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

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Manager
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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}
[Reveal] Spoiler: OA

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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27 Oct 2009, 23:39
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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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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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27 Oct 2009, 23:59
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KocharRohit wrote:
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.

Hope it's clear.
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28 Oct 2009, 00:01
Thanks Buddy..much clear now..
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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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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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12 Jan 2010, 05:21
chetan2u wrote:
C..

Why
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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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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
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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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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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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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05 Aug 2016, 23:17
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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
Bunuel
Hello, 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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26 Dec 2016, 02:53
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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.
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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.

Now it is clear,thanks!
Re: a, b, and c are integers and a < b < c. S is the set of all   [#permalink] 28 Dec 2016, 23:59
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