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a, b, and c are integers and a < b < c. S is the set

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New post 18 Oct 2006, 09:34
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

Could anyone explain these question?

Thanks
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New post 18 Oct 2006, 12:03
Answer is C

the explanation I have is (1/4+7/8)/2...
Can anyone explain this in a better way?
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New post 18 Oct 2006, 12:07
Explanation

For consecutive integer series, Mean = Median = (First Term + last term)/ 2 because the entire series behaves as a number line.

So from set S, (a + b) / 2 = ¾ b => a = b / 2
From other set, ( b + c ) / 2 = 7/8 c => b = ¾ c

A = b/2 = 3/8 c

For the new set Median is (a + c) / 2 = (3/8 c + c) / 2 = 11/16 c
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New post 18 Oct 2006, 12:41
I got the answer right by chance..thanks aditya for a nice explanantion :)
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New post 18 Oct 2006, 21:29
ummm. got it. Thanks for your explanation.
  [#permalink] 18 Oct 2006, 21:29
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a, b, and c are integers and a < b < c. S is the set

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