CAT 2025

Slot 1

DILR

Question & SolutionQues & Sol

Setup: We convert each clue into a diameter inequality and chain them together to build a single master ordering of all balls and hoops. Once this ordering is established, counting pings for B1, B2, and B3 is mechanical — a ball pings a hoop if and only if the ball's diameter is less than or equal to the hoop's diameter, i.e., the ball appears to the left of the hoop in the master sequence.

Steps:

1. Clue 1 — B1 and B6 passed H4, but B5 did not → diameters satisfy: B1 < H4, B6 < H4, and B5 > H4, giving us B1, B6 < H4 < B5.
2. Clue 2 — B4 passed H3, but B1 did not → B4 < H3 and B1 > H3, giving us B4 < H3 < B1.
3. Clue 3 — All balls except B3 passed H1 → B1, B2, B4, B5, B6 < H1 and B3 > H1, giving us B5 < H1 < B3.
4. Clue 4 — Only B2 passed H2 → B2 < H2 and all others (B1, B3, B4, B5, B6) > H2, giving us B2 < H2 < B1, B4.
5. Chaining all inequalities: From Clues 2 and 4: B2 < H2 < B4 < H3 < B1. From Clue 1: B1 < H4 < B5. From Clue 3: B5 < H1 < B3. Full master sequence: B2 < H2 < B4 < H3 < B1 < H4 < B5 < H1 < B3. (B6 is somewhere between H2 and H4, but this does not affect B1, B2, or B3.)
6. Pings by B1: B1 < H4 ✓ and B1 < H1 ✓; B1 > H3 ✗ and B1 > H2 ✗ → 2 pings (on H4 and H1).
7. Pings by B2: B2 is the smallest element in the sequence — B2 < H2 < H3 < H4 < H1 — so B2 passes all four hoops → 4 pings (on H1, H2, H3, H4).
8. Pings by B3: B3 is larger than H1, which is the largest hoop, so B3 cannot pass any hoop → 0 pings.
9. Total pings = 2 + 4 + 0 = 6.

Final Answer: 6

Setup: After building the master diameter ordering from the four clues, we identify which ball's position is ambiguous (not fully pinned). We then test each option to see whether it holds in ALL valid configurations or only in some — the option that fails in at least one valid configuration is the answer.

Steps:

1. Master ordering (fixed part): B2 < H2 < B4 < H3 < B1 < H4 < B5 < H1 < B3. Every ball except B6 has a fully determined position in the chain.
2. B6's ambiguity: We know B6 < H4 (from Clue 1) and B6 > H2 (from Clue 4, since only B2 passed H2). So B6 must lie somewhere in the range (H2, H4). This gives two distinct cases:
   • Case A: H2 < B6 < H3, meaning B6 < H3, so B6 is also smaller than B1 → B6 < B4 or H2 < B6 < H3 < B1.
   • Case B: H3 < B6 < H4, meaning B6 is between H3 and H4 — B6 could be larger or smaller than B1 within this window, so B6 > H3 and specifically B6 may be greater than B1 or less than B1 (both B1 and B6 lie between H3 and H4).
3. Check Option A (B4 < B5 < B3): From the fixed sequence: B4 < H3 < B1 < H4 < B5 < H1 < B3. So B4 < B5 < B3 holds in both cases. Always true.
4. Check Option B (B2 < B1 < B5): B2 is the smallest ball; B1 < H4 < B5 is fixed. Always true.
5. Check Option C (B1 < B6 < B3): In Case A, B6 < H3 < B1, so B1 < B6 is false — this statement does NOT hold in Case A. Therefore, B1 < B6 < B3 is NOT necessarily true.
6. Check Option D (B1 < B5 < B3): B1 < H4 < B5 < H1 < B3 is fixed in both cases. Always true.
7. Only Option C fails in at least one valid configuration, making it the statement that is NOT necessarily true.

Final Answer: Option C (B1 < B6 < B3)

Setup: The sizes of the hoops relative to each other are directly encoded in the master diameter ordering built from the four clues. Once the full sequence is established, we simply extract the positions of the four hoops within that sequence to determine their size ranking.

Steps:

1. Reconstruct the master ordering: Chaining all four clues gives: B2 < H2 < B4 < H3 < B1 < H4 < B5 < H1 < B3.
2. Read off hoop positions: Within the master sequence, the hoops appear in this left-to-right order: H2, then H3, then H4, then H1. Since leftmost means smallest diameter, the hoop size ranking is H2 < H3 < H4 < H1.
3. Cross-verify each hoop's role: H2 is the smallest hoop — only B2 (the smallest ball) can pass it (consistent with Clue 4). H1 is the largest hoop — every ball except B3 (the largest ball) can pass it (consistent with Clue 3). H3 is intermediate — B4 passes it but B1 does not (consistent with Clue 2). H4 is larger than H3 — B1 and B6 pass it but B5 does not (consistent with Clue 1).
4. Match with options: H2 < H3 < H4 < H1 corresponds exactly to Option A. Options B and C incorrectly place H1 as the smallest hoop, and Option D incorrectly swaps H3 and H4.

Final Answer: Option A (H2 < H3 < H4 < H1)

Setup: Since B6's exact position in the master ordering is ambiguous (it can lie in one of two windows), the total ping count is not uniquely determined. We calculate pings for all balls whose positions are fixed, then handle B6's two cases separately and sum up both possible totals.

Steps:

1. Fixed master ordering (excluding B6): B2 < H2 < B4 < H3 < B1 < H4 < B5 < H1 < B3. B6 lies somewhere between H2 and H4 (since B6 > H2 from Clue 4 and B6 < H4 from Clue 1).
2. B6's two valid placements:
   • Case 1: H2 < B6 < H3 → B6 passes H3, H4, and H1 → 3 pings for B6.
   • Case 2: H3 < B6 < H4 → B6 passes H4 and H1 only → 2 pings for B6.
3. Pings for B1 (fixed): B1 passes H4 and H1 (B1 > H2 and H3, so no ping there) → 2 pings.
4. Pings for B2 (fixed): B2 is smaller than all hoops → passes H2, H3, H4, H1 → 4 pings.
5. Pings for B3 (fixed): B3 is larger than all hoops → passes none → 0 pings.
6. Pings for B4 (fixed): B4 passes H3, H4, H1 (B4 < H3 < H4 < H1; B4 > H2) → 3 pings.
7. Pings for B5 (fixed): B5 passes only H1 (B5 > H4, H3, H2) → 1 ping.
8. Fixed subtotal (B1+B2+B3+B4+B5) = 2 + 4 + 0 + 3 + 1 = 10 pings.
9. Total in Case 1: 10 + 3 = 13 pings. Total in Case 2: 10 + 2 = 12 pings.
10. The total is either 12 or 13. Option A (12 or 13) is the most precise correct answer. Option B (13 or 14) is wrong since 14 is impossible. Option C (12 or 13 or 14) is imprecise. Option D (at least 9) is too vague.

Final Answer: Option A (12 or 13)