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Statistics [#permalink] New post 19 Aug 2009, 11:04
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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

Explanations welcome. Found it worth sharing
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Re: Statistics [#permalink] New post 19 Aug 2009, 11:25
is it C?

I kinda drawn it out like:

3< 4.5(M) < 6 < 7(M) <8
I assumed C was 8 and since M(median) is 7, b has to be 6 and then figured out a being 3. Now add them all and found to be 33/6 = 5.5 average and then found which answer multiplies by C(8) gives me 5.5 Obviously not an efficient way of doing it but at least it got some kinda of result...

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Re: Statistics [#permalink] New post 18 Sep 2009, 00:15
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The answer is C: 11/16.

The key to this problem is remembering that the median for a consecutive set of numbers is equivalent to its mean. For example, the mean and median of a set consisting of x, x+1, x+2, ... ,y will always be (x+y)/2.

For set S, consisting of numbers (a, a+1,...,b), the median is given to be 3/4*b:

(a+b)/2 = (3/4)*b
a = b/2

For set Q, consisting of numbers (b, b+1,...,c), the median is given to be 7/8*c:

(b+c)/2 = (7/8)*c
b = (3/4)*c

For set R, consisting of numbers (a, a+1,...c), the median needs to be found:
a = b/2 = (3/4*c)/2 = (3/8)*c

Median = (a + c)/2 = (3/8*c + c)/2 = (11/8)*c/2 = (11/16)*c
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Re: Statistics [#permalink] New post 18 Sep 2009, 05:08
very good question, goldgoldandgold!
Clear and easy explanation, AKProdigy87!
Thank you too, guys. Kudos.
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Re: Statistics [#permalink] New post 05 Nov 2010, 00:31
tough question to solve in real-time... looks easy when u check soln
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Re: Statistics [#permalink] New post 15 Dec 2010, 02:57
I have a doubt here...the question does not say anywhere that the numbers are consecutive...then is it feasible to assume the mean=median property? or is plugging numbers a better option?
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Re: Statistics [#permalink] New post 15 Dec 2010, 03:19
Expert's post
rishabh26m wrote:
I have a doubt here...the question does not say anywhere that the numbers are consecutive...then is it feasible to assume the mean=median property? or is plugging numbers a better option?


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 given:
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}).
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Re: Statistics   [#permalink] 15 Dec 2010, 03:19
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