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

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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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New post 18 Sep 2009, 00:15
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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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New post 19 Aug 2009, 11:25
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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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New post 18 Sep 2009, 05:08
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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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New post 05 Nov 2010, 00:31
tough question to solve in real-time... 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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New post 15 Dec 2010, 02:57
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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New post 15 Dec 2010, 03:19
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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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New post 07 Mar 2016, 22:00
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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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New post 26 Mar 2016, 04:08
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Hi Bunuel, VeritasPrepKarishma


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
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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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New post 28 Mar 2016, 23:05
kzivrev wrote:
Hi Bunuel, VeritasPrepKarishma


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 from  [#permalink]

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New post 04 Jul 2017, 17:29
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)

Answer (C)


Where does 28 come from?
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Re: a, b, and c are integers and a [#permalink]

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New post 04 Jul 2017, 19:40
Consider set S =(a.. 3/4b...b) because 3/4b is the median of the set S
Since the integers in a given set would be consecutive, median, (a+b) /2=3/4b

Now consider set Q=(b... 7/8c...c) because 7/8c is the median of the set Q
Since the integers in a given set would be consecutive, median, (b+c)/2=7/8c

set R=(a..... c).
Thus median =(a+c) /2

Adding the median equations of set S and set Q and simplifying we get,
7/8c + 1/4b = (a+c) /2
But b = 3/4c.

Thus (a+c) /2 = 11/16c.

Answer is Option C.

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Shreya

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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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New post 05 Jul 2017, 08:50
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 from  [#permalink]

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New post 05 Jul 2017, 09:29
When I was doing the task I tried just to make a common numbers.
3/4 and 7/8
3/4 = 21/28 (x7)
7/8 = 28/32 (x4)
Hence, 21/32 = 10,5/16 or 11/16
Tell me please is that coincidence???
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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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New post 06 Jul 2017, 03:34
Jahfors wrote:
When I was doing the task I tried just to make a common numbers.
3/4 and 7/8
3/4 = 21/28 (x7)
7/8 = 28/32 (x4)
Hence, 21/32 = 10,5/16 or 11/16
Tell me please is that coincidence???


What logic are you using here?
And 21/32 is not the same as 11/16.
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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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New post 06 Jul 2017, 09:49
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...

Kudos for the question!


Hi VeritasPrepKarishma

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.
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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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New post 07 Jul 2017, 01:57
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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...

Kudos for the question!


Hi VeritasPrepKarishma

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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