CAT 2025 DILR Slot 3 Question Paper With Detailed PDF Solutions

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CAT 2025 DILR Slot 3 Paper With Answers & Explanation

Directions for questions 1 to 5: Answer the questions on the basis of the information given below.

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.

Q. 1 How much is exported from C to X, in IC?

Explanation
Correct Answer

48

Explanation

Let total exports of P, X, C, ROW be Ep, Ex, Ec,
Er respectively and imports Ip, Ix, Ic, Ir respectively.
From normalized trade balances:
For P: (Ep – Ip)/(Ep + Ip) = 0 ⇒ Ep = Ip
For X: (Ex – Ix)/(Ex + Ix) = 10%
⇒ Ex – Ix = 0.1(Ex + Ix) ⇒ 0.9Ex = 1.1Ix ⇒ Ex/Ix = 11/9
For C: (Ec – Ic)/(Ec + Ic) = –20%
⇒ Ec – Ic = –0.2(Ec + Ic)
⇒ 1.2Ec = 0.8Ic ⇒ Ic = 1.5Ec
Imports and exports of C:
As per condition (5): P is the only country that
exports to C.
C’s only import comes from P and the import volume, in IC, is 1200.
Therefore, Ic = 1200.
Since Ic = 1.5Ec, therefore, Ec = 800.
Exports of C:
90% to P ⇒ 0.9Ec = 720
4% to ROW ⇒ 0.04Ec = 32
Remaining 6% to X ⇒ C → X = 0.06 × 800 = 48

Q. 2 How much is exported from P to ROW, in IC?

Explanation
Correct Answer

200

Explanation

Step 1:
Let total exports of P, X, C, ROW be Ep, Ex, Ec, Er respectively and imports Ip, Ix, Ic, Ir respectively.
From normalized trade balances:
For P: (Ep – Ip)/(Ep + Ip) = 0 ⇒ Ep = Ip
For X: (Ex – Ix)/(Ex + Ix) = 10%
⇒ Ex – Ix = 0.1(Ex + Ix) ⇒ 0.9Ex = 1.1Ix ⇒ Ex/Ix = 11/9
For C: (Ec – Ic)/(Ec + Ic) = –20%
⇒ Ec – Ic = –0.2(Ec + Ic) ⇒ 1.2Ec = 0.8Ic ⇒ Ic = 1.5Ec
Imports and exports of C:
As per condition (5): P is the only country that exports to C.
C’s only import comes from P and the import volume, in IC, is 1200.
Therefore, Ic = 1200.
Since Ic = 1.5Ec, therefore, Ec = 800.
Exports of C:
90% to P ⇒ 0.9Ec = 720
4% to ROW ⇒ 0.04Ec = 32
Remaining 6% to X ⇒ C ⇒ X = 0.06 × 800 = 48
Step 2:
From condition (2):
40% of exports of X to P (X → P) is same as 22% of
imports of P from X.
So 0.4Ex = 0.22Ip or, Ex = 11Ip/20.
But Ep = Ip.
⇒ 0.4Ex = 0.22Ep ⇒ Ex = 0.55Ep
And similarly, 20Ix = 9Ip.
Imports and exports of P:
Imports of P come from X, C, and ROW.
From X = 0.22Ip
From C = 720
From ROW = Ip – (0.22Ip + 720) = 0.78Ip – 720
Exports of ROW to P = 40% of Er
⇒ 0.4Er = 0.78Ip – 720 ...(1)
Imports and Exports of X:
12% of exports of ROW go to X
⇒ ROW → X = 0.12Er
Imports of X:
From P = 600
From C = 48
From ROW = 0.12Er
So Ix = 648 + 0.12Er. Since Ix = 0.45Ep, therefore, ⇒ 0.45Ep = 648 + 0.12Er ...(2)
Step 3:
From equations (1) and (2):
From (1): Er = (0.78Ep – 720)/0.4 = 1.95Ep – 1800 Substitute into (2):
⇒ 0.45Ep = 648 + 0.12(1.95Ep – 1800)
⇒ 0.45Ep = 648 + 0.234Ep – 216
⇒ 0.216Ep = 432 ⇒ Ep = 2000
So Ip = 2000.
Hence, the final table looks like:


200 IC is exported from P to ROW.

Q. 3 How much is exported from ROW to ROW, in IC?

Explanation
Correct Answer

1008

Explanation

Step 1:
Let total exports of P, X, C, ROW be Ep, Ex, Ec, Er respectively and imports Ip, Ix, Ic, Ir respectively.
From normalized trade balances:
For P: (Ep – Ip)/(Ep + Ip) = 0 ⇒ Ep = Ip
For X: (Ex – Ix)/(Ex + Ix) = 10%
⇒ Ex – Ix = 0.1(Ex + Ix) ⇒ 0.9Ex = 1.1Ix ⇒ Ex/Ix = 11/9
For C: (Ec – Ic)/(Ec + Ic) = –20%
⇒ Ec – Ic = –0.2(Ec + Ic) ⇒ 1.2Ec = 0.8Ic ⇒ Ic = 1.5Ec
Imports and exports of C:
As per condition (5): P is the only country that exports to C.
C’s only import comes from P and the import volume, in IC, is 1200.
Therefore, Ic = 1200.
Since Ic = 1.5Ec, therefore, Ec = 800.
Exports of C:
90% to P ⇒ 0.9Ec = 720
4% to ROW ⇒ 0.04Ec = 32
Remaining 6% to X ⇒ C ⇒ X = 0.06 × 800 = 48
Step 2:
From condition (2):
40% of exports of X to P (X → P) is same as 22% of
imports of P from X.
So 0.4Ex = 0.22Ip or, Ex = 11Ip/20.
But Ep = Ip.
⇒ 0.4Ex = 0.22Ep ⇒ Ex = 0.55Ep
And similarly, 20Ix = 9Ip.
Imports and exports of P:
Imports of P come from X, C, and ROW.
From X = 0.22Ip
From C = 720
From ROW = Ip – (0.22Ip + 720) = 0.78Ip – 720
Exports of ROW to P = 40% of Er
⇒ 0.4Er = 0.78Ip – 720 ...(1)
Imports and Exports of X:
12% of exports of ROW go to X
⇒ ROW → X = 0.12Er
Imports of X:
From P = 600
From C = 48
From ROW = 0.12Er
So Ix = 648 + 0.12Er. Since Ix = 0.45Ep, therefore, ⇒ 0.45Ep = 648 + 0.12Er ...(2)
Step 3:
From equations (1) and (2):
From (1): Er = (0.78Ep – 720)/0.4 = 1.95Ep – 1800 Substitute into (2):
⇒ 0.45Ep = 648 + 0.12(1.95Ep – 1800)
⇒ 0.45Ep = 648 + 0.234Ep – 216
⇒ 0.216Ep = 432 ⇒ Ep = 2000
So Ip = 2000.
Hence, the final table looks like:


1008 IC is exported from ROW to ROW.

Q. 4 What is the trade balance of ROW?

Explanation
Correct Answer

4

Explanation

Step 1:
Let total exports of P, X, C, ROW be Ep, Ex, Ec, Er respectively and imports Ip, Ix, Ic, Ir respectively.
From normalized trade balances:
For P: (Ep – Ip)/(Ep + Ip) = 0 ⇒ Ep = Ip
For X: (Ex – Ix)/(Ex + Ix) = 10%
⇒ Ex – Ix = 0.1(Ex + Ix) ⇒ 0.9Ex = 1.1Ix ⇒ Ex/Ix = 11/9
For C: (Ec – Ic)/(Ec + Ic) = –20%
⇒ Ec – Ic = –0.2(Ec + Ic) ⇒ 1.2Ec = 0.8Ic ⇒ Ic = 1.5Ec
Imports and exports of C:
As per condition (5): P is the only country that exports to C.
C’s only import comes from P and the import volume, in IC, is 1200.
Therefore, Ic = 1200.
Since Ic = 1.5Ec, therefore, Ec = 800.
Exports of C:
90% to P ⇒ 0.9Ec = 720
4% to ROW ⇒ 0.04Ec = 32
Remaining 6% to X ⇒ C ⇒ X = 0.06 × 800 = 48
Step 2:
From condition (2):
40% of exports of X to P (X → P) is same as 22% of
imports of P from X.
So 0.4Ex = 0.22Ip or, Ex = 11Ip/20.
But Ep = Ip.
⇒ 0.4Ex = 0.22Ep ⇒ Ex = 0.55Ep
And similarly, 20Ix = 9Ip.
Imports and exports of P:
Imports of P come from X, C, and ROW.
From X = 0.22Ip
From C = 720
From ROW = Ip – (0.22Ip + 720) = 0.78Ip – 720
Exports of ROW to P = 40% of Er
⇒ 0.4Er = 0.78Ip – 720 ...(1)
Imports and Exports of X:
12% of exports of ROW go to X
⇒ ROW → X = 0.12Er
Imports of X:
From P = 600
From C = 48
From ROW = 0.12Er
So Ix = 648 + 0.12Er. Since Ix = 0.45Ep, therefore, ⇒ 0.45Ep = 648 + 0.12Er ...(2)
Step 3:
From equations (1) and (2):
From (1): Er = (0.78Ep – 720)/0.4 = 1.95Ep – 1800 Substitute into (2):
⇒ 0.45Ep = 648 + 0.12(1.95Ep – 1800)
⇒ 0.45Ep = 648 + 0.234Ep – 216
⇒ 0.216Ep = 432 ⇒ Ep = 2000
So Ip = 2000.
Hence, the final table looks like:


The trade balance of ROW = Er – Ir = 2100 – 1900 = 200 IC.

Q. 5 Which among the countries P, X, and C has/have the least total trade?

Explanation
Correct Answer

3

Explanation

Step 1:
Let total exports of P, X, C, ROW be Ep, Ex, Ec, Er respectively and imports Ip, Ix, Ic, Ir respectively.
From normalized trade balances:
For P: (Ep – Ip)/(Ep + Ip) = 0 ⇒ Ep = Ip
For X: (Ex – Ix)/(Ex + Ix) = 10%
⇒ Ex – Ix = 0.1(Ex + Ix) ⇒ 0.9Ex = 1.1Ix ⇒ Ex/Ix = 11/9
For C: (Ec – Ic)/(Ec + Ic) = –20%
⇒ Ec – Ic = –0.2(Ec + Ic) ⇒ 1.2Ec = 0.8Ic ⇒ Ic = 1.5Ec
Imports and exports of C:
As per condition (5): P is the only country that exports to C.
C’s only import comes from P and the import volume, in IC, is 1200.
Therefore, Ic = 1200.
Since Ic = 1.5Ec, therefore, Ec = 800.
Exports of C:
90% to P ⇒ 0.9Ec = 720
4% to ROW ⇒ 0.04Ec = 32
Remaining 6% to X ⇒ C ⇒ X = 0.06 × 800 = 48
Step 2:
From condition (2):
40% of exports of X to P (X → P) is same as 22% of
imports of P from X.
So 0.4Ex = 0.22Ip or, Ex = 11Ip/20.
But Ep = Ip.
⇒ 0.4Ex = 0.22Ep ⇒ Ex = 0.55Ep
And similarly, 20Ix = 9Ip.
Imports and exports of P:
Imports of P come from X, C, and ROW.
From X = 0.22Ip
From C = 720
From ROW = Ip – (0.22Ip + 720) = 0.78Ip – 720
Exports of ROW to P = 40% of Er
⇒ 0.4Er = 0.78Ip – 720 ...(1)
Imports and Exports of X:
12% of exports of ROW go to X
⇒ ROW → X = 0.12Er
Imports of X:
From P = 600
From C = 48
From ROW = 0.12Er
So Ix = 648 + 0.12Er. Since Ix = 0.45Ep, therefore, ⇒ 0.45Ep = 648 + 0.12Er ...(2)
Step 3:
From equations (1) and (2):
From (1): Er = (0.78Ep – 720)/0.4 = 1.95Ep – 1800 Substitute into (2):
⇒ 0.45Ep = 648 + 0.12(1.95Ep – 1800)
⇒ 0.45Ep = 648 + 0.234Ep – 216
⇒ 0.216Ep = 432 ⇒ Ep = 2000
So Ip = 2000.
Hence, the final table looks like:


Total trade:
P = Ep + Ip = 4000
X = Ex + Ix = (1100 + 900) = 2000
C = Ec + Ic = (800 + 1200) = 2000
Hence, the least total trade = X and C.

Directions for questions 6 to 9: Answer the questions on the basis of the information given below.

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 '?'.

Q. 6 Who is sitting immediately to the right of Bina?

Explanation
Correct Answer

1

Explanation


In Round 8, Diya receives the Buck. In Round 9, she passes it in such a way that Farhan receives it.
If Farhan sits to the immediate left of Diya, then there is no way Farhan passes it in a way receives it.
Therefore, Diya must pass second to the right and then Farhan must pass immediate to the right and Chirag received it.
Final arrangement looks like:


Eshan is sitting immediately to the right of Bina.

Q. 7 Who is sitting third to the left of Eshan?

Explanation
Correct Answer

1

Explanation


In Round 8, Diya receives the Buck. In Round 9, she passes it in such a way that Farhan receives it.
If Farhan sits to the immediate left of Diya, then there is no way Farhan passes it in a way receives it.
Therefore, Diya must pass second to the right and then Farhan must pass immediate to the right and Chirag received it.
Final arrangement looks like:


Chirag is sitting third to the left of Eshan.

Q. 8 For which of the following pass types can the total number of occurrences be uniquely determined?

Explanation
Correct Answer

3

Explanation


In Round 8, Diya receives the Buck. In Round 9, she passes it in such a way that Farhan receives it.
If Farhan sits to the immediate left of Diya, then there is no way Farhan passes it in a way receives it.
Therefore, Diya must pass second to the right and then Farhan must pass immediate to the right and Chirag received it.
Final arrangement looks like:


For Immediately to the right pass type the total number of occurrences can be uniquely determined.

Q. 9 For which of the following children is it possible to determine how many times they received the Buck?

Explanation
Correct Answer

2

Explanation


In Round 8, Diya receives the Buck. In Round 9, she passes it in such a way that Farhan receives it.
If Farhan sits to the immediate left of Diya, then there is no way Farhan passes it in a way receives it.
Therefore, Diya must pass second to the right and then Farhan must pass immediate to the right and Chirag received it.
Final arrangement looks like:


For Gaurav it is possible to determine the number of times he received the Buck due to round 4.

Directions for questions 10 to 13: 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.

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.

Q. 10 What was the duration of calls (in minutes) from Bijay to Anu?

Explanation
Correct Answer

50

Explanation

Step 1:
Four mobile operator combinations are possible:
Xitel operator to Xitel (Orange area)
Xitel to Yocel (Purple area)
Yocel to Yocel (Blue area)
Yocel to Xitel (Green area)

Step 2:
Note: Bijay is Xitel user and makes an outgoing call to a Xitel user and that must be Anu only. Therefore, a = 50
and e = 100.
Similarly, equate the areas coloured by same colour.
Green: 50 + 100 + b = 225 + 125 or, b = 200.
Purple: c + 200 = 250 + 275 + 200 or, c = 525
Blue: 175 + 150 + 100 + d = 150 + 100 + 375 + 150 or, d = 350


The duration of calls (in minutes) from Bijay to Anu was 50 minutes.

Q. 11 What was the total duration of calls (in minutes) made by Anu to friends having mobile numbers from Operator Yocel?

Explanation
Correct Answer

525

Explanation

Step 1:
Four mobile operator combinations are possible:
Xitel operator to Xitel (Orange area)
Xitel to Yocel (Purple area)
Yocel to Yocel (Blue area)
Yocel to Xitel (Green area)

Step 2:
Note: Bijay is Xitel user and makes an outgoing call to a Xitel user and that must be Anu only. Therefore, a = 50
and e = 100.
Similarly, equate the areas coloured by same colour.
Green: 50 + 100 + b = 225 + 125 or, b = 200.
Purple: c + 200 = 250 + 275 + 200 or, c = 525
Blue: 175 + 150 + 100 + d = 150 + 100 + 375 + 150 or, d = 350


The total duration of calls (in minutes) made by Anu to friends who have mobile numbers from Operator Yocel was 525 minutes.

Q. 12 What was the total duration of calls (in minutes) made by Faruq to friends having mobile numbers from Operator Yocel?

Explanation
Correct Answer

350

Explanation

Step 1:
Four mobile operator combinations are possible:
Xitel operator to Xitel (Orange area)
Xitel to Yocel (Purple area)
Yocel to Yocel (Blue area)
Yocel to Xitel (Green area)

Step 2:
Note: Bijay is Xitel user and makes an outgoing call to a Xitel user and that must be Anu only. Therefore, a = 50
and e = 100.
Similarly, equate the areas coloured by same colour.
Green: 50 + 100 + b = 225 + 125 or, b = 200.
Purple: c + 200 = 250 + 275 + 200 or, c = 525
Blue: 175 + 150 + 100 + d = 150 + 100 + 375 + 150 or, d = 350


The total duration of calls (in minutes) made by Faruq to friends who had mobile numbers from Operator Yocel was 350 minutes.

Q. 13 What was the duration of calls (in minutes) from Deepak to Chetan?

Explanation
Correct Answer

4

Explanation

Step 1:
Four mobile operator combinations are possible:
Xitel operator to Xitel (Orange area)
Xitel to Yocel (Purple area)
Yocel to Yocel (Blue area)
Yocel to Xitel (Green area)

Step 2:
Note: Bijay is Xitel user and makes an outgoing call to a Xitel user and that must be Anu only. Therefore, a = 50
and e = 100.
Similarly, equate the areas coloured by same colour.
Green: 50 + 100 + b = 225 + 125 or, b = 200.
Purple: c + 200 = 250 + 275 + 200 or, c = 525
Blue: 175 + 150 + 100 + d = 150 + 100 + 375 + 150 or, d = 350


Directions for questions 14 to 17: 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.

Q. 14 Who had solved the largest number of puzzles by the 20-th minute from the start of the competition?

Explanation
Correct Answer

1

Explanation

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 one 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.

Q. 15 How many minutes did Suranjan take to solve the third visual puzzle in the competition?

Explanation
Correct Answer

2

Explanation

From graph 1, we know Suranjan completed puzzle 1 at 5th minute, puzzle 2 at 7th minute, puzzle 3 at 12th
minute, puzzle 4 at 18th minute, puzzle 5 at 21st minute, puzzle 6 at 23rd minute, puzzle 7 at 26th minute, puzzle 8 at 28th minute, puzzle 9 at 30th minute and puzzle 10 at 35th minute.
From graph 2, we know Suranjan completed visual puzzle 1 at 5th minute, visual puzzle 2 at 18th minute, visual puzzle 3 at 28th minute and visual puzzle 4 at 30th minute.
So if we consider both information together, puzzle 8 is visual puzzle 3. And time taken to complete that would be puzzle 8 – puzzle 7 = 28 – 26 = 2 minutes.

Q. 16 At what number in the sequence was the fourth number-based puzzle?

Explanation
Correct Answer

6

Explanation

From graph 1, we know Suranjan completed puzzle 1 at 5th minute, puzzle 2 at 7th minute, puzzle 3 at 12th minute, puzzle 4 at 18th minute, puzzle 5 at 21st minute, puzzle 6 at 23rd minute, puzzle 7 at 26th minute, puzzle 8 at 28th minute, puzzle 9 at 30th minute and puzzle 10 at 35th minute.
From graph 2, we know Suranjan completed visual puzzle 1 at 5th minute, visual puzzle 2 at 18th minute, visual puzzle 3 at 28th minute and visual puzzle 4 at 30th minute.
So if we consider both information together:
Puzzle 1 is visual puzzle 1
Puzzle 4 is visual puzzle 4
puzzle 8 is visual puzzle 3
puzzle 9 was visual puzzle 4.
Therefore, puzzle 2, 3, 5, 6, 7 and 10 are number based puzzle. Hence, puzzle 6 was the fourth number-based puzzle.

Q. 17 Which of the following is the closest to the average time taken by Anirbid to solve the number-based puzzles in the competition?

Explanation
Correct Answer

1

Explanation

The average time taken by Anirbid to solve the number-based puzzles can be calculated as follows:
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 = (5 + 2 + 5 + 3 + 5 + 4) / 6 = 24 / 6 = 4 minutes
The average time taken by Anirbid to solve the number-based puzzles is 4 minutes.

Directions for questions 18 to 22: 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.

Q. 18 What is the sum of Travel Costs for all travelers in Zentars?

Explanation
Correct Answer

41000

Explanation

Step 1:
As per the information given to us:
Let us denote the local currencies Aurels as A, Brins as B, Crowns as C and Zentars as Z.
And 1 C = 0.5 Z.

Lian’s Calculation:
Lian’s Cost = 36000 C = 18000 Z.
18000 Z = (6000/5000) Z + 3000 B + 2000 C In Zentars: 18000 = (6000/5000) + 3000b + 1000 If Flight = 5000:
3000b = 12000 ⇒ b = 4 If Flight = 6000:
3000b = 11000 ⇒ b = 3.66 Step 2:
Kira’s Calculation:
In Brins, her cost is 4000.
Total Zentars = 4000 × b.
If b = 4, Total = 16000 Z.
16000 = Flight + 1000a + 2000(4) 8000 = Flight + 1000a If Flight = 6000:
1000a = 2000 ⇒ a = 2 If b = 11/3, Total = 44000/3 Z.
16000 = 5000 + 1000a + 2000(11/3) ⇒ a = 11/3.
Checking Jano: it will only satisfy for a = 2 and not for a = 11/3.


Sum of Travel Costs for all travelers (Zentars):
Jano: 7000 Z
Kira: 8000 × 2 = 16000 Z
Lian: 36000/2 = 18000 Z
Total = 7000 + 16000 + 18000 = 41000 Z.

Q. 19 How many Zentars did Lian spend in the two countries he visited?

Explanation
Correct Answer

13000

Explanation

Step 1:
As per the information given to us:
Let us denote the local currencies Aurels as A, Brins as B, Crowns as C and Zentars as Z.
And 1 C = 0.5 Z.

Lian’s Calculation:
Lian’s Cost = 36000 C = 18000 Z.
18000 Z = (6000/5000) Z + 3000 B + 2000 C In Zentars: 18000 = (6000/5000) + 3000b + 1000 If Flight = 5000:
3000b = 12000 ⇒ b = 4 If Flight = 6000:
3000b = 11000 ⇒ b = 3.66 Step 2:
Kira’s Calculation:
In Brins, her cost is 4000.
Total Zentars = 4000 × b.
If b = 4, Total = 16000 Z.
16000 = Flight + 1000a + 2000(4) 8000 = Flight + 1000a If Flight = 6000:
1000a = 2000 ⇒ a = 2 If b = 11/3, Total = 44000/3 Z.
16000 = 5000 + 1000a + 2000(11/3) ⇒ a = 11/3.
Checking Jano: it will only satisfy for a = 2 and not for a = 11/3.


Lian spend = 3000 × 4 + 2000/2 = 12000 + 1000 = 13000 Z.

Q. 20 What was Jano's total spend in the two countries he visited, in Aurels?

Explanation
Correct Answer

1500

Explanation

Step 1:
As per the information given to us:
Let us denote the local currencies Aurels as A, Brins as B, Crowns as C and Zentars as Z.
And 1 C = 0.5 Z.

Lian’s Calculation:
Lian’s Cost = 36000 C = 18000 Z.
18000 Z = (6000/5000) Z + 3000 B + 2000 C In Zentars: 18000 = (6000/5000) + 3000b + 1000 If Flight = 5000:
3000b = 12000 ⇒ b = 4 If Flight = 6000:
3000b = 11000 ⇒ b = 3.66 Step 2:
Kira’s Calculation:
In Brins, her cost is 4000.
Total Zentars = 4000 × b.
If b = 4, Total = 16000 Z.
16000 = Flight + 1000a + 2000(4) 8000 = Flight + 1000a If Flight = 6000:
1000a = 2000 ⇒ a = 2 If b = 11/3, Total = 44000/3 Z.
16000 = 5000 + 1000a + 2000(11/3) ⇒ a = 11/3.
Checking Jano: it will only satisfy for a = 2 and not for a = 11/3.


Jano’s total spend = 1000 × 2 + 2000/2 = 3000 Z
In Aurels: 3000/2 = 1500 A.

Q. 21 One Brin is equivalent to how many Crowns?

Explanation
Correct Answer

4

Explanation

Step 1:
As per the information given to us:
Let us denote the local currencies Aurels as A, Brins as B, Crowns as C and Zentars as Z.
And 1 C = 0.5 Z.

Lian’s Calculation:
Lian’s Cost = 36000 C = 18000 Z.
18000 Z = (6000/5000) Z + 3000 B + 2000 C In Zentars: 18000 = (6000/5000) + 3000b + 1000 If Flight = 5000:
3000b = 12000 ⇒ b = 4 If Flight = 6000:
3000b = 11000 ⇒ b = 3.66 Step 2:
Kira’s Calculation:
In Brins, her cost is 4000.
Total Zentars = 4000 × b.
If b = 4, Total = 16000 Z.
16000 = Flight + 1000a + 2000(4) 8000 = Flight + 1000a If Flight = 6000:
1000a = 2000 ⇒ a = 2 If b = 11/3, Total = 44000/3 Z.
16000 = 5000 + 1000a + 2000(11/3) ⇒ a = 11/3.
Checking Jano: it will only satisfy for a = 2 and not for a = 11/3.


1 Brin = 4 Zentars and 1 Crown = 0.5 Zentars
Hence, one Brin is equivalent to = 4 / 0.5 = 8 Crowns.

Q. 22 Which of the following statements is NOT true about money spent in the local currency?

Explanation
Correct Answer

1

Explanation

Step 1:
As per the information given to us:
Let us denote the local currencies Aurels as A, Brins as B, Crowns as C and Zentars as Z.
And 1 C = 0.5 Z.

Lian’s Calculation:
Lian’s Cost = 36000 C = 18000 Z.
18000 Z = (6000/5000) Z + 3000 B + 2000 C In Zentars: 18000 = (6000/5000) + 3000b + 1000 If Flight = 5000:
3000b = 12000 ⇒ b = 4 If Flight = 6000:
3000b = 11000 ⇒ b = 3.66 Step 2:
Kira’s Calculation:
In Brins, her cost is 4000.
Total Zentars = 4000 × b.
If b = 4, Total = 16000 Z.
16000 = Flight + 1000a + 2000(4) 8000 = Flight + 1000a If Flight = 6000:
1000a = 2000 ⇒ a = 2 If b = 11/3, Total = 44000/3 Z.
16000 = 5000 + 1000a + 2000(11/3) ⇒ a = 11/3.
Checking Jano: it will only satisfy for a = 2 and not for a = 11/3.


1. Jano spent 2000 in Aurevia (False)
2. Lian spent 2000 in Cyrenia (True)
3. Jano spent 2000 in Cyrenia (True)
4. Kira spent 1000 in Aurevia (True)