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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}
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Last edited by Bunuel on 20 Aug 2012, 01:50, edited 1 time in total.
Edited the question and added the OA.



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Re: Statistics.. [#permalink]
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KocharRohit wrote: 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 \(R=\frac{a+c}{2}=\frac{c*\frac{3}{8}+c}{2}=c*\frac{11}{16}\) Answer: C (\(\frac{11}{16}\))
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Re: Statistics.. [#permalink]
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27 Oct 2009, 23:46
Set R has three numbers..a , b , c..so why did we take median as (a+c)/2 shudnt it be only b??



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Re: Statistics.. [#permalink]
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27 Oct 2009, 23:59



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Re: Statistics.. [#permalink]
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28 Oct 2009, 00:01
Thanks Buddy..much clear now..



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Re: Statistics.. [#permalink]
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28 Oct 2009, 17:27
got the same C Though, i am afraid, how i will manage it under stress, and under time constraints
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Median advanced and tricky [#permalink]
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12 Jan 2010, 04:31
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?
(A) 3/8 (B) 1/2 (C) 11/16 (D) 5/7 (E) 3/4



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Re: Median advanced and tricky [#permalink]
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12 Jan 2010, 05:21
chetan2u wrote: C.. Why



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Re: Median advanced and tricky [#permalink]
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12 Jan 2010, 05:46
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..
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Re: Median advanced and tricky [#permalink]
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16 Jan 2010, 12:54
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



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Re: Median advanced and tricky [#permalink]
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16 Jan 2010, 13:43
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



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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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26 Dec 2016, 01:25
BunuelHello, can you please clarify how it is clear from basic condition that sets are evenly spaced or consecutive? Thanks.



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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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Alexey1989x wrote: BunuelHello, 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: BunuelHello, 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. Now it is clear,thanks!




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