Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

How many odd integers are greater than the integer x and less than the integer y?

(1) There are 12 even integers greater than x and less than y
(2) There are 24 integers greater than x and less than y

There are 2 variables (x,y) and 2 equations from the question and the 2 conditions, so there is high chance (C) will be our answer.
Looking at the conditions together, if there are 12 even numbers out of the 24 integers, there are of course 12 odd integers, so the answer becomes (C). But this is an integer question which is one the the key questions, if we apply 4(A) mistake types,
Looking at condition 1, the number of odd integers becomes (35-11)/2+1=13 when x=10, y=36, and (33-13)/2+1=11 when x=11, y=35; this does not give unique answer, so this is insufficient.
From condition 2, if there are 24 integers, there has to be 12 even and 12 odd. This is sufficient, making the answer (B).

For cases where we need 2 more equation, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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(2) says that: there are 24 integers greater than x and less than y. Naturally those 24 integers between x and y are consecutive, how else? Consider x=1 and y=26: there are following 24 integers between them: 2, 3, 4, ..., 25.

Also check this post about the same issue: how-many-odd-integers-are-greater-than-the-integer-x-and-100521.html#p809821

Hope it's clear.
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In statement 2- If the number of integers between X and Y was odd, say 23 or 21. Would the statement be sufficient?
What would have been the answer in that case?

Answer would be insufficient - eg. take smaller nos.
if x=1 & y=5, so we have 1 <2,3,4<5 so only 1 odd no
if x=2 & y=6, so we have 2<3,4,5<6 , so now we have 2 odd nos.

I have posted a similar query to bunuel for clarification regarding generalization of this counting principle.But from what I can gather, when we have an even no of consecutive integers, we will have have an equal no of odd & equal no of even nos between them. & when we have even no or odd number of integers to count, then it will depend on the 2 numbers encapsulating them.

Hello

If you are taking X=1 and Y=1000, then there are not just '24' integers between them. Between 1 and 1000 there are '998' integers (2, 3, 4, ...999). So this example (of 1 and 1000) cannot be taken considering the second statement.

When the statement says, "there are 24 integers between X and Y", it means there are '24 integers only'. That is only possible if Y = X+25. And out of those 24 integers, 12 will be odd and 12 will be even (no matter whether X is odd or even, or whether Y is even or odd)

Method: Solving an easier Q and testing cases

1)
Let x = 1, y = 5
#s in middle: 2,3,4 => # of Odd ints = 1 (i.e. 3)
Let x = 1, y = 6
#s in middle: 2,3,4,5 => # of Odd ints = 2 (i.e. 2)
Not sufficient

2)
Let x = 1, y = 6
#s in middle: 2,3,4,5 => # of Odd ints = 2 (i.e. 3,5)
Let x = 1, y = 7
#s in middle: 2,3,4,5,6 => # of Odd ints = 2 (i.e. 3,5)
Sufficient

ANSWER: B