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a, b, and c are integers and a < b < c. S is the set of all [#permalink]
27 Oct 2009, 22:18

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Question Stats:

57% (05:15) correct
43% (02:50) wrong based on 69 sessions

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

\sqrt{I solved it lke this... for set S (a+b)/ 2 = (3/4 ) B for set Q (b+c)/2 = (7c)/8 after solving we got 3c = 4b = 8a median for set R is b so 4b = 3c b = (3c)/4..

got E ..plasde tell me am i wrong somewhere? Rohit}

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

I solved it lke this... for set S (a+b)/ 2 = (3/4 ) B for set Q (b+c)/2 = (7c)/8 after solving we got 3c = 4b = 8a median for set R is b so 4b = 3c b = (3c)/4..

got E ..plasde tell me am i wrong somewhere? Rohit

Given: Median of S=\frac{a+b}{2}=b*\frac{3}{4} --> b=2a;

Median of Q=\frac{b+c}{2}=c*\frac{7}{8} --> b=c*\frac{3}{4} --> 2a=c*\frac{3}{4} --> a=c*\frac{3}{8};

Median of R=\frac{a+c}{2}=\frac{c*\frac{3}{8}+c}{2}=c*\frac{11}{16}

Set R has three numbers..a , b , c..so why did we take median as (a+c)/2 shudnt it be only b??

Not so. We are told that a<b<c, but not that they are consecutive integers, it could be 3<55<79. So set R in that case would be all integers from 3 to 79 inclusive. And b not necessarily would be the median. Median would be (3+79)/2=41.

Median advanced and tricky [#permalink]
12 Jan 2010, 03:31

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?

Re: Median advanced and tricky [#permalink]
12 Jan 2010, 04:46

shalva wrote:

chetan2u wrote:

C..

Why

for nos between b and c ,7c/8 is the median..means c has to be a multiple of 4 or 8... but for 3b/4 to be a median, c has to be a multiple of 8 ..... lets take it as 8.. median=7*8/7=7.... set is 6,7,8...this makes b=6...median=6*3/4= 9/2.... so the set is 3,4,5,6.. set Q is 3,4,5,6,7,8.. the median is 11/2,which is 11/16 of 8 or c.... we can try the same with other multiple of 8 also..

Re: Median advanced and tricky [#permalink]
16 Jan 2010, 11:54

i fails to understand how can the median between two number a and b can be a+b/2.It depends on number of integer between the range so median between 1 6 and 7 cannot be 4 it will be 6 .so please calcify the approach in details

Re: Median advanced and tricky [#permalink]
16 Jan 2010, 12:43

arghya05 wrote:

i fails to understand how can the median between two number a and b can be a+b/2.It depends on number of integer between the range so median between 1 6 and 7 cannot be 4 it will be 6 .so please calcify the approach in details

Well, from some point you are right: median divides the sample in two parts: Equal number of observations in each. It doesn't necessarily mean (a+b)/2

But you should also consider that, in set S, there are "all integers from a to b, inclusive", and in set Q there are "all integers from b to c, inclusive"

So these sets are sets of consecutive integers. In this case Mean=Median=(a+b)/2

Re: a, b, and c are integers and a < b < c. S is the set of all [#permalink]
23 Feb 2014, 02:51

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