If S is a finite set of consecutive even numbers, is the median of S

(1) The mean of set S is an even number.
Set S consists of consecutive even numbers hence mean=median
Mean is even then Median also even
Suff

(2) The range of set S is divisible by 4.

Let a be first number in set
then
last number = a+(n-1)d
since d=2
Last Number=a+(n-1)2

Range = a+(n-1)2 - a= 2(n-1)
Range = 4k = 2(n-1)
n=2k+1
n= Odd number

As we know now, set S contains odd number of consecutive even numbers median will be one of this numbers and
hence is EVEN
Suff

IMO D

S is an evenly spaced set thus median=mean. The median/mean of an evenly spaced set of consecutive even numbers is odd when the number of elements in the set is EVEN. For instance
(2; 4; 6) median=mean=4=even; (2; 4; 6; 8) median=mean=5=odd.

Every statement that helps us determine the number of elements in S is sufficient.

statement 1: Sufficient. If mean=median=even number of element in S=odd.

statement 2: Not sufficient, we have no clue about how many elements are in S.

Answer is A.
_________________

D

2nd Stmnt says Max-Min is divisible by 4.
Consider ( 2n-2, 2n, 2n +2) or (2n, 2n+2, 2n +4) or ( 2n-4, 2n-2, 2n, 2n+2, 2n+4) ..... all these cases satisfy the stmnt
Hence Median is even (i.e. not odd.) ... Sufficient

If you have even number of terms, it will not satisfy the Divisibility by 4 constraint

Whats the OA?

Statement 1
The question says S is a finite set of consecutive even integers, which means it is an equally spaced set. In equally spaced set mean = median so median is even ... Sufficient

Statement 2
Range of set S will be divisible by 4 only when the number of values is odd, which means the middle value is even. So median is even ... Sufficient

We can do this either with number picking or algebra

Lets suppose set S is [2,4,6] or [2,4,6,8,10] or [2,4,6,8,10,12,14]
Range in first set is 4 in second set and in last set 12. For range to be 4 we need minimum of three consecutive even integers from there on you can keep on adding two more values and we ll get multiples of 4 as range. So the number of values in the set will always be odd.

Correct Ans D

first of all: are negative numbers included in sets like this?
(1) set: {-4;-2;0;2;4;6;8} so the mean is 3 and the median is 2 -> suff

(2) set: {-4;-2;0;2;4;6;8} the range is 12, the median is 2
set: {2;4;..;10} -> range is 8, median is 6. the median is NEVER odd -> suff

both sets on their own are sufficient, because the answer the question with no.
-> D

The Rule that you have to know for this problem:
- "If the number of terms in a set of consecutive even integers is even, then the median is odd"
- "If the number of terms in a set of consecutive even integers is odd, then the median is even"

You can try this with the following sets:
2, 4, 6, 8 (Median = 5)
2, 4, 6, 8, 10 (Median = 6)

As you know, mean and median are equal in evenly spaced sets.


Statement 1: Sufficient
If the mean is even, then the median is even.

Statement 2: Sufficient
You can try different scenarios and you´ll always yield to the right answer. However, I wouldn´t know how to explain the algebraic approach.


OA = D

Good question! I also go with D.
Basically median of S is odd if # terms in S is even and vice versa. So we really only need to know# terms in S.

1) Mean of set is even --> the #terms is odd. Just visualize with simple examples, (2,4,6) and (2,4,6,8). Only if #terms is odd, mean is even. SUFFICIENT.
2) Range is divisible by 4. Again this would mean the #terms is odd (using same examples as above). SUFFICIENT.

Answer = D.

Just go with sample data for this kind of problem
Consecutive even numbers {2,4,6,8...}
From statement1--> mean is an even number -->{2,4,6,8}-->Median is odd number => Statement1 is sufficient
From statement2--> Range is divisble by 4 --> {2,4,6,8,10} --> Median is even number => Statement 2 is sufficient
Answer should D -->Both the statement alone is sufficient

{0, 2, 4, 8} is not a set of consecutive even integers. You are missing number 6 there.
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I don't think it will matter how many terms are in S, because we know they are consecutive and we know that the range is 4. This combination only works with an odd number of terms I reckon.

So there's a couple of housekeeping rules with stat ds question- the mean of a set of consecutive integers will always be the median. This applies to consecutive even sets and consecutive odd sets too.

St 1

Clearly gives us info about the median

St 2

In order to have a range that is a multiple of 4 the number of terms in the set must be odd: 3,5,7, etc. And because of this the median will always be even.

D

I know this comment is quite old but just to correct the tiny highlighted part to avoid confusion for anybody reading your reply in the future. Any evenly spaced numbers whether positive or negative or a mix included in a set, median is equal to mean.

If we actually do the long calculation for the mean instead of doing the short version for a set of evenly spaced numbers. (-4-2+0+2+4+6+8)/7=2 and not 3.

As for the shortest way to do this, the rules needed for this were mentioned earlier by minwoswoh.

Good Luck!

Helpful formula for evaluating statement 2)

The number of terms (n) in a set of consecutive even integers is given by:

\(n=\frac{(L-F)}{2}+1\)

Where;
\(L\) - Last term
\(F\) - First term
i.e. \((L-F)\) is the range

If the range is divisible by 4, then:
\(n=\frac{(L-F)}{2}\)=even number, and
\(n=\frac{(L-F)}{2}+1\) = odd number (because Odd+Even=Odd)

So our median will be even

For statement 2 ; what of if the set of numbers are (0,2,4,8) . Won’t this satisfy statement 2 and the median will be odd ?why is no one talking about this possibility. Or did I miss something .

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Thank you 👍 for the correction.