Correct Answer: 6

Explanation: From the conditions, B2 is the only ball that pings H2, so B2’s diameter is the smallest. All balls except B3 ping H1, so B3 is the largest ball. From H4, B1 and B6 ping but B5 does not, implying B5 is larger than both B1 and B6, and H4 lies between them in size. From H3, B4 pings but B1 does not, so B4 is smaller than B1, and H3 lies between them. Using all constraints, B1 pings exactly on H1 and H4 (2 pings), B2 pings on all four hoops (4 pings), and B3 pings on none (0 pings). Hence total pings by B1, B2, and B3 = 2 + 4 + 0 = 6.

Steps to fill the table

(Ping means ball diameter ≤ hoop diameter)
(Source: Question statement)

Step 1
All balls except B3 made a ping on H1 → B1, B2, B4, B5, B6 ≤ H1 and B3 > H1
(Source: Clue 3)

Step 2
None of the balls except B2 made a ping on H2 → only B2 ≤ H2, all others > H2
(Source: Clue 4)

From Steps 1 and 2:
H2 < B1, B3, B4, B5, B6 < H1
(Source: Direct comparison)

Step 3
B4 made a ping on H3, but B1 did not → B4 ≤ H3 < B1
(Source: Clue 2)

Step 4
B1 and B6 made pings on H4, but B5 did not → B1 ≤ H4, B6 ≤ H4 < B5
(Source: Clue 1)

This gives B1 < B5 and B6 < B5
(Source: Diameter implication)

Step 5
From earlier steps:
B2 is the smallest ball (only one passing H2)
B3 is the largest ball (fails H1)
Also B4 < B1 < B5

So partial order becomes:
B2 < B4 < B1 < B5 < B3, with B6 between B1 and B5 or between B4 and B1
(Source: Combined deductions)

Step 6
Since B4 passes H3 and B1 fails H3 → H3 < B1
Since B1 passes H4 → B1 ≤ H4
Thus hoops satisfy:
H2 < H3 < H4 < H1
(Source: Steps 1, 3, and 4)

Step 7
Filling all forced pings based on final orderings:

(Source: All constraints applied)

Step 8
Pings required for Q10:
B1 = 2, B2 = 4, B3 = 0 → total = 6
(Source: Step 7)

Step 9
Total guaranteed pings = B1(2) + B2(4) + B3(0) + B5(1) = 7
B4 can have 2 or 3, B6 can have 2 or 3
So total pings = 12 or 13
(Source: Step 7 optional passes)

Final Deductions
Ball order (small → large):
B2 < B4 < B1 < B6 < B5 < B3

Hoop order (small → large):
H2 < H3 < H4 < H1

Correct Answer: 2

Explanation: From the deductions, B2 is the smallest and B3 is the largest ball. We also have B4 < B1 and B1 < B5, giving B4 < B1 < B5 < B3 as a necessary chain. The relative order of B1 and B6, however, is not fixed: both B1 and B6 ping H4 and fail H3, so both lie on the same side of those hoops without a strict comparison between them. Therefore, the statement “B1 < B6 < B3” is not necessarily true, while the other options are consistent with all valid configurations.

Steps to fill the table

(Ping means ball diameter ≤ hoop diameter)
(Source: Question statement)

Step 1
All balls except B3 made a ping on H1 → B1, B2, B4, B5, B6 ≤ H1 and B3 > H1
(Source: Clue 3)

Step 2
None of the balls except B2 made a ping on H2 → only B2 ≤ H2, all others > H2
(Source: Clue 4)

From Steps 1 and 2:
H2 < B1, B3, B4, B5, B6 < H1
(Source: Direct comparison)

Step 3
B4 made a ping on H3, but B1 did not → B4 ≤ H3 < B1
(Source: Clue 2)

Step 4
B1 and B6 made pings on H4, but B5 did not → B1 ≤ H4, B6 ≤ H4 < B5
(Source: Clue 1)

This gives B1 < B5 and B6 < B5
(Source: Diameter implication)

Step 5
From earlier steps:
B2 is the smallest ball (only one passing H2)
B3 is the largest ball (fails H1)
Also B4 < B1 < B5

So partial order becomes:
B2 < B4 < B1 < B5 < B3, with B6 between B1 and B5 or between B4 and B1
(Source: Combined deductions)

Step 6
Since B4 passes H3 and B1 fails H3 → H3 < B1
Since B1 passes H4 → B1 ≤ H4
Thus hoops satisfy:
H2 < H3 < H4 < H1
(Source: Steps 1, 3, and 4)

Step 7
Filling all forced pings based on final orderings:

(Source: All constraints applied)

Step 8
Pings required for Q10:
B1 = 2, B2 = 4, B3 = 0 → total = 6
(Source: Step 7)

Step 9
Total guaranteed pings = B1(2) + B2(4) + B3(0) + B5(1) = 7
B4 can have 2 or 3, B6 can have 2 or 3
So total pings = 12 or 13
(Source: Step 7 optional passes)

Final Deductions
Ball order (small → large):
B2 < B4 < B1 < B6 < B5 < B3

Correct Answer: 1

Explanation: From H2, only B2 pings, so H2 is the smallest hoop. From H1, all except B3 ping, and since B3 is the largest ball, H1 must be the largest hoop. From H3 and H4, we know H3 allows B4 but not B1, while H4 allows B1 and B6 but not B5. This places the hoops in increasing order as H2 < H3 < H4 < H1. Hence option 1 is correct.

Steps to fill the table

(Ping means ball diameter ≤ hoop diameter)
(Source: Question statement)

Step 1
All balls except B3 made a ping on H1 → B1, B2, B4, B5, B6 ≤ H1 and B3 > H1
(Source: Clue 3)

Step 2
None of the balls except B2 made a ping on H2 → only B2 ≤ H2, all others > H2
(Source: Clue 4)

From Steps 1 and 2:
H2 < B1, B3, B4, B5, B6 < H1
(Source: Direct comparison)

Step 3
B4 made a ping on H3, but B1 did not → B4 ≤ H3 < B1
(Source: Clue 2)

Step 4
B1 and B6 made pings on H4, but B5 did not → B1 ≤ H4, B6 ≤ H4 < B5
(Source: Clue 1)

This gives B1 < B5 and B6 < B5
(Source: Diameter implication)

Step 5
From earlier steps:
B2 is the smallest ball (only one passing H2)
B3 is the largest ball (fails H1)
Also B4 < B1 < B5

So partial order becomes:
B2 < B4 < B1 < B5 < B3, with B6 between B1 and B5 or between B4 and B1
(Source: Combined deductions)

Step 6
Since B4 passes H3 and B1 fails H3 → H3 < B1
Since B1 passes H4 → B1 ≤ H4
Thus hoops satisfy:
H2 < H3 < H4 < H1
(Source: Steps 1, 3, and 4)

Step 7
Filling all forced pings based on final orderings:

(Source: All constraints applied)

Step 8
Pings required for Q10:
B1 = 2, B2 = 4, B3 = 0 → total = 6
(Source: Step 7)

Step 9
Total guaranteed pings = B1(2) + B2(4) + B3(0) + B5(1) = 7
B4 can have 2 or 3, B6 can have 2 or 3
So total pings = 12 or 13
(Source: Step 7 optional passes)

Final Deductions
Ball order (small → large):
B2 < B4 < B1 < B6 < B5 < B3

Correct Answer: 3

Explanation: From the full ordering, pings can be counted ball by ball. B1 pings on H1 and H4 (2). B2 pings on all four hoops (4). B3 pings on none (0). B4 pings on H1, H3, and possibly H4 (2 or 3). B5 pings on H1 only (1). B6 pings on H1 and H4, and may or may not ping H3 (2 or 3). Adding these, the total number of pings is either 12 or 13 depending on whether B6 (or B4) fits through H3. No configuration allows fewer than 12 or more than 13. Therefore, the best statement is “12 or 13”.

Steps to fill the table

(Ping means ball diameter ≤ hoop diameter)
(Source: Question statement)

Step 1
All balls except B3 made a ping on H1 → B1, B2, B4, B5, B6 ≤ H1 and B3 > H1
(Source: Clue 3)

Step 2
None of the balls except B2 made a ping on H2 → only B2 ≤ H2, all others > H2
(Source: Clue 4)

From Steps 1 and 2:
H2 < B1, B3, B4, B5, B6 < H1
(Source: Direct comparison)

Step 3
B4 made a ping on H3, but B1 did not → B4 ≤ H3 < B1
(Source: Clue 2)

Step 4
B1 and B6 made pings on H4, but B5 did not → B1 ≤ H4, B6 ≤ H4 < B5
(Source: Clue 1)

This gives B1 < B5 and B6 < B5
(Source: Diameter implication)

Step 5
From earlier steps:
B2 is the smallest ball (only one passing H2)
B3 is the largest ball (fails H1)
Also B4 < B1 < B5

So partial order becomes:
B2 < B4 < B1 < B5 < B3, with B6 between B1 and B5 or between B4 and B1
(Source: Combined deductions)

Step 6
Since B4 passes H3 and B1 fails H3 → H3 < B1
Since B1 passes H4 → B1 ≤ H4
Thus hoops satisfy:
H2 < H3 < H4 < H1
(Source: Steps 1, 3, and 4)

Step 7
Filling all forced pings based on final orderings:

(Source: All constraints applied)

Step 8
Pings required for Q10:
B1 = 2, B2 = 4, B3 = 0 → total = 6
(Source: Step 7)

Step 9
Total guaranteed pings = B1(2) + B2(4) + B3(0) + B5(1) = 7
B4 can have 2 or 3, B6 can have 2 or 3
So total pings = 12 or 13
(Source: Step 7 optional passes)

Final Deductions
Ball order (small → large):
B2 < B4 < B1 < B6 < B5 < B3