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: Jul 4, 2026
Show Solution

Correct Answer: 6

Solution and Explanation

Step 1: B1's status: pings H1 and H4 (both stated directly), fails H2 (only B2 pings H2) and fails H3 (stated directly). That's 2 pings.
Step 2: B2's status: since B2 is the only ball small enough for H2, it is the smallest ball overall, so it also fits the comparatively larger hoops H1, H3 and H4. That's 4 pings.
Step 3: B3's status: since B3 is the only ball too big for H1, it is the largest ball overall, so it is too big for every hoop, including H2, H3 and H4. That's 0 pings.
\[ \boxed{2+4+0 = 6} \]
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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: Jul 2, 2026
  • B4 < B5 < B3
  • B2 < B1 < B5
  • B1 < B5 < B3
  • B1 < B6 < B3
Show Solution

The Correct Option is D

Solution and Explanation

Approach: Instead of one big chain, place each ball on a number line using the hoop thresholds, then look for the ball whose slot is ambiguous.

Setting up: Ping = ball fits, i.e. ball $\le$ hoop. The hoop thresholds from the data are $H2 < H3 < H4 < H1$. Each ball sits in the gap defined by the largest hoop it still fits through and the smallest it fails.

Placing balls by which hoops they clear:

B2 clears H2 (smallest) so it clears all four $\to$ smallest ball.

B4 clears H3 but fails H2 $\to$ sits in $(H2, H3]$.

B1 fails H3 but clears H4 $\to$ sits in $(H3, H4]$, above B4.

B5 fails H4 $\to$ sits above H4.

B3 fails H1 $\to$ biggest.

B6 clears H4 but fails H2, and is never tested against H3 $\to$ it sits somewhere in the whole band $(H2, H4]$, overlapping both B4's slot and B1's slot.

Reading off: The fixed order is $B2 < B4 < B1 < B5 < B3$. B6 is pinned only as $B2 < B6 < B5$; it can land below or above B1. So any statement that does not involve B6's loose position is forced, but a claim like $B1 < B6$ can break. Among the options, the only one asserting $B1 < B6$ is $B1 < B6 < B3$.

Final answer: B1 < B6 < B3
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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: Jul 2, 2026
  • H1 < H4 < H3 < H2
  • H2 < H3 < H4 < H1
  • H1 < H3 < H4 < H2
  • H2 < H4 < H3 < H1
Show Solution

The Correct Option is B

Solution and Explanation

Approach: Rank hoops by counting how many balls each one lets through \(-\) a wider hoop passes more balls.

Counting balls passed (ping = ball $\le$ hoop):

H2: only B2 passes $\to$ 1 ball. Fewest passes $=$ smallest hoop.

H3: B4 passes, B1 fails. Since B2 (smallest) also passes and B1, B5, B3 are above B1, H3 passes B2 and B4 $\to$ 2 balls.

H4: B1 and B6 pass (and B2, B4 which are smaller), but B5 fails $\to$ passes 4 balls (B1, B2, B4, B6).

H1: every ball except B3 passes $\to$ 5 balls. Most passes $=$ largest hoop.

Ordering by passes: $1 < 2 < 4 < 5$ gives $H2 < H3 < H4 < H1$. More balls cleared means a wider hoop, so this count order is exactly the size order.

Final answer: H2 < H3 < H4 < H1
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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: Jul 2, 2026
  • 13 or 14
  • At least 9
  • 12 or 13
  • 12 or 13 or 14
Show Solution

The Correct Option is C

Solution and Explanation

Approach: Count from the hoop side \(-\) for each hoop add up how many balls clear it; total pings is the same whether you sum across balls or across hoops.

Per-hoop counts (ball $\le$ hoop clears it):

H2 (smallest): only B2 fits $\to$ 1.

H3: clears B2 and B4 for sure; B1 fails. B6 versus H3 is unknown $\to$ 2 or 3 (the $+1$ if B6 clears H3).

H4: clears B2, B4, B1, B6; B5 fails $\to$ 4.

H1 (largest): clears everyone except B3 $\to$ 5.

Summing: $1 + (2\text{ or }3) + 4 + 5 = 12 \text{ or } 13$.

Why a range: Every comparison is forced except B6 against H3 (B6 is the one ball never sized against that hoop), so the single ambiguous test shifts the total by exactly one.

Final answer: 12 or 13
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