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Anu, Bijay, Chetan, Deepak, Eshan, and Faruq are six friends. Each of them uses a mobile number from exactly one of the two mobile operators - Xitel and Yocel. During the last month, the six friends made several calls to each other. Each call was made by one of these six friends to another. The table below summarizes the number of minutes of calls that each of the six made to (outgoing minutes) and received from (incoming minutes) these friends, grouped by the operators. Some of the entries are missing.
| Friend | Operator | Outgoing minutes to | Incoming minutes from | ||
|---|---|---|---|---|---|
| Operator Xitel | Operator Yocel | Operator Xitel | Operator Yocel | ||
| Anu | Xitel | 100 | 50 | 225 | |
| Bijay | Xitel | 200 | 125 | ||
| Chetan | Yocel | 50 | 175 | 250 | 150 |
| Deepak | Yocel | 100 | 150 | 275 | 100 |
| Eshan | Yocel | 100 | 100 | 375 | |
| Faruq | Yocel | 0 | 100 | 150 | |
It is known that the duration of calls from Faruq to Eshan was 200 minutes.
Also, there were no calls from:
i. Bijay to Eshan,
ii. Chetan to Anu and Chetan to Deepak,
iii. Deepak to Bijay and Deepak to Faruq,
iv. Eshan to Chetan and Eshan to Deepak.
What was the duration of calls (in minutes) from Bijay to Anu?
Answer: 50
Operators: Anu, Bijay use Xitel; Chetan, Deepak, Eshan, Faruq use Yocel. So 'to/from Xitel' means to/from Anu or Bijay, and 'to/from Yocel' means to/from C, D, E, F. Write A→B for minutes of calls from A to B.
Anu's 'incoming from Xitel' is 50. The only other Xitel user is Bijay, so all of it came from Bijay: B→A = 50.
Hence, the answer is 50.
What was the total duration of calls (in minutes) made by Anu to friends having mobile numbers from Operator Yocel?
Answer: 525
Operators: Anu, Bijay use Xitel; Chetan, Deepak, Eshan, Faruq use Yocel. So 'to/from Xitel' means to/from Anu or Bijay, and 'to/from Yocel' means to/from C, D, E, F. Write A→B for minutes of calls from A to B.
Step 1 (direct from the table and the 'no call' list):
Anu's outgoing to Xitel = 100 = A→B.
Chetan→Xitel = 50 and C→A = 0, so C→B = 50.
Deepak→Xitel = 100 and D→B = 0, so D→A = 100.
Eshan→Yocel = 100 and E→C = E→D = 0, so E→F = 100.
Faruq→Xitel = 0, so F→A = F→B = 0. Given F→E = 200.
Anu's incoming from Xitel = 50 = B→A.
Eshan's incoming from Xitel = 100 and B→E = 0, so A→E = 100.
Step 2 (incoming from Yocel):
Anu: C→A + D→A + E→A + F→A = 0 + 100 + E→A + 0 = 225, so E→A = 125.
Bijay: 50 + 0 + E→B + 0 = 125, so E→B = 75.
Deepak: C→D + E→D + F→D = 0 + 0 + F→D = 100, so F→D = 100.
Faruq: C→F + D→F + E→F = C→F + 0 + 100 = 150, so C→F = 50.
Chetan's outgoing to Yocel = C→E + C→F = 175, so C→E = 125.
Eshan's incoming from Yocel: C→E + D→E + F→E = 125 + D→E + 200 = 375, so D→E = 50.
Deepak's outgoing to Yocel: D→C + D→E = 150, so D→C = 100.
Chetan's incoming from Yocel: D→C + E→C + F→C = 100 + 0 + F→C = 150, so F→C = 50.
Calls determined (minutes, row = caller, column = receiver):
| From \ To | Anu | Bijay | Chetan | Deepak | Eshan | Faruq |
|---|---|---|---|---|---|---|
| Anu | - | 100 | ? | ? | 100 | ? |
| Bijay | 50 | - | ? | ? | 0 | ? |
| Chetan | 0 | 50 | - | 0 | 125 | 50 |
| Deepak | 100 | 0 | 100 | - | 50 | 0 |
| Eshan | 125 | 75 | 0 | 0 | - | 100 |
| Faruq | 0 | 0 | 50 | 100 | 200 | - |
(The '?' cells are not individually fixed, but A→C + B→C = 250, A→D + B→D = 275, A→F + B→F = 100 from the incoming-from-Xitel column, and B→C + B→D + B→F = 200.)
Anu's calls to Yocel = A→C + A→D + A→E + A→F. A→E = 100. From the incoming-from-Xitel column: (A→C + B→C) + (A→D + B→D) + (A→F + B→F) = 250 + 275 + 100 = 625. Bijay's outgoing to Yocel is B→C + B→D + B→F = 200 (B→E = 0). So A→C + A→D + A→F = 625 − 200 = 425.
Total = 425 + 100 = 525 minutes.
Hence, the answer is 525.
What was the total duration of calls (in minutes) made by Faruq to friends having mobile numbers from Operator Yocel?
Answer: 350
Operators: Anu, Bijay use Xitel; Chetan, Deepak, Eshan, Faruq use Yocel. So 'to/from Xitel' means to/from Anu or Bijay, and 'to/from Yocel' means to/from C, D, E, F. Write A→B for minutes of calls from A to B.
Step 1 (direct from the table and the 'no call' list):
Anu's outgoing to Xitel = 100 = A→B.
Chetan→Xitel = 50 and C→A = 0, so C→B = 50.
Deepak→Xitel = 100 and D→B = 0, so D→A = 100.
Eshan→Yocel = 100 and E→C = E→D = 0, so E→F = 100.
Faruq→Xitel = 0, so F→A = F→B = 0. Given F→E = 200.
Anu's incoming from Xitel = 50 = B→A.
Eshan's incoming from Xitel = 100 and B→E = 0, so A→E = 100.
Step 2 (incoming from Yocel):
Anu: C→A + D→A + E→A + F→A = 0 + 100 + E→A + 0 = 225, so E→A = 125.
Bijay: 50 + 0 + E→B + 0 = 125, so E→B = 75.
Deepak: C→D + E→D + F→D = 0 + 0 + F→D = 100, so F→D = 100.
Faruq: C→F + D→F + E→F = C→F + 0 + 100 = 150, so C→F = 50.
Chetan's outgoing to Yocel = C→E + C→F = 175, so C→E = 125.
Eshan's incoming from Yocel: C→E + D→E + F→E = 125 + D→E + 200 = 375, so D→E = 50.
Deepak's outgoing to Yocel: D→C + D→E = 150, so D→C = 100.
Chetan's incoming from Yocel: D→C + E→C + F→C = 100 + 0 + F→C = 150, so F→C = 50.
Calls determined (minutes, row = caller, column = receiver):
| From \ To | Anu | Bijay | Chetan | Deepak | Eshan | Faruq |
|---|---|---|---|---|---|---|
| Anu | - | 100 | ? | ? | 100 | ? |
| Bijay | 50 | - | ? | ? | 0 | ? |
| Chetan | 0 | 50 | - | 0 | 125 | 50 |
| Deepak | 100 | 0 | 100 | - | 50 | 0 |
| Eshan | 125 | 75 | 0 | 0 | - | 100 |
| Faruq | 0 | 0 | 50 | 100 | 200 | - |
(The '?' cells are not individually fixed, but A→C + B→C = 250, A→D + B→D = 275, A→F + B→F = 100 from the incoming-from-Xitel column, and B→C + B→D + B→F = 200.)
Faruq to Yocel users = F→C + F→D + F→E = 50 + 100 + 200 = 350 minutes.
Hence, the answer is 350.
What was the duration of calls (in minutes) from Deepak to Chetan?
Answer: C
Operators: Anu, Bijay use Xitel; Chetan, Deepak, Eshan, Faruq use Yocel. So 'to/from Xitel' means to/from Anu or Bijay, and 'to/from Yocel' means to/from C, D, E, F. Write A→B for minutes of calls from A to B.
Step 1 (direct from the table and the 'no call' list):
Anu's outgoing to Xitel = 100 = A→B.
Chetan→Xitel = 50 and C→A = 0, so C→B = 50.
Deepak→Xitel = 100 and D→B = 0, so D→A = 100.
Eshan→Yocel = 100 and E→C = E→D = 0, so E→F = 100.
Faruq→Xitel = 0, so F→A = F→B = 0. Given F→E = 200.
Anu's incoming from Xitel = 50 = B→A.
Eshan's incoming from Xitel = 100 and B→E = 0, so A→E = 100.
Step 2 (incoming from Yocel):
Anu: C→A + D→A + E→A + F→A = 0 + 100 + E→A + 0 = 225, so E→A = 125.
Bijay: 50 + 0 + E→B + 0 = 125, so E→B = 75.
Deepak: C→D + E→D + F→D = 0 + 0 + F→D = 100, so F→D = 100.
Faruq: C→F + D→F + E→F = C→F + 0 + 100 = 150, so C→F = 50.
Chetan's outgoing to Yocel = C→E + C→F = 175, so C→E = 125.
Eshan's incoming from Yocel: C→E + D→E + F→E = 125 + D→E + 200 = 375, so D→E = 50.
Deepak's outgoing to Yocel: D→C + D→E = 150, so D→C = 100.
Chetan's incoming from Yocel: D→C + E→C + F→C = 100 + 0 + F→C = 150, so F→C = 50.
Calls determined (minutes, row = caller, column = receiver):
| From \ To | Anu | Bijay | Chetan | Deepak | Eshan | Faruq |
|---|---|---|---|---|---|---|
| Anu | - | 100 | ? | ? | 100 | ? |
| Bijay | 50 | - | ? | ? | 0 | ? |
| Chetan | 0 | 50 | - | 0 | 125 | 50 |
| Deepak | 100 | 0 | 100 | - | 50 | 0 |
| Eshan | 125 | 75 | 0 | 0 | - | 100 |
| Faruq | 0 | 0 | 50 | 100 | 200 | - |
(The '?' cells are not individually fixed, but A→C + B→C = 250, A→D + B→D = 275, A→F + B→F = 100 from the incoming-from-Xitel column, and B→C + B→D + B→F = 200.)
D→C = 100 minutes.
Hence, the answer is (C) 100.
Aurevia, Brelosia, Cyrenia and Zerathania are four countries with their currencies being Aurels, Brins, Crowns, and Zentars, respectively. The currencies have different exchange values. Crown's currency exchange rate with Zentars = 0.5, i.e., 1 Crown is worth 0.5 Zentars.
Three travelers, Jano, Kira, and Lian set out from Zerathania visiting exactly two of the countries. Each country is visited by exactly two travelers. Each traveler has a unique Flight Cost, which represents the total cost of airfare in traveling to both the countries and back to Zerathania. The Flight Cost of Jano was 4000 Zentars, while that of the other two travelers were 5000 and 6000 Zentars, not necessarily in that order.
When visiting a country, a traveler spent either 1000, 2000 or 3000 in the country's local currency. Each traveler had different spends (in the country's local currency) in the two countries he/she visited. Across all the visits, there were exactly two spends of 1000 and exactly one spend of 3000 (in the country's local currency).
The total "Travel Cost" for a traveler is the sum of his/her Flight Cost and the money spent in the countries visited.
The citizens of the four countries with knowledge of these travels made a few observations, with spends measured in their respective local currencies:
i. Aurevia citizen: Jano and Kira visited our country, and their Travel Costs were 3500 and 8000, respectively.
ii. Brelosia citizen: Kira and Lian visited our country, spending 2000 and 3000, respectively. Kira's Travel Cost was 4000.
iii. Cyrenia citizen: Lian visited our country and her Travel Cost was 36000.
What is the sum of Travel Costs for all travelers in Zentars?
Answer: 41000
Who visited where: Each traveler visits two of Aurevia, Brelosia, Cyrenia (6 visits), and each of these is visited by exactly two. Aurevia: Jano, Kira. Brelosia: Kira, Lian. So Cyrenia: Lian and Jano. Hence Jano: A, C; Kira: A, B; Lian: B, C.
Exchange rates (value of 1 unit in Zentars): Aurel = \(a\), Brin = \(b\), Crown = 0.5.
Kira's Travel Cost is 8000 Aurels = 4000 Brins, so \(8000a = 4000b\), i.e. \(b = 2a\).
Spends: Lian spent 3000 in Brelosia (the only 3000), Kira 2000 in Brelosia. Kira's Aurevia spend must differ from 2000 and cannot be 3000, so it is 1000. The one remaining 1000 must go to Jano (his two spends differ and cannot include 3000, so they are 1000 and 2000). Hence Lian spent 2000 in Cyrenia.
Kira: \(F_K + 2000b + 1000a = 8000a \Rightarrow F_K + 5000a = 8000a \Rightarrow F_K = 3000a\).
Lian: Travel Cost = 36000 Crowns = 18000 Z, so \(F_L + 3000b + 2000(0.5) = 18000 \Rightarrow F_L = 17000 - 6000a\).
\(F_K, F_L\) are 5000 and 6000. \(F_K = 5000\) gives \(a = 5/3\), \(F_L = 7000\) (impossible). So \(F_K = 6000\), \(a = 2\), \(F_L = 5000\), \(b = 4\).
Jano: Travel Cost = 3500 Aurels = 7000 Z: \(4000 + 2\,s_A + 0.5\,s_C = 7000\). With \(\{s_A, s_C\} = \{1000, 2000\}\), only \(s_A = 1000, s_C = 2000\) works (2000 + 1000 = 3000).
| Traveler | Flight (Z) | Aurevia (Aurels) | Brelosia (Brins) | Cyrenia (Crowns) | Travel Cost (Z) |
|---|---|---|---|---|---|
| Jano | 4000 | 1000 | - | 2000 | 4000 + 2000 + 1000 = 7000 |
| Kira | 6000 | 1000 | 2000 | - | 6000 + 2000 + 8000 = 16000 |
| Lian | 5000 | - | 3000 | 2000 | 5000 + 12000 + 1000 = 18000 |
Rates: 1 Aurel = 2 Z, 1 Brin = 4 Z, 1 Crown = 0.5 Z.
Sum of Travel Costs = 7000 + 16000 + 18000 = 41000 Zentars.
Hence, the answer is 41000.
How many Zentars did Lian spend in the two countries he visited?
Answer: 13000
Who visited where: Each traveler visits two of Aurevia, Brelosia, Cyrenia (6 visits), and each of these is visited by exactly two. Aurevia: Jano, Kira. Brelosia: Kira, Lian. So Cyrenia: Lian and Jano. Hence Jano: A, C; Kira: A, B; Lian: B, C.
Exchange rates (value of 1 unit in Zentars): Aurel = \(a\), Brin = \(b\), Crown = 0.5.
Kira's Travel Cost is 8000 Aurels = 4000 Brins, so \(8000a = 4000b\), i.e. \(b = 2a\).
Spends: Lian spent 3000 in Brelosia (the only 3000), Kira 2000 in Brelosia. Kira's Aurevia spend must differ from 2000 and cannot be 3000, so it is 1000. The one remaining 1000 must go to Jano (his two spends differ and cannot include 3000, so they are 1000 and 2000). Hence Lian spent 2000 in Cyrenia.
Kira: \(F_K + 2000b + 1000a = 8000a \Rightarrow F_K + 5000a = 8000a \Rightarrow F_K = 3000a\).
Lian: Travel Cost = 36000 Crowns = 18000 Z, so \(F_L + 3000b + 2000(0.5) = 18000 \Rightarrow F_L = 17000 - 6000a\).
\(F_K, F_L\) are 5000 and 6000. \(F_K = 5000\) gives \(a = 5/3\), \(F_L = 7000\) (impossible). So \(F_K = 6000\), \(a = 2\), \(F_L = 5000\), \(b = 4\).
Jano: Travel Cost = 3500 Aurels = 7000 Z: \(4000 + 2\,s_A + 0.5\,s_C = 7000\). With \(\{s_A, s_C\} = \{1000, 2000\}\), only \(s_A = 1000, s_C = 2000\) works (2000 + 1000 = 3000).
| Traveler | Flight (Z) | Aurevia (Aurels) | Brelosia (Brins) | Cyrenia (Crowns) | Travel Cost (Z) |
|---|---|---|---|---|---|
| Jano | 4000 | 1000 | - | 2000 | 4000 + 2000 + 1000 = 7000 |
| Kira | 6000 | 1000 | 2000 | - | 6000 + 2000 + 8000 = 16000 |
| Lian | 5000 | - | 3000 | 2000 | 5000 + 12000 + 1000 = 18000 |
Rates: 1 Aurel = 2 Z, 1 Brin = 4 Z, 1 Crown = 0.5 Z.
Lian spent 3000 Brins = 3000 × 4 = 12000 Z and 2000 Crowns = 2000 × 0.5 = 1000 Z, a total of 13000 Zentars.
Hence, the answer is 13000.
What was Jano's total spend in the two countries he visited, in Aurels?
Answer: 1500
Who visited where: Each traveler visits two of Aurevia, Brelosia, Cyrenia (6 visits), and each of these is visited by exactly two. Aurevia: Jano, Kira. Brelosia: Kira, Lian. So Cyrenia: Lian and Jano. Hence Jano: A, C; Kira: A, B; Lian: B, C.
Exchange rates (value of 1 unit in Zentars): Aurel = \(a\), Brin = \(b\), Crown = 0.5.
Kira's Travel Cost is 8000 Aurels = 4000 Brins, so \(8000a = 4000b\), i.e. \(b = 2a\).
Spends: Lian spent 3000 in Brelosia (the only 3000), Kira 2000 in Brelosia. Kira's Aurevia spend must differ from 2000 and cannot be 3000, so it is 1000. The one remaining 1000 must go to Jano (his two spends differ and cannot include 3000, so they are 1000 and 2000). Hence Lian spent 2000 in Cyrenia.
Kira: \(F_K + 2000b + 1000a = 8000a \Rightarrow F_K + 5000a = 8000a \Rightarrow F_K = 3000a\).
Lian: Travel Cost = 36000 Crowns = 18000 Z, so \(F_L + 3000b + 2000(0.5) = 18000 \Rightarrow F_L = 17000 - 6000a\).
\(F_K, F_L\) are 5000 and 6000. \(F_K = 5000\) gives \(a = 5/3\), \(F_L = 7000\) (impossible). So \(F_K = 6000\), \(a = 2\), \(F_L = 5000\), \(b = 4\).
Jano: Travel Cost = 3500 Aurels = 7000 Z: \(4000 + 2\,s_A + 0.5\,s_C = 7000\). With \(\{s_A, s_C\} = \{1000, 2000\}\), only \(s_A = 1000, s_C = 2000\) works (2000 + 1000 = 3000).
| Traveler | Flight (Z) | Aurevia (Aurels) | Brelosia (Brins) | Cyrenia (Crowns) | Travel Cost (Z) |
|---|---|---|---|---|---|
| Jano | 4000 | 1000 | - | 2000 | 4000 + 2000 + 1000 = 7000 |
| Kira | 6000 | 1000 | 2000 | - | 6000 + 2000 + 8000 = 16000 |
| Lian | 5000 | - | 3000 | 2000 | 5000 + 12000 + 1000 = 18000 |
Rates: 1 Aurel = 2 Z, 1 Brin = 4 Z, 1 Crown = 0.5 Z.
Jano spent 1000 Aurels in Aurevia and 2000 Crowns = 1000 Z = 500 Aurels in Cyrenia. Total = 1000 + 500 = 1500 Aurels.
Hence, the answer is 1500.
One Brin is equivalent to how many Crowns?
Answer: C
Who visited where: Each traveler visits two of Aurevia, Brelosia, Cyrenia (6 visits), and each of these is visited by exactly two. Aurevia: Jano, Kira. Brelosia: Kira, Lian. So Cyrenia: Lian and Jano. Hence Jano: A, C; Kira: A, B; Lian: B, C.
Exchange rates (value of 1 unit in Zentars): Aurel = \(a\), Brin = \(b\), Crown = 0.5.
Kira's Travel Cost is 8000 Aurels = 4000 Brins, so \(8000a = 4000b\), i.e. \(b = 2a\).
Spends: Lian spent 3000 in Brelosia (the only 3000), Kira 2000 in Brelosia. Kira's Aurevia spend must differ from 2000 and cannot be 3000, so it is 1000. The one remaining 1000 must go to Jano (his two spends differ and cannot include 3000, so they are 1000 and 2000). Hence Lian spent 2000 in Cyrenia.
Kira: \(F_K + 2000b + 1000a = 8000a \Rightarrow F_K + 5000a = 8000a \Rightarrow F_K = 3000a\).
Lian: Travel Cost = 36000 Crowns = 18000 Z, so \(F_L + 3000b + 2000(0.5) = 18000 \Rightarrow F_L = 17000 - 6000a\).
\(F_K, F_L\) are 5000 and 6000. \(F_K = 5000\) gives \(a = 5/3\), \(F_L = 7000\) (impossible). So \(F_K = 6000\), \(a = 2\), \(F_L = 5000\), \(b = 4\).
Jano: Travel Cost = 3500 Aurels = 7000 Z: \(4000 + 2\,s_A + 0.5\,s_C = 7000\). With \(\{s_A, s_C\} = \{1000, 2000\}\), only \(s_A = 1000, s_C = 2000\) works (2000 + 1000 = 3000).
| Traveler | Flight (Z) | Aurevia (Aurels) | Brelosia (Brins) | Cyrenia (Crowns) | Travel Cost (Z) |
|---|---|---|---|---|---|
| Jano | 4000 | 1000 | - | 2000 | 4000 + 2000 + 1000 = 7000 |
| Kira | 6000 | 1000 | 2000 | - | 6000 + 2000 + 8000 = 16000 |
| Lian | 5000 | - | 3000 | 2000 | 5000 + 12000 + 1000 = 18000 |
Rates: 1 Aurel = 2 Z, 1 Brin = 4 Z, 1 Crown = 0.5 Z.
1 Brin = 4 Zentars and 1 Crown = 0.5 Zentar, so 1 Brin = 4 / 0.5 = 8 Crowns.
Hence, the answer is (C) 8.
Which of the following statements is NOT true about money spent in the local currency?
Answer: B
Who visited where: Each traveler visits two of Aurevia, Brelosia, Cyrenia (6 visits), and each of these is visited by exactly two. Aurevia: Jano, Kira. Brelosia: Kira, Lian. So Cyrenia: Lian and Jano. Hence Jano: A, C; Kira: A, B; Lian: B, C.
Exchange rates (value of 1 unit in Zentars): Aurel = \(a\), Brin = \(b\), Crown = 0.5.
Kira's Travel Cost is 8000 Aurels = 4000 Brins, so \(8000a = 4000b\), i.e. \(b = 2a\).
Spends: Lian spent 3000 in Brelosia (the only 3000), Kira 2000 in Brelosia. Kira's Aurevia spend must differ from 2000 and cannot be 3000, so it is 1000. The one remaining 1000 must go to Jano (his two spends differ and cannot include 3000, so they are 1000 and 2000). Hence Lian spent 2000 in Cyrenia.
Kira: \(F_K + 2000b + 1000a = 8000a \Rightarrow F_K + 5000a = 8000a \Rightarrow F_K = 3000a\).
Lian: Travel Cost = 36000 Crowns = 18000 Z, so \(F_L + 3000b + 2000(0.5) = 18000 \Rightarrow F_L = 17000 - 6000a\).
\(F_K, F_L\) are 5000 and 6000. \(F_K = 5000\) gives \(a = 5/3\), \(F_L = 7000\) (impossible). So \(F_K = 6000\), \(a = 2\), \(F_L = 5000\), \(b = 4\).
Jano: Travel Cost = 3500 Aurels = 7000 Z: \(4000 + 2\,s_A + 0.5\,s_C = 7000\). With \(\{s_A, s_C\} = \{1000, 2000\}\), only \(s_A = 1000, s_C = 2000\) works (2000 + 1000 = 3000).
| Traveler | Flight (Z) | Aurevia (Aurels) | Brelosia (Brins) | Cyrenia (Crowns) | Travel Cost (Z) |
|---|---|---|---|---|---|
| Jano | 4000 | 1000 | - | 2000 | 4000 + 2000 + 1000 = 7000 |
| Kira | 6000 | 1000 | 2000 | - | 6000 + 2000 + 8000 = 16000 |
| Lian | 5000 | - | 3000 | 2000 | 5000 + 12000 + 1000 = 18000 |
Rates: 1 Aurel = 2 Z, 1 Brin = 4 Z, 1 Crown = 0.5 Z.
Kira spent 1000 in Aurevia (true); Jano spent 1000, not 2000, in Aurevia (false); Jano spent 2000 in Cyrenia (true); Lian spent 2000 in Cyrenia (true).
Hence, the answer is (B) Jano spent 2000 in Aurevia.
Anirbid, Chandranath, Koushik, and Suranjan participated in a puzzle solving competition. The competition comprised 10 puzzles that had to be solved in the same sequence, i.e., a competitor got access to a puzzle as soon as they solved the previous puzzle. Some of the puzzles were visual puzzles and the others were number-based puzzles. The winner of the competition was the one who solved all puzzles in the least time.
The following charts describe their progress in the competition. The chart on the left shows the number of puzzles solved by each competitor at a given time (in minutes) after the start of the competition. The chart on the right shows the number of visual puzzles solved by each competitor at a given time (in minutes) after the start of the competition.
Who had solved the largest number of puzzles by the 20-th minute from the start of the competition?
Answer: B
Reading the charts (completion time, in minutes, of each puzzle):
| Puzzle | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| Anirbid | 4 | 9 | 11 | 14 | 19 | 22 | 27 | 29 | 33 | 37 |
| Chandranath | 4 | 6 | 10 | 13 | 17 | 21 | 23 | 26 | 28 | 30 |
| Koushik | 2 | 4 | 7 | 12 | 14 | 16 | 19 | 25 | 29 | 31 |
| Suranjan | 5 | 7 | 12 | 18 | 21 | 23 | 26 | 28 | 30 | 35 |
From the visual-puzzle chart, Koushik's visual puzzles were completed at 2, 12, 25, 29, i.e. puzzles 1, 4, 8, 9 (Anirbid: 4, 14, 29, 33; Suranjan: 5, 18, 28, 30; the same positions). So the visual puzzles are #1, #4, #8, #9 and the number-based puzzles are #2, #3, #5, #6, #7, #10.
Puzzles solved by minute 20: Koushik 7 (7th at minute 19), Anirbid 5, Chandranath 5 (6th only at 21), Suranjan 4 (5th at 21). Koushik leads.
Hence, the answer is (B) Koushik.
How many minutes did Suranjan take to solve the third visual puzzle in the competition?
Answer: 2
Reading the charts (completion time, in minutes, of each puzzle):
| Puzzle | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| Anirbid | 4 | 9 | 11 | 14 | 19 | 22 | 27 | 29 | 33 | 37 |
| Chandranath | 4 | 6 | 10 | 13 | 17 | 21 | 23 | 26 | 28 | 30 |
| Koushik | 2 | 4 | 7 | 12 | 14 | 16 | 19 | 25 | 29 | 31 |
| Suranjan | 5 | 7 | 12 | 18 | 21 | 23 | 26 | 28 | 30 | 35 |
From the visual-puzzle chart, Koushik's visual puzzles were completed at 2, 12, 25, 29, i.e. puzzles 1, 4, 8, 9 (Anirbid: 4, 14, 29, 33; Suranjan: 5, 18, 28, 30; the same positions). So the visual puzzles are #1, #4, #8, #9 and the number-based puzzles are #2, #3, #5, #6, #7, #10.
The third visual puzzle is puzzle #8. Suranjan solved #7 at minute 26 and #8 at minute 28 (his visual count rises from 2 to 3 at 28), so he took 28 − 26 = 2 minutes.
Hence, the answer is 2.
At what number in the sequence was the fourth number-based puzzle?
Answer: 6
Reading the charts (completion time, in minutes, of each puzzle):
| Puzzle | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| Anirbid | 4 | 9 | 11 | 14 | 19 | 22 | 27 | 29 | 33 | 37 |
| Chandranath | 4 | 6 | 10 | 13 | 17 | 21 | 23 | 26 | 28 | 30 |
| Koushik | 2 | 4 | 7 | 12 | 14 | 16 | 19 | 25 | 29 | 31 |
| Suranjan | 5 | 7 | 12 | 18 | 21 | 23 | 26 | 28 | 30 | 35 |
From the visual-puzzle chart, Koushik's visual puzzles were completed at 2, 12, 25, 29, i.e. puzzles 1, 4, 8, 9 (Anirbid: 4, 14, 29, 33; Suranjan: 5, 18, 28, 30; the same positions). So the visual puzzles are #1, #4, #8, #9 and the number-based puzzles are #2, #3, #5, #6, #7, #10.
The number-based puzzles in order are #2, #3, #5, #6, ..., so the fourth one is puzzle #6.
Hence, the answer is 6.
Which of the following is the closest to the average time taken by Anirbid to solve the number-based puzzles in the competition?
Answer: C
Reading the charts (completion time, in minutes, of each puzzle):
| Puzzle | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| Anirbid | 4 | 9 | 11 | 14 | 19 | 22 | 27 | 29 | 33 | 37 |
| Chandranath | 4 | 6 | 10 | 13 | 17 | 21 | 23 | 26 | 28 | 30 |
| Koushik | 2 | 4 | 7 | 12 | 14 | 16 | 19 | 25 | 29 | 31 |
| Suranjan | 5 | 7 | 12 | 18 | 21 | 23 | 26 | 28 | 30 | 35 |
From the visual-puzzle chart, Koushik's visual puzzles were completed at 2, 12, 25, 29, i.e. puzzles 1, 4, 8, 9 (Anirbid: 4, 14, 29, 33; Suranjan: 5, 18, 28, 30; the same positions). So the visual puzzles are #1, #4, #8, #9 and the number-based puzzles are #2, #3, #5, #6, #7, #10.
Anirbid's times on number-based puzzles: #2: 9 − 4 = 5, #3: 11 − 9 = 2, #5: 19 − 14 = 5, #6: 22 − 19 = 3, #7: 27 − 22 = 5, #10: 37 − 33 = 4. Total = 24 minutes for 6 puzzles, average = 4.0 minutes.
Hence, the answer is (C) 4.0 minutes.
Seven children, Aarav, Bina, Chirag, Diya, Eshan, Farhan, and Gaurav, are sitting in a circle facing inside (not necessarily in the same order) and playing a game of 'Passing the Buck'.
The game is played over 10 rounds. In each round, the child holding the Buck must pass it directly to a child sitting in one of the following positions:
• Immediately to the left;
• Immediate to the right;
• Second to the left; or
• Second to the right.
The game starts with Bina passing the Buck and ends with Chirag receiving the Buck. The table below provides some information about the pass types and the child receiving the Buck. Some information is missing and labelled as '?'.
| Round | Position | Name |
|---|---|---|
| 1 | Immediately to the left | Aarav |
| 2 | Second to the right | ? |
| 3 | Immediately to the right | Diya |
| 4 | ? | ? |
| 5 | ? | Aarav |
| 6 | Second to the left | ? |
| 7 | Immediately to the left | Gaurav |
| 8 | Immediately to the left | ? |
| 9 | ? | Farhan |
| 10 | ? | Chirag |
Who is sitting immediately to the right of Bina?
Answer: D
Number the seats around the circle so that 'immediately to the left' of a seat is +1 and 'immediately to the right' is −1 (mod 7). Put Bina at seat 0.
Round 1 (left 1): Aarav = 1. Round 2 (right 2): receiver at seat 6. Round 3 (right 1): Diya = 5. Round 5 receiver is Aarav (1); Round 6 (left 2): seat 3; Round 7 (left 1): Gaurav = 4; Round 8 (left 1): seat 5 = Diya.
Chirag, Eshan, Farhan occupy seats 2, 3, 6. Round 9 goes from Diya (5) to Farhan, so Farhan is within two seats of 5: seat 3 or 6. Round 10 goes from Farhan to Chirag. If Farhan = 6, Chirag would need seat 4, 5, 0 or 1, none free. So Farhan = 3, Chirag = 2, Eshan = 6.
Seating (going leftwards from Bina): Bina(0), Aarav(1), Chirag(2), Farhan(3), Gaurav(4), Diya(5), Eshan(6).
Round 4 receiver must be within two seats of Diya (5) and of Aarav (1): seat 0, 3 or 6 (Bina, Farhan or Eshan).
| Round | Pass | Receiver |
|---|---|---|
| 1 | Left 1 | Aarav |
| 2 | Right 2 | Eshan |
| 3 | Right 1 | Diya |
| 4 | Left 2 / Right 2 / Left 1 | Bina / Farhan / Eshan |
| 5 | Left 1 / Right 2 / Left 2 | Aarav |
| 6 | Left 2 | Farhan |
| 7 | Left 1 | Gaurav |
| 8 | Left 1 | Diya |
| 9 | Right 2 | Farhan |
| 10 | Right 1 | Chirag |
Immediately to the right of Bina (seat 0) is seat 6, Eshan.
Hence, the answer is (D) Eshan.
Who is sitting third to the left of Eshan?
Answer: B
Number the seats around the circle so that 'immediately to the left' of a seat is +1 and 'immediately to the right' is −1 (mod 7). Put Bina at seat 0.
Round 1 (left 1): Aarav = 1. Round 2 (right 2): receiver at seat 6. Round 3 (right 1): Diya = 5. Round 5 receiver is Aarav (1); Round 6 (left 2): seat 3; Round 7 (left 1): Gaurav = 4; Round 8 (left 1): seat 5 = Diya.
Chirag, Eshan, Farhan occupy seats 2, 3, 6. Round 9 goes from Diya (5) to Farhan, so Farhan is within two seats of 5: seat 3 or 6. Round 10 goes from Farhan to Chirag. If Farhan = 6, Chirag would need seat 4, 5, 0 or 1, none free. So Farhan = 3, Chirag = 2, Eshan = 6.
Seating (going leftwards from Bina): Bina(0), Aarav(1), Chirag(2), Farhan(3), Gaurav(4), Diya(5), Eshan(6).
Round 4 receiver must be within two seats of Diya (5) and of Aarav (1): seat 0, 3 or 6 (Bina, Farhan or Eshan).
| Round | Pass | Receiver |
|---|---|---|
| 1 | Left 1 | Aarav |
| 2 | Right 2 | Eshan |
| 3 | Right 1 | Diya |
| 4 | Left 2 / Right 2 / Left 1 | Bina / Farhan / Eshan |
| 5 | Left 1 / Right 2 / Left 2 | Aarav |
| 6 | Left 2 | Farhan |
| 7 | Left 1 | Gaurav |
| 8 | Left 1 | Diya |
| 9 | Right 2 | Farhan |
| 10 | Right 1 | Chirag |
Eshan is at seat 6; third to his left is seat 6 + 3 = 9 ≡ 2, which is Chirag.
Hence, the answer is (B) Chirag.
For which of the following pass types can the total number of occurrences be uniquely determined?
Answer: D
Number the seats around the circle so that 'immediately to the left' of a seat is +1 and 'immediately to the right' is −1 (mod 7). Put Bina at seat 0.
Round 1 (left 1): Aarav = 1. Round 2 (right 2): receiver at seat 6. Round 3 (right 1): Diya = 5. Round 5 receiver is Aarav (1); Round 6 (left 2): seat 3; Round 7 (left 1): Gaurav = 4; Round 8 (left 1): seat 5 = Diya.
Chirag, Eshan, Farhan occupy seats 2, 3, 6. Round 9 goes from Diya (5) to Farhan, so Farhan is within two seats of 5: seat 3 or 6. Round 10 goes from Farhan to Chirag. If Farhan = 6, Chirag would need seat 4, 5, 0 or 1, none free. So Farhan = 3, Chirag = 2, Eshan = 6.
Seating (going leftwards from Bina): Bina(0), Aarav(1), Chirag(2), Farhan(3), Gaurav(4), Diya(5), Eshan(6).
Round 4 receiver must be within two seats of Diya (5) and of Aarav (1): seat 0, 3 or 6 (Bina, Farhan or Eshan).
| Round | Pass | Receiver |
|---|---|---|
| 1 | Left 1 | Aarav |
| 2 | Right 2 | Eshan |
| 3 | Right 1 | Diya |
| 4 | Left 2 / Right 2 / Left 1 | Bina / Farhan / Eshan |
| 5 | Left 1 / Right 2 / Left 2 | Aarav |
| 6 | Left 2 | Farhan |
| 7 | Left 1 | Gaurav |
| 8 | Left 1 | Diya |
| 9 | Right 2 | Farhan |
| 10 | Right 1 | Chirag |
Fixed passes: Left 1 in rounds 1, 7, 8; Right 2 in rounds 2, 9; Right 1 in rounds 3, 10; Left 2 in round 6. Rounds 4 and 5 are (Left 2, Left 1), (Right 2, Right 2) or (Left 1, Left 2), all valid. None of these options adds a 'Right 1' pass, so 'Immediately to the right' occurs exactly 2 times in every case. The other three counts change with the case.
Hence, the answer is (D) Immediately to the right.
For which of the following children is it possible to determine how many times they received the Buck?
Answer: B
Number the seats around the circle so that 'immediately to the left' of a seat is +1 and 'immediately to the right' is −1 (mod 7). Put Bina at seat 0.
Round 1 (left 1): Aarav = 1. Round 2 (right 2): receiver at seat 6. Round 3 (right 1): Diya = 5. Round 5 receiver is Aarav (1); Round 6 (left 2): seat 3; Round 7 (left 1): Gaurav = 4; Round 8 (left 1): seat 5 = Diya.
Chirag, Eshan, Farhan occupy seats 2, 3, 6. Round 9 goes from Diya (5) to Farhan, so Farhan is within two seats of 5: seat 3 or 6. Round 10 goes from Farhan to Chirag. If Farhan = 6, Chirag would need seat 4, 5, 0 or 1, none free. So Farhan = 3, Chirag = 2, Eshan = 6.
Seating (going leftwards from Bina): Bina(0), Aarav(1), Chirag(2), Farhan(3), Gaurav(4), Diya(5), Eshan(6).
Round 4 receiver must be within two seats of Diya (5) and of Aarav (1): seat 0, 3 or 6 (Bina, Farhan or Eshan).
| Round | Pass | Receiver |
|---|---|---|
| 1 | Left 1 | Aarav |
| 2 | Right 2 | Eshan |
| 3 | Right 1 | Diya |
| 4 | Left 2 / Right 2 / Left 1 | Bina / Farhan / Eshan |
| 5 | Left 1 / Right 2 / Left 2 | Aarav |
| 6 | Left 2 | Farhan |
| 7 | Left 1 | Gaurav |
| 8 | Left 1 | Diya |
| 9 | Right 2 | Farhan |
| 10 | Right 1 | Chirag |
Receivers: Aarav (R1, R5), Eshan (R2), Diya (R3, R8), Farhan (R6, R9), Gaurav (R7), Chirag (R10), and the Round 4 receiver is Bina, Farhan or Eshan. So Bina receives 0 or 1 times, Farhan 2 or 3, Eshan 1 or 2, while Gaurav receives exactly once in every case.
Hence, the answer is (B) Gaurav.
Three countries — Pumpland (P), Xiland (X) and Cheeseland (C) — trade among themselves and with the (other countries in) Rest of World (ROW). All trade volumes are given in IC (international currency). The following terminology is used:
• Trade balance = Exports – Imports
• Total trade = Exports + Imports
• Normalized trade balance = Trade balance / Total trade, expressed in percentage terms
The following information is known.
1. The normalized trade balances of P, X and C are 0%, 10%, and –20%, respectively.
2. 40% of exports of X are to P. 22% of imports of P are from X.
3. 90% of exports of C are to P; 4% are to ROW.
4. 12% of exports of ROW are to X, 40% are to P.
5. The export volumes of P, in IC, to X and C are 600 and 1200, respectively. P is the only country that exports to C.
How much is exported from C to X, in IC?
Answer: 48
C: P is the only exporter to C, so C's imports = 1200. Normalized balance −20%: \(\frac{E_C - 1200}{E_C + 1200} = -0.2 \Rightarrow E_C = 800\). C exports 90% = 720 to P, 4% = 32 to ROW, and the remaining 6% = 48 to X.
Let ROW's total exports = \(R\) (12% to X, 40% to P, nothing to C, so 48% to ROW itself), X's exports = \(E_X\) (40% to P, rest 60% to ROW), P's imports = \(I_P\).
P's imports: \(I_P = 0.22I_P + 720 + 0.4R \Rightarrow 0.78I_P = 720 + 0.4R\).
X to P: \(0.4E_X = 0.22I_P \Rightarrow E_X = 0.55I_P\).
X's imports = 600 + 48 + 0.12R; balance 10% gives \(0.9E_X = 1.1I_X\), so \(E_X = \frac{11}{9}(648 + 0.12R)\).
Equating: \(\frac{0.55(720 + 0.4R)}{0.78} = \frac{11}{9}(648 + 0.12R) \Rightarrow 3564 + 1.98R = 5559.84 + 1.0296R \Rightarrow R = 2100\).
Then \(I_P = 1560/0.78 = 2000\), \(E_X = 1100\), \(I_X = 900\). P's balance is 0, so P's exports = 2000 = 600 + 1200 + (P to ROW), giving P to ROW = 200.
| Exporter \ Importer | P | X | C | ROW | Total exports |
|---|---|---|---|---|---|
| P | - | 600 | 1200 | 200 | 2000 |
| X | 440 | - | 0 | 660 | 1100 |
| C | 720 | 48 | - | 32 | 800 |
| ROW | 840 | 252 | 0 | 1008 | 2100 |
| Total imports | 2000 | 900 | 1200 | 1900 |
C to X = 6% of 800 = 48 IC.
Hence, the answer is 48.
How much is exported from P to ROW, in IC?
Answer: 200
C: P is the only exporter to C, so C's imports = 1200. Normalized balance −20%: \(\frac{E_C - 1200}{E_C + 1200} = -0.2 \Rightarrow E_C = 800\). C exports 90% = 720 to P, 4% = 32 to ROW, and the remaining 6% = 48 to X.
Let ROW's total exports = \(R\) (12% to X, 40% to P, nothing to C, so 48% to ROW itself), X's exports = \(E_X\) (40% to P, rest 60% to ROW), P's imports = \(I_P\).
P's imports: \(I_P = 0.22I_P + 720 + 0.4R \Rightarrow 0.78I_P = 720 + 0.4R\).
X to P: \(0.4E_X = 0.22I_P \Rightarrow E_X = 0.55I_P\).
X's imports = 600 + 48 + 0.12R; balance 10% gives \(0.9E_X = 1.1I_X\), so \(E_X = \frac{11}{9}(648 + 0.12R)\).
Equating: \(\frac{0.55(720 + 0.4R)}{0.78} = \frac{11}{9}(648 + 0.12R) \Rightarrow 3564 + 1.98R = 5559.84 + 1.0296R \Rightarrow R = 2100\).
Then \(I_P = 1560/0.78 = 2000\), \(E_X = 1100\), \(I_X = 900\). P's balance is 0, so P's exports = 2000 = 600 + 1200 + (P to ROW), giving P to ROW = 200.
| Exporter \ Importer | P | X | C | ROW | Total exports |
|---|---|---|---|---|---|
| P | - | 600 | 1200 | 200 | 2000 |
| X | 440 | - | 0 | 660 | 1100 |
| C | 720 | 48 | - | 32 | 800 |
| ROW | 840 | 252 | 0 | 1008 | 2100 |
| Total imports | 2000 | 900 | 1200 | 1900 |
P to ROW = 2000 − 600 − 1200 = 200 IC.
Hence, the answer is 200.
How much is exported from ROW to ROW, in IC?
Answer: 1008
C: P is the only exporter to C, so C's imports = 1200. Normalized balance −20%: \(\frac{E_C - 1200}{E_C + 1200} = -0.2 \Rightarrow E_C = 800\). C exports 90% = 720 to P, 4% = 32 to ROW, and the remaining 6% = 48 to X.
Let ROW's total exports = \(R\) (12% to X, 40% to P, nothing to C, so 48% to ROW itself), X's exports = \(E_X\) (40% to P, rest 60% to ROW), P's imports = \(I_P\).
P's imports: \(I_P = 0.22I_P + 720 + 0.4R \Rightarrow 0.78I_P = 720 + 0.4R\).
X to P: \(0.4E_X = 0.22I_P \Rightarrow E_X = 0.55I_P\).
X's imports = 600 + 48 + 0.12R; balance 10% gives \(0.9E_X = 1.1I_X\), so \(E_X = \frac{11}{9}(648 + 0.12R)\).
Equating: \(\frac{0.55(720 + 0.4R)}{0.78} = \frac{11}{9}(648 + 0.12R) \Rightarrow 3564 + 1.98R = 5559.84 + 1.0296R \Rightarrow R = 2100\).
Then \(I_P = 1560/0.78 = 2000\), \(E_X = 1100\), \(I_X = 900\). P's balance is 0, so P's exports = 2000 = 600 + 1200 + (P to ROW), giving P to ROW = 200.
| Exporter \ Importer | P | X | C | ROW | Total exports |
|---|---|---|---|---|---|
| P | - | 600 | 1200 | 200 | 2000 |
| X | 440 | - | 0 | 660 | 1100 |
| C | 720 | 48 | - | 32 | 800 |
| ROW | 840 | 252 | 0 | 1008 | 2100 |
| Total imports | 2000 | 900 | 1200 | 1900 |
ROW to ROW = 48% of 2100 = 1008 IC.
Hence, the answer is 1008.
What is the trade balance of ROW?
Answer: A
C: P is the only exporter to C, so C's imports = 1200. Normalized balance −20%: \(\frac{E_C - 1200}{E_C + 1200} = -0.2 \Rightarrow E_C = 800\). C exports 90% = 720 to P, 4% = 32 to ROW, and the remaining 6% = 48 to X.
Let ROW's total exports = \(R\) (12% to X, 40% to P, nothing to C, so 48% to ROW itself), X's exports = \(E_X\) (40% to P, rest 60% to ROW), P's imports = \(I_P\).
P's imports: \(I_P = 0.22I_P + 720 + 0.4R \Rightarrow 0.78I_P = 720 + 0.4R\).
X to P: \(0.4E_X = 0.22I_P \Rightarrow E_X = 0.55I_P\).
X's imports = 600 + 48 + 0.12R; balance 10% gives \(0.9E_X = 1.1I_X\), so \(E_X = \frac{11}{9}(648 + 0.12R)\).
Equating: \(\frac{0.55(720 + 0.4R)}{0.78} = \frac{11}{9}(648 + 0.12R) \Rightarrow 3564 + 1.98R = 5559.84 + 1.0296R \Rightarrow R = 2100\).
Then \(I_P = 1560/0.78 = 2000\), \(E_X = 1100\), \(I_X = 900\). P's balance is 0, so P's exports = 2000 = 600 + 1200 + (P to ROW), giving P to ROW = 200.
| Exporter \ Importer | P | X | C | ROW | Total exports |
|---|---|---|---|---|---|
| P | - | 600 | 1200 | 200 | 2000 |
| X | 440 | - | 0 | 660 | 1100 |
| C | 720 | 48 | - | 32 | 800 |
| ROW | 840 | 252 | 0 | 1008 | 2100 |
| Total imports | 2000 | 900 | 1200 | 1900 |
ROW's exports = 2100, imports = 200 (from P) + 660 (from X) + 32 (from C) + 1008 (from ROW) = 1900. Trade balance = 2100 − 1900 = 200 (the same if the 1008 internal trade is left out of both sides: 1092 − 892).
Hence, the answer is (A) 200.
Which among the countries P, X, and C has/have the least total trade?
Answer: A
C: P is the only exporter to C, so C's imports = 1200. Normalized balance −20%: \(\frac{E_C - 1200}{E_C + 1200} = -0.2 \Rightarrow E_C = 800\). C exports 90% = 720 to P, 4% = 32 to ROW, and the remaining 6% = 48 to X.
Let ROW's total exports = \(R\) (12% to X, 40% to P, nothing to C, so 48% to ROW itself), X's exports = \(E_X\) (40% to P, rest 60% to ROW), P's imports = \(I_P\).
P's imports: \(I_P = 0.22I_P + 720 + 0.4R \Rightarrow 0.78I_P = 720 + 0.4R\).
X to P: \(0.4E_X = 0.22I_P \Rightarrow E_X = 0.55I_P\).
X's imports = 600 + 48 + 0.12R; balance 10% gives \(0.9E_X = 1.1I_X\), so \(E_X = \frac{11}{9}(648 + 0.12R)\).
Equating: \(\frac{0.55(720 + 0.4R)}{0.78} = \frac{11}{9}(648 + 0.12R) \Rightarrow 3564 + 1.98R = 5559.84 + 1.0296R \Rightarrow R = 2100\).
Then \(I_P = 1560/0.78 = 2000\), \(E_X = 1100\), \(I_X = 900\). P's balance is 0, so P's exports = 2000 = 600 + 1200 + (P to ROW), giving P to ROW = 200.
| Exporter \ Importer | P | X | C | ROW | Total exports |
|---|---|---|---|---|---|
| P | - | 600 | 1200 | 200 | 2000 |
| X | 440 | - | 0 | 660 | 1100 |
| C | 720 | 48 | - | 32 | 800 |
| ROW | 840 | 252 | 0 | 1008 | 2100 |
| Total imports | 2000 | 900 | 1200 | 1900 |
Total trade: P = 2000 + 2000 = 4000, X = 1100 + 900 = 2000, C = 800 + 1200 = 2000. X and C tie for the least.
Hence, the answer is (A) Both X and C.
