Question 1 · Data Interpretation and Logical Reasoning NAT

Common Description: Instructions [42 - 46]
Eight gymnastics players numbered 1 through 8 underwent a training camp where they were coached by three coaches - Xena, Yuki, and Zara. Each coach trained at least two players. Yuki trained only even numbered players, while Zara trained only odd numbered players. After the camp, the coaches evaluated the players and gave integer ratings to the respective players trained by them on a scale of 1 to 7, with 1 being the lowest rating and 7 the highest.
The following additional information is known.
1. Xena trained more players than Yuki.
2. Player-1 and Player-4 were trained by the same coach, while the coaches who trained Player-2, Player-3 and Player-5 were all different.
3. Player-5 and Player-7 were trained by the same coach and got the same rating. All other players got a unique rating.
4. The average of the ratings of all the players was 4.
5. Player-2 got the highest rating.
6. The average of the ratings of the players trained by Yuki was twice that of the players trained by Xena and two more than that of the players trained by Zara.
7. Player-4's rating was double of Player-8's and less than Player-5's. For how many players the ratings can be determined with certainty?

Correct Answer:

6

Step-by-Step Solution

Key idea: This is a constrained integer partition and group-assignment problem. We recognise it because players must be assigned to coaches under parity restrictions, then rated with integers subject to sum, average, and distinctness constraints.

Step 1: Determine group sizes and averages.

Let $n_{X}, n_{Y}, n_{Z}$ be the number of players trained by Xena, Yuki, and Zara respectively.

Total players = 8. Each coach trains at least 2 players ( $n \geq 2$ ).

Clue 1 says $n_{X} > n_{Y}$ .

The only integer partition of 8 into three parts $\geq 2$ where one part is strictly greater than another is ${4, 2, 2}$ .

Since $n_{X} > n_{Y}$ , we must have $n_{X} = 4$ . The remaining two are 2. So $n_{Y} = 2$ and $n_{Z} = 2$ .

Now for averages. Let $A_{X}, A_{Y}, A_{Z}$ be the average ratings.

Clue 6: $A_{Y} = 2 A_{X}$ and $A_{Y} = A_{Z} + 2 ⟹ A_{Z} = 2 A_{X} - 2$ .

Clue 4: Overall average is 4, so Total Sum = $8 \times 4 = 32$ .

Equation: $4 A_{X} + 2 A_{Y} + 2 A_{Z} = 32$ .

Substitute: $4 A_{X} + 2 (2 A_{X}) + 2 (2 A_{X} - 2) = 32$ .

$4 A_{X} + 4 A_{X} + 4 A_{X} - 4 = 32 ⟹ 12 A_{X} = 36 ⟹ A_{X} = 3$ .

Therefore, $A_{Y} = 6$ and $A_{Z} = 4$ .

Group Sums: $S_{X} = 4 \times 3 = 12$ , $S_{Y} = 2 \times 6 = 12$ , $S_{Z} = 2 \times 4 = 8$ .

Step 2: Assign players to coaches.

Yuki (Y) trains only even numbers. Zara (Z) trains only odd numbers.

Clue 3: P5 and P7 are same coach, same rating. Both are odd, so they belong to Z.

Since $S_{Z} = 8$ and $R_{5} = R_{7}$ , we have $2 R_{5} = 8 ⟹ R_{5} = R_{7} = 4$ .

Clue 2: P1 and P4 are same coach. P1 is odd, P4 is even. Only Xena can train both. So ${P 1, P 4} \subset X$ .

Clue 2 also says coaches for P2, P3, P5 are all different. P5 is Z.

Remaining coaches for P2, P3 are X and Y.

P2 is even, so P2 cannot be Z. P2 could be Y or X.

P3 is odd, so P3 cannot be Y. P3 could be X or Z.

But P5 is already Z. Since coaches for {P2, P3, P5} are distinct, P3 cannot be Z.

Therefore, P3 must be X.

This leaves P2 to be Y.

Current assignment: Z={5,7}, X={1,3,4}, Y={2}.

Y needs 2 players total. Remaining evens: {6, 8}. One goes to Y, one to X (to complete X's 4).

Clue 7: $R_{4} = 2 \times R_{8}$ and $R_{4} < R_{5} (= 4)$ .

Possible integers: If $R_{8} = 1 ⟹ R_{4} = 2$ . If $R_{8} = 2 ⟹ R_{4} = 4$ (invalid as $< 4$ ).

So $R_{8} = 1$ and $R_{4} = 2$ .

Where is P8? If P8 were in Y, then $S_{Y} = R_{2} + R_{8} = R_{2} + 1 = 12 ⟹ R_{2} = 11$ . Impossible (max 7).

So P8 must be in X. This forces P6 into Y.

Final Groups: Z={5,7}, Y={2,6}, X={1,3,4,8}.

Step 3: Calculate certain ratings.

Known: $R_{5} = 4, R_{7} = 4, R_{4} = 2, R_{8} = 1$ .

Y sum = 12. Clue 5 says P2 got highest rating. Max possible is 7. So $R_{2} = 7$ .

Then $R_{6} = 12 - 7 = 5$ .

X sum = 12. Current X sum = $R_{1} + R_{3} + 2 + 1 = R_{1} + R_{3} + 3$ .

So $R_{1} + R_{3} = 9$ .

Used ratings: {1, 2, 4, 5, 7}. Remaining available: {3, 6}.

Since $3 + 6 = 9$ , the set ${R_{1}, R_{3}}$ is definitely ${3, 6}$ , but no clue distinguishes them.

Certain ratings: P2(7), P4(2), P5(4), P6(5), P7(4), P8(1).

Uncertain: P1, P3.

Count of certain ratings = 6.

Answer: 6

Question 2 · Data Interpretation and Logical Reasoning MCQ

Common Description: Instructions [33 - 38]
10 players - P1, P2, ... , P10 - competed in an international javelin throw event. The number (after P) of a player reflects his rank at the beginning of the event, with rank 1 going to the topmost player. There were two phases in the event with the first phase consisting of rounds 1, 2, and 3, and the second phase consisting of rounds 4, 5, and 6. A throw is measured in terms of the distance it covers (in meters, up to one decimal point accuracy), only if the throw is a ‘valid’ one. For an invalid throw, the distance is taken as zero. A player’s score at the end of a round is the maximum distance of all his throws up to that round. Players are re-ranked after every round based on their current scores. In case of a tie in scores, the player with a prevailing higher rank retains the higher rank. This ranking determines the order in which the players go for their throws in the next round.
In each of the rounds in the first phase, the players throw in increasing order of their latest rank, i.e. the player ranked 1 at that point throws first, followed by the player ranked 2 at that point and so on. The top six players at the end of the first phase qualify for the second phase. In each of the rounds in the second phase, the players throw in decreasing order of their latest rank i.e. the player ranked 6 at that point throws first, followed by the player ranked 5 at that point and so on. The players ranked 1, 2, and 3 at the end of the sixth round receive gold, silver, and bronze medals respectively.
All the valid throws of the event were of distinct distances (as per stated measurement accuracy). The tables below show distances (in meters) covered by all valid throws in the first and the third round in the event.
Distances covered by all the valid throws in the first round

Player	Distance(in m)
P1	82.9
P3	81.5
P5	86.4
P6	82.5
P7	87.2
P9	84.1

Distances covered by all the valid throws in the third round

Player	Distance(in m)
P1	88.6
P3	79.0
P9	81.4

The following facts are also known.
i. Among the throws in the second round, only the last two were valid. Both the throws enabled these players to qualify for the second phase, with one of them qualifying with the least score. None of these players won any medal.
ii. If a player throws first in a round AND he was also the last (among the players in the current round) to throw in the previous round, then the player is said to get a double. Two players got a double.
iii. In each round of the second phase, exactly one player improved his score. Each of these improvements was by the same amount.
iv. The gold and bronze medalists improved their scores in the fifth and the sixth rounds respectively. One medal winner improved his score in the fourth round.
v. The difference between the final scores of the gold medalist and the silver medalist, as well as the difference between the final scores of the silver medalist and the bronze medalist was 1.0 m. Who threw the last javelin in the event?

A.
P7
B.
P1
C.
P9
D.
P10

Correct Answer:

A

Step-by-Step Solution

Key idea: This is a Sports Ranking Simulation problem. It is recognisable because players are re-ranked after every round, throwing order depends on current rank, and the question asks about a specific positional event (who threw last).

Step 1: Understand throwing order in Phase 2.

Phase 2 (Rounds 4, 5, 6) uses DECREASING order of latest rank. Rank 6 throws first, Rank 1 throws LAST. The question asks who threw last in Round 6, which means we need the player holding Rank 1 at the end of Round 5.

Step 2: Determine Phase 2 improvements.

Clue (iii) says exactly one player improves per round by the same amount $d$ . Clue (iv) says Gold improved in R5, Bronze in R6, and one medalist in R4.

Step 3: Identify medalists and gaps.

End of R3 scores: P1=88.6, P7=87.2, P5=86.4. Final gaps are 1.0m.

If Silver = P1 (no improvement), then Gold = 89.6, Bronze = 87.6.

Gold (P7) needs +2.4, so $2 d = 2.4 ⟹ d = 1.2$ .

Bronze (P5) needs +1.2, so $d = 1.2$ .

This perfectly matches: P7 improves in R4 and R5, P5 improves in R6.

Step 4: Track Rank 1.

After R3, P1 is Rank 1 (88.6).

In R4, P7 improves to 88.4 (P1 remains 1st).

In R5, P7 improves to 89.6 (P7 becomes 1st).

Step 5: Conclusion.

Since P7 is Rank 1 at the end of R5, P7 throws last in R6.

Answer: A

Question 3 · Data Interpretation and Logical Reasoning MCQ

Common Description: Instructions [42 - 46]
Eight gymnastics players numbered 1 through 8 underwent a training camp where they were coached by three coaches - Xena, Yuki, and Zara. Each coach trained at least two players. Yuki trained only even numbered players, while Zara trained only odd numbered players. After the camp, the coaches evaluated the players and gave integer ratings to the respective players trained by them on a scale of 1 to 7, with 1 being the lowest rating and 7 the highest.
The following additional information is known.
1. Xena trained more players than Yuki.
2. Player-1 and Player-4 were trained by the same coach, while the coaches who trained Player-2, Player-3 and Player-5 were all different.
3. Player-5 and Player-7 were trained by the same coach and got the same rating. All other players got a unique rating.
4. The average of the ratings of all the players was 4.
5. Player-2 got the highest rating.
6. The average of the ratings of the players trained by Yuki was twice that of the players trained by Xena and two more than that of the players trained by Zara.
7. Player-4's rating was double of Player-8's and less than Player-5's. What best can be concluded about the number of players coached by Zara?

A.
Either 2 or 3 or 4
B.
Exactly 2
C.
Either 2 or 3
D.
Either 3

Correct Answer:

B

Step-by-Step Solution

Key idea: This is an evaluation-matrix question with group counts and average constraints requiring integer validation. It is recognisable because it combines partition of players into groups with average score equations.

Step 1: Determine the number of players per coach.

Total players = 8. Coaches X, Y, Z each train $\geq 2$ players.

Y trained only even numbered players (2, 4, 6, 8), so $n_{Y} \leq 4$ .

X trained more players than Y ( $n_{X} > n_{Y}$ ).

If $n_{Y} = 3$ , then $n_{X} \geq 4$ and $n_{Z} \geq 2$ , total $\geq 9$ . Impossible since only 8 players exist.

So $n_{Y} = 2$ .

Then $n_{X} + n_{Z} = 6$ . Since $n_{X} > 2$ , $n_{X}$ can be 3 or 4.

Step 2: Use the average constraints.

Let $a_{X}, a_{Y}, a_{Z}$ be the average ratings for X, Y, Z.

Total sum of ratings = $8 \times 4 = 32$ .

Clue 6: $a_{Y} = 2 a_{X}$ and $a_{Z} = a_{Y} - 2 = 2 a_{X} - 2$ .

Sum equation: $n_{X} a_{X} + n_{Y} a_{Y} + n_{Z} a_{Z} = 32$ .

Substitute $n_{Y} = 2$ : $n_{X} a_{X} + 2 (2 a_{X}) + n_{Z} (2 a_{X} - 2) = 32$ .

Simplify: $(n_{X} + 4 + 2 n_{Z}) a_{X} - 2 n_{Z} = 32$ .

Step 3: Test the possible values for $n_{X}$ and $n_{Z}$ .

Case 1: $n_{X} = 3, n_{Z} = 3$ .

$(3 + 4 + 6) a_{X} - 6 = 32 \Rightarrow 13 a_{X} = 38$ . No clean solution. Invalid.

Case 2: $n_{X} = 4, n_{Z} = 2$ .

$(4 + 4 + 4) a_{X} - 4 = 32 \Rightarrow 12 a_{X} = 36 \Rightarrow a_{X} = 3$ .

This gives integer averages: $a_{X} = 3, a_{Y} = 6, a_{Z} = 4$ . Valid.

Step 4: Conclusion.

The only valid configuration is $n_{X} = 4, n_{Y} = 2, n_{Z} = 2$ .

Zara trained exactly 2 players.

Answer: B

Question 4 · Data Interpretation and Logical Reasoning MCQ

Common Description: Instructions [39 - 44]
Ravi works in an online food-delivery company. After each delivery, customers rate Ravi on each of four parameters - Behaviour, Packaging, Hygiene, and Timeliness, on a scale from 1 to 9. If the total of the four rating points is 25 or more, then Ravi gets a bonus of ₹20 for that delivery. Additionally, a customer may or may not give Ravi a tip. If the customer gives a tip, it is either ₹30 or ₹50.
One day, Ravi made four deliveries - one to each of Atal, Bihari, Chirag, and Deepak, and received a total of ₹120 in bonus and tips. He did not get both a bonus and a tip from the same customer.
The following additional facts are also known.
1. In Timeliness, Ravi received a total of 21 points, and three of the customers gave him the same rating points in this parameter. Atal gave higher rating points than Bihari and Chirag in this parameter.
2. Ravi received distinct rating points in Packaging from the four customers adding up to 29 points. Similarly, Ravi received distinct rating points in Hygiene from the four customers adding up to 26 points.
3. Chirag gave the same rating points for Packaging and Hygiene.
4. Among the four customers, Bihari gave the highest rating points in Packaging, and Chirag gave the highest rating points in Hygiene.
5. Everyone rated Ravi between 5 and 7 in Behaviour. Unique maximum and minimum ratings in this parameter were given by Atal and Deepak respectively.
6. If the customers are ranked based on ratings given by them in individual parameters, then Atal’s rank based on Packaging is the same as that based on Hygiene. This is also true for Deepak. What rating did Deepak give on Packaging?

A.
7
B.
8
C.
5
D.
6

Correct Answer:

A

Step-by-Step Solution

Key idea: This is an evaluation-matrix cell-finding question. It is recognisable because a specific cell value is asked after several column, rank, and money constraints.

Step 1: Fix Behaviour.

Behaviour ratings are between 5 and 7. Atal gave the unique maximum and Deepak gave the unique minimum.

Therefore Atal = 7, Deepak = 5, and Bihari and Chirag both get 6.

Step 2: Fix Packaging values.

Packaging has four distinct ratings from 1 to 9 summing to 29.

The only possible distinct set is ${9, 8, 7, 5}$ .

Bihari gave the highest Packaging rating, so Bihari = 9.

Step 3: Fix Hygiene values using Chirag.

Chirag gave the same rating in Packaging and Hygiene, and Chirag gave the highest Hygiene rating.

Chirag cannot be 9 in Packaging because Bihari is 9.

If Chirag's common value were 7 or lower, the maximum possible distinct Hygiene sum would be too small to reach 26.

Therefore Chirag = 8 in both Packaging and Hygiene.

Hygiene values must be ${8, 7, 6, 5}$ .

Step 4: Use ranks to create two cases.

Packaging ranks: Bihari rank 1 with 9, Chirag rank 2 with 8.

Hygiene ranks: Chirag rank 1 with 8.

Atal and Deepak must have the same rank in Packaging and Hygiene.

This forces Bihari to be rank 2 in Hygiene, so Bihari = 7 in Hygiene.

The remaining values give two cases for Atal and Deepak:

Case I: Atal Packaging 7, Hygiene 6; Deepak Packaging 5, Hygiene 5.

Case II: Atal Packaging 5, Hygiene 5; Deepak Packaging 7, Hygiene 6.

Step 5: Use Timeliness and money to select the valid case.

Timeliness can only be ${9, 4, 4, 4}$ or ${6, 5, 5, 5}$ to sum to 21 with three identical values, and Atal > Bihari, Chirag.

Base sums (Beh + Pkg + Hyg):

In Case I: Atal = 20, Bihari = 22, Chirag = 22, Deepak = 15.

In Case II: Atal = 17, Bihari = 22, Chirag = 22, Deepak = 18.

Bihari and Chirag will always reach $\geq 25$ and get the ₹20 bonus.

Total money is ₹120. Since Bihari and Chirag get bonuses (₹40), the remaining ₹80 must come from tips (₹30 + ₹50).

This means exactly two customers gave tips, and they cannot be Bihari or Chirag.

Thus, Atal and Deepak must be the tippers, meaning their final totals must be $< 25$ .

In Case I, Atal's base is 20. Even with the minimum Timeliness of 4, his total is 24 (valid). But if Timeliness is ${6, 5, 5, 5}$ , Atal gets 6, total 26 (bonus, invalid).

In Case II, Atal's base is 17. With Timeliness ${6, 5, 5, 5}$ , Atal gets 6, total 23 (valid). Deepak's base is 18, gets 5, total 23 (valid).

So Case II is the only valid configuration.

In Case II, Deepak's Packaging rating is 7.

Answer: A

Question 5 · Data Interpretation and Logical Reasoning MCQ

Common Description: Instructions [33 - 38]
10 players - P1, P2, ... , P10 - competed in an international javelin throw event. The number (after P) of a player reflects his rank at the beginning of the event, with rank 1 going to the topmost player. There were two phases in the event with the first phase consisting of rounds 1, 2, and 3, and the second phase consisting of rounds 4, 5, and 6. A throw is measured in terms of the distance it covers (in meters, up to one decimal point accuracy), only if the throw is a ‘valid’ one. For an invalid throw, the distance is taken as zero. A player’s score at the end of a round is the maximum distance of all his throws up to that round. Players are re-ranked after every round based on their current scores. In case of a tie in scores, the player with a prevailing higher rank retains the higher rank. This ranking determines the order in which the players go for their throws in the next round.
In each of the rounds in the first phase, the players throw in increasing order of their latest rank, i.e. the player ranked 1 at that point throws first, followed by the player ranked 2 at that point and so on. The top six players at the end of the first phase qualify for the second phase. In each of the rounds in the second phase, the players throw in decreasing order of their latest rank i.e. the player ranked 6 at that point throws first, followed by the player ranked 5 at that point and so on. The players ranked 1, 2, and 3 at the end of the sixth round receive gold, silver, and bronze medals respectively.
All the valid throws of the event were of distinct distances (as per stated measurement accuracy). The tables below show distances (in meters) covered by all valid throws in the first and the third round in the event.
Distances covered by all the valid throws in the first round

Player	Distance(in m)
P1	82.9
P3	81.5
P5	86.4
P6	82.5
P7	87.2
P9	84.1

Distances covered by all the valid throws in the third round

Player	Distance(in m)
P1	88.6
P3	79.0
P9	81.4

The following facts are also known.
i. Among the throws in the second round, only the last two were valid. Both the throws enabled these players to qualify for the second phase, with one of them qualifying with the least score. None of these players won any medal.
ii. If a player throws first in a round AND he was also the last (among the players in the current round) to throw in the previous round, then the player is said to get a double. Two players got a double.
iii. In each round of the second phase, exactly one player improved his score. Each of these improvements was by the same amount.
iv. The gold and bronze medalists improved their scores in the fifth and the sixth rounds respectively. One medal winner improved his score in the fourth round.
v. The difference between the final scores of the gold medalist and the silver medalist, as well as the difference between the final scores of the silver medalist and the bronze medalist was 1.0 m. Who won the silver medal?

A.
P5
B.
P7
C.
P9
D.
P1

Correct Answer:

D

Step-by-Step Solution

Key idea: This is a logical-deduction question built on an arithmetic structure, recognisable because the final medal scores have a fixed gap (1.0 m) and every improvement is by the same amount $d$ , so the medal scores form a rigid arithmetic template.

Step 1: Establish the medalist pool and baselines.

Qualifiers for Phase 2: P1, P7, P5, P9, P10, P8.

Clue (i) says P8 and P10 win no medal.

Scores at end of R3: P1 = 88.6, P7 = 87.2, P5 = 86.4, P9 = 84.1.

P9 is too far behind. Medalists come from {P1, P7, P5}.

Step 2: Model improvements. Let common improvement = $d$ .

Clue (iii): Exactly one player improves per round in Phase 2 (R4, R5, R6), each by $d$ .

Clue (iv): Gold improved in R5, Bronze improved in R6, one medal winner improved in R4.

Three improvement slots: R4 = some medalist, R5 = Gold, R6 = Bronze.

Step 3: Apply the gap condition.

$G_{f ina l} - S_{f ina l} = 1.0$ and $S_{f ina l} - B_{f ina l} = 1.0$ .

Step 4: Test Case - Silver = P1 (base 88.6), Silver does NOT improve.

Then $G_{f ina l} = 89.6$ , $B_{f ina l} = 87.6$ .

Gold must reach 89.6. Try Gold = P7 (base 87.2): needs $+ 2.4$ .

If Gold improves in R4 and R5: $2 d = 2.4$ , so $d = 1.2$ .

Bronze must reach 87.6. Try Bronze = P5 (base 86.4): needs $+ 1.2 = d$ . Bronze improves in R6 only.

This perfectly matches all conditions: Gold (P7) improves in R4, R5. Bronze (P5) improves in R6. Silver (P1) does not improve.

Thus, P1 is the Silver medalist.

Answer: D

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