Comprehension
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?

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For logic puzzles involving relative ordering, the first step is always to establish the relationship between the items. Create a single inequality chain if possible (e.g., A > B > C > D). This makes answering specific questions much easier. Always double-check your initial deductions as all subsequent answers will depend on them.
Updated On: Dec 4, 2025
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Correct Answer: 6

Solution and Explanation

Step 1: Understanding the Question and Initial Deductions:
We need to find the total number of pings for balls B1, B2, and B3 across all four hoops. A ping occurs if Diameter(Ball) $\le$ Diameter(Hoop). Let's use Bx and Hx to denote the diameters.
From the rules, we can deduce the relative sizes.
- Rule 4 implies B2 $\le$ H2 and (B1, B3, B4, B5, B6) > H2. This makes B2 the smallest ball.
- Rule 3 implies B3 > H1 and (B1, B2, B4, B5, B6) $\le$ H1. This makes B3 the largest ball.
- From Rule 2, B1 > H3 and B4 $\le$ H3. This implies B1 > B4.
- From Rule 1, B5 > H4, while B1 $\le$ H4 and B6 $\le$ H4. This implies B5 > B1 and B5 > B6.
- Combining these, we get a partial order for balls: B3 > B5 > B1 > B4 > B2.
- For hoops: H1 $\ge$ B5 > H4 $\ge$ B1 > H3 $\ge$ B4 > H2 $\ge$ B2. This gives a clear order: H1 > H4 > H3 > H2.
Step 2: Calculating Pings for Each Ball:
- Pings for B1:
- vs H1: B1 $\le$ H1 (Rule 3) $\implies$ Ping.
- vs H2: B1 > H2 (Rule 4) $\implies$ No Ping.
- vs H3: B1 > H3 (Rule 2) $\implies$ No Ping.
- vs H4: B1 $\le$ H4 (Rule 1) $\implies$ Ping.
- Total for B1 = 2 pings.
- Pings for B2:
- B2 is the smallest ball. It is smaller than H2, which is the smallest hoop. Therefore, B2 is smaller than all hoops.
- vs H1: B2 $\le$ H1 (Rule 3) $\implies$ Ping.
- vs H2: B2 $\le$ H2 (Rule 4) $\implies$ Ping.
- vs H3: Since H3 > H2 $\ge$ B2, B2 $\le$ H3 $\implies$ Ping.
- vs H4: Since H4 > H2 $\ge$ B2, B2 $\le$ H4 $\implies$ Ping.
- Total for B2 = 4 pings.
- Pings for B3:
- B3 is the largest ball. It is larger than H1, which is the largest hoop. Therefore, B3 is larger than all hoops.
- vs H1: B3 > H1 (Rule 3) $\implies$ No Ping.
- It will not ping on any other smaller hoop either.
- Total for B3 = 0 pings.
Step 3: Final Answer:
Total number of pings for B1, B2, and B3 is the sum of their individual pings.
Total = Pings(B1) + Pings(B2) + Pings(B3) = 2 + 4 + 0 = 6.
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Question: 2

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

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In "Not Necessarily True" questions, you are looking for ambiguity. If you can construct a valid scenario where the statement is false, then it is not necessarily true. The key here was realizing that two items being smaller than a third item (B1 $\le$ H4 and B6 $\le$ H4) doesn't define the relationship between those two items.
Updated On: Dec 4, 2025
  • B4 < B5 < B3
  • B2 < B1 < B5
  • B1 < B5 < B3
  • B1 < B6 < B3
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The Correct Option is D

Solution and Explanation

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. 
 

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Question: 3

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

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The most effective way to solve ordering problems is to use a "bridge." Find an element (in this case, a ball) that connects two other elements (hoops) to establish their relative order. For example, Ball B1 was the bridge to prove H4 > H3. Systematically finding these bridges will reveal the complete order.
Updated On: Dec 4, 2025
  • H1 < H4 < H3 < H2
  • H2 < H3 < H4 < H1
  • H1 < H3 < H4 < H2
  • H2 < H4 < H3 < H1
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The Correct Option is B

Solution and Explanation

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).
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Question: 4

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

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When a question asks what "BEST" can be said, look for the most precise answer that is logically certain. A vague but true statement (like "At least 9") is usually not the best answer if a more specific range or value (like "12 or 13") can be proven. Identify any uncertainties and calculate the range of possible outcomes based on them.
Updated On: Dec 4, 2025
  • 13 or 14
  • At least 9
  • 12 or 13
  • 12 or 13 or 14
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The Correct Option is C

Solution and Explanation

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).
\textit{(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.)}
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