How many odd integers are greater than the integer x and

"There are 24 integers greater than x and less than y" means that exactly 24 integers are greater than x and less than y.
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Once again, don't blindly justify the GMAT answer even when it is wrong.

The answer is clearly C. Very happy to be proven otherwise.

There is no indication anywhere the 24 integers are consecutive.

Let's shrunk it to 4 integers for the purpose of example. Statement two can be x,2,3,4,y or x,3,8,7,y = inconclusive.

I don't know why you've chosen to adopt such a patronizing and categorical tone in your response. No-one is blindly justifying the answer. If you read the thread you'll be surprised to find elaborate responses there (by the way your doubt is also addressed in the thread!).

"There are 24 integers greater than x and less than y" means that there are exactly 24 integers between x and y, not inclusive. Those integers will obviously be consecutive.

Not sure what your examples shows there but if I ask you, how many integers are greater than 1 and less than 6, what would be the answer? The answer would be four integers: 2, 3, 4, and 5. Half are odd and half are even.

How many integers are greater than -8 and less than -3? Four: -7, -6, -5, and -4. Half are odd and half are even.

Hope it's clear!
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Hey Bunuel, thanks for the reply, I didn’t mean that way at all sorry it came across that way, I was just being confident in my statement. (also text doesn't translate my tone well)
I am solely addressing the consecutive point which is the crux of this question, all I am saying is no where in the question specifies all the numbers appeared once or have to appear at least once. It makes perfect sense if you assume that, but the rule of GMAT is don’t assume anything that is not stated; and definition of integer does not entail consecutiveness. I believe we are talking about the same thing here but just poor wording by the question

EDIT just to be clear, for example: how many integers between 1 and 100.

I can equally say 98 - counting all number once

or two (just for example) - if the set is {1, 4, 88, 100} the two are 4 and 88 (just random number picked for demonstration)

The answer to the question how many integers are there between 1 and 100, not inclusive? is 98: 2, 3, 4, ..., 99. You count ALL integers which are between 1 and 100, not inclusive. We don't have some specific set there and there is no ambiguity whatsoever about this in the question.
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We are wasting time here debating this, when you say there are 3 integers, you don't automatically group them into a consecutive cloud. I did not assume they are a set, in fact I did not mak any presumption, but you did (assuming inclusive and count all numbers while the Q didn't state it)

There is another question here coincidentally I come across just now as we speak, which correctly phrase "list of consecutive integers":
https://gmatclub.com/forum/list-k-consi ... 47855.html

Basically if my boss as me to buy three numbers for him for lottery between 0-9, I don't just start saying it should be 1,2,3. Anyway, I think you are clouding your judgement a bit by the preestablished position, it is a time wasting here I will move on. Thanks again.

Yes, I agree that you are wasting time here.
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jabhatta2 and others: I created an analogous problem - please give it a try and let me know whether you found it helpful:

How many Wednesdays are there in the month of February in a certain year?
(1) There are four Thursdays in the month of February of that year
(2) It is not a leap year

*how would your answer change if statement (2) said that it is a leap year?

Wow avigutman - the analogy certainly helped but i think i may be going down the wrong rabbit hole with this analogy

I think you are using Feb because 28 days is divisible by 7

• If not a leap year, the answer is B (answer is 4)
• If leap year -- the answer is E (answer could be 4 or 5)

Going down the rabbit hole with my analysis of "why"
Above seems sufficient when the TOTAL number of days (=28 in Feb) is EXACTLY DIVISIBLE by the number of items in the cyclity cycle (=7 in the case of a week)

In the original question, the number of items in the cyclicity is 2 (even and odd)

But whats the total number of elements in the set to divide by 2 ?
(From S1) We dont know the total number of elements to divide by 2 --> hence INS
(From S2) The total number to elements is 24 (x and y themselves are NOT included in this total). Given 24 is EXACTLY divisible by 2, S2 is sufficient

I think thats why S2 is sufficient but i maybe wrong

Deeper logic -- Same condition - if the number of items in the cyclicty were 3 instead or 4 or 8 instead, i think (S2) would be sufficient given 24 is divisible by 2 / 3 , 4 and 8

I maybe am going down the wrong rabbit hole

notes for JD only

-- how small tweaks can give different DS results

Hi Bunuel avigutman - From the wording in S2 -- i think duplicates should be allowed

You mention the following



I dont believe your question posted above (for which the answer is 98) is analgous to S2

S2 - just says -- there are 4 integers greater than x and less than y

Duplicates should be allowed to be counted twice.

Check out the same question in the context of height

example - How many people are taller than 6 feet ?
- Person A - 6'1
- Person B - 6'1
- Person C - 6'1
- Person D - 6'1

All 4 people will be counted (duplication in exact heights is still to be counted as 4 people)

You didn't go down a rabbit hole, jabhatta2. This was a perfect analysis!

I'm not sure I'm following your point here, jabhatta2, but your height example is very different from there are 4 integers greater than x and less than y
Some examples:
There are 4 integers greater than 5 and less than 10 (they are 6, 7, 8, and 9)
There are 4 integers greater than -20 and less than -15 (they are -19, -18, -17, and -16)
There are 4 integers greater than 31.3 and less than 35.9 (they are 32, 33, 34, and 35)

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

Statement 1: There are 12 even integers greater than x and less than y
There are many scenarios that satisfy statement 1. Here are two:
Case a: x = 1 and y = 25. In this case, the answer to the target question is there are 11 odd integers greater than the integer x and less than the integer y
Case b: x = 1 and y = 26. In this case, the answer to the target question is there are 12 odd integers greater than the integer x and less than the integer y
Since we can’t answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: There are 24 integers greater than x and less than y
ASIDE: Many students will read the statement 2 and conclude that, since we aren't told whether those 24 integers are CONSECUTIVE integers, there's no way to tell how many how many odd integers there are.
However, the integers between x and y will always be consecutive.
For example, the integers between 5 and 11 are 6, 7, 8, 9 and 10 (consecutive)
Similarly, the integers between -2 and 8 are -1, 0, 1, 2, 3, 4, 5, 6 and 7 (consecutive)

Since even and odd integers alternate in consecutive integers (e.g., 1 is odd, 2 is even, 3 is odd, 4 is even,... etc.) , we know that 12 of the 24 integers must be even, and the other 12 must be odd
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer: B

Cheers,
Brent

Answer: Option B

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IanStewart please explain this question? In a simpler way

I haven't read the other explanations, but looking first at Statement 2:

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

Since even and odd integers alternate, 12 of these integers must be even, and the other 12 must be odd. So Statement 2 is sufficient. If instead Statement 2 said "There are 25 integers between x and y" (or any other odd number), it would not be sufficient, because it might be that 13 are even, 12 are odd, or it might be that 13 are odd and 12 are even.

Looking at the harder Statement, Statement 1:

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

It doesn't matter which integers these are, so we can just imagine they are the twelve integers 2, 4, 6, 8, ..., 22, 24. These need to be the only even integers between the two integers x and y. So it might be that x = 1 and y = 25, and then there are eleven odd integers between x and y. But it might be that x = 1 and y = 26, in which case we'll have one more odd integer in our range, namely 25, so we could have twelve odd integers in total. Or it might even be that x = 0 and y = 26, in which case we'll also have '1' in our range, and we'll have thirteen odd integers between x and y. So from Statement 1, all we know is that there are 11, 12 or 13 odd integers between x and y, and Statement 1 is not sufficient. So the answer is B.
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