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
Plugging in values is the easiest and fastest approach.
Here's another approach that employs some algebra:
Set S:
The largest value in the set = b
The median = 3b/4
USEFUL TRICK: Rewrite b as 4b/4
So,
4b/4 is the largest number in the set
3b/4 is the median
Since the median is in the MIDDLE of the numbers of set S, we know that the smallest number in set S is
2b/4
Notice that
3b/4 (the median) lies directly in the middle of the set {
2b/4,
3b/4,
4b/4}
So, set S ranges form
2b/4 to
4b/4, with the median,
3b/4, in the middle.
Set Q:
The largest value in the set = c
The median = 7c/8
USEFUL TRICK: Rewrite c as 8c/8
So,
8c/8 is the largest number in the set
7c/8 is the median
sing similar logic from above, we can conclude that set Q ranges from
6c/8 to
8c/8, with
7c/8 squarely in the middle.
Here's the important step.
6c/8 is the SMALLEST number in set Q. This value is equivalent to b (the GREATEST value in set S)
So, we can say that
6c/8 = b
From here, we can find two INTEGERS VALUES for b and c that satisfy this equation and use these values tp determine how sets S and Q might look.
One solution to the equation is b = 6 and c = 8
This means that set S = {3,4,5,6} and set Q = {6,7,8}
The median of {3,4,5,6,7,8} is 5.5, which means the median is 11/16 of c
Answer:
C
We could have chosen OTHER numbers for b and c but the fraction remains 11/16
For example, b = 12 and c = 16 gives us S = {6,7,8,9,10,11,12} and Q = {12,13,14,15,16}
The median of {6, 7, . . . 15, 16} is 11, making it 11/16 of c
Answer:
still C
Cheers,
Brent
