Question

Rishi and Swathi are students of Class 5. Pavan and Tanvi are students of Class 4. Rishi and Pavan are boys. Swathi and Tanvi are girls. The four students played a total of three games of chess. The games were played one after another. A player who lost a game did not participate in any more games. It was observed that:

1. the first game was the only game where two students of the same class played against each other,
2. the students of Class 5 won more games than the students of Class 4, and
3. the boys won two games and the girls won one game.

The student who did not lose any game is __________.

• A. Pavan
• B. Rishi
• C. Swathi
• D. Tanvi

Hint:

With four players and three knockout-style games, the win totals split as Class 5 = 2 wins, Class 4 = 1 win; work out who can hold those wins without creating a second same-class game.

Answer 1

The Correct Option is D

Step 1: Set up win-count variables.
Let $R, S, P, T$ be the number of games won by Rishi, Swathi, Pavan and Tanvi. Three games are played in total, so $R + S + P + T = 3$. Class 5 (Rishi, Swathi) wins more than Class 4 (Pavan, Tanvi), and with only 3 games to split, the only way to keep Class 5 strictly ahead is $R + S = 2$ and $P + T = 1$. Also, boys (Rishi, Pavan) win 2 games total and girls (Swathi, Tanvi) win 1: $R + P = 2$, $S + T = 1$.

Step 2: Solve the equations.
From $R + S = 2$ and $R + P = 2$, subtracting gives $S = P$. From $P + T = 1$ and $S + T = 1$, subtracting gives $P = S$ again, so this alone does not fix the values yet. Since $P + T = 1$ with $P, T$ non-negative integers, either $P = 0, T = 1$ or $P = 1, T = 0$.
If $P = 1, T = 0$: then $S = P = 1$, and $R = 2 - S = 1$. This gives $(R,S,P,T) = (1,1,1,0)$.
If $P = 0, T = 1$: then $S = P = 0$, and $R = 2 - S = 2$. This gives $(R,S,P,T) = (2,0,0,1)$.
Both satisfy the arithmetic, so the win totals alone give two candidate patterns; the tournament structure is needed to eliminate one.

Step 3: Test (R,S,P,T) = (1,1,1,0) against the structure.
Here Rishi and Swathi both need exactly 1 win each. But in the elimination format, whichever of them loses game 1 (if game 1 is Rishi vs Swathi) is out forever with 0 wins, so they cannot both end up with 1 win, one of them is stuck at 0. This forces game 1 to be Pavan vs Tanvi instead, with Tanvi needing $T = 0$, so Tanvi loses game 1 and Pavan wins it (1 win, matching $P=1$, so Pavan must lose every later game he plays). Pavan then plays game 2 against a Class 5 student (Rishi or Swathi), and must lose it, so that Class 5 student wins game 2. That same Class 5 student then plays game 3 against the last remaining student, who is the other Class 5 student, since both Rishi and Swathi get used up in games 1 and 2, making game 3 a same-class match. This breaks condition (i), which says only game 1 pairs same-class students. So this whole pattern is structurally impossible.

Step 4: Confirm (R,S,P,T) = (2,0,0,1) works.
With $S = 0$ and $P = 0$, both Swathi and Pavan never win a single game, meaning each of them loses the one game they play and is eliminated immediately. Rishi wins 2 games and Tanvi wins 1. Placing Rishi as the winner of games 1 and 2 (beating Swathi, then Pavan) and Tanvi as the winner of game 3 (beating Rishi) satisfies condition (i): game 1 (Rishi vs Swathi) is the only same-class pairing, since games 2 (Rishi vs Pavan) and 3 (Rishi vs Tanvi) are both cross-class. This is consistent, so this is the valid pattern.

Step 5: Read off who never lost.
In this valid pattern, Swathi loses game 1, Pavan loses game 2, and Rishi loses game 3 (after having won games 1 and 2). Tanvi only plays game 3 and wins it, so Tanvi is the only student with zero losses.
\[ \boxed{\text{Tanvi}} \]