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.

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

The correct option is Both X and C.

Let us solve this step-by-step by defining the variables for exports and imports of each country, and using the given trade relationships to calculate their total trades.

Let the total exports of Pumpland (P), Xiland (X), and Cheeseland (C) be denoted by _EP, _EX, and _EC, respectively. Let their total imports be _MP, _MX, and _MC, respectively.
The trade balance is given by _{Exports − Imports}, and the total trade is _{Exports + Imports}.
The normalized trade balance (NTB) is defined as:
$$\text{NTB}=\frac{\text{Exports}-\text{Imports}}{\text{Exports}+\text{Imports}}\times 100\%$$

Step 1: Calculate the relations from Normalized Trade Balances
1. For Pumpland (P): NTB is 0%.
$E_P=M_P$
Let this common value be T. So, _{EP = MP = T}.
2. For Xiland (X): NTB is 10%.
$$\frac{E_X-M_X}{E_X+M_X}=0.10\Rightarrow E_X-M_X=0.10(E_X+M_X)\Rightarrow 0.9E_X=1.1M_X\Rightarrow E_X=\frac{11}{9}{M}_X$$
3. For Cheeseland (C): NTB is -20%.
$$\frac{E_C-M_C}{E_C+M_C}=-0.20\Rightarrow E_C-M_C=-0.20(E_C+M_C)\Rightarrow 1.2E_C=0.8M_C\Rightarrow M_C=1.5E_C$$

Step 2: Calculate Cheeseland (C) Trade Volumes
From statement 5, P is the only country that exports to C, and P's export to C is 1200 IC. Therefore, C's total imports are:
_{MC = 1200} IC.
Using the relation _{MC = 1.5 EC}:
$$E_C=\frac{1200}{1.5}=800$$
Thus, the total trade of C is:
_{Total TradeC = EC + MC = 800 + 1200 = 2000} IC.

Step 3: Analyze C's exports and X's exports
From statement 3: 90% of C's exports go to P, and 4% go to ROW. The remaining exports of C must go to X.
• Export from C to P: _{0.90 × 800 = 720} IC.
• Export from C to ROW: _{0.04 × 800 = 32} IC.
• Export from C to X: _{800 − 720 − 32 = 48} IC.

From statement 2: 40% of X's exports go to P, and this amount constitutes 22% of P's imports.
Since _{MP = T}, we have:
_{Export from X to P = 0.22 T}
Also, since 40% of X's exports go to P:
_{0.40 EX = 0.22 T ⇒ EX = 0.55 T}.
Using the relation from Step 1 where _{EX = (11 / 9) MX}, we get:
$$M_X=\frac{9}{11}{E}_X=\frac{9}{11}\left(0.55T\right)=0.45T$$

Step 4: Formulate and Solve for T
The imports of X (_MX) come from P, C, and ROW:
_{MX = EP→X + EC→X + EROW→X}
We are given _{EP→X = 600} (from statement 5) and we found _{EC→X = 48}. Thus:
_{0.45 T = 600 + 48 + EROW→X ⇒ EROW→X = 0.45 T − 648}

From statement 4: 12% of ROW's exports are to X, and 40% are to P.
This means:
$$E_{\text{ROW→P}}=\frac{40}{12}{E}_{\text{ROW→X}}=\frac{10}{3}{E}_{\text{ROW→X}}$$
The imports of P (_{MP = T}) come from X, C, and ROW:
_{T = EX→P + EC→P + EROW→P}
Substitute the known values:
_{T = 0.22 T + 720 + (10 / 3) (0.45 T − 648)
0.78 T = 720 + 1.5 T − 2160
1440 = 0.72 T
T = 2000}

Step 5: Compare Total Trades
Now we calculate the total trade for each country:
• For Pumpland (P):
_{EP = 2000}, _{MP = 2000}.
_{Total TradeP = EP + MP = 2000 + 2000 = 4000} IC.

• For Xiland (X):
_{EX = 0.55 × 2000 = 1100}, _{MX = 0.45 × 2000 = 900}.
_{Total TradeX = 1100 + 900 = 2000} IC.

• For Cheeseland (C):
_{Total TradeC = 2000} IC.

Comparing the values, both X and C have the minimum total trade of 2000 IC.