A list contains n distinct integers. Are all n integers cons

[color=#ffff00]

PerfectScores wrote:

UdaySJha wrote:

A list contains n distinct integers. Are all n integers consecutive?

(1) The average (arithmetic mean) of the list with the lowest number removed is 1 more than the average (arithmetic mean) of the list with the highest number removed.

(2) The positive difference between any two numbers in the list is always less than n

(1) The average (arithmetic mean) of the list with the lowest number removed is 1 more than the average (arithmetic mean) of the list with the highest number removed.

(2) The positive difference between any two numbers in the list is always less than n

Statement I is sufficient

Let us say n = 4 and the numbers are 4, 5, 6 and 7

If we remove 4 then the average is 6
If we remove 7 then the average is 5 which satisfies with the statement

Now lets go algebra:

If the numbers are {x, x+1, x+2.....x+n-1} with average (2x + n - 1)/2 (First + last)/2

If we remove x then the average will become (x + 1 + x + n - 1)/2 = (2x + n)/2 increasing by 1/2
If we remove x + n - 1 the average will become (x + x + n -2)/2 = (2x + n -2)/2 decreasing by 1/2

Hence the difference will always be the 1 between them.

Statement II is sufficient:

If the numbers are {x, x+1, x+2.....x+n-1}

If we subtract the first and the last we will get = |x + n - 1 - x| = n - 1 which is always less than n. Now is there any other set which can have that. Since the members of the set are distinct it is not possible to have that.

Answer is D.[/quote][/color]

Hi Perfect score!

Nice solution. I have little confusion in understanding the language used in the question stem . Here it is mentioned as list of number so whether we are suppose to assume numbers in the list always follow sequence or constant increment or decrease? Pls advise.

[/color]

Hi Perfect score!

Nice solution. I have little confusion in understanding the language used in the question stem . Here it is mentioned as list of number so whether we are suppose to assume numbers in the list always follow sequence or constant increment or decrease? Pls advise.[/quote]

You have to look at both the scenarios. Since the numbers are distinct in the set it is not possible to have any other set apart from numbers which are consecutive.

What if the set includes {1,2,4,5,6}? Then n = 5 and the difference between 2 and 4 is less than 5 but the set is not consecutive.

Notice that the second statement says "the positive difference between ANY two numbers in the list is always less than n". In your example the difference between 6 and 1 is 5, not less than 5, therefore your set is not valid.

Does this make sense?

Thanks Bunuel! Makes perfect sense.

Why are we assuming that the set will always contain positive integers?

We did not assume that. Check here: a-list-contains-n-distinct-integers-are-all-n-integers-cons-165873.html#p1316160

What if my set is -1,0,1,2. Then the diff 2-(-1) >2
What do we mean by positive difference? (Is it |a-b| or |a|-|b|?)

The positive difference between a and b means |a - b|.

In your example, if set is {-1, 0, 1, 2}, then n = 4 (the number of elements in the set) and the positive difference between any two numbers in the list is less than 4.

Is there a propery like " if the list consists of concecutive integers, then with any one number removed from the set, the mean of the set will vary maximum (+- 0.5) .

Hi there, still i have concerns
for me statement 1 is only sufficient
for B let's assume 2 different scenarios
1- N contains {1,2,3,4,5} which means n=5 and if we deduct 5-1 = 4 so far so good
2- N contains {2,4,6,8} which means N =4 and if we deduct 8-2 = 6 so B should be insufficient

Hi,
the statement says ...
(2) The positive difference between any two numbers in the list is always less than n..

in the highlighted portion the positive difference is 6 which is more than 4(n here)... so the example taken here is not correct..
the least possible diff is when the number is consecutive and it is always one less than n, so no other scenario is possible where the diff is less than n in n distinct integers

chetan2u Thank you

VeritasPrepKarishma somewhere wrote that in a set of consecutive n integers, if we remove one integer, the average of the new set can decrease/increase MAX by 0.5.
1. if we remove the lowest, our new average would be +0.5. if we remove the highest number - the new average would be -0.5.
the statement tells that the difference is 1. this works in consecutive N integers only.
2. since we have distinctive integers and since the difference between any must be less than n, there is no way to create a list of N integers so that integers would not be consecutive.

D

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.

A list contains n distinct integers. Are all n integers consecutive?

(1) The average (arithmetic mean) of the list with the lowest number removed is 1 more than the average (arithmetic mean) of the list with the highest number removed.
(2) The positive difference between any two numbers in the list is always less than n


In the original condition, the number of variables is n, which should match with the number of equations. So you need n equations. For 1) 1 equation, for 2) 1 equation, which is likely to make E the answer. When 1) & 2),
For 1), if m,m+1,m+2,,,,,,,,m+n-2, m+n-1, the number of these consecutive integers is n, which starts with m. The average excluding m(the smallest integer) is ={(m+1)+.......+(m+n-2)+(m+n-1)}/(n-1)={m+(m+1)+.......+(m+n-2)+(n-1)}/(n-1)={m+(m+1)+.......+(m+n-2)}/(n-1)+{(n-1)/(n-1)}
={m+(m+1)+.......+(m+n-2)}/(n-1)+1=m+n-1(the biggest integer) average+1 excluding the biggest integer, which is yes and suffcient.
For 2), it goes same for 1). That is, 1)=2).
Therefore, the answer is D.


 For cases where we need 3 more equations, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80% chance that E is the answer (especially about 90% of 2 by 2 questions where there are more than 3 variables), while C has 15% 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 E 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, C or D.

Solution from "Thursdays With Ron"

Hi All,

This DS question has a complex 'feel' to it, so you might find it useful to deal with the information in a slightly different "order"

We're told that we have N DISTINCT integers (meaning that the numbers are all different - NO duplicates). We're asked if the N integers are consecutive. This is a YES/NO question.

To start, if we ARE dealing with consecutive integers, then there will be a variety of built-in patterns that will define the integers as consecutive (and if we can spot any of those patterns, then we might be able to reverse-engineer that the group is comprised of consecutive integers). Fact 2 is the easier of the two options, so I'm going to start there...

2) The positive difference between ANY two numbers in the list is always LESS than N.

IF.... N = 2, then the difference between those two integers has to be LESS than 2. Since we already know that the numbers are DISTINCT, the two values would have to have a difference of 1. For example, the list could be {0,1}, {3,4}, {-9, -8}, etc. This means that they ARE consecutive integers and the answer to the question is YES.

IF... N = 3, then the difference between any two of those three integers has to be LESS than 3. Since we already know that the numbers are DISTINCT, then any two values would have to have a difference of 1 or 2. The group is 3 integers though, so again - they will have to be consecutive. For example {1, 2, 3}, {7, 8, 9}, {-2, -1, 0}, etc. The answer to the question is also YES.

Increasing the number of integers will NOT change the outcome (try an example with 4 or 5 integers and you'll see). The answer to the question will ALWAYS be YES.
Fact 2 is SUFFICIENT.

1) The average (arithmetic mean) of the list with the lowest number removed is 1 more than the average (arithmetic mean) of the list with the highest number removed.

Since Fact 2 found a 'quirky' way to define that the list of integers was consecutive, it gets me thinking that maybe Fact 1 has also done that. However, we need proof before we make our assessment, so let's start off with a consecutive list of integers and see if it 'fits' what Fact 1 describes.

IF... we're dealing with {1, 2, 3}, then N = 3.
Remove the lowest value and the average becomes (2+3)/2 = 2.5
Remove the largest value and the average becomes (1+2)/2 = 1.5
The difference in averages IS exactly 1, so this example 'fits' what we're told and the answer to the question is YES. (incidentally, if you try any other group of consecutive integers, you'll find that this pattern holds true - you should try a few and prove it for yourself).

Now let's try a group of numbers that is NOT consecutive....
IF... we're dealing with {1, 2, 4}, then N = 3.
Remove the lowest value and the average becomes (2+4)/2 = 3
Remove the largest value and the average becomes (1+2)/2 = 1.5
The difference here is NOT 1 though, so this does NOT 'fit' what we were told and is NOT a valid example. Looking at this example, it seems that if the largest number is "too far" from the smallest number, and there are 'missing' integers in between, then the difference in averages will NOT equal 1 (it will be LARGER every time). With a few additional examples, you can prove it.

Thus, the only groups of numbers that will fit Fact 1 are consecutive integers and the answer to the question will always be YES.
Fact 1 is SUFFICIENT.

Final Answer:

GMAT assassins aren't born, they're made,
Rich