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a, b, and c are integers and a < b < c. S is the set of all [#permalink]

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27 Oct 2009, 23:18

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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

\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.

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?

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..
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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

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]

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23 Feb 2014, 03:51

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

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15 Apr 2015, 07:20

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

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Bunuel Hello, can you please clarify how it is clear from basic condition that sets are evenly spaced or consecutive? Thanks.

We are told that S is the set of all integers from a to b, inclusive. For example, S can be from 1 to 10, inclusive: 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10. In any way S will be set of consecutive integers.
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Re: a, b, and c are integers and a < b < c. S is the set of all [#permalink]

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28 Dec 2016, 23:59

Bunuel wrote:

Alexey1989x wrote:

Bunuel Hello, can you please clarify how it is clear from basic condition that sets are evenly spaced or consecutive? Thanks.

We are told that S is the set of all integers from a to b, inclusive. For example, S can be from 1 to 10, inclusive: 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10. In any way S will be set of consecutive integers.

a, b, and c are integers and a < b < c. S is the set of all [#permalink]

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04 Sep 2017, 08:42

Ans is C

S : a,........,b Q: b,........,c R: a,.........,c

median of Q= 7/8 c , let us say c=8 ,16,... median of Q= 7 if c = 8 ___________________________________and median Q = 14 if c = 16 and so on... if c is 8 median is 7 then b = 6 and ________________________on other hand c=14 means series will be 12,13,14,15,16 , b=12 if b=6 then median 3/4xb = 18/4 =4.5 ____________________if b=12 then median of S = 9 if median =4.5 then series is 3,4,5,6 if a=3 then series is 3,4,5,6,7,8____________________________if median S=9 then series of S = 6,7,8,9,10,11,12 which implies a =6 so its median is 11/2 ______________________________________if a=6 then series R: 6,7,8,9..........,15,16 and median = 11 and 16 and c x fraction=11/2______________________________________if median =11 and c=16 then fraction is 11/16 8X=11/2___________________________________________________fraction x 11 =16 X=\(\frac{11}{16}\)__________________________________________________fraction = \(\frac{11}{16}\)
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