CATSprint

CAT 2025Slot 2DILRQues & Sol

Logical ReasoningHard

Data Set

There are six spherical balls, B1, B2, B3, B4, B5, and B6, and four circular hoops H1, H2, H3, and H4.

Each ball was tested on each hoop once, by attempting to pass the ball through the hoop. If the diameter of a ball is not larger than the diameter of the hoop, the ball passes through the hoop and makes a "ping". Any ball having a diameter larger than that of the hoop gets stuck on that hoop and does not make a ping.

The following additional information is known:
1. B1 and B6 each made a ping on H4, but B5 did not.
2. B4 made a ping on H3, but B1 did not.
3. All balls, except B3, made pings on H1.
4. None of the balls, except B2, made a ping on H2.

Question 1

What was the total number of pings made by B1, B2, and B3?

Solution

Setup: We build a master ordering of ball and hoop diameters by translating each clue into a size inequality. Once the ordering is established, we count pings for B1, B2, and B3 individually — a ping happens only when a ball's diameter ≤ hoop's diameter.

Steps:

  1. Clue 1 → B1, B6 passed H4; B5 did not → (B1, B6) < H4 < B5.
  2. Clue 2 → B4 passed H3; B1 did not → B4 < H3 < B1.
  3. Clue 3 → All except B3 passed H1 → (B1, B2, B4, B5, B6) < H1 < B3.
  4. Clue 4 → Only B2 passed H2 → B2 < H2 < (B1, B3, B4, B5, B6).
  5. Combining clues 1 & 2: B4 < H3 < B1 < H4 < B5. Merging all clues: B2 < H2 < B4 < H3 < B1 < H4 < B5 < H1 < B3. B6 fits either between H2 and H3, or between H3 and H4 — but this ambiguity does NOT affect B1, B2, or B3.
  6. Pings by B1: B1 < H4 ✓ and B1 < H1 ✓; B1 > H3 ✗ and B1 > H2 ✗ → 2 pings (H4, H1).
  7. Pings by B2: B2 < H2 < H3 < H4 < H1, so B2 passes all four hoops → 4 pings (H1, H2, H3, H4).
  8. Pings by B3: B3 > H1 > all other hoops, so B3 passes none → 0 pings.
  9. Total = 2 + 4 + 0 = 6.

Final Answer: 6

Question 2

Which of the following statements about the relative sizes of the balls is NOT NECESSARILY true?

Solution

Setup: After establishing the master diameter ordering from the four clues, we identify which ball's position is ambiguous and then test each option against all possible orderings.

Steps:

  1. Master ordering (fixed part): B2 < H2 < B4 < H3 < B1 < H4 < B5 < H1 < B3.
  2. B6's position is ambiguous: B6 passed H4 but we don't know its relation to H3. Two cases exist:
    • Case 1: B2 < H2 < (B4, B6) < H3 < B1 < H4 < B5 < H1 < B3 → B6 < H3 < B1, so B6 < B1.
    • Case 2: B2 < H2 < B4 < H3 < (B1, B6) < H4 < B5 < H1 < B3 → B6 could be > B1 or < B1 (both are between H3 and H4).
  3. Check Option A (B4 < B5 < B3): In both cases, B4 < H3 < B1 < H4 < B5 < H1 < B3, so B4 < B5 < B3 is always true.
  4. Check Option B (B2 < B1 < B5): B2 is smallest; B1 < H4 < B5 — always true.
  5. Check Option C (B1 < B6 < B3): In Case 1, B6 < B1, so B1 < B6 is FALSE. This statement is NOT necessarily true.
  6. Check Option D (B1 < B5 < B3): B1 < H4 < B5 < H1 < B3 in both cases — always true.

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

Question 3

Which of the following statements about the relative sizes of the hoops is true?

Solution

Setup: The relative sizes of the hoops can be directly read off the master diameter ordering established from the four clues. We simply extract the hoop positions from that sequence.

Steps:

  1. Master ordering: B2 < H2 < B4 < H3 < B1 < H4 < B5 < H1 < B3.
  2. Hoop positions in the sequence: H2 appears first among hoops, then H3, then H4, then H1 — giving H2 < H3 < H4 < H1.
  3. Verify against clues:
    • H1 is the largest hoop (only B3 can't pass it → H1 is just smaller than B3 but larger than B5).
    • H2 is the smallest hoop (only B2, the smallest ball, can pass it).
    • H3 is between H2 and H4 (B4 passes H3 but not H4; B1 passes H4 but not H3 — consistent).
  4. Match with options: H2 < H3 < H4 < H1 matches Option A exactly. All other options contradict the established ordering.

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

Question 4

What BEST can be said about the total number of pings from all the tests undertaken?

Solution

Setup: Since B6's position in the diameter ordering has two valid placements, the total ping count depends on which case holds. We calculate the total pings in each case separately and report both possible values.

Steps:

  1. Fixed ordering (excluding B6): B2 < H2 < B4 < H3 < B1 < H4 < B5 < H1 < B3.
  2. B6 has two possible placements:
    • Case 1: B6 sits between H2 and H3 (i.e., B6 < H3 < H4 < H1) → B6 pings H3, H4, H1 = 3 pings.
    • Case 2: B6 sits between H3 and H4 (i.e., H3 < B6 < H4 < H1) → B6 pings H4, H1 = 2 pings.
  3. Pings for fixed balls (same in both cases):
    • B1: passes H4, H1 → 2 pings.
    • B2: passes H2, H3, H4, H1 → 4 pings.
    • B3: passes none → 0 pings.
    • B4: passes H3, H4, H1 → 3 pings.
    • B5: passes H1 only → 1 ping.
  4. Total in Case 1: 2 + 4 + 0 + 3 + 1 + 3 (B6) = 13 pings.
  5. Total in Case 2: 2 + 4 + 0 + 3 + 1 + 2 (B6) = 12 pings.
  6. Therefore, the total number of pings is either 12 or 13 — Option A is the most precise answer. Options B, C, and D are either incorrect or less precise.

Final Answer: Option A (12 or 13)