Set A consists of 8 distinct prime numbers. If x is equal to the range

Question

Set A consists of 8 distinct prime numbers. If x is equal to the range of set A and y is equal to the median of set A, is the product xy even?

(1) The smallest integer in the set is 5.
(2) The largest integer in the set is 101

DHRJ0032 · Answer

y will always be odd.
We have to look out whether x is odd or even !!

1) since 5 is smallest in the set..then X will always be even as (odd-odd=even)
So xy=even 
.
2) if first element is 2 then 
X=odd and xy=odd
But if first element is 3 then 
X=even and xy=even.
INSUFFICIENT

A it is
Suffiicient

Posted from my mobile device

Samuel0709 · Answer

Y will not always certainly be odd.  Y can be odd or even   as for calculating the median the set needs to be arranged first. ( for eg if 4th and 5th term are 11 and 29, y is 20 and if middle two terms are 7 and 11 , y=9 ) 
So the question drills down to is X odd or even.

1. Smallest is 5 : odd-odd is always even hence x will always be even 
this gives us xy= even x odd= even
or
xy= even x even = even
 Therefore A is sufficient.

2.Largest is 101 : 
if the smallest number is 2 then 101-2=99 which is odd
Therefore we can have xy= odd x odd= odd
or
xy= odd x even= even 
Since we are getting two contradictory answers  B is clearly NOT SUFFICIENT.

TestPrepUnlimited · Answer

Note the median can be odd, even if all the numbers are odd. For example, let 19 and 23 be the 4th and 5th biggest number in the list, and the median would be 21.

Statement 1:

We know the range must be even now since the biggest integer must also be odd. Then x*y must be even. Sufficient.

Statement 2:

We don't know if the median is odd or even. We don't know if the range is odd or even (depends on if we include 2). Insufficient.

Ans: A

yasharoraa · Answer

Points to note:
  The set has 8  distinct prime  numbers   x is the range of set   y is the median of set of 8 numbers  
Let's consider possibilities for x to be even or odd. 
x can be  odd  only when set A consists of 2. Since 2 is the only even prime number, all other numbers in the set will be odd. Also 2 is the smallest prime number, and hence range will always be odd:  odd - 2  (odd - even = odd)
If A does not contain 2, then x will always be  even , since both the biggest and the smallest number in the set will be odd and odd - odd = even.

We don't know anything about y, y could be even or odd. 
Eg. if numbers in set A were {2,3,5,7,11,13,17}, then y = (5+7)/2 = 6, hence even.
Now if numbers in set A were {2,3,7,11,13,17,19}, then y = (7+11)/2 = 9, hence odd.

Now let's take the two statements-
 (1) The smallest integer in the set is 5. 
Since, the smallest integer in the set is 5, x, i.e., the range of the set will always be even. Hence product xy will always be even (even * any number = even).
Hence statement 1 is  SUFFICIENT .

(2) The largest integer in the set is 101 
This doesn't give us much information. It's quicker to plug in to make sure that we are correct
 Let's say set A were {2,3,5,7,11,13,101}, then y is even, x is odd, and the product xy is even. Answer is  yes , xy is even.
Now to get another answer as no, let's try to make y odd as well.
Let's assume A to be {2,3,7,11,13,17,101}, now y is odd, as well as x is odd. Hence, xy is odd. Thus, the answer is  no , xy is not even.
Hence  INSUFFICIENT .

Thus answer is A) Statement 1 alone is sufficient and statement 2 is insufficient.

760Abhi · Answer

I don't think that y will always be odd because

y == odd+odd/2 == even/2
Now it can be even as well as odd depending on even no (even/2).

Bunuel  : Please correct me if I am wrong

tiwariii · Answer

760Abhi

You are correct, the Median (Y) can be both Even or Odd. But in case 1 we know for sure that X (range) is even. Because Odd- 5 = even. Hence Irrespective of the median the outcome of xy will be even. 
But in case 2, we do not know if Range (X) is even or odd. because we have no idea of the lowest value (which can be 2 or 3). Hence in case 2, we have X which can be either even or odd, and Y which too can be either even or odd. Hence there can be a case when X = odd and Y = odd which will make xy odd. Hence this makes case 2 not sufficient.

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Set A consists of 8 distinct prime numbers. If x is equal to the range

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