How much is exported from P to ROW, in IC?

From	P	X	C	ROW	Exports
P	—	600	1200
X		—	0
C			—
ROW			0
Imports			1200

From clue 5, $P\to X = 600$ and $P\to C = 1200$ are filled directly.

---

Clue 5 also states P is the only country exporting to C. Hence $X\to C = 0$ and $ROW\to C = 0$, and all of C's imports come from P:

Imports of C $= 1200$.

From	P	X	C	ROW	Exports
P	—	600	1200
X		—	0
C	720	48	—	32	800
ROW			0
Imports			1200

C's row is filled. For C, $M_C = 1.5\,E_C$ and $M_C = 1200$, so $E_C = 800$.

---

Clue 3 splits C's exports: $90\%$ to P, $4\%$ to ROW, leaving $6\%$ to X.

$C\to P = 0.90\times 800 = 720$.

$C\to ROW = 0.04\times 800 = 32$.

$C\to X = 0.06\times 800 = 48$.

Exports of C $= 800$.

From	P	X	C	ROW	Exports
P	—	600	1200	200	2000
X		—	0
C	720	48	—	32	800
ROW			0
Imports	2000		1200

Let $M_P$ denote imports of P. Since P's balance is $0\%$, exports of P also equal $M_P$.

Exports of P $= 600 + 1200 + (P\to ROW) = 1800 + (P\to ROW)$, so $P\to ROW = M_P - 1800$.

---

Clue 2: $X\to P = 0.22\,M_P$ and $X\to P = 0.40\,E_X$, giving $E_X = 0.55\,M_P$.

Imports of P $= (X\to P)+(C\to P)+(ROW\to P) = 0.22M_P + 720 + (ROW\to P) = M_P$,

so $ROW\to P = 0.78M_P - 720$.

Clue 4: $ROW\to P = 0.40\,E_{ROW}$, so $E_{ROW} = 1.95M_P - 1800$ and $ROW\to X = 0.12E_{ROW} = 0.234M_P - 216$.

Imports of X $= 600 + 48 + (0.234M_P - 216) = 432 + 0.234M_P$.

X's $10\%$ balance gives imports $= \tfrac{9}{11}E_X = 0.45M_P$.

So $432 + 0.234M_P = 0.45M_P \Rightarrow 0.216M_P = 432 \Rightarrow M_P = 2000$.

Exports of P $=$ Imports of P $= 2000$, and $P\to ROW = 2000 - 1800 = 200$.

From	P	X	C	ROW	Exports
P	—	600	1200	200	2000
X	440	—	0	660	1100
C	720	48	—	32	800
ROW			0
Imports	2000		1200

With $M_P = 2000$, X's row follows.

$E_X = 0.55\times 2000 = 1100$.

$X\to P = 0.22\times 2000 = 440$.

Since $X\to C = 0$, the remainder goes to ROW: $X\to ROW = 1100 - 440 = 660$.

Exports of X $= 1100$.

From	P	X	C	ROW	Exports
P	—	600	1200	200	2000
X	440	—	0	660	1100
C	720	48	—	32	800
ROW	840	252	0	1008	2100
Imports	2000	900	1200

The ROW row is filled.

$ROW\to P = 0.78\times 2000 - 720 = 1560 - 720 = 840$.

$E_{ROW} = 840 / 0.40 = 2100$.

---

Clue 4 gives the split of ROW exports: $40\%$ to P, $12\%$ to X, $0\%$ to C, leaving $48\%$ internal.

$ROW\to X = 0.12\times 2100 = 252$.

$ROW\to ROW = 0.48\times 2100 = 1008$.

Exports of ROW $= 2100$.

Column X now sums to $600 + 48 + 252 = 900$, matching X's imports.

From	P	X	C	ROW	Exports
P	—	600	1200	200	2000
X	440	—	0	660	1100
C	720	48	—	32	800
ROW	840	252	0	1008	2100
Imports	2000	900	1200	1900

The last column total is computed:

Imports of ROW $= 200 + 660 + 32 + 1008 = 1900$.

---

Summary of totals and balances:

P: $E = 2000$, $M = 2000$, balance $0$, total trade $4000$.

X: $E = 1100$, $M = 900$, balance $+200$, normalized $+10\%$, total trade $2000$.

C: $E = 800$, $M = 1200$, balance $-400$, normalized $-20\%$, total trade $2000$.

ROW: $E = 2100$, $M = 1900$, balance $+200$, total trade $4000$.

The world balances sum to $0 + 200 - 400 + 200 = 0$, as required. All values are uniquely determined.

---

P has a $0\%$ balance, so exports $=$ imports $= M_P$. Solving the system (clues 2 and 4 with X's $10\%$ balance) gives $M_P = 2000$. Since P's exports $= 600 + 1200 + (P\to ROW) = 1800 + (P\to ROW) = 2000$, we get $P\to ROW = 200$.