CAT Previous Year Question Paper 2023 (June 9) - CAT MCQ

Option C is the correct answer because it contradicts the idea of "remapping" discussed in the passage. The passage emphasizes that the novels under consideration challenge the common representations found in English fiction, particularly those centered in the West. Option C, suggesting that cosmopolitanism originated in the West and traveled to the East through globalization, aligns with the conventional Western-centric narrative rather than the passage's argument of reshaping perspectives and centralizing the interconnected global south, particularly the Indian Ocean world, as a key space in the reimagined literary landscape.

Option A aligns with the passage's discussion of the novels focusing on the Indian Ocean world, contributing to the "remapping" beyond national concerns.

Option B aligns with the passage's emphasis on the interconnected Indian Ocean world, challenging the Eurocentric perspective on trade and commerce.

Option D supports the passage's claim that historical evidence suggests that globalization first appeared in the Indian Ocean, contributing to the "remapping" of the world's historical and geographical perspectives.

Option C cannot be directly inferred from the passage. The passage does discuss the infiuence of both geographic factors (such as biogeography) and non-geographic factors (culture, history, individual decisions) on human phenomena. However, the passage does not explicitly quantify or compare the prevalence of these infiuences by stating that "most human phenomena result from culture and individual choice."

Option A: The author discusses the development of warm fur clothes among the Inuit living north of the Arctic Circle and asserts that it was not due to a specific individual decision or historical contingency in 1783 but rather a response to environmental factors.

Option B: The author discusses the current state of agricultural practices in Australia, stating that the crops and domestic animals that make Australia a food and wool exporter today are all non-native species (mainly Eurasian) brought to Australia by overseas colonists. The use of the term "non-native" implies a change in agricultural practices from what was originally present in the Australian continent.

Option D: The author mentions that some geographic explanations advanced a century ago were racist, causing all geographic explanations to become tainted by racist associations in the minds of many scholars.

Option B is the correct answer because the passage uses examples like the Inuit and Aboriginal Australians to illustrate the influence of physical circumstances, particularly environmental factors, on human behavior and cultural practices. The discussion about the development of warm fur clothes among the Inuit due to the Arctic environment and the absence of indigenous farming in Aboriginal Australia because of the lack of domesticable native species underscores how physical circumstances dictate certain aspects of human behavior and shape cultural adaptations. Therefore, Option B accurately captures the main idea conveyed by the examples provided in the passage.
Option A is not explicitly emphasized in the passage; the focus is more on how environmental factors influence behavior and cultures.
Option C: The passage doesn't explicitly highlight self-suflciency but rather the impact of speciflc environmental factors on the development of societies.
Option D is not entirely incorrect, but Option B more precisely captures the emphasis on physical circumstances dictating human behavior and cultures in the context of the examples provided in the passage.

The passage mentions that as early as the ninth century, the royal oflce of the Luparii, or wolf-catchers, was created in France to tackle the predators. However, this oflce became redundant as it had flnished it’s job (kill the last wolf). So the resurgence of the wolfs can’t be attributed to it shutting down. The other options on the other hand, can be clearly inferred.
Option A: “Various factors explain the changes of the past few decades. Rural depopulation is part of the story.
In Lozère, for example, farming and a once-flourishing mining industry supported a population of over 140,000 residents in the mid-19th century. Today the department has fewer than 80,000 people, many in its towns. “ Option B: “As humans withdraw, forests are expanding. In France, between 1990 and 2015, forest cover increased by an average of 102,000 hectares each year, as more flelds were given over to trees. Now, nearly one-third of mainland France is covered by woodland of some sort. “ Option D: “The mostly protected status of the wolf in Europe—hunting them is now forbidden, other than when occasional culls are sanctioned by the state—plus the efforts of NGOs to track and count the animals, also contribute to the recovery of wolf populations.”

Considering the first paragraph: “RESIDENTS of Lozère, a hilly department in southern France, recite complaints familiar to many rural corners of Europe. In remote hamlets and villages, with names such as Le Bacon and Le Bacon Vieux, mayors grumble about a lack of local schools, jobs, or phone and internet connections. Farmers of grazing animals add another concern: the return of wolves. Eradicated from France last century, the predators are gradually creeping back to more forests and hillsides. “The wolf must be taken in hand,” said an aspiring parliamentarian, Francis Palombi, when pressed by voters in an election campaign early this summer.
Tourists enjoy visiting a wolf park in Lozère, but farmers fret over their livestock and their livelihoods. .” Options A, B and C can be clearly inferred from the highlighted part.
The passage mentions that the number of people holding hunting licenses is still high but the number of people who still actively hunt is low. So Option D which states that there is decline in the number of hunting licences is incorrect.

It is given that the applications are scheduled for processing in twenty 15-minute slots starting at 9:00 am and ending at 2:00 pm. Ten applications are scheduled in each slot.
Hence, the total number of applicants = (20*10) = 200. It is also known that 50% of the applications were US applications, and the number of US applications was the same in all the slots. The same was true for the other three categories.
Hence, the number of total number of US applicants = (200*50%) = 100, and the number of US applicants in each slot = (100/20) = 5 It is also known that Ira, Vijay, and Nandini were scheduled for Schengen visa processing in that order. They had a 9:15 am slot. Since the number of Schengen applicants was the same in all the slots, it implies the number of Schengen applicants in each slot is at least 3.
Similarly, it is given that Mahira and Osman were scheduled in the 9:30 am slot on that day for visa processing in the Others category, which implies the number of other category applicants in each slot is at least 2. Since the number of total applicants in each slot is 10, this implies the number of Schengen and other applicants in each slot is 3, and 2, respectively. Hence, the number of UK applicants is 0 in each slot.

It is also known that the number of total counters is 10, among which four are dedicated to US applications, and two each for UK applications, Schengen applications, and Others applications. It is given that each US and UK application requires 10 minutes of processing time, and Vijay was called to a counter at 9:25 am. (Who is 5th in the queue). It can only be possible when the processing time of Schengen applications is 12.5 minutes.

On a particular day, Ira, Vijay, and Nandini were scheduled for Schengen visa processing in that order. They had a 9:15 am slot but entered the VPO at 9:20 am. When they entered the ofice, exactly six out of the ten counters were either processing applications, or had finished processing one and ready to start processing the next.
Hence, at 9.20 am, there are exactly four free counters. Out of these 4, 2 is the UK counter, and the other two are other counters. (Since the US counters and Schengen Counters were either processing applications, or had finished processing one and were ready to start processing the next.) Nandini's position was sixth in the queue in the Schengen Applications. From the table, we can see that her process will end at 9.45 am.

The correct option is D

It is given that the applications are scheduled for processing in twenty 15-minute slots starting at 9:00 am and ending at 2:00 pm. Ten applications are scheduled in each slot.
Hence, the total number of applicants = (20*10) = 200. It is also known that 50% of the applications were US applications, and the number of US applications was the same in all the slots. The same was true for the other three categories.
Hence, the number of total number of US applicants = (200*50%) = 100, and the number of US applicants in each slot = (100/20) = 5 It is also known that Ira, Vijay, and Nandini were scheduled for Schengen visa processing in that order. They had a 9:15 am slot. Since the number of Schengen applicants was the same in all the slots, it implies the number of Schengen applicants in each slot is at least 3.
Similarly, it is given that Mahira and Osman were scheduled in the 9:30 am slot on that day for visa processing in the Others category, which implies the number of other category applicants in each slot is at least 2. Since the number of total applicants in each slot is 10, this implies the number of Schengen and other applicants in each slot is 3, and 2, respectively. Hence, the number of UK applicants is 0 in each slot.
It is also known that the number of total counters is 10, among which four are dedicated to US applications, and two each for UK applications, Schengen applications, and Others applications. It is given that each US and UK application requires 10 minutes of processing time, and Vijay was called to a counter at 9:25 am. (Who is 5th in the queue). It can only be possible when the processing time of Schengen applications is 12.5 minutes.

On a particular day, Ira, Vijay, and Nandini were scheduled for Schengen visa processing in that order. They had a 9:15 am slot but entered the VPO at 9:20 am. When they entered the office, exactly six out of the ten counters were either processing applications, or had On a particular day, Ira, Vijay, and Nandini were scheduled for Schengen visa processing in that order. They had a 9:15 am slot but entered the VPO at 9:20 am. When they entered the office, exactly six out of the ten counters were either processing applications, or had finished processing one and ready to start processing the next.
Hence, at 9.20 am, there are exactly four free counters. Out of these 4, 2 is the UK counter, and the other two are other counters. (Since the US counters and Schengen Counters were either processing applications, or had finished processing one and were ready to start processing the next.) Let's check the options.
Option A: The application process of Osman was completed before 9:45 am. => True (Since he is 5th in the queue, his process will end at 9.35 am) Option B: The application process of Mahira started after Nandini’s. => The application process for Mahira starts at 9.30 am, and the application process for Nandini starts at 9.32.30 am => False.
The correct option is Bnished processing one and ready to start processing the next.
Hence, at 9.20 am, there are exactly four free counters. Out of these 4, 2 is the UK counter, and the other two are other counters. (Since the US counters and Schengen Counters were either processing applications, or had
nished processing one and were ready to start processing the next.) Let's check the options.
Option A: The application process of Osman was completed before 9:45 am. => True (Since he is 5th in the queue, his process will end at 9.35 am) Option B: The application process of Mahira started after Nandini’s. => The application process for Mahira starts at 9.30 am, and the application process for Nandini starts at 9.32.30 am => False.
The correct option is B

It is given that the applications are scheduled for processing in twenty 15-minute slots starting at 9:00 am and ending at 2:00 pm. Ten applications are scheduled in each slot.
Hence, the total number of applicants = (20*10) = 200. It is also known that 50% of the applications were US applications, and the number of US applications was the same in all the slots. The same was true for the other three categories.
Hence, the number of total number of US applicants = (200*50%) = 100, and the number of US applicants in each slot = (100/20) = 5

It is also known that Ira, Vijay, and Nandini were scheduled for Schengen visa processing in that order. They had a 9:15 am slot. Since the number of Schengen applicants was the same in all the slots, it implies the number of Schengen applicants in each slot is at least 3.
Similarly, it is given that Mahira and Osman were scheduled in the 9:30 am slot on that day for visa processing in the Others category, which implies the number of other category applicants in each slot is at least 2. Since the number of total applicants in each slot is 10, this implies the number of Schengen and other applicants in each slot is 3, and 2, respectively. Hence, the number of UK applicants is 0 in each slot.
It is also known that the number of total counters is 10, among which four are dedicated to US applications, and two each for UK applications, Schengen applications, and Others applications. It is given that each US and UK application requires 10 minutes of processing time, and Vijay was called to a counter at 9:25 am. (Who is 5th in the queue). It can only be possible when the processing time of Schengen applications is 12.5 minutes.

On a particular day, Ira, Vijay, and Nandini were scheduled for Schengen visa processing in that order. They had a 9:15 am slot but entered the VPO at 9:20 am. When they entered the o
ce, exactly six out of the ten counters were either processing applications, or had
nished processing one and ready to start processing the next.
Hence, at 9.20 am, there are exactly four free counters. Out of these 4, 2 is the UK counter, and the other two are other counters. (Since the US counters and Schengen Counters were either processing applications, or had
nished processing one and were ready to start processing the next.) From the table, we can see that the
rst slot takes 20 minutes to complete, and after that remaining 19 slots take 15 minutes each to complete the US application process.
Hence, the total time taken = 20+15*19 = 305 minutes = 5 hrs 5 minutes. Hence, the time will be (9 am + 5hrs 5 minutes) = 2.05 pm

The correct option is A

It is given that some of the houses are occupied. The remaining ones are vacant and are the only ones available for sale.
The base price of a vacant house is Rs. 10 lakhs if the house does not have a parking space, and Rs. 12 lakhs if it does. The quoted price (in lakhs of Rs.) of a vacant house is calculated as (base price) + 5 × (road adjacency value) + 3 × (neighbor count).
It is also known that the maximum quoted price of a house in Block XX is Rs. 24 lakhs Hence, there can be two cases for the maximum quoted price of a house in block XX.
Case 1: House with parking space: => 12+5a+3b = 24 => 5a+3b = 12 ( a = road adjacency value, b= neighbor count) The only value for which the equation satisfies is (a = 0, and b=4). But the value of b can't be 4 because the maximum neighbor count can be at most 3.
Hence, case 1 is invalid.

Case 2: House without parking space: => 10+5a+3b = 24 => 5a+3b = 14 => (a, b) = (1, 3) Hence, the house must have 3 neighbors and 1 road connected to it. Hence, the only possible case is B2.
Therefore, the neighbor houses of B2, which are (B1, A2, and C2) are occupied.
It is known that Row 1 has two occupied houses, one in each block. Since B1 is already occupied, it implies A1, and C1 are vacant.
Hence, the configuration of block XX is given below: (Where U = Unoccupied/ Vacant, and U = Occupied)

Now for block YY, we know that both houses in Column E are vacant. Each of Column-D and Column-F has at least one occupied house. There is only one house with parking space in Block YY.
It is also known that the minimum quoted price of a house in block YY is Rs. 15 lakhs, and one such house is in Column E.
Case 1: The minimum quoted house is E2: We know that the road adjacency of E2 is 1, hence we can calculate whether the house has parking space or not, and the neighbor count (b) If the house has parking space, then: 12+5*1+3*b = 15 => 3b = -2 (which is not possible) Hence, the house has no parking space => 10+5*1+3b = 15 => b = 0 b = 0 implies all the neighbor house of E2 is vacant, which are (E1, D2, and F2).
It is known that each of Column-D and Column-F has at least one occupied house, which implies D1, and F1 must be occupied.
But D1 and F1 can't be occupied together since the total number of occupied houses in Row 1 is 2 (one in each block).
Hence, This case is invalid.

Case 2: The minimum quoted house is E1: We know that the road adjacency of E1 is 0, hence we can calculate whether the house has parking space or not, and the neighbor count (b). i) If the house has no working space, then: 10+5*0+3b = 15 => b = 5/3 (this is not possible since b has to be an integer value) Hence, the house has parking space => 12+5*0+3b = 15 => b = 1 => One neighbor house is occupied among D1 and F1.
Let's take the case that house D1 is occupied and F1 is empty. In that case, the value of house F1 would be 10(there is no parking space)+ (5*0) + (3*the number of neighbours) Here, even if we take the number of neighbours to be 1, which is maximum for F1 in this case, the value of F1 would be a maximum of 13. This is lower than the lowest value house in block YY. Therefore, F1 cannot be empty.
Let us see the other scenario of D1 being unoccupied.
Here, the value of D1 can be 15 or 18 depending on if D2 is unoccupied or occupied respectively.
We do not know the status of houses D2 and F2.
Therefore, the final diagram is given below:

From the diagram, we can see that 3 houses are vacant in block XX.

It is given that some of the houses are occupied. The remaining ones are vacant and are the only ones available for sale.
The base price of a vacant house is Rs. 10 lakhs if the house does not have a parking space, and Rs. 12 lakhs if it does. The quoted price (in lakhs of Rs.) of a vacant house is calculated as (base price) + 5 × (road adjacency value) + 3 × (neighbor count).
It is also known that the maximum quoted price of a house in Block XX is Rs. 24 lakhs Hence, there can be two cases for the maximum quoted price of a house in block XX.
Case 1: House with parking space: => 12+5a+3b = 24 => 5a+3b = 12 ( a = road adjacency value, b= neighbor count) The only value for which the equation satisfies is (a = 0, and b=4). But the value of b can't be 4 because the maximum neighbor count can be at most 3.
Hence, case 1 is invalid.

Case 2: House without parking space: => 10+5a+3b = 24 => 5a+3b = 14 => (a, b) = (1, 3) Hence, the house must have 3 neighbors and 1 road connected to it. Hence, the only possible case is B2.
Therefore, the neighbor houses of B2, which are (B1, A2, and C2) are occupied.
It is known that Row 1 has two occupied houses, one in each block. Since B1 is already occupied, it implies A1, and C1 are vacant.
Hence, the configuration of block XX is given below: (Where U = Unoccupied/ Vacant, and U = Occupied)

Now for block YY, we know that both houses in Column E are vacant. Each of Column-D and Column-F has at least one occupied house. There is only one house with parking space in Block YY.
It is also known that the minimum quoted price of a house in block YY is Rs. 15 lakhs, and one such house is in Column E.
Case 1: The minimum quoted house is E2: We know that the road adjacency of E2 is 1, hence we can calculate whether the house has parking space or not, and the neighbor count (b) If the house has parking space, then: 12+5*1+3*b = 15 => 3b = -2 (which is not possible) Hence, the house has no parking space => 10+5*1+3b = 15 => b = 0 b = 0 implies all the neighbor house of E2 is vacant, which are (E1, D2, and F2).
It is known that each of Column-D and Column-F has at least one occupied house, which implies D1, and F1 must be occupied.
But D1 and F1 can't be occupied together since the total number of occupied houses in Row 1 is 2 (one in each block).
Hence, This case is invalid.

Case 2: The minimum quoted house is E1: We know that the road adjacency of E1 is 0, hence we can calculate whether the house has parking space or not, and the neighbor count (b). i) If the house has no working space, then: 10+5*0+3b = 15 => b = 5/3 (this is not possible since b has to be an integer value) Hence, the house has parking space => 12+5*0+3b = 15 => b = 1 => One neighbor house is occupied among D1 and F1.
Let's take the case that house D1 is occupied and F1 is empty. In that case, the value of house F1 would be 10(there is no parking space)+ (5*0) + (3*the number of neighbours) Here, even if we take the number of neighbours to be 1, which is maximum for F1 in this case, the value of F1 would be a maximum of 13. This is lower than the lowest value house in block YY. Therefore, F1 cannot be empty.
Let us see the other scenario of D1 being unoccupied.
Here, the value of D1 can be 15 or 18 depending on if D2 is unoccupied or occupied respectively.
We do not know the status of houses D2 and F2.
Therefore, the final diagram is given below:

From the diagram, we can see that B1 is definitely occupied. The rest opinions are not de
nitely correct.
The correct option is C

Given that -2 is a root of the given cubic equation.
=> Dividing the given equation by (x + 2), Using the Horners method of synthetic division:

coeficient of x² is 1, and coeficient of x is (2r+1)-2 = 2r-1 and the constant term = (4r-1)-2(2r-1) = 1.
=> The quadratic obtained by dividing the cubic = x²+(2r-1)x+ = 0.
Since, this equation has 2 real roots => Discriminant should be greater than 0
=> (2r-1)² > 4 . => 2r-1 > 2 or 2r - 1 <- 2 = > r >3/2 or r <-1/2

Minimum possible non-negative integer value of r is 2.

The given time is 8:48 AM.
Angle made by hours hand w.r.t 12 is 8 * 30 (30 degrees in 1 hour) + 0.5 * 48 (0.5 degree in 1 minute) = 240 + 24 = 264 degrees.
Angle made by minutes hands w.r.t 12 is 48 * 6 = 288 degrees.
=> The angle between them is 288 - 264 = 24 degrees.
This should further increase by 12 degrees (50% of 24)

After m minutes, the further increase in angle

Let us assume the initial selling prices of A and B is p.

Given, she made profit of 20% on A => 1.2 * c = p => c = 5p/6 => cost of A is 5/6p

Given, she made a loss of 10% on B => 0.9 * c = p => c = 10p/9 => cost of B is 10/9p

Now, she sold them at a price such that a 10% profit is made on B

=> Selling price = s = 11/10 * 10/9 p =>11/9p

Given the total number of kilometres travelled, including the trip = is 26862 Km, and the duration of the trip is 8 hrs.
If avg. speed of the car during the trip is 's'
=> the km travelled till just before the trip is 26862 - 8s, which should also be a palindrome.
=> From the options if s = 110
=> The reading will be 26862 - 110*8 = 25982 (Not a palindrome) => If s = 100 => The reading will be 26862 - 100*8 = 26062 => It is a palindrome.
=> s = 100 is the correct option.