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

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rishabh26m
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Re: a, b, and c are integers and a < b < c. S is the set of all integers [#permalink] New post   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 [#permalink] New post   06 Feb 2012, 04:01
Another quick question, so for this question we're assuming that the medium is equal to mean. I thought the only way for that to happen is if there is no skewness in the set, but it doesn't say that anywhere. Is there any sort of general rule to tell if medium = mean?

Thanks so much Bunuel
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a, b, and c are integers and a < b < c. S is the set of all integers [#permalink] New post   06 Feb 2012, 04:34
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nonameee wrote:
Median of a combined interval will be in the middle between the median of Q and the median of S:

(\(3/4\) b + \(7/8\) c) * \(1/2\) (1)

From the formula for median of Q we get:

(b+c)/2 = \(7/8\) c ==> b = \(3/4\) c (2)

Substituting b from (2) into (1) we get:

(\(3/4\) *\(3/4\)c + \(7/8\) c) * 1/2 ==> \(23/32\) c


Please help.

Thank you.


Red part is not correct: we cannot assume that as we don't know that a and c are equidistant from b. If it were so then the median would simply be b.

It should be as shown in my post: (a+c)/2.
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Re: a, b, and c are integers and a < b < c. S is the set of all integers [#permalink] New post   06 Feb 2012, 05:11
Bunuel, so in order to determine a median of two intervals of integers (a,b) and (b,c) (where a<b<c), you should always use the formula: (a+c)/2?
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Re: a, b, and c are integers and a < b < c. S is the set of all integers [#permalink] New post   10 Jul 2013, 02:58
Bunnel, if i take set S as 3,6,8 and set Q as 8,14,16 whats wrong with it? satisfy questions requirement and are not in AP.
We cant apply consecutive integers formula then.
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Re: a, b, and c are integers and a < b < c. S is the set of all integers [#permalink] New post   09 Feb 2016, 23:39
Hi.. So for this question only way to solve is by knowing the fact that median will lie between the 2 and eliminating the options... right?
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Re: a, b, and c are integers and a < b < c. S is the set of all integers [#permalink] New post   29 Mar 2016, 00:05
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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 [#permalink] New post   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 integers [#permalink] New post   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)

Answer (C)


Where does 28 come from?
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Re: a, b, and c are integers and a < b < c. S is the set of all integers [#permalink] New post   05 Jul 2017, 09:50
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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 [#permalink] New post   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...

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 [#permalink] New post   15 Sep 2020, 11:43
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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


Q is the set of all integers from b to c, inclusive.
The median of set Q is (7/8)c.
Let \(c=16\).
Median of Q \(= \frac{7}{8}*c = \frac{7}{8}*6 = 14\).
Since the median of 14 must be halfway between \(b\) and c=16, \(b=12\).

S is the set of all integers from a to b, inclusive.
The median of set S is (3/4)b.
Median of S \(= \frac{3}{4}*b = \frac{3}{4}*12 = 9\).
Since the median of 9 must be halfway between \(a\) and b=12, \(a=6\).

The median of set R -- which is composed of all of the integers from a=6 to c=16, inclusive -- is equal to the the average of 6 and 16:
\(\frac{6+16}{2} = 11\)

What fraction of c is the median of set R?
\(\frac{median-of-R}{c} = \frac{11}{16}\)

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C
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Re: a, b, and c are integers and a < b < c. S is the set of all integers [#permalink] New post   15 Sep 2020, 13:21
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--> correct: median and mean for consecutive integers are same. so \(\frac{a+b}{2} =\frac{3c}{4} => b=2a \) & \(\frac{b+c}{2} =\frac{7c}{8} => 4b=3c => c = \frac{8a}{3}\), a, b & c are integers, so let's say a=3, so b = 2*3=6 & c = 8, so median of \(R = \frac{a+c}{2} =\frac{3+8}{2} =\frac{11}{2} = \frac{11}{16}*8 = \frac{11}{16}*c\)
(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 [#permalink] New post   18 Dec 2020, 00:18
VeritasKarishma
Hello,

Can we take the series as S= (1,3,4), Q= (4,7,8) and R= (1,3,4,7,8),
In this case the median of R = 4 is 1/2 of the value of 8.
Also all given conditions are fulfilled by the set of numbers.
How can we consider the series is in A.P?
Please help.
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Re: a, b, and c are integers and a < b < c. S is the set of all integers [#permalink] New post   24 Jan 2022, 09:29
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

Show Spoiler
\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}


median of consecutive integers a to b = (a+b)/2
median of consecutive integers b to c = (b+c)/2
median of consecutive integers a to c = (a+c)/2

(a+b)/2 = 3/4 b => b = 2a

(b+c)/2 = 7/8c => 4b = 3c

=> 8a = 3c

median of consecutive integers a to c = (a+c)/2 = (3c/8+c)/2 = 11c/16

Answer = 11/16
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Re: a, b, and c are integers and a < b < c. S is the set of all integers [#permalink] New post   03 Dec 2023, 12:28
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Re: a, b, and c are integers and a < b < c. S is the set of all integers [#permalink]
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