Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 27 Aug 2016, 17:30

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Manager
Status: Berkeley Haas 2013
Joined: 23 Jul 2009
Posts: 191
Followers: 1

Kudos [?]: 36 [3] , given: 16

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

### Show Tags

19 Aug 2009, 12:04
3
KUDOS
4
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

53% (04:10) correct 47% (02:38) wrong based on 73 sessions

### HideShow timer Statistics

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
[Reveal] Spoiler: OA
Current Student
Joined: 12 Jun 2009
Posts: 1847
Location: United States (NC)
Concentration: Strategy, Finance
Schools: UNC (Kenan-Flagler) - Class of 2013
GMAT 1: 720 Q49 V39
WE: Programming (Computer Software)
Followers: 23

Kudos [?]: 236 [0], given: 52

Re: a, b, and c are integers and a<b<c. S is the set of all integers from [#permalink]

### Show Tags

19 Aug 2009, 12:25
1
This post was
BOOKMARKED
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!
_________________

Manager
Joined: 11 Sep 2009
Posts: 129
Followers: 5

Kudos [?]: 317 [5] , given: 6

Re: a, b, and c are integers and a<b<c. S is the set of all integers from [#permalink]

### Show Tags

18 Sep 2009, 01:15
5
KUDOS
2
This post was
BOOKMARKED

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
Manager
Joined: 10 Jul 2009
Posts: 129
Location: Ukraine, Kyiv
Followers: 4

Kudos [?]: 121 [0], given: 60

Re: a, b, and c are integers and a<b<c. S is the set of all integers from [#permalink]

### Show Tags

18 Sep 2009, 06:08
very good question, goldgoldandgold!
Clear and easy explanation, AKProdigy87!
Thank you too, guys. Kudos.
_________________

Never, never, never give up

Intern
Joined: 11 Oct 2010
Posts: 27
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: a, b, and c are integers and a<b<c. S is the set of all integers from [#permalink]

### Show Tags

05 Nov 2010, 01:31
tough question to solve in real-time... looks easy when u check soln
Intern
Joined: 22 Nov 2010
Posts: 1
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: a, b, and c are integers and a<b<c. S is the set of all integers from [#permalink]

### Show Tags

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?
Math Expert
Joined: 02 Sep 2009
Posts: 34457
Followers: 6279

Kudos [?]: 79672 [1] , given: 10022

Re: a, b, and c are integers and a<b<c. S is the set of all integers from [#permalink]

### Show Tags

15 Dec 2010, 04:19
1
KUDOS
Expert's post
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 ($$\frac{11}{16}$$).
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 11100
Followers: 511

Kudos [?]: 134 [0], given: 0

Re: a, b, and c are integers and a<b<c. S is the set of all integers from [#permalink]

### Show Tags

07 Mar 2016, 20:42
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 6830
Location: Pune, India
Followers: 1926

Kudos [?]: 11956 [1] , given: 221

Re: a, b, and c are integers and a<b<c. S is the set of all integers from [#permalink]

### Show Tags

07 Mar 2016, 23:00
1
KUDOS
Expert's post
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)

_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 06 Jun 2014 Posts: 45 Followers: 0 Kudos [?]: 3 [0], given: 100 Re: a, b, and c are integers and a<b<c. S is the set of all integers from [#permalink] ### Show Tags 26 Mar 2016, 05:08 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 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 6830 Location: Pune, India Followers: 1926 Kudos [?]: 11956 [0], given: 221 Re: a, b, and c are integers and a<b<c. S is the set of all integers from [#permalink] ### Show Tags 29 Mar 2016, 00:05 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. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

Veritas Prep Reviews

Re: a, b, and c are integers and a<b<c. S is the set of all integers from   [#permalink] 29 Mar 2016, 00:05
Similar topics Replies Last post
Similar
Topics:
10 Set A consists of the integers from 4 to 12, inclusive, while set B 15 21 Jul 2015, 03:18
8 In the figure, BD and CD are angular bisectors of <B and <C in an 1 25 Oct 2014, 06:12
8 Two integers will be randomly selected from sets A and B 7 24 Oct 2012, 22:05
37 a, b, and c are integers and a < b < c. S is the set of all 19 06 Feb 2012, 03:37
16 a, b, and c are integers and a < b < c. S is the set of all 13 27 Oct 2009, 23:18
Display posts from previous: Sort by