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I 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.

I 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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Re: a, b, and c are integers and a < b < c. S is the set of all integers
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30 Aug 2016, 17:11

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nonameee 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

Another option is to PLUG in numbers.

We're told that the median of set Q is (7/8)c Let's let c = 8 This means the median = (7/8)(8) = 7 For the median to equal 7, the other integer in set Q must be 6 So, set Q = {6, 7, 8}

This means that b = 6

We're told that the median of set S is (3/4)b This means the median of set S = (3/4)(6) = 4.5 Starting with b = 6, let's keep adding integers LESS THAN 6 to set S until we get a median of 4.5 Set S = {5, 6} (nope, the median is not 4.5) Set S = {4, 5, 6} (nope, the median is not 4.5) Set S = {3, 4, 5, 6} (YES, the median is 4.5!) So, set S = {3, 4, 5, 6}

This means set R = {3, 4, 5, 6, 7, 8} Here, c = 8 and the median = 5.5 What fraction of c is the median of set R? 5.5/8 = 11/16

Re: a, b, and c are integers and a < b < c. S is the set of all integers
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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 integers
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04 Jul 2017, 18:29

VeritasPrepKarishma wrote:

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)

Re: a, b, and c are integers and a < b < c. S is the set of all integers
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05 Jul 2017, 09:50

azelastine wrote:

VeritasPrepKarishma wrote:

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)

Where does 28 come from?

We assumed c = 32 Median of Q = (7/8)*c = (7/8)*32 = 28

Since all data is given in fractions and we are asked to give the answer in fraction too, assuming any suitable value of c will help get the answer.
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Re: a, b, and c are integers and a < b < c. S is the set of all integers
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06 Jul 2017, 10:49

shaselai wrote:

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...

Is the way above where the person does not rely on the medians given but simply assumes numbers that fit the set correct?

Also, as long as the values are integers, the median does not have to be a whole number, right? I can have set S consist of [3,4,5,6] which would meet the requirements but the median in this case would be 4.5, i.e. not an integer.

Re: a, b, and c are integers and a < b < c. S is the set of all integers
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07 Jul 2017, 02:57

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azelastine wrote:

shaselai wrote:

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...

Is the way above where the person does not rely on the medians given but simply assumes numbers that fit the set correct?

Also, as long as the values are integers, the median does not have to be a whole number, right? I can have set S consist of [3,4,5,6] which would meet the requirements but the median in this case would be 4.5, i.e. not an integer.

Thanks.

Yes, it is correct. The person does rely on the medians to get the elements of the set. You start with c = 8, get median as (7/8)*c so median is 7. This would mean that the set Q would be {6, 7, 8}. This means b = 6. So median of (3/4)b = 4.5. Set S would need median of 4.5 which means its elements will be {3, 4, 5, 6}. Median of even number of consecutive integers will be a non-integer. Set R = {3, 4, 5, 6, 7, 8} Its median is 5.5. Fraction = 5.5/8 = 11/16

Yes, median of a set of integers can certainly be a non integer.
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Re: a, b, and c are integers and a < b < c. S is the set of all integers
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