GATE CS 2021 Set 2 Q10 — General Aptitude

Let us arrange the six students in descending order of their heights from 1 (tallest) to 6 (shortest):

1 (Tallest) > 2 > 3 > 4 > 5 > 6 (Shortest)

Using the given observations, we can determine their positions step-by-step:

1. Observation II: $Q$ is the shortest of all.
   $$\text{Position 6} = Q$$
2. Observation III: $U$ is taller than only one student (which must be $Q$).
   $$\text{Position 5} = U$$
3. Observation I & IV: $T$ is taller than $S$, and $S$ is taller than $R$. This gives the relative order: $T > S > R$.
   We also know $T$ is not the tallest (so $T$ cannot be at Position 1).
   The remaining positions for $T, S,$ and $R$ are 1, 2, 3, and 4. Since $T$ cannot be at 1, the only possible assignment to maintain the order $T > S > R$ is:
   $$\text{Position 2} = T, \quad \text{Position 3} = S, \quad \text{Position 4} = R$$
4. The only remaining position (Position 1) must be occupied by $P$:
   $$\text{Position 1} = P$$

Thus, the unique height order from tallest to shortest is:

P > T > S > R > U > Q

Now, let's answer the question:

• The number of students taller than $R$ (who is at position 4) is 3 (namely $P, T,$ and $S$).
• We need to find the student who has exactly 3 students shorter than them.
• Looking at our order, the students shorter than $S$ are $R, U,$ and $Q$ (exactly 3 students).

Therefore, the blank should be filled with S.