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

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Manager
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a, b, and c are integers and a < b < c. S is the set [#permalink] New post 17 Jun 2007, 11:51
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

Please explain the approach
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Re: Mean of the 3rd set [#permalink] New post 17 Jun 2007, 14:17
iamba 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) 1/2
(C) 11/16
(D) 5/7
(E) 3/4

Please explain the approach


Note that for a set of consecutive integers, the median is the the average of the first and the last integer
Median of S =(a+b)/2 therefore a=b/2
Median of Q=(b+c)/2 therefore b= (3/4)c
Thus a= (3/8)c
Median of R = (a+c)/2 = (11/16)c
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 [#permalink] New post 17 Jun 2007, 16:25
Excellent solution by kevincan. I would guess such a question in exam.
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 [#permalink] New post 17 Jun 2007, 17:37
Great explaination. Knowing such systematic approaches helps a lot
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 [#permalink] New post 17 Jun 2007, 17:49
Good one. Excellent solution
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reply [#permalink] New post 19 Jun 2007, 12:27
Kevin, how did you decipher that 'they were consective integers'. It didn't mention in the problem?
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Re: reply [#permalink] New post 19 Jun 2007, 13:36
alimad wrote:
Kevin, how did you decipher that 'they were consective integers'. It didn't mention in the problem?


The stem says - ALL INTEGERS between a and b. It means integers are consecutive.
Re: reply   [#permalink] 19 Jun 2007, 13:36
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a, b, and c are integers and a < b < c. S is the set

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