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RenB
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List A is split into two sub-lists B and C. All the elements of list B 11 Aug 2023, 21:58
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List A is split into two sub-lists B and C. All the elements of list B are less than or equal to the elements of List C. What is the mode of list A?

(1) Both list B and list C are bimodal.
(2) 11 is a mode of list B and list C.
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B
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List A is split into two sub-lists B and C. All the elements of list B 21 Nov 2023, 07:26
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RenB
List A is split into two sub-lists B and C. All the elements of list B are less than or equal to the elements of List C. What is the mode of list A?

(1) Both list B and list C are bimodal.
(2) 11 is a mode of list B and list C.
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OA is B

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List A is split into two sub-lists B and C. All the elements of list B 25 Jun 2026, 10:52
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I don't agree with the answer choice (B) and the simple reasoning is as follows.

We know that any element in List B cannot be more than List C. At max it can be equal or less, but not more.

Choice (B) tells us that 11 is "A" mode in both List B & List C. "a" mode indicates that 11 is one mode common to List B & List C. We do not know if there is any other element which is also a mode in List B & C as it is statistically possible for a list/set to have 2 or more modes at one time.

Since, the question asks for "The" mode of List A, Answer Choice B is clearly incorrect.

Example as follows:

List B = (1, 2, 5, 5, 5, 9, 11, 11, 11, 12, 12, 12)

List C = (11, 11, 12, 12 15, 19, 21, 21)

Therefore, we satisfy both conditions presented i.e. both list B & list C are bi-modal and 11 is a mode in both List B & C.

However, combining this into List A which is (1, 2, 5, 5, 5, 9, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 15, 19, 21, 21) - we see that List A is also bimodal & has 2 modes i.e. 11 & 12 - both of which are repeated 5 times each. Therefore there is not one single mode and accordingly we cannot determine "The Mode" which is asked in the question.

Now, if the question asked for "a" mode of List A - then Choice (B) is correct but since it asks for "the" mode, I feel Choice (E) here is the most appropriate answer.
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List A is split into two sub-lists B and C. All the elements of list B 01 Jul 2026, 00:39
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In the example you have provided, List B contains the element 12, which is greater than an element in List C (11), and hence violates the condition in the question stem. If you remove the elements 12 from List B, 12 ceases to be a mode in List A, and 11 becomes "the mode".
Tinytrotter
I don't agree with the answer choice (B) and the simple reasoning is as follows.

We know that any element in List B cannot be more than List C. At max it can be equal or less, but not more.

Choice (B) tells us that 11 is "A" mode in both List B & List C. "a" mode indicates that 11 is one mode common to List B & List C. We do not know if there is any other element which is also a mode in List B & C as it is statistically possible for a list/set to have 2 or more modes at one time.

Since, the question asks for "The" mode of List A, Answer Choice B is clearly incorrect.

Example as follows:

List B = (1, 2, 5, 5, 5, 9, 11, 11, 11, 12, 12, 12)

List C = (11, 11, 12, 12 15, 19, 21, 21)

Therefore, we satisfy both conditions presented i.e. both list B & list C are bi-modal and 11 is a mode in both List B & C.

However, combining this into List A which is (1, 2, 5, 5, 5, 9, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 15, 19, 21, 21) - we see that List A is also bimodal & has 2 modes i.e. 11 & 12 - both of which are repeated 5 times each. Therefore there is not one single mode and accordingly we cannot determine "The Mode" which is asked in the question.

Now, if the question asked for "a" mode of List A - then Choice (B) is correct but since it asks for "the" mode, I feel Choice (E) here is the most appropriate answer.
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List A is split into two sub-lists B and C. All the elements of list B 01 Jul 2026, 00:54
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The question stem says "All the elements of List B are less than or equal to the elements of List C". So the wording is not clear to cover the case you have presented, in my view at least. Because 11 & 12 which are present in both List B & List C and thus satisfy the criteria as that the elements of List B are equal to the Elements of List C. My reading of the question stem is that as long as both lists contain the same element - they meet the "equal" criteria even if they don't miss the less than criteria. And since the wording is "OR" (Less Than OR equal), I can proceed with only the equality condition.

I may be wrong but the wording is confusing from that extent.
DawnHub
In the example you have provided, List B contains the element 12, which is greater than an element in List C (11), and hence violates the condition in the question stem. If you remove the elements 12 from List B, 12 ceases to be a mode in List A, and 11 becomes "the mode".

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List A is split into two sub-lists B and C. All the elements of list B 01 Jul 2026, 10:25
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Hi Tinytrotter,

Your instinct about the "or equal" wording is worth taking seriously, so let's put it under the microscope, because that's exactly where the setup breaks.

The condition "all elements of List B are less than or equal to the elements of List C" is not a per-element pairing - it's a single blanket rule between the two lists: every value in B must sit at or below every value in C. In plain terms, the largest thing in B can't exceed the smallest thing in C. So the "or equal" branch only rescues values that land right on that shared boundary; it does not give a value free rein to appear on both sides.

Where the 12 goes wrong

The moment 11 shows up in List C (from statement 2), every element of B is pinned to be less than or equal to 11 - because each B element must be less than or equal to that 11. A 12 sitting in List B would be greater than an element of List C (the 11), which breaks the rule outright. It doesn't matter that 12 also appears in C; the illegal comparison is "12 in B vs. 11 in C," and that one alone is fatal. So the example with 12 in both lists isn't a valid case to test - it never satisfies the stem.

Here's the clean way to see what statement 2 forces:

- 11 is a mode of B, and every other B value is less than or equal to 11.
- 11 is a mode of C, and every other C value is greater than or equal to 11.
- So 11 is the only value that can live in both lists (it's the boundary). Every other value belongs to just one side.

When you pool B and C into A, 11 collects its count from both lists, while any other value can only draw from one list - and in its own list it never out-appeared 11. So nothing can catch 11 once merged. 11 is the mode of A, uniquely. That's why statement (2) alone is sufficient.

A quick way to lock the rule in: take List B = {2, 5, 5} and List C = {5, 8, 8}. Legal, because max(B) = 5 is less than or equal to min(C) = 5, meeting on the shared value 5. Now try to slip a 6 into B - instantly 6 > 5 in C, illegal. The boundary is the only place the two lists may touch.

Answer: B

Tinytrotter
I don't agree with the answer choice (B) and the simple reasoning is as follows.

We know that any element in List B cannot be more than List C. At max it can be equal or less, but not more.

Choice (B) tells us that 11 is "A" mode in both List B & List C. "a" mode indicates that 11 is one mode common to List B & List C. We do not know if there is any other element which is also a mode in List B & C as it is statistically possible for a list/set to have 2 or more modes at one time.

Since, the question asks for "The" mode of List A, Answer Choice B is clearly incorrect.

Example as follows:

List B = (1, 2, 5, 5, 5, 9, 11, 11, 11, 12, 12, 12)

List C = (11, 11, 12, 12 15, 19, 21, 21)

Therefore, we satisfy both conditions presented i.e. both list B & list C are bi-modal and 11 is a mode in both List B & C.

However, combining this into List A which is (1, 2, 5, 5, 5, 9, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 15, 19, 21, 21) - we see that List A is also bimodal & has 2 modes i.e. 11 & 12 - both of which are repeated 5 times each. Therefore there is not one single mode and accordingly we cannot determine "The Mode" which is asked in the question.

Now, if the question asked for "a" mode of List A - then Choice (B) is correct but since it asks for "the" mode, I feel Choice (E) here is the most appropriate answer.
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