CAT 2019 Question Paper Slot 1

Option A: It has been mentioned that food metabolism is used to maintain body temperature. But it cannot be inferred that heat is lost due to food metabolism.

Option B: The colder temperature of plumage results in slight heat gain from the surrounding air. Hence this option is incorrect.

Option C: In the last paragraph of the passage, it has been mentioned that heat is very important for the breeding of Emperor Penguins. So it can be inferred that this conserved heat might be used in the reproductive process of Emperor Penguins. Hence C is the answer.

Option D: Consider the line: "Since their outer plumage is..........................thermal convection—the transfer of heat via the movement of a fluid (in this case, the air)." It is clear that the process of thermal convection is responsible for heat gain and not heat loss. Hence D is incorrect.

The primary findings of the study conclude that Emperor Penguins reduce the heat loss by keeping the temperature of the outer surface of their plumage lower than the surrounding air. In fact, they gain a little heat from the surrounding air through thermal convection.

Option A: If the plumage did not allow thermal convection, it would contradict the findings of the study. Hence A is not the answer.

Option B: Since the transfer of heat takes place through the plumage, variation in the average temperature of feet will not affect the conclusions of the study. Hence B is the answer.

Option C: The average temperature of plumage should be lower than that of the air. It has been mentioned in the passage that the temperatures of the plumage on their heads, chests and backs were -1.84, -7.24 and -9.76 degrees Fahrenheit respectively. If the temperature of the air is -10 degrees Fahrenheit, Penguins would not be able to gain the heat. Hence, this will negate the study findings. 

Option D: All the temperatures mentioned in this option are higher than the temperature of the air, but the study assumes the surrounding air temperature to be higher. This option will also negate the study findings. 

The word "paradoxically" has been used by the author to indicate the two contradictory characteristics mentioned in the statement. 

Option A: This option states the exact opposite conclusion mentioned in the passage. As per the passage, penguins keep their plumage colder to keep their body warmer. Hence A is incorrect.

Option B: It has been mentioned that the penguins lose heat through radiation and gain heat through convection. Hence B is incorrect.

Option C: Although this statement is true, it does not contain self-contradictory parts. It has been mentioned that the heat loss and heat gain happen through the given processes but one has no relation to the other. Hence C is not the answer.

Option D: This statement combines two contradictory qualities. The penguins are keeping their plumage colder, which is responsible for the heat gain from the surrounding air and making their body warmer. Hence D is the answer.

Option A: Consider the sentence: "As the cold Antarctic air cycles around their bodies, slightly warmer air comes into contact with the plumage and donates minute amounts of heat back to the penguins, then cycles away at a slightly colder temperature." It has been mentioned in the passage that the plumage temperature is lower than the surrounding air temperature. Hence, "slightly warmer air" refers to the Antarctica air that surrounds the plumage and "at a slightly colder temperature" refers to the fall in temperature due to heat loss.

Option B: The process of convections and not radiation is involved in this case. Hence the first part of the option is incorrect. B is not the answer.

Option C: The passage does not mention air trapped in plumage. Hence this option is rejected.

Option D: "slightly warmer air" refers to the Antarctica air and not the air inside the penguins' bodies. Hence D is incorrect.

Option A: The startups Casper and Glossier are certainly breaking the trend of choice anxiety. Yet, the author argues that they are turning into something that they intended to disrupt. Hence, this does not capture the purpose of the author.

Option B: The author argues that even these startups are targeting select few mid-range customers rather than the lower classes. Hence, this option directly contradicts the author's claim.

Option C: These startups initially started as an exception to offering a wide variety of choices. Yet, due to limited customers, and want of steep growth, they might transform into a type of company that they intended to disrupt. Hence, this option correctly resounds the authors fear and captures his purpose of argument. Hence C is correct

Option D: This option is largely vague and can have multiple interpretations. One interpretation can be that these startups are targeting a selected band of customers and do not have offering for lower-class customers. Hence, there is no uniform distribution.

Option A: Paragraph 1 says "choice fatigue is one reason so many people gravitate toward lifestyle influencers on Instagram". Hence, as per the passage, a company with wide range of products and a lifestyle influencer is likely to perform better than a company with only wide range of products. Hence, this statement negates the claim of the author.

Option B: "As options have expanded for people with disposable income, the opportunity to buy even basic things such as fresh food or quality diapers has contracted for much of America’s lower classes." The author argues that variety of products are offered only for a certain class of consumers other than the lower class. If variety of options indeed helped the poor, then his argument is weakened.

Option C: "Research has consistently held that people who are presented with a few options make better, easier decisions than those presented with many". "Americans have lost the ability to sort through the sheer volume". Clearly, people are overwhelmed by options and prefer lesser variety. Hence, option C is contradictory.

Option D: This option is largely vague and leaves unanswered questions behind. Also, the author doesn't make any comparison between the growth of these two type of companies. The author only says that, as the company targets only few consumers, for the want of growth they are likely to expand to variety of products. As there is no information about their growths, this option neither strengthens nor weakens the claim.

Option A: Paragraph 1 says "Since Americans have lost the ability to sort through the sheer volume of the consumer choices available to them" Since the product options are overwhelming, they are unable to sort through the options. Hence, option A can be inferred from the passage.

Option B: Paragraph 1 says "Research has consistently ..... industry unto itself." As people experience choice anxiety due to overwhelming options, they are unable to trust products while selecting. Hence, they look-out for celebrities and curators to make a decision. 

Option C: There is no such comparison in the passage that shows people's preference towards products by startups. Hence, option C cannot be inferred.

Option D: Paragraph 1 says "a ghost now has to be in the retail machine, whether it’s an algorithm, an influencer, or some snazzy ad tech to help a product follow you around the internet". Due to our inability to sort, we depend on influencers or we are vulnerable to snazzy ads to purchase products. Hence, D can be inferred.

The author principally argues for lesser choices. He says that choice anxiety is overwhelming and people make better decisions with lesser choices.

He is also critical about companies targeting only certain band of well-off customers and critiques them for not offering products for consumers of lower classes.

Hence, a product group with lesser variety, and targeted to lower class customers would be most acceptable to the author.

By "Depth", the author suggests a scenario that adds value or supplies additional information which supports his claim.

Option A: If the startup products grow exponentially and are self-sufficient and do not expand to other products, this scenario directly contradicts the author's probable prediction of these companies. Hence, it would add the least depth to the author's argument. A is the correct answer.

Option B: Lets consider that startups with few product options already exist. In such a case, these startups are no exceptions. For the sake of steep growth and surviving, they might have to expand into different product categories. Hence it adds some depth to the author's prediction.

Option C: "There may be no way to opt-out of stuff by buying into the right thing." The author is clearly displeased with startups ending up with overwhelming variety. Losing regular customers for better growth further invigorates the author's claim against numerous choices. Hence, it adds some value to his criticism.

Option D: If the government doubles their tax rates, as these startups are dependent on select customers for income, they might have to venture into other products and varieties to accentuate their returns and keep the company afloat. Hence, their fate would likely end up the way author predicted it to be.

"Residents of upscale residential developments have disclosed how important it is to maintain their community’s distinct identity, often by casting themselves in a superior social position and by reinforcing class and racial differences."

Option A: The option implies that the clients are made to feel at home. While the phrase “Residents of upscale residential developments” is used to capture the intent of social dominance of a particular class. Hence this option is incorrect.

Option B: The option implies that jingoism of a certain class might lead to topophobia. The option is yet again unrelated.

Option C: Residents of upscale residential developments intend to promote their community by reinforcing sectarian differences. This exclusivism(Practice of being exclusive/important) is clearly captured in the option. Hence C is correct.

Option D: Sensitive response indicates a considerate response where other's sentiments are considered. While these residents are inconsiderate and consider themself superior. Also, the option doesn't capture the purpose clearly. Hence, incorrect

Option A: The entire passage deals with "TOPOPHILIA" and "TOPOGRAPHY" is unrelated. Also, the author says that we experience topophilia in three forms and that we are not born with it.

Option C: An illustration of topophobia doesn't represent the author's view on topophilia

Option D: The option speaks about glossophilia(Love of language) and is unrelated to topophilia

Option B: "Topophilia connotes a positive relationship, but it often is useful to explore the darker affiliations between people and place. Patriotism, literally meaning the love of one’s terra patria or homeland". 
Despite a negative tone, the author says that one form of topophilia is patriotism. Even though not wholesome, it comes "closest" to the author's understanding of topophilia among the given options. Hence B is correct.

"As Tuan noted, purely aesthetic responses often are suddenly revealed, but their intensity rarely is longlasting. Topophilia is difficult to design for and impossible to quantify". The author says that people's response to aesthetics is shortlived and usually subsides overtime. Hence, it is difficult to design or quantify.

Option A: "Amomie" means lack of morals or ethics. It is unrelated to the passage.

Option B: An objective analysis by architects does not explain the reason as to why it is difficult to quantify topophilia.

Option C: This statement is in the form of an opinion and does not explain the above statement.

Option D: Since every person has different topophilic attractions and have different responses to aesthetics. Capturing topophilia in the form of design is impossible. This option elaborates and explains the reason for quantifying topophilia. Hence option D is correct.

"And just as a beloved landscape is suddenly revealed, so too may landscapes of fear cast a dark shadow over a place that makes one feel a sense of dread or anxiety—or topophobia."

Option B speaks about topography, while Option C speaks about dread towards people.
Option D is unrelated to topophobia. Hence, all of them are incorrect

Option A clearly captures the essence of the last sentence in the passage. 

Option A: "new urbanism seeks to... Although motivated by good intentions, such attempts to create places rich in meaning are perhaps bound to disappoint." The author says new urbanism that tries to induce sense of place is bound to fail. Since there is no mention of clients, irrespectively new urbanism is going to fail. Hence, it is contradicting the author.

Option B: "His 1974 book set forth a wide-ranging exploration of how the emotive ties with the material environment vary greatly from person to person and in intensity, subtlety, and mode of expression." This option is contradicting the passage yet again.

Option C: The author lists out three ways of experiencing topophilia but doesn't emphasize about any one way. Hence, even though not contradictory, this option is factually misquoting the passage.

Option D: "Topophilia connotes a positive relationship, but it often is useful to explore the darker affiliations between people and place. Patriotism, literally meaning the love of one’s terra patria or homeland.." Clearly, the author has a negative intonation when he says "darker affiliation". He presents patriotism as a darker manifestation of topophilia. Hence, this statement is correct and does not contradict the author. Hence option D is correct.

Cecil Sharp says "One man sings a song, and then others sing it after him, changing what they do not like". This signifies that folk music is constantly evolving. Hence, this adaptability contributes to its plurality. Hence the author is going to agree with option B

"Just as the effusive ..... on countless record labels" This indicates that - "Just as the radical views of Morris became popular and mainstream, similarly folk music which is considered parochial is becoming popular and conformist. This popularity is being rejoiced by media as "folk is hip again". Hence, option C correctly captures this sentiment.

"For the early-20th-century composers .... tradition itself." This line captures the idea that folk is also inspired by various philosophies and schools of thought. Hence, we can infer that folk is intellectually relevant in contemporary times. Option D is in coherence with the author's views.

Option A says that folk forms exhibit homogeneity. The author in the entire passage describes the diversity of folk and says it paves way for vivid imagination. "The very obscurity and anonymity of folk music's origins open up space for rampant imaginative fancies". Cecil Sharp cites an analogy of an oak tree to show the constant transformation of folk. Hence, this option is contradicting author's opinion and he is least likely to agree with it.

"Just as the effusive floral prints of the radical William Morris now cover genteel sofas, so the revolutionary intentions of many folk historians and revivalists have led to music that is commonly regarded as parochial and conservative.

Here the author compares two aspects. We know that William Morris is a radical conservationist as per para 1. (genteel refers to respectable/gentlemanly, genteel sofas refers to people in respectable place in life) As the footprints/views of William Morris becomes more popular i.e. as conservative folk forms once considered radical became more mainstream, similarly folk music which is considered parochial is now being revived by revivalists. The primary purpose is to show an analogy that a radical folk form became more mainstream/acceptable over time. This expression is best captured in option C

"Free of the taint of manufacture" ...... been contested territory."

The phrase "Free of the taint of manufacture" is likely to have emerged post-industrialisation when conservationinsts(conserving status quo especially natural resources) fancied a pre-industrial age and expressed nostalgic attachment towards it. Hence the main point of the first paragraph can be summarised as "Conservationists envisioned a cosy folk form inspired by preindustrial times".

Hence option B is the correct answer

" In the late 1960s, purists were suspicious of folk songs recast in rock idioms." Purists oppose any altercation or adaptation of original folk from and they criticized the adaptations by rock too. Hence, option B can be inferred

Folk music was inspired by revolutionary intentions in 1940s, various philosophies and school of thoughts in 1960s, Freedom of expression(Bob dylan) and psychedalia.

This shows a constant transformation of folk reinforcing the beliefs of Cecil Sharp. Option C is correct.
Option D can be rightly inferred from the lyrical freedom of bob dylan and revolutionary intentions in 1940s. 

Option A : "In the late 1960s, purists were suspicious of folk songs recast in rock idioms. Electrification, however, comes in many forms." Even though the electrification of folk by rock was rejected by purists, electrification came in many forms and not individually by Rock alone. Hence, electrification of folk music is not causated by rock alone. Hence option A cannot be inferred.

Option A and option D both signifies the inspiration of folk music from two different times. Folk is not limited to immediate past or to any specific time-line. This contributes to the plurality of folk music.

Option C talks about "fluidity". Fluidity indicates flexibility or different modes of oral rendition. For example different vocal styles can be generated by pitch, rhythm, style of rendition. Hence, the variance in oral transmission of music can lead to various iterations of one original form. Hence, this again contributes to the diversity and plurality of folk music.

Option B: Popularity or unpopularity is an opinion. An opinion in no way contributes to the diversity of a folk form. It might be regarded as diverse opinions but does not inherently contribute to the diversity of folk itself. Hence, option B does not contribute to folk's plurality.

The passage says "describes his own hard-knocks upbringing and the way he marvelled at the extravagance of Versailles. The descriptions he uses were very similar to the descriptions of the lavish palace that ended up in Galland’s version of the Aladdin story."
Hence, option B and option C depicts the similarities of Hanna Diyab's life and Aladdin's character

Option D:“Diyab was ideally placed to embody the overlapping world of East and West, blending the storytelling traditions of his homeland with his youthful observations of the wonder of 18th-century France.”. Since Diyab is a middle eastern man who came to France, his cross-culture experience would make him ideal to embody the character of Aladdin. Hence, option D gives evidence for the claim aladdin is based on hana diyab's life.

Option A: Even though Diyab narrated the story, he might have read it somewhere or heard it from someone else, it doesn't necessarily give any insight about its relationship with his life.

It reflects not only “a history of the French and the Middle East, but also [a story about] Middle Easterners coming to Paris and that speaks to our world today,”

The above statement indicates that the primary reason for scholars to go back to aladdin is the intrigue about middle easteners coming to paris.

Option B and D have references from third paragragh which is unrelated to the context.

While option C is one of the reasons, the primary reason as per the author is option A(traveller's experience indicates midleeasteners experience in france). 

Option A can be considered as one of primary importance as he says "that speaks to the world today" highlighting the importance of middle-easteners coming to paris.

The narrative sensibility of Diyab’s travelogue indicates similarity in characters of Aladdin and Diyab in terms of sensibility, being considerate. Hence, option A strengthens the passage's claim about Aladdin's character having ties with that of Diyab

Option B, Galland's acknowledgement again indicates that Aladdin might be predated than 1712 and might have some roots associated with Diyab

Option D, The affluence in the story of Aladdin and Diyab's travelogue have major similarities. This suggests that Aladdin maybe based on Diyab's life experiences. Hence, it supports the passage's claim about authorship.

As per option C, the french fairy tales inform us about the probable cause/motive behind writing Aladdin but doesn't lead to information regarding its authorship.

'"Galland wrote in his diary that he first heard the tale from a Syrian storyteller from Aleppo named Hanna Diyab

Since he heard the story, option B is incorrect.

Also, the first paragraph implies that - While Arabian nights predates to medieval times, the earliest appearance of Aladdin is in 1712. Hence, option C and D are incorrect.

"Transmit" means passing from one person to another. This can imply that Diyab told it to Galland. Also, Aladdin is one of the stories of "Arabian Nights" (Others include Alibaba & 40 thieves, Sindbad). Hence option A is correct.

The second paragraph says that there are 2 possible motivations for writing the story of Aladdin, first being french fairy tales and second being French orientalism. He goes on to say, if aladdin is actually based on the life of hanna Diyab, then the idea of french orientalism is inversed.

"or that the story was invented in that 18th century period as a byproduct of French Orientalism, a fascination with stereotypical exotic Middle Eastern luxuries that was prevalent then. The idea that Diyab might have based it on his own life — the experiences of a Middle Eastern man encountering the French, not vice-versa — flips the script."

French Orientalism implies an intrigue of French towards Middle-eastern luxuries, while Diyab coming to france shows an interest of middle-easteners in France. Hence, the script is inversed if Aladdin's story is based on Hanna Diyab.
The question is looking for option which invalidates the inversion. This implies that the script shouldn't be inversed. This occurs when Aladdin's story is not based on Diyab.

Option C says that Diyab's travellogue doesn't bear any resemblance to Galland's Aladdin. This implies that Aladdin is not based on Diyab. Hence, the inversion of script doesn't occur/invalidated. None of the other options invalidate the script.

Hence, option C is correct

Option 4 and option 5 are related as both statement start with a suffix.
While option 3 is a continuation of the idea in option 5

Option 5 says that the suffix signifies pride, while option 3 elaborates on this and explains how it is displayed as pride to friends and families alike. Hence 53 is a logical block.

Among all the statements, 4 is the only one which doesn't have a pronoun or a tone indicating the presence of a preceding statement.

While 4 opens the statement, it must be succeeded by 1 as the terms cannot be explained at the end. 

The logical coherence of this para jumble is 4(Introduction of terms)-1(Explanation of terms)-5(Consequence of terms(Pride))-3(Elaboration of consequence)

Statement 2 speaks about modernism and that every phenomenon can be easily grasped. It is unrelated to the context of the passage and a misfit.

Hence, option 2 is the odd one out

Statement 1 displays a contrast of 2 kinds of reading. Logically these reading types must be defined before their negatives are discussed. This indicates that both statement 2 and 4 necessarily precede statement 1

Statement 4 and 1 form a logical block as statement 4 introduces the idea of developmental disorders while statement 1 cites example of such disorders.  

Statement 4 cannot be opening as it has a phrase "like mind reading". While, Statement 3 can neither be an opening nor closing statement. 

Hence, statement 2 is a good introductory statement as it starts by describing mind reading and statement 3 extends the idea of mindreading. As 3 doesn't fit anywhere else, it has to necessarily follow statement 2

Hence, the correct order is 2341

Statement 3 and 2 form a natural block. While statement 3 describes "carpe diem", statement 4 explains how every language has subtle differences in its essence and interpretation. Statement 4 says "its an example of" suggesting to the logical block 32 which shows misinterpretation of metaphor. Hence, 324 forms a block. While statement 1 can serve as both opening as concluding closing statement. Both the sequesnces 3241 and 1324 seem accurate. Although CAT 2019, considered 3241 as the final answer to this parajumble with statement 1 concluding the paragragh.

Statement 1 suggests that we will live under mob-rule until a specific event occurs. Hence, it cannot be the opening statement.
Statement 2 is a suggestion to deal with criticism.
Statement 3 is a good opening statement as it sets the agenda for the passage.
Statement 4 says that in the upcoming future, repetitive criticism will not be permitted.

41 makes a logical block as statement 4 talks about a specific event, where mob frenzy attitude is curtailed by eliminating repetitive criticism. Until this occurs, we will live under mob-rule.

Statement 3 opens the passage that we need to learn to deal with criticism. While statement 2 extends the idea as to how one has to deal with criticism on a personal level. Then, the author speaks about curtailing mob culture by censoring repetitive criticism.

Hence, the correct logical order is 3241

In this context, "Psychoanalytical analytical technique" implies that the advertising agencies are adapting methods to tap into the unconscious mind of the consumers. They are conducting detailed interviews to identify hidden motivations.

Here, subliminal advertising represents some portion of the ad being difficult to comprehend or simply put, when one of the motives of the ad is so subtle that it is difficult to be understood by a layman.

While supraliminal advertising can be clearly conceived by most people.

Packard claims that the 'Hidden persuaders' use supraliminal advertising to entice customers by tapping into consumers without their knowledge.  (....can be persuaded by supraliminal messages without their knowledge.)

Option B and D say that the method is subliminal, hence, it is incorrect
Option C says that people are well aware about being persuaded, hence incorrect.
Option A is a wholesome summary of the method of persuation.

The paragraph says that humans think about past occurrences suddenly without any immediate stimuli.
The author also says that thinking/thoughts about a certain distant past is a necessity before one can speak about it. 
He says that thoughts are a pre-requisite before one talks about it. He also gives an example that various human-like symbols might have emerged without any immediate stimuli.

Option A and C: There is no mention of specificity to humans in the passage
Option B : "All speech acts" is a false generalisation. The passage says that speaking about distant past requires thinking about it first
Option D : It clearly captures the essence of the passage and says that one needs to think about distant past events before talking about them
Hence option D is correct.

Statement 1 says "this collaborative filtering". Here "this" refers to a likely preceding statement which explains about collaborative filtering. While statement 1 is an example of collaborative filtering.
Statement 2 starts with "these algorithms". This doesn't look like an opening statement as it describes a certain algorithm ("these").
Statement 3 is explaining about the flaw in the algorithm
Statement 4 makes for a good opening statement
Statement 41 makes a logical block as the example in 1 refers to collaborative filtering of statement 4
Statement 2 must appear soon after the logical block of 41 as it contains the phrase "these algorithms". While statement 2 explains the problem in the algorithm, statement 3 extends the idea of the problem.

Statement 4 : Introduces collaborative filtering
Statement 1 : Given an example of collaborative filtering
Statement 2 : Speaks about the drawback in the algorithm
Statement 3 : Gives an example of the drawback

Hence the logically coherent order is 4123

Statement 34 makes a logical block as statement 4 speaks about "this prejudice" against deaf people. While statement 3 highlights the prejudice by saying that they were considered to be dumb and not allowed to sign a will.

The idea of sign language is continued in statement 1 where the pronoun "he" refers to pedro and the statement discusses about the origin of the sign language.

Statement 5 is a good opening statement as it introduces the idea of discrimination against deaf and this idea is continued by an example in statement 3. Hence these statements can be arranged in the order 5341.

Statement 2 is a good standalone opening statement. Yet, this statement cannot be succeeded by any of the other statements as it is a misfit in the paragraph. 

Hence statement 2 is the odd one out

The passage says that pure science intends to discover without any end-goal in mind. While engineers use these benefits for practical applications. The author says that the science behind these practical applications are often considered false by pure science since they are approximated or not applied as per ideal conditions. In any case, even though they are rejected, these approximated science theories find lot of practical applications in everyday life.

Option A is correct. By diluting science, these theories are put into practical benefits. Hence, option A is correct

Option B is incorrect as no such implication can be drawn from the passage

Option C is incorrect. Linear relationship indicates that, if a certain theory is rejected by pure science, it is bound to be rejected by applied science too. This is clearly not the case as engineers use rejected theories for practical benefits.

Option D speaks only about engineers and has no reference to sciences or the main point of the paragraph. The paragraph intends to compare the functionalities of scientists and engineers while option D is specific to engineers and does not encapsulate the essence of the paragraph.

Hence, by way of elimination Option A is the most suitable summary

Statement 1 says that actors who do not fit within a certain category such as "humour", "action" etc will face certain difficulty

Statement 2 has a negative tone. It starts with "others"  indicating that the preceding statement is likely to have a positive tone.

Statement 4 is a good opening statement as it sets the agenda for the passage by saying that there are opposing views with regard to effects of identities in organisations.

Statement 5 is a favourable view to complex identities, a positive tone. Hence 52 is a logical block.

Statement 3 cannot open the paragraph as the succeeding statement 4 will be disconnected to the central idea of statement 1. Statement 3 can neither be a conclusion as it is too generic. Hence, statement 3 can logically occur only after statement 4 and before the logical block 52.

4 - Idea of organisational identity introduced
3 - One of the features of organisational identities explained
52 - Two opposing views expressed

Statement 1 is unrelated to organisational identity as it speaks about stereotyping the actors.
Hence statement 1 is odd one out

The total number of biscuits = 5, the total number of candies =3 and the total number of savouries = 12-(3+5)=4

Representing the candies as C, biscuits as B and savories as S. K is to be placed in shelf number 16. D, E and F are savouries and are to be placed in consecutively numbered shelves in increasing order after all the biscuits and candies. Since there is no empty shelf between the items of same type, D,E,F and K are savouries and placed at 13,14,15 and 16 respectively. This can be tabulated as follows:

The shelf 12 will be empty.

It is given that items are to be placed such that all items of same type are clustered together.

From 1, A and B are to be placed in consecutively numbered shelves in increasing order.

From 6, C is a candy and is to be placed in a shelf preceded by two empty shelves and from 7, L is to be placed in a shelf preceded by exactly one empty shelf.

Hence C and L are items of different types. Since C is a candy, L will be a biscuit.

From 5, L and J are items of the same type, while H is an item of a different type.

Since I and J are clustered together, I, J and L are biscuits and H is a candy.

So C,H are candies and I,J,L are biscuits. It is given that A, B are place consecutively. Hence A and B are items of same types. So A, B should be biscuits because if they are candies, there will be 4 candies.

Hence, I,J,L,A,B are biscuits and C,H and G are candies.

Now there are two empty shelves before C and exactly one empty shelf before L, then the different cases can be tabulated as follows:

Case 1:

Case 2:

The number of arrangements for the first case = 2*2=4 

The number of arrangements for the second case = 2*2=4

The total number of arrangements = 4+4=8

The total number of biscuits = 5, the total number of candies =3 and the total number of savouries = 12-(3+5)=4

Representing the candies as C, biscuits as B and savories as S. K is to be placed in shelf number 16. D, E and F are savouries and are to be placed in consecutively numbered shelves in increasing order after all the biscuits and candies. Since there is no empty shelf between the items of same type, D,E,F and K are savouries and placed at 13,14,15 and 16 respectively. This can be tabulated as follows:

The shelf 12 will be empty.

It is given that items are to be placed such that all items of same type are clustered together.

From 1, A and B are to be placed in consecutively numbered shelves in increasing order.

From 6, C is a candy and is to be placed in a shelf preceded by two empty shelves and from 7, L is to be placed in a shelf preceded by exactly one empty shelf.

Hence C and L are items of different types. Since C is a candy, L will be a biscuit.

From 5, L and J are items of the same type, while H is an item of a different type.

Since I and J are clustered together, I, J and L are biscuits and H is a candy.

So C,H are candies and I,J,L are biscuits. It is given that A, B are place consecutively. Hence A and B are items of same types. So A, B should be biscuits because if they are candies, there will be 4 candies.

Hence, I,J,L,A,B are biscuits and C,H and G are candies.

Now there are two empty shelves before C and exactly one empty shelf before L, then the different cases can be tabulated as follows:

Case 1:

Case 2:

G is a candy. Hence D is the answer.

The total number of biscuits = 5, the total number of candies =3 and the total number of savouries = 12-(3+5)=4

Representing the candies as C, biscuits as B and savories as S. K is to be placed in shelf number 16. D, E and F are savouries and are to be placed in consecutively numbered shelves in increasing order after all the biscuits and candies. Since there is no empty shelf between the items of same type, D,E,F and K are savouries and placed at 13,14,15 and 16 respectively. This can be tabulated as follows:

The shelf 12 will be empty.

It is given that items are to be placed such that all items of same type are clustered together.

From 1, A and B are to be placed in consecutively numbered shelves in increasing order.

From 6, C is a candy and is to be placed in a shelf preceded by two empty shelves and from 7, L is to be placed in a shelf preceded by exactly one empty shelf.

Hence C and L are items of different types. Since C is a candy, L will be a biscuit.

From 5, L and J are items of the same type, while H is an item of a different type.

Since I and J are clustered together, I, J and L are biscuits and H is a candy.

So C,H are candies and I,J,L are biscuits. It is given that A, B are place consecutively. Hence A and B are items of same types. So A, B should be biscuits because if they are candies, there will be 4 candies.

Hence, I,J,L,A,B are biscuits and C,H and G are candies.

Now there are two empty shelves before C and exactly one empty shelf before L, then the different cases can be tabulated as follows:

Case 1:

Case 2:

From the table(case 2), only 1,2,6 and 12 are empty in the same arrangement. Hence, C is the answer.

The total number of biscuits = 5, the total number of candies =3 and the total number of savouries = 12-(3+5)=4

Representing the candies as C, biscuits as B and savories as S. K is to be placed in shelf number 16. D, E and F are savouries and are to be placed in consecutively numbered shelves in increasing order after all the biscuits and candies. Since there is no empty shelf between the items of same type, D,E,F and K are savouries and placed at 13,14,15 and 16 respectively. This can be tabulated as follows:

The shelf 12 will be empty.

It is given that items are to be placed such that all items of same type are clustered together.

From 1, A and B are to be placed in consecutively numbered shelves in increasing order.

From 6, C is a candy and is to be placed in a shelf preceded by two empty shelves and from 7, L is to be placed in a shelf preceded by exactly one empty shelf.

Hence C and L are items of different types. Since C is a candy, L will be a biscuit.

From 5, L and J are items of the same type, while H is an item of a different type.

Since I and J are clustered together, I, J and L are biscuits and H is a candy.

So C,H are candies and I,J,L are biscuits. It is given that A, B are place consecutively. Hence A and B are items of same types. So A, B should be biscuits because if they are candies, there will be 4 candies.

Hence, I,J,L,A,B are biscuits and C,H and G are candies.

Now there are two empty shelves before C and exactly one empty shelf before L, then the different cases can be tabulated as follows:

Case 1:

Case 2:

Option A and C are wrong as candies can come before biscuits and vice versa. B is not necessarily true as there can be one empty shelf too as shown in the table. Option D is true as there are at least 4 shelves between B and C. Hence D is the answer.

It is given that every bull’s eye score in the first three rounds gave a player one additional chance to shoot in the bonus rounds, Rounds 4 to 6, which means Tanzi scored Bull's eye only once in the first 3 rounds because she participated only once in round 4 to 6. Similarly, Umeza scored Bull's eye exactly 2 times in the first 3 rounds. Wangdu did not score Bull's eye in the first three rounds and so on. 

Now from 1, Tanzi, Umeza and Yonita had the same total score.

So, Total score of Tanzi will be 4+5+5+a=14+a,  (She scored Bull's eye(a score of 5) in exactly one round and a is the unknown score)

Total score of Umeza = 1+2+5+5+b = 13+b    (She scored Bull's eye(a score of 5) in exactly 2 rounds and b is the unknown score)

Total score of Yonita = 3+5+5+c=13+c (She scored Bull's eye(a score of 5) in exactly one round and c is the unknown score)

Now 14+a=13+b=13+c,

Also it is given that total scores for all players, except one, were in multiples of three, so these three will have to be a multiple of 3.

So, (a,b,c) can be either (1,2,2) or (4,5,5) in the same order. But the value (5,5) for b and c is not possible.  (Umeza scored Bull's eye in exactly 2 rounds and Yonita in exactly 1 round)

Hence, a=1,b=2 and c=2. So each of Tanzi, Umeza and Yonita had total score of 15.

Tabulating the data, we have

From 5, Tanzi and Zeneca had the same score in Round 1 but different scores in Round 3.

Zeneca score Bull's eye 2 times in round 1 to 3. If Tanzi scored 1 in round 1, then Zeneca also has to score 1 in round 1, which means both Tanzi and Zeneca scores in round 3 will be 5, which violates 5. Hence Tanzi scored 5 in round 1 and Zeneca also scored the same in round 1.So the new table is:

From 4, the number of players hitting bull’s eye in Round 2 was double of that in Round 3.

So, in round 3 either 1 or 2 Bull's eye can be scored and in round 2, 2 or 4 Bull's eye can be scored.

Case 1: If only 1 Bull's eye is scored in the round 3, then in round 3 Umeza will score 2 and Zeneca will score 2/3/4 in round 3, which means both will score 5 in round 2. So minimum Bull's eye in round 2 will be 3. (Umeza, Zeneca and Xyla)

Hence this case is rejected.

Case 2: 2 Bull's eye were scored in round 3 and 4 Bull's eye were scored in round 2. So in round 2 Umeza, Yonita and Zeneca scored 5. This can be tabulated as:

In round 3, 2 Bull's eye can only be scored by Xyla and Umeza.

The highest scorer can be either Xyla or Zeneca. The lowest scorer will be Wangdu.

1.Consider Zeneca is the highest scorer. 

From 3, the highest total score was one more than double of the lowest total score. So the only possible score for Zeneca is 23 and that for Wangdu is 11. (11*2+1=23)

But this will violate condition 2, since both Zeneca and Wangdu do not have their scores as multiples of three in this case.

Hence, Xyla will be the highest scorer. The only possible total score for Xyla will be 25, and that for Wangdu is 12(4+4+4). (12*2+1=25) 

Since Xyla already has non-multiple of 3 as total score. Zeneca will have 24 as the total score. The complete table is:

The highest score is 25.

It is given that every bull’s eye score in the first three rounds gave a player one additional chance to shoot in the bonus rounds, Rounds 4 to 6, which means Tanzi scored Bull's eye only once in the first 3 rounds because she participated only once in round 4 to 6. Similarly, Umeza scored Bull's eye exactly 2 times in the first 3 rounds. Wangdu did not score Bull's eye in the first three rounds and so on. 

Now from 1, Tanzi, Umeza and Yonita had the same total score.

So, Total score of Tanzi will be 4+5+5+a=14+a,  (She scored Bull's eye(a score of 5) in exactly one round and a is the unknown score)

Total score of Umeza = 1+2+5+5+b = 13+b    (She scored Bull's eye(a score of 5) in exactly 2 rounds and b is the unknown score)

Total score of Yonita = 3+5+5+c=13+c (She scored Bull's eye(a score of 5) in exactly one round and c is the unknown score)

Now 14+a=13+b=13+c,

Also it is given that total scores for all players, except one, were in multiples of three, so these three will have to be a multiple of 3.

So, (a,b,c) can be either (1,2,2) or (4,5,5) in the same order. But the value (5,5) for b and c is not possible.  (Umeza scored Bull's eye in exactly 2 rounds and Yonita in exactly 1 round)

Hence, a=1,b=2 and c=2. So each of Tanzi, Umeza and Yonita had total score of 15.

Tabulating the data, we have

From 5, Tanzi and Zeneca had the same score in Round 1 but different scores in Round 3.

Zeneca score Bull's eye 2 times in round 1 to 3. If Tanzi scored 1 in round 1, then Zeneca also has to score 1 in round 1, which means both Tanzi and Zeneca scores in round 3 will be 5, which violates 5. Hence Tanzi scored 5 in round 1 and Zeneca also scored the same in round 1.So the new table is:

From 4, the number of players hitting bull’s eye in Round 2 was double of that in Round 3.

So, in round 3 either 1 or 2 Bull's eye can be scored and in round 2, 2 or 4 Bull's eye can be scored.

Case 1: If only 1 Bull's eye is scored in the round 3, then in round 3 Umeza will score 2 and Zeneca will score 2/3/4 in round 3, which means both will score 5 in round 2. So minimum Bull's eye in round 2 will be 3. (Umeza, Zeneca and Xyla)

Hence this case is rejected.

Case 2: 2 Bull's eye were scored in round 3 and 4 Bull's eye were scored in round 2. So in round 2 Umeza, Yonita and Zeneca scored 5. This can be tabulated as:

In round 3, 2 Bull's eye can only be scored by Xyla and Umeza.

The highest scorer can be either Xyla or Zeneca. The lowest scorer will be Wangdu.

1.Consider Zeneca is the highest scorer. 

From 3, the highest total score was one more than double of the lowest total score. So the only possible score for Zeneca is 23 and that for Wangdu is 11. (11*2+1=23)

But this will violate condition 2, since both Zeneca and Wangdu do not have their scores as multiples of three in this case.

Hence, Xyla will be the highest scorer. The only possible total score for Xyla will be 25, and that for Wangdu is 12(4+4+4). (12*2+1=25) 

Since Xyla already has non-multiple of 3 as total score. Zeneca will have 24 as the total score. The complete table is:

Zeneca total score is 24.

It is given that every bull’s eye score in the first three rounds gave a player one additional chance to shoot in the bonus rounds, Rounds 4 to 6, which means Tanzi scored Bull's eye only once in the first 3 rounds because she participated only once in round 4 to 6. Similarly, Umeza scored Bull's eye exactly 2 times in the first 3 rounds. Wangdu did not score Bull's eye in the first three rounds and so on. 

Now from 1, Tanzi, Umeza and Yonita had the same total score.

So, Total score of Tanzi will be 4+5+5+a=14+a,  (She scored Bull's eye(a score of 5) in exactly one round and a is the unknown score)

Total score of Umeza = 1+2+5+5+b = 13+b    (She scored Bull's eye(a score of 5) in exactly 2 rounds and b is the unknown score)

Total score of Yonita = 3+5+5+c=13+c (She scored Bull's eye(a score of 5) in exactly one round and c is the unknown score)

Now 14+a=13+b=13+c,

Also it is given that total scores for all players, except one, were in multiples of three, so these three will have to be a multiple of 3.

So, (a,b,c) can be either (1,2,2) or (4,5,5) in the same order. But the value (5,5) for b and c is not possible.  (Umeza scored Bull's eye in exactly 2 rounds and Yonita in exactly 1 round)

Hence, a=1,b=2 and c=2. So each of Tanzi, Umeza and Yonita had total score of 15.

Tabulating the data, we have

From 5, Tanzi and Zeneca had the same score in Round 1 but different scores in Round 3.

Zeneca score Bull's eye 2 times in round 1 to 3. If Tanzi scored 1 in round 1, then Zeneca also has to score 1 in round 1, which means both Tanzi and Zeneca scores in round 3 will be 5, which violates 5. Hence Tanzi scored 5 in round 1 and Zeneca also scored the same in round 1.So the new table is:

From 4, the number of players hitting bull’s eye in Round 2 was double of that in Round 3.

So, in round 3 either 1 or 2 Bull's eye can be scored and in round 2, 2 or 4 Bull's eye can be scored.

Case 1: If only 1 Bull's eye is scored in the round 3, then in round 3 Umeza will score 2 and Zeneca will score 2/3/4 in round 3, which means both will score 5 in round 2. So minimum Bull's eye in round 2 will be 3. (Umeza, Zeneca and Xyla)

Hence this case is rejected.

Case 2: 2 Bull's eye were scored in round 3 and 4 Bull's eye were scored in round 2. So in round 2 Umeza, Yonita and Zeneca scored 5. This can be tabulated as:

In round 3, 2 Bull's eye can only be scored by Xyla and Umeza.

The highest scorer can be either Xyla or Zeneca. The lowest scorer will be Wangdu.

1.Consider Zeneca is the highest scorer. 

From 3, the highest total score was one more than double of the lowest total score. So the only possible score for Zeneca is 23 and that for Wangdu is 11. (11*2+1=23)

But this will violate condition 2, since both Zeneca and Wangdu do not have their scores as multiples of three in this case.

Hence, Xyla will be the highest scorer. The only possible total score for Xyla will be 25, and that for Wangdu is 12(4+4+4). (12*2+1=25) 

Since Xyla already has non-multiple of 3 as total score. Zeneca will have 24 as the total score. The complete table is:

Xyla was the highest scorer.

It is given that every bull’s eye score in the first three rounds gave a player one additional chance to shoot in the bonus rounds, Rounds 4 to 6, which means Tanzi scored Bull's eye only once in the first 3 rounds because she participated only once in round 4 to 6. Similarly, Umeza scored Bull's eye exactly 2 times in the first 3 rounds. Wangdu did not score Bull's eye in the first three rounds and so on. 

Now from 1, Tanzi, Umeza and Yonita had the same total score.

So, Total score of Tanzi will be 4+5+5+a=14+a,  (She scored Bull's eye(a score of 5) in exactly one round and a is the unknown score)

Total score of Umeza = 1+2+5+5+b = 13+b    (She scored Bull's eye(a score of 5) in exactly 2 rounds and b is the unknown score)

Total score of Yonita = 3+5+5+c=13+c (She scored Bull's eye(a score of 5) in exactly one round and c is the unknown score)

Now 14+a=13+b=13+c,

Also it is given that total scores for all players, except one, were in multiples of three, so these three will have to be a multiple of 3.

So, (a,b,c) can be either (1,2,2) or (4,5,5) in the same order. But the value (5,5) for b and c is not possible.  (Umeza scored Bull's eye in exactly 2 rounds and Yonita in exactly 1 round)

Hence, a=1,b=2 and c=2. So each of Tanzi, Umeza and Yonita had total score of 15.

Tabulating the data, we have

From 5, Tanzi and Zeneca had the same score in Round 1 but different scores in Round 3.

Zeneca score Bull's eye 2 times in round 1 to 3. If Tanzi scored 1 in round 1, then Zeneca also has to score 1 in round 1, which means both Tanzi and Zeneca scores in round 3 will be 5, which violates 5. Hence Tanzi scored 5 in round 1 and Zeneca also scored the same in round 1.So the new table is:

From 4, the number of players hitting bull’s eye in Round 2 was double of that in Round 3.

So, in round 3 either 1 or 2 Bull's eye can be scored and in round 2, 2 or 4 Bull's eye can be scored.

Case 1: If only 1 Bull's eye is scored in the round 3, then in round 3 Umeza will score 2 and Zeneca will score 2/3/4 in round 3, which means both will score 5 in round 2. So minimum Bull's eye in round 2 will be 3. (Umeza, Zeneca and Xyla)

Hence this case is rejected.

Case 2: 2 Bull's eye were scored in round 3 and 4 Bull's eye were scored in round 2. So in round 2 Umeza, Yonita and Zeneca scored 5. This can be tabulated as:

In round 3, 2 Bull's eye can only be scored by Xyla and Umeza.

The highest scorer can be either Xyla or Zeneca. The lowest scorer will be Wangdu.

1.Consider Zeneca is the highest scorer. 

From 3, the highest total score was one more than double of the lowest total score. So the only possible score for Zeneca is 23 and that for Wangdu is 11. (11*2+1=23)

But this will violate condition 2, since both Zeneca and Wangdu do not have their scores as multiples of three in this case.

Hence, Xyla will be the highest scorer. The only possible total score for Xyla will be 25, and that for Wangdu is 12(4+4+4). (12*2+1=25) 

Since Xyla already has non-multiple of 3 as total score. Zeneca will have 24 as the total score. The complete table is:

Tanzi scored 1 in round 3.

The value of F can only be 0 as F+F=F can only hold if F=0.

Also, A can only be 1(in the second column) because to get a carry of more than 1, B has to be a double-digit number which is not possible. (A carry is a digit that is transferred from one column of digits to another column of more significant digits.)

So the data can be tabulated as follows:

Since the last row in the third column is 0, the carry to the second column must have been 1, Hence B+1+1=11  => B=9

In the 4th column, H+H = 10 since a carry 1 has gone to the 3rd column. Hence H=5.

G+K must be 11 and the carry 1 goes to the next column, so C=1+1=2.

Now, G,K can take values (3,8), (4,7) and (5,6) in any order.

From 5th column G=J+1  => J=G-1

Case: G=3 and K=8, here J =2 which is not possible as C =2

Case: G=8 and K=3, J=7, a possible case.

Case: G=4 and K=7, J=3 possible

Case: G=7 and K=4, J=6 possible

Case: G=5 and K=6, J=4 not possible as H =5.

Case: G=6 and K=5, J=5 both J and K are same, not possible.

Hence the cases can be tabulated as follows:

The letter A represents 1.

The value of F can only be 0 as F+F=F can only hold if F=0.

Also, A can only be 1(in the second column) because to get a carry of more than 1, B has to be a double-digit number which is not possible. (A carry is a digit that is transferred from one column of digits to another column of more significant digits.)

So the data can be tabulated as follows:

Since the last row in the third column is 0, the carry to the second column must have been 1, Hence B+1+1=11  => B=9

In the 4th column, H+H = 10 since a carry 1 has gone to the 3rd column. Hence H=5.

G+K must be 11 and the carry 1 goes to the next column, so C=1+1=2.

Now, G,K can take values (3,8), (4,7) and (5,6) in any order.

From 5th column G=J+1  => J=G-1

Case: G=3 and K=8, here J =2 which is not possible as C =2

Case: G=8 and K=3, J=7, a possible case.

Case: G=4 and K=7, J=3 possible

Case: G=7 and K=4, J=6 possible

Case: G=5 and K=6, J=4 not possible as H =5.

Case: G=6 and K=5, J=5 both J and K are same, not possible.

Hence the cases can be tabulated as follows:

The letter B represents 9.

The value of F can only be 0 as F+F=F can only hold if F=0.

Also, A can only be 1(in the second column) because to get a carry of more than 1, B has to be a double-digit number which is not possible. (A carry is a digit that is transferred from one column of digits to another column of more significant digits.)

So the data can be tabulated as follows:

Since the last row in the third column is 0, the carry to the second column must have been 1, Hence B+1+1=11  => B=9

In the 4th column, H+H = 10 since a carry 1 has gone to the 3rd column. Hence H=5.

G+K must be 11 and the carry 1 goes to the next column, so C=1+1=2.

Now, G,K can take values (3,8), (4,7) and (5,6) in any order.

From 5th column G=J+1  => J=G-1

Case: G=3 and K=8, here J =2 which is not possible as C =2

Case: G=8 and K=3, J=7, a possible case.

Case: G=4 and K=7, J=3 possible

Case: G=7 and K=4, J=6 possible

Case: G=5 and K=6, J=4 not possible as H =5.

Case: G=6 and K=5, J=5 both J and K are same, not possible.

Hence the cases can be tabulated as follows:

In all possible cases 7 is already represented by a letter other than D. Hence 7 is the answer.

The value of F can only be 0 as F+F=F can only hold if F=0.

Also, A can only be 1(in the second column) because to get a carry of more than 1, B has to be a double-digit number which is not possible. (A carry is a digit that is transferred from one column of digits to another column of more significant digits.)

So the data can be tabulated as follows:

Since the last row in the third column is 0, the carry to the second column must have been 1, Hence B+1+1=11  => B=9

In the 4th column, H+H = 10 since a carry 1 has gone to the 3rd column. Hence H=5.

G+K must be 11 and the carry 1 goes to the next column, so C=1+1=2.

Now, G,K can take values (3,8), (4,7) and (5,6) in any order.

From 5th column G=J+1  => J=G-1

Case: G=3 and K=8, here J =2 which is not possible as C =2

Case: G=8 and K=3, J=7, a possible case.

Case: G=4 and K=7, J=3 possible

Case: G=7 and K=4, J=6 possible

Case: G=5 and K=6, J=4 not possible as H =5.

Case: G=6 and K=5, J=5 both J and K are same, not possible.

Hence the cases can be tabulated as follows:

From the table it is clear that 6 cannot be represented by G.

The data can be tabulated as follows(approximately):

Customer Services: 28,41,50,55,70 (The median is 50)

Cost: 50,71,77,81,90 (The median is 77)

Reliability: 26, 40, 52, 60, 75 (The median is 52)

Quality: 40, 48, 62, 69, 72 (The median is 62)

Features: 40, 45, 56, 75, 90 (The median is 56)

Reach: 46, 58, 63, 70, 80 (The median is 63)

Hence the customer services has the lowest median.

The data can be tabulated as follows(approximately):

The average of the vendor will be highest which has highest total score. Hence vendor 3 has the highest average.

The data can be tabulated as follows(approximately):

Top 3 on Reliability: Vendor 3, Vendor 5

Top 3 on Reach: Vendor 1, Vendor 5

Top 3 on Quality: Vendor 1, Vendor 2

Top 3 on Features: Vendor 4, Vendor 5

Top 3 on Customer Services: Vendor 4, Vendor 1

Top 3 on Cost: Vendor 3, Vendor 2

Vendor 1: 3 times         Vendor 2: Only once           Vendor 3: 2 times              Vendor 4:  2 times                    Vendor 5: 3 times

Here 1 and 5 comes 3 times. Hence B is the answer.

The data can be tabulated as follows(approximately): 

Top 3 on Reliability: Vendor 3, Vendor 5, Vendor 1

Top 3 on Reach: Vendor 1, Vendor 5, Vendor 3

Top 3 on Quality: Vendor 1, Vendor 2, Vendor 3

Top 3 on Features: Vendor 4, Vendor 5, Vendor 3

Top 3 on Customer Services: Vendor 4, Vendor 1, Vendor 3

Top 3 on Cost: Vendor 3,  Vendor 2, Vendor 1

Only Vendor 3 ranks among top 3 in all the six parameters.

The data can be tabulated as follows(approximately):

Rank of Delhi in IPC crimes category = 1, The rank of Karnataka and Maharashtra is 3(from table), then the rank of Goa can only be 2.

The rank of Telangana is 6 which has less |IPC crimes than Kerala, which means the rank of Kerala can be less than or equal to 5.

Now, there are two states with 3 ranks, so there will be no rank 4, there can only be rank 5 which is Kerala.

The data can be tabulated as follows(approximately):

The highest cases are registered in West Bengal and Delhi.

The total number of IPC crimes = 63-64

The total number of SLL crimes = 520+35-36 = 555-556

Hence the ratio = (63-64)/(555-556)  = 0.11  (Approximately)   = 1:9

The data can be tabulated as follows(approximately):

From the table, the rank of Tamilnadu in other crimes is 2. The states which are not in the table will have crimes less than Telangana(i.e 24-25)

From the table the rank of Pudducherry in other crimes is 3.

The data can be tabulated as follows(approximately): 

The data can be tabulated as follows(approximately):

The rank of Delhi in IPC crimes should be 1 because the states which are not in table cannot crime more than that of Telangana which is 24-25.

Similarly Delhi Rank in Other crimes will be 1.

Now in SLL crimes clearly West Bengal has rank 1. It is given that Karnataka has rank 2. The rank 3 can go to either Goa, Delhi and Maharashtra but Goa and Maharashtra already have rank 4. So Delhi will have rank 3. Also no state outside of the table can be ranked 3 in SLL crimes as maximum number of crime should be less than that of Telangana(24-25). Here the number of SLL crimes is 35-36.

Hence the sum of the ranks = 1+3+1=5

From 1, X, U, and Z are standing at the three corners of a triangle formed by three street segments.

From 2, X can see only U and Z.

From 4, U sees V standing in the next intersection behind Z. Also, no one among the six is standing at intersection d.

Only cases possible are:

1.

W cannot see V or Z. So W can only be at the intersection a. Since Y can see only U and W, Y can only be at c where X can see him. Hence this case is rejected.

2.

Y can only see U and W. Y cannot be placed anywhere. Hence this case is also rejected.

3.

Y can only see U and W. Y cannot be placed anywhere. Hence this case is also rejected.

4.

W cannot see V or Z. W can only be placed at i. Y can see only U and W. Y can only be placed at j or e, where he can see more people than U and W. Hence this case is also rejected.

5.

W cannot see V or Z. Y can only see U and W. Hence W and Y can only be placed as shown:

No one is standing at the intersection A. Hence C is the answer.

From 1, X, U, and Z are standing at the three corners of a triangle formed by three street segments.

From 2, X can see only U and Z.

From 4, U sees V standing in the next intersection behind Z. Also, no one among the six is standing at intersection d.

Only cases possible are:

1.

W cannot see V or Z. So W can only be at the intersection a. Since Y can see only U and W, Y can only be at c where X can see him. Hence this case is rejected.

2.

Y can only see U and W. Y cannot be placed anywhere. Hence this case is also rejected.

3.

Y can only see U and W. Y cannot be placed anywhere. Hence this case is also rejected.

4.

W cannot see V or Z. W can only be placed at i. Y can see only U and W. Y can only be placed at j or e, where he can see more people than U and W. Hence this case is also rejected.

5.

W cannot see V or Z. Y can only see U and W. Hence W and Y can only be placed as shown:

V can see U and Z only. Hence C is the answer.

From 1, X, U, and Z are standing at the three corners of a triangle formed by three street segments.

From 2, X can see only U and Z.

From 4, U sees V standing in the next intersection behind Z. Also, no one among the six is standing at intersection d.

Only cases possible are:

1.

W cannot see V or Z. So W can only be at the intersection a. Since Y can see only U and W, Y can only be at c where X can see him. Hence this case is rejected.

2.

Y can only see U and W. Y cannot be placed anywhere. Hence this case is also rejected.

3.

Y can only see U and W. Y cannot be placed anywhere. Hence this case is also rejected.

4.

W cannot see V or Z. W can only be placed at i. Y can see only U and W. Y can only be placed at j or e, where he can see more people than U and W. Hence this case is also rejected.

5.

W cannot see V or Z. Y can only see U and W. Hence W and Y can only be placed as shown:

To reach Y, X has to go from b to g and g to k, i.e. 2 streets.

From 1, X, U, and Z are standing at the three corners of a triangle formed by three street segments.

From 2, X can see only U and Z.

From 4, U sees V standing in the next intersection behind Z. Also, no one among the six is standing at intersection d.

Only cases possible are:

1.

W cannot see V or Z. So W can only be at the intersection a. Since Y can see only U and W, Y can only be at c where X can see him. Hence this case is rejected.

2.

Y can only see U and W. Y cannot be placed anywhere. Hence this case is also rejected.

3.

Y can only see U and W. Y cannot be placed anywhere. Hence this case is also rejected.

4.

W cannot see V or Z. W can only be placed at i. Y can see only U and W. Y can only be placed at j or e, where he can see more people than U and W. Hence this case is also rejected.

5.

W cannot see V or Z. Y can only see U and W. Hence W and Y can only be placed as shown:

If a new person stands at d(left down corner), they can see W and X only. Hence A is the answer.

Since the dancers performed their second items in the same sequence of their performance of their first items. The table will be as follows:

The items assigned by Ashman were performed consecutively. The number of performances between items assigned by each of the remaining composers was the same.

Also, the first items performed by the four dancers were all assigned by different music composers. Badal can come only at the place as shown in the table.

Then Ashman can only compose for the following performances.

Hence Dyu will compose for the following performances:

From (i) No composer who assigned item to Princess, assigned any item to Queen.
From (ii) No composer who assigned item to Rani, assigned any item to Samragni.

Hence Dyu will compose for Samragni 1st Performance and Gagan will compose for Queen 1st Performance. Also, Badal will compose for Samragni 2nd Performance and Dyu will compose for Queens 2nd Performance.

Hence, the complete table is as follows:

The second performance was composed by Dyu. Hence A is the answer.

Since the dancers performed their second items in the same sequence of their performance of their first items. The table will be as follows:

The items assigned by Ashman were performed consecutively. The number of performances between items assigned by each of the remaining composers was the same.

Also, the first items performed by the four dancers were all assigned by different music composers. Badal can come only at the place as shown in the table.

Then Ashman can only compose for the following performances.

Hence Dyu will compose for the following performances:

From (i) No composer who assigned item to Princess, assigned any item to Queen.
From (ii) No composer who assigned item to Rani, assigned any item to Samragni.

Hence Dyu will compose for Samragni 1st Performance and Gagan will compose for Queen 1st Performance. Also, Badal will compose for Samragni 2nd Performance and Dyu will compose for Queens 2nd Performance.

Hence, the complete table is as follows:

Option A: Samragni did not perform in any item composed by Ashman. This statement is true.

Option B: Princess did not perform in any item composed by Dyu. This is also true.

Option C: Rani did not perform in any item composed by Badal. This statement is true.

Option D: Queen did not perform in any item composed by Gagan. This statement is false.

Hence D is the answer.

Since the dancers performed their second items in the same sequence of their performance of their first items. The table will be as follows:

The items assigned by Ashman were performed consecutively. The number of performances between items assigned by each of the remaining composers was the same.

Also, the first items performed by the four dancers were all assigned by different music composers. Badal can come only at the place as shown in the table.

Then Ashman can only compose for the following performances.

Hence Dyu will compose for the following performances:

From (i) No composer who assigned item to Princess, assigned any item to Queen.
From (ii) No composer who assigned item to Rani, assigned any item to Samragni.

Hence Dyu will compose for Samragni 1st Performance and Gagan will compose for Queen 1st Performance. Also, Badal will compose for Samragni 2nd Performance and Dyu will compose for Queens 2nd Performance.

Hence, the complete table is as follows:

The sixth performance was composed by Badal. Hence C is the answer.

Since the dancers performed their second items in the same sequence of their performance of their first items. The table will be as follows:

The items assigned by Ashman were performed consecutively. The number of performances between items assigned by each of the remaining composers was the same.

Also, the first items performed by the four dancers were all assigned by different music composers. Badal can come only at the place as shown in the table.

Then Ashman can only compose for the following performances.

Hence Dyu will compose for the following performances:

From (i) No composer who assigned item to Princess, assigned any item to Queen.
From (ii) No composer who assigned item to Rani, assigned any item to Samragni.

Hence Dyu will compose for Samragni 1st Performance and Gagan will compose for Queen 1st Performance. Also, Badal will compose for Samragni 2nd Performance and Dyu will compose for Queens 2nd Performance.

Hence, the complete table is as follows:

The first and the sixth items were composed by Badal. Hence D is the answer.

It is given that the most expensive item is a diamond ring of type a and there is exactly one of these. Since the item b should be at least twice. The minimum number of items will be obtained when a=1 and b=99, which means there are only two different types of items.

It is given that the most expensive item is a diamond ring of type a and there is exactly one of these. Since the number of items of type b should be at least twice of that of a and the number of items of type c should be at least twice of that of b and so on. So the maximum number of different types of items of a, b and c will be obtained when a=1, b=2, c=4, d=8, e=16, f=69. Hence the maximum number of different types of items will be 6.

If the number of items is 7, then the minimum number of prizes should be 1+2+4+8+16+32+64=127 which is more than 100.

Hence 6 is the answer.

Option A: There are exactly 75 items of type e.

a=1,b=2,c=4,d=8, e=85. Here the maximum value of e= 85. Hence it can take the value 75.

An example of such case is a=1,b=2,c=4,d=18, e=75

Option B: There are exactly 30 items of type b.

a=1 b=30 and c=69. Hence this case is also possible.

Option C: There are exactly 45 items of type c.

Since the value of d should be at least 90, it means that d is not present because 45+90 will be more than 100(maximum number of items). Only a,b and c are present.

The maximum value of b = 22 and a =1, but 45+22+1=68, which is less than 100. So this case is not possible.

Option D: There are exactly 60 items of type d.

d=60, c=30, b=9 and a=1. a+b+c+d=100. Hence this case is possible.

C is the answer.

The total number of items from 1 to 100, which are of same type as in box 45 = 31+1+43=75

Now to maximize the number of items, a=1, b=2, c=4, d=18 and e=75(given)

There can be maximum 5 types of items.

If we consider number of items to be 6, then minimum number of items of 5th type will be 16, 1+2+4+8+16+75=106 which is more than 100.

Let the speed of cars be a and b and the distance =d

Minimum time taken by 1st car = 6 hours,

For maximum difference in time taken by both of them, car 1 has to start at 10:00 AM and car 2 has to start at 11:00 AM. 

Hence, car 2 will take 5 hours. 

Hence a= $$\ \frac{\ d}{6}$$ and b = $$\ \frac{\ d}{5}$$

Hence the speed of car 2 will exceed the speed of car 1 by $$\ \dfrac{\ \ \frac{\ d}{5}-\ \frac{\ d}{6}}{\ \frac{\ d}{6}}\times\ 100$$ = $$\ \dfrac{\ \ \frac{\ d}{30}}{\ \frac{\ d}{6}}\times\ 100$$ = 20

We have, $$\frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + ....... + \frac{1}{\sqrt{a_n} + \sqrt{a_{n + 1}}}$$

Now, $$\frac{1}{\sqrt{a_1} + \sqrt{a_2}}$$ = $$\frac{\sqrt{a_2} - \sqrt{a_1}}{(\sqrt{a_2} + \sqrt{a_1})(\sqrt{a_2} - \sqrt{a_1})}$$   (Multiplying numerator and denominator by $$\sqrt{a_2} - \sqrt{a_1}$$)

= $$\frac{\sqrt{a_2} - \sqrt{a_1}}{({a_2} - {a_1}}$$

=$$\frac{\sqrt{a_2} - \sqrt{a_1}}{d}$$   (where d is the common difference)

Similarly, $$ \frac{1}{\sqrt{a_2} + \sqrt{a_3}}$$ = $$\frac{\sqrt{a_3} - \sqrt{a_2}}{d}$$ and so on.

Then the expression $$\frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + ....... + \frac{1}{\sqrt{a_n} + \sqrt{a_{n + 1}}}$$

can be written as $$\ \frac{\ 1}{d}(\sqrt{a_2}-\sqrt{a_1}+\sqrt{a_3}-\sqrt{a_3}+..........................\sqrt{a_{n+1}} - \sqrt{a_{n}}$$

= $$\ \frac{\ n}{nd}(\sqrt{a_{n+1}}-\sqrt{a_1})$$ (Multiplying both numerator and denominator by n)

= $$\ \frac{n(\sqrt{a_{n+1}}-\sqrt{a_1})}{{a_{n+1}} - {a_1}}$$     $$(a_{n+1} - {a_1} =nd)$$

= $$\frac{n}{\sqrt{a_1} + \sqrt{a_{n + 1}}}$$

Since AB is a diameter,  AQB and APB will right angles.

In right triangle APB, AP = $$\sqrt{10^2-6^2}=8$$

Now, 2AQ=AP  => AQ= 8/2=4

In right triangle AQB, AP = $$\sqrt{10^2-4^2}=9.165$$ =9.1 (Approx)

We have, $$(5.55)^x = (0.555)^y = 1000$$

Taking log in base 10 on both sides,

x($$\log_{10}555$$-2) = y($$\log_{10}555$$-3) = 3

Then, x($$\log_{10}555$$-2) = 3.....(1)

y($$\log_{10}555$$-3) = 3 .....(2)

From (1) and (2)

=> $$\log_{10}555$$=$$\ \frac{\ 3}{x}$$+2=$$\ \frac{\ 3}{y}+3$$

=> $$\frac{1}{x} - \frac{1}{y}$$ = $$\frac{1}{3}$$

Assuming the income of Bimla = 100a, then the income of Amala will be 120a.

And the income of Kamala will be 120a*100/80=150a

If Kamala's income goes down by 4%, then new income of Kamala = 150a-150a(4/100) = 150a-6a=144a

If Bimla's income goes up by 10 percent, her new income will be 100a+100a(10/100)=110a

=> Hence the Kamala income will exceed Bimla income by (144a-110a)*100/110a=31

Distance covered by A in 1 revolution = 2$$\pi\ $$*30 = 60$$\pi\ $$

Distance covered by B in 1 revolution = 2$$\pi\ $$*40 = 80$$\pi\ $$

Now, (5000+n)60$$\pi\ $$ = 80$$\pi\ $$n

=> 15000= 4n-3n   =>n=15000

Then distance travelled by B = 15000*80$$\pi\ $$ cm = 12$$\pi\ $$ km

Hence, the speed = $$\ \frac{\ 12\pi\times\ 60\ }{45}$$ = 16$$\pi\ $$

We have, $$\mid x^2 - x - 6 \mid = x + 2$$

=> |(x-3)(x+2)|=x+2

For x<-2, (3-x)(-x-2)=x+2

=> x-3=1   =>x=4 (Rejected as x<-2)

For -2$$\le\ $$x<3, (3-x)(x+2)=x+2    =>x=2,-2

For x$$\ge\ $$3, (x-3)(x+2)=x+2   =>x=4

Hence the product =4*-2*2=-16

Assuming the length of race course = x and the speed of three horses be a,b and c respectively.

Hence, $$\ \frac{\ x}{a}=\ \frac{\ x-11}{b}$$......(1)

and $$\ \frac{\ x}{a}=\ \frac{\ x-90}{c}$$......(2)

Also, $$\ \frac{\ x}{b}=\ \frac{\ x-80}{c}$$......(3)

From 1 and 2, we get, $$\ \frac{\ x-11}{b}=\ \frac{\ x-90}{c}$$ .....(4)

Dividing (3) by (4), we get, $$\ \frac{\ x-11}{x}=\ \frac{\ x-90}{x-80}$$  

=> (x-11)(x-80)=x(x-90)

=> 91x-90x=880  => x=880

The population of town at the beginning of 1st year = p

The population of town at the beginning of 2nd year = 3+2p

The population of town at the beginning of 3rd year = 2(3+2p)+3 = 2*2p+2*3+3 =4p+3(1+2)

The population of town at the beginning of 4th year = 2(2*2p+2*3+3)+3 = 8p+3(1+2+4)

Similarly population at the beginning of the nth year = $$2^{n-1}$$p+3($$2^{n-1}-1$$) = $$2^{n-1}\left(p+3\right)$$-3 

The population in the beginning of 2019 is 1000, then the population in the beginning of 2034 will be $$(2^{2034-2019})\left(1000+3\right)$$-3 = $$2^{15}\left(1003\right)$$-3

f (x + y) = f (x) f (y)

Hence, f(2)=f(1+1)=f(1)*f(1)=2*2=4

f(3)=f(2+1)=f(2)*f(1)=4*2=8

f(4)=f(3+1)=f(3)*f(1)=8*2=16

.......=> f(x)=$$2^x$$

Now, f(a + 1) +f (a + 2) + ... + f(a + n) = 16 (2$$^n$$ - 1)

On putting n=1 in the equation we get, f(a+1)=16   => f(a)*f(1)=16  (It is given that f (x + y) = f (x) f (y))

=> $$2^a$$*2=16

=> a=3

Assuming the investment of Amala, Bina, and Gouri be 300x, 400x and 500x, hence the interest incomes will be 300x*6/100=18x, 400x*5/100=20x and 500x*4/100 = 20x

Given, Bina's interest income exceeds Amala by 20x-18x=2x=250  => x=125

Now, total interest income = 18x+20x+20x=58x = 58*125 = 7250

Assuming m is even, then 8f(m+1)-f(m)=2

m+1 will be odd

So, 8(m+1+3)-m(m+1)=2

=> 8m+32-$$m^2-m$$=2

=> $$m^2-7m-30=0$$

=> m=10,-3 

Rejecting the negative value, we get m=10

Assuming m is odd, m+1 will be even.

then, 8(m+1)(m+2)-m-3=2

=> 8($$m^2+3m+2$$)-m-3=2

=> $$8m^2+23m+11=0$$

Solving this, m = -2.26 and -0.60

Hence, the value of m is not integral. Hence this case will be rejected.

Assume the numbers are a and b, then ab=616

We have, $$\ \ \frac{\ a^3-b^3}{\left(a-b\right)^3}$$ = $$\ \frac{\ 157}{3}$$

=> $$\ 3\left(a^3-b^3\right)\ =\ 157\left(a^3-b^3+3ab\left(b-a\right)\right)$$

=> $$154\left(a^3-b^3\right)+3*157*ab\left(b-a\right)$$ = 0

=> $$154\left(a^3-b^3\right)+3*616*157\left(b-a\right)$$ = 0        (ab=616)

=>$$a^3-b^3+\left(3\times\ 4\times\ 157\left(b-a\right)\right)$$    (154*4=616)

=> $$\left(a-b\right)\left(a^2+b^2+ab\right)\ =\ 3\times\ 4\times\ 157\left(a-b\right)$$

=> $$a^2+b^2+ab\ =\ 3\times\ 4\times\ 157$$

Adding ab=616 on both sides, we get

$$a^2+b^2+ab\ +ab=\ 3\times\ 4\times\ 157+616$$

=> $$\left(a+b\right)^2=\ 3\times\ 4\times\ 157+616$$ = 2500

=> a+b=50

Assume the total distance between A and B as d and time taken by Amal = t

Since Amal travelled $$\frac{1}{3}^{rd}$$ of his total journey time in different speeds
d = $$\ \frac{\ t}{3}\times\ 10+\ \frac{\ t}{3}\times\ 20+\frac{\ t}{3}\times\ 30\ \ =\ 20t$$

$$\text{Total time taken by Bimal} = \ \frac{d_1}{s_1}+\frac{d_2}{s_2}+\frac{d_3}{s_3}$$

$$=\ \frac{20t}{3}\times\ \frac{1}{10}+\frac{20t}{3}\times\ \frac{1}{20}+\frac{20t}{3}\times\ \frac{1}{30}\ \ =\frac{20t\left(6+3+2\right)}{3\ \times30}\ =\frac{11}{9}t$$

Hence, the ratio of time taken by Bimal to time taken by Amal = $$\frac{\frac{11t}{9}}{t}=\frac{11}{9}$$
Therefore, Bimal will exceed Amal's time by $$\ \ \ \frac{\ \ \frac{\ 11t}{9}-t}{t}\times\ 100 = 22.22%$$

Assuming the maximum marks =100a, then Meena got 40a

After increasing her score by 50%, she will get 40a(1+50/100)=60a

Passing score = 60a+35

Post review score after 20% increase = 60a*1.2=72a

=>Hence, 60a+35+7=72a

=>12a=42   =>a=3.5

=> maximum marks = 350 and passing marks = 210+35=245

=> Passing percentage = 245*100/350 = 70

Assuming the amount invested in the ratio 2:1 was 200x and 100x, then the fixed deposit investment = 1500000-300x

Hence, the interest = 200x*4/100 = 8x and 100x*3/100=3x

Interest from the fixed deposit = (1500000-300x)*6/100 = 90000-18x

Hence the total interest  = 90000-18x+8x+3x=90000-7x =76000

=> 7x=14000   => x=2000

Hence, the fixed deposit investment = 1500000-300*2000 = 900000 = 9 lakhs

Assume the number of members who can play exactly 1 game = I

The number of members who can play exactly 2 game = II

The number of members who can play exactly 3 game = III

I+2II+3III=144+123+132=399....(1)

I+II+III=256......(2)

subtracting (1) from (2), we get

=> II+2III=143.....(3)

Also, II+3III=58+25+63=146 ......(4)

subtracting (3) from (4), we get

=> III = 3 (From 3 and 4)

=> II =137

=> I = 116

The members who play only tennis = 123-58-25+3 = 43

11th term of series = $$a_{11}$$ = Sum of 11 terms - Sum of 10 terms = $$3(2^{11 + 1} - 2)$$-3$$(2^{10 + 1} - 2)$$ 

= 3$$(2^{12} - 2-2^{11} +2)$$=3$$(2^{11})(2-1)$$= 3*$$2^{11}$$ = 6144

$$2 \cos (x(x + 1)) = 2^x + 2^{-x}$$

The maximum value of LHS is 2 when $$\cos (x(x + 1))$$ is 1 and the minimum value of RHS is 2 using AM $$\geq$$ GM 

Hence LHS and RHS can only be equal when both sides are 2. For LHS, cosx(x+1)=1   => x(x+1)=0   => x=0,-1

For RHS minimum value, x=0

Hence only one solution x=0

Assuming A completes a units of work in a day and B completes B units of work in a day and the total work = 1 unit

Hence, 12(a+b)=1.........(1)

Also, 9($$\ \frac{\ a}{2}$$+3b)=1  .........(2)

Using both equations, we get, 12(a+b)= 9($$\ \frac{\ a}{2}$$+3b)

=> 4a+4b=$$\ \frac{\ 3a}{2}$$+9b

=> $$\ \frac{\ 5a}{2}$$=5b

=> a=2b

Substituting the value of b in equation (1),

12($$\ \frac{\ 3a}{2}$$)=1

=> a=$$\ \frac{\ 1}{18}$$

Hence, the number of days required = 1/($$\ \frac{\ 1}{18}$$)=18

Assuming the number of students =100x

Hence, the number of girls = 60x and the number of boys = 40x

We have, 60x-40x=30  => x=1.5

The number of girls = 60*1.5=90

Number of girls that pass = 68x-30=68*1.5-30 = 102-30=72

The number of girls who do not pass = 90-72=18

Hence the percentage of girls who do not pass = 1800/90=20

In figure AE*BE=CE*DE    (The   intersecting chords theorem)

=> 7*15= x(20.5-x)                  (Assuming AE=x)

=> 210=x(41-2x)

=> $$2x^2$$-41x+210=0

=> x=10 or x=10.5   => AE=10 or AE=10.5      Hence BE = 20.5-10=10.5  or BE = 20.5-10.5=10

Required difference= 10.5-10=0.5

Assuming the cost price of pen = 100p and the cost price of book = 100b

So, on selling a pen at 5% loss and a book at 15% gain, net gain =  -5p+15b = 7 ....1

On selling the pen at 5% gain and the book at 10% gain, net gain = 5p+10b = 13  .....2

Adding 1 and 2 we get, 25b=20

Hence 100b= 20*4=80, 

C is the answer. 

The weight/volume(g/L) for liquid 1 = 1000

The weight/volume(g/L) for liquid 2 = 800

The weight/volume(g/L) of the mixture = 480/(1/2) = 960

Using alligation the ratio of liquid 1 and liquid 2 in the mixture = (960-800)/(1000-960) = 160/40 = 4:1

Hence the percentage of liquid 1 in the mixture = 4*100/(4+1)=80

Assume the average of 21 students other than Ramesh = a

Sum of the scores of 21 students other than Ramesh = 21a

Hence the average of 22 students = a+1

Sum of the scores of all 22 students = 22(a+1)

The score of Ramesh = Sum of scores of all 22 students - Sum of the scores of 21 students other than Ramesh = 22(a+1)-21a=a+22  = 82.5   (Given)

=> a = 60.5

Hence, sum of the scores of all 22 students = 22(a+1) = 22*61.5 = 1353

Now the sum of the scores of students other than Gautam = 21*62 = 1302

Hence the score of Gautam = 1353-1302=51

We have, $$(\surd2)^{19} 3^4 4^2 9^m 8^n = 3^n 16^m (\sqrt[4]{64})$$

Converting both sides in powers of 2 and 3, we get

$$2^{\ \frac{19\ }{2}}3^42^43^{2m}2^{3n}$$ = $$3^n2^{4m}2^{\frac{\ 6}{4}}$$

Comparing the power of 2 we get, $$\ \frac{\ 19}{2}+4+3n\ =4m+\frac{\ 6}{4}\ $$

=> 4m=3n+12 .....(1)

Comparing the power of 3 we get, $$4+2m=n$$

Substituting the value of n in (1), we get

4m=3(4+2m)+12

=> m=-12

Consider the work done by a man in a day = a and that by a machine = b

Since, three men and eight machines can finish a job in half the time taken by three machines and eight men to finish the same job, hence the efficiency will be double.

=> 3a+8b = 2(3b+8a)

=> 13a=2b

Hence work done by 13 men in a day = work done by 2 machines in a day.

=> If two machines can finish the job in 13 days, then same work will be done 13 men in 13 days. 

Hence the required number of men = 13

The given figure can be divided into 9 regions or equilateral triangles of equal areas as shown below,

Now the hexagon consists of 6 regions and the triangle consists of 9 regions.

Hence the ratio of areas = 6/9 =2:3

We have, $$\log_{5}{(x + y)} + \log_{5}{(x - y)} = 3$$

=> $$x^2-y^2=125$$......(1)

$$\log_{2}{y} - \log_{2}{x} = 1 - \log_{2}{3}$$

=>$$\ \frac{\ y}{x}$$ = $$\ \frac{\ 2}{3}$$

=> 2x=3y   => x=$$\ \frac{\ 3y}{2}$$

On substituting the value of x in 1, we get

$$\ \frac{\ 5x^2}{4}$$=125

=>y=10, x=15

Hence xy=150

Sum of the area of region I and II is the required area.

Now, required area =  $$\ 4\times\frac{\ 1}{2}\times\ 1\times\ 1$$ = 2

The number of paths from (1, 1) to (8, 10) via (4, 6) = The number of paths from (1,1) to (4,6) * The number of paths from (4,6) to (8,10)

To calculate the number of paths from (1,1) to (4,6), 4-1 =3 steps in x-directions and 6-1=5 steps in y direction

Hence the number of paths from (1,1) to (4,6) = $$^{(3+5)}C_3$$ = 56

To calculate the number of paths from (4,6) to (8,10), 8-4 =4 steps in x-directions and 10-6=4 steps in y direction

Hence the number of paths from (4,6) to (8,10) = $$^{(4+4)}C_4$$ = 70

The number of paths from (1, 1) to (8, 10) via (4, 6) = 56*70=3920

Assuming the dimensions of the brick are a, b and c and the diagonals are 3, 2 $$\surd3$$ and $$\surd{15}$$

Hence, $$a^{2\ }+\ b^2$$ = $$3^2$$   ......(1)

$$b^{2\ }+\ c^2$$ = $$(2\sqrt{3})^2$$ ......(2)

$$c^{2\ }+\ a^2$$ = $$(\sqrt{15})^2$$ ......(3)

Adding the three equations, 2($$a^2+b^2+c^2$$) = 9+12+15=36

=>$$a^2+b^2+c^2$$ = 18......(4)

Subtracting (1) from (4), we get $$c^2$$ = 9    =>c=3

Subtracting (2) from (4), we get $$a^2$$ = 6    =>a=$$\sqrt{6}$$

Subtracting (3) from (4), we get $$b^2$$ = 3    =>b=$$\sqrt{3}$$

The ratio of the length of the shortest edge of the brick to that of its longest edge is = $$\ \frac{\ \sqrt{3}}{3}$$ = $$1 : \sqrt{3}$$

For x <0, -x($$6x^2+1$$) = $$5x^2$$

=> ($$6x^2+1$$) = -5x

=> ($$6x^2 + 5x+ 1$$) = 0 

=>($$6x^2 + 3x+2x+ 1$$) = 0

=> (3x+1)(2x+1)=0    =>x=$$\ -\frac{\ 1}{3}$$  or x=$$\ -\frac{\ 1}{2}$$

For x=0, LHS=RHS=0    (Hence, 1 solution)

For x >0, x($$6x^2+1$$) = $$5x^2$$

=> ($$6x^2 - 5x+ 1$$) = 0 

=>(3x-1)(2x-1)=0    =>x=$$\ \frac{\ 1}{3}$$   or   x=$$\ \frac{\ 1}{2}$$

Hence, the total number of solutions = 5

In any right triangle, the circumradius is half of the hypotenuse. Here,L=$$\ \frac{\ 1}{2}\ $$* the length of the hypotenuse = $$\ \frac{\ 1}{2}$$($$\sqrt{\ 15^2+9^2}$$) = $$\ \frac{\ 1}{2}\ $$*$$\sqrt{\ 306}$$ = $$\ \frac{\ 1}{\ 2}\times\ $$17.49 = 8.74

Hence, the integer close to L = 9