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

Step 1: C's imports \( =1200 \) (only from P). Using \( \text{NTB}_C=-20\%=\frac{E_C-1200}{E_C+1200} \), solve to get \( E_C=800 \).
Step 2: C's exports: 90% to P \( =720 \), 4% to ROW \( =32 \).
Step 3: Remaining exports go to X (only other destination): \( 800-720-32=48 \).
\[ \boxed{48 \text{ IC}} \]

Answer 2

Step 1: P's exports \( =600(\text{to X})+1200(\text{to C})+x(\text{to ROW}) \). Since \( \text{NTB}_P=0 \), exports \( = \) imports \( =Y \).
Step 2: Solving the full set of trade equations, using the X and ROW percentage clues and \( \text{NTB}_X=10\% \), gives \( Y=2000 \).
Step 3: So \( x = 2000-600-1200=200 \).
\[ \boxed{200 \text{ IC}} \]

Answer 3

Step 1: Let ROW's total exports be \( R \). It sends 40% to P and 12% to X, and (since only P exports to C) none to C.
Step 2: Solving the system of trade-balance equations for all four parties gives \( R=2100 \).
Step 3: The remaining share, \( 100-40-12=48\% \), stays within ROW itself: \( 0.48\times2100=1008 \).
\[ \boxed{1008 \text{ IC}} \]

Answer 4

Approach: Use a conservation shortcut: for the whole world, all trade balances must sum to zero. So ROW's balance is the negative of (P + X + C) balances combined — no need to tally ROW's own legs.

Step 1: Compute each country's balance in IC. P: $0\%\Rightarrow 0$. X: $E_X-I_X=1100-900=+200$. C: $E_C-I_C=800-1200=-400$.

Step 2: Sum of all four balances over a closed world equals $0$ (every export is someone's import). So $\text{Bal}_P+\text{Bal}_X+\text{Bal}_C+\text{Bal}_{ROW}=0$.

Step 3: Substitute: $0+200+(-400)+\text{Bal}_{ROW}=0\Rightarrow \text{Bal}_{ROW}=+200$ IC.

Final answer: $+200$ IC (Option 3).

Answer 5

Approach: Skip the full matrix. The NTB formula directly links a country's exports and imports as a fixed ratio, and once you know either the exports or the imports of a country, its total trade pops out in one step. I anchor each country on the single number that is easiest to read off.

Setup: Write the NTB as $\dfrac{E-I}{E+I}=r$. A neat rearrangement is $\dfrac{I}{E}=\dfrac{1-r}{1+r}$, so total trade $E+I=E\left(1+\dfrac{1-r}{1+r}\right)=E\cdot\dfrac{2}{1+r}$. That single relation does all the work.

Country C ($r=-0.20$): The cleanest anchor is its imports. Since P is the only exporter to C and P sends $1200$ to C, $I_C=1200$. With $r=-0.20$, $\dfrac{E}{I}=\dfrac{1+r}{1-r}=\dfrac{0.8}{1.2}=\dfrac{2}{3}$, so $E_C=\tfrac{2}{3}(1200)=800$. Total trade $=800+1200=2000$.

Country P ($r=0$): $r=0$ forces exports $=$ imports, so total trade is just twice the exports. P exports $600+1200=1800$ to X and C, plus a flow to ROW; matching the X-side constraints fixes that flow at $200$, giving exports $=2000$. Total trade $=2\times 2000=4000$.

Country X ($r=0.10$): Its exports come to $1100$. Using total trade $=E\cdot\dfrac{2}{1+r}=1100\cdot\dfrac{2}{1.1}=2000$.

Read off: P is at $4000$, while X and C are each at $2000$. The smallest belongs jointly to X and C.

Final answer: Both X and C.