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 P to ROW, in IC?

The correct answer is 200.

To find the volume of exports from Pumpland (P) to the Rest of the World (ROW), we can define variables and systematically solve the system of equations based on the given information.

Step 1: Understand the relationships for each country's total trade

Let $E_P$, $E_X$, and $E_C$ be the total exports of Pumpland, Xiland, and Cheeseland, respectively.
Let $I_P$, $I_X$, and $I_C$ be the total imports of Pumpland, Xiland, and Cheeseland, respectively.

Using the normalized trade balance (NTB) formula:
$$\text{NTB}=\frac{\text{Exports}-\text{Imports}}{\text{Exports}+\text{Imports}}\times 100\%$$

1. For Pumpland (P), the NTB is 0%:
$$\frac{E_P-I_P}{E_P+I_P}=0\Rightarrow E_P=I_P$$

2. For Xiland (X), the NTB is 10%:
$$\frac{E_X-I_X}{E_X+I_X}=0.10\Rightarrow E_X-I_X=0.10E_X+0.10I_X\Rightarrow 0.90E_X=1.10I_X\Rightarrow E_X=\frac{11}{9}{I}_X$$

3. For Cheeseland (C), the NTB is -20%:
$$\frac{E_C-I_C}{E_C+I_C}=-0.20\Rightarrow E_C-I_C=-0.20E_C-0.20I_C\Rightarrow 1.20E_C=0.80I_C\Rightarrow E_C=\frac{2}{3}{I}_C$$

Step 2: Determine Cheeseland's (C) exports and imports

From Information 5, we know that the export volume of P to C is 1200 IC, and P is the only country that exports to C. Therefore, all of Cheeseland's imports come from Pumpland:
$$I_C=1200\text{ IC}$$

Using the relation for Cheeseland from Step 1:
$$E_C=\frac{2}{3}\times 1200=800\text{ IC}$$

From Information 3:
- 90% of C's exports are to P: $E_{C\to P}=0.90\times 800=720\text{ IC}$.
- 4% of C's exports are to ROW: $E_{C\to \text{ROW}}=0.04\times 800=32\text{ IC}$.
- The rest of C's exports go to Xiland (X): $E_{C\to X}=(1-0.90-0.04)\times 800=0.06\times 800=48\text{ IC}$.

Step 3: Analyze Pumpland's and Xiland's trade links

From Information 2:
- 40% of exports of X are to P: $E_{X\to P}=0.40E_X$.
- 22% of imports of P are from X: $E_{X\to P}=0.22I_P$.
Thus:
$$0.40E_X=0.22I_P$$

Since we established in Step 1 that $E_P=I_P$:
$$E_P=\frac{0.40}{0.22}{E}_X=\frac{20}{11}{E}_X$$

Step 4: Establish equations using ROW trade relationships

From Information 4, ROW exports are distributed such that 12% go to X and 40% go to P. Let $E_{\text{ROW}}$ be the total exports of ROW.
- ROW exports to X: $E_{\text{ROW}\to X}=0.12E_{\text{ROW}}$.
- ROW exports to P: $E_{\text{ROW}\to P}=0.40E_{\text{ROW}}$.
This gives:
$$E_{\text{ROW}\to P}=\frac{0.40}{0.12}{E}_{\text{ROW}\to X}=\frac{10}{3}{E}_{\text{ROW}\to X}$$

Now, let us write the total import equation for Xiland (X):
$$I_X=E_{P\to X}+E_{C\to X}+E_{\text{ROW}\to X}$$

Substitute the known exports into X: $E_{P\to X}=600$ and $E_{C\to X}=48$:
$$I_X=600+48+E_{\text{ROW}\to X}=648+E_{\text{ROW}\to X}$$

Using $E_X=\frac{11}{9}{I}_X$:
$$E_X=\frac{11}{9}(648+E_{\text{ROW}\to X})=792+\frac{11}{9}{E}_{\text{ROW}\to X}$$

Since $I_P=\frac{20}{11}{E}_X$, we have:
$$I_P=\frac{20}{11}(792+\frac{11}{9}{E}_{\text{ROW}\to X})=1440+\frac{20}{9}{E}_{\text{ROW}\to X}$$

Step 5: Solve for the trade volumes

The total import equation for Pumpland (P) is:
$$I_P=E_{X\to P}+E_{C\to P}+E_{\text{ROW}\to P}$$

Substitute the expressions and values we know:
$$I_P=0.22I_P+720+\frac{10}{3}{E}_{\text{ROW}\to X}$$

Subtracting $0.22I_P$ from both sides:
$$0.78I_P=720+\frac{10}{3}{E}_{\text{ROW}\to X}$$

Now, substitute the expression for $I_P$ from Step 4 into this equation:
$$0.78(1440+\frac{20}{9}{E}_{\text{ROW}\to X})=720+\frac{10}{3}{E}_{\text{ROW}\to X}$$

Distributing $0.78$:
$$1123.2+\frac{15.6}{9}{E}_{\text{ROW}\to X}=720+\frac{10}{3}{E}_{\text{ROW}\to X}$$

Since $\frac{15.6}{9}=\frac{5.2}{3}$:
$$1123.2-720=(\frac{10}{3}-\frac{5.2}{3})E_{\text{ROW}\to X}$$

$$403.2=1.6E_{\text{ROW}\to X}\Rightarrow E_{\text{ROW}\to X}=252$$

Now, calculate Xiland's total exports ($E_X$):
$$E_X=792+\frac{11}{9}\times 252=792+308=1100\text{ IC}$$

Then, find Pumpland's total exports ($E_P$):
$$E_P=\frac{20}{11}\times 1100=2000\text{ IC}$$

Step 6: Find Pumpland's exports to ROW

Pumpland's total exports ($E_P$) consist of its exports to Xiland, Cheeseland, and the Rest of the World:
$$E_P=E_{P\to X}+E_{P\to C}+E_{P\to \text{ROW}}$$

Substitute the known values ($E_P=2000$, $E_{P\to X}=600$, and $E_{P\to C}=1200$):
$$2000=600+1200+E_{P\to \text{ROW}}$$

$$E_{P\to \text{ROW}}=2000-1800=200\text{ IC}$$

Therefore, Pumpland exports 200 IC to the Rest of the World (ROW).