a, b, and c are integers and a<b<c. S is the set of all integers from a to b, inclusive. Q is the set of all integers from b to c, inclusive. The median of set S is (3/4)*b. The median of set Q is (7.5/8)*c. If R is the set of all integers from a to c, inclusive, what fraction of c is the median of set R?	Pick one
A) 3/8 B) 1/2 C) 11/16 D) 23/32 E) 3/4
The answer isC: 11/16.   The key to this problem is remembering that the median for a consecutive set of numbers is equivalent to its mean. For example, the mean and median of a set consisting of x, x+1, x+2, ... ,y will always be (x+y)/2.   For set S, consisting of numbers (a, a+1,...,b), the median is given to be 3/4*b:   (a+b)/2 = (3/4)*b   a = b/2   For set Q, consisting of numbers (b, b+1,...,c), the median is given to be 7.5/8*c:   (b+c)/2 = (7.5/8)*c   b = (7/8)*c   For set R, consisting of numbers (a, a+1,...c), the median needs to be found:   a = b/2 = (7/8*c)/2 = (7/16)*c   Median = (a + c)/2 = (7/16*c + c)/2 = (23/16)*c/2 = (23/32)*c (ANSWER 23/32)
<result>D</result>
a, b, and c are integers and a<b<c. S is the set of all integers from a to b, inclusive. Q is the set of all integers from b to c, inclusive. The median of set S is (3/4)*b. The median of set Q is (7.5/8)*c. If R is the set of all integers from a to c, inclusive, what fraction of c is the median of set R?