Teacher distributes ‘N’ chocolates to ‘3x’ students. Find the value of x.

A. If he distributes 6 chocolates to each student, then teacher left with 24 chocolates.

B. If number of students were 1/3rd of original number, then no chocolate was left with teacher.

C. 30 < number of students < 40 and 200 < number of chocolates < 300.

To find the value of $x$ (and hence the number of students, which is $3x$), let us analyze the information given in the question and the statements.

Let the total number of chocolates be $N$.
Let the total number of students be $S=3x$, where $x$ is an integer.

Let's evaluate each statement individually and then in combination:

Statement A: "If he distributes 6 chocolates to each student, then teacher left with 24 chocolates."
This gives the equation:
$N=6\left(3x\right)+24=18x+24$
Here, we have two variables ($N$ and $x$) and only one equation. Statement A alone is not sufficient.

Statement B: "If number of students were 1/3rd of original number, then no chocolate was left with teacher."
The new number of students is $\frac{1}{3}\left(3x\right)=x$.
Let $k$ be the number of chocolates each of these $x$ students receives. Then:
$N=kx$ (where $k$ is some positive integer).
Since $k$ is unknown, Statement B alone is not sufficient.

Statement C: "30 < number of students < 40 and 200 < number of chocolates < 300."
This gives the ranges:
$30<3x<40$ which simplifies to $10<x<13.33$.
Since $x$ is an integer, $x$ could be 11, 12, or 13. Thus, the number of students $3x$ could be 33, 36, or 39.
Also, $200<N<300$.
Statement C alone is clearly not sufficient to find a unique value for $x$.

Combining Statements A and C:
From Statement A, we have $N=18x+24$.
From Statement C, the possible values for $x$ are 11, 12, or 13.
- If $x=11$, then $N=18\left(11\right)+24=198+24=222$ (which is between 200 and 300).
- If $x=12$, then $N=18\left(12\right)+24=216+24=240$ (which is between 200 and 300).
- If $x=13$, then $N=18\left(13\right)+24=234+24=258$ (which is between 200 and 300).
Since we have multiple valid pairs for $(x,N)$, namely $(11,222)$, $(12,240)$, and $(13,258)$, Statements A and C together are not sufficient.

Combining Statements B and C:
From B, $N=kx$ where $k$ is an integer.
From C, $x\in \{11,12,13\}$ and $200<N<300$.
This gives many possibilities for $x$ and $N$, so B and C together are not sufficient.

Combining Statements A, B, and C:
We have:
1) $N=18x+24$
2) $N=kx$ (meaning $N$ is a multiple of $x$)
3) $x\in \{11,12,13\}$ and $200<N<300$.
Let us test the possible values of $x$:
- If $x=11$, then $N=222$. We check if 222 is divisible by 11: $222/11=20.18$ (not an integer). So $x=11$ is not possible.
- If $x=12$, then $N=240$. We check if 240 is divisible by 12: $240/12=20$ (which is an integer). So $x=12$ is a possible solution.
- If $x=13$, then $N=258$. We check if 258 is divisible by 13: $258/13=19.84$ (not an integer). So $x=13$ is not possible.

However, the options provided only combine at most two statements together (e.g., "Either A & B together or B & C together", "Any two of them", etc.). Since no combination of two statements is sufficient to find a unique value for $x$, none of the options representing combinations of two statements can answer the question.

Therefore, the correct choice is: None of the given statements can answer the question.