nonameee wrote:
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
OA:
Bunuel or someone else, where am I going wrong with this one?
Median of a combined interval will be in the middle between the median of Q and the median of S:
(\(3/4\) b + \(7/8\) c) * \(1/2\) (1)
From the formula for median of Q we get:
(b+c)/2 = \(7/8\) c ==> b = \(3/4\) c (2)
Substituting b from (2) into (1) we get:
(\(3/4\) *\(3/4\)c + \(7/8\) c) * 1/2 ==> \(23/32\) c
Please help.
Thank you.
Given that S is the
set of all integers from a to b, inclusive, Q is the
set of all integers from b to c, inclusive and R is the set of
all integers from a to c, inclusive, so sets S, Q and R have to be consecutive integers sets. For any set of consecutive integers (generally for any evenly spaced set) median (also the mean) equals to the average of the first and the last terms.
So we have:
Median of \(S=\frac{a+b}{2}=b*\frac{3}{4}\) --> \(b=2a\);
Median of \(Q=\frac{b+c}{2}=c*\frac{7}{8}\) --> \(b=c*\frac{3}{4}\) --> \(2a=c*\frac{3}{4}\) --> \(a=c*\frac{3}{8}\);
Median of \(R=\frac{a+c}{2}=\frac{c*\frac{3}{8}+c}{2}=c*\frac{11}{16}\)
Answer: C (\(\frac{11}{16}\)).