abhijeetsinghai 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.11/16
c. 1/2
d.5/7
[spoiler]OA: b[/spoiler]
I don't knw how to approach this question. Please throw some light. Thanx !!
Here's one approach:
Set S: We know that the largest number is b and the median is 3b/4. 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
You can see that 3b/4 (the median) lies directly in the middle 2b/4 and b (i.e., 2b/4, 3b/4, 4b/4)
So, set S ranges from b/2 to b
Set Q: Here the largest number is c and the median is 7c/8. Using 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 equivalant to b (the largest value in set S)
So, now we have a relationship between c and b. We know that 6c/8 = b
From here, we can find two integer values for b and c that satisfy this equation amd determine how sets S and Q might look.
One solution is b=6 and c=8
For set S, we know that the largest value is 6, so keep adding numbers until we meet the criterion that the median is 3/4 b. We do the same with set S, beginning with c=8.
From here we can conclude 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
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
