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.

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

The correct answer is 1008.

1. Understanding the Trade Definitions and Initial Data
Let the three countries be Pumpland (P), Xiland (X), and Cheeseland (C), and the rest of the world be ROW.
For any country, we define:
• Normalized Trade Balance (NTB) = $$\frac{\text{Exports}-\text{Imports}}{\text{Exports}+\text{Imports}}\times 100\%$$

From the given information:
• Pumpland (P): NTB_P = 0%, which means:
$${\mathrm{Exports}}_P={\mathrm{Imports}}_P$$
Let $E_P=I_P$.

• Xiland (X): NTB_X = 10%, which means:
$$\frac{E_X-I_X}{E_X+I_X}=0.1\Rightarrow E_X-I_X=0.1E_X+0.1I_X\Rightarrow 0.9E_X=1.1I_X$$
Therefore, $I_X=\frac{9}{11}{E}_X$.

• Cheeseland (C): NTB_C = -20%, which means:
$$\frac{E_C-I_C}{E_C+I_C}=-0.2\Rightarrow E_C-I_C=-0.2E_C-0.2I_C\Rightarrow 1.2E_C=0.8I_C$$
Therefore, $I_C=1.5E_C$.

2. Analyzing Cheeseland (C) Trade
From Information 5, P's exports to C are 1200 IC, and P is the only country that exports to C.
This means Cheeseland's total imports are:
$$I_C=1200$$
Using $I_C=1.5E_C$:
$$1200=1.5E_C\Rightarrow E_C=800$$

From Information 3, C's exports are distributed as follows:
• To P: 90% of C's exports = $0.9\times 800=720$
• To ROW: 4% of C's exports = $0.04\times 800=32$
• To X (the remainder): $100\%-90\%-4\%=6\%$ of C's exports = $0.06\times 800=48$

3. Connecting Xiland (X) and Pumpland (P)
From Information 2:
• 40% of exports of X are to P:
$$E_{X\to P}=0.4E_X$$
• 22% of imports of P are from X:
$$I_{}P\leftarrow X=0.22I_P$$
Since the volume exported from X to P equals the volume imported by P from X:
$$0.4E_X=0.22I_P\Rightarrow E_X=0.55I_P$$
Substituting this into our equation for X's imports:
$$I_X=\frac{9}{11}{E}_X=\frac{9}{11}\left(0.55I_P\right)=0.45I_P$$

4. Formulating the Total Imports of Xiland (X) and Pumpland (P)
Let the total exports of ROW be $E_{\mathrm{ROW}}$.
From Information 4:
• ROW exports to X = $0.12E_{\mathrm{ROW}}$
• ROW exports to P = $0.4E_{\mathrm{ROW}}$

Now, X's total imports are composed of imports from P (600, from Information 5), C (48, calculated above), and ROW:
$$I_X=600+48+0.12E_{\mathrm{ROW}}=648+0.12E_{\mathrm{ROW}}$$
Since $I_X=0.45I_P$, we can write:
$$I_P=\frac{648+0.12E_{\mathrm{ROW}}}{0.45}$$

Next, P's total imports are composed of imports from X (which is 22% of $I_P$), C (720, calculated above), and ROW ( $0.4E_{\mathrm{ROW}}$):
$$I_P=0.22I_P+720+0.4E_{\mathrm{ROW}}$$
$$0.78I_P=720+0.4E_{\mathrm{ROW}}$$

Substitute $I_P$ in terms of $E_{\mathrm{ROW}}$ into this equation:
$$0.78\times \left(\frac{648+0.12E_{\mathrm{ROW}}}{0.45}\right)=720+0.4E_{\mathrm{ROW}}$$
Multiplying both sides by 0.45:
$$0.78\times (648+0.12E_{\mathrm{ROW}})=0.45\times (720+0.4E_{\mathrm{ROW}})$$
$$505.44+0.0936E_{\mathrm{ROW}}=324+0.18E_{\mathrm{ROW}}$$
$$505.44-324=0.18E_{\mathrm{ROW}}-0.0936E_{\mathrm{ROW}}$$
$$181.44=0.0864E_{\mathrm{ROW}}$$
$$E_{\mathrm{ROW}}=\frac{181.44}{0.0864}=2100$$

5. Calculating the Export Volume from ROW to ROW
The total exports of ROW are $E_{\mathrm{ROW}}=2100$ IC.
These exports go to Xiland (12%), Pumpland (40%), Cheeseland (0%, as P is the only country exporting to C), and the remainder goes to the Rest of the World (ROW) itself.
The percentage of ROW exports that go to ROW is:
$$100\%-12\%-40\%=48\%$$

Therefore, the volume of exports from ROW to ROW is:
$$E_{\mathrm{ROW}\to \mathrm{ROW}}=0.48\times 2100=1008$$