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

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a , b, and c are integers and a < b < c. S is the set [#permalink] New post 17 Jun 2007, 05:56
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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
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@grad_mba [#permalink] New post 17 Jun 2007, 15:20
If you take b=8 and c=16 then how come the median of set Q is (7/8)C ??
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C... [#permalink] New post 24 Jun 2007, 10:24
Since we are calculating the median of all the integers from a to b & from b to c. Median will be same as average in both cases (even or odd number of numbers between any range).

From the Media of Set S

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

=> a = b/2

similarly, From the Median of Set Q

b = (3/4)c

& hence median of integers from a to c is (11/16)c
:wink:
C...   [#permalink] 24 Jun 2007, 10:24
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a , b, and c are integers and a < b < c. S is the set

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