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

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

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03 Jul 2005, 08:55
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a, b, and c are integers and a < b < c. S is the set of all integers k such that a <= k <=b . Q is the set of all integers l such that b <= l <= c. 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 m such that a <= m <=c, 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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03 Jul 2005, 09:33
I will go with C

S = {6,7,8,9,10,11,12}
Q = {12,13,14,15,16}
R = {6,7,8,9,10,11,12,13,14,15,16}

The median of the above set is 11 and 11 is 11/16 of 16, so C
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05 Jul 2005, 13:41
I will go with C

S = {6,7,8,9,10,11,12}
Q = {12,13,14,15,16}
R = {6,7,8,9,10,11,12,13,14,15,16}

The median of the above set is 11 and 11 is 11/16 of 16, so C

can you give me a more detailled explanation
how did you find that S has an odd number of terms and which one to include in it , for Q and R as well
thanks

regards

Mandy
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05 Jul 2005, 15:34
mandy wrote:
can you give me a more detailled explanation
how did you find that S has an odd number of terms and which one to include in it , for Q and R as well
thanks

regards

Mandy

Since the median of set S is 3/4b & set Q is 7/8C. Pick a number that will be divisble by 8. So I chose 16.
So let C be 16 because this one has the maximum among a,b,c.
If C is 16, set Q's median number should be 7/8 * 8 = 14.
So this is how I arrived at set Q.
Similarly, I chose 12 for b and came up with set S.
Set R includes everything from Set S & Q. After building set R one can find the median and arrive at the answer.

AJB, Do you want to wait for some participants before posting the OA
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06 Jul 2005, 14:00
mandy wrote:
can you give me a more detailled explanation
how did you find that S has an odd number of terms and which one to include in it , for Q and R as well
thanks

regards

Mandy

Since the median of set S is 3/4b & set Q is 7/8C. Pick a number that will be divisble by 8. So I chose 16.
So let C be 16 because this one has the maximum among a,b,c.
If C is 16, set Q's median number should be 7/8 * 8 = 14.
So this is how I arrived at set Q.
Similarly, I chose 12 for b and came up with set S.
Set R includes everything from Set S & Q. After building set R one can find the median and arrive at the answer.

AJB, Do you want to wait for some participants before posting the OA

Thanks just a least question how did u determine A AND 6 TO BE the beggining number in set S
Thanks again,
regards
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06 Jul 2005, 15:53
mandy wrote:
Thanks just a least question how did u determine A AND 6 TO BE the beggining number in set S
Thanks again,
regards

I started with the last number 12 and since the median k is an integer and 'a' has to be less than 'b' I equally spaced out the numbers
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06 Jul 2005, 21:39
The median of the combined set should be somewhere between 3/4 and 7/8 . Only 11/16 matches this range.
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07 Jul 2005, 07:10
BG wrote:
The median of the combined set should be somewhere between 3/4 and 7/8 . Only 11/16 matches this range.

Plz BG can you explain this further
thanks

regards

mandy
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08 Jul 2005, 06:13
(a+b)/2=3/4b => a=1/2b
(b+c)/2=7/8c => b=3/4c

=> a=1/2*3/4c=3/8c => insert in (a+c)/2=median of R

=> (3/8c+c)/2=11/16c

C)...
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08 Jul 2005, 14:50
christoph wrote:
(a+b)/2=3/4b => a=1/2b
(b+c)/2=7/8c => b=3/4c

=> a=1/2*3/4c=3/8c => insert in (a+c)/2=median of R

=> (3/8c+c)/2=11/16c

C)...

Thanks Christoph, This is a much more shorter & sweeter version than the one that I had.
08 Jul 2005, 14:50
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