Question

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

• A. B4 < B5 < B3
• B. B2 < B1 < B5
• C. B1 < B5 < B3
• D. B1 < B6 < B3
• A. H1 < H4 < H3 < H2
• B. H2 < H3 < H4 < H1
• C. H1 < H3 < H4 < H2
• D. H2 < H4 < H3 < H1
• A. 13 or 14
• B. At least 9
• C. 12 or 13
• D. 12 or 13 or 14

Answer 1

Correct Answer: 6

Approach: Using the size orders from the set — balls \(B3 > B5 > B1 > B4 > B2\), hoops \(H1 > H4 > H3 > H2\) — each ball pings exactly the hoops that are at least as big as it. So just count, for B1, B2, B3, how many hoops sit above the ball.

Step 1: B1. \(B1 \le H1\) (ping), \(B1 \le H4\) (ping, given directly in clue 1), \(B1 > H3\) (no ping, clue 2), \(B1 > H2\) (no ping). So \[ B1 \to 2 \text{ pings}. \]

Step 2: B2. B2 is the smallest ball — it even fits H2, the smallest hoop — so it fits all four hoops. \[ B2 \to 4 \text{ pings}. \]

Step 3: B3. B3 is the largest ball, bigger than even H1, the largest hoop, so it sticks everywhere. \[ B3 \to 0 \text{ pings}. \]

Step 4: Add. \[ 2 + 4 + 0 = 6. \]

Final Answer: 6.

Answer 2

Using the size order. B2 turns out to be the smallest ball of all six, so it pings every hoop: H1 (given), H2 (given, it's the only one), H3 (since it is smaller than B4, which already pings H3), and H4 (since it is smaller than B1, which already pings H4) — 4 pings in total.

B3 turns out to be the largest ball of all six (it is the only one that fails H1, and H1 already admits the biggest of the other five), so it fails every hoop: H1 (given), H2 (only B2 pings it), H3 (since it is bigger than B1, which already fails H3), and H4 (since it is bigger than B5, which already fails H4) — 0 pings.

B1 pings H1 (given) and H4 (given), but fails H2 (only B2 pings H2) and H3 (given) — 2 pings.

Total \(=4+0+2=6\).

Answer: 6.

Answer 3

Approach: A "ping" means the ball passes through the hoop, so ping happens exactly when ball diameter \(\le\) hoop diameter. Convert every test result into a ball-vs-ball inequality, then see which option is the only one not forced.

Step 1 \(-\) read the clues as inequalities. B1, B6 ping H4 but B5 does not: so \(B1, B6 \le H4 < B5\). B4 pings H3 but B1 does not: so \(B4 \le H3 < B1\). Every ball except B3 pings H1: so all others \(\le H1 < B3\) (B3 is the biggest ball). Only B2 pings H2: so \(B2 \le H2 <\) everyone else (B2 is the smallest ball).

Step 2 \(-\) build the ball order. From \(B2\) smallest, \(B4 \le H3 < B1 \le H4 < B5\), and \(B3\) largest, the certain chain is \[ B2 < B4 < B1 < B5 < B3. \] B6 only satisfies \(H2 < B6 \le H4\) (it pings H4, not H2). Crucially, B6 is never compared with H3, so its position relative to B4 and B1 is not fixed \(-\) all we know is \(B6 < B5 < B3\).

Step 3 \(-\) test the options. \(B4 < B5 < B3\): forced (true). \(B2 < B1 < B5\): forced. \(B1 < B5 < B3\): forced. \(B1 < B6 < B3\): the part \(B6 < B3\) is forced, but \(B1 < B6\) is NOT \(-\) B6 could be smaller than B1.

Step 4 \(-\) conclude. Only \(B1 < B6 < B3\) is not necessarily true.

Final answer: B1 < B6 < B3

Answer 4

Step 1: Understanding the Question:
We need to identify which of the given inequalities about the ball sizes cannot be definitively proven from the information given. We will use the size relationships derived in the previous question. 
Step 2: Reviewing Ball Size Deductions: 
- B3 is the largest ball.
- B2 is the smallest ball.
- B5 > H4 and B1 $\le$ H4 $\implies$ B5 > B1.
- B1 > H3 and B4 $\le$ H3 $\implies$ B1 > B4.
- B5 > H4 and B6 $\le$ H4 $\implies$ B5 > B6.
- The definite order is: B3 > B5 > B1 > B4 > B2.
- The position of B6 is uncertain. We only know B5 > B6 and B6 > H2 > B2. The relationship between B6 and B1, and B6 and B4 is not determined by the given rules. 
Step 3: Evaluating the Options: 
- (A) B4 < B5 < B3:
- Is B5 < B3? Yes, B3 is the largest.
- Is B4 < B5? Yes, we established B5 > B1 and B1 > B4, so B5 > B4.
- This statement is necessarily true. 
- (B) B2 < B1 < B5:
- Is B1 < B5? Yes, we established this.
- Is B2 < B1? Yes, B2 is the smallest.
- This statement is necessarily true. 
- (C) B1 < B5 < B3:
- Is B5 < B3? Yes.
- Is B1 < B5? Yes.
- This statement is necessarily true. 
- (D) B1 < B6 < B3:
- Is B6 < B3? Yes, B3 is the largest.
- Is B1 < B6? This is unknown. From Rule 1, we have B1 $\le$ H4 and B6 $\le$ H4. This does not allow us to compare B1 and B6. It is possible that B1 < B6, B1 > B6, or B1 = B6.
- Since we cannot prove B1 < B6, the entire statement is not necessarily true. 
Step 4: Final Answer: 
The relationship between B1 and B6 cannot be determined from the given information. Therefore, the statement "B1 < B6 < B3" is not necessarily true. 
 

Answer 5

Approach: The same ping data that orders the balls also pins down the hoops \(-\) read each clue for what it says about hoop size.

Step 1 \(-\) H2 is smallest. Only B2 pings H2, so H2 is too small for every other ball: \(H2 < B1, B3, B4, B5, B6\). That is the tightest hoop.

Step 2 \(-\) H1 is largest. Every ball except the biggest (B3) pings H1, so H1 clears all of them: H1 is the widest hoop.

Step 3 \(-\) order H3 and H4. B4 pings H3 but B1 does not, so \(B4 \le H3 < B1\). B1 pings H4, so \(B1 \le H4\). Chaining: \(H3 < B1 \le H4\), hence \(H3 < H4\).

Step 4 \(-\) assemble. Smallest to largest: \[ H2 < H3 < H4 < H1. \]

Final answer: H2 < H3 < H4 < H1

Answer 6

Step 1: Understanding the Question:
We need to find the correct ascending or descending order of the hoop sizes based on the deductions from the problem statement.
Step 2: Establishing the Hoop Order:
We will use the relationships between balls and hoops to determine the relative sizes of the hoops.
- From Rule 3, B3 > H1 and B5 $\le$ H1. From Rule 1, B5 > H4.
- Combining these: H1 $\ge$ B5 > H4. Therefore, H1 > H4.
- From Rule 1, B1 $\le$ H4. From Rule 2, B1 > H3.
- Combining these: H4 $\ge$ B1 > H3. Therefore, H4 > H3.
- From Rule 2, B4 $\le$ H3. From Rule 4, B4 > H2.
- Combining these: H3 $\ge$ B4 > H2. Therefore, H3 > H2.
Step 3: Combining the Inequalities and Final Answer:
Putting all the derived inequalities together, we get:
H1 > H4 > H3 > H2.
This can be written in ascending order as:
H2 < H3 < H4 < H1.
This matches option (B).

Answer 7

Approach: Count pings ball by ball using the fixed orders \(B2 < B4 < B1 < B5 < B3\) and \(H2 < H3 < H4 < H1\); the only loose piece is B6 versus H3, which creates a small range.

Step 1 \(-\) count the certain balls (ping = ball \(\le\) hoop).

B2 (smallest) clears all 4 hoops \(\to\) 4 pings.

B3 (largest) clears none \(\to\) 0 pings.

B4 sits in \((H2, H3]\): clears H3, H4, H1, fails H2 \(\to\) 3 pings.

B1 sits in \((H3, H4]\): clears H4, H1, fails H2, H3 \(\to\) 2 pings.

B5 fails H4 (and H2, H3), clears only H1 \(\to\) 1 ping.

Step 2 \(-\) the uncertain ball B6. B6 clears H4 and H1, fails H2. Whether it clears H3 is unknown, so B6 gives 2 pings (if it fails H3) or 3 pings (if it clears H3).

Step 3 \(-\) total. \[ 4 + 0 + 3 + 2 + 1 + (2 \text{ or } 3) = 10 + (2 \text{ or } 3) = 12 \text{ or } 13. \]

Step 4 \(-\) conclude. The best that can be said is the total is 12 or 13.

Final answer: 12 or 13

Answer 8

Step 1: Understanding the Question: 
We need to find the total number of pings across all 24 tests (6 balls x 4 hoops). The answer might be a specific number or a range if there is uncertainty. 
Step 2: Calculating Pings for Each Ball: 
We use the established size orders: B3 > B5 > B1 > B4 > B2 and H1 > H4 > H3 > H2. The position of B6 is B5 > B6 > B2. 
- B1 Pings: On H1(Y), H4(Y). On H2(N), H3(N). Total = 2.
- B2 Pings: Smallest ball, pings on all hoops. Total = 4.
- B3 Pings: Largest ball, pings on no hoops. Total = 0.
- B4 Pings:
- vs H1 (Y, since H1 is largest)
- vs H2 (N, rule 4)
- vs H3 (Y, rule 2)
- vs H4 (Y, since H4 > H3 $\ge$ B4)
- Total = 3.
- B5 Pings:
- vs H1 (Y, rule 3)
- vs H2 (N, rule 4)
- vs H3 (N, since B5 > B1 > H3)
- vs H4 (N, rule 1)
- Total = 1.
- B6 Pings:
- vs H1 (Y, rule 3)
- vs H2 (N, rule 4)
- vs H4 (Y, rule 1)
- vs H3: This is unknown. We know B1 > H3 and B4 $\le$ H3. We do not have information to place B6 relative to H3. B6 could be larger or smaller than H3.
- So, B6 pings on H3 if B6 $\le$ H3, and does not ping if B6 > H3. - Total = 2 or 3.
Step 3: Calculating the Total Number of Pings: 
Summing the pings for all balls: 
Total Pings = Pings(B1) + Pings(B2) + Pings(B3) + Pings(B4) + Pings(B5) + Pings(B6) 
Total Pings = 2 + 4 + 0 + 3 + 1 + (2 or 3) 
Total Pings = 10 + (2 or 3) 
- If B6 does not ping on H3, Total = 10 + 2 = 12.
- If B6 pings on H3, Total = 10 + 3 = 13.
Step 4: Final Answer: 
The total number of pings can be either 12 or 13, depending on the size of B6 relative to H3. The statement that best captures this is "12 or 13". This corresponds to option (C).
(Note: While "At least 9" is technically true, it is not the BEST description of the total, as we can prove the total must be at least 12. "12 or 13" is the most precise and accurate statement.)