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what fraction of c

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what fraction of c

by sanju09 » Mon Mar 30, 2009 3:40 am
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 ¾ 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. ½
C. 11/16
D. 5/7
E. ¾

OA C
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Re: what fraction of c

by Vemuri » Mon Mar 30, 2009 4:18 am
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
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by lav » Mon Mar 30, 2009 4:50 am
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
Last edited by lav on Mon Mar 30, 2009 4:52 am, edited 1 time in total.
Kid in Verbal :(
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Re: what fraction of c

by gabriel » Mon Mar 30, 2009 4:52 am
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.
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by gabriel » Mon Mar 30, 2009 5:01 am
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.
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by lav » Mon Mar 30, 2009 5:04 am
@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 ?
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by gabriel » Mon Mar 30, 2009 5:08 am
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.

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by sanju09 » Mon Mar 30, 2009 5:12 am
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.
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by gabriel » Mon Mar 30, 2009 5:16 am
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.

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by Vemuri » Mon Mar 30, 2009 8:04 am
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).
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by sanju09 » Wed Feb 17, 2010 4:15 am
you're all WELCOME, mates
The mind is everything. What you think you become. -Lord Buddha



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