CAT 2025 Slot 2 Question Paper

If all the two-author-type books are counted for both authors separately, then we get the sum to be 4 * 2 = 8.

If all the three-author-type books are counted for all three authors separately, then we get the sum to be 3 * 3 = 9.

If all the four-author-type books are counted for all four authors separately, then we get the sum to be 4 * 2 = 8.

We are given that there are 2 four-author books, so there are exactly two books for each that are of the four-author type.

Putting the values in the table, we get,

We are given that each author has at least one book of each type. In the case of Arman, there are three books left to be assigned, and there are 3 types of books left, so each of them has to be equal to 1 for the above condition to be satisfied.

In the case of Brajen, we are told that he had an equal number of single-author and two-author books.

Case 1: If the equal number is 1, then the three-author type books become 4, and we are given that both Chintan and Devon had more three-author books than Brajen, which is not possible in this case, as both of them combined will be left with 5 books together and whaterver we split 5 books there wont be a case of both of them having more books than Brajen.

Case 2: If the equal number is 2, then the three-author type books become 2, and we are given that both Chintan and Devon had more three-author books than Brajen, which is possible in this case, as both of them combined will be left with 6 books together and the only way to split 6 books such that both of them are greater than 2 is to split them into 3, 3 and it is the only possible way.

Case 3: If the equal number is greater than or equal to 3, then the three-author type books become 0 or negative, which can directly be eliminated, as we are given that all authors wrote at least 1 book of each type.

So, we can eliminate both cases 1 and 3, and we will be left with the case of Brajen having 2 books each of Single-author, Two-author, and Three-author types. Chintan and Devon have three books each of the three-author type.

Filling up the table with these values, we get,

We are given that each author has a different number of single-author type books. The only possibilities for the single-author books of Chintan and Devon are 3 and 4 in some order.

If the single-author books of Chintan are 3, then the double-author books of Chintan become 4, the single-author books of Devon become 4, and the double-author books of Devon become 1.

If the single-author books of Chintan are 4, then the double-author books of Chintan become 3, the single-author books of Devon become 3, and the double-author books of Devon become 2.

Placing both possibilities in the table, we get,

Statement 1: Chintan wrote exactly three two-author papers.

This is not necessarily true, as he could have written four books as well.

Statement 2: Chintan wrote more single-author papers than Devon.

This is not necessarily true, as the opposite can be possible.

So, neither I nor 2 is necessarily true.

Hence, the correct answer is option A.

If all the two-author-type books are counted for both authors separately, then we get the sum to be 4 * 2 = 8.

If all the three-author-type books are counted for all three authors separately, then we get the sum to be 3 * 3 = 9.

If all the four-author-type books are counted for all four authors separately, then we get the sum to be 4 * 2 = 8.

We are given that there are 2 four-author books, so there are exactly two books for each that are of the four-author type.

Putting the values in the table, we get,

We are given that each author has at least one book of each type. In the case of Arman, there are three books left to be assigned, and there are 3 types of books left, so each of them has to be equal to 1 for the above condition to be satisfied.

In the case of Brajen, we are told that he had an equal number of single-author and two-author books.

Case 1: If the equal number is 1, then the three-author type books become 4, and we are given that both Chintan and Devon had more three-author books than Brajen, which is not possible in this case, as both of them combined will be left with 5 books together and whaterver we split 5 books there wont be a case of both of them having more books than Brajen.

Case 2: If the equal number is 2, then the three-author type books become 2, and we are given that both Chintan and Devon had more three-author books than Brajen, which is possible in this case, as both of them combined will be left with 6 books together and the only way to split 6 books such that both of them are greater than 2 is to split them into 3, 3 and it is the only possible way.

Case 3: If the equal number is greater than or equal to 3, then the three-author type books become 0 or negative, which can directly be eliminated, as we are given that all authors wrote at least 1 book of each type.

So, we can eliminate both cases 1 and 3, and we will be left with the case of Brajen having 2 books each of Single-author, Two-author, and Three-author types. Chintan and Devon have three books each of the three-author type.

Filling up the table with these values, we get,

We are given that each author has a different number of single-author type books. The only possibilities for the single-author books of Chintan and Devon are 3 and 4 in some order.

If the single-author books of Chintan are 3, then the double-author books of Chintan become 4, the single-author books of Devon become 4, and the double-author books of Devon become 1.

If the single-author books of Chintan are 4, then the double-author books of Chintan become 3, the single-author books of Devon become 3, and the double-author books of Devon become 2.

Placing both possibilities in the table, we get,

Statement 1: Arman wrote three-author papers only with Chintan and Devon.

There are in total 3 three-author books, and Chintan and Devon are authors of all three of them according to the table. So, Arman wrote the three-author papers only with Chintan and Devon. So, the statement is necessarily true

Statement 2: Brajen wrote three-author papers only with Chintan and Devon.

There are in total 3 three-author books, and Chintan and Devon are authors of all three of them according to the table. So, Brajen wrote the three-author papers only with Chintan and Devon. So, the statement is necessarily true.

So, both I and 2 are necessarily true.

Hence, the correct answer is option B.

If all the two-author-type books are counted for both authors separately, then we get the sum to be 4 * 2 = 8.

If all the three-author-type books are counted for all three authors separately, then we get the sum to be 3 * 3 = 9.

If all the four-author-type books are counted for all four authors separately, then we get the sum to be 4 * 2 = 8.

We are given that there are 2 four-author books, so there are exactly two books for each that are of the four-author type.

Putting the values in the table, we get,

We are given that each author has at least one book of each type. In the case of Arman, there are three books left to be assigned, and there are 3 types of books left, so each of them has to be equal to 1 for the above condition to be satisfied.

In the case of Brajen, we are told that he had an equal number of single-author and two-author books.

Case 1: If the equal number is 1, then the three-author type books become 4, and we are given that both Chintan and Devon had more three-author books than Brajen, which is not possible in this case, as both of them combined will be left with 5 books together and whaterver we split 5 books there wont be a case of both of them having more books than Brajen.

Case 2: If the equal number is 2, then the three-author type books become 2, and we are given that both Chintan and Devon had more three-author books than Brajen, which is possible in this case, as both of them combined will be left with 6 books together and the only way to split 6 books such that both of them are greater than 2 is to split them into 3, 3 and it is the only possible way.

Case 3: If the equal number is greater than or equal to 3, then the three-author type books become 0 or negative, which can directly be eliminated, as we are given that all authors wrote at least 1 book of each type.

So, we can eliminate both cases 1 and 3, and we will be left with the case of Brajen having 2 books each of Single-author, Two-author, and Three-author types. Chintan and Devon have three books each of the three-author type.

Filling up the table with these values, we get,

We are given that each author has a different number of single-author type books. The only possibilities for the single-author books of Chintan and Devon are 3 and 4 in some order.

If the single-author books of Chintan are 3, then the double-author books of Chintan become 4, the single-author books of Devon become 4, and the double-author books of Devon become 1.

If the single-author books of Chintan are 4, then the double-author books of Chintan become 3, the single-author books of Devon become 3, and the double-author books of Devon become 2.

Placing both possibilities in the table, we get,

If Devon wrote more than one two-author paper, he would have written 2 papers, making the number of two-author papers written by Chintan 3.

Hence, the correct answer is 3.

Let us try to arrange the hoops and the spheres in increasing order of their diameter according to the information given,

In clue 1, we are given that B1 and B6 made a ping on H4, and B5 did not.

So, we can definitely say that,

(B1, B6) < H4 < B5  --(1)

In clue 2, we are given that B4 made a ping on H3, but B1 did not.

So, we can definitely say that,

B4 < H3 < B1 --(2)

In clue 3, we are given that all balls, except B3, made pings on H1.

So, we can definitely say that,

(B1, B2, B4, B5, B6) < H1 < B3 --(3)

In clue 4, we are given that none of the balls, except B2, made a ping on H2.

So, we can definitely say that,

B2 < H2 < (B1, B3, B4, B5, B6) --(4)

Combining (1) and (2), we can definitely say that

B4 < H3 < B1 < H4 < B5 --(5)

The only ball that we do not have enough information is about B6, as it can be either less than H3 or greater than H3 and less than H4.

Combining (3), (4) and (5), we get,

B2 < H2 < B4 < H3 < B1 < H4 < B5 < H1 < B3

The sphere B6 can be placed in 2 positions, giving us two possible inequalities, which are

B2 < H2 < (B4, B6) < H3 < B1 < H4 < B5 < H1 < B3

B2 < H2 < B4 < H3 < (B1,B6) < H4 < B5 < H1 < B3

The position of B6 is not clear, and except for that, all the positions are fixed.

Pings made by B1 = H4, H1 = 2

Pings made by B2 = H2, H3, H4, H1 = 4

Pings made by B3 = 0

Total number of pings by B1, B2 and B3 = 2 + 4 + 0 = 6.

Hence, the correct answer is 6.

Let us try to arrange the hoops and the spheres in increasing order of their diameter according to the information given,

In clue 1, we are given that B1 and B6 made a ping on H4, and B5 did not.

So, we can definitely say that,

(B1, B6) < H4 < B5 --(1)

In clue 2, we are given that B4 made a ping on H3, but B1 did not.

So, we can definitely say that,

B4 < H3 < B1 --(2)

In clue 3, we are given that all balls, except B3, made pings on H1.

So, we can definitely say that,

(B1, B2, B4, B5, B6) < H1 < B3 --(3)

In clue 4, we are given that none of the balls, except B2, made a ping on H2.

So, we can definitely say that,

B2 < H2 < (B1, B3, B4, B5, B6) --(4)

Combining (1) and (2), we can definitely say that

B4 < H3 < B1 < H4 < B5 --(5)

The only ball for which we do not have enough information is about B6, as it can be either less than H3 or greater than H3 and less than H4.

Combining (3), (4) and (5), we get,

B2 < H2 < B4 < H3 < B1 < H4 < B5 < H1 < B3

The sphere B6 can be placed in 2 positions, giving us two possible inequalities, which are

B2 < H2 < (B4, B6) < H3 < B1 < H4 < B5 < H1 < B3

B2 < H2 < B4 < H3 < (B1,B6) < H4 < B5 < H1 < B3

The position of B6 is unclear, and except for that, all the other positions are fixed.

Option A) B4 < B5 < B3. This is definitely true.

Option B) B2 < B1 < B5. This is definitely true.

Option C) B1 < B6 < B3. This need not be true, as we do not know which of B1 and B6 has the larger diameter.

Option D) B1 < B5 < B3. This is definitely true.

Hence, the correct answer is option C.

Let us try to arrange the hoops and the spheres in increasing order of their diameter according to the information given,

In clue 1, we are given that B1 and B6 made a ping on H4, and B5 did not.

So, we can definitely say that,

(B1, B6) < H4 < B5 --(1)

In clue 2, we are given that B4 made a ping on H3, but B1 did not.

So, we can definitely say that,

B4 < H3 < B1 --(2)

In clue 3, we are given that all balls, except B3, made pings on H1.

So, we can definitely say that,

(B1, B2, B4, B5, B6) < H1 < B3 --(3)

In clue 4, we are given that none of the balls, except B2, made a ping on H2.

So, we can definitely say that,

B2 < H2 < (B1, B3, B4, B5, B6) --(4)

Combining (1) and (2), we can definitely say that

B4 < H3 < B1 < H4 < B5 --(5)

The only ball for which we do not have enough information is about B6, as it can be either less than H3 or greater than H3 and less than H4.

Combining (3), (4) and (5), we get,

B2 < H2 < B4 < H3 < B1 < H4 < B5 < H1 < B3

The sphere B6 can be placed in 2 positions, giving us two possible inequalities, which are

B2 < H2 < (B4, B6) < H3 < B1 < H4 < B5 < H1 < B3

B2 < H2 < B4 < H3 < (B1,B6) < H4 < B5 < H1 < B3

The position of B6 is unclear, and except for that, all the other positions are fixed.

We know the correct order of the diameters of the hoops is H2 < H3 < H4 < H1.

Hence, the correct answer is option A.

Let us try to arrange the hoops and the spheres in increasing order of their diameter according to the information given,

In clue 1, we are given that B1 and B6 made a ping on H4, and B5 did not.

So, we can definitely say that,

(B1, B6) < H4 < B5 --(1)

In clue 2, we are given that B4 made a ping on H3, but B1 did not.

So, we can definitely say that,

B4 < H3 < B1 --(2)

In clue 3, we are given that all balls, except B3, made pings on H1.

So, we can definitely say that,

(B1, B2, B4, B5, B6) < H1 < B3 --(3)

In clue 4, we are given that none of the balls, except B2, made a ping on H2.

So, we can definitely say that,

B2 < H2 < (B1, B3, B4, B5, B6) --(4)

Combining (1) and (2), we can definitely say that

B4 < H3 < B1 < H4 < B5 --(5)

The only ball for which we do not have enough information is about B6, as it can be either less than H3 or greater than H3 and less than H4.

Combining (3), (4) and (5), we get,

B2 < H2 < B4 < H3 < B1 < H4 < B5 < H1 < B3

The sphere B6 can be placed in 2 positions, giving us two possible inequalities, which are

B2 < H2 < (B4, B6) < H3 < B1 < H4 < B5 < H1 < B3

B2 < H2 < B4 < H3 < (B1,B6) < H4 < B5 < H1 < B3

The position of B6 is unclear, and except for that, all the other positions are fixed.

In the first case, the number of pings possible can be calculated as 4 for B2, 3 for B4, 3 for B6, 2 for B1, 1 for B5 and 0 for B3. In total, the number of pings possible in the 1st case is 4 + 3 + 3 + 2 + 1 = 13.

In the second case, the number of pings possible can be calculated as 4 for B2, 3 for B4, 2 for B6, 2 for B1, 1 for B5 and 0 for B3. In total, the number of pings possible in the 2nd case is 4 + 3 + 2 + 2 + 1 = 12.

So, the total number of pings can be 12 or 13.

Hence, the correct answer is Option A.

Let us assume the values of A, B, C, D, E and F to be a, b, c, d, e and f in 2016.

For A:

Value in 2020 = increased by 25% from 2016 = $$\left(1\ +\ \dfrac{25}{100}\right)a$$ = $$\dfrac{5}{4}a$$

Value in 2024 = increased by 50% from 2020 = $$\left(1\ +\ \dfrac{50}{100}\right)\times\ \dfrac{5a}{4}$$ = $$\dfrac{15}{8}a$$

For B:

Value in 2020 = decreased by 25% from 2016 = $$\left(1\ -\ \dfrac{25}{100}\right)b$$ = $$\dfrac{3}{4}b$$

Value in 2024 = decreased by 25% from 2020 = $$\left(1\ -\ \dfrac{25}{100}\right)\times\ \dfrac{3b}{4}$$ = $$\dfrac{9}{16}b$$

For C:

Value in 2020 = decreased by 20% from 2016 = $$\left(1\ -\ \dfrac{20}{100}\right)c$$ = $$\dfrac{4}{5}c$$

Value in 2024 = increased by 40% from 2020 = $$\left(1\ +\ \dfrac{40}{100}\right)\times\ \dfrac{4c}{5}$$ = $$\dfrac{28}{25}c$$

For D:

Value in 2020 = increased by 100% from 2016 = $$\left(1\ +\ \dfrac{100}{100}\right)d$$ = $$2d$$

Value in 2024 = increased by 20% from 2020 = $$\left(1\ +\ \dfrac{20}{100}\right)\times\ 2d$$ = $$\dfrac{12}{5}d$$

For E:

Value in 2020 = increased by 25% from 2016 = $$\left(1\ +\ \dfrac{25}{100}\right)e$$ = $$\dfrac{5}{4}e$$

Value in 2024 = increased by 20% from 2020 = $$\left(1\ +\ \dfrac{20}{100}\right)\times\ \dfrac{5e}{4}$$ = $$\dfrac{3}{2}e$$

For F:

Value in 2020 = increased by 100% from 2016 = $$\left(1\ +\ \dfrac{100}{100}\right)f$$ = $$2f$$

Value in 2024 = decreased by 25% from 2020 = $$\left(1\ -\ \dfrac{25}{100}\right)\times\ 2f$$ = $$\dfrac{3}{2}f$$

We are given that the value of E is 90 in 2024. We can calculate the value of e as,

$$\dfrac{3e}{2}\ =\ 90$$

$$e\ =\ 60$$

The value of E in 2016 is 60, and its value in 2020 can be calculated as,

$$\dfrac{5e}{4}\ =\ \dfrac{5}{4}\times\ 60\ =\ 75$$

We are given that B, C, E, and A had ranks 1, 2, 3, and 4, respectively, in 2016, and F has the lowest rank(6) in all the years.

Putting the values in the table, we get,

SI of E in 2016 is 60.

Hence, the correct answer is 60.

Let us assume the values of A, B, C, D, E and F to be a, b, c, d, e and f in 2016.

For A:

Value in 2020 = increased by 25% from 2016 = $$\left(1\ +\ \dfrac{25}{100}\right)a$$ = $$\dfrac{5}{4}a$$

Value in 2024 = increased by 50% from 2020 = $$\left(1\ +\ \dfrac{50}{100}\right)\times\ \dfrac{5a}{4}$$ = $$\dfrac{15}{8}a$$

For B:

Value in 2020 = decreased by 25% from 2016 = $$\left(1\ -\ \dfrac{25}{100}\right)b$$ = $$\dfrac{3}{4}b$$

Value in 2024 = decreased by 25% from 2020 = $$\left(1\ -\ \dfrac{25}{100}\right)\times\ \dfrac{3b}{4}$$ = $$\dfrac{9}{16}b$$

For C:

Value in 2020 = decreased by 20% from 2016 = $$\left(1\ -\ \dfrac{20}{100}\right)c$$ = $$\dfrac{4}{5}c$$

Value in 2024 = increased by 40% from 2020 = $$\left(1\ +\ \dfrac{40}{100}\right)\times\ \dfrac{4c}{5}$$ = $$\dfrac{28}{25}c$$

For D:

Value in 2020 = increased by 100% from 2016 = $$\left(1\ +\ \dfrac{100}{100}\right)d$$ = $$2d$$

Value in 2024 = increased by 20% from 2020 = $$\left(1\ +\ \dfrac{20}{100}\right)\times\ 2d$$ = $$\dfrac{12}{5}d$$

For E:

Value in 2020 = increased by 25% from 2016 = $$\left(1\ +\ \dfrac{25}{100}\right)e$$ = $$\dfrac{5}{4}e$$

Value in 2024 = increased by 20% from 2020 = $$\left(1\ +\ \dfrac{20}{100}\right)\times\ \dfrac{5e}{4}$$ = $$\dfrac{3}{2}e$$

For F:

Value in 2020 = increased by 100% from 2016 = $$\left(1\ +\ \dfrac{100}{100}\right)f$$ = $$2f$$

Value in 2024 = decreased by 25% from 2020 = $$\left(1\ -\ \dfrac{25}{100}\right)\times\ 2f$$ = $$\dfrac{3}{2}f$$

We are given that the value of E is 90 in 2024. We can calculate the value of e as,

$$\dfrac{3e}{2}\ =\ 90$$

$$e\ =\ 60$$

The value of E in 2016 is 60, and its value in 2020 can be calculated as,

$$\dfrac{5e}{4}\ =\ \dfrac{5}{4}\times\ 60\ =\ 75$$

We are given that B, C, E, and A had ranks 1, 2, 3, and 4, respectively, in 2016, and F has the lowest rank(6) in all the years.

Putting the values in the table, we get,

We know that F is ranked 6 and B is ranked 1 in 2016, and we are also given that the range of SI in 2016 and 2024 is 60. This means that the value of b-f = 60, and we can conclude that b is definitely greater than 60, as we know that f is an integer greater than 0.

We know that all the SI values in all the years are integers, and for SI of B to be an integer in 2024, the value of b must be a multiple of 16. The multiples of 16 that are greater than 60 and less than 100 are 64, 80 and 96.

CASE 1: b = 64

If b = 64, then the value of f becomes 4, as we know b - f = 60.

If f = 4, then the value of F in 2024 is 3f/2 = 12/2 = 6.

We already know the value of E in 2024 is 90.

In that case, the range of SI in 2024 is at least 90 - 6 = 84, but we are given that the range is 60 in 2024, which is not possible in this case.

So, we can eliminate this case.

CASE 2: b = 96

If b = 96, then the value of f becomes 36, as we know b - f = 60.

If f = 36, then the value of F in 2024 is 3f/2 = 108/2 = 54.

The value of B in 2024 = 9b/16 = (9 * 96)/16 = 54

We calculated the value of B and F to be 54 in 2024, which is not possible, as we are given that F had a lower SI than any other country in 2024, and we obtained its value to be the same as B.

So, we can eliminate this case.

CASE 3: b = 80

If b = 80, then the value of f becomes 20, as we know b - f = 60.

If f = 20, then the value of F in 2024 is 3f/2 = 60/2 = 30.

The value of B in 2024 = 9b/16 = (9 * 80)/16 = 45

We can see that no condition is violated in this case, as F is less than B in 2024, and the range in 2024 from the obtained values as of now is 90 - 30 = 60.

So, we can conclude that the value of b = 80 and f = 20.

Substituting the values in the table, we get,

So, SI of F in 2020 is 40.

Hence, the correct answer is 40.

Let us assume the values of A, B, C, D, E and F to be a, b, c, d, e and f in 2016.

For A:

Value in 2020 = increased by 25% from 2016 = $$\left(1\ +\ \dfrac{25}{100}\right)a$$ = $$\dfrac{5}{4}a$$

Value in 2024 = increased by 50% from 2020 = $$\left(1\ +\ \dfrac{50}{100}\right)\times\ \dfrac{5a}{4}$$ = $$\dfrac{15}{8}a$$

For B:

Value in 2020 = decreased by 25% from 2016 = $$\left(1\ -\ \dfrac{25}{100}\right)b$$ = $$\dfrac{3}{4}b$$

Value in 2024 = decreased by 25% from 2020 = $$\left(1\ -\ \dfrac{25}{100}\right)\times\ \dfrac{3b}{4}$$ = $$\dfrac{9}{16}b$$

For C:

Value in 2020 = decreased by 20% from 2016 = $$\left(1\ -\ \dfrac{20}{100}\right)c$$ = $$\dfrac{4}{5}c$$

Value in 2024 = increased by 40% from 2020 = $$\left(1\ +\ \dfrac{40}{100}\right)\times\ \dfrac{4c}{5}$$ = $$\dfrac{28}{25}c$$

For D:

Value in 2020 = increased by 100% from 2016 = $$\left(1\ +\ \dfrac{100}{100}\right)d$$ = $$2d$$

Value in 2024 = increased by 20% from 2020 = $$\left(1\ +\ \dfrac{20}{100}\right)\times\ 2d$$ = $$\dfrac{12}{5}d$$

For E:

Value in 2020 = increased by 25% from 2016 = $$\left(1\ +\ \dfrac{25}{100}\right)e$$ = $$\dfrac{5}{4}e$$

Value in 2024 = increased by 20% from 2020 = $$\left(1\ +\ \dfrac{20}{100}\right)\times\ \dfrac{5e}{4}$$ = $$\dfrac{3}{2}e$$

For F:

Value in 2020 = increased by 100% from 2016 = $$\left(1\ +\ \dfrac{100}{100}\right)f$$ = $$2f$$

Value in 2024 = decreased by 25% from 2020 = $$\left(1\ -\ \dfrac{25}{100}\right)\times\ 2f$$ = $$\dfrac{3}{2}f$$

We are given that the value of E is 90 in 2024. We can calculate the value of e as,

$$\dfrac{3e}{2}\ =\ 90$$

$$e\ =\ 60$$

The value of E in 2016 is 60, and its value in 2020 can be calculated as,

$$\dfrac{5e}{4}\ =\ \dfrac{5}{4}\times\ 60\ =\ 75$$

We are given that B, C, E, and A had ranks 1, 2, 3, and 4, respectively, in 2016, and F has the lowest rank(6) in all the years.

Putting the values in the table, we get,

We know that F is ranked 6 and B is ranked 1 in 2016, and we are also given that the range of SI in 2016 and 2024 is 60. This means that the value of b-f = 60, and we can conclude that b is definitely greater than 60, as we know that f is an integer greater than 0.

We know that all the SI values in all the years are integers, and for SI of B to be an integer in 2024, the value of b must be a multiple of 16. The multiples of 16 that are greater than 60 and less than 100 are 64, 80 and 96.

CASE 1: b = 64

If b = 64, then the value of f becomes 4, as we know b - f = 60.

If f = 4, then the value of F in 2024 is 3f/2 = 12/2 = 6.

We already know the value of E in 2024 is 90.

In that case, the range of SI in 2024 is at least 90 - 6 = 84, but we are given that the range is 60 in 2024, which is not possible in this case.

So, we can eliminate this case.

CASE 2: b = 96

If b = 96, then the value of f becomes 36, as we know b - f = 60.

If f = 36, then the value of F in 2024 is 3f/2 = 108/2 = 54.

The value of B in 2024 = 9b/16 = (9 * 96)/16 = 54

We calculated the value of B and F to be 54 in 2024, which is not possible, as we are given that F had a lower SI than any other country in 2024, and we obtained its value to be the same as B.

So, we can eliminate this case.

CASE 3: b = 80

If b = 80, then the value of f becomes 20, as we know b - f = 60.

If f = 20, then the value of F in 2024 is 3f/2 = 60/2 = 30.

The value of B in 2024 = 9b/16 = (9 * 80)/16 = 45

We can see that no condition is violated in this case, as F is less than B in 2024, and the range in 2024 from the obtained values as of now is 90 - 30 = 60.

So, we can conclude that the value of b = 80 and f = 20.

Substituting the values in the table, we get,

We know that C is ranked 2nd in 2016, and we know the values of 1st and 3rd to be 80 and 60. We also know that SI of C has to be integers in 2016, 2020 and 2024. So, for the value of SI of C to be an integer in 2024, the value of c has to be a multiple of 25.

So, we know that c lies between 80 and 60, and is also a multiple of 25. The only multiple of 25 that lies between 60 and 80 is 75. So, the value of c has to be 75, and the value of C in 2024 can be calculated as,

C in 2024 = 28c/25 = (28 * 75)/25 = 84.

Hence, the correct answer is 84.