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a, b, and c

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by ektamatta » Sun May 25, 2008 7:14 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 (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

OA is C
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Source: — Problem Solving |
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by ksh » Sun May 25, 2008 7:37 am
Hi ektamatta,

Answer is C

median of set S=(a+b)/2
3b/4=(a+b)/2=> a=b/2

Now median of set Q= (b+c)/2
7c/8= (b+c)/2 => c=4b/3

Median of set R=(a+c)/2
=b/2+4b/3
=11b/12
=11/12*3c/4
=11c/16 ->answer C
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by airan » Sun May 25, 2008 8:45 am
Hi Ksh,
median of set S=(a+b)/2
3b/4=(a+b)/2=> a=b/2
Can u help to explain how 3b/4=(a+b)/2
Thanks
Airan
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by netigen » Sun May 25, 2008 11:17 am
In a set of consecutive integers the median of the set in equal to the mean of the set

https://www.beatthegmat.com/set-rules-pl ... 11062.html
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