Bunuel wrote:
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
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 101This 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.