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by k.pankaj.r » Mon Dec 05, 2011 6:26 pm
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 the set S is (3/4)b. the median of the 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 C
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by GMATGuruNY » Mon Dec 05, 2011 6:52 pm
I posted a solution here:

https://www.beatthegmat.com/integers-nic ... 82287.html
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by chieftang » Mon Dec 05, 2011 6:54 pm
Median of a -> b = (a+b)/2 = (3/4)*b
Median of b -> c = (b+c)/2 = (7/8)*c
Median of a -> c = (a+c)/2 = ?

Solve for a in terms of c:
a = (3/8)*c

Plug it in:
((3/8)*c + c)/2 = (11/16)*c

ANSWER: C
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by k.pankaj.r » Mon Dec 05, 2011 7:35 pm
thanks mitch..the post helped a lot..
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by [email protected] » Sat Dec 24, 2011 4:57 am
chieftang's solution is better!!!
IT IS TIME TO BEAT THE GMAT

LEARNING, APPLICATION AND TIMING IS THE FACT OF GMAT AND LIFE AS WELL... KEEP PLAYING!!!

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