM sharp and then the admission co-ordinators would send in a certain number of students to different waiting rooms- X, Y and Z prior to their one on one interviews with the panellists. There were 3 panellists- A, B and C who were conducting interviews for the candidates waiting in the rooms X, Y and Z respectively. The capacity for the waiting rooms X, Y and Z were 5, 10 and 15 respectively.
The 60 students were given a rank from 1 to 60 based on their composite score and it is known that no two students got the same score. To start off the interviews, the ones with top 5 ranks are sent to panellist A. The 10 below them are sent to B and students ranked 16 to 30 were sent to C. When the last person from the waiting room X, Y and Z gives an interview, the next set of students are immediately sent to the waiting room as soon as the interview gets over and the priority is maintained as students with higher ranks are sent to X, Y and Z respectively if the interviews of the previous batch end simultaneously.
It is known that:
1. Panellist A takes an interview for 15 minutes and takes a break of 5 minutes only after interviewing 3 candidates 
2. Panellist B takes an interview for 10 minutes and takes a break of 5 minutes after interviewing 5 students.
3. Panellist C takes an interview of 5 minutes and takes a break of 5 minutes after interviewing 4 students.
4. Lunch break is scheduled between 12:00 noon to 12:30 PM and students are not cut abruptly if their interview began before 12:00 noon, i.e. the interviewers take the break only after finishing the interview if it started before 12:00 noon.
5. Batches of 5, 10 and 15 students enter the waiting room together such that students who enter the waiting room at a time have continuous ranks. For example, if the interviewer X is ending up the interview of the last student from the previous batch, 5 students with continuous ranks (say 21-25) enter the waiting room X together.
6. Students allotted to a particular waiting room is only interviewed by the designated interviewer and is not interviewed by anyone else even if the other interviewer is free.
7. If panellist A goes into the lunch break after having interviewed 1 student, he will again take his 5 minute break after interviewing 2 students post the lunch break so as to satisfy the condition 1 always. The same applies for panellist B and C who follow conditions 2 and 3 respectively.

At what time does Interviewer A finish all his interviews?

Detailed Solution for CAT Mock Test- 10 (November 12) - Question 15

Students ranked from 1 to 5 will be interviewed by A and thus are moved to the waiting room X.
Similarly, students ranked 6 to 15 (10 students) will be interviewed by B and are moved to waiting room Y.
And students ranked 16 to 30 (15 students) will be interviewed by C and are moved to waiting room Z.
A takes a break after every 3 interviews. B and C take breaks after every 5 and 4 interviews respectively.
So, a map of the first few students will look like:

Here, M, N, X and Y are the ranks of the students. M to M+4 are the set of 5 students who will be interviewed once the interview of the student ranked 5th gets over and similarly for others.
We see that interviewer A finishes his first batch of interviews first (at 10:20) and hence M must be 31. Students ranked 31 to 35 move to waiting room X.
Next to finish interviewing his/her first batch of interviewees is B (at 10:30) and hence Y=36. Students ranked from 36 to 50 will move into waiting room Z.
Finally, interviewer Y will get the next 10 students ranked from 51 to 60 and hence X=51.
We do not require N.
Interviewer A interviews students ranked from 1 to 5, 31 to 35.
Interviewer B interviews students ranked from 6 to 15 and 51 to 60.
Interviewer C interviews students ranked from 16 to 30 and 36 to 50.

An updated table will look like:

We can break down the interviews of students ranked 41 to 50 for interviewer C as:
Each set of 4 consecutively ranked students will be interviewed in 20 minutes. The fifth student will be interviewed after 20+5(break) minutes= 25 minutes.
So, the interview of student ranked 41 starts at 11:05 am and that of student ranked 45 starts at 11:05+25= 11:30 am. Similarly the interview of student ranked 49 starts at 11:55 am and so on.
Interviewer A finished his interviews at 11:45 AM

CAT Mock Test- 10 (November 12) - Question 16

5 persons- P1, P2, P3, P4 and P5 took part in a unique high jump competition where each one was given 10 chances to maximise their points. Each one of them started with a height of 150 cm to clear. If a particular participant cleared the height, in the next jump he was to clear a height of 5 cm more than the previous one. For eg. if a participant cleared 150 cm in his first jump, he will need to clear the height of 155 cm in his next jump. A foul is said to be committed if the participant fails to clear the barrier height. 3 continuous fouls would lead to straight elimination. 10 points are given to a participant who clears the given height and 2 points are deducted for every foul. Also, a participant is said to face an elimination jump if he had committed fouls on both of his two previous attempts. The following additional information is known about the jumps made by the 5 participants-
1. P5 was eliminated in his 5th jump and managed to score a total of points in single digit.
2. P1 and P2 faced elimination jumps thrice but managed to escape elimination. P1 never cleared two jumps continuously without a foul but P2 did this exactly once. Both of them scored 28 points each and cleared the 10th jump.
3. P4 was the only participant who made a maximum of 3 jumps continuously twice and he faced elimination jump once and escaped elimination.
4. P3 was eliminated in his 8th jump and he never faced the elimination jump before his 8th jump. Also, he cleared two different heights in two jumps continuously exactly once. 
5. P2 failed in his 4th and 7th jumps.
6. P4 made a foul in his second jump.

How many participants did not commit a foul in their 5th attempt?

Detailed Solution for CAT Mock Test- 10 (November 12) - Question 16

We will start off by using the information given in point 1.
It says that P5 was eliminated in his 5th jump. We will try to represent the jump in which he cleared the height successfully with 'Y' and the jumps in which he fouled using 'N'. Since he was eliminated in his fifth jump, he must have made fouls in his 3rd, 4th and 5th jumps and cleared the 2nd jump successfully. Had this not been the case, he would have been eliminated in his 4th jump with fouls in 2nd, 3rd and 4th jumps.

So,

In his first jump, he cannot have made the jump successful, else his points would then become 20-6=14 which is a double-digit.

.'. P5 must have had the following set of jumps:

Moving on to point 3 and 6 for P4.
It is said that he made a maximum of 3 continuous correct attempts twice and failed in his second jump.
We can have the following possibilities:

Next, we will look into point 4. P3 was eliminated in his 8th jump. So, we can have:

We know the logic behind putting YNNN in the last 4 jumps as he was eliminated in his 8th jump. Also, he never faced elimination before his 8th jump. So, 2 Ns cannot be together. A N has to be preceded and followed by a Y. If J4 is an N, J2 and J3 must be the continuous jumps. Similarly for the other case, if J4 is a Y, we have the continuous jumps and all the previous jumps will alternate between Y and N.
Now, for P1 using the second point, we can see that he scored 28 points with facing the elimination jump thrice. He must have cleared the heights successfully 4 times (40 points) and committed a foul 6 times (-12 points) to have scored 28 points. Also, all his 6 fouls are part of 2 continuous fouls because of which he faced the elimination jump thrice. The only possible way to do this without any two successful jumps is:


For P2, he also has 4 successful jumps and 6 fouls. Also, using point 5, we can infer that:

Now, jumps 3 and 4 cannot be the first set of fouls, because in that case, we will have only two elimination jumps for P2- 5th jump after 3rd and 4th fouls and 8th jump after 6th and 7th foul. Jump 10 is already successful. Therefore, jump 4 must be part of the second continuous fouls. Hence, we have:

P3 and P4 did not commit any foul in every case in the 5th jump. The rest of the participants did commit the foul.

CAT Mock Test- 10 (November 12) - Question 17

Directions: Read the passage and answer the following question.

Bill, Cody, Finch, Harper, Joana and Nick are playing a card game which had blue and red coloured cards. At the end of the game, each person was left with each of the two colored cards ranging from 1 to 6 in number.
No two persons have the same number of cards of the same colour.

Further, it is also known that:
(1) For any person, the sum of the number of cards with that person in both the colours is an odd number.
(2) Cody had more blue cards than Finch.
(3) The difference in the number of blue cards of Bill and red cards of Harper is 2.
(4) Finch, who has three red cards, has less blue cards than at least three persons.
(5) Harper, who has more blue cards than at least three persons, has more red cards than Joana.
(6) Nick has four blue cards and more red cards than those with Harper.

Q. How many blue cards are there with Cody?

Detailed Solution for CAT Mock Test- 10 (November 12) - Question 17

Given that the sum of the number of cards with any person is an odd number, then, for each person, the number of one of the two colored cards with him must be an odd one and the other an even one.

From (4), Finch has 3 red cards. Since he has odd number of red cards, then he must be having even number of blue cards. Hence, his blue cards can be 2/4/6. However, he has less blue cards than at least 3 persons. Hence, he must have 2 blue cards.
From (6), Nick has 4 blue cards. Hence, his red cards can be 1/3/5. Since Finch has 3 red cards, Nick can have 1/5 red cards. From (6), he has more red cards than Harper. Hence, he cannot have 1 red card. Hence, Nick must have 5 red cards.
From (2), Cody cannot have 1 or 2 blue cards (since Finch has 2 blue cards).
From (3), if Bill has 1 blue card, Harper must have 3 red cards. This is not possible as Finch has 3 red cards. Hence, Bill cannot have 1 blue card.
From (5), Harper can have 4/5/6 blue cards. Since Nick has 4 blue cards, Harper can have 5/6 blue cards.
Joana is the only person who can have 1 blue card.
From (5), Harper cannot have 1 red card and from (6), Harper cannot have 6 red cards. He also cannot have 3 or 5 red cards because Finch and Nick have 3 and 5 red cards, respectively. Hence, Harper can have 2/4 red cards. He must have odd number of blue cards (since he has even number of red cards). Therefore, Harper must have 5 blue cards.
If Harper has 2 red cards, Bill must have 4 blue cards, which is not possible as Nick has 4 blue cards. If Harper has 4 red cards, Bill must be having 6 blue cards (as Bill cannot have 4). This is the only possibility. Hence, Harper has 4 red cards and Bill has 6 blue cards.
Since Joana has 1 blue card, she must be having 2/4/6 red cards. From (5), Joana must have 2 red cards (since Harper has more red cards than Joana).
Since Bill has 6 blue cards, he must have 1 red card.
Cody has 3 blue cards and must have 6 red cards.

The following table provides the cards of each person and the cards they have.

Cody has 3 blue cards.

*Answer can only contain numeric values
CAT Mock Test- 10 (November 12) - Question 18

Directions: Read the passage and answer the following question.

Bill, Cody, Finch, Harper, Joana and Nick are playing a card game which had blue and red coloured cards. At the end of the game, each person was left with each of the two colored cards ranging from 1 to 6 in number.
No two persons have the same number of cards of the same colour.

Further, it is also known that:
(1) For any person, the sum of the number of cards with that person in both the colours is an odd number.
(2) Cody had more blue cards than Finch.
(3) The difference in the number of blue cards of Bill and red cards of Harper is 2.
(4) Finch, who has three red cards, has less blue cards than at least three persons.
(5) Harper, who has more blue cards than at least three persons, has more red cards than Joana.
(6) Nick has four blue cards and more red cards than those with Harper.

Q. How many people have less cards than Harper in both blue and red colours?


Detailed Solution for CAT Mock Test- 10 (November 12) - Question 18

Given that the sum of the number of cards with any person is an odd number, then, for each person, the number of one of the two colored cards with him must be an odd one and the other an even one.

From (4), Finch has 3 red cards. Since he has odd number of red cards, then he must be having even number of blue cards. Hence, his blue cards can be 2/4/6. However, he has less blue cards than at least 3 persons. Hence, he must have 2 blue cards.
From (6), Nick has 4 blue cards. Hence, his red cards can be 1/3/5. Since Finch has 3 red cards, Nick can have 1/5 red cards. From (6), he has more red cards than Harper. Hence, he cannot have 1 red card. Hence, Nick must have 5 red cards.
From (2), Cody cannot have 1 or 2 blue cards (since Finch has 2 blue cards).
From (3), if Bill has 1 blue card, Harper must have 3 red cards. This is not possible as Finch has 3 red cards. Hence, Bill cannot have 1 blue card.
From (5), Harper can have 4/5/6 blue cards. Since Nick has 4 blue cards, Harper can have 5/6 blue cards.
Joana is the only person who can have 1 blue card.
From (5), Harper cannot have 1 red card and from (6), Harper cannot have 6 red cards. He also cannot have 3 or 5 red cards because Finch and Nick have 3 and 5 red cards, respectively. Hence, Harper can have 2/4 red cards. He must have odd number of blue cards (since he has even number of red cards). Therefore, Harper must have 5 blue cards.
If Harper has 2 red cards, Bill must have 4 blue cards, which is not possible as Nick has 4 blue cards. If Harper has 4 red cards, Bill must be having 6 blue cards (as Bill cannot have 4). This is the only possibility. Hence, Harper has 4 red cards and Bill has 6 blue cards.
Since Joana has 1 blue card, she must be having 2/4/6 red cards. From (5), Joana must have 2 red cards (since Harper has more red cards than Joana).
Since Bill has 6 blue cards, he must have 1 red card.
Cody has 3 blue cards and must have 6 red cards.

The following table provides the cards of each person.

Only two persons have less cards than Harper in both blue and red colours.

CAT Mock Test- 10 (November 12) - Question 19

Directions: Read the passage and answer the following question.

Bill, Cody, Finch, Harper, Joana and Nick are playing a card game which had blue and red coloured cards. At the end of the game, each person was left with each of the two colored cards ranging from 1 to 6 in number.
No two persons have the same number of cards of the same colour.

Further, it is also known that:
(1) For any person, the sum of the number of cards with that person in both the colours is an odd number.
(2) Cody had more blue cards than Finch.
(3) The difference in the number of blue cards of Bill and red cards of Harper is 2.
(4) Finch, who has three red cards, has less blue cards than at least three persons.
(5) Harper, who has more blue cards than at least three persons, has more red cards than Joana.
(6) Nick has four blue cards and more red cards than those with Harper.

Q. Who has the same number of blue cards as the number of red cards with Bill?

Detailed Solution for CAT Mock Test- 10 (November 12) - Question 19

Given that the sum of the number of cards with any person is an odd number, then, for each person, the number of one of the two colored cards with him must be an odd one and the other an even one.

From (4), Finch has 3 red cards. Since he has odd number of red cards, then he must be having even number of blue cards. Hence, his blue cards can be 2/4/6. However, he has less blue cards than at least 3 persons. Hence, he must have 2 blue cards.
From (6), Nick has 4 blue cards. Hence, his red cards can be 1/3/5. Since Finch has 3 red cards, Nick can have 1/5 red cards. From (6), he has more red cards than Harper. Hence, he cannot have 1 red card. Hence, Nick must have 5 red cards.
From (2), Cody cannot have 1 or 2 blue cards (since Finch has 2 blue cards).
From (3), if Bill has 1 blue card, Harper must have 3 red cards. This is not possible as Finch has 3 red cards. Hence, Bill cannot have 1 blue card.
From (5), Harper can have 4/5/6 blue cards. Since Nick has 4 blue cards, Harper can have 5/6 blue cards.
Joana is the only person who can have 1 blue card.
From (5), Harper cannot have 1 red card and from (6), Harper cannot have 6 red cards. He also cannot have 3 or 5 red cards because Finch and Nick have 3 and 5 red cards, respectively. Hence, Harper can have 2/4 red cards. He must have odd number of blue cards (since he has even number of red cards). Therefore, Harper must have 5 blue cards.
If Harper has 2 red cards, Bill must have 4 blue cards, which is not possible as Nick has 4 blue cards. If Harper has 4 red cards, Bill must be having 6 blue cards (as Bill cannot have 4). This is the only possibility. Hence, Harper has 4 red cards and Bill has 6 blue cards.
Since Joana has 1 blue card, she must be having 2/4/6 red cards. From (5), Joana must have 2 red cards (since Harper has more red cards than Joana).
Since Bill has 6 blue cards, he must have 1 red card.
Cody has 3 blue cards and must have 6 red cards.

The following table provides the cards of each person and the cards they have.

Joana has the same number of blue cards as number of red cards with Bill.

*Answer can only contain numeric values
CAT Mock Test- 10 (November 12) - Question 20

Directions: Read the passage and answer the following question.

Bill, Cody, Finch, Harper, Joana and Nick are playing a card game which had blue and red coloured cards. At the end of the game, each person was left with each of the two colored cards ranging from 1 to 6 in number.
No two persons have the same number of cards of the same colour.

Further, it is also known that:
(1) For any person, the sum of the number of cards with that person in both the colours is an odd number.
(2) Cody had more blue cards than Finch.
(3) The difference in the number of blue cards of Bill and red cards of Harper is 2.
(4) Finch, who has three red cards, has less blue cards than at least three persons.
(5) Harper, who has more blue cards than at least three persons, has more red cards than Joana.
(6) Nick has four blue cards and more red cards than those with Harper.

Q. How many red cards more/less does Harper have than Bill? Key in the number.


Detailed Solution for CAT Mock Test- 10 (November 12) - Question 20

Given that the sum of the number of cards with any person is an odd number, then, for each person, the number of one of the two colored cards with him must be an odd one and the other an even one.

From (4), Finch has 3 red cards. Since he has odd number of red cards, then he must be having even number of blue cards. Hence, his blue cards can be 2/4/6. However, he has less blue cards than at least 3 persons. Hence, he must have 2 blue cards.
From (6), Nick has 4 blue cards. Hence, his red cards can be 1/3/5. Since Finch has 3 red cards, Nick can have 1/5 red cards. From (6), he has more red cards than Harper. Hence, he cannot have 1 red card. Hence, Nick must have 5 red cards.
From (2), Cody cannot have 1 or 2 blue cards (since Finch has 2 blue cards).
From (3), if Bill has 1 blue card, Harper must have 3 red cards. This is not possible as Finch has 3 red cards. Hence, Bill cannot have 1 blue card.
From (5), Harper can have 4/5/6 blue cards. Since Nick has 4 blue cards, Harper can have 5/6 blue cards.
Joana is the only person who can have 1 blue card.
From (5), Harper cannot have 1 red card and from (6), Harper cannot have 6 red cards. He also cannot have 3 or 5 red cards because Finch and Nick have 3 and 5 red cards, respectively. Hence, Harper can have 2/4 red cards. He must have odd number of blue cards (since he has even number of red cards). Therefore, Harper must have 5 blue cards.
If Harper has 2 red cards, Bill must have 4 blue cards, which is not possible as Nick has 4 blue cards. If Harper has 4 red cards, Bill must be having 6 blue cards (as Bill cannot have 4). This is the only possibility. Hence, Harper has 4 red cards and Bill has 6 blue cards.
Since Joana has 1 blue card, she must be having 2/4/6 red cards. From (5), Joana must have 2 red cards (since Harper has more red cards than Joana).
Since Bill has 6 blue cards, he must have 1 red card.
Cody has 3 blue cards and must have 6 red cards.

The following table provides the cards of each person and the cards they have.

Harper has 3 more red cards than Bill.

CAT Mock Test- 10 (November 12) - Question 21

C-Biz is a network marketing platform wherein a pyramid scheme is followed. The member who joins initially gets 25% of the revenue generated as commission. He is also liable to get an additional 25% of the commissions generated by a member who was referred by him. And this scheme goes on till infinity. If the revenue generated by each of the members of C-Biz is Rs. 20,000; then what amount is earned by the first member who referred another member only once, who in turn referred another only once and so on?

Detailed Solution for CAT Mock Test- 10 (November 12) - Question 21

The person, say A, gets 25% of revenue as a commission directly, which is
​Let B be the person referred by him. A gets 25% of B's commision as well, which is equal to
Let C be the person referred by B. C's commision=
​B's commission because of C = 
​And A's commission because of C= 
​Therefore, A is seen to earn in the following pattern- 

Total earnings for A= Sum to infinite G.P. with a= 5000 and 
​Total earnings = 

CAT Mock Test- 10 (November 12) - Question 22

What is the number of integers in the domain of (15−2x) C (x+3) ?

Detailed Solution for CAT Mock Test- 10 (November 12) - Question 22

The domain of nCr is such that r > 0 and n > r.
So, x+3 ≥ 0  and 15-2x≥x+3 are the only two conditions which needs to be satisfied.
x > -3 and 12> 3x x > -3 x < 4  are the two conditions.
.'. The possible integer values of x are -3, -2, -1, 0, 1, 2, 3 and 4.

CAT Mock Test- 10 (November 12) - Question 23

Grapes when dried up turn into raisins. A raisin has only 5% of water and grapes have 80% of water. 2 faulty weighing machines- A and B are used to measure grapes and raisins respectively. Faulty weighing machine A shows 10% less weight than the correct weight and faulty weighing machine B shows 10% more weight than the correct weight. A certain amount of grapes and raisins was measured, and according to the readings displayed by the balances, the mixture had 20% water content. What approximately was the true water content of the mixture of raisins and grapes that were measured?

Detailed Solution for CAT Mock Test- 10 (November 12) - Question 23

Let the amount of grapes be x kg and the amount of raisins be y kg.
Weight displayed by A = 0.9x
Weight displayed by B = 1.1y
Now,
% of water in the mixture = 20%

72x+5.5y=18x+22y
54x=16.5y54x=16.5y


To find out the original water content,

Replacing x with 11/36 y, we get the following

*Answer can only contain numeric values
CAT Mock Test- 10 (November 12) - Question 24

A city is classified as a green city when its forest cover is greater than or equal to 20%. The mayor of Hyderabad wanted the city to fall under this category. When calculated, it was seen that the forest area cannot increase by more than 4% annually. If the current forest area of Hyderabad is 17%, then after 'x' years, Hyderabad falls under the category of Green city, where x is the smallest possible integer. Then x=? 


Detailed Solution for CAT Mock Test- 10 (November 12) - Question 24

Let the area of Hyderabad be x sq. Kms
Current forest area= 0.17x
Every year the forest area can increase by 4% max.
So, we will compound 4% on 0.17x to check the value of 'n', the number of years after which the forest area increases to 20%.
0.17x(1.04) n =0.2x


By checking the value of n=2, we get L.H.S.= 1.0816 <1.1765
When n=4, we get L.H.S.= 1.1698< 1.1765
When n=5, we get L.H.S.= 1.2167>1.1765
Therefore min. value of n has to be 5.

*Answer can only contain numeric values
CAT Mock Test- 10 (November 12) - Question 25

A right-angled triangle ABC has an inradius and circumradius of 4 cm and 10 cm respectively. What is the area of triangle ABC in sq. cm?


Detailed Solution for CAT Mock Test- 10 (November 12) - Question 25

For any right-angled triangle with hypotenuse=h and the other two sides= 'a' and 'b', we know that:
Circumradius= h/2 and inradius= 
So, h= 2(10) cm= 20 cm.
Also, (a+b-20)=2(4)
⇒ a+b-20=8
.'. a+b= 28      ....(1)
Also, we have: a+b= h2
 From equation (1),we have a2+ (28-a)2=202
⇒ a2 + 784 + a2 - 56a = 400
⇒ 2a2 + 384 - 56a = 0
⇒ a2 - 28a + 192 = 0.
The two roots of this equation will be a and (28-a) which is nothing but a and b. Also, the product of the roots then become 192.
Area of a right angled triangle with non-hypotenuse sides 'a' and 'b'= 

CAT Mock Test- 10 (November 12) - Question 26

The exact time after 12:00 noon :after which the minute hand and the hour hand are exactly opposite to each other for the second time is?

Detailed Solution for CAT Mock Test- 10 (November 12) - Question 26

When the hour hand and the minute hands are opposite to each other for the first time, relative angle covered = 180
But when the two hands are opposite to each other for the second time, the relative angle covered = 180∘ + 360 = 540.
The hour hand covers 360 in 12 hours and 30 in 1 hour.
So, it covers 0.5 in 1 minute.
The minute hand on the other hand covers 360∘ in 60 minutes or 6 in 1 minute.
Therefore the relative speed = (6 - 0.5) = 5.5∘ per minute.
To cover 540, it will take minutes after 12 i.e

CAT Mock Test- 10 (November 12) - Question 27

If A, B, C and D are distinct natural numbers, the sum of which adds up to 50. Then, the maximum value of their product is __?

Detailed Solution for CAT Mock Test- 10 (November 12) - Question 27

Given, A, B, C and D are 4 distinct integers and A+B+C+D= 50.
The products of four numbers will be maximum if the numbers are as close to each other as possible.
The A.M. of the four numbers= 
.'. The four numbers are around 12 and continuous. A, B, C and D are 11, 12, 13 and 14.
And hence, the maximum value of the products of the four number=11×12×13×14=24024. 

*Answer can only contain numeric values
CAT Mock Test- 10 (November 12) - Question 28

Then what is the value of log2 n?


Detailed Solution for CAT Mock Test- 10 (November 12) - Question 28


.'. n=8 
And log2n= log28 = 3

CAT Mock Test- 10 (November 12) - Question 29

In the diagram, square ABCD has a side length of 6 units. Circular arcs of radius 6 units are drawn with centres B and D. What is the area of the shaded region?

Detailed Solution for CAT Mock Test- 10 (November 12) - Question 29

Shaded Area = 2 [Area of Sector ACB - Area of Right ΔABC]

= 2 [Area of Sector ACD - Area of Right ΔADC]

CAT Mock Test- 10 (November 12) - Question 30

Alok took a loan of Rs. 24000 at 5% simple interest per annum from Bhanu. He immediately lent out one-third of the money to Clara at 4% compound interest for two years and the rest to Dhruv at 8% simple interest. How much did Alok earn (in rupees) from the entire transactions if the money was returned back by Clara and Dhruv at the end of the second year? Consider Alok paying back the loan to Bhanu as soon as he got the money back from the other two.

Detailed Solution for CAT Mock Test- 10 (November 12) - Question 30

Initially, Alok lent out all the money he borrowed from Bhanu. Alok will earn only when the amount received by him by lending out to Clara and Dhruv is greater than the amount payable to Bhanu at the end of 2 years.
Amount payable to Bhanu at the end of two years= Principal+ S.I.= 24000+ = 24000+2400= Rs. 26,400.
Amount receivable from Clara= 
Principal amount for Dhruv= Rs. 24000- Rs. 8000= Rs. 16000.
Amount receivable from Dhruv= 16000+ = 16000+2560= Rs. 18560.
Total amount receivable by Alok= Rs. (8652.8+ 18,560)= Rs. 27212.8
Profit for Alok= Rs. (27212.8- 26400)= Rs. 812.8

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