You can solve this problem algebraically, or you can plug in numbers.
For the algebraic approach, you need to recognize that sets S and Q consist of consecutive integers. We know that the median of a set of consecutive integers is equal to the mean, which can be found by averaging the first and last integers in the set.
Since S runs from a to b, its mean/median = (a+b)/2
Since Q runs from b to c, its mean/median=(b+c)/2
So now we can use the provided information about the medians to create two equations:
From here we just need to do some manipulations to get to our answer:
Since R runs from a to c, let’s combine these results to find a in terms of c:
Now we can plug these values in to find the median of R. The mean/median of R=(a+c)/2=(3c/8 + c)/2=(11c/8)/2=11c/16
The answer is C.
However, it might be easier to plug in numbers here. We just have to think it over a little bit. If these sets consist of integers, and the medians are fractions of the end numbers, there are only so many possibilities.
Let’s look at set S first. If the median is (3/4)b, we know that b must be a multiple of 4. What are our possibilities?
b=4, median=3, set S runs from 2-4 (there is one term after the median, so there must be one term before)
b=8, median=6, set S runs from 4-8 (there are two terms after the median, so there must be two terms before)
If we want additional possibilities, we can just keep expanding this pattern:
a/last term/b = 2/3/4 or 4/6/8 or 6/9/12 or 9/12/16, etc. Each valid set is just a multiple of the first one.
We can do the same thing for set Q. Since the median is (7/8)c, we know that c must be a multiple of 8. So our possibilities for b/median/c are as follows:
Since both sets contain b, we need values for sets S and Q that have the same value for b. The first overlapping value is 12. So we can choose 6-12 for set S and 12-16 for set Q. Now we have our set R.
Since set R runs from a to c, or 6 to 16, its median is 11. c=16, so median/c=11/16.
Dmitry Farber | Manhattan GMAT Instructor | New York
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