CAT 2025 Slot 3 Question Paper

Let us assume units of Aurels, Brins, Crowns, and Zentars to be A, B, C and D, respectively.

We are given the travel cost of Jano to be 3500A in Statement 1.

We are also given the travel cost of Kira to be 8000A in Statement 1 and 4000B in Statement 2.

We are also given the travel cost of Lian to be 36000C in statement 3.

We know that the travel cost has to be constant throughout, so by equating the travel costs of Kira, we get,

8000A = 4000B

B = 2A ---(1)

We are also given the value of C to be 0.5 Z.

C = 0.5Z ---(2)

So, the travel cost of Lian = 36000C = 36000*0.5Z = 18000Z

Let us put the known information in the table, and we get,

We are given that the spending cost of any individual is different in different cities.

We are given that the spending amounts are amongst {1000, 2000, 3000} in their local currencies and also told that there were two spends of 1000 and one spend of 3000, which makes the total spending amount 2000, to be 3 because there are 6 spending amounts in total.

One 2000 and one 3000 are already assigned in the above table, so we will be left with two spends of 1000 and two spends of 2000 to be assigned.

Kira has already spent 2000, so the only amount that she must have spent is 1000, as she cannot spend 2000 twice in different countries.

Lian has already spent 3000, so the only amount that she must have spent is 2000. If she spent 1000, then Jano will be left with two 2000s, which is not possible.

We can also conclude that Jano spend 1000 and 2000 in some order in the two countries.

The flight costs are given as 4000Z, 5000Z, and 6000Z, out of which 4000Z is for Jano and 5000Z and 6000Z are for Kira and Lian, in some order.

Filling up the table with these values, we get,

Travel Cost = Spending Cost + Flight Cost

For Kira,

Travel Cost = 8000A

Spending Cost = 1000A + 4000A = 5000A

Flight Cost = 5000Z/6000Z

CASE 1: Flight cost of Kira = 5000Z and Flight cost of Lian = 6000Z

If we assume the Flight cost to be 5000Z, then we get,

8000A = 5000A + 5000Z

3000A = 5000Z

A = $$\frac{5}{3}$$Z

For Lian,

Travel Cost = 18000Z

Spending Cost = 6000A + 1000Z = 6000$$\times\ \frac{5}{3}$$Z + 1000Z = 11000Z

Flight Cost = Travel Cost - Spending Cost = 18000Z - 11000Z = 7000Z

But in our assumption, the Flight Cost of Lian is 6000Z, which does not match the above answer.

So, we can eliminate this case.

CASE 2: Flight cost of Kira = 6000Z and Flight cost of Lian = 5000Z

If we assume the Flight cost to be 6000Z, then we get,

8000A = 5000A + 6000Z

3000A = 6000Z

A = 2Z

For Lian,

Travel Cost = 18000Z

Spending Cost = 6000A + 1000Z = 6000*2Z + 1000Z = 13000Z

Flight Cost = Travel Cost - Spending Cost = 18000Z - 13000Z = 5000Z

In our assumption, the Flight Cost of Lian is 5000Z, which matches the above answer.

So, Flight cost of Kira = 6000Z, Flight cost of Lian = 5000Z and A = 2Z.

Filling the table with calculated values all in the currency of Z, we get,

In the case of Zano,

Spending Cost = Travel Cost - Flight Cost = 7000Z - 4000Z = 3000Z

In the first case, the Spending Cost = 2000Z + 1000Z = 3000Z

In the second case, the Spending Cost = 4000Z + 500Z = 4500Z

We obtained a spending cost of 3000Z only in the first case, so we can eliminate the second case.

The final table looks like,

Total Spending Cost of Jano = 2000Z + 1000Z = 3000Z

We know that A = 2Z, so the above value in A can be calculated as,

3000Z = 3000*A/2 = 1500A

So, the spending of Jano in Aurels is 1500A.

Hence, the correct answer is 1500.

Let us assume units of Aurels, Brins, Crowns, and Zentars to be A, B, C and D, respectively.

We are given the travel cost of Jano to be 3500A in Statement 1.

We are also given the travel cost of Kira to be 8000A in Statement 1 and 4000B in Statement 2.

We are also given the travel cost of Lian to be 36000C in statement 3.

We know that the travel cost has to be constant throughout, so by equating the travel costs of Kira, we get,

8000A = 4000B

B = 2A ---(1)

We are also given the value of C to be 0.5 Z.

C = 0.5Z ---(2)

So, the travel cost of Lian = 36000C = 36000*0.5Z = 18000Z

Let us put the known information in the table, and we get,

We are given that the spending cost of any individual is different in different cities.

We are given that the spending amounts are amongst {1000, 2000, 3000} in their local currencies and also told that there were two spends of 1000 and one spend of 3000, which makes the total spending amount 2000, to be 3 because there are 6 spending amounts in total.

One 2000 and one 3000 are already assigned in the above table, so we will be left with two spends of 1000 and two spends of 2000 to be assigned.

Kira has already spent 2000, so the only amount that she must have spent is 1000, as she cannot spend 2000 twice in different countries.

Lian has already spent 3000, so the only amount that she must have spent is 2000. If she spent 1000, then Jano will be left with two 2000s, which is not possible.

We can also conclude that Jano spent 1000 and 2000 in some order in the two countries.

The flight costs are given as 4000Z, 5000Z, and 6000Z, out of which 4000Z is for Jano and 5000Z and 6000Z are for Kira and Lian, in some order.

Filling up the table with these values, we get,

Travel Cost = Spending Cost + Flight Cost

For Kira,

Travel Cost = 8000A

Spending Cost = 1000A + 4000A = 5000A

Flight Cost = 5000Z/6000Z

CASE 1: Flight cost of Kira = 5000Z and Flight cost of Lian = 6000Z

If we assume the Flight cost to be 5000Z, then we get,

8000A = 5000A + 5000Z

3000A = 5000Z

A = $$\frac{5}{3}$$Z

For Lian,

Travel Cost = 18000Z

Spending Cost = 6000A + 1000Z = 6000$$\times\ \frac{5}{3}$$Z + 1000Z = 11000Z

Flight Cost = Travel Cost - Spending Cost = 18000Z - 11000Z = 7000Z

But in our assumption, the Flight Cost of Lian is 6000Z, which does not match the above answer.

So, we can eliminate this case.

CASE 2: Flight cost of Kira = 6000Z and Flight cost of Lian = 5000Z

If we assume the Flight cost to be 6000Z, then we get,

8000A = 5000A + 6000Z

3000A = 6000Z

A = 2Z

For Lian,

Travel Cost = 18000Z

Spending Cost = 6000A + 1000Z = 6000*2Z + 1000Z = 13000Z

Flight Cost = Travel Cost - Spending Cost = 18000Z - 13000Z = 5000Z

In our assumption, the Flight Cost of Lian is 5000Z, which matches the above answer.

So, Flight cost of Kira = 6000Z, Flight cost of Lian = 5000Z and A = 2Z.

Filling the table with calculated values all in the currency of Z, we get,

In the case of Zano,

Spending Cost = Travel Cost - Flight Cost = 7000Z - 4000Z = 3000Z

In the first case, the Spending Cost = 2000Z + 1000Z = 3000Z

In the second case, the Spending Cost = 4000Z + 500Z = 4500Z

We obtained a spending cost of 3000Z only in the first case, so we can eliminate the second case.

The final table looks like,

We know that B = 2A and A = 2Z, so the value of B in terms of Z is 

B = 2*2Z = 4Z  --(3)

We also know that 

C = 0.5Z  --(4)

Dividing (3) by (4), we get,

$$\dfrac{B}{C}\ =\ \dfrac{4Z}{0.5Z}\ =\ 8$$

B = 8C

So, one Brin is equivalent to 8 crowns.

Hence, the correct answer is option A.

Let us assume units of Aurels, Brins, Crowns, and Zentars to be A, B, C and D, respectively.

We are given the travel cost of Jano to be 3500A in Statement 1.

We are also given the travel cost of Kira to be 8000A in Statement 1 and 4000B in Statement 2.

We are also given the travel cost of Lian to be 36000C in statement 3.

We know that the travel cost has to be constant throughout, so by equating the travel costs of Kira, we get,

8000A = 4000B

B = 2A ---(1)

We are also given the value of C to be 0.5 Z.

C = 0.5Z ---(2)

So, the travel cost of Lian = 36000C = 36000*0.5Z = 18000Z

Let us put the known information in the table, and we get,

We are given that the spending cost of any individual is different in different cities.

We are given that the spending amounts are amongst {1000, 2000, 3000} in their local currencies and also told that there were two spends of 1000 and one spend of 3000, which makes the total spending amount 2000, to be 3 because there are 6 spending amounts in total.

One 2000 and one 3000 are already assigned in the above table, so we will be left with two spends of 1000 and two spends of 2000 to be assigned.

Kira has already spent 2000, so the only amount that she must have spent is 1000, as she cannot spend 2000 twice in different countries.

Lian has already spent 3000, so the only amount that she must have spent is 2000. If she spent 1000, then Jano will be left with two 2000s, which is not possible.

We can also conclude that Jano spend 1000 and 2000 in some order in the two countries.

The flight costs are given as 4000Z, 5000Z, and 6000Z, out of which 4000Z is for Jano and 5000Z and 6000Z are for Kira and Lian, in some order.

Filling up the table with these values, we get,

Travel Cost = Spending Cost + Flight Cost

For Kira,

Travel Cost = 8000A

Spending Cost = 1000A + 4000A = 5000A

Flight Cost = 5000Z/6000Z

CASE 1: Flight cost of Kira = 5000Z and Flight cost of Lian = 6000Z

If we assume the Flight cost to be 5000Z, then we get,

8000A = 5000A + 5000Z

3000A = 5000Z

A = $$\frac{5}{3}$$Z

For Lian,

Travel Cost = 18000Z

Spending Cost = 6000A + 1000Z = 6000$$\times\ \frac{5}{3}$$Z + 1000Z = 11000Z

Flight Cost = Travel Cost - Spending Cost = 18000Z - 11000Z = 7000Z

But in our assumption, the Flight Cost of Lian is 6000Z, which does not match the above answer.

So, we can eliminate this case.

CASE 2: Flight cost of Kira = 6000Z and Flight cost of Lian = 5000Z

If we assume the Flight cost to be 6000Z, then we get,

8000A = 5000A + 6000Z

3000A = 6000Z

A = 2Z

For Lian,

Travel Cost = 18000Z

Spending Cost = 6000A + 1000Z = 6000*2Z + 1000Z = 13000Z

Flight Cost = Travel Cost - Spending Cost = 18000Z - 13000Z = 5000Z

In our assumption, the Flight Cost of Lian is 5000Z, which matches the above answer.

So, Flight cost of Kira = 6000Z, Flight cost of Lian = 5000Z and A = 2Z.

Filling the table with calculated values all in the currency of Z, we get,

In the case of Zano,

Spending Cost = Travel Cost - Flight Cost = 7000Z - 4000Z = 3000Z

In the first case, the Spending Cost = 2000Z + 1000Z = 3000Z

In the second case, the Spending Cost = 4000Z + 500Z = 4500Z

We obtained a spending cost of 3000Z only in the first case, so we can eliminate the second case.

The final table looks like,

Statement 1) Jano spent 2000 in Aurevia

Jano spent 2000Z in Aurevia = 1000A in Aurevia.

So, this option is incorrect.

Statement 2) Lian spent 2000 in Cyrenia

Lian spent 1000Z in Cyrenia = 2000C in Cyrenia.

So, this option is correct.

Statement 3) Jano spent 2000 in Cyrenia

Jano spent 1000Z in Cyrenia = 2000C in Cyrenia.

So, this option is correct.

Statement 4) Kira spent 1000 in Aurevia

Kira spent 2000Z in Aurevia = 1000A in Aurevia.

So, this option is correct.

Hence, the correct answer is option A.

The left plot shows, for each competitor, a step (staircase) curve whose height at any time t equals the number of puzzles that competitor has completed by minute t.

To find who solved the most puzzles by the 20th minute, draw a vertical line at t=20 (or simply read the height of each competitor’s colored stair at the vertical line x=20).

Reading the four curves at t=20:

The green staircase (Koushik) is at the highest horizontal level at t=20.

The yellow (Chandranath) and red (Anirbid) staircases are lower than the green ones at that time.

The blue (Suranjan) staircase is the lowest of the four at t=20.

Since Koushik’s value is greater than every other competitor’s at t=20, Koushik has solved the largest number of puzzles by the 20th minute.

Hence, the correct answer is option B.

Comparing the 1st and 2nd graphs, we should be able to determine the type of puzzle.

From the second graph, we can say that there are 4 visual-based puzzles in total, which makes the number-based puzzles 6.

Let us follow the green line in both graphs:

The 1st visual-based puzzle was completed by Green in 2 minutes from the 2nd graph. From the 1st graph at the 2nd minute, the puzzle that was solved by Green was 1. So, we can say that 1 is a visual-based puzzle.

The 2nd visual-based puzzle was completed by Green in 12 minutes from the 2nd graph. From the 1st graph at the 12th minute, the puzzle that was solved by Green was 4. So, we can say that 4 is a visual-based puzzle.

The 3rd visual-based puzzle was completed by Green in 25 minutes from the 2nd graph. From the 1st graph at the 25th minute, the puzzle that was solved by Green was 8. So, we can say that 8 is a visual-based puzzle.

The 4th visual-based puzzle was completed by Green in 29 minutes from the 2nd graph. From the 1st graph at the 29th minute, the puzzle that was solved by Green was 9. So, we can say that 9 is a visual-based puzzle.

We can conclude that puzzles 1, 4, 8, and 9 are visual-based, while puzzles 2, 3, 5, 6, 7, and 10 are number-based.

So, the third visual puzzle is 8, and the time taken by Suranjan(blue) to complete the 3rd visual-based puzzle, which is the 8th puzzle, can be calculated as,

The time taken by Suranjan to solve the 8th puzzle from graph 1 = 28 - 26 = 2 minutes.

Hence, the correct answer is 2.

Comparing the 1st and 2nd graphs, we should be able to determine the type of puzzle.

From the second graph, we can say that there are 4 visual-based puzzles in total, which makes the number-based puzzles 6.

Let us follow the green line in both graphs:

The 1st visual-based puzzle was completed by Green in 2 minutes from the 2nd graph. From the 1st graph at the 2nd minute, the puzzle that was solved by Green was 1. So, we can say that 1 is a visual-based puzzle.

The 2nd visual-based puzzle was completed by Green in 12 minutes from the 2nd graph. From the 1st graph at the 12th minute, the puzzle that was solved by Green was 4. So, we can say that 4 is a visual-based puzzle.

The 3rd visual-based puzzle was completed by Green in 25 minutes from the 2nd graph. From the 1st graph at the 25th minute, the puzzle that was solved by Green was 8. So, we can say that 8 is a visual-based puzzle.

The 4th visual-based puzzle was completed by Green in 29 minutes from the 2nd graph. From the 1st graph at the 29th minute, the puzzle that was solved by Green was 9. So, we can say that 9 is a visual-based puzzle.

We can conclude that puzzles 1, 4, 8, and 9 are visual-based, while puzzles 2, 3, 5, 6, 7, and 10 are number-based.

The fourth number-based puzzle is 6th puzzle in the order.

Hence, the correct answer is 6.

Comparing the 1st and 2nd graphs, we should be able to determine the type of puzzle.

From the second graph, we can say that there are 4 visual-based puzzles in total, which makes the number-based puzzles 6.

Let us follow the green line in both graphs:

The 1st visual-based puzzle was completed by Green in 2 minutes from the 2nd graph. From the 1st graph at the 2nd minute, the puzzle that was solved by Green was 1. So, we can say that 1 is a visual-based puzzle.

The 2nd visual-based puzzle was completed by Green in 12 minutes from the 2nd graph. From the 1st graph at the 12th minute, the puzzle that was solved by Green was 4. So, we can say that 4 is a visual-based puzzle.

The 3rd visual-based puzzle was completed by Green in 25 minutes from the 2nd graph. From the 1st graph at the 25th minute, the puzzle that was solved by Green was 8. So, we can say that 8 is a visual-based puzzle.

The 4th visual-based puzzle was completed by Green in 29 minutes from the 2nd graph. From the 1st graph at the 29th minute, the puzzle that was solved by Green was 9. So, we can say that 9 is a visual-based puzzle.

We can conclude that puzzles 1, 4, 8, and 9 are visual-based, while puzzles 2, 3, 5, 6, 7, and 10 are number-based.

The average time taken by Anirbid to solve the number-based puzzles can be calculated as,

Time for 1st number-based puzzle(2nd puzzle) from 1st graph = 9 - 4 = 5 minutes.

Time for 2nd number-based puzzle(3rd puzzle) from 1st graph = 11 - 9 = 2 minutes.

Time for 3rd number-based puzzle(5th puzzle) from 1st graph = 19 - 14 = 5 minutes.

Time for 4th number-based puzzle(6th puzzle) from 1st graph = 22 - 19 = 3 minutes.

Time for 5th number-based puzzle(7th puzzle) from 1st graph = 27 - 22 = 5 minutes.

Time for 6th number-based puzzle(10th puzzle) from 1st graph = 37 - 33 = 4 minutes.

Average $$=\ \dfrac{5\ +\ 2\ +\ 5\ +\ 3\ +\ 5\ +\ 4}{6}\ =\ \dfrac{24}{6}\ =\ 4$$ minutes

The average time taken by Anirbid to solve the number-based puzzles is 4 minutes.

Hence, the correct answer is option D.

In clue 5, we are given that export volumes of P, in IC, to X and C are 600 and 1200, respectively, and P is the only country that exports to C. So, the export from X to C and ROW to C is 0.

In clue 1, we are given that normalised trade balances of P, X and C are 0%, 10%, and -20%, respectively.

For P, since the normalised trade balance is 0% the exports have to be equal to the imports according to the given formula. So, let us assume them to be 10a and 10a.

For X,

$$\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ 10$$

$$10E\ -\ 10I\ =\ E\ +\ I$$

$$11I\ =\ 9E$$

So, let us assume export to be 11b and import to be 9b.

For C,

$$\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ -20$$

$$5I\ -\ 5E\ =\ E\ +\ I$$

$$2I\ =\ 3E$$

So, let us assume export to be 2c and import to be 3c.

In clue 2, we are given that 40% of exports of X are to P and 22% of imports of P are from X. 

40% of the exports of X $$=\dfrac{40}{100}\ \times\ 11b\ =\ 4.4b\ =$$ exports from X to P

22% of the imports of P $$=\dfrac{22}{100}\ \times\ 10a\ =\ 2.2a\ =$$ imports to P from X

Equating both, we get,

4.4b = 2.2a

a = 2b.

We are also given that imports for C are only from P, and we can set the rest of the imports for C to 0.

The total imports for C = 3c = 1200

c = 400

This makes the exports for C = 2c = 800

In clue 3, we are given that 90% of exports of C are to P and 4% are to ROW.

90% of the exports of C $$=\dfrac{90}{100}\ \times\ 2c\ =\ 1.8c\ =1.8\ \times\ 400\ =\ 720\ =\ $$exports from C to P

4% of the exports of C $$=\dfrac{4}{100}\ \times\ 2c\ =\ 0.08c\ =\ 0.08\ \times\ 400\ =\ 32\ =$$ exports from C to ROW

The exports from C to X = 800 - 720 - 32 = 48

In clue 4, we are given that 12% of exports of ROW are to X, 40% are to P.

If we assume that Imports of ROW are m and exports of ROW are n.

12% of the exports of ROW $$=\dfrac{12}{100}\ \times\ n\ =\ 0.12n\ =$$ exports from ROW to X

40% of the exports of ROW $$=\dfrac{40}{100}\ \times\ n\ =\ 0.4n\ =$$ exports from ROW to P

Placing all the values in the table, we get,

Equating the imports of P and X to the Totals, we get,

For P,

4.4b + 720 + 0.4n = 20b

15.6b = 720 + 0.4n  --(1)

For X,

600 + 48 + 0.12n = 9b

9b = 648 + 0.12n  --(2)

Solving (1) and (2), we get

$$\dfrac{648\ +\ 0.12n}{9}\ =\ \dfrac{720\ +\ 0.4n}{15.6}$$

$$10108.8+1.872n=6480+3.6n$$

$$3628.8=1.728n$$

$$n=2100$$

From (1),

15.6b = 720 + 0.4(2100)

15.6b = 1560

b = 100

Exports of P to ROW = 20b - 600 - 1200 = 2000 - 1800 = 200

Exports of X to ROW = 11b - 4.4b = 6.6b = 660

Substituting the values of b and n, we can figure out all the values except for Imports and Exports of ROW to ROW.

The value of exports and imports of ROW to ROW has to be equal.

We calculated the value of n to be 2100.

The exports of ROW to ROW can be calculated as,

Exports of ROW to ROW = 2100 - 840 - 252 = 1008 = Imports of ROW to ROW.

The value of m can be calculated as,

m = 200 + 660 + 32 + 1008 = 1900.

Filling up the table with all the values, we get,

Exports from C to X are 48 IC.

Hence, the correct answer is 48. 

In clue 5, we are given that export volumes of P, in IC, to X and C are 600 and 1200, respectively, and P is the only country that exports to C. So, the export from X to C and ROW to C is 0.

In clue 1, we are given that normalised trade balances of P, X and C are 0%, 10%, and -20%, respectively.

For P, since the normalised trade balance is 0% the exports have to be equal to the imports according to the given formula. So, let us assume them to be 10a and 10a.

For X,

$$\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ 10$$

$$10E\ -\ 10I\ =\ E\ +\ I$$

$$11I\ =\ 9E$$

So, let us assume export to be 11b and import to be 9b.

For C,

$$\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ -20$$

$$5I\ -\ 5E\ =\ E\ +\ I$$

$$2I\ =\ 3E$$

So, let us assume export to be 2c and import to be 3c.

In clue 2, we are given that 40% of exports of X are to P and 22% of imports of P are from X.

40% of the exports of X $$=\dfrac{40}{100}\ \times\ 11b\ =\ 4.4b\ =$$ exports from X to P

22% of the imports of P $$=\dfrac{22}{100}\ \times\ 10a\ =\ 2.2a\ =$$ imports to P from X

Equating both, we get,

4.4b = 2.2a

a = 2b.

We are also given that imports for C are only from P, and we can set the rest of the imports for C to 0.

The total imports for C = 3c = 1200

c = 400

This makes the exports for C = 2c = 800

In clue 3, we are given that 90% of exports of C are to P and 4% are to ROW.

90% of the exports of C $$=\dfrac{90}{100}\ \times\ 2c\ =\ 1.8c\ =1.8\ \times\ 400\ =\ 720\ =\ $$exports from C to P

4% of the exports of C $$=\frac{4}{100}\ \times\ 2c\ =\ 0.08c\ =\ 0.08\ \times\ 400\ =\ 32\ =$$ exports from C to ROW

The exports from C to X = 800 - 720 - 32 = 48

In clue 4, we are given that 12% of exports of ROW are to X, 40% are to P.

If we assume that Imports of ROW are m and exports of ROW are n.

12% of the exports of ROW $$=\dfrac{12}{100}\ \times\ n\ =\ 0.12n\ =$$ exports from ROW to X

40% of the exports of ROW $$=\dfrac{40}{100}\ \times\ n\ =\ 0.4n\ =$$ exports from ROW to P

Placing all the values in the table, we get,

Equating the imports of P and X to the Totals, we get,

For P,

4.4b + 720 + 0.4n = 20b

15.6b = 720 + 0.4n --(1)

For X,

600 + 48 + 0.12n = 9b

9b = 648 + 0.12n --(2)

Solving (1) and (2), we get

$$\dfrac{648\ +\ 0.12n}{9}\ =\ \dfrac{720\ +\ 0.4n}{15.6}$$

$$10108.8+1.872n=6480+3.6n$$

$$3628.8=1.728n$$

$$n=2100$$

From (1),

15.6b = 720 + 0.4(2100)

15.6b = 1560

b = 100

Exports of P to ROW = 20b - 600 - 1200 = 2000 - 1800 = 200

Exports of X to ROW = 11b - 4.4b = 6.6b = 660

Substituting the values of b and n, we can figure out all the values except for Imports and Exports of ROW to ROW.

The value of exports and imports of ROW to ROW has to be equal.

We calculated the value of n to be 2100.

The exports of ROW to ROW can be calculated as,

Exports of ROW to ROW = 2100 - 840 - 252 = 1008 = Imports of ROW to ROW.

The value of m can be calculated as,

m = 200 + 660 + 32 + 1008 = 1900.

Filling up the table with all the values, we get,

The exports from P to ROW 200 IC.

Hence, the correct answer is 200.

In clue 5, we are given that export volumes of P, in IC, to X and C are 600 and 1200, respectively, and P is the only country that exports to C. So, the export from X to C and ROW to C is 0.

In clue 1, we are given that normalised trade balances of P, X and C are 0%, 10%, and -20%, respectively.

For P, since the normalised trade balance is 0% the exports have to be equal to the imports according to the given formula. So, let us assume them to be 10a and 10a.

For X,

$$\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ 10$$

$$10E\ -\ 10I\ =\ E\ +\ I$$

$$11I\ =\ 9E$$

So, let us assume export to be 11b and import to be 9b.

For C,

$$\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ -20$$

$$5I\ -\ 5E\ =\ E\ +\ I$$

$$2I\ =\ 3E$$

So, let us assume export to be 2c and import to be 3c.

In clue 2, we are given that 40% of exports of X are to P and 22% of imports of P are from X.

40% of the exports of X $$=\dfrac{40}{100}\ \times\ 11b\ =\ 4.4b\ =$$ exports from X to P

22% of the imports of P $$=\dfrac{22}{100}\ \times\ 10a\ =\ 2.2a\ =$$ imports to P from X

Equating both, we get,

4.4b = 2.2a

a = 2b.

We are also given that imports for C are only from P, and we can set the rest of the imports for C to 0.

The total imports for C = 3c = 1200

c = 400

This makes the exports for C = 2c = 800

In clue 3, we are given that 90% of exports of C are to P and 4% are to ROW.

90% of the exports of C $$=\dfrac{90}{100}\ \times\ 2c\ =\ 1.8c\ =1.8\ \times\ 400\ =\ 720\ =\ $$exports from C to P

4% of the exports of C $$=\frac{4}{100}\ \times\ 2c\ =\ 0.08c\ =\ 0.08\ \times\ 400\ =\ 32\ =$$ exports from C to ROW

The exports from C to X = 800 - 720 - 32 = 48

In clue 4, we are given that 12% of exports of ROW are to X, 40% are to P.

If we assume that Imports of ROW are m and exports of ROW are n.

12% of the exports of ROW $$=\dfrac{12}{100}\ \times\ n\ =\ 0.12n\ =$$ exports from ROW to X

40% of the exports of ROW $$=\dfrac{40}{100}\ \times\ n\ =\ 0.4n\ =$$ exports from ROW to P

Placing all the values in the table, we get,

Equating the imports of P and X to the Totals, we get,

For P,

4.4b + 720 + 0.4n = 20b

15.6b = 720 + 0.4n --(1)

For X,

600 + 48 + 0.12n = 9b

9b = 648 + 0.12n --(2)

Solving (1) and (2), we get

$$\dfrac{648\ +\ 0.12n}{9}\ =\ \dfrac{720\ +\ 0.4n}{15.6}$$

$$10108.8+1.872n=6480+3.6n$$

$$3628.8=1.728n$$

$$n=2100$$

From (1),

15.6b = 720 + 0.4(2100)

15.6b = 1560

b = 100

Exports of P to ROW = 20b - 600 - 1200 = 2000 - 1800 = 200

Exports of X to ROW = 11b - 4.4b = 6.6b = 660

Substituting the values of b and n, we can figure out all the values except for Imports and Exports of ROW to ROW.

The value of exports and imports of ROW to ROW has to be equal.

We calculated the value of n to be 2100.

The exports of ROW to ROW can be calculated as,

Exports of ROW to ROW = 2100 - 840 - 252 = 1008 = Imports of ROW to ROW.

The value of m can be calculated as,

m = 200 + 660 + 32 + 1008 = 1900.

Filling up the table with all the values, we get,

The exports from ROW to ROW are 1008 IC.

Hence, the correct answer is 1008.