(i) the first game was the only game where two students of the same class played against each other,
(ii) the students of Class 5 won more games than the students of Class 4, and
(iii) the boys won two games and the girls won one game.
The student who did not lose any game is __________.
Problem Setup
We have four students:
- Class 5: Rishi (Boy), Swathi (Girl)
- Class 4: Pavan (Boy), Tanvi (Girl)
Game Rules:
- 3 games played sequentially.
- The loser of a game is eliminated from further participation.
Observations:
- (i) Only the first game featured players from the same class.
- (ii) Class 5 students won more games than Class 4 students (implying 2 wins for Class 5, 1 win for Class 4).
- (iii) Boys won 2 games, Girls won 1 game.
Objective: Identify the student who did not lose any game.
Logical Deduction
There are 3 games and 3 losers. The student who did not lose any game must be the winner of the final game.
Step 1: Analyze Game 1 Pairings (Condition i)
Condition (i) states that only the first game had students from the same class. This means Game 2 and Game 3 must have students from different classes.
Therefore, Game 1 must be either:
- Rishi (Class 5) vs Swathi (Class 5)
- Pavan (Class 4) vs Tanvi (Class 4)
Step 2: Analyze Win Conditions (Conditions ii & iii)
Total Wins:
- Class 5 wins = 2
- Class 4 wins = 1
- Boys wins = 2
- Girls wins = 1
Step 3: Evaluate Scenarios
Scenario A: Game 1 is Class 5 vs Class 5
Assume Game 1 is Rishi (C5, B) vs Swathi (C5, G).
- Game 1: Rishi vs Swathi.
- Let Rishi win. Swathi loses (eliminated).
- Wins: C5=1, Boys=1. Remaining players: Rishi (C5, B), Pavan (C4, B), Tanvi (C4, G).
- Game 2: Must be different classes. Let's assume Rishi (C5, B) vs Pavan (C4, B).
- Let Rishi win. Pavan loses (eliminated).
- Wins: C5=1, Boys=1. Cumulative Wins: C5=2, C4=0, Boys=2, Girls=0.
- Remaining players: Rishi (C5, B), Tanvi (C4, G).
- Game 3: Must be different classes. Rishi (C5, B) vs Tanvi (C4, G).
- To satisfy the final win counts (C5=2, C4=1, Boys=2, Girls=1), Tanvi must win this game.
- Tanvi wins. Rishi loses (eliminated).
- Wins: C4=1, Girls=1.
Final Check for Scenario A:
- Total Wins: C5=2 (Rishi G1, G2), C4=1 (Tanvi G3), Boys=2 (Rishi G1, G2), Girls=1 (Tanvi G3). Conditions (ii) and (iii) are met.
- Class Pairings: G1 (C5 vs C5 - Same), G2 (C5 vs C4 - Different), G3 (C5 vs C4 - Different). Condition (i) is met.
- Losers: Swathi (G1), Pavan (G2), Rishi (G3).
- Undefeated Student: Tanvi.
This scenario fits all conditions.
Scenario B: Game 1 is Class 4 vs Class 4
Assume Game 1 is Pavan (C4, B) vs Tanvi (C4, G).
If Tanvi wins G1 (C4=1, G=1), players are Tanvi(C4,G), Rishi(C5,B), Swathi(C5,G). Game 2 could be Rishi vs Swathi (C5 vs C5). This violates condition (i) because Game 2 would also be same-class.
If Pavan wins G1 (C4=1, B=1), players are Pavan(C4,B), Rishi(C5,B), Swathi(C5,G). Game 2 could be Rishi vs Swathi (C5 vs C5). This also violates condition (i).
Therefore, Scenario B is impossible.
Conclusion
Based on the logical deduction, the only scenario fitting all conditions is Scenario A, where Tanvi wins the final game and does not lose any game.
Important Questions from Puzzles
In the 4 x 4 array shown below, each cell of the first three rows has either a cross (X) or a number.
1 X 4 3 X 5 5 4 3 X 6 X
The number in a cell represents the count of the immediate neighboring cells (left, right, top, bottom, diagonals) NOT having a cross (X). Given that the last row has no crosses (X), the sum of the four numbers to be filled in the last row is- Three children P, Q, R and two grown-ups X, Y play a badminton doubles tournament. X and Y are parents to two of the children playing. The child of X is not the same as the child of Y. Exactly one of the children does not have a parent playing in the tournament. The following rules are followed:
(i) A parent and his/her child cannot be on the same team.
(ii) A match can feature at most one parent and his/her child, that is, a maximum of one parent-child pair can play in a match.
The following matches were played:TEAM 1 TEAM 2 MATCH 1 P and X Q and R MATCH 2 P and R X and Y MATCH 3 R and X Q and Y
Which one of the following options is correct? - In the 2020 summer Olympics’ Javelin throw finals, Neeraj Chopra exhibited a spectacular performance to win the gold medal. The silver medal was won by Jakub Vadlejch and the bronze medal was won by Vitezlav Vesely. There were six rounds of throws with each athlete having one throw per round. The best of all the throws of each athlete is considered for the medal. Following were the observations about the throws:
i. The first and second rounds were dominated by Neeraj Chopra with a gold medal performance in his second throw, while the other two athletes did not have any medal winning throws in these rounds.
ii. The throws in the last round by both Jakub Vadlejch and Vitezlav Vesely were fouls and were not considered for scoring.
iii. After four rounds, Vitezlav Vesely was in the second position and could not improve upon his best throw in the succeeding rounds.
iv. In the fourth round, the throw by Jakub Vadlejch was the best in that round.
In which round did Vitezlav Vesely have his best throw? Students applying for hostel rooms are allotted rooms in order of seniority. Students already staying in a room will move if they get a room in their preferred list. Preferences of lower ranked applicants are ignored during allocation.
Given the data below, which room will Ajit stay in?Names Student seniority Current room Room preference list Amar 1 P R, S, Q Akbar 2 None R, S Anthony 3 Q P Ajit 4 S Q, P, R The diagram below shows a river system consisting of 7 segments, marked P, Q, R, S, T, U, and V. It splits the land into 5 zones, marked Z1, Z2, Z3, Z4, and Z5. We need to connect these zones using the least number of bridges. Out of the following options, which one is correct?
Note: The figure shown is representative.
