CAT 2025 Slot 3

Let us place Bina at position 1 and arrange the numbers in a clockwise direction for better understanding, and then position the rest of the people on the table.

Round 1:

The holder of the buck is Bina. The buck is passed immediately to the left, which means it is given to position 2 from 1, and it is received by Aarav. So, Aarav's position is 2.

Round 2:

The holder of the buck is Aarav. The buck is passed second to the right, which means it is given to the position 7 from 2, and the person receiving it is unknown. 

Round 3:

The holder of the buck is unknown, but it is at position 7. The buck is passed immediately to the right, which means it is given to the position 6 from 7, and the person receiving it is Diya. Therefore, Diya's position is 6.

Round 4 and 5:

The buck is at position 6 at the start of round 4 with Diya, and the pass types of rounds 4 and 5 are not provided to us. The only information known to us is that after round 5, the buck is with Aarav, who is at position 2.

The possible ways for Aarav to receive the buck after rounds 4 and 5 are:

1) Passing the buck second to the right in round 4 and second to the right in round 5.
2) Passing the buck second to the left in round 4 and to the left in round 5.
3) Passing the buck to the left in round 4 and second to the left in round 5.

These are the only possible ways for the buck to reach Aarav after round 6.

Round 6:

The holder of the buck is Aarav. The buck is passed second to the left, which means it is given to the position 4 from 2, and the person receiving it is unknown.

Round 7:

The holder of the buck is unknown, but it is at position 4. The buck is passed immediately to the left, which means it is given to position 5 from 4, and the person receiving it is Gaurav. Therefore, Gaurav's position is 5.

Round 8: 

The holder of the buck is Gaurav. The buck is passed immediately to the left, which means it is given to position 6 from 5, and the person receiving it is unknown. But we know that the person in position 6 is Diya. So, the buck is currently at position 6 with Diya.

Round 9 and 10:

The pass types of rounds 9 and 10 are unknown to us, but we are told that the buck is passed to Farhan in round 9 and to Chirag in round 10. The positions of Farhan and Chirag are currently not assigned, and they are out of 3, 4, and 7. From position 6, the only positions that Buck can reach out of the three are 4 or 7.
If the buck is passed to position 7 in round 9, then it has to reach either 3 or 4 in round 10 to get to Chirag, but it is not possible to pass the buck from 7 to 3 or 4 in a single round. So, we can eliminate the case of Buck going to 7 in round 9.
The only case left is for the buck to go to position 4 in round 9 and to position 3 in round 10, as from 4, the buck can only be passed to 3 out of 3 and 7 in one round. Hence, we can conclude that Farhan is placed at position 4 and Chirag is placed at position 3. This leaves out only position 7 and the only person left to be assigned is Eshan making the person at position 7 to be Eshan.

This is the final arrangement. 

The person immediately to the right of Bina is Eshan.

Hence, the correct answer is option B.

Let us place Bina at position 1 and arrange the numbers in a clockwise direction for better understanding, and then position the rest of the people on the table.

Round 1:

The holder of the buck is Bina. The buck is passed immediately to the left, which means it is given to position 2 from 1, and it is received by Aarav. So, Aarav's position is 2.

Round 2:

The holder of the buck is Aarav. The buck is passed second to the right, which means it is given to the position 7 from 2, and the person receiving it is unknown.

Round 3:

The holder of the buck is unknown, but it is at position 7. The buck is passed immediately to the right, which means it is given to the position 6 from 7, and the person receiving it is Diya. Therefore, Diya's position is 6.

Round 4 and 5:

The buck is at position 6 at the start of round 4 with Diya, and the pass types of rounds 4 and 5 are not provided to us. The only information known to us is that after round 5, the buck is with Aarav, who is at position 2.

The possible ways for Aarav to receive the buck after rounds 4 and 5 are:

1) Passing the buck second to the right in round 4 and second to the right in round 5.
2) Passing the buck second to the left in round 4 and to the left in round 5.
3) Passing the buck to the left in round 4 and second to the left in round 5.

These are the only possible ways for the buck to reach Aarav after round 6.

Round 6:

The holder of the buck is Aarav. The buck is passed second to the left, which means it is given to the position 4 from 2, and the person receiving it is unknown.

Round 7:

The holder of the buck is unknown, but it is at position 4. The buck is passed immediately to the left, which means it is given to position 5 from 4, and the person receiving it is Gaurav. Therefore, Gaurav's position is 5.

Round 8:

The holder of the buck is Gaurav. The buck is passed immediately to the left, which means it is given to position 6 from 5, and the person receiving it is unknown. But we know that the person in position 6 is Diya. So, the buck is currently at position 6 with Diya.

Round 9 and 10:

The pass types of rounds 9 and 10 are unknown to us, but we are told that the buck is passed to Farhan in round 9 and to Chirag in round 10. The positions of Farhan and Chirag are currently not assigned, and they are out of 3, 4, and 7. From position 6, the only positions that Buck can reach out of the three are 4 or 7.
If the buck is passed to position 7 in round 9, then it has to reach either 3 or 4 in round 10 to get to Chirag, but it is not possible to pass the buck from 7 to 3 or 4 in a single round. So, we can eliminate the case of Buck going to 7 in round 9.
The only case left is for the buck to go to position 4 in round 9 and to position 3 in round 10, as from 4, the buck can only be passed to 3 out of 3 and 7 in one round. Hence, we can conclude that Farhan is placed at position 4 and Chirag is placed at position 3. This leaves out only position 7, and the only person left to be assigned is Eshan, making the person at position 7 Eshan.

This is the final arrangement.

The person third to the left of Eshan is Chirag.

Hence, the correct answer is option C.

Let us place Bina at position 1 and arrange the numbers in a clockwise direction for better understanding, and then position the rest of the people on the table.

Round 1:

The holder of the buck is Bina. The buck is passed immediately to the left, which means it is given to position 2 from 1, and it is received by Aarav. So, Aarav's position is 2.

Round 2:

The holder of the buck is Aarav. The buck is passed second to the right, which means it is given to the position 7 from 2, and the person receiving it is unknown.

Round 3:

The holder of the buck is unknown, but it is at position 7. The buck is passed immediately to the right, which means it is given to the position 6 from 7, and the person receiving it is Diya. Therefore, Diya's position is 6.

Round 4 and 5:

The buck is at position 6 at the start of round 4 with Diya, and the pass types of rounds 4 and 5 are not provided to us. The only information known to us is that after round 5, the buck is with Aarav, who is at position 2.

The possible ways for Aarav to receive the buck after rounds 4 and 5 are:

1) Passing the buck second to the right in round 4 and second to the right in round 5.
2) Passing the buck second to the left in round 4 and to the left in round 5.
3) Passing the buck to the left in round 4 and second to the left in round 5.

These are the only possible ways for the buck to reach Aarav after round 6.

Round 6:

The holder of the buck is Aarav. The buck is passed second to the left, which means it is given to the position 4 from 2, and the person receiving it is unknown.

Round 7:

The holder of the buck is unknown, but it is at position 4. The buck is passed immediately to the left, which means it is given to position 5 from 4, and the person receiving it is Gaurav. Therefore, Gaurav's position is 5.

Round 8:

The holder of the buck is Gaurav. The buck is passed immediately to the left, which means it is given to position 6 from 5, and the person receiving it is unknown. But we know that the person in position 6 is Diya. So, the buck is currently at position 6 with Diya.

Round 9 and 10:

The pass types of rounds 9 and 10 are unknown to us, but we are told that the buck is passed to Farhan in round 9 and to Chirag in round 10. The positions of Farhan and Chirag are currently not assigned, and they are out of 3, 4, and 7. From position 6, the only positions that Buck can reach out of the three are 4 or 7.
If the buck is passed to position 7 in round 9, then it has to reach either 3 or 4 in round 10 to get to Chirag, but it is not possible to pass the buck from 7 to 3 or 4 in a single round. So, we can eliminate the case of Buck going to 7 in round 9.
The only case left is for the buck to go to position 4 in round 9 and to position 3 in round 10, as from 4, the buck can only be passed to 3 out of 3 and 7 in one round. Hence, we can conclude that Farhan is placed at position 4 and Chirag is placed at position 3. This leaves out only position 7, and the only person left to be assigned is Eshan, making the person at position 7 Eshan.

In rounds 4 and 5, the possibilities were

1) Passing the buck second to the right in round 4 and second to the right in round 5.
2) Passing the buck second to the left in round 4 and to the left in round 5.
3) Passing the buck to the left in round 4 and second to the left in round 5.

Here, Buck could have been passed second to the right, second to the left or immediately to the left in rounds 4 and 5, which is unknown to us. The only pass type that we are certain to have happened is immediately to the right.

Hence, the correct answer is option C.

Let us place Bina at position 1 and arrange the numbers in a clockwise direction for better understanding, and then position the rest of the people on the table.

Round 1:

The holder of the buck is Bina. The buck is passed immediately to the left, which means it is given to position 2 from 1, and it is received by Aarav. So, Aarav's position is 2.

Round 2:

The holder of the buck is Aarav. The buck is passed second to the right, which means it is given to the position 7 from 2, and the person receiving it is unknown.

Round 3:

The holder of the buck is unknown, but it is at position 7. The buck is passed immediately to the right, which means it is given to the position 6 from 7, and the person receiving it is Diya. Therefore, Diya's position is 6.

Round 4 and 5:

The buck is at position 6 at the start of round 4 with Diya, and the pass types of rounds 4 and 5 are not provided to us. The only information known to us is that after round 5, the buck is with Aarav, who is at position 2.

The possible ways for Aarav to receive the buck after rounds 4 and 5 are:

1) Passing the buck second to the right in round 4 and second to the right in round 5.
2) Passing the buck second to the left in round 4 and to the left in round 5.
3) Passing the buck to the left in round 4 and second to the left in round 5.

These are the only possible ways for the buck to reach Aarav after round 6.

Round 6:

The holder of the buck is Aarav. The buck is passed second to the left, which means it is given to the position 4 from 2, and the person receiving it is unknown.

Round 7:

The holder of the buck is unknown, but it is at position 4. The buck is passed immediately to the left, which means it is given to position 5 from 4, and the person receiving it is Gaurav. Therefore, Gaurav's position is 5.

Round 8:

The holder of the buck is Gaurav. The buck is passed immediately to the left, which means it is given to position 6 from 5, and the person receiving it is unknown. But we know that the person in position 6 is Diya. So, the buck is currently at position 6 with Diya.

Round 9 and 10:

The pass types of rounds 9 and 10 are unknown to us, but we are told that the buck is passed to Farhan in round 9 and to Chirag in round 10. The positions of Farhan and Chirag are currently not assigned, and they are out of 3, 4, and 7. From position 6, the only positions that Buck can reach out of the three are 4 or 7.
If the buck is passed to position 7 in round 9, then it has to reach either 3 or 4 in round 10 to get to Chirag, but it is not possible to pass the buck from 7 to 3 or 4 in a single round. So, we can eliminate the case of Buck going to 7 in round 9.
The only case left is for the buck to go to position 4 in round 9 and to position 3 in round 10, as from 4, the buck can only be passed to 3 out of 3 and 7 in one round. Hence, we can conclude that Farhan is placed at position 4 and Chirag is placed at position 3. This leaves out only position 7, and the only person left to be assigned is Eshan, making the person at position 7 Eshan.

In rounds 4 and 5, the possibilities were

1) Passing the buck second to the right in round 4 and second to the right in round 5. Here, the buck would have gone to Farhan in round 4.
2) Passing the buck second to the left in round 4 and to the left in round 5. Here, the buck would have gone to Bina in round 4.
3) Passing the buck to the left in round 4 and second to the left in round 5. Here, the buck would have gone to Eshan in round 4.

In round 4, the buck might have gone to either Farhan, Bina or Eshan, which is not known to us. Therefore, the count of the number of times the buck was received by them cannot be determined uniquely, whereas it can be uniquely determined in the case of Gaurav, which is 1 in round 7.

Hence, the correct answer is option D.

Let us assume units of Aurels, Brins, Crowns, and Zentars to be A, B, C and D, respectively. 

We are given the travel cost of Jano to be 3500A in Statement 1.

We are also given the travel cost of Kira to be 8000A in Statement 1 and 4000B in Statement 2.

We are also given the travel cost of Lian to be 36000C in statement 3.

We know that the travel cost has to be constant throughout, so by equating the travel costs of Kira, we get,

8000A = 4000B

B = 2A ---(1)

We are also given the value of C to be 0.5 Z.

C = 0.5Z  ---(2)

So, the travel cost of Lian = 36000C = 36000*0.5Z = 18000Z

Let us put the known information in the table, and we get,

We are given that the spending cost of any individual is different in different cities.

We are given that the spending amounts are amongst {1000, 2000, 3000} in their local currencies and also told that there were two spends of 1000 and one spend of 3000, which makes the total spending amount 2000, to be 3 because there are 6 spending amounts in total.

One 2000 and one 3000 are already assigned in the above table, so we will be left with two spends of 1000 and two spends of 2000 to be assigned.

Kira has already spent 2000, so the only amount that she must have spent is 1000, as she cannot spend 2000 twice in different countries.

Lian has already spent 3000, so the only amount that she must have spent is 2000. If she spent 1000, then Jano will be left with two 2000s, which is not possible.

We can also conclude that Jano spend 1000 and 2000 in some order in the two countries.

The flight costs are given as 4000Z, 5000Z, and 6000Z, out of which 4000Z is for Jano and 5000Z and 6000Z are for Kira and Lian, in some order.

Filling up the table with these values, we get,

Travel Cost = Spending Cost + Flight Cost

For Kira,

Travel Cost = 8000A

Spending Cost = 1000A + 4000A = 5000A

Flight Cost = 5000Z/6000Z

CASE 1: Flight cost of Kira = 5000Z and Flight cost of Lian = 6000Z

If we assume the Flight cost to be 5000Z, then we get,

8000A = 5000A + 5000Z

3000A = 5000Z

A = $$\frac{5}{3}$$Z

For Lian,

Travel Cost = 18000Z

Spending Cost = 6000A + 1000Z = 6000$$\times\ \frac{5}{3}$$Z + 1000Z = 11000Z

Flight Cost = Travel Cost - Spending Cost = 18000Z - 11000Z = 7000Z

But in our assumption, the Flight Cost of Lian is 6000Z, which does not match the above answer.

So, we can eliminate this case.

CASE 2: Flight cost of Kira = 6000Z and Flight cost of Lian = 5000Z

If we assume the Flight cost to be 6000Z, then we get,

8000A = 5000A + 6000Z

3000A = 6000Z

A = 2Z

For Lian,

Travel Cost = 18000Z

Spending Cost = 6000A + 1000Z = 6000*2Z + 1000Z = 13000Z

Flight Cost = Travel Cost - Spending Cost = 18000Z - 13000Z = 5000Z

In our assumption, the Flight Cost of Lian is 5000Z, which matches the above answer.

So, Flight cost of Kira = 6000Z, Flight cost of Lian = 5000Z and A = 2Z.

Filling the table with calculated values all in the currency of Z, we get,

In the case of Zano,

Spending Cost = Travel Cost - Flight Cost = 7000Z - 4000Z = 3000Z

In the first case, the Spending Cost = 2000Z + 1000Z = 3000Z

In the second case, the Spending Cost = 4000Z + 500Z = 4500Z

We obtained a spending cost of 3000Z only in the first case, so we can eliminate the second case.

The final table looks like,

The total Travel Cost = 7000Z + 16000Z + 18000Z = 41000Z

Hence, the correct answer is 41000.

Let us assume units of Aurels, Brins, Crowns, and Zentars to be A, B, C and D, respectively.

We are given the travel cost of Jano to be 3500A in Statement 1.

We are also given the travel cost of Kira to be 8000A in Statement 1 and 4000B in Statement 2.

We are also given the travel cost of Lian to be 36000C in statement 3.

We know that the travel cost has to be constant throughout, so by equating the travel costs of Kira, we get,

8000A = 4000B

B = 2A ---(1)

We are also given the value of C to be 0.5 Z.

C = 0.5Z ---(2)

So, the travel cost of Lian = 36000C = 36000*0.5Z = 18000Z

Let us put the known information in the table, and we get,

We are given that the spending cost of any individual is different in different cities.

We are given that the spending amounts are amongst {1000, 2000, 3000} in their local currencies and also told that there were two spends of 1000 and one spend of 3000, which makes the total spending amount 2000, to be 3 because there are 6 spending amounts in total.

One 2000 and one 3000 are already assigned in the above table, so we will be left with two spends of 1000 and two spends of 2000 to be assigned.

Kira has already spent 2000, so the only amount that she must have spent is 1000, as she cannot spend 2000 twice in different countries.

Lian has already spent 3000, so the only amount that she must have spent is 2000. If she spent 1000, then Jano will be left with two 2000s, which is not possible.

We can also conclude that Jano spent 1000 and 2000 in some order in the two countries.

The flight costs are given as 4000Z, 5000Z, and 6000Z, out of which 4000Z is for Jano and 5000Z and 6000Z are for Kira and Lian, in some order.

Filling up the table with these values, we get,

Travel Cost = Spending Cost + Flight Cost

For Kira,

Travel Cost = 8000A

Spending Cost = 1000A + 4000A = 5000A

Flight Cost = 5000Z/6000Z

CASE 1: Flight cost of Kira = 5000Z and Flight cost of Lian = 6000Z

If we assume the Flight cost to be 5000Z, then we get,

8000A = 5000A + 5000Z

3000A = 5000Z

A = $$\frac{5}{3}$$Z

For Lian,

Travel Cost = 18000Z

Spending Cost = 6000A + 1000Z = 6000$$\times\ \frac{5}{3}$$Z + 1000Z = 11000Z

Flight Cost = Travel Cost - Spending Cost = 18000Z - 11000Z = 7000Z

But in our assumption, the Flight Cost of Lian is 6000Z, which does not match the above answer.

So, we can eliminate this case.

CASE 2: Flight cost of Kira = 6000Z and Flight cost of Lian = 5000Z

If we assume the Flight cost to be 6000Z, then we get,

8000A = 5000A + 6000Z

3000A = 6000Z

A = 2Z

For Lian,

Travel Cost = 18000Z

Spending Cost = 6000A + 1000Z = 6000*2Z + 1000Z = 13000Z

Flight Cost = Travel Cost - Spending Cost = 18000Z - 13000Z = 5000Z

In our assumption, the Flight Cost of Lian is 5000Z, which matches the above answer.

So, Flight cost of Kira = 6000Z, Flight cost of Lian = 5000Z and A = 2Z.

Filling the table with calculated values all in the currency of Z, we get,

In the case of Zano,

Spending Cost = Travel Cost - Flight Cost = 7000Z - 4000Z = 3000Z

In the first case, the Spending Cost = 2000Z + 1000Z = 3000Z

In the second case, the Spending Cost = 4000Z + 500Z = 4500Z

We obtained a spending cost of 3000Z only in the first case, so we can eliminate the second case.

The final table looks like,

Spending Cost of Lian = 12000Z + 1000Z = 13000Z

Hence, the correct answer is 13000.

Let us assume units of Aurels, Brins, Crowns, and Zentars to be A, B, C and D, respectively.

We are given the travel cost of Jano to be 3500A in Statement 1.

We are also given the travel cost of Kira to be 8000A in Statement 1 and 4000B in Statement 2.

We are also given the travel cost of Lian to be 36000C in statement 3.

We know that the travel cost has to be constant throughout, so by equating the travel costs of Kira, we get,

8000A = 4000B

B = 2A ---(1)

We are also given the value of C to be 0.5 Z.

C = 0.5Z ---(2)

So, the travel cost of Lian = 36000C = 36000*0.5Z = 18000Z

Let us put the known information in the table, and we get,

We are given that the spending cost of any individual is different in different cities.

We are given that the spending amounts are amongst {1000, 2000, 3000} in their local currencies and also told that there were two spends of 1000 and one spend of 3000, which makes the total spending amount 2000, to be 3 because there are 6 spending amounts in total.

One 2000 and one 3000 are already assigned in the above table, so we will be left with two spends of 1000 and two spends of 2000 to be assigned.

Kira has already spent 2000, so the only amount that she must have spent is 1000, as she cannot spend 2000 twice in different countries.

Lian has already spent 3000, so the only amount that she must have spent is 2000. If she spent 1000, then Jano will be left with two 2000s, which is not possible.

We can also conclude that Jano spend 1000 and 2000 in some order in the two countries.

The flight costs are given as 4000Z, 5000Z, and 6000Z, out of which 4000Z is for Jano and 5000Z and 6000Z are for Kira and Lian, in some order.

Filling up the table with these values, we get,

Travel Cost = Spending Cost + Flight Cost

For Kira,

Travel Cost = 8000A

Spending Cost = 1000A + 4000A = 5000A

Flight Cost = 5000Z/6000Z

CASE 1: Flight cost of Kira = 5000Z and Flight cost of Lian = 6000Z

If we assume the Flight cost to be 5000Z, then we get,

8000A = 5000A + 5000Z

3000A = 5000Z

A = $$\frac{5}{3}$$Z

For Lian,

Travel Cost = 18000Z

Spending Cost = 6000A + 1000Z = 6000$$\times\ \frac{5}{3}$$Z + 1000Z = 11000Z

Flight Cost = Travel Cost - Spending Cost = 18000Z - 11000Z = 7000Z

But in our assumption, the Flight Cost of Lian is 6000Z, which does not match the above answer.

So, we can eliminate this case.

CASE 2: Flight cost of Kira = 6000Z and Flight cost of Lian = 5000Z

If we assume the Flight cost to be 6000Z, then we get,

8000A = 5000A + 6000Z

3000A = 6000Z

A = 2Z

For Lian,

Travel Cost = 18000Z

Spending Cost = 6000A + 1000Z = 6000*2Z + 1000Z = 13000Z

Flight Cost = Travel Cost - Spending Cost = 18000Z - 13000Z = 5000Z

In our assumption, the Flight Cost of Lian is 5000Z, which matches the above answer.

So, Flight cost of Kira = 6000Z, Flight cost of Lian = 5000Z and A = 2Z.

Filling the table with calculated values all in the currency of Z, we get,

In the case of Zano,

Spending Cost = Travel Cost - Flight Cost = 7000Z - 4000Z = 3000Z

In the first case, the Spending Cost = 2000Z + 1000Z = 3000Z

In the second case, the Spending Cost = 4000Z + 500Z = 4500Z

We obtained a spending cost of 3000Z only in the first case, so we can eliminate the second case.

The final table looks like,

Total Spending Cost of Jano = 2000Z + 1000Z = 3000Z

We know that A = 2Z, so the above value in A can be calculated as,

3000Z = 3000*A/2 = 1500A

So, the spending of Jano in Aurels is 1500A.

Hence, the correct answer is 1500.

Let us assume units of Aurels, Brins, Crowns, and Zentars to be A, B, C and D, respectively.

We are given the travel cost of Jano to be 3500A in Statement 1.

We are also given the travel cost of Kira to be 8000A in Statement 1 and 4000B in Statement 2.

We are also given the travel cost of Lian to be 36000C in statement 3.

We know that the travel cost has to be constant throughout, so by equating the travel costs of Kira, we get,

8000A = 4000B

B = 2A ---(1)

We are also given the value of C to be 0.5 Z.

C = 0.5Z ---(2)

So, the travel cost of Lian = 36000C = 36000*0.5Z = 18000Z

Let us put the known information in the table, and we get,

We are given that the spending cost of any individual is different in different cities.

We are given that the spending amounts are amongst {1000, 2000, 3000} in their local currencies and also told that there were two spends of 1000 and one spend of 3000, which makes the total spending amount 2000, to be 3 because there are 6 spending amounts in total.

One 2000 and one 3000 are already assigned in the above table, so we will be left with two spends of 1000 and two spends of 2000 to be assigned.

Kira has already spent 2000, so the only amount that she must have spent is 1000, as she cannot spend 2000 twice in different countries.

Lian has already spent 3000, so the only amount that she must have spent is 2000. If she spent 1000, then Jano will be left with two 2000s, which is not possible.

We can also conclude that Jano spent 1000 and 2000 in some order in the two countries.

The flight costs are given as 4000Z, 5000Z, and 6000Z, out of which 4000Z is for Jano and 5000Z and 6000Z are for Kira and Lian, in some order.

Filling up the table with these values, we get,

Travel Cost = Spending Cost + Flight Cost

For Kira,

Travel Cost = 8000A

Spending Cost = 1000A + 4000A = 5000A

Flight Cost = 5000Z/6000Z

CASE 1: Flight cost of Kira = 5000Z and Flight cost of Lian = 6000Z

If we assume the Flight cost to be 5000Z, then we get,

8000A = 5000A + 5000Z

3000A = 5000Z

A = $$\frac{5}{3}$$Z

For Lian,

Travel Cost = 18000Z

Spending Cost = 6000A + 1000Z = 6000$$\times\ \frac{5}{3}$$Z + 1000Z = 11000Z

Flight Cost = Travel Cost - Spending Cost = 18000Z - 11000Z = 7000Z

But in our assumption, the Flight Cost of Lian is 6000Z, which does not match the above answer.

So, we can eliminate this case.

CASE 2: Flight cost of Kira = 6000Z and Flight cost of Lian = 5000Z

If we assume the Flight cost to be 6000Z, then we get,

8000A = 5000A + 6000Z

3000A = 6000Z

A = 2Z

For Lian,

Travel Cost = 18000Z

Spending Cost = 6000A + 1000Z = 6000*2Z + 1000Z = 13000Z

Flight Cost = Travel Cost - Spending Cost = 18000Z - 13000Z = 5000Z

In our assumption, the Flight Cost of Lian is 5000Z, which matches the above answer.

So, Flight cost of Kira = 6000Z, Flight cost of Lian = 5000Z and A = 2Z.

Filling the table with calculated values all in the currency of Z, we get,

In the case of Zano,

Spending Cost = Travel Cost - Flight Cost = 7000Z - 4000Z = 3000Z

In the first case, the Spending Cost = 2000Z + 1000Z = 3000Z

In the second case, the Spending Cost = 4000Z + 500Z = 4500Z

We obtained a spending cost of 3000Z only in the first case, so we can eliminate the second case.

The final table looks like,

We know that B = 2A and A = 2Z, so the value of B in terms of Z is 

B = 2*2Z = 4Z  --(3)

We also know that 

C = 0.5Z  --(4)

Dividing (3) by (4), we get,

$$\dfrac{B}{C}\ =\ \dfrac{4Z}{0.5Z}\ =\ 8$$

B = 8C

So, one Brin is equivalent to 8 crowns.

Hence, the correct answer is option A.

Let us assume units of Aurels, Brins, Crowns, and Zentars to be A, B, C and D, respectively.

We are given the travel cost of Jano to be 3500A in Statement 1.

We are also given the travel cost of Kira to be 8000A in Statement 1 and 4000B in Statement 2.

We are also given the travel cost of Lian to be 36000C in statement 3.

We know that the travel cost has to be constant throughout, so by equating the travel costs of Kira, we get,

8000A = 4000B

B = 2A ---(1)

We are also given the value of C to be 0.5 Z.

C = 0.5Z ---(2)

So, the travel cost of Lian = 36000C = 36000*0.5Z = 18000Z

Let us put the known information in the table, and we get,

We are given that the spending cost of any individual is different in different cities.

We are given that the spending amounts are amongst {1000, 2000, 3000} in their local currencies and also told that there were two spends of 1000 and one spend of 3000, which makes the total spending amount 2000, to be 3 because there are 6 spending amounts in total.

One 2000 and one 3000 are already assigned in the above table, so we will be left with two spends of 1000 and two spends of 2000 to be assigned.

Kira has already spent 2000, so the only amount that she must have spent is 1000, as she cannot spend 2000 twice in different countries.

Lian has already spent 3000, so the only amount that she must have spent is 2000. If she spent 1000, then Jano will be left with two 2000s, which is not possible.

We can also conclude that Jano spend 1000 and 2000 in some order in the two countries.

The flight costs are given as 4000Z, 5000Z, and 6000Z, out of which 4000Z is for Jano and 5000Z and 6000Z are for Kira and Lian, in some order.

Filling up the table with these values, we get,

Travel Cost = Spending Cost + Flight Cost

For Kira,

Travel Cost = 8000A

Spending Cost = 1000A + 4000A = 5000A

Flight Cost = 5000Z/6000Z

CASE 1: Flight cost of Kira = 5000Z and Flight cost of Lian = 6000Z

If we assume the Flight cost to be 5000Z, then we get,

8000A = 5000A + 5000Z

3000A = 5000Z

A = $$\frac{5}{3}$$Z

For Lian,

Travel Cost = 18000Z

Spending Cost = 6000A + 1000Z = 6000$$\times\ \frac{5}{3}$$Z + 1000Z = 11000Z

Flight Cost = Travel Cost - Spending Cost = 18000Z - 11000Z = 7000Z

But in our assumption, the Flight Cost of Lian is 6000Z, which does not match the above answer.

So, we can eliminate this case.

CASE 2: Flight cost of Kira = 6000Z and Flight cost of Lian = 5000Z

If we assume the Flight cost to be 6000Z, then we get,

8000A = 5000A + 6000Z

3000A = 6000Z

A = 2Z

For Lian,

Travel Cost = 18000Z

Spending Cost = 6000A + 1000Z = 6000*2Z + 1000Z = 13000Z

Flight Cost = Travel Cost - Spending Cost = 18000Z - 13000Z = 5000Z

In our assumption, the Flight Cost of Lian is 5000Z, which matches the above answer.

So, Flight cost of Kira = 6000Z, Flight cost of Lian = 5000Z and A = 2Z.

Filling the table with calculated values all in the currency of Z, we get,

In the case of Zano,

Spending Cost = Travel Cost - Flight Cost = 7000Z - 4000Z = 3000Z

In the first case, the Spending Cost = 2000Z + 1000Z = 3000Z

In the second case, the Spending Cost = 4000Z + 500Z = 4500Z

We obtained a spending cost of 3000Z only in the first case, so we can eliminate the second case.

The final table looks like,

Statement 1) Jano spent 2000 in Aurevia

Jano spent 2000Z in Aurevia = 1000A in Aurevia.

So, this option is incorrect.

Statement 2) Lian spent 2000 in Cyrenia

Lian spent 1000Z in Cyrenia = 2000C in Cyrenia.

So, this option is correct.

Statement 3) Jano spent 2000 in Cyrenia

Jano spent 1000Z in Cyrenia = 2000C in Cyrenia.

So, this option is correct.

Statement 4) Kira spent 1000 in Aurevia

Kira spent 2000Z in Aurevia = 1000A in Aurevia.

So, this option is correct.

Hence, the correct answer is option A.

The left plot shows, for each competitor, a step (staircase) curve whose height at any time t equals the number of puzzles that competitor has completed by minute t.

To find who solved the most puzzles by the 20th minute, draw a vertical line at t=20 (or simply read the height of each competitor’s colored stair at the vertical line x=20).

Reading the four curves at t=20:

The green staircase (Koushik) is at the highest horizontal level at t=20.

The yellow (Chandranath) and red (Anirbid) staircases are lower than the green ones at that time.

The blue (Suranjan) staircase is the lowest of the four at t=20.

Since Koushik’s value is greater than every other competitor’s at t=20, Koushik has solved the largest number of puzzles by the 20th minute.

Hence, the correct answer is option B.

Comparing the 1st and 2nd graphs, we should be able to determine the type of puzzle.

From the second graph, we can say that there are 4 visual-based puzzles in total, which makes the number-based puzzles 6.

Let us follow the green line in both graphs:

The 1st visual-based puzzle was completed by Green in 2 minutes from the 2nd graph. From the 1st graph at the 2nd minute, the puzzle that was solved by Green was 1. So, we can say that 1 is a visual-based puzzle.

The 2nd visual-based puzzle was completed by Green in 12 minutes from the 2nd graph. From the 1st graph at the 12th minute, the puzzle that was solved by Green was 4. So, we can say that 4 is a visual-based puzzle.

The 3rd visual-based puzzle was completed by Green in 25 minutes from the 2nd graph. From the 1st graph at the 25th minute, the puzzle that was solved by Green was 8. So, we can say that 8 is a visual-based puzzle.

The 4th visual-based puzzle was completed by Green in 29 minutes from the 2nd graph. From the 1st graph at the 29th minute, the puzzle that was solved by Green was 9. So, we can say that 9 is a visual-based puzzle.

We can conclude that puzzles 1, 4, 8, and 9 are visual-based, while puzzles 2, 3, 5, 6, 7, and 10 are number-based.

So, the third visual puzzle is 8, and the time taken by Suranjan(blue) to complete the 3rd visual-based puzzle, which is the 8th puzzle, can be calculated as,

The time taken by Suranjan to solve the 8th puzzle from graph 1 = 28 - 26 = 2 minutes.

Hence, the correct answer is 2.

Comparing the 1st and 2nd graphs, we should be able to determine the type of puzzle.

From the second graph, we can say that there are 4 visual-based puzzles in total, which makes the number-based puzzles 6.

Let us follow the green line in both graphs:

The 1st visual-based puzzle was completed by Green in 2 minutes from the 2nd graph. From the 1st graph at the 2nd minute, the puzzle that was solved by Green was 1. So, we can say that 1 is a visual-based puzzle.

The 2nd visual-based puzzle was completed by Green in 12 minutes from the 2nd graph. From the 1st graph at the 12th minute, the puzzle that was solved by Green was 4. So, we can say that 4 is a visual-based puzzle.

The 3rd visual-based puzzle was completed by Green in 25 minutes from the 2nd graph. From the 1st graph at the 25th minute, the puzzle that was solved by Green was 8. So, we can say that 8 is a visual-based puzzle.

The 4th visual-based puzzle was completed by Green in 29 minutes from the 2nd graph. From the 1st graph at the 29th minute, the puzzle that was solved by Green was 9. So, we can say that 9 is a visual-based puzzle.

We can conclude that puzzles 1, 4, 8, and 9 are visual-based, while puzzles 2, 3, 5, 6, 7, and 10 are number-based.

The fourth number-based puzzle is 6th puzzle in the order.

Hence, the correct answer is 6.

Comparing the 1st and 2nd graphs, we should be able to determine the type of puzzle.

From the second graph, we can say that there are 4 visual-based puzzles in total, which makes the number-based puzzles 6.

Let us follow the green line in both graphs:

The 1st visual-based puzzle was completed by Green in 2 minutes from the 2nd graph. From the 1st graph at the 2nd minute, the puzzle that was solved by Green was 1. So, we can say that 1 is a visual-based puzzle.

The 2nd visual-based puzzle was completed by Green in 12 minutes from the 2nd graph. From the 1st graph at the 12th minute, the puzzle that was solved by Green was 4. So, we can say that 4 is a visual-based puzzle.

The 3rd visual-based puzzle was completed by Green in 25 minutes from the 2nd graph. From the 1st graph at the 25th minute, the puzzle that was solved by Green was 8. So, we can say that 8 is a visual-based puzzle.

The 4th visual-based puzzle was completed by Green in 29 minutes from the 2nd graph. From the 1st graph at the 29th minute, the puzzle that was solved by Green was 9. So, we can say that 9 is a visual-based puzzle.

We can conclude that puzzles 1, 4, 8, and 9 are visual-based, while puzzles 2, 3, 5, 6, 7, and 10 are number-based.

The average time taken by Anirbid to solve the number-based puzzles can be calculated as,

Time for 1st number-based puzzle(2nd puzzle) from 1st graph = 9 - 4 = 5 minutes.

Time for 2nd number-based puzzle(3rd puzzle) from 1st graph = 11 - 9 = 2 minutes.

Time for 3rd number-based puzzle(5th puzzle) from 1st graph = 19 - 14 = 5 minutes.

Time for 4th number-based puzzle(6th puzzle) from 1st graph = 22 - 19 = 3 minutes.

Time for 5th number-based puzzle(7th puzzle) from 1st graph = 27 - 22 = 5 minutes.

Time for 6th number-based puzzle(10th puzzle) from 1st graph = 37 - 33 = 4 minutes.

Average $$=\ \dfrac{5\ +\ 2\ +\ 5\ +\ 3\ +\ 5\ +\ 4}{6}\ =\ \dfrac{24}{6}\ =\ 4$$ minutes

The average time taken by Anirbid to solve the number-based puzzles is 4 minutes.

Hence, the correct answer is option D.

In clue 5, we are given that export volumes of P, in IC, to X and C are 600 and 1200, respectively, and P is the only country that exports to C. So, the export from X to C and ROW to C is 0.

In clue 1, we are given that normalised trade balances of P, X and C are 0%, 10%, and -20%, respectively.

For P, since the normalised trade balance is 0% the exports have to be equal to the imports according to the given formula. So, let us assume them to be 10a and 10a.

For X,

$$\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ 10$$

$$10E\ -\ 10I\ =\ E\ +\ I$$

$$11I\ =\ 9E$$

So, let us assume export to be 11b and import to be 9b.

For C,

$$\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ -20$$

$$5I\ -\ 5E\ =\ E\ +\ I$$

$$2I\ =\ 3E$$

So, let us assume export to be 2c and import to be 3c.

In clue 2, we are given that 40% of exports of X are to P and 22% of imports of P are from X. 

40% of the exports of X $$=\dfrac{40}{100}\ \times\ 11b\ =\ 4.4b\ =$$ exports from X to P

22% of the imports of P $$=\dfrac{22}{100}\ \times\ 10a\ =\ 2.2a\ =$$ imports to P from X

Equating both, we get,

4.4b = 2.2a

a = 2b.

We are also given that imports for C are only from P, and we can set the rest of the imports for C to 0.

The total imports for C = 3c = 1200

c = 400

This makes the exports for C = 2c = 800

In clue 3, we are given that 90% of exports of C are to P and 4% are to ROW.

90% of the exports of C $$=\dfrac{90}{100}\ \times\ 2c\ =\ 1.8c\ =1.8\ \times\ 400\ =\ 720\ =\ $$exports from C to P

4% of the exports of C $$=\dfrac{4}{100}\ \times\ 2c\ =\ 0.08c\ =\ 0.08\ \times\ 400\ =\ 32\ =$$ exports from C to ROW

The exports from C to X = 800 - 720 - 32 = 48

In clue 4, we are given that 12% of exports of ROW are to X, 40% are to P.

If we assume that Imports of ROW are m and exports of ROW are n.

12% of the exports of ROW $$=\dfrac{12}{100}\ \times\ n\ =\ 0.12n\ =$$ exports from ROW to X

40% of the exports of ROW $$=\dfrac{40}{100}\ \times\ n\ =\ 0.4n\ =$$ exports from ROW to P

Placing all the values in the table, we get,

Equating the imports of P and X to the Totals, we get,

For P,

4.4b + 720 + 0.4n = 20b

15.6b = 720 + 0.4n  --(1)

For X,

600 + 48 + 0.12n = 9b

9b = 648 + 0.12n  --(2)

Solving (1) and (2), we get

$$\dfrac{648\ +\ 0.12n}{9}\ =\ \dfrac{720\ +\ 0.4n}{15.6}$$

$$10108.8+1.872n=6480+3.6n$$

$$3628.8=1.728n$$

$$n=2100$$

From (1),

15.6b = 720 + 0.4(2100)

15.6b = 1560

b = 100

Exports of P to ROW = 20b - 600 - 1200 = 2000 - 1800 = 200

Exports of X to ROW = 11b - 4.4b = 6.6b = 660

Substituting the values of b and n, we can figure out all the values except for Imports and Exports of ROW to ROW.

The value of exports and imports of ROW to ROW has to be equal.

We calculated the value of n to be 2100.

The exports of ROW to ROW can be calculated as,

Exports of ROW to ROW = 2100 - 840 - 252 = 1008 = Imports of ROW to ROW.

The value of m can be calculated as,

m = 200 + 660 + 32 + 1008 = 1900.

Filling up the table with all the values, we get,

Exports from C to X are 48 IC.

Hence, the correct answer is 48. 

In clue 5, we are given that export volumes of P, in IC, to X and C are 600 and 1200, respectively, and P is the only country that exports to C. So, the export from X to C and ROW to C is 0.

In clue 1, we are given that normalised trade balances of P, X and C are 0%, 10%, and -20%, respectively.

For P, since the normalised trade balance is 0% the exports have to be equal to the imports according to the given formula. So, let us assume them to be 10a and 10a.

For X,

$$\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ 10$$

$$10E\ -\ 10I\ =\ E\ +\ I$$

$$11I\ =\ 9E$$

So, let us assume export to be 11b and import to be 9b.

For C,

$$\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ -20$$

$$5I\ -\ 5E\ =\ E\ +\ I$$

$$2I\ =\ 3E$$

So, let us assume export to be 2c and import to be 3c.

In clue 2, we are given that 40% of exports of X are to P and 22% of imports of P are from X.

40% of the exports of X $$=\dfrac{40}{100}\ \times\ 11b\ =\ 4.4b\ =$$ exports from X to P

22% of the imports of P $$=\dfrac{22}{100}\ \times\ 10a\ =\ 2.2a\ =$$ imports to P from X

Equating both, we get,

4.4b = 2.2a

a = 2b.

We are also given that imports for C are only from P, and we can set the rest of the imports for C to 0.

The total imports for C = 3c = 1200

c = 400

This makes the exports for C = 2c = 800

In clue 3, we are given that 90% of exports of C are to P and 4% are to ROW.

90% of the exports of C $$=\dfrac{90}{100}\ \times\ 2c\ =\ 1.8c\ =1.8\ \times\ 400\ =\ 720\ =\ $$exports from C to P

4% of the exports of C $$=\frac{4}{100}\ \times\ 2c\ =\ 0.08c\ =\ 0.08\ \times\ 400\ =\ 32\ =$$ exports from C to ROW

The exports from C to X = 800 - 720 - 32 = 48

In clue 4, we are given that 12% of exports of ROW are to X, 40% are to P.

If we assume that Imports of ROW are m and exports of ROW are n.

12% of the exports of ROW $$=\dfrac{12}{100}\ \times\ n\ =\ 0.12n\ =$$ exports from ROW to X

40% of the exports of ROW $$=\dfrac{40}{100}\ \times\ n\ =\ 0.4n\ =$$ exports from ROW to P

Placing all the values in the table, we get,

Equating the imports of P and X to the Totals, we get,

For P,

4.4b + 720 + 0.4n = 20b

15.6b = 720 + 0.4n --(1)

For X,

600 + 48 + 0.12n = 9b

9b = 648 + 0.12n --(2)

Solving (1) and (2), we get

$$\dfrac{648\ +\ 0.12n}{9}\ =\ \dfrac{720\ +\ 0.4n}{15.6}$$

$$10108.8+1.872n=6480+3.6n$$

$$3628.8=1.728n$$

$$n=2100$$

From (1),

15.6b = 720 + 0.4(2100)

15.6b = 1560

b = 100

Exports of P to ROW = 20b - 600 - 1200 = 2000 - 1800 = 200

Exports of X to ROW = 11b - 4.4b = 6.6b = 660

Substituting the values of b and n, we can figure out all the values except for Imports and Exports of ROW to ROW.

The value of exports and imports of ROW to ROW has to be equal.

We calculated the value of n to be 2100.

The exports of ROW to ROW can be calculated as,

Exports of ROW to ROW = 2100 - 840 - 252 = 1008 = Imports of ROW to ROW.

The value of m can be calculated as,

m = 200 + 660 + 32 + 1008 = 1900.

Filling up the table with all the values, we get,

The exports from P to ROW 200 IC.

Hence, the correct answer is 200.

In clue 5, we are given that export volumes of P, in IC, to X and C are 600 and 1200, respectively, and P is the only country that exports to C. So, the export from X to C and ROW to C is 0.

In clue 1, we are given that normalised trade balances of P, X and C are 0%, 10%, and -20%, respectively.

For P, since the normalised trade balance is 0% the exports have to be equal to the imports according to the given formula. So, let us assume them to be 10a and 10a.

For X,

$$\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ 10$$

$$10E\ -\ 10I\ =\ E\ +\ I$$

$$11I\ =\ 9E$$

So, let us assume export to be 11b and import to be 9b.

For C,

$$\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ -20$$

$$5I\ -\ 5E\ =\ E\ +\ I$$

$$2I\ =\ 3E$$

So, let us assume export to be 2c and import to be 3c.

In clue 2, we are given that 40% of exports of X are to P and 22% of imports of P are from X.

40% of the exports of X $$=\dfrac{40}{100}\ \times\ 11b\ =\ 4.4b\ =$$ exports from X to P

22% of the imports of P $$=\dfrac{22}{100}\ \times\ 10a\ =\ 2.2a\ =$$ imports to P from X

Equating both, we get,

4.4b = 2.2a

a = 2b.

We are also given that imports for C are only from P, and we can set the rest of the imports for C to 0.

The total imports for C = 3c = 1200

c = 400

This makes the exports for C = 2c = 800

In clue 3, we are given that 90% of exports of C are to P and 4% are to ROW.

90% of the exports of C $$=\dfrac{90}{100}\ \times\ 2c\ =\ 1.8c\ =1.8\ \times\ 400\ =\ 720\ =\ $$exports from C to P

4% of the exports of C $$=\frac{4}{100}\ \times\ 2c\ =\ 0.08c\ =\ 0.08\ \times\ 400\ =\ 32\ =$$ exports from C to ROW

The exports from C to X = 800 - 720 - 32 = 48

In clue 4, we are given that 12% of exports of ROW are to X, 40% are to P.

If we assume that Imports of ROW are m and exports of ROW are n.

12% of the exports of ROW $$=\dfrac{12}{100}\ \times\ n\ =\ 0.12n\ =$$ exports from ROW to X

40% of the exports of ROW $$=\dfrac{40}{100}\ \times\ n\ =\ 0.4n\ =$$ exports from ROW to P

Placing all the values in the table, we get,

Equating the imports of P and X to the Totals, we get,

For P,

4.4b + 720 + 0.4n = 20b

15.6b = 720 + 0.4n --(1)

For X,

600 + 48 + 0.12n = 9b

9b = 648 + 0.12n --(2)

Solving (1) and (2), we get

$$\dfrac{648\ +\ 0.12n}{9}\ =\ \dfrac{720\ +\ 0.4n}{15.6}$$

$$10108.8+1.872n=6480+3.6n$$

$$3628.8=1.728n$$

$$n=2100$$

From (1),

15.6b = 720 + 0.4(2100)

15.6b = 1560

b = 100

Exports of P to ROW = 20b - 600 - 1200 = 2000 - 1800 = 200

Exports of X to ROW = 11b - 4.4b = 6.6b = 660

Substituting the values of b and n, we can figure out all the values except for Imports and Exports of ROW to ROW.

The value of exports and imports of ROW to ROW has to be equal.

We calculated the value of n to be 2100.

The exports of ROW to ROW can be calculated as,

Exports of ROW to ROW = 2100 - 840 - 252 = 1008 = Imports of ROW to ROW.

The value of m can be calculated as,

m = 200 + 660 + 32 + 1008 = 1900.

Filling up the table with all the values, we get,

The exports from ROW to ROW are 1008 IC.

Hence, the correct answer is 1008.

In clue 5, we are given that export volumes of P, in IC, to X and C are 600 and 1200, respectively, and P is the only country that exports to C. So, the export from X to C and ROW to C is 0.

In clue 1, we are given that normalised trade balances of P, X and C are 0%, 10%, and -20%, respectively.

For P, since the normalised trade balance is 0% the exports have to be equal to the imports according to the given formula. So, let us assume them to be 10a and 10a.

For X,

$$\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ 10$$

$$10E\ -\ 10I\ =\ E\ +\ I$$

$$11I\ =\ 9E$$

So, let us assume export to be 11b and import to be 9b.

For C,

$$\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ -20$$

$$5I\ -\ 5E\ =\ E\ +\ I$$

$$2I\ =\ 3E$$

So, let us assume export to be 2c and import to be 3c.

In clue 2, we are given that 40% of exports of X are to P and 22% of imports of P are from X.

40% of the exports of X $$=\dfrac{40}{100}\ \times\ 11b\ =\ 4.4b\ =$$ exports from X to P

22% of the imports of P $$=\dfrac{22}{100}\ \times\ 10a\ =\ 2.2a\ =$$ imports to P from X

Equating both, we get,

4.4b = 2.2a

a = 2b.

We are also given that imports for C are only from P, and we can set the rest of the imports for C to 0.

The total imports for C = 3c = 1200

c = 400

This makes the exports for C = 2c = 800

In clue 3, we are given that 90% of exports of C are to P and 4% are to ROW.

90% of the exports of C $$=\dfrac{90}{100}\ \times\ 2c\ =\ 1.8c\ =1.8\ \times\ 400\ =\ 720\ =\ $$exports from C to P

4% of the exports of C $$=\frac{4}{100}\ \times\ 2c\ =\ 0.08c\ =\ 0.08\ \times\ 400\ =\ 32\ =$$ exports from C to ROW

The exports from C to X = 800 - 720 - 32 = 48

In clue 4, we are given that 12% of exports of ROW are to X, 40% are to P.

If we assume that Imports of ROW are m and exports of ROW are n.

12% of the exports of ROW $$=\dfrac{12}{100}\ \times\ n\ =\ 0.12n\ =$$ exports from ROW to X

40% of the exports of ROW $$=\dfrac{40}{100}\ \times\ n\ =\ 0.4n\ =$$ exports from ROW to P

Placing all the values in the table, we get,

Equating the imports of P and X to the Totals, we get,

For P,

4.4b + 720 + 0.4n = 20b

15.6b = 720 + 0.4n --(1)

For X,

600 + 48 + 0.12n = 9b

9b = 648 + 0.12n --(2)

Solving (1) and (2), we get

$$\dfrac{648\ +\ 0.12n}{9}\ =\ \dfrac{720\ +\ 0.4n}{15.6}$$

$$10108.8+1.872n=6480+3.6n$$

$$3628.8=1.728n$$

$$n=2100$$

From (1),

15.6b = 720 + 0.4(2100)

15.6b = 1560

b = 100

Exports of P to ROW = 20b - 600 - 1200 = 2000 - 1800 = 200

Exports of X to ROW = 11b - 4.4b = 6.6b = 660

Substituting the values of b and n, we can figure out all the values except for Imports and Exports of ROW to ROW.

The value of exports and imports of ROW to ROW has to be equal.

We calculated the value of n to be 2100.

The exports of ROW to ROW can be calculated as,

Exports of ROW to ROW = 2100 - 840 - 252 = 1008 = Imports of ROW to ROW.

The value of m can be calculated as,

m = 200 + 660 + 32 + 1008 = 1900.

Filling up the table with all the values, we get,

Trade balance of ROW = Exports - Imports = 2100 - 1900 = 200.

Hence, the correct answer is option D.

In clue 5, we are given that export volumes of P, in IC, to X and C are 600 and 1200, respectively, and P is the only country that exports to C. So, the export from X to C and ROW to C is 0.

In clue 1, we are given that normalised trade balances of P, X and C are 0%, 10%, and -20%, respectively.

For P, since the normalised trade balance is 0% the exports have to be equal to the imports according to the given formula. So, let us assume them to be 10a and 10a.

For X,

$$\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ 10$$

$$10E\ -\ 10I\ =\ E\ +\ I$$

$$11I\ =\ 9E$$

So, let us assume export to be 11b and import to be 9b.

For C,

$$\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ -20$$

$$5I\ -\ 5E\ =\ E\ +\ I$$

$$2I\ =\ 3E$$

So, let us assume export to be 2c and import to be 3c.

In clue 2, we are given that 40% of exports of X are to P and 22% of imports of P are from X.

40% of the exports of X $$=\dfrac{40}{100}\ \times\ 11b\ =\ 4.4b\ =$$ exports from X to P

22% of the imports of P $$=\dfrac{22}{100}\ \times\ 10a\ =\ 2.2a\ =$$ imports to P from X

Equating both, we get,

4.4b = 2.2a

a = 2b.

We are also given that imports for C are only from P, and we can set the rest of the imports for C to 0.

The total imports for C = 3c = 1200

c = 400

This makes the exports for C = 2c = 800

In clue 3, we are given that 90% of exports of C are to P and 4% are to ROW.

90% of the exports of C $$=\dfrac{90}{100}\ \times\ 2c\ =\ 1.8c\ =1.8\ \times\ 400\ =\ 720\ =\ $$exports from C to P

4% of the exports of C $$=\frac{4}{100}\ \times\ 2c\ =\ 0.08c\ =\ 0.08\ \times\ 400\ =\ 32\ =$$ exports from C to ROW

The exports from C to X = 800 - 720 - 32 = 48

In clue 4, we are given that 12% of exports of ROW are to X, 40% are to P.

If we assume that Imports of ROW are m and exports of ROW are n.

12% of the exports of ROW $$=\dfrac{12}{100}\ \times\ n\ =\ 0.12n\ =$$ exports from ROW to X

40% of the exports of ROW $$=\dfrac{40}{100}\ \times\ n\ =\ 0.4n\ =$$ exports from ROW to P

Placing all the values in the table, we get,

Equating the imports of P and X to the Totals, we get,

For P,

4.4b + 720 + 0.4n = 20b

15.6b = 720 + 0.4n --(1)

For X,

600 + 48 + 0.12n = 9b

9b = 648 + 0.12n --(2)

Solving (1) and (2), we get

$$\dfrac{648\ +\ 0.12n}{9}\ =\ \dfrac{720\ +\ 0.4n}{15.6}$$

$$10108.8+1.872n=6480+3.6n$$

$$3628.8=1.728n$$

$$n=2100$$

From (1),

15.6b = 720 + 0.4(2100)

15.6b = 1560

b = 100

Exports of P to ROW = 20b - 600 - 1200 = 2000 - 1800 = 200

Exports of X to ROW = 11b - 4.4b = 6.6b = 660

Substituting the values of b and n, we can figure out all the values except for Imports and Exports of ROW to ROW.

The value of exports and imports of ROW to ROW has to be equal.

We calculated the value of n to be 2100.

The exports of ROW to ROW can be calculated as,

Exports of ROW to ROW = 2100 - 840 - 252 = 1008 = Imports of ROW to ROW.

The value of m can be calculated as,

m = 200 + 660 + 32 + 1008 = 1900.

Filling up the table with all the values, we get,

Total trade of P = 2000 + 2000 = 4000

Total trade of X = 900 + 1100 = 2000

Total trade of C = 1200 + 800 = 2000

So, both countries X and C have the least total trade.

Hence, the correct answer is option C.

This is the table given in the question with unknown values of a, b, c, d and e.

We are told that Anu and Bijay use the Xitel operator, and Chetan, Deepak, Esthan and Faruq use the Yocel operator.

The outgoing call from Anu to the Xitel operator has to be the incoming call from Xitel to Bijay because there are only two friends using Xitel, and if one of the Xitel users is calling another Xitel user, then those two have to be Anu and Bijay.

The outgoing to Xitel by Anu = 100

The incoming from Xitel for Bijay = c = The outgoing to Xitel by Anu = 100

So, the value of c in the above table is 100.

The outgoing to Xitel by Bijay = b

The incoming from Xitel for Anu = 50 = The outgoing to Xitel by Bijay = b

So, the value of b in the above table is 50.

The sum of Outgoing from Anu and Bijay to operator Yocel must be equal to the sum of Incoming minutes from Xitel for Chetan, Deepak, Esthan and Faruq.

The sum of Outgoing from Anu and Bijay to operator Yocel = a + 200 

The sum of Incoming minutes from Xitel for Chetan, Deepak, Esthan and Faruq = 250 + 275 + 100 + 100 = 725

Equating them, we get,

a + 200 = 725

a = 525

So, the value of a in the above table is 525.

The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Xitel must be equal to the sum of Incoming minutes from Yocel for Anu and Bijay.

The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Xitel = 50 + 100 + d + 0 = 150 + d

The sum of Incoming minutes from Yocel for Anu and Bijay = 225 + 125 = 350

Equating them, we get,

150 + d = 350

d = 200

So, the value of d in the above table is 200.

The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Yocel must be equal to the sum of Incoming minutes from Yocel for Chetan, Deepak, Esthan, and Faruq.

The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Yocel = 175 + 150 + 100 + e = 425 + e

The sum of Incoming minutes from Yocel for Chetan, Deepak, Esthan, and Faruq = 150 + 100 + 375 + 150 = 775

Equating them, we get,

425 + e = 775

e = 350

So, the value of e in the above table is 350.

The final table is,

The duration of calls from Bijay to Ajay is 50 minutes, which is the outgoing minutes to operator Xitel of Bijay.

Hence, the correct answer is 50.

This is the table given in the question with unknown values of a, b, c, d and e.

We are told that Anu and Bijay use the Xitel operator, and Chetan, Deepak, Esthan and Faruq use the Yocel operator.

The outgoing call from Anu to the Xitel operator has to be the incoming call from Xitel to Bijay because there are only two friends using Xitel, and if one of the Xitel users is calling another Xitel user, then those two have to be Anu and Bijay.

The outgoing to Xitel by Anu = 100

The incoming from Xitel for Bijay = c = The outgoing to Xitel by Anu = 100

So, the value of c in the above table is 100.

The outgoing to Xitel by Bijay = b

The incoming from Xitel for Anu = 50 = The outgoing to Xitel by Bijay = b

So, the value of b in the above table is 50.

The sum of Outgoing from Anu and Bijay to operator Yocel must be equal to the sum of Incoming minutes from Xitel for Chetan, Deepak, Esthan and Faruq.

The sum of Outgoing from Anu and Bijay to operator Yocel = a + 200

The sum of Incoming minutes from Xitel for Chetan, Deepak, Esthan and Faruq = 250 + 275 + 100 + 100 = 725

Equating them, we get,

a + 200 = 725

a = 525

So, the value of a in the above table is 525.

The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Xitel must be equal to the sum of Incoming minutes from Yocel for Anu and Bijay.

The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Xitel = 50 + 100 + d + 0 = 150 + d

The sum of Incoming minutes from Yocel for Anu and Bijay = 225 + 125 = 350

Equating them, we get,

150 + d = 350

d = 200

So, the value of d in the above table is 200.

The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Yocel must be equal to the sum of Incoming minutes from Yocel for Chetan, Deepak, Esthan, and Faruq.

The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Yocel = 175 + 150 + 100 + e = 425 + e

The sum of Incoming minutes from Yocel for Chetan, Deepak, Esthan, and Faruq = 150 + 100 + 375 + 150 = 775

Equating them, we get,

425 + e = 775

e = 350

So, the value of e in the above table is 350.

The final table is,

The total duration of calls made by Anu to friends having mobile numbers from Operator Yocel is Anu's outgoing minutes to Operator Yocel, which is 525 minutes.

Hence, the correct answer is 525.

This is the table given in the question with unknown values of a, b, c, d and e.

We are told that Anu and Bijay use the Xitel operator, and Chetan, Deepak, Esthan and Faruq use the Yocel operator.

The outgoing call from Anu to the Xitel operator has to be the incoming call from Xitel to Bijay because there are only two friends using Xitel, and if one of the Xitel users is calling another Xitel user, then those two have to be Anu and Bijay.

The outgoing to Xitel by Anu = 100

The incoming from Xitel for Bijay = c = The outgoing to Xitel by Anu = 100

So, the value of c in the above table is 100.

The outgoing to Xitel by Bijay = b

The incoming from Xitel for Anu = 50 = The outgoing to Xitel by Bijay = b

So, the value of b in the above table is 50.

The sum of Outgoing from Anu and Bijay to operator Yocel must be equal to the sum of Incoming minutes from Xitel for Chetan, Deepak, Esthan and Faruq.

The sum of Outgoing from Anu and Bijay to operator Yocel = a + 200

The sum of Incoming minutes from Xitel for Chetan, Deepak, Esthan and Faruq = 250 + 275 + 100 + 100 = 725

Equating them, we get,

a + 200 = 725

a = 525

So, the value of a in the above table is 525.

The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Xitel must be equal to the sum of Incoming minutes from Yocel for Anu and Bijay.

The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Xitel = 50 + 100 + d + 0 = 150 + d

The sum of Incoming minutes from Yocel for Anu and Bijay = 225 + 125 = 350

Equating them, we get,

150 + d = 350

d = 200

So, the value of d in the above table is 200.

The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Yocel must be equal to the sum of Incoming minutes from Yocel for Chetan, Deepak, Esthan, and Faruq.

The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Yocel = 175 + 150 + 100 + e = 425 + e

The sum of Incoming minutes from Yocel for Chetan, Deepak, Esthan, and Faruq = 150 + 100 + 375 + 150 = 775

Equating them, we get,

425 + e = 775

e = 350

So, the value of e in the above table is 350.

The final table is,

The total duration of calls made by Faruq to friends having mobile numbers from Operator Yocel is the outgoing minutes from Faruq to Operator Yocel, which is 350 minutes.

Hence, the correct answer is 350.

This is the table given in the question with unknown values of a, b, c, d and e.

We are told that Anu and Bijay use the Xitel operator, and Chetan, Deepak, Esthan and Faruq use the Yocel operator.

The outgoing call from Anu to the Xitel operator has to be the incoming call from Xitel to Bijay because there are only two friends using Xitel, and if one of the Xitel users is calling another Xitel user, then those two have to be Anu and Bijay.

The outgoing to Xitel by Anu = 100

The incoming from Xitel for Bijay = c = The outgoing to Xitel by Anu = 100

So, the value of c in the above table is 100.

The outgoing to Xitel by Bijay = b

The incoming from Xitel for Anu = 50 = The outgoing to Xitel by Bijay = b

So, the value of b in the above table is 50.

The sum of Outgoing from Anu and Bijay to operator Yocel must be equal to the sum of Incoming minutes from Xitel for Chetan, Deepak, Esthan and Faruq.

The sum of Outgoing from Anu and Bijay to operator Yocel = a + 200

The sum of Incoming minutes from Xitel for Chetan, Deepak, Esthan and Faruq = 250 + 275 + 100 + 100 = 725

Equating them, we get,

a + 200 = 725

a = 525

So, the value of a in the above table is 525.

The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Xitel must be equal to the sum of Incoming minutes from Yocel for Anu and Bijay.

The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Xitel = 50 + 100 + d + 0 = 150 + d

The sum of Incoming minutes from Yocel for Anu and Bijay = 225 + 125 = 350

Equating them, we get,

150 + d = 350

d = 200

So, the value of d in the above table is 200.

The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Yocel must be equal to the sum of Incoming minutes from Yocel for Chetan, Deepak, Esthan, and Faruq.

The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Yocel = 175 + 150 + 100 + e = 425 + e

The sum of Incoming minutes from Yocel for Chetan, Deepak, Esthan, and Faruq = 150 + 100 + 375 + 150 = 775

Equating them, we get,

425 + e = 775

e = 350

So, the value of e in the above table is 350.

The final table is,

Label the unknown value we need as

$$x=\text{minutes from Deepak to Chetan}$$

Use the operator groups from the table:

Xitel users: Anu, Bijay.

Yocel users: Chetan, Deepak, Eshan, Faruq.

Chetan’s incoming from Yocel = 150 minutes.

Yocel$$\to$$Chetan can come only from Deepak, Eshan, or Faruq. But the problem states Eshan $$\to$$ Chetan = 0, so

$$(\text{Deepak}\to\text{Chetan}) + (\text{Faruq}\to\text{Chetan}) = 150$$

Hence

$$x + (\text{Faruq}\to\text{Chetan}) = 150 \qquad(1)$$

Faruq’s outgoing to Yocel = 350 minutes (table). Those 350 minutes are split among Yocel recipients Chetan, Deepak and Eshan. We are told Faruq$$\to$$Eshan = 200 (given). So

$$(\text{Faruq}\to\text{Chetan}) + (\text{Faruq}\to\text{Deepak}) + 200 = 350,$$

hence

$$(\text{Faruq}\to\text{Chetan}) + (\text{Faruq}\to\text{Deepak}) = 150 \qquad(2)$$

Consider Deepak’s incoming from Yocel (table) = 100 minutes. The ones in Yocel that can call Deepak are Chetan, Eshan, and Faruq. But we are told Chetan $$\to$$ Deepak = 0 and Eshan $$\to$$ Deepak = 0, so the only Yocel caller to Deepak is Faruq. Therefore

$$(\text{Faruq}\to\text{Deepak}) = 100 \qquad(3)$$

Substitute (3) into (2):

$$(\text{Faruq}\to\text{Chetan}) + 100 = 150 \quad\Rightarrow\quad (\text{Faruq}\to\text{Chetan}) = 50.$$

Now use (1): 

$$x + (\text{Faruq}\to\text{Chetan}) = 150$$. 

With $$\text{Faruq}\to\text{Chetan} = 50$$, we get,

$$x + 50 = 150$$

$$x = 100$$

Thus Deepak $$\to$$ Chetan = 100 minutes.

Hence, the correct answer is option A.