CAT 2025 DILR Questions — Slot 3

Setting up the full arrangement

This is a multi-layer puzzle: we must figure out who visited which country, how much each traveler spent (in local currency), and the exchange rates — all from the citizens' observations. Let's build it step by step.

Step 1 — Who visited where

Each traveler visits exactly 2 of the 3 foreign countries (Aurevia, Brelosia, Cyrenia), and each country is visited by exactly 2 travelers. The citizens tell us directly:

• Aurevia: visited by Jano and Kira
• Brelosia: visited by Kira and Lian
• Cyrenia: visited by Lian and …?

Kira already has 2 countries (Aurevia + Brelosia), so she cannot visit Cyrenia. Lian already has 2 countries (Brelosia + Cyrenia). Cyrenia still needs a second visitor — that must be Jano.

Step 2 — Spends in local currency

There are 6 spends total (3 travelers × 2 countries). The constraints say: exactly two spends of 1000, exactly one spend of 3000, and the remaining three are 2000. Also, each traveler's two spends must differ.

The Brelosia citizen tells us: Kira spent 2000 Brins and Lian spent 3000 Brins in Brelosia. That 3000 is the only 3000 across all visits.

• Kira's two spends must differ. She already spent 2000 in Brelosia, so her Aurevia spend cannot be 2000. It also can't be 3000 (the only 3000 is used). So Kira spent 1000 in Aurevia.

Used so far: one 1000 (Kira–Aurevia), one 3000 (Lian–Brelosia), one 2000 (Kira–Brelosia). Remaining: one 1000 and two 2000s, to be distributed among Jano's two spends and Lian's Cyrenia spend.

• Lian already has 3000, so her Cyrenia spend must differ → it is 2000.
• That leaves {1000, 2000} for Jano's two spends (Aurevia and Cyrenia), which do differ. ✓

Step 3 — Exchange rates

Let 1 Aurel $= a$ Zentars, 1 Brin $= b$ Zentars, and 1 Crown $= 0.5$ Zentars (given).

Kira's Travel Cost is quoted by two citizens in their own currencies: 8000 Aurels (Aurevia) and 4000 Brins (Brelosia). These are the same real amount: $8000a = 4000b \quad\Rightarrow\quad b = 2a$

Kira's Travel Cost in Zentars $= 8000a$. This equals Flight + spends: $8000a = \text{Flight}_K + 1000a + 2000b = \text{Flight}_K + 1000a + 4000a = \text{Flight}_K + 5000a$ $\Rightarrow \text{Flight}_K = 3000a$

Kira's flight is either 5000 or 6000 Z. So $a = \frac{5000}{3}$ or $a = 2$.

Now use Lian. The Cyrenia citizen says Lian's Travel Cost $= 36000$ Crowns $= 36000 \times 0.5 = 18000$ Zentars. $\text{Flight}_L + 3000b + 2000 \times 0.5 = 18000$ $\text{Flight}_L + 3000(2a) + 1000 = 18000 \quad\Rightarrow\quad \text{Flight}_L + 6000a = 17000$

• If $a = 2$: $\text{Flight}_L + 12000 = 17000 \Rightarrow \text{Flight}_L = 5000$ ✓ (valid — 5000 is an available flight)
• If $a = \frac{5}{3}$: $\text{Flight}_L + 10000 = 17000 \Rightarrow \text{Flight}_L = 7000$ ✗ (not 5000 or 6000)

So $a = 2$, $b = 4$, and flights are: Jano 4000, Kira 6000, Lian 5000.

Step 4 — Jano's specific spends

Jano's Travel Cost $= 3500$ Aurels $= 3500 \times 2 = 7000$ Zentars. $4000 + 2 \times j_A + 0.5 \times j_C = 7000 \quad\Rightarrow\quad 2j_A + 0.5j_C = 3000$

Jano's spends are {1000, 2000} in some order:

• $j_A = 1000,\; j_C = 2000$: $2(1000) + 0.5(2000) = 2000 + 1000 = 3000$ ✓
• $j_A = 2000,\; j_C = 1000$: $2(2000) + 0.5(1000) = 4500$ ✗

So Jano spent 1000 in Aurevia and 2000 in Cyrenia.

Solved table (spends in local currency; Travel Cost in Zentars)

Exchange rates: 1 Aurel = 2 Z, 1 Brin = 4 Z, 1 Crown = 0.5 Z.

Answering this question

The sum of all Travel Costs in Zentars: $7000 + 16000 + 18000 = \boxed{41000} \text{ Zentars}$

Answer: 41000

Setting up the full arrangement

We need to solve the entire puzzle to answer any question. Here is the complete deduction.

Step 1 — Who visited where

Each traveler visits exactly 2 of the 3 foreign countries, and each country gets exactly 2 visitors. From the citizens:

• Aurevia: Jano and Kira
• Brelosia: Kira and Lian
• Cyrenia: Lian and …?

Kira already has 2 countries (Aurevia + Brelosia). Lian already has 2 (Brelosia + Cyrenia). Cyrenia needs a second visitor → Jano.

Step 2 — Spends in local currency

Six spends: exactly two 1000s, exactly one 3000, three 2000s. Each traveler's two spends differ.

Brelosia citizen: Kira spent 2000 Brins, Lian spent 3000 Brins (the only 3000).

• Kira's other spend (Aurevia) can't be 2000 (must differ) or 3000 (already used) → Kira spent 1000 in Aurevia.
• Remaining: one 1000, two 2000s for Jano's two spends and Lian's Cyrenia spend.
• Lian already has 3000, so her Cyrenia spend must differ → 2000.
• Jano's two spends are the remaining {1000, 2000}.

Step 3 — Exchange rates

Let 1 Aurel $= a$ Z, 1 Brin $= b$ Z, 1 Crown $= 0.5$ Z.

Kira's Travel Cost is 8000 Aurels and also 4000 Brins (same real amount): $8000a = 4000b \;\Rightarrow\; b = 2a$

Kira's TC in Z: $8000a = \text{Flight}_K + 1000a + 2000b = \text{Flight}_K + 5000a$, so $\text{Flight}_K = 3000a$.

Lian's TC $= 36000$ Crowns $= 18000$ Z: $\text{Flight}_L + 3000b + 2000(0.5) = 18000 \;\Rightarrow\; \text{Flight}_L + 6000a = 17000$

Testing $a = 2$: $\text{Flight}_L = 5000$ ✓. Testing $a = \frac{5}{3}$: $\text{Flight}_L = 7000$ ✗.

So $a = 2$, $b = 4$. Flights: Jano 4000, Kira 6000, Lian 5000.

Step 4 — Jano's spends

Jano's TC $= 3500$ Aurels $= 7000$ Z: $4000 + 2j_A + 0.5j_C = 7000 \Rightarrow 2j_A + 0.5j_C = 3000$. With {1000, 2000}: $j_A = 1000,\; j_C = 2000$ works.

Solved table

Rates: 1 Aurel = 2 Z, 1 Brin = 4 Z, 1 Crown = 0.5 Z.

Answering this question

Lian visited Brelosia and Cyrenia. Her spends were:

• Brelosia: 3000 Brins $= 3000 \times 4 = 12000$ Zentars
• Cyrenia: 2000 Crowns $= 2000 \times 0.5 = 1000$ Zentars

$12000 + 1000 = \boxed{13000} \text{ Zentars}$

Answer: 13000

Setting up the full arrangement

Step 1 — Who visited where

Each traveler visits exactly 2 of the 3 foreign countries; each country gets exactly 2 visitors. From the citizens' remarks:

• Aurevia: Jano and Kira
• Brelosia: Kira and Lian
• Cyrenia: Lian and Jano (the only remaining option)

Step 2 — Spends in local currency

Six spends: exactly two 1000s, one 3000, three 2000s. Each traveler's two spends differ.

Brelosia citizen: Kira = 2000 Brins, Lian = 3000 Brins (the only 3000).

• Kira's Aurevia spend can't be 2000 (must differ) or 3000 (used) → 1000 in Aurevia.
• Lian's Cyrenia spend can't be 3000 (used) → 2000.
• Jano's two spends are the remaining {1000, 2000}.

Step 3 — Exchange rates

Let 1 Aurel $= a$ Z, 1 Brin $= b$ Z, 1 Crown $= 0.5$ Z.

Kira's TC is 8000 Aurels = 4000 Brins → $b = 2a$.

Kira: $\text{Flight}_K = 3000a$. Lian: $\text{Flight}_L + 6000a = 17000$.

Only $a = 2$ works (giving $\text{Flight}_L = 5000$). So $a = 2$, $b = 4$. Flights: Jano 4000, Kira 6000, Lian 5000.

Step 4 — Jano's specific spends

Jano's TC $= 3500$ Aurels $= 7000$ Z: $4000 + 2j_A + 0.5j_C = 7000 \;\Rightarrow\; 2j_A + 0.5j_C = 3000$ With {1000, 2000}: $j_A = 1000,\; j_C = 2000$ is the only solution.

Solved table

Rates: 1 Aurel = 2 Z, 1 Brin = 4 Z, 1 Crown = 0.5 Z.

Answering this question

💡 Teacher tip: "In Aurels" means we must convert all of Jano's spending into a single currency — Aurels. Since his spends are in two different local currencies (Aurels and Crowns), convert both to Zentars first, then divide by 2 to get Aurels.

Jano spent:

• Aurevia: 1000 Aurels $= 1000 \times 2 = 2000$ Zentars
• Cyrenia: 2000 Crowns $= 2000 \times 0.5 = 1000$ Zentars

Total in Zentars: $2000 + 1000 = 3000$ Z.

Converting to Aurels ($1$ Aurel $= 2$ Z): $\frac{3000}{2} = \boxed{1500} \text{ Aurels}$

Answer: 1500

Setting up the full arrangement

Step 1 — Who visited where

Each traveler visits exactly 2 of the 3 foreign countries; each country gets exactly 2 visitors.

• Aurevia: Jano and Kira
• Brelosia: Kira and Lian
• Cyrenia: Lian and Jano

Step 2 — Spends in local currency

Six spends: exactly two 1000s, one 3000, three 2000s. Each traveler's two spends differ.

Brelosia citizen: Kira = 2000 Brins, Lian = 3000 Brins (the only 3000).

• Kira's Aurevia spend → 1000 (can't be 2000 or 3000).
• Lian's Cyrenia spend → 2000 (can't be 3000).
• Jano's two spends → {1000, 2000}.

Step 3 — Exchange rates

Let 1 Aurel $= a$ Z, 1 Brin $= b$ Z, 1 Crown $= 0.5$ Z.

Kira's TC is 8000 Aurels = 4000 Brins → $b = 2a$.

Kira: $\text{Flight}_K = 3000a$. Lian: $\text{Flight}_L + 6000a = 17000$.

Only $a = 2$ works (giving $\text{Flight}_L = 5000$). So $a = 2$, $b = 4$.

Step 4 — Jano's spends

Jano's TC $= 3500$ Aurels $= 7000$ Z: $2j_A + 0.5j_C = 3000$ with {1000, 2000} gives $j_A = 1000,\; j_C = 2000$.

Solved table

Rates: 1 Aurel = 2 Z, 1 Brin = 4 Z, 1 Crown = 0.5 Z.

Answering this question

We need to find how many Crowns equal 1 Brin. We know both in terms of Zentars:

• 1 Brin $= 4$ Zentars
• 1 Crown $= 0.5$ Zentars

So: $1 \text{ Brin} = \frac{4}{0.5} = \boxed{8} \text{ Crowns}$

Answer: 8 (Choice A)

Setting up the full arrangement

Step 1 — Who visited where

Each traveler visits exactly 2 of the 3 foreign countries; each country gets exactly 2 visitors.

• Aurevia: Jano and Kira
• Brelosia: Kira and Lian
• Cyrenia: Lian and Jano

Step 2 — Spends in local currency

Six spends: exactly two 1000s, one 3000, three 2000s. Each traveler's two spends differ.

Brelosia citizen: Kira = 2000 Brins, Lian = 3000 Brins (the only 3000).

• Kira's Aurevia spend → 1000 (can't be 2000 or 3000).
• Lian's Cyrenia spend → 2000 (can't be 3000).
• Jano's two spends → {1000, 2000}.

Step 3 — Exchange rates

Let 1 Aurel $= a$ Z, 1 Brin $= b$ Z, 1 Crown $= 0.5$ Z.

Kira's TC is 8000 Aurels = 4000 Brins → $b = 2a$.

Kira: $\text{Flight}_K = 3000a$. Lian: $\text{Flight}_L + 6000a = 17000$.

Only $a = 2$ works (giving $\text{Flight}_L = 5000$). So $a = 2$, $b = 4$.

Step 4 — Jano's specific spends

Jano's TC $= 3500$ Aurels $= 7000$ Z: $2j_A + 0.5j_C = 3000$ with {1000, 2000} gives $j_A = 1000$ (Aurevia), $j_C = 2000$ (Cyrenia).

Solved table (spends in local currency)

Answering this question

We check each statement against the solved table:

The statement that is NOT true is A.

Answer: A — "Jano spent 2000 in Aurevia" is the false statement.

Finding who solved the most puzzles by minute 20

The left chart is a cumulative step-line: each step up means one more puzzle finished. So the number of puzzles a competitor has solved by minute 20 is simply the count of their completion times that are ≤ 20.

Reading each competitor's puzzle-by-puzzle finish times off the left chart:

Koushik has finished 7 puzzles by minute 20 — two more than the next-best competitor.

💡 Teacher tip: In a cumulative step chart, "how many by time t" is just "how many step-times are ≤ t" — you don't need to estimate the y-value; count the risers.

Answer: Koushik (Choice B)

Finding Suranjan's time on the third visual puzzle

First we must identify which puzzles in the sequence are visual. The right chart gives each competitor's four visual-puzzle finish times. Matching those against the left chart's all-puzzle finish times tells us which sequence positions are visual.

For Suranjan:

So the visual puzzles sit at sequence positions 1, 4, 8, 9. The third visual puzzle is therefore the one at position 8.

The time taken on a single puzzle = (finish time of that puzzle) − (finish time of the previous puzzle). From the left chart, Suranjan finished puzzle 7 at minute 26 and puzzle 8 at minute 28:

$28 - 26 = 2 \text{ minutes}$

💡 Teacher tip: "Time to solve puzzle k" is the gap between consecutive cumulative finish times, not the finish time itself — puzzle 1's time is measured from minute 0.

Answer: 2 minutes

Finding the position of the fourth number-based puzzle

We first determine which of the 10 sequence positions are visual. Cross-reading the two charts, every competitor's four visual-puzzle finish times line up with their 1st, 4th, 8th, and 9th overall completions:

Since all four competitors agree, the visual puzzles are at positions 1, 4, 8, 9, and the remaining six are number-based:

The number-based puzzles in order are: 2 (1st), 3 (2nd), 5 (3rd), 6 (4th), 7 (5th), 10 (6th).

💡 Teacher tip: List the number-based positions in ascending order and simply count to the fourth entry — no arithmetic needed.

Answer: 6

Finding Anirbid's average time on the number-based puzzles

Step 1 — Identify the number-based positions. Cross-reading the two charts, the visual puzzles fall at sequence positions 1, 4, 8, 9 for every competitor. Therefore the number-based puzzles are at positions 2, 3, 5, 6, 7, 10.

Step 2 — Compute Anirbid's per-puzzle times. The time on each puzzle is the gap between consecutive cumulative finish times (puzzle 1 measured from minute 0). Anirbid's finish times from the left chart are: 4, 9, 11, 14, 19, 22, 27, 29, 33, 37.

Step 3 — Average the six number-based times (positions 2, 3, 5, 6, 7, 10 → times 5, 2, 5, 3, 5, 4):

$\text{Average} = \frac{5 + 2 + 5 + 3 + 5 + 4}{6} = \frac{24}{6} = 4.0 \text{ minutes}$

💡 Teacher tip: Each individual puzzle time comes from differences of consecutive cumulative values, not the cumulative values themselves. Get the six gaps first, then average.

Answer: 4.0 minutes (Choice D)

Set up the two operators first. Xitel = {Anu, Bijay}; Yocel = {Chetan, Deepak, Eshan, Faruq}. Each person's table row gives group totals: minutes they sent to Xitel users, to Yocel users, and received from each group. The "no-call" rules plus Faruq→Eshan = 200 and Faruq's outgoing-to-Xitel = 0 (so Faruq never called Anu or Bijay) let us pin every needed call.

Key deductions (in order):

1. Anu's incoming from Xitel = 50, and only Bijay (the other Xitel user) can call Anu ⇒ Bijay→Anu = 50.
2. Anu's outgoing to Xitel = 100 can only go to Bijay ⇒ Anu→Bijay = 100.
3. Chetan's outgoing to Xitel = 50, and Chetan never calls Anu ⇒ Chetan→Bijay = 50.
4. Bijay's incoming from Yocel = 125. Deepak and Faruq never call Bijay, so it is Chetan→Bijay + Eshan→Bijay $= 50 + \text{Eshan→Bijay} = 125$ ⇒ Eshan→Bijay = 75.
5. Deepak's outgoing to Xitel = 100, and Deepak never calls Bijay ⇒ Deepak→Anu = 100.
6. Anu's incoming from Yocel = 225. Chetan and Faruq never call Anu, so it is Deepak→Anu + Eshan→Anu $= 100 + \text{Eshan→Anu} = 225$ ⇒ Eshan→Anu = 125.
7. Faruq→Eshan = 200 is given, and the Yocel columns fix Faruq→Chetan + Faruq→Deepak = 150.

Solved directed-call table (minutes; blank = no call). The last column is each person's outgoing total to Xitel / Yocel:

This question

Anu's incoming from Xitel is 50 minutes, and among Xitel users only Bijay can call Anu. So all of it is Bijay's call:

$\text{Bijay→Anu} = 50 \text{ minutes}$

Answer: 50

Set up the two operators first. Xitel = {Anu, Bijay}; Yocel = {Chetan, Deepak, Eshan, Faruq}. Each person's table row gives group totals: minutes they sent to Xitel users, to Yocel users, and received from each group. The "no-call" rules plus Faruq→Eshan = 200 and Faruq's outgoing-to-Xitel = 0 (so Faruq never called Anu or Bijay) let us pin every needed call.

Key deductions (in order):

1. Anu's incoming from Xitel = 50, and only Bijay (the other Xitel user) can call Anu ⇒ Bijay→Anu = 50.
2. Anu's outgoing to Xitel = 100 can only go to Bijay ⇒ Anu→Bijay = 100.
3. Chetan's outgoing to Xitel = 50, and Chetan never calls Anu ⇒ Chetan→Bijay = 50.
4. Bijay's incoming from Yocel = 125. Deepak and Faruq never call Bijay, so it is Chetan→Bijay + Eshan→Bijay $= 50 + \text{Eshan→Bijay} = 125$ ⇒ Eshan→Bijay = 75.
5. Deepak's outgoing to Xitel = 100, and Deepak never calls Bijay ⇒ Deepak→Anu = 100.
6. Anu's incoming from Yocel = 225. Chetan and Faruq never call Anu, so it is Deepak→Anu + Eshan→Anu $= 100 + \text{Eshan→Anu} = 225$ ⇒ Eshan→Anu = 125.
7. Faruq→Eshan = 200 is given, and the Yocel columns fix Faruq→Chetan + Faruq→Deepak = 150.

Solved directed-call table (minutes; blank = no call). The last column is each person's outgoing total to Xitel / Yocel:

This question

We need Anu's outgoing to Yocel total. Since Bijay never calls Eshan, Eshan's incoming from Xitel = 100 must all come from Anu, so Anu→Eshan = 100. For Chetan, Deepak and Faruq, Anu's call equals their incoming from Xitel minus Bijay's call to them; summed, that is $(250 + 275 + 100) - 200 = 425$. Total:

$425 + 100 = 525 \text{ minutes}$

Answer: 525

Set up the two operators first. Xitel = {Anu, Bijay}; Yocel = {Chetan, Deepak, Eshan, Faruq}. Each person's table row gives group totals: minutes they sent to Xitel users, to Yocel users, and received from each group. The "no-call" rules plus Faruq→Eshan = 200 and Faruq's outgoing-to-Xitel = 0 (so Faruq never called Anu or Bijay) let us pin every needed call.

Key deductions (in order):

1. Anu's incoming from Xitel = 50, and only Bijay (the other Xitel user) can call Anu ⇒ Bijay→Anu = 50.
2. Anu's outgoing to Xitel = 100 can only go to Bijay ⇒ Anu→Bijay = 100.
3. Chetan's outgoing to Xitel = 50, and Chetan never calls Anu ⇒ Chetan→Bijay = 50.
4. Bijay's incoming from Yocel = 125. Deepak and Faruq never call Bijay, so it is Chetan→Bijay + Eshan→Bijay $= 50 + \text{Eshan→Bijay} = 125$ ⇒ Eshan→Bijay = 75.
5. Deepak's outgoing to Xitel = 100, and Deepak never calls Bijay ⇒ Deepak→Anu = 100.
6. Anu's incoming from Yocel = 225. Chetan and Faruq never call Anu, so it is Deepak→Anu + Eshan→Anu $= 100 + \text{Eshan→Anu} = 225$ ⇒ Eshan→Anu = 125.
7. Faruq→Eshan = 200 is given, and the Yocel columns fix Faruq→Chetan + Faruq→Deepak = 150.

Solved directed-call table (minutes; blank = no call). The last column is each person's outgoing total to Xitel / Yocel:

This question

Faruq only calls Yocel friends (his outgoing to Xitel is 0). He called Eshan for 200 minutes (given), and the Yocel incoming columns fix Faruq→Chetan + Faruq→Deepak = 150. So:

$200 + 150 = 350 \text{ minutes}$

Answer: 350

Set up the two operators first. Xitel = {Anu, Bijay}; Yocel = {Chetan, Deepak, Eshan, Faruq}. Each person's table row gives group totals: minutes they sent to Xitel users, to Yocel users, and received from each group. The "no-call" rules plus Faruq→Eshan = 200 and Faruq's outgoing-to-Xitel = 0 (so Faruq never called Anu or Bijay) let us pin every needed call.

Key deductions (in order):

1. Anu's incoming from Xitel = 50, and only Bijay (the other Xitel user) can call Anu ⇒ Bijay→Anu = 50.
2. Anu's outgoing to Xitel = 100 can only go to Bijay ⇒ Anu→Bijay = 100.
3. Chetan's outgoing to Xitel = 50, and Chetan never calls Anu ⇒ Chetan→Bijay = 50.
4. Bijay's incoming from Yocel = 125. Deepak and Faruq never call Bijay, so it is Chetan→Bijay + Eshan→Bijay $= 50 + \text{Eshan→Bijay} = 125$ ⇒ Eshan→Bijay = 75.
5. Deepak's outgoing to Xitel = 100, and Deepak never calls Bijay ⇒ Deepak→Anu = 100.
6. Anu's incoming from Yocel = 225. Chetan and Faruq never call Anu, so it is Deepak→Anu + Eshan→Anu $= 100 + \text{Eshan→Anu} = 225$ ⇒ Eshan→Anu = 125.
7. Faruq→Eshan = 200 is given, and the Yocel columns fix Faruq→Chetan + Faruq→Deepak = 150.

Solved directed-call table (minutes; blank = no call). The last column is each person's outgoing total to Xitel / Yocel:

This question

Chetan's incoming from Yocel = 150 comes from Deepak, Eshan and Faruq, but Eshan never calls Chetan. So it is Deepak→Chetan + Faruq→Chetan = 150. Solving the remaining Yocel balances fixes Faruq→Chetan, leaving:

$\text{Deepak→Chetan} = 100 \text{ minutes}$

Answer: 100

We need the volume of Cheeseland's exports to Xiland. This can be found using only Cheeseland's data — no need to solve the entire system.

Step 1 — Find C's total imports.

Clue 5 tells us P is the only country that exports to C. We are also told P exports 1200 to C. Therefore:

$I_C = 1200 \text{ IC}$

Step 2 — Use C's normalized trade balance to find C's total exports.

C's normalized trade balance is $-20\%$:

$\frac{E_C - I_C}{E_C + I_C} = -0.20$

Substituting $I_C = 1200$:

$\frac{E_C - 1200}{E_C + 1200} = -0.20$

$E_C - 1200 = -0.20\,E_C - 240$

$1.20\,E_C = 960$

$E_C = 800 \text{ IC}$

Step 3 — Split C's exports among destinations.

From clue 3: $90\%$ of C's exports go to P, and $4\%$ go to ROW. Since a country cannot export to itself, the remaining percentage must go to X:

$100\% - 90\% - 4\% = 6\%$

$C \to X = 0.06 \times 800 = 48 \text{ IC}$

Answer: 48

To find P's exports to ROW, we need P's total exports, which requires solving the full trade system.

Step 1 — Cheeseland's exports.

P is the only exporter to C, and P→C = 1200, so $I_C = 1200$. With C's normalized balance at $-20\%$:

$\frac{E_C - 1200}{E_C + 1200} = -0.20 \implies 1.2\,E_C = 960 \implies E_C = 800$

C's exports split: $90\% \to P = 720$, $4\% \to \text{ROW} = 32$, remaining $6\% \to X = 48$.

Step 2 — Xiland's relationships.

X's normalized balance is $10\%$:

$\frac{E_X - I_X}{E_X + I_X} = 0.10 \implies E_X = \frac{11}{9}\,I_X$

X's exports go to P ($40\%$) and ROW ($60\%$) — not to C, since P is the only exporter to C.

X's imports come from P (600), C (48), and ROW ($12\%$ of $E_{\text{ROW}}$):

$I_X = 648 + 0.12\,E_{\text{ROW}}$

Step 3 — Pumpland's relationships.

P's normalized balance is $0\%$, so $E_P = I_P$.

P's exports: $600 + 1200 + P{\to}\text{ROW}$

P's imports: $X{\to}P + C{\to}P + \text{ROW}{\to}P = 0.4\,E_X + 720 + 0.4\,E_{\text{ROW}}$

From clue 2, $X{\to}P$ is $22\%$ of P's imports:

$0.4\,E_X = 0.22\,I_P \implies I_P = \frac{20}{11}\,E_X$

Step 4 — Solve the system.

Since $E_P = I_P = \frac{20}{11}E_X$ and also $I_P = 0.4\,E_X + 720 + 0.4\,E_{\text{ROW}}$:

$\frac{20}{11}E_X = 0.4\,E_X + 720 + 0.4\,E_{\text{ROW}}$

$\frac{78}{55}E_X = 720 + 0.4\,E_{\text{ROW}} \quad \cdots (1)$

From X's balance and imports:

$E_X = \frac{11}{9}(648 + 0.12\,E_{\text{ROW}}) = 792 + \frac{11}{75}E_{\text{ROW}} \quad \cdots (2)$

Substituting (2) into (1):

$\frac{78}{55}\!\left(792 + \frac{11}{75}E_{\text{ROW}}\right) = 720 + 0.4\,E_{\text{ROW}}$

$1123.2 + 0.208\,E_{\text{ROW}} = 720 + 0.4\,E_{\text{ROW}}$

$403.2 = 0.192\,E_{\text{ROW}} \implies E_{\text{ROW}} = 2100$

Back-substituting: $E_X = 1100$, $I_P = E_P = 2000$.

Step 5 — Find P → ROW.

P's total exports = 2000. Subtracting the known flows to X (600) and C (1200):

$P{\to}\text{ROW} = 2000 - 600 - 1200 = 200 \text{ IC}$

Solved export matrix (rows = exporter, columns = importer), in IC:

Answer: 200

To find ROW's exports to itself, we need ROW's total exports, which requires solving the full trade system.

Step 1 — Cheeseland's exports.

P is the only exporter to C, and P→C = 1200, so $I_C = 1200$. With C's normalized balance at $-20\%$:

$\frac{E_C - 1200}{E_C + 1200} = -0.20 \implies 1.2\,E_C = 960 \implies E_C = 800$

C's exports split: $90\% \to P = 720$, $4\% \to \text{ROW} = 32$, remaining $6\% \to X = 48$.

Step 2 — Xiland's relationships.

X's normalized balance is $10\%$:

$\frac{E_X - I_X}{E_X + I_X} = 0.10 \implies E_X = \frac{11}{9}\,I_X$

X's exports go to P ($40\%$) and ROW ($60\%$) — not to C, since P is the only exporter to C.

X's imports: $I_X = 600 + 48 + 0.12\,E_{\text{ROW}} = 648 + 0.12\,E_{\text{ROW}}$

Step 3 — Pumpland's relationships.

P's normalized balance is $0\%$, so $E_P = I_P$.

From clue 2: $0.4\,E_X = 0.22\,I_P \implies I_P = \frac{20}{11}E_X$

Also: $I_P = 0.4\,E_X + 720 + 0.4\,E_{\text{ROW}}$

Step 4 — Solve the system.

Equating the two expressions for $I_P$:

$\frac{20}{11}E_X = 0.4\,E_X + 720 + 0.4\,E_{\text{ROW}} \implies \frac{78}{55}E_X = 720 + 0.4\,E_{\text{ROW}} \quad \cdots (1)$

From X's balance: $E_X = \frac{11}{9}(648 + 0.12\,E_{\text{ROW}}) = 792 + \frac{11}{75}E_{\text{ROW}} \quad \cdots (2)$

Substituting (2) into (1):

$\frac{78}{55}\!\left(792 + \frac{11}{75}E_{\text{ROW}}\right) = 720 + 0.4\,E_{\text{ROW}}$

$1123.2 + 0.208\,E_{\text{ROW}} = 720 + 0.4\,E_{\text{ROW}}$

$403.2 = 0.192\,E_{\text{ROW}} \implies E_{\text{ROW}} = 2100$

Step 5 — Find ROW → ROW.

From clue 4, ROW's exports go to P ($40\%$) and X ($12\%$). The remaining share stays within ROW:

$100\% - 40\% - 12\% = 48\%$

$\text{ROW}{\to}\text{ROW} = 0.48 \times 2100 = 1008 \text{ IC}$

Solved export matrix (rows = exporter, columns = importer), in IC:

Answer: 1008

To find ROW's trade balance, we need both ROW's total exports and total imports, which requires solving the full trade system.

Step 1 — Cheeseland's exports.

P is the only exporter to C, and P→C = 1200, so $I_C = 1200$. With C's normalized balance at $-20\%$:

$\frac{E_C - 1200}{E_C + 1200} = -0.20 \implies 1.2\,E_C = 960 \implies E_C = 800$

C's exports split: $90\% \to P = 720$, $4\% \to \text{ROW} = 32$, remaining $6\% \to X = 48$.

Step 2 — Xiland's relationships.

X's normalized balance is $10\%$:

$\frac{E_X - I_X}{E_X + I_X} = 0.10 \implies E_X = \frac{11}{9}\,I_X$

X's exports go to P ($40\%$) and ROW ($60\%$) — not to C, since P is the only exporter to C.

X's imports: $I_X = 600 + 48 + 0.12\,E_{\text{ROW}} = 648 + 0.12\,E_{\text{ROW}}$

Step 3 — Pumpland's relationships.

P's normalized balance is $0\%$, so $E_P = I_P$.

From clue 2: $0.4\,E_X = 0.22\,I_P \implies I_P = \frac{20}{11}E_X$

Also: $I_P = 0.4\,E_X + 720 + 0.4\,E_{\text{ROW}}$

Step 4 — Solve the system.

Equating the two expressions for $I_P$:

$\frac{20}{11}E_X = 0.4\,E_X + 720 + 0.4\,E_{\text{ROW}} \implies \frac{78}{55}E_X = 720 + 0.4\,E_{\text{ROW}} \quad \cdots (1)$

From X's balance: $E_X = \frac{11}{9}(648 + 0.12\,E_{\text{ROW}}) = 792 + \frac{11}{75}E_{\text{ROW}} \quad \cdots (2)$

Substituting (2) into (1):

$\frac{78}{55}\!\left(792 + \frac{11}{75}E_{\text{ROW}}\right) = 720 + 0.4\,E_{\text{ROW}}$

$1123.2 + 0.208\,E_{\text{ROW}} = 720 + 0.4\,E_{\text{ROW}}$

$403.2 = 0.192\,E_{\text{ROW}} \implies E_{\text{ROW}} = 2100$

Back-substituting: $E_X = 1100$, $E_P = I_P = 2000$, $P{\to}\text{ROW} = 200$.

Step 5 — Compute ROW's trade balance.

ROW's total exports = 2100.

ROW's imports come from every exporter that sends goods to ROW: P (200), X (660), C (32), and ROW itself (1008):

$I_{\text{ROW}} = 200 + 660 + 32 + 1008 = 1900$

💡 Teacher tip: ROW's imports are the sum of the entire ROW column in the trade matrix — including the diagonal entry where ROW exports to itself.

$\text{Trade Balance}_{\text{ROW}} = E_{\text{ROW}} - I_{\text{ROW}} = 2100 - 1900 = 200$

Solved export matrix (rows = exporter, columns = importer), in IC:

Answer: D. 200

To compare total trade across P, X, and C, we need each country's exports and imports, which requires solving the full trade system.

Step 1 — Cheeseland's exports.

P is the only exporter to C, and P→C = 1200, so $I_C = 1200$. With C's normalized balance at $-20\%$:

$\frac{E_C - 1200}{E_C + 1200} = -0.20 \implies 1.2\,E_C = 960 \implies E_C = 800$

C's exports split: $90\% \to P = 720$, $4\% \to \text{ROW} = 32$, remaining $6\% \to X = 48$.

Step 2 — Xiland's relationships.

X's normalized balance is $10\%$:

$\frac{E_X - I_X}{E_X + I_X} = 0.10 \implies E_X = \frac{11}{9}\,I_X$

X's exports go to P ($40\%$) and ROW ($60\%$) — not to C, since P is the only exporter to C.

X's imports: $I_X = 600 + 48 + 0.12\,E_{\text{ROW}} = 648 + 0.12\,E_{\text{ROW}}$

Step 3 — Pumpland's relationships.

P's normalized balance is $0\%$, so $E_P = I_P$.

From clue 2: $0.4\,E_X = 0.22\,I_P \implies I_P = \frac{20}{11}E_X$

Also: $I_P = 0.4\,E_X + 720 + 0.4\,E_{\text{ROW}}$

Step 4 — Solve the system.

Equating the two expressions for $I_P$:

$\frac{20}{11}E_X = 0.4\,E_X + 720 + 0.4\,E_{\text{ROW}} \implies \frac{78}{55}E_X = 720 + 0.4\,E_{\text{ROW}} \quad \cdots (1)$

From X's balance: $E_X = \frac{11}{9}(648 + 0.12\,E_{\text{ROW}}) = 792 + \frac{11}{75}E_{\text{ROW}} \quad \cdots (2)$

Substituting (2) into (1):

$\frac{78}{55}\!\left(792 + \frac{11}{75}E_{\text{ROW}}\right) = 720 + 0.4\,E_{\text{ROW}}$

$1123.2 + 0.208\,E_{\text{ROW}} = 720 + 0.4\,E_{\text{ROW}}$

$403.2 = 0.192\,E_{\text{ROW}} \implies E_{\text{ROW}} = 2100$

Back-substituting: $E_X = 1100$, $I_X = 900$, $E_P = I_P = 2000$.

Step 5 — Compare total trade.

Total trade = Exports + Imports for each country:

X and C are tied at 2000 IC each, which is the lowest among the three.

Solved export matrix (rows = exporter, columns = importer), in IC:

Answer: C. Both X and C

Finding the circular seating

The children face inwards, so each child's immediately to the left neighbour is the next seat in one fixed rotational direction, and immediately to the right is the opposite direction. Second to the left/right means skipping one seat in that direction.

The holder of each round is the receiver of the previous round; Bina holds in Round 1, and Chirag receives in Round 10.

Step-by-step deduction of the seating order

1. Round 1: Bina passes immediately to the left → Aarav. So Aarav sits immediately to Bina's left.
2. Round 2: Aarav passes second to the right. From Aarav, going right one seat reaches Bina, and a second seat reaches Eshan — so Eshan receives in Round 2.
3. Round 3: Eshan passes immediately to the right → Diya. So Diya sits immediately to Eshan's right.
4. Round 6: Aarav passes second to the left → Farhan. So Farhan sits second to Aarav's left.
5. Round 7: Farhan passes immediately to the left → Gaurav. So Gaurav sits immediately to Farhan's left.
6. Round 8: Gaurav passes immediately to the left → Diya (confirmed by the table). So Diya sits immediately to Gaurav's left.
7. Round 9: Diya passes to Farhan. From Diya, Farhan is second to the right — this fixes Round 9's pass type.
8. Round 10: Farhan passes to Chirag. From Farhan, Chirag is immediately to the right — this fixes Round 10's pass type.

Putting all the neighbour relationships together, the circular order (going in the immediately-to-the-left direction) is:

$\textbf{Bina} \to \textbf{Aarav} \to \textbf{Chirag} \to \textbf{Farhan} \to \textbf{Gaurav} \to \textbf{Diya} \to \textbf{Eshan} \to (\text{back to Bina})$

💡 Teacher tip: Each child's immediate-left neighbour is the next name in the loop above; each child's immediate-right neighbour is the previous name in the loop.

Answering this question

We need Bina's immediate-right neighbour — the name just before Bina in the left-going loop. Reading the loop, the name before Bina is Eshan.

Answer: Eshan (Choice B)

The solved circular seating

From the round-by-round pass analysis (Bina starts, Chirag receives last, and each round's holder is the previous round's receiver), the forced circular order going in the immediately-to-the-left direction is:

$\textbf{Bina} \to \textbf{Aarav} \to \textbf{Chirag} \to \textbf{Farhan} \to \textbf{Gaurav} \to \textbf{Diya} \to \textbf{Eshan} \to (\text{back to Bina})$

Key relationships:

• Each child's immediate-left neighbour = next name in the loop.
• Each child's immediate-right neighbour = previous name in the loop.

Answering this question

We need the child sitting third to the left of Eshan. Starting at Eshan and stepping three seats in the immediately-to-the-left direction:

1. 1st to the left of Eshan → Bina
2. 2nd to the left of Eshan → Aarav
3. 3rd to the left of Eshan → Chirag

Answer: Chirag (Choice C)

The solved circular seating and round-by-round passes

The forced circular order (going in the immediately-to-the-left direction) is:

$\textbf{Bina} \to \textbf{Aarav} \to \textbf{Chirag} \to \textbf{Farhan} \to \textbf{Gaurav} \to \textbf{Diya} \to \textbf{Eshan} \to (\text{back to Bina})$

Working through all 10 rounds (holder = previous round's receiver):

The only flexibility is in Rounds 4–5: Diya must reach Aarav in exactly two legal passes. There are three possible ways:

Tallying pass-type counts

Fixed rounds (1, 2, 3, 6, 7, 8, 9, 10):

Adding Rounds 4–5 from each scenario:

Final totals across scenarios:

Only Immediately to the right has the same count (2) in every possible scenario.

Answer: Immediately to the right (Choice C)

The solved circular seating and round-by-round passes

The forced circular order (going in the immediately-to-the-left direction) is:

$\textbf{Bina} \to \textbf{Aarav} \to \textbf{Chirag} \to \textbf{Farhan} \to \textbf{Gaurav} \to \textbf{Diya} \to \textbf{Eshan} \to (\text{back to Bina})$

The full round-by-round table (holder = previous round's receiver):

Rounds 4–5 are the only flexible part: Diya must reach Aarav in two legal passes, and the Round-4 receiver can be Eshan, Bina, or Farhan.

Counting receptions per child

Ignoring Round 4 (the one variable receiver), the fixed receptions are:

Now, Round 4's receiver is Eshan, Bina, or Farhan — so only those three children's counts can change:

Gaurav is never the Round-4 receiver, so his reception count is fixed at 1 regardless of which scenario plays out. Among the four options, only Gaurav's count can be uniquely determined.

💡 Teacher tip: When part of a set is flexible, check whether each answer choice is ever affected by the flexible part. Only a child who cannot appear in the variable slot has a fixed count.

Answer: Gaurav (Choice D)