CAT 2023 Slot 3 Question Paper (July 31) - CAT MCQ

The primary purpose of patrimony laws is stated in the passage as being "aimed at protecting cultural property," implying that the intention behind these laws is to preserve and safeguard a country's cultural heritage.

However, the paradox lies in the unintended consequence of these laws, as highlighted in the passage. The author argues that, despite the good intentions of protecting cultural property, the strict implementation of patrimony laws has led to a reduction in new archaeological discoveries. This reduction is attributed to diminished incentives for foreign entities, such as governments, NGOs, and educational institutions, to invest in overseas archaeological exploration. In other words, the very laws designed to protect cultural property end up hindering the process of making new archaeological discoveries. This underscores the tension between preserving cultural heritage and the potential negative impact on the exploration and understanding of that heritage. Option D aptly captures this point.

The author suggests that archaeological sites are important to some source countries because they can reap bene
ts from new archaeological discoveries, and one of the mentioned bene
ts is that such discoveries typically increase tourism. The passage emphasizes the economic and cultural advantages associated with tourism, which includes enhancing cultural pride and potentially attracting visitors to explore archaeological sites. Option D correctly presents this point. None of the other choices can be considered as valid inferences.

The central idea of the passage is that strict cultural property laws, although popular, may reduce incentives for foreign entities to invest in overseas archaeological exploration. The passage suggests that this reduction in incentives could be detrimental to archaeological discoveries and, consequently, to the tourism and cultural pride of source countries.

Among the given options, the only statement that would undermine the central idea is presented in Option C - it introduces the idea that there is external financial support for archaeological research in these countries. If true, then the lack of discoveries could potentially be attributed to a completely different factor/variable that the author might have failed to account for.

Based on the discussion, the option that is NOT associated with Pinker's view of rational thinking (as well as that of the ancient Greek philosophers) is Option C - the passage suggests that while sequential reasoning is valuable, many profound human achievements come from moments of epiphany or insight rather than solely from conscious, sequential reasoning.

In relation to this thought, we are told that the emphasis on rational thought involves an understanding of the gaps in one’s own knowledge [Option A] and also ‘arriving at independent conclusions’ [Option D]: {“Even Plato’s Socrates — who anticipated many of Pinker’s points by nearly 2,500 years, showing the virtue of knowing what you do not know and examining all premises in arguments, not simply trusting speakers’ authority or charisma...”}

Towards the end of the passage, we are informed of an ethical and moral dimension [Option B] to rationality, which the author asserts that Pinker considers but does not elaborate on.
Hence, Option C is the correct choice.

The passage emphasises Pinker's focus on sequential reasoning and the tools of rationality, suggesting that greater mastery of these tools can improve decision-making in various practical contexts where individuals must act on ‘uncertain and shifting information.’ The author’s endorsement or support for Pinker’s work is centred on the idea that logical reasoning “equips people with the ability to tackle challenging practical problems” [Option C].

Option A is incorrect - while the author acknowledges that rationality is seen by Pinker as a moral virtue, he adds that this role of moral and ethical education is underexplored in Pinker's work. Option B presents a very specific use case of Pinker’s views and fails to capture the broader message. Option D is similarly limited in scope - the emphasis is more on the broader applicability of rationality in decision-making.
Hence, Option C is the correct choice. 

Let us assume, A is the total number of AC's sold
=> From the information that the total number of ACs sold in the city, 25% were of Window variant => Window AC's = A/4 and Split AC's = 3A/4
Now, let us assume B is the total number of inverter ACs
=> From the information that among the Inverter ACs sold, 20% were of Window variant.=> Window Inverter AC's = B/5 and Window Non-Inverter AC's = 4B/5

From - Condition-3
=> A/4 - B/5 = 6 and 4B/5 = 36 => B = 46 and A = 60.

Now, from condition-6
a) D1 & D4 sold "0" window Non-inverter ACs => D2 & D3 sold 6 window non-inverter ACs, it is given that D2 sold twice as many as D3 => D2 sold 4 and D3 sold 2 ACs of this type.
From condition-2
b) Let us assume, D1 sold "x" window inverter ACs => Number of split inverter ACs sold is 13-x From condition-4
c) Number of split ACs sold by D1 will be "2x"
From condition-5
d) Let us assume 'y' is the number of window ACs sold by D3 & D4 => D2 sold 3y ACs of this type.
From condition-7
e) Let us assume 'z' is the number of split inverter ACs sold by D3 and D4 => D2 sold 2z ACs of this type.
Let us use a, b, c, d, and e make a table:

We know that the total number of window ACs is 15
=> x + 3y + y + y = 15
=> x + 5y = 15, also x and y should be greater than or equal to 2 from condition-1
=> x = 5 and y = 2 is the only solution.
Filling this in the table:

Now, Number of split inverter ACs is 36
=> 8 + 2z + z + z = 36 => 4z = 28 => z = 7.

Filling this and using (5), the number of split AC's sold by D1 is 2*5 = 10.

From the table, we see that 14 split inverter ACs are sold.

Let us assume, A is the total number of AC's sold
=> From the information that the total number of ACs sold in the city, 25% were of Window variant => Window AC's = A/4 and Split AC's = 3A/4
Now, let us assume B is the total number of inverter ACs
=> From the information that among the Inverter ACs sold, 20% were of Window variant.=> Window Inverter AC's = B/5 and Window Non-Inverter AC's = 4B/5

From - Condition-3
=> A/4 - B/5 = 6 and 4B/5 = 36 => B = 46 and A = 60.

Now, from condition-6
a) D1 & D4 sold "0" window Non-inverter ACs => D2 & D3 sold 6 window non-inverter ACs, it is given that D2 sold twice as many as D3 => D2 sold 4 and D3 sold 2 ACs of this type.
From condition-2
b) Let us assume, D1 sold "x" window inverter ACs => Number of split inverter ACs sold is 13-x From condition-4
c) Number of split ACs sold by D1 will be "2x"
From condition-5
d) Let us assume 'y' is the number of window ACs sold by D3 & D4 => D2 sold 3y ACs of this type.
From condition-7
e) Let us assume 'z' is the number of split inverter ACs sold by D3 and D4 => D2 sold 2z ACs of this type.
Let us use a, b, c, d, and e make a table:

We know that the total number of window ACs is 15
=> x + 3y + y + y = 15
=> x + 5y = 15, also x and y should be greater than or equal to 2 from condition-1
=> x = 5 and y = 2 is the only solution.
Filling this in the table:

Now, Number of split inverter ACs is 36
=> 8 + 2z + z + z = 36 => 4z = 28 => z = 7.

Filling this and using (5), the number of split AC's sold by D1 is 2*5 = 10.

From this table, we see that total number of non-inverter ACs is 9 + 6 = 15.
Required percentage is 15 out of 60 => 25%.

Let us assume, A is the total number of AC's sold
=> From the information that the total number of ACs sold in the city, 25% were of Window variant => Window AC's = A/4 and Split AC's = 3A/4
Now, let us assume B is the total number of inverter ACs
=> From the information that among the Inverter ACs sold, 20% were of Window variant.=> Window Inverter AC's = B/5 and Window Non-Inverter AC's = 4B/5

From - Condition-3
=> A/4 - B/5 = 6 and 4B/5 = 36 => B = 46 and A = 60.

Now, from condition-6
a) D1 & D4 sold "0" window Non-inverter ACs => D2 & D3 sold 6 window non-inverter ACs, it is given that D2 sold twice as many as D3 => D2 sold 4 and D3 sold 2 ACs of this type.
From condition-2
b) Let us assume, D1 sold "x" window inverter ACs => Number of split inverter ACs sold is 13-x From condition-4
c) Number of split ACs sold by D1 will be "2x"
From condition-5
d) Let us assume 'y' is the number of window ACs sold by D3 & D4 => D2 sold 3y ACs of this type.
From condition-7
e) Let us assume 'z' is the number of split inverter ACs sold by D3 and D4 => D2 sold 2z ACs of this type.
Let us use a, b, c, d, and e make a table:

We know that the total number of window ACs is 15
=> x + 3y + y + y = 15
=> x + 5y = 15, also x and y should be greater than or equal to 2 from condition-1
=> x = 5 and y = 2 is the only solution.
Filling this in the table:

Now, Number of split inverter ACs is 36
=> 8 + 2z + z + z = 36 => 4z = 28 => z = 7.

Filling this and using (5), the number of split AC's sold by D1 is 2*5 = 10.

Total number of ACs sold by D2 and D4 = 60 - D1 - D3 = 60 - 15 - 12 = 33.

Let us assume, A is the total number of AC's sold
=> From the information that the total number of ACs sold in the city, 25% were of Window variant => Window AC's = A/4 and Split AC's = 3A/4
Now, let us assume B is the total number of inverter ACs
=> From the information that among the Inverter ACs sold, 20% were of Window variant.=> Window Inverter AC's = B/5 and Window Non-Inverter AC's = 4B/5

From - Condition-3
=> A/4 - B/5 = 6 and 4B/5 = 36 => B = 46 and A = 60.

Now, from condition-6
a) D1 & D4 sold "0" window Non-inverter ACs => D2 & D3 sold 6 window non-inverter ACs, it is given that D2 sold twice as many as D3 => D2 sold 4 and D3 sold 2 ACs of this type.
From condition-2
b) Let us assume, D1 sold "x" window inverter ACs => Number of split inverter ACs sold is 13-x From condition-4
c) Number of split ACs sold by D1 will be "2x"
From condition-5
d) Let us assume 'y' is the number of window ACs sold by D3 & D4 => D2 sold 3y ACs of this type.
From condition-7
e) Let us assume 'z' is the number of split inverter ACs sold by D3 and D4 => D2 sold 2z ACs of this type.
Let us use a, b, c, d, and e make a table:

We know that the total number of window ACs is 15
=> x + 3y + y + y = 15
=> x + 5y = 15, also x and y should be greater than or equal to 2 from condition-1
=> x = 5 and y = 2 is the only solution.
Filling this in the table:

Now, Number of split inverter ACs is 36
=> 8 + 2z + z + z = 36 => 4z = 28 => z = 7.

Filling this and using (5), the number of split AC's sold by D1 is 2*5 = 10.

We see that D1 & D3 sold 27 ACs together which is less than 60 - 27 = 33 sold by D2 & D4.
=> Option-D is definitely false.

It is given that none of the streets has more than one team traveling along it in any direction at any point in time (point 1), which implies at 9.00 hrs, all 4 teams have chosen different roots from the starting point.
It is also known that Teams 2 and 3 are the only ones in stations E and D respectively at 10:00 hrs, and Team 1 and Team 4 are the only teams that patrol the street connecting stations A and E.
It is only possible when Team 2 traveled (A-E) via F, and Team 3 reached station D via station C.
It is also known that Teams 1 and 3 are the only ones in Station E at 10:30 hrs, and Team 4 never passes through Stations B, D, or F. Hence, Team 1 must have chosen the (A-B) root at the starting point, and Team 4 has chosen the (A-E) root at 9.00 hrs.
Hence, Team 1 will reach B at 9.30, and come to A at 10.00 hrs. After that, they will go to E at 10.30 hrs. Since Team 4 never passes through stations B, D, or F. Team 4 only can pass through stations A, E, and C.
Hence, the roots of team 4 to reach station E at 11.30 will be (A-E-A-C-A-E) or (A-E-A-E-A-E).
Since team 1 is already traveling to E from A at 10.00 hrs, at that time team 4 can't choose the same route.
Hence, the final route for team 4 to reach E at 11.30 is (A-E-A-C-A-E), and at 12.00 hrs, team 4 will come back to station A.
Hence, the complete route diagram for team 4 is (A-E-A-C-A-E-A)

We can see that team 1 is at station E at 10.30 hrs, and they will reach station B at 11.30 hrs, which is only possible when they travel to B via A.
Hence, the complete route diagram for team 1 is (A-B-A-E-A-B-A). It is also known that Teams 1 and 3 are the only ones in station E at 10:30 hrs.

The only possible root for Team 2 at 10.00 hrs is from E to F since they can't choose E to D because Team 3 is already on this route. Since team 3 has to reach A at 12.00. The only possible combination for team 3 is E-D-C-A

Now the roots for team 2 going back to A is from F at 10.30 hrs (F-A-F-A) or (F-E-F-A).
Hence, the final table is given below:

From the table, we can see that among the options station E is visited the largest number of times.

It is given that there are exactly 41 numbers, which can be expressed as the power of two, and exist between 8^m and 8ⁿ, (where m, and n are positive integers, and m<n)
Hence, 2^3m < 41 numbers < 2³ⁿ
Since, m is a positive integer, the least value of m is 1. Therefore, 2^3m = 2³, hence, the 41 numbers between them are 2⁴, 2⁵, 2⁶, ..., 2⁴⁴
Then the lowest possible value of 8ⁿ is 2⁴⁵. Hence, the smallest value of n is 2⁴⁵ = 8ⁿ=>2³ⁿ =  2⁴⁵ =>n= 15
Hence, the smallest value of m+n is (15+1) = 16
The correct option is D

It is given that for some real numbers a and b, the system of equations x+y = 4 and (a+5)x (b² - 15)y = 8b has infinitely many solutions for x and y.
Hence, we can say that

This equation can be used to find the value of a, and b.

Hence, the values of b are 5, and -3, respectively.
The value of a can be expressed in terms of b, which is

The values of $$2^$$ can't be 4 (log will be undefined), which implies The value of 2^x is 16.
Therefore, the common difference is log₄ (2^x-9) - log₄2

It is given that 5^n-1 < 3ⁿ⁺¹, where n is a natural number. By inspection, we can say that the inequality holds when n = 1, 2, 3 4, and 5.
Now, we need to find the least integer value of m that satisfies 

For, n =1, the least integer value of m is 2.
For, n = 2, the least integer value of m is 3
For, n = 3, the least integer value of m is 4.
For, n = 4, the least integer value of m is 4.
For, n= 5, the least integer value of m is 5.
Hence, the least integer value of m such that for all the values of n, the equation holds is 5.