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a, b, and c are integers and a<b<c. S is the set of all integers from [#permalink]
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19 Aug 2009, 12:04
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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
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Re: a, b, and c are integers and a<b<c. S is the set of all integers from [#permalink]
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19 Aug 2009, 12:25
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is it C? I kinda drawn it out like: 3< 4.5(M) < 6 < 7(M) <8 I assumed C was 8 and since M(median) is 7, b has to be 6 and then figured out a being 3. Now add them all and found to be 33/6 = 5.5 average and then found which answer multiplies by C(8) gives me 5.5 Obviously not an efficient way of doing it but at least it got some kinda of result... Kudos for the question!
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Re: a, b, and c are integers and a<b<c. S is the set of all integers from [#permalink]
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18 Sep 2009, 01:15
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The answer is C: 11/16.
The key to this problem is remembering that the median for a consecutive set of numbers is equivalent to its mean. For example, the mean and median of a set consisting of x, x+1, x+2, ... ,y will always be (x+y)/2.
For set S, consisting of numbers (a, a+1,...,b), the median is given to be 3/4*b:
(a+b)/2 = (3/4)*b a = b/2
For set Q, consisting of numbers (b, b+1,...,c), the median is given to be 7/8*c:
(b+c)/2 = (7/8)*c b = (3/4)*c
For set R, consisting of numbers (a, a+1,...c), the median needs to be found: a = b/2 = (3/4*c)/2 = (3/8)*c
Median = (a + c)/2 = (3/8*c + c)/2 = (11/8)*c/2 = (11/16)*c



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Re: a, b, and c are integers and a<b<c. S is the set of all integers from [#permalink]
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18 Sep 2009, 06:08
very good question, goldgoldandgold! Clear and easy explanation, AKProdigy87! Thank you too, guys. Kudos.
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Re: a, b, and c are integers and a<b<c. S is the set of all integers from [#permalink]
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05 Nov 2010, 01:31
tough question to solve in realtime... looks easy when u check soln



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Re: a, b, and c are integers and a<b<c. S is the set of all integers from [#permalink]
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15 Dec 2010, 03:57
I have a doubt here...the question does not say anywhere that the numbers are consecutive...then is it feasible to assume the mean=median property? or is plugging numbers a better option?



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Re: a, b, and c are integers and a<b<c. S is the set of all integers from [#permalink]
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rishabh26m wrote: I have a doubt here...the question does not say anywhere that the numbers are consecutive...then is it feasible to assume the mean=median property? or is plugging numbers a better option? Given that S is the set of all integers from a to b, inclusive, Q is the set of all integers from b to c, inclusive and R is the set of all integers from a to c, inclusive, so sets S, Q and R have to be consecutive integers sets. For any set of consecutive integers (generally for any evenly spaced set) median (also the mean) equals to the average of the first and the last terms. So 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: a, b, and c are integers and a<b<c. S is the set of all integers from [#permalink]
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07 Mar 2016, 23:00
goldgoldandgold 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 Since the question and options give all data in terms of fractions, we can assume values. The point is for which variable we should assume a value? a < b < c S = { a, ..., b } Median = (3/4)b Q = {b, ..., c } Median = (7/8)c Assume a value of c (which will lead to a value of b) such that c will get divided by 8 and 4 successively so that both medians are integers. Say c = 32 Median of Q = (7/8)*32 = 28 b = 28  4 = 24 Median of S = (3/4)*24 = 18 a = 18  6 = 12 R = { 12, 13, 14, ... 32} Median of R = 22 which as a fraction of c is 22/32 = (11/16) Answer (C)
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Re: a, b, and c are integers and a<b<c. S is the set of all integers from [#permalink]
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26 Mar 2016, 05:08
Hi Bunuel, VeritasPrepKarishmaI think that assumption in this question is to have distinct numbers, numbers can not duplicate, for example: 6 , 7 , 8 , 9 , 10 , 11 , 12 , 13 , 14 , 15 , 16 here a=6 b= 12 c= 16 and conditions are fulfiled for 3/4*b and 7/8*c so final answer would be 11/16. BUT if we include more 9's in set S and more 14's in set Q we will still have all numbers inclusive from a to c in the joint set, but than the median would not be 11, can be different numebr depending how many 9's and 14' are duplicating. Do you think that is possable or I'm wrong Thanks in advance for your respnce



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Re: a, b, and c are integers and a<b<c. S is the set of all integers from [#permalink]
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29 Mar 2016, 00:05
kzivrev wrote: Hi Bunuel, VeritasPrepKarishmaI think that assumption in this question is to have distinct numbers, numbers can not duplicate, for example: 6 , 7 , 8 , 9 , 10 , 11 , 12 , 13 , 14 , 15 , 16 here a=6 b= 12 c= 16 and conditions are fulfiled for 3/4*b and 7/8*c so final answer would be 11/16. BUT if we include more 9's in set S and more 14's in set Q we will still have all numbers inclusive from a to c in the joint set, but than the median would not be 11, can be different numebr depending how many 9's and 14' are duplicating. Do you think that is possable or I'm wrong Thanks in advance for your respnce By definition, a set is a collection of distinct objects. So we can easily say that all integers in the sets must be distinct. Sometimes, some questions do specify sets with identical elements but strictly speaking, sets have distinct elements.
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