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prachi18oct
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One option is to PLUG in numbers.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
We're told that the median of set Q is (7/8)c
Let's let c = 8
This means the median = (7/8)(8) = 7
For the median to equal 7, the other integer in set Q must be 6
So, set Q = {6, 7, 8}
This means that b = 6
We're told that the median of set S is (3/4)b
This means the median of set S = (3/4)(6) = 4.5
Starting with b = 6, let's keep adding integers LESS THAN 6 to set S until we get a median of 4.5
Set S = {5, 6} (nope, the median is not 4.5)
Set S = {4, 5, 6} (nope, the median is not 4.5)
Set S = {3, 4, 5, 6} (YES, the median is 4.5!)
So, set S = {3, 4, 5, 6}
This means set R = {3, 4, 5, 6, 7, 8}
Here, c = 8 and the median = 5.5
What fraction of c is the median of set R?
5.5/8 = [spoiler]11/16 = C[/spoiler]
Cheers,
Brent
