CAT 2025 Slot 3 DILR Question & Solution

Data InterpretationHard

Data Set

Three countries — Pumpland (P), Xiland (X) and Cheeseland (C) — trade among themselves and with the (other countries in) Rest of World (ROW). All trade volumes are given in IC (international currency). The following terminology is used:
• Trade balance = Exports - Imports
• Total trade = Exports + Imports
• Normalized trade balance = Trade balance / Total trade, expressed in percentage terms

The following information is known.
1. The normalized trade balances of P, X and C are 0%, 10%, and -20%, respectively.
2. 40% of exports of X are to P. 22% of imports of P are from X.
3. 90% of exports of C are to P; 4% are to ROW.
4. 12% of exports of ROW are to X, 40% are to P.
5. The export volumes of P, in IC, to X and C are 600 and 1200, respectively. P is the only country that exports to C.

Question 1

How much is exported from C to X, in IC?

Solution:

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.

Question 2

How much is exported from P to ROW, in IC?

Solution:

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.

Question 3

How much is exported from ROW to ROW, in IC?

Solution:

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.

Question 4

What is the trade balance of ROW?

-200

100

200

Solution:

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,

Trade balance of ROW = Exports - Imports = 2100 - 1900 = 200.

Hence, the correct answer is option D.

Question 5

Which among the countries P, X, and C has/have the least total trade?

Only X

Only C

Both X and C

Only p

Solution:

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,

Total trade of P = 2000 + 2000 = 4000

Total trade of X = 900 + 1100 = 2000

Total trade of C = 1200 + 800 = 2000

So, both countries X and C have the least total trade.

Hence, the correct answer is option C.