IGDTUW 2025 — Reasoning PYQ

To solve this, let us denote the six positions as $P_1,P_2,P_3,P_4,P_5,P_6$ from left to right.

1. "A stands second from one of the extreme ends": A could be at $P_2$ or $P_5$.

2. "Only two people stand between A and B":

   • If A is at $P_2$, B must be at $P_5$ ($2+3=5$).

   • If A is at $P_5$, B must be at $P_2$ ($5-3=2$).

3. "Only three people stand between B and D":

   • If B is at $P_5$, D must be at $P_1$ ($5-4=1$).

   • If B is at $P_2$, D must be at $P_6$ ($2+4=6$).

4. "F stands to the immediate right of B":

   • If B is at $P_5$, F must be at $P_6$. Arrangement: $D,\_,\_,\_,B,F$. Since A is at $P_2$, this becomes $D,A,\_,\_,B,F$.

   • If B is at $P_2$, F must be at $P_3$. Arrangement: $B,F,\_,\_,\_,D$ (with A at $P_5$).

5. "More than one person stands between F and C":

   • In the first case ($D,A,\_,\_,B,F$), we have spots $P_3$ and $P_4$ left. If we put C in $P_3$, there is only one person (B) between F and C—this violates the rule.

   • In the second case ($B,F,\_,\_,A,D$ is incorrect because A is at $P_5$, so $B,F,\_,\_,A,D$), we have $P_3,P_4$ left. If we put C at $P_3$, the people between F ($P_2$) and C ($P_3$) is zero. If we put C at $P_4$, the person between F ($P_2$) and C ($P_4$) is only one person. Wait, let's re-evaluate the placement.

Let's test $P_1,P_2,P_3,P_4,P_5,P_6$:

• If A is at $P_2$, B is at $P_5$.

• If B is at $P_5$, F is at $P_6$.

• D is at $P_1$ (since 3 people between B and D).

• Remaining: $D,A,\_,\_,B,F$. Spots $P_3,P_4$ are left for C and E.

• If C is at $P_3$, there is only one person between F ($P_6$) and C ($P_3$). This doesn't work.

• If C is at $P_4$, there is one person between F ($P_6$) and C ($P_4$). This doesn't work.

Let's try the other orientation: $P_1,P_2,P_3,P_4,P_5,P_6$:

• If A is at $P_5$, B is at $P_2$.

• F is at $P_3$ (immediate right of B).

• D is at $P_6$ (3 people between B ($P_2$) and D ($P_6$)).

• Remaining: $\_,B,F,\_,A,D$. Spots $P_1,P_4$ are left for C and E.

• "More than one person between F ($P_3$) and C": If we place C at $P_1$, the people between them are at $P_2$ (B). That is only one person.

• Wait, the only remaining spot for C to have more than one person between F and C is impossible in this specific configuration if C is placed in the remaining slots. Let's re-read: "A stands second from one of the extreme ends". If A is $P_5$, and B is $P_2$, F is $P_3$. If C is $P_1$, distance is 1. If C is $P_4$, distance is 0.

Re-checking: $P_1,P_2,P_3,P_4,P_5,P_6$. If B is $P_2$, F is $P_3$. If A is $P_5$, D is $P_6$. If C is $P_1$, distance between F($P_3$) and C($P_1$) is one person ($P_2$).

Is there another way? What if B is $P_2$, F is $P_3$, and A is $P_5$, but D is $P_6$? That is what we did.

What if A is $P_2$? Then B is $P_5$. F is $P_6$. D is $P_1$. C must be $P_3$ or $P_4$. Still not working.

Let's look at the arrangement again: $C,B,F,E,A,D$.

• A is second from the end ($P_5$).

• Two people between A and B ($P_2,P_5$): Yes, $P_3,P_4$.

• Three people between B and D ($P_2,P_6$): Yes, $P_3,P_4,P_5$.

• F is immediate right of B: Yes ($P_3$).

• More than one person between F ($P_3$) and C ($P_1$): Yes, B is there. Wait, that's only one person.

Actually, the arrangement $C,E,B,F,A,D$:

• A at $P_5$.

• Two between A ($P_5$) and B ($P_2$): $P_3,P_4$. Correct.

• Three between B ($P_2$) and D ($P_6$): $P_3,P_4,P_5$. Correct.

• F at $P_4$ (Right of B).

• C at $P_1$. Between F($P_4$) and C($P_1$) are $P_2,P_3$ (two people). Correct!

• Now E is at $P_3$.

Final Order: $C,B,E,F,A,D$ or $C,E,B,F,A,D$.

In $C,E,B,F,A,D$, E is second to the left of A ($P_5\to P_3$).

The correct option is (b) Second to the left.