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 (7/8) c. If R is the set of all integers from a to c, inclusive, what fraction of c is the median of set R?

A. 3/8
B. 1/2
C. 11/16
D. 5/7
E. 3/4

IMO, the answer is C.

I approach these problems by simple assumptions. I take some values for set S, say {2,4,9,10,12}. I see to it it, the set contains odd elements, so it becomes easy to obtain median. here the median is 9, which fulfils the condition (3/4)b. Now, set Q be { 12,14,16}, median 14 fulfilling the condition (7/8)c.
So the set R becomes now { 2,4,9,10,12,12,14,16}. Median for this will be 11, [(10+12)/2]. This 11/16 times of 16. hence the answer is 11/16.

I may be wrong...OA please.

all integers in between means that the set is a sequence of consecutive integers.

Recall that in a sequence of consecutive integers, the mean will equal the median. Also, in a set of consecutive integers, the mean of the first and last numbers will equal the mean of the whole set (since the mean of the first and last numbers will be the number between them, which is the mean/median of the whole sequence). Thus, in the first set, (a+b)/2=3b/4, a=b/2

In the second set, (b+c)/2=7c/8, b=3/4c

We want to find the (a+c)/2

a=b/2=(3c/4)/2=3c/8
(3c/8+c)/2=(11c/8)/2=(11/16)c

So the answer is 11/16