Question

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

• A. Only P
• B. Only X
• C. Both X and C
• D. Only C

Hint:

When comparing trade volumes, add exports and imports for each country to determine the total trade value.

Answer 1

The Correct Option is C

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.