This is an interesting question & I am sure if such a question comes I will panic in the real exam watching the clock ticking.

Anyway, here's how I would approach the problem when I am sane & calm. Lets give some numbers to a, b & c.
a=1
b=16 (I chose 16 for easy computation with the fraction)
c=32 (I chose 32 for easy computation with the fraction)

The median between a & b is 3/4b, i.e. 12
The median between b & c is 7/8c, i.e. 28

Now, we have our numbers, so lets lay them down:

1_12_16_28_32 (totally 9 numbers). The median for the entire set is 16. Now, the question is asking what fraction of c is the median.

i.e. 32x=16
==> x=1/2. So, the fraction is 1/2. My answer is B

thats a nice try vemuri
but i am less convinced if that is right way to find answer , may get lucky ni saying B

what if i take ( for easy calc )
a=1
b=48
c=64

The median between a & b is 3/4b, i.e. 36
The median between b & c is 7/8c, i.e. 56

Now, we have our numbers, so lets lay them down:

1_36_48_56_64 (totally 9 numbers). The median for the entire set is 48. Now, the question is asking what fraction of c is the median.

i.e. 64x=48
==> x=3/4. So, the fraction is 3/4. ans E ??

In short taking speciifc values of b and c is not right way according to me
I can choose b and c to get other ans choices , It all depends on b and c as with the logic above

bx=c always

Vemuri wrote:This is an interesting question & I am sure if such a question comes I will panic in the real exam watching the clock ticking.

Anyway, here's how I would approach the problem when I am sane & calm. Lets give some numbers to a, b & c.
a=1
b=16 (I chose 16 for easy computation with the fraction)
c=32 (I chose 32 for easy computation with the fraction)

The median between a & b is 3/4b, i.e. 12
The median between b & c is 7/8c, i.e. 28

Now, we have our numbers, so lets lay them down:

1_12_16_28_32 (totally 9 numbers). The median for the entire set is 16. Now, the question is asking what fraction of c is the median.

i.e. 32x=16
==> x=1/2. So, the fraction is 1/2. My answer is B

Well I am pretty sure if you change the numbers the answer would be different. The thing is that the median is a positional average and hence does not have a lot of mathematical value and cannot be easily interpreted mathematically. I am really curious to read Sanju's answer, maybe I am missing something.

Actually just read the question again. It is pretty simple. The only thing to note is that for a set of consecutive integers the median is same as the mean. Use this and the solution is pretty simple.

@gabriel
But the question never says that we have a set of consecutive integers. it just says that

a, b, and c are integers and a < b < c.

right ? or am i missing something ?

It says S is the set of all inetgers from a to c inclusive, similarly for Q it says it is the set of all inetgers from b to c, inclusive.

Cheers

I am really curious to read Sanju's answer, maybe I am missing something.

Sure gabriel!

Since S contains consecutive integers only, its median will be the arithmetic mean of the extremes; hence we can take

(a + b)/2 = 3 b/4 => 2 a = b _________(i)

same take with set Q enables us to have

(b + c)/2 = 7 c/8 => 4 b = 3 c ________ (ii)

Median of set R will be (a + c)/2, take a in terms of c by the help of (i) & (ii) to get a = 3 c/8, substitution gets us the median in terms of c, as here [3 c/8 + c]/2 = 11/16 of c.

Thanks a lot Sanju !!.

I did get the answer. I had not read the question properly initially and missed the fact that the question is talking about consecutive integers. Anyway good question, thanks for sharing.

Cheers.

Lav, Thanks for pointing the flaw in my logic. I guess that's what happens in real time as well. I did not give a second thought to my logic because I had already taken 3-4 mins trying to figure out how to solve this problem.

Thanks a lot Sanju for posting this question. Lesson Learnt: Every word in the question needs to be comprehended properly otherwise it will be nightmare. In this scenario, I never bothered to understand the hint the question was trying to give me (the consecutive numbers).

you're all WELCOME, mates