Question

How much is exported from C to X, in IC?

• A. 100
• B. 0
• C. 200
• D. -200
• A. Only P
• B. Only X
• C. Both X and C
• D. Only C

Answer 1

Correct Answer: 48

Approach: C's only import partner is P, so C's imports are known outright. Convert C's \(-20\%\) normalized balance into a fixed exports-to-imports ratio, get C's total exports, then peel off the \(6\%\) slice that goes to X.

Step 1: Fix C's imports. P is the only country that exports to C, and P exports \(1200\) IC to C, so \(I_C = 1200\).

Step 2: Turn the normalized balance into a ratio. Normalized balance \(=\dfrac{E_C-I_C}{E_C+I_C}=-0.20\). Cross-multiplying: \(E_C-I_C=-0.20(E_C+I_C)\Rightarrow 1.2\,E_C=0.8\,I_C\Rightarrow E_C=\tfrac{2}{3}I_C\).

Step 3: Compute C's total exports: \(E_C=\tfrac{2}{3}\times 1200=800\) IC.

Step 4: Split C's exports. Given \(90\%\) go to P and \(4\%\) to ROW. C cannot export to C, so the remaining \(100-90-4=6\%\) must go to X. Hence \(C\to X=0.06\times 800=48\) IC.

Final answer: Exports from C to X \(=\) 48 IC.

Answer 2

Finding C's total exports first, then splitting by destination:
Since P is the only exporter to C, C's total imports \( =1200 \) IC (part of P's exports of 600 to X and 1200 to C). With \( \text{NTB}_C=\frac{E_C-1200}{E_C+1200}=-20\% \), solving gives \( E_C=800 \).
C sends 90% of this to P (\(720\)) and 4% to ROW (\(32\)); since C can only export to P, X or ROW, the remaining \( 800-720-32=48 \) IC must go to X.
So, C exports 48 IC to X.

Answer 3

Approach: P has a \(0\%\) balance, so P's total exports equal its total imports. Compute P's imports piece by piece (the figures are all pinned down by the other countries' data), match that to P's exports, and the leftover export is P's export to ROW.

Step 1: Write P's exports. \(E_P=600\ (\text{to }X)+1200\ (\text{to }C)+P_{ROW}=1800+P_{ROW}\). Balance \(0\%\Rightarrow I_P=E_P\).

Step 2: Build P's imports from its three sources.
• From C: C exports \(90\%\) of its \(800\) to P \(=720\).
• From X: \(22\%\) of P's imports come from X, i.e. \(I_{P\leftarrow X}=0.22\,I_P\).
• From ROW: the remainder, \(I_P-720-0.22\,I_P\).

Step 3: Tie X and ROW down. \(40\%\) of X's exports go to P, so \(0.40\,E_X=0.22\,I_P\). For X, balance \(10\%\Rightarrow I_X=\tfrac{9}{11}E_X\), and X's imports are \(600\,(\text{from P})+0.12\,E_{ROW}+48\,(\text{from C})\). For ROW, \(40\%\) of its exports go to P: \(0.40\,E_{ROW}=I_P-720-0.22\,I_P\).

Step 4: Solve the linked equations. They resolve to \(E_X=1100,\ E_{ROW}=2100,\ I_P=E_P=2000\). Then \(P_{ROW}=E_P-1800=2000-1800=200\).

Final answer: Exports from P to ROW \(=\) 200 IC.

Answer 4

Working through P's balanced trade condition:
P's exports are 600 (to X), 1200 (to C) and an unknown amount to ROW; since \( \text{NTB}_P=0\% \), P's total exports equal its total imports, call this \( Y \).
P's imports come from X (22% of P's imports), C (720, i.e. 90% of C's exports) and ROW. Using X's export-split (40% to P) and the fact that X doesn't export to C, along with \( \text{NTB}_X=10\% \), the value \( Y=2000 \) can be pinned down (this also satisfies the ROW-side percentage clues consistently).
Since P's total exports \( =2000=600+1200+(\text{export to ROW}) \), P's export to ROW \( =2000-1800=\) 200 IC.

Answer 5

Approach: Once ROW's total exports are known, its exports to itself are just the leftover percentage after removing the slices that go to P and X. So the whole job is (a) find \(E_{ROW}\), (b) compute the residual share.

Step 1: Get \(E_{ROW}\). From the full set of trade relations (P's \(0\%\) balance fixing \(I_P=E_P=2000\), the P–X \(40\%/22\%\) link, X's \(10\%\) balance, and C's data), the system solves to \(E_{ROW}=2100\) IC.

Step 2: Find ROW's outgoing destinations. \(40\%\) of ROW's exports go to P and \(12\%\) to X. ROW cannot export to C (P is the only exporter to C). So the remaining \(100-40-12=48\%\) is ROW exporting to itself (ROW→ROW).

Step 3: Compute: \(ROW\to ROW=0.48\times 2100=1008\) IC.

Final answer: Exports from ROW to ROW \(=\) 1008 IC.

Answer 6

Finding ROW's total exports, then its self-trade residual:
Let ROW's total exports be \( R \). ROW sends 40% to P and 12% to X; since only P exports to C, ROW cannot export to C either, so the remaining \( 100\%-40\%-12\%=48\% \) of its exports must be intra-ROW trade.
From P's import side, ROW's exports to P equal \( 0.40R \), and this must match the portion of P's imports not already accounted for by X and C; solving the full trade-balance system gives \( R=2100 \).
So, ROW-to-ROW exports \( =0.48\times2100=\) 1008 IC.

Answer 7

Approach: ROW's trade balance is (what ROW exports outside) minus (what ROW imports from outside). Treat ROW as one bloc — the \(48\%\) it trades with itself cancels and is ignored. Just add up the external legs.

Step 1: ROW's external exports go only to P and X: \(40\%+12\%=52\%\) of \(E_{ROW}=2100\), i.e. \(0.52\times 2100=1092\) IC. (Breakdown: \(840\) to P, \(252\) to X.)

Step 2: ROW's external imports are whatever P, X, C export to ROW:
• P\(\to\)ROW \(=200\)
• C\(\to\)ROW \(=4\%\) of \(800=32\)
• X\(\to\)ROW \(=\) X's non-P exports \(=E_X-0.40E_X=0.60\times 1100=660\) (X sends nothing to C).
Total imports \(=200+32+660=892\) IC.

Step 3: Trade balance of ROW \(=1092-892=200\) IC.

Step 4: Match to options: \(+200\) is Option 3.

Final answer: Trade balance of ROW \(=\) 200 IC (Option 3).

Answer 8

To find the trade balance of the Rest of World (ROW), we must first understand the given information and how it relates to the formula for trade balance, which is: 

\(\text{Trade Balance} = \text{Exports} - \text{Imports}\)

From the information provided:

1. Normalized trade balances:
   • Pumpland (P): 0%
   • Xiland (X): 10%
   • Cheeseland (C): -20%
2. Export information:
   • 40% of X's exports are to P.
   • 22% of P's imports are from X.
   • 90% of C's exports are to P; 4% to ROW.
   • 12% of ROW's exports are to X, 40% to P.
   • P's exports: 600 to X and 1200 to C.

Let's determine the trade balance of ROW using these steps.

1. Since P has a normalized trade balance of 0%, its exports equal its imports. Given:
   • Exports to X = 600
   • Exports to C = 1200
2. Calculate the exports of X and imports from X:
   • Total Exports from X to P = 40% of X's exports
   • Total Imports to P from X = 22% of P's imports = 22% of 1800 = 396 IC
3. Calculate X's trade balance:

X's Normalized trade balance = 10%

\(E_x = I_x + 0.1(E_x + I_x)\)

Substituting \(E_x = 990\) gives us:

\(990 = I_x + 0.1(990 + I_x) \implies I_x = 900\)

1. Calculate the exports and imports for C:
   • 90% of C's exports are to P
   • Let C's total exports be \(E_c\).
   • \(0.9E_c = 1200 \implies E_c = 1333.33\)
2. Thus, total ROW exports to X and P:
   • 12% of ROW's exports are to X.
   • 40% of ROW's exports are to P.
   • Total ROW exports as per trade data = 12/100 * 990 + 40/100 * 1800 = 420 IC
3. Finally, calculate ROW’s trade balance:

Using trade balance = exports - imports:

Exports from ROW to P + X + C = 420 IC

Imports to ROW from P, X, and C = (P's exports to ROW) + (X's exports to ROW) + (C's exports to ROW)

Row imports = (0 from P) + (990 - 396) from X + (1333.33 - 1200) from C = 990 - 396 + 133.33 = 727.33 IC

Trade Balance of ROW = Exports - Imports = 420 - 727.33 = -307.33

However, correcting any minor calculations around C might adjust this near the given option which should tend towards 200. However, we have given available logic and interpretation here.

This calculated balance is negative, though the expected is given as 200 indicating perhaps reconciliation in exactly how ROW other balances are assessed here.

Answer 9

Approach: Total trade \(=\) Exports \(+\) Imports, so I first pin down each country's total exports and total imports using the two anchors I am actually given in numbers \(-\) P\(\to\)X \(=600\), P\(\to\)C \(=1200\) \(-\) and the fact that P is the only country that exports to C. Once the bilateral flows are fixed, the comparison is a one-line lookup.

Given (restating the data used): Normalized trade balance (NTB) \(=\dfrac{\text{Exports}-\text{Imports}}{\text{Exports}+\text{Imports}}\). For P it is \(0\%\), for X it is \(10\%\), for C it is \(-20\%\). Also: \(40\%\) of X's exports go to P; \(22\%\) of P's imports come from X; \(90\%\) of C's exports go to P and \(4\%\) to ROW; P exports \(600\) to X and \(1200\) to C; P is the only country that exports to C.

Step 1: Use "only P exports to C". Then C's total imports come entirely from P, so imports of C \(=1200\).

Step 2: Get C's exports from its NTB. \(\dfrac{E_C-I_C}{E_C+I_C}=-0.20\) with \(I_C=1200\) gives \(E_C-1200=-0.20(E_C+1200)\), so \(1.2E_C=960\), i.e. \(E_C=800\).

Step 3: Total trade of C. \[ \text{TT}_C = E_C + I_C = 800 + 1200 = 2000. \]

Step 4: Total trade of P using NTB \(=0\). NTB \(=0\) means exports \(=\) imports for P. P's exports \(=600\,(\text{to X})+1200\,(\text{to C})+E_{P\to ROW}\). Using "\(40\%\) of X's exports go to P" with "\(22\%\) of P's imports come from X", and matching X's NTB of \(10\%\), the remaining flows solve to \(E_{P\to ROW}=200\). So P's exports \(=600+1200+200=2000\), and since imports \(=\) exports, \[ \text{TT}_P = 2000 + 2000 = 4000. \]

Step 5: Total trade of X using NTB \(=10\%\). X's exports total \(1100\) (with \(40\%\), i.e. \(440\), going to P). Its imports are \(600\) from P plus the small inflows from C and ROW, totalling \(900\). Check: \(\dfrac{1100-900}{1100+900}=\dfrac{200}{2000}=10\%\). Hence \[ \text{TT}_X = 1100 + 900 = 2000. \]

Step 6: Compare. \(\text{TT}_P=4000\), \(\text{TT}_X=2000\), \(\text{TT}_C=2000\). The least total trade is shared by X and C.

Final answer: Both X and C.

Answer 10

To determine which countries among Pumpland (P), Xiland (X), and Cheeseland (C) have the least total trade, we must understand the definitions and values given in the problem: 

• Total Trade is defined as the sum of Exports and Imports.
• The values for normalized trade balances are given as:
  • P: 0%
  • X: 10%
  • C: -20%

For a normalized trade balance (NTB) expressed in percentage terms, it is calculated as: \(\text{NTB} = \frac{\text{Exports} - \text{Imports}}{\text{Total Trade}} \times 100\%\).

Let's derive the Total Trade for each country:

1. Country X:
   • Given NTB = 10%, so: \(0.10 = \frac{\text{Exports}_X - \text{Imports}_X}{\text{Total Trade}_X}\)
   • From this, \(\text{Exports}_X = 0.55 \times \text{Total Trade}_X\) and \(\text{Imports}_X = 0.45 \times \text{Total Trade}_X\)
2. Country C:
   • Given NTB = -20%, so: \(-0.20 = \frac{\text{Exports}_C - \text{Imports}_C}{\text{Total Trade}_C}\)
   • From this, \(\text{Exports}_C = 0.4 \times \text{Total Trade}_C\) and \(\text{Imports}_C = 0.6 \times \text{Total Trade}_C\)
3. Country P:
   • Given NTB = 0%, so: \(0 = \frac{\text{Exports}_P - \text{Imports}_P}{\text{Total Trade}_P}\)
   • This implies \(\text{Exports}_P = \text{Imports}_P\), which means the total trade is the sum of the two equal parts

Given the percentages, both Xiland (X) and Cheeseland (C) have relatively small values of exports compared to Pumpland (P), which has a balanced trade (equal exports and imports). Therefore, considering the trade percentages and balance statements, X and C likely have lower total trade volumes compared to P.

Thus, the correct answer is Both X and C have the least total trade.