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 (4/8)*c. If R is the set of all integers from a to c, inclusive, what fraction of c is the median of set R?	Choose the correct choice from the following options
A) 2/5	B) 1/2	C) 11/16	D) 5/7	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 4/8*c:   (b+c)/2 = (4/8)*c   b = 0   For set R, consisting of numbers (a, a+1,...c), the median needs to be found:   a = b/2 = (0)/2= 
<gadget id="calculator">(0)/2</gadget>
<output>0</output>
0   Median= (a + c)/2 = (0 + c)/2 = c/2 = (1/2)*c (ANSWER 1/2)
<result>B</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 (4/8)*c. If R is the set of all integers from a to c, inclusive, what fraction of c is the median of set R?