CAT 2023 Slot 2 Question Paper (July 30) - CAT MCQ

Option C is the correct answer because it aligns with the passage's perspective that the interpretation of facts is subjective and can be influenced by different perspectives. The passage argues that historians play a crucial role in selecting and interpreting facts, and Option C supports this idea by suggesting that facts, like truth, can be relative.
If facts are relative, it means that what one person considers a fact may not be viewed the same way by another person. This relativity of facts supports the notion that the historian's interpretation and perspective play a significant role in determining what is considered a fact. Therefore, Option C, if true, reinforces the passage's claim that facts are not entirely objective and independent of the historian's perspective, and it does not weaken the passage's argument.

The passage suggests that historians can rely on disciplines such as archaeology, among others, to establish basic facts. The relevant portion of the passage is:
"But [to] praise a historian for his accuracy is like praising an architect for using well-seasoned timber or properly mixed concrete in his building. It is a necessary condition of his work, but not his essential function. It is precisely for matters of this kind that the historian is entitled to rely on what have been called the 'auxiliary sciences' of history—archaeology, epigraphy, numismatics, chronology, and so forth."

In this context, the "auxiliary sciences" are mentioned as tools that historians can use to ensure the accuracy of basic facts. Archaeology is included in this list, suggesting that it helps historians in ascertaining factual accuracy by providing evidence from material remains, artifacts, and other archaeological findings. Therefore, Option B correctly reflects the role of archaeology in supporting historians in their pursuit of factual accuracy.

The passage suggests that while establishing basic facts is necessary, the essential function of historians goes beyond this. It emphasizes the selective and interpretive nature of historical writing, where historians are expected to go deeper into understanding the context and motivations behind historical events.

Option A, which involves exploring the socio-political and economic factors that led to the Battle of Hastings, is in line with the idea of providing a nuanced interpretation. This option indicates a focus on understanding the underlying causes and in
uences that shaped the historical event, re
ecting a more comprehensive and contextual approach to historical writing.
Option B: While timelines are important, the author suggests that the historian's essential function goes beyond establishing basic chronological facts.

Option C: The author acknowledges the importance of basic facts but argues that historians must go beyond mere fact-
nding.

Option D: While the author acknowledges the use of auxiliary sciences, it also suggests that the historian's focus should go beyond relying solely on these sciences for basic facts, emphasizing the historian's selective and interpretive role in presenting historical events.

Option B is not supported by the information in the passage. While the passage mentions that Netflix has offices across Europe, it also notes that the big decisions still rest with American executives. The passage suggests that Netflix's content might still exhibit a somewhat mid-Atlantic quality, and the company's executive decisions remain under the control of Americans, making it less accurate to claim that Netflix has completely transformed into a truly European entity.
Option A is supported by the passage, which mentions shows like "Lupin," a French crime caper on Netflix, becoming global hits.
Option C: The passage specifically mentions that, according to Ampere, a media-analysis company, in 2015, about 75% of Netflix’s original content was American, but now the figure is half. This indicates a shift in the geographical distribution of Netflix's original programming, with a decrease in the proportion of American content.
Option D is true as the passage mentions that streaming services like Netflix account for about a third of all viewing hours, challenging the dominance of national broadcasters.

Option A is the correct answer because the passage emphasizes that the rise of Netflix in Europe is seen as a unifying force. The author notes that Netflix and similar streaming services, by pumping the same content into homes across the continent, contribute to making culture a cross-border endeavor. This is described as a shared experience among Europeans, as they binge-watch the same series. The idea is that having a common cultural experience, facilitated by platforms like Netflix, can be a unifying factor among the diverse populations of Europe. Therefore, the rise of Netflix is portrayed in a positive light as a force that brings people together through shared cultural consumption.

We are given that each box contains three sacks. Each sack has a certain number of coins, between 1 and 9, both inclusive.
The average number of coins per sack in the boxes are all distinct integers. The total number of coins in each row is the same. The total number of coins in each column is also the same.
=> The total number of coins in a box range from 3 (1 + 1 + 1) to 27 (9 + 9 + 9)
Since, it is given that the average number of coins per sack in the boxes are all distinct integers => The total number of coins in a box would be 3, 6, 9, 12, 15, 18, 21, 24, 27 => averages of 1, 2, 3, 4,....,9 => Sum = 45.
=> Sum of averages coins in a box in a row or column = 45/3 = 15 [The total number of coins in each row is the same. The total number of coins in each column is also the same.] ==> (1)
Let us represent the final configuration of the sacks in boxes as follows:

Also a bag (x,y) => bag in xth row and yth column.
We are given 2 clues => Table-1 & Table-2
Consider bag (3,1)
=> From Table-1
=> Median = 8 & From Table-2 all 3 sacks have more than 5 coins. Also *=> There is a 9 in one of the sacks.
=> c, 8, 9 are the coins in bag (3,1), now c > 5 & c + 8 + 9 should be a multiple of 3 => c = 7 is the only possiblility. => bag (3,1) has 7, 8, 9 coins with average = 8.
Consider bag (2,1)
Median = 2 and 1 sack has more than 5 coins. Also ** => conditions i & iii should be satisfied.
=> 1, 2, 9 are the coins in bag (2,1) with average = 4
Consider bag (1,2) Median = 9 and 2 elements are more than 5. Also * => (9 is present & 1 is not present)
=> c, 9, 9 are the coins in bag (1,2) and c is not equal to 1 and less than 5 => c = 3 for c + 18 to be a multiple of 3.
=> 3, 9, 9 are the coins in bag (1,2) with average = 7.
Capturing this info. in the table:

From (1), The average in bag (1,1) is 15 - 4 - 8 = 3.
From (1), The average in bag (1,3) is 15 - 3 - 7 = 5.

Consider bag (1,1)
Avg = 3, 1 sack has more than 5 and ** => 2 conditions are being satisfied. => (can't be condition-3 => 9 coins as the total sum of coins is it self 3*3 = 9)
=> bag (1,1) has 1, 1, 7 coins with average = 3.
Consider bag (1,3)
Avg. = 5 => Sum = 15.
Median = 6 and 2 sacks have more than 5 and * => (1 condition is satisfied)
Not condition ii as the median is 6 & Not condition iii as the sum of 2 sacks itself will become 6 + 9 = 15
=> 1, 6, c are the coins => For sum = 15 => c = 15 - 1 - 6 = 8
=> bag (1,3) has 1, 6, 8 coins with average = 5.

Consider bag (3,3)

0 sacks have more than 5 coins and ** => conditions i & ii are being satisfied.
=> 1,1,c are the coins. Now c = 1 or 2 or 3 or 4 => c = 1 or 4 for number of coins to be a multiple of 3.
But c = 1 as no other bag has the possibility to get avg. = 1 as bag (2,2) should have 1, b, c coins and b and c should be more than 1 as only 1*
=> bag (3,3) has 1, 1, 1 coins with average = 1.
Now, we can fill the averages in all the bags.

In bag (2,3) Avg. = 9 => 9, 9, 9 are the coins.
In bag (2,2) => Avg. = 2 => Sum = 6 and only 1* => smallest elemens=t should be 1.
=> 1, b, c are the coins where b + c = 5 and b,c can't be equal to 1 and less than 5 => 2 + 3 = 5 is the only possibility.
=> 1, 2, 3 are the coins with average = 2.

Considering bag (3,2)
Avg. = 6 => Sum = 18.
2 sacks more than 5 coins and ** => 2 sacks have 1 and 9 coins.
=> bag (3,2) has 1, c, 9 coins and c = 18 - 1 - 9 = 8
=> bag (3,2) has 1, 8, 9 coins with average = 6 coins.
==> Final required table, bracket number => average coins per sack in the bag.

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

We are given that each box contains three sacks. Each sack has a certain number of coins, between 1 and 9, both inclusive.
The average number of coins per sack in the boxes are all distinct integers. The total number of coins in each row is the same. The total number of coins in each column is also the same.
=> The total number of coins in a box range from 3 (1 + 1 + 1) to 27 (9 + 9 + 9)
Since, it is given that the average number of coins per sack in the boxes are all distinct integers => The total number of coins in a box would be 3, 6, 9, 12, 15, 18, 21, 24, 27 => averages of 1, 2, 3, 4,....,9 => Sum = 45.
=> Sum of averages coins in a box in a row or column = 45/3 = 15 [The total number of coins in each row is the same. The total number of coins in each column is also the same.] ==> (1)
Let us represent the final configuration of the sacks in boxes as follows:

Also a bag (x,y) => bag in xth row and yth column.
We are given 2 clues => Table-1 & Table-2
Consider bag (3,1)
=> From Table-1
=> Median = 8 & From Table-2 all 3 sacks have more than 5 coins. Also *=> There is a 9 in one of the sacks.
=> c, 8, 9 are the coins in bag (3,1), now c > 5 & c + 8 + 9 should be a multiple of 3 => c = 7 is the only possiblility. => bag (3,1) has 7, 8, 9 coins with average = 8.
Consider bag (2,1)
Median = 2 and 1 sack has more than 5 coins. Also ** => conditions i & iii should be satisfied.
=> 1, 2, 9 are the coins in bag (2,1) with average = 4
Consider bag (1,2) Median = 9 and 2 elements are more than 5. Also * => (9 is present & 1 is not present)
=> c, 9, 9 are the coins in bag (1,2) and c is not equal to 1 and less than 5 => c = 3 for c + 18 to be a multiple of 3.
=> 3, 9, 9 are the coins in bag (1,2) with average = 7.
Capturing this info. in the table:

From (1), The average in bag (1,1) is 15 - 4 - 8 = 3.
From (1), The average in bag (1,3) is 15 - 3 - 7 = 5.

Consider bag (1,1)
Avg = 3, 1 sack has more than 5 and ** => 2 conditions are being satisfied. => (can't be condition-3 => 9 coins as the total sum of coins is it self 3*3 = 9)
=> bag (1,1) has 1, 1, 7 coins with average = 3.
Consider bag (1,3)
Avg. = 5 => Sum = 15.
Median = 6 and 2 sacks have more than 5 and * => (1 condition is satisfied)
Not condition ii as the median is 6 & Not condition iii as the sum of 2 sacks itself will become 6 + 9 = 15
=> 1, 6, c are the coins => For sum = 15 => c = 15 - 1 - 6 = 8
=> bag (1,3) has 1, 6, 8 coins with average = 5.

Consider bag (3,3)

0 sacks have more than 5 coins and ** => conditions i & ii are being satisfied.
=> 1,1,c are the coins. Now c = 1 or 2 or 3 or 4 => c = 1 or 4 for number of coins to be a multiple of 3.
But c = 1 as no other bag has the possibility to get avg. = 1 as bag (2,2) should have 1, b, c coins and b and c should be more than 1 as only 1*
=> bag (3,3) has 1, 1, 1 coins with average = 1.
Now, we can fill the averages in all the bags.

In bag (2,3) Avg. = 9 => 9, 9, 9 are the coins.
In bag (2,2) => Avg. = 2 => Sum = 6 and only 1* => smallest elemens=t should be 1.
=> 1, b, c are the coins where b + c = 5 and b,c can't be equal to 1 and less than 5 => 2 + 3 = 5 is the only possibility.
=> 1, 2, 3 are the coins with average = 2.

Considering bag (3,2)
Avg. = 6 => Sum = 18.
2 sacks more than 5 coins and ** => 2 sacks have 1 and 9 coins.
=> bag (3,2) has 1, c, 9 coins and c = 18 - 1 - 9 = 8
=> bag (3,2) has 1, 8, 9 coins with average = 6 coins.
==> Final required table, bracket number => average coins per sack in the bag.

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

We are given that each box contains three sacks. Each sack has a certain number of coins, between 1 and 9, both inclusive.
The average number of coins per sack in the boxes are all distinct integers. The total number of coins in each row is the same. The total number of coins in each column is also the same.
=> The total number of coins in a box range from 3 (1 + 1 + 1) to 27 (9 + 9 + 9)
Since, it is given that the average number of coins per sack in the boxes are all distinct integers => The total number of coins in a box would be 3, 6, 9, 12, 15, 18, 21, 24, 27 => averages of 1, 2, 3, 4,....,9 => Sum = 45.
=> Sum of averages coins in a box in a row or column = 45/3 = 15 [The total number of coins in each row is the same. The total number of coins in each column is also the same.] ==> (1)
Let us represent the final configuration of the sacks in boxes as follows:

Also a bag (x,y) => bag in xth row and yth column.
We are given 2 clues => Table-1 & Table-2
Consider bag (3,1)
=> From Table-1
=> Median = 8 & From Table-2 all 3 sacks have more than 5 coins. Also *=> There is a 9 in one of the sacks.
=> c, 8, 9 are the coins in bag (3,1), now c > 5 & c + 8 + 9 should be a multiple of 3 => c = 7 is the only possiblility. => bag (3,1) has 7, 8, 9 coins with average = 8.
Consider bag (2,1)
Median = 2 and 1 sack has more than 5 coins. Also ** => conditions i & iii should be satisfied.
=> 1, 2, 9 are the coins in bag (2,1) with average = 4
Consider bag (1,2) Median = 9 and 2 elements are more than 5. Also * => (9 is present & 1 is not present)
=> c, 9, 9 are the coins in bag (1,2) and c is not equal to 1 and less than 5 => c = 3 for c + 18 to be a multiple of 3.
=> 3, 9, 9 are the coins in bag (1,2) with average = 7.
Capturing this info. in the table:

From (1), The average in bag (1,1) is 15 - 4 - 8 = 3.
From (1), The average in bag (1,3) is 15 - 3 - 7 = 5.

Consider bag (1,1)
Avg = 3, 1 sack has more than 5 and ** => 2 conditions are being satisfied. => (can't be condition-3 => 9 coins as the total sum of coins is it self 3*3 = 9)
=> bag (1,1) has 1, 1, 7 coins with average = 3.
Consider bag (1,3)
Avg. = 5 => Sum = 15.
Median = 6 and 2 sacks have more than 5 and * => (1 condition is satisfied)
Not condition ii as the median is 6 & Not condition iii as the sum of 2 sacks itself will become 6 + 9 = 15
=> 1, 6, c are the coins => For sum = 15 => c = 15 - 1 - 6 = 8
=> bag (1,3) has 1, 6, 8 coins with average = 5.

Consider bag (3,3)

0 sacks have more than 5 coins and ** => conditions i & ii are being satisfied.
=> 1,1,c are the coins. Now c = 1 or 2 or 3 or 4 => c = 1 or 4 for number of coins to be a multiple of 3.
But c = 1 as no other bag has the possibility to get avg. = 1 as bag (2,2) should have 1, b, c coins and b and c should be more than 1 as only 1*
=> bag (3,3) has 1, 1, 1 coins with average = 1.
Now, we can fill the averages in all the bags.

In bag (2,3) Avg. = 9 => 9, 9, 9 are the coins.
In bag (2,2) => Avg. = 2 => Sum = 6 and only 1* => smallest elemens=t should be 1.
=> 1, b, c are the coins where b + c = 5 and b,c can't be equal to 1 and less than 5 => 2 + 3 = 5 is the only possibility.
=> 1, 2, 3 are the coins with average = 2.

Considering bag (3,2)
Avg. = 6 => Sum = 18.
2 sacks more than 5 coins and ** => 2 sacks have 1 and 9 coins.
=> bag (3,2) has 1, c, 9 coins and c = 18 - 1 - 9 = 8
=> bag (3,2) has 1, 8, 9 coins with average = 6 coins.
==> Final required table, bracket number => average coins per sack in the bag.

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

In this set, we are told that the amount each firm raised every year increased until it reached a maximum, and then decreased until the firm closed down and no firm raised the same amount of money in two consecutive years.
The increase or decrease can be ±1  or ±2 . => (1)
We are also told that each firm raised Rs. 1 crore in its first and last year of existence
Consider A:
It raised money for 8 years => The raising pattern looks like follows: 1, a, b, c, d, e, f, 1 => where a, b, c,..,, f are the unknown amounts raised.
Also a + b + c + d + e + f = 21 - 2 = 19.
We can observe that 19/6 is slightly greater than 3 => The average amount raised should be around 3.
If a = 3 and f = 3 => b + c + d + e = 13 (not possible) as the minimum case would be (4, 5, 6, 4) => Not possible.
If a = 3 and f = 2 => b + c + d + e = 14 (not possible) as the minimum case would be (4, 5, 4, 3) => Not possible.
=> a = 2 and f = 2 => b + c + d + e = 15 the minimum case is (3, 4, 5, 3) or (3, 5, 4, 3) which gives a sum of 15.
So, the possible cases for A are:

Consider B:
The patterns looks as follows:
1, a, b, 1
If a = 2, b has to be equal to 3 to satisfy (1)
if a = 3, b has to be equal to 2 to satisfy (1)
=> The possible cases for B are:

Consider C: The pattern looks as follows:
1, ..., 1
Let us assume there are 2 gaps between => a + b = 7 (Not possible) as maximum case would be 1, 3, 2, 1
Let us assume there are 3 gaps between => a + b + c = 7, the minimum case possible is 1, 2, 3, 2, 1 => Satisfies.
Now, if there are 4 gaps => a + b + c + d = 7 => The average value is 7/4 which is less than 2
=> Not possible.
=> The possible cases for C are:

Consider D:
The pattern looks as follows:
1, a, b, c, 1
=> a + b + c = 8 When a = 2 and c = 2 => b = 4 => 2, 4, 2 => Satisfies.
When a = 2 and c = 3, b should be 3 (Not satisfying (1))
When a = 3 and c = 3, b should be 2 (Not satisfying (1))
=> The possible cases for D are:

Consider E:
The pattern looks as follows: 1,.....,1
For 1 or 2 gaps, we can't get a sum of 11.
Assume 3 gaps => a + b + c = 11, the maximum case is 3, 5, 3 => Satis
es.
Now, assume 4 gaps
=> a + b + c + d = 11, the minimum case is 2, 3, 4, 2 or 2, 4, 3, 2 which satisfies (1) and 2 + 3 + 4 + 2 = 11.
=> The possible cases for E are:

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

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

In this set, we are told that the amount each firm raised every year increased until it reached a maximum, and then decreased until the firm closed down and no firm raised the same amount of money in two consecutive years.
The increase or decrease can be ±1  or ±2 . => (1)
We are also told that each firm raised Rs. 1 crore in its first and last year of existence
Consider A:
It raised money for 8 years => The raising pattern looks like follows: 1, a, b, c, d, e, f, 1 => where a, b, c,..,, f are the unknown amounts raised.
Also a + b + c + d + e + f = 21 - 2 = 19.
We can observe that 19/6 is slightly greater than 3 => The average amount raised should be around 3.
If a = 3 and f = 3 => b + c + d + e = 13 (not possible) as the minimum case would be (4, 5, 6, 4) => Not possible.
If a = 3 and f = 2 => b + c + d + e = 14 (not possible) as the minimum case would be (4, 5, 4, 3) => Not possible.
=> a = 2 and f = 2 => b + c + d + e = 15 the minimum case is (3, 4, 5, 3) or (3, 5, 4, 3) which gives a sum of 15.
So, the possible cases for A are:

Consider B:
The patterns looks as follows:
1, a, b, 1
If a = 2, b has to be equal to 3 to satisfy (1)
if a = 3, b has to be equal to 2 to satisfy (1)
=> The possible cases for B are:

Consider C: The pattern looks as follows:
1, ..., 1
Let us assume there are 2 gaps between => a + b = 7 (Not possible) as maximum case would be 1, 3, 2, 1
Let us assume there are 3 gaps between => a + b + c = 7, the minimum case possible is 1, 2, 3, 2, 1 => Satisfies.
Now, if there are 4 gaps => a + b + c + d = 7 => The average value is 7/4 which is less than 2
=> Not possible.
=> The possible cases for C are:

Consider D:
The pattern looks as follows:
1, a, b, c, 1
=> a + b + c = 8 When a = 2 and c = 2 => b = 4 => 2, 4, 2 => Satisfies.
When a = 2 and c = 3, b should be 3 (Not satisfying (1))
When a = 3 and c = 3, b should be 2 (Not satisfying (1))
=> The possible cases for D are:

Consider E:
The pattern looks as follows: 1,.....,1
For 1 or 2 gaps, we can't get a sum of 11.
Assume 3 gaps => a + b + c = 11, the maximum case is 3, 5, 3 => Satis
es.
Now, assume 4 gaps
=> a + b + c + d = 11, the minimum case is 2, 3, 4, 2 or 2, 4, 3, 2 which satisfies (1) and 2 + 3 + 4 + 2 = 11.
=> The possible cases for E are:

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

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

It is given that  ^mbⁿ = 144¹⁴⁵ where a> 1 and b>1
144 can be written as 144 = 24 × 32
Hence a^m.bⁿ can be written as 144^145. can be written as

We know that 3²⁹⁰ is a natural number, which implies it can be written as a¹ , where a>1
Hence, the least possible value of m is 1. Similarly, the largest value of n is 580.
Hence, the largest value of (n-m) is (580-1) = 579
The correct option is D

From this inequality, we can say that, when y<0  = > y(y-3) > 0.
Now to satisfy the given equation 

(x+y) must be greater than zero Hence, x>0 and |x| > |y|

Therefore, the magnitude of  is greater than the magnitude of y

Hence x>y, and |x| > |y|=> -x < y (Since the magnitude of x is greater than the magnitude of y.)
The correct option is B.

It is given that k divides m+2n and 3m+4n.
Since k divides (m+2n), it implies k will also divide 3(m+2n). Therefore, k divides 3m+6n.
Similarly, we know that k divides 3m+4n.
We know that if two numbers a, and b both are divisible by c, then their difference (a-b) is also divisible by c.
By the same logic, we can say that {(3m+6n)-(3m+4n)} is divisible by k. Hence, 2n is also divisible by k.
Now, (m+2n) is divisible by k, it implies 2(m+2n) =2m+4n is also divisible by k.
Hence, {(3m+4n)-(2m+4n)} = m is also divisible by k.
Therefore, m, and 2n are also divisible by k.
The correct option is C

It is given that  which can be written as:

Hence, the possible values of x are -5/2, and 3, respectively.
Therefore, the sum of the possible values is 

The correct option is D

Since there are two distinct factors other than 1, and itself, which implies the total number of factors of N is 4.
It can be done in two ways.
First case:

Second case:

From case 1, we can see that the numbers which is a cube of prime and less than 50 are 8, and 27 (2 numbers).
From case 2, we will get the numbers in the form (2*3), (2*5), (2*7), (2*11), (2*13), (2*17), (2*19), (2*23), (3*5), (3*7), (3*11), (3*13), (5*7) {(13 numbers)}
Hence, the total number of numbers having two distinct factors is (13+2) = 15.