Instructions [38 - 42 ] 

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.

What is the trade balance of ROW?

The correct answer is 200.

Let us break down the problem step-by-step to find the export and import volumes of each country and ultimately determine the trade balance of the Rest of the World (ROW).

Let:
$E_P$, $E_X$, $E_C$, and $E_R$ be the total exports of Pumpland (P), Xiland (X), Cheeseland (C), and ROW, respectively.
$I_P$, $I_X$, $I_C$, and $I_R$ be the total imports of P, X, C, and ROW, respectively.
Let $T_{A\to B}$ denote the trade volume exported from country A to country B.

Step 1: Analyze Cheeseland (C)
From Statement 5, we are told that P is the only country that exports to C. Since only P exports to C, the total imports of C are equal to the exports from P to C:
$I_C=T_{P\to C}=1200$.

From Statement 1, the normalized trade balance of C is -20% (-0.20):
$\frac{E_C-I_C}{E_C+I_C}=-0.20$
Substituting $I_C=1200$:
$E_C-1200=-0.20(E_C+1200)$
$E_C-1200=-0.20E_C-240$
$1.20E_C=960$
$E_C=800$.

From Statement 3, the distribution of C's exports is:
• To P: 90% of C's exports = $0.90\times 800=720$
• To ROW: 4% of C's exports = $0.04\times 800=32$
• The remaining 6% of C's exports must go to X: $0.06\times 800=48$.

Step 2: Analyze Pumpland (P)
From Statement 1, the normalized trade balance of P is 0%, which means:
$E_P=I_P$.

From Statement 5, we know P's exports to X and C are 600 and 1200, respectively. Let $T_{P\to R}$ be P's exports to ROW. Thus:
$E_P=600+1200+T_{P\to R}=1800+T_{P\to R}$.

Now, let us represent P's total imports ($I_P$). P imports from X, C, and ROW. We know:
• Imports from C: $T_{C\to P}=720$
• From Statement 2, imports from X make up 22% of P's total imports: $T_{X\to P}=0.22I_P$
• Since $E_P=I_P$, we have $T_{X\to P}=0.22E_P$.

From Statement 2, we also know that 40% of X's exports are to P:
$T_{X\to P}=0.40E_X$.
Equating the two expressions for $T_{X\to P}$:
$0.40E_X=0.22E_P\Rightarrow E_X=0.55E_P$.

Step 3: Analyze Xiland (X)
From Statement 1, the normalized trade balance of X is 10% (0.10):
$\frac{E_X-I_X}{E_X+I_X}=0.10$
$E_X-I_X=0.10E_X+0.10I_X$
$0.90E_X=1.10I_X\Rightarrow I_X=\frac{9}{11}{E}_X$.

Since $E_X=0.55E_P$:
$I_X=\frac{9}{11}\times 0.55E_P=0.45E_P$.

Let us express $I_X$ as the sum of its imports from P, C, and ROW:
• From P: $T_{P\to X}=600$ (from Statement 5)
• From C: $T_{C\to X}=48$ (from Step 1)
• From ROW: From Statement 4, 12% of ROW exports are to X: $T_{R\to X}=0.12E_R$.
Thus:
$I_X=600+48+0.12E_R=648+0.12E_R$.

Step 4: Connect P and ROW
We know that P's total imports are:
$I_P=T_{X\to P}+T_{C\to P}+T_{R\to P}$.
Using the values:
• $T_{X\to P}=0.22I_P=0.22E_P$
• $T_{C\to P}=720$
• From Statement 4, 40% of ROW exports go to P: $T_{R\to P}=0.40E_R$.
Since $I_P=E_P$, we have:
$E_P=0.22E_P+720+0.40E_R$
$0.78E_P-0.40E_R=720$ (Equation A).

From Step 3, we also have:
$I_X=0.45E_P=648+0.12E_R$
$0.45E_P-0.12E_R=648$ (Equation B).

Step 5: Solve for $E_P$ and $E_R$
From Equation B, we can express $E_R$ in terms of $E_P$:
$0.12E_R=0.45E_P-648$
$E_R=3.75E_P-5400$.

Substitute this into Equation A:
$0.78E_P-0.40(3.75E_P-5400)=720$
$0.78E_P-1.5E_P+2160=720$
$-0.72E_P=-1440$
$E_P=2000$.

Now, calculate $E_R$:
$E_R=3.75\left(2000\right)-5400=7500-5400=2100$.

Calculate other values based on $E_P=2000$:
• $E_X=0.55\times 2000=1100$.
• $I_X=0.45\times 2000=900$.
• Since $E_P=1800+T_{P\to R}$, we have $T_{P\to R}=2000-1800=200$.
• Since $E_X=1100$ and $T_{X\to P}=0.40E_X=440$, and exports to C are 0, the exports of X to ROW are:
$T_{X\to R}=E_X-T_{X\to P}=1100-440=660$.

Step 6: Calculate the trade balance of ROW
The imports of ROW ($I_R$) consist of the exports from all other countries to ROW:
$I_R=T_{P\to R}+T_{X\to R}+T_{C\to R}$
Substituting the calculated values:
$I_R=200+660+32=892$
By accounting consistency, the sum of all trade balances in a closed world economy must equal zero. Let us verify:
• Trade balance of P: $E_P-I_P=2000-2000=0$.
• Trade balance of X: $E_X-I_X=1100-900=200$.
• Trade balance of C: $E_C-I_C=800-1200=-400$.
• Let $B$ be the trade balance of ROW:
$0+200-400+B=0\Rightarrow B=200$.

Thus, the trade balance of ROW is 200.