Trade Network | DILR CAT 2025 Slot 3 | Caselet | Hard
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.
How much is exported from C to X, in IC?
Answer & Explanation
Correct Answer: 48
Explanation: Pumpland is the only country exporting to Cheeseland, and exports 1200 to it. So total imports of Cheeseland are 1200. Cheeseland has a normalized trade balance of minus 20 percent. That means (total exports − total imports) divided by (total exports + total imports) equals −0.2. Substituting total imports = 1200, we get: (total exports − 1200) / (total exports + 1200) = −0.2. Solving this, we obtain total exports of Cheeseland equal to 800. The problem states that 90 percent of Cheeseland’s exports go to Pumpland and 4 percent go to Rest of World. So Cheeseland exports 0.9 × 800 = 720 to Pumpland and 0.04 × 800 = 32 to Rest of World. The remaining exports must go to Xiland. Hence exports from Cheeseland to Xiland are 800 − 720 − 32 = 48.
Step wide Explanation
Step 0 – Blank structure
| To \ From | P | X | C | ROW | Total Imports |
| P | ? | ? | ? | ? | ? |
| X | ? | ? | ? | ? | ? |
| C | ? | ? | ? | ? | ? |
| ROW | ? | ? | ? | ? | ? |
| Total Exports | ? | ? | ? | ? | ? |
Step 1 – Use direct information for P’s exports
Given P → X = 600 and P → C = 1200. No country trades with itself.
| To \ From | P | X | C | ROW | Total Imports |
| P | 0 | ? | ? | ? | ? |
| X | 600 | 0 | ? | ? | ? |
| C | 1200 | ? | 0 | ? | ? |
| ROW | ? | ? | ? | 0 | ? |
| Total Exports | ? | ? | ? | ? | ? |
Step 2 – Fix Cheeseland completely
C imports only from P and I_C = 1200.
Normalized trade balance of C = −20% gives E_C = 800.
Export split of C gives: C → P = 720, C → X = 48, C → ROW = 32.
| To \ From | P | X | C | ROW | Total Imports |
| P | 0 | ? | 720 | ? | ? |
| X | 600 | 0 | 48 | ? | ? |
| C | 1200 | ? | 0 | ? | 1200 |
| ROW | ? | ? | 32 | 0 | ? |
| Total Exports | ? | ? | 800 | ? | ? |
Step 3 – Solve for total trade of P and ROW (algebra stage)
Using normalized balances of P and X and ROW export percentages, we obtain:
I_P = E_P = 2000
E_ROW = 2100
From totals of P:
P → ROW = 2000 − 600 − 1200 = 200.
| To \ From | P | X | C | ROW | Total Imports |
| P | 0 | ? | 720 | ? | ? |
| X | 600 | 0 | 48 | ? | ? |
| C | 1200 | ? | 0 | ? | 1200 |
| ROW | 200 | ? | 32 | 0 | ? |
| Total Exports | 2000 | ? | 800 | ? | ? |
Step 4 – Fill ROW exports
40% of ROW exports go to P and 12% to X.
ROW → P = 840
ROW → X = 252
ROW → C = 0
ROW → ROW = 2100 − 840 − 252 = 1008
| To \ From | P | X | C | ROW | Total Imports |
| P | 0 | ? | 720 | 840 | ? |
| X | 600 | 0 | 48 | 252 | ? |
| C | 1200 | ? | 0 | 0 | 1200 |
| ROW | 200 | ? | 32 | 1008 | ? |
| Total Exports | 2000 | ? | 800 | 2100 | ? |
Step 5 – Complete X and close the table
From earlier results:
E_X = 1100
I_X = 900
Given 40% of X exports go to P:
X → P = 440
X → ROW = 1100 − 440 = 660
Final Table
| To \ From | P | X | C | ROW | Total Imports |
| P | 0 | 440 | 720 | 840 | 2000 |
| X | 600 | 0 | 48 | 252 | 900 |
| C | 1200 | 0 | 0 | 0 | 1200 |
| ROW | 200 | 660 | 32 | 1008 | 1900 |
| Total Exports | 2000 | 1100 | 800 | 2100 | 6200 |
How much is exported from P to ROW, in IC?
Answer & Explanation
Correct Answer: 200
Explanation: Pumpland’s normalized trade balance is 0 percent, so its total exports equal its total imports. From the equations involving Xiland’s 10 percent surplus and the export percentages of Rest of World, we get total exports and total imports of Pumpland both equal to 2000. We already know that Pumpland exports 600 to Xiland and 1200 to Cheeseland. Let exports from Pumpland to Rest of World be p. Then total exports of Pumpland are 600 to Xiland, 1200 to Cheeseland, and p to Rest of World. This sum must equal 2000. So 600 + 1200 + p = 2000. That gives 1800 + p = 2000, hence p = 200. Therefore, exports from Pumpland to Rest of World are 200.
Step wide Explanation
Step 0 – Blank structure
| To \ From | P | X | C | ROW | Total Imports |
| P | ? | ? | ? | ? | ? |
| X | ? | ? | ? | ? | ? |
| C | ? | ? | ? | ? | ? |
| ROW | ? | ? | ? | ? | ? |
| Total Exports | ? | ? | ? | ? | ? |
Step 1 – Use direct information for P’s exports
Given P → X = 600 and P → C = 1200. No country trades with itself.
| To \ From | P | X | C | ROW | Total Imports |
| P | 0 | ? | ? | ? | ? |
| X | 600 | 0 | ? | ? | ? |
| C | 1200 | ? | 0 | ? | ? |
| ROW | ? | ? | ? | 0 | ? |
| Total Exports | ? | ? | ? | ? | ? |
Step 2 – Fix Cheeseland completely
C imports only from P and I_C = 1200.
Normalized trade balance of C = −20% gives E_C = 800.
Export split of C gives: C → P = 720, C → X = 48, C → ROW = 32.
| To \ From | P | X | C | ROW | Total Imports |
| P | 0 | ? | 720 | ? | ? |
| X | 600 | 0 | 48 | ? | ? |
| C | 1200 | ? | 0 | ? | 1200 |
| ROW | ? | ? | 32 | 0 | ? |
| Total Exports | ? | ? | 800 | ? | ? |
Step 3 – Solve for total trade of P and ROW (algebra stage)
Using normalized balances of P and X and ROW export percentages, we obtain:
I_P = E_P = 2000
E_ROW = 2100
From totals of P:
P → ROW = 2000 − 600 − 1200 = 200.
| To \ From | P | X | C | ROW | Total Imports |
| P | 0 | ? | 720 | ? | ? |
| X | 600 | 0 | 48 | ? | ? |
| C | 1200 | ? | 0 | ? | 1200 |
| ROW | 200 | ? | 32 | 0 | ? |
| Total Exports | 2000 | ? | 800 | ? | ? |
Step 4 – Fill ROW exports
40% of ROW exports go to P and 12% to X.
ROW → P = 840
ROW → X = 252
ROW → C = 0
ROW → ROW = 2100 − 840 − 252 = 1008
| To \ From | P | X | C | ROW | Total Imports |
| P | 0 | ? | 720 | 840 | ? |
| X | 600 | 0 | 48 | 252 | ? |
| C | 1200 | ? | 0 | 0 | 1200 |
| ROW | 200 | ? | 32 | 1008 | ? |
| Total Exports | 2000 | ? | 800 | 2100 | ? |
Step 5 – Complete X and close the table
From earlier results:
E_X = 1100
I_X = 900
Given 40% of X exports go to P:
X → P = 440
X → ROW = 1100 − 440 = 660
Final Table
| To \ From | P | X | C | ROW | Total Imports |
| P | 0 | 440 | 720 | 840 | 2000 |
| X | 600 | 0 | 48 | 252 | 900 |
| C | 1200 | 0 | 0 | 0 | 1200 |
| ROW | 200 | 660 | 32 | 1008 | 1900 |
| Total Exports | 2000 | 1100 | 800 | 2100 | 6200 |
How much is exported from ROW to ROW, in IC?
Answer & Explanation
Correct Answer: 1008
Explanation: From the balance conditions of Pumpland and Xiland and the percentage breakdown of exports from Rest of World, we find that total exports of Rest of World are 2100. The data say that 40 percent of Rest of World’s exports go to Pumpland and 12 percent go to Xiland. So exports from Rest of World to Pumpland are 0.40 × 2100 = 840, and exports from Rest of World to Xiland are 0.12 × 2100 = 252. There are no exports from Rest of World to Cheeseland. Let exports from Rest of World to itself be r. The sum of all exports from Rest of World must be 2100. So 840 to Pumpland, 252 to Xiland, 0 to Cheeseland, and r to itself must add up to 2100. That gives 840 + 252 + r = 2100, so r = 2100 − 1092 = 1008. Therefore, exports from Rest of World to Rest of World are 1008.
Step wide Explanation
Step 0 – Blank structure
| To \ From | P | X | C | ROW | Total Imports |
| P | ? | ? | ? | ? | ? |
| X | ? | ? | ? | ? | ? |
| C | ? | ? | ? | ? | ? |
| ROW | ? | ? | ? | ? | ? |
| Total Exports | ? | ? | ? | ? | ? |
Step 1 – Use direct information for P’s exports
Given P → X = 600 and P → C = 1200. No country trades with itself.
| To \ From | P | X | C | ROW | Total Imports |
| P | 0 | ? | ? | ? | ? |
| X | 600 | 0 | ? | ? | ? |
| C | 1200 | ? | 0 | ? | ? |
| ROW | ? | ? | ? | 0 | ? |
| Total Exports | ? | ? | ? | ? | ? |
Step 2 – Fix Cheeseland completely
C imports only from P and I_C = 1200.
Normalized trade balance of C = −20% gives E_C = 800.
Export split of C gives: C → P = 720, C → X = 48, C → ROW = 32.
| To \ From | P | X | C | ROW | Total Imports |
| P | 0 | ? | 720 | ? | ? |
| X | 600 | 0 | 48 | ? | ? |
| C | 1200 | ? | 0 | ? | 1200 |
| ROW | ? | ? | 32 | 0 | ? |
| Total Exports | ? | ? | 800 | ? | ? |
Step 3 – Solve for total trade of P and ROW (algebra stage)
Using normalized balances of P and X and ROW export percentages, we obtain:
I_P = E_P = 2000
E_ROW = 2100
From totals of P:
P → ROW = 2000 − 600 − 1200 = 200.
| To \ From | P | X | C | ROW | Total Imports |
| P | 0 | ? | 720 | ? | ? |
| X | 600 | 0 | 48 | ? | ? |
| C | 1200 | ? | 0 | ? | 1200 |
| ROW | 200 | ? | 32 | 0 | ? |
| Total Exports | 2000 | ? | 800 | ? | ? |
Step 4 – Fill ROW exports
40% of ROW exports go to P and 12% to X.
ROW → P = 840
ROW → X = 252
ROW → C = 0
ROW → ROW = 2100 − 840 − 252 = 1008
| To \ From | P | X | C | ROW | Total Imports |
| P | 0 | ? | 720 | 840 | ? |
| X | 600 | 0 | 48 | 252 | ? |
| C | 1200 | ? | 0 | 0 | 1200 |
| ROW | 200 | ? | 32 | 1008 | ? |
| Total Exports | 2000 | ? | 800 | 2100 | ? |
Step 5 – Complete X and close the table
From earlier results:
E_X = 1100
I_X = 900
Given 40% of X exports go to P:
X → P = 440
X → ROW = 1100 − 440 = 660
Final Table
| To \ From | P | X | C | ROW | Total Imports |
| P | 0 | 440 | 720 | 840 | 2000 |
| X | 600 | 0 | 48 | 252 | 900 |
| C | 1200 | 0 | 0 | 0 | 1200 |
| ROW | 200 | 660 | 32 | 1008 | 1900 |
| Total Exports | 2000 | 1100 | 800 | 2100 | 6200 |
What is the trade balance of ROW?
1. 200 2. 0 3. 100 4. -200
Answer & Explanation
Correct Answer: 1. 200
Explanation: Trade balance is defined as total exports minus total imports. For Rest of World, total exports are 2100. To find total imports of Rest of World, we sum all flows going into Rest of World. From Pumpland, exports to Rest of World are 200. From Xiland, first note total exports of Xiland are 1100, and 40 percent of that (440) go to Pumpland. The remaining 660 go to Rest of World. From Cheeseland, exports to Rest of World are 32. Finally, Rest of World also “imports” from itself through its own internal trade, which is 1008. So total imports of Rest of World are 200 (from Pumpland) + 660 (from Xiland) + 32 (from Cheeseland) + 1008 (from itself) = 1900. Therefore, the trade balance of Rest of World is 2100 − 1900 = 200, which corresponds to option 1.
Step wide Explanation
Step 0 – Blank structure
| To \ From | P | X | C | ROW | Total Imports |
| P | ? | ? | ? | ? | ? |
| X | ? | ? | ? | ? | ? |
| C | ? | ? | ? | ? | ? |
| ROW | ? | ? | ? | ? | ? |
| Total Exports | ? | ? | ? | ? | ? |
Step 1 – Use direct information for P’s exports
Given P → X = 600 and P → C = 1200. No country trades with itself.
| To \ From | P | X | C | ROW | Total Imports |
| P | 0 | ? | ? | ? | ? |
| X | 600 | 0 | ? | ? | ? |
| C | 1200 | ? | 0 | ? | ? |
| ROW | ? | ? | ? | 0 | ? |
| Total Exports | ? | ? | ? | ? | ? |
Step 2 – Fix Cheeseland completely
C imports only from P and I_C = 1200.
Normalized trade balance of C = −20% gives E_C = 800.
Export split of C gives: C → P = 720, C → X = 48, C → ROW = 32.
| To \ From | P | X | C | ROW | Total Imports |
| P | 0 | ? | 720 | ? | ? |
| X | 600 | 0 | 48 | ? | ? |
| C | 1200 | ? | 0 | ? | 1200 |
| ROW | ? | ? | 32 | 0 | ? |
| Total Exports | ? | ? | 800 | ? | ? |
Step 3 – Solve for total trade of P and ROW (algebra stage)
Using normalized balances of P and X and ROW export percentages, we obtain:
I_P = E_P = 2000
E_ROW = 2100
From totals of P:
P → ROW = 2000 − 600 − 1200 = 200.
| To \ From | P | X | C | ROW | Total Imports |
| P | 0 | ? | 720 | ? | ? |
| X | 600 | 0 | 48 | ? | ? |
| C | 1200 | ? | 0 | ? | 1200 |
| ROW | 200 | ? | 32 | 0 | ? |
| Total Exports | 2000 | ? | 800 | ? | ? |
Step 4 – Fill ROW exports
40% of ROW exports go to P and 12% to X.
ROW → P = 840
ROW → X = 252
ROW → C = 0
ROW → ROW = 2100 − 840 − 252 = 1008
| To \ From | P | X | C | ROW | Total Imports |
| P | 0 | ? | 720 | 840 | ? |
| X | 600 | 0 | 48 | 252 | ? |
| C | 1200 | ? | 0 | 0 | 1200 |
| ROW | 200 | ? | 32 | 1008 | ? |
| Total Exports | 2000 | ? | 800 | 2100 | ? |
Step 5 – Complete X and close the table
From earlier results:
E_X = 1100
I_X = 900
Given 40% of X exports go to P:
X → P = 440
X → ROW = 1100 − 440 = 660
Final Table
| To \ From | P | X | C | ROW | Total Imports |
| P | 0 | 440 | 720 | 840 | 2000 |
| X | 600 | 0 | 48 | 252 | 900 |
| C | 1200 | 0 | 0 | 0 | 1200 |
| ROW | 200 | 660 | 32 | 1008 | 1900 |
| Total Exports | 2000 | 1100 | 800 | 2100 | 6200 |
Which among the countries P, X, and C has/have the least total trade?
1. Only P 2. Only C 3. Only X 4. Both X and C
Answer & Explanation
Correct Answer: 4. Both X and C
Explanation: Total trade of a country is defined as total exports plus total imports. For Pumpland, we already know total exports are 2000 and, since its normalized balance is zero, total imports are also 2000. So total trade of Pumpland is 2000 + 2000 = 4000. For Cheeseland, we previously found total exports equal to 800 and total imports equal to 1200. So total trade of Cheeseland is 800 + 1200 = 2000. For Xiland, total exports are 1100. Its normalized trade balance is +10 percent, which implies total imports of Xiland are 900. So total trade of Xiland is 1100 + 900 = 2000. Comparing total trades, Pumpland has 4000, while Xiland and Cheeseland each have 2000. Therefore the least total trade is shared by Xiland and Cheeseland, giving option 4.
Step wide Explanation
Step 0 – Blank structure
| To \ From | P | X | C | ROW | Total Imports |
| P | ? | ? | ? | ? | ? |
| X | ? | ? | ? | ? | ? |
| C | ? | ? | ? | ? | ? |
| ROW | ? | ? | ? | ? | ? |
| Total Exports | ? | ? | ? | ? | ? |
Step 1 – Use direct information for P’s exports
Given P → X = 600 and P → C = 1200. No country trades with itself.
| To \ From | P | X | C | ROW | Total Imports |
| P | 0 | ? | ? | ? | ? |
| X | 600 | 0 | ? | ? | ? |
| C | 1200 | ? | 0 | ? | ? |
| ROW | ? | ? | ? | 0 | ? |
| Total Exports | ? | ? | ? | ? | ? |
Step 2 – Fix Cheeseland completely
C imports only from P and I_C = 1200.
Normalized trade balance of C = −20% gives E_C = 800.
Export split of C gives: C → P = 720, C → X = 48, C → ROW = 32.
| To \ From | P | X | C | ROW | Total Imports |
| P | 0 | ? | 720 | ? | ? |
| X | 600 | 0 | 48 | ? | ? |
| C | 1200 | ? | 0 | ? | 1200 |
| ROW | ? | ? | 32 | 0 | ? |
| Total Exports | ? | ? | 800 | ? | ? |
Step 3 – Solve for total trade of P and ROW (algebra stage)
Using normalized balances of P and X and ROW export percentages, we obtain:
I_P = E_P = 2000
E_ROW = 2100
From totals of P:
P → ROW = 2000 − 600 − 1200 = 200.
| To \ From | P | X | C | ROW | Total Imports |
| P | 0 | ? | 720 | ? | ? |
| X | 600 | 0 | 48 | ? | ? |
| C | 1200 | ? | 0 | ? | 1200 |
| ROW | 200 | ? | 32 | 0 | ? |
| Total Exports | 2000 | ? | 800 | ? | ? |
Step 4 – Fill ROW exports
40% of ROW exports go to P and 12% to X.
ROW → P = 840
ROW → X = 252
ROW → C = 0
ROW → ROW = 2100 − 840 − 252 = 1008
| To \ From | P | X | C | ROW | Total Imports |
| P | 0 | ? | 720 | 840 | ? |
| X | 600 | 0 | 48 | 252 | ? |
| C | 1200 | ? | 0 | 0 | 1200 |
| ROW | 200 | ? | 32 | 1008 | ? |
| Total Exports | 2000 | ? | 800 | 2100 | ? |
Step 5 – Complete X and close the table
From earlier results:
E_X = 1100
I_X = 900
Given 40% of X exports go to P:
X → P = 440
X → ROW = 1100 − 440 = 660
Final Table
| To \ From | P | X | C | ROW | Total Imports |
| P | 0 | 440 | 720 | 840 | 2000 |
| X | 600 | 0 | 48 | 252 | 900 |
| C | 1200 | 0 | 0 | 0 | 1200 |
| ROW | 200 | 660 | 32 | 1008 | 1900 |
| Total Exports | 2000 | 1100 | 800 | 2100 | 6200 |









