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

 It is currently 24 Oct 2016, 17:32

### 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

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Manager
Joined: 03 Oct 2009
Posts: 93
Followers: 1

Kudos [?]: 31 [1] , given: 2

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

### Show Tags

27 Oct 2009, 23:18
1
KUDOS
16
This post was
BOOKMARKED
00:00

Difficulty:

75% (hard)

Question Stats:

55% (04:46) correct 45% (02:55) wrong based on 159 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

\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}
[Reveal] Spoiler: OA

Last edited by Bunuel on 20 Aug 2012, 01:50, edited 1 time in total.
Edited the question and added the OA.
Math Expert
Joined: 02 Sep 2009
Posts: 35275
Followers: 6636

Kudos [?]: 85571 [2] , given: 10237

### Show Tags

27 Oct 2009, 23:39
2
KUDOS
Expert's post
6
This post was
BOOKMARKED
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

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

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 ($$\frac{11}{16}$$)
_________________
Manager
Joined: 03 Oct 2009
Posts: 93
Followers: 1

Kudos [?]: 31 [0], given: 2

### Show Tags

27 Oct 2009, 23:46
Set R has three numbers..a , b , c..so why did we take median as
(a+c)/2 shudnt it be only b??
Math Expert
Joined: 02 Sep 2009
Posts: 35275
Followers: 6636

Kudos [?]: 85571 [1] , given: 10237

### Show Tags

27 Oct 2009, 23:59
1
KUDOS
Expert's post
KocharRohit wrote:
Set R has three numbers..a , b , c..so why did we take median as
(a+c)/2 shudnt it be only b??

Not so. We are told that a<b<c, but not that they are consecutive integers, it could be 3<55<79. So set R in that case would be all integers from 3 to 79 inclusive. And b not necessarily would be the median. Median would be (3+79)/2=41.

Hope it's clear.
_________________
Manager
Joined: 03 Oct 2009
Posts: 93
Followers: 1

Kudos [?]: 31 [0], given: 2

### Show Tags

28 Oct 2009, 00:01
Thanks Buddy..much clear now..
Senior Manager
Joined: 18 Aug 2009
Posts: 435
Schools: UT at Austin, Indiana State University, UC at Berkeley
WE 1: 5.5
WE 2: 5.5
WE 3: 6.0
Followers: 8

Kudos [?]: 120 [0], given: 16

### Show Tags

28 Oct 2009, 17:27
got the same C

Though, i am afraid, how i will manage it under stress, and under time constraints
_________________

Never give up,,,

Manager
Joined: 25 Dec 2009
Posts: 99
Followers: 1

Kudos [?]: 143 [0], given: 3

### Show Tags

12 Jan 2010, 04:31
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
Joined: 20 Aug 2009
Posts: 311
Location: Tbilisi, Georgia
Schools: Stanford (in), Tuck (WL), Wharton (ding), Cornell (in)
Followers: 17

Kudos [?]: 137 [0], given: 69

### Show Tags

12 Jan 2010, 05:21
chetan2u wrote:
C..

Why
Verbal Forum Moderator
Joined: 02 Aug 2009
Posts: 4098
Followers: 287

Kudos [?]: 2986 [0], given: 100

### Show Tags

12 Jan 2010, 05:46
shalva wrote:
chetan2u wrote:
C..

Why

for nos between b and c ,7c/8 is the median..means c has to be a multiple of 4 or 8... but for 3b/4 to be a median, c has to be a multiple of 8 .....
lets take it as 8.. median=7*8/7=7.... set is 6,7,8...this makes b=6...median=6*3/4= 9/2.... so the set is 3,4,5,6..
set Q is 3,4,5,6,7,8.. the median is 11/2,which is 11/16 of 8 or c.... we can try the same with other multiple of 8 also..
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

Intern
Joined: 18 Mar 2009
Posts: 17
Followers: 0

Kudos [?]: 10 [0], given: 1

### Show Tags

16 Jan 2010, 12:54
i fails to understand how can the median between two number a and b can be
a+b/2.It depends on number of integer between the range so median between
1 6 and 7 cannot be 4 it will be 6 .so please calcify the approach in details
Joined: 20 Aug 2009
Posts: 311
Location: Tbilisi, Georgia
Schools: Stanford (in), Tuck (WL), Wharton (ding), Cornell (in)
Followers: 17

Kudos [?]: 137 [0], given: 69

### Show Tags

16 Jan 2010, 13:43
arghya05 wrote:
i fails to understand how can the median between two number a and b can be
a+b/2.It depends on number of integer between the range so median between
1 6 and 7 cannot be 4 it will be 6 .so please calcify the approach in details

Well, from some point you are right: median divides the sample in two parts: Equal number of observations in each. It doesn't necessarily mean (a+b)/2

But you should also consider that, in set S, there are "all integers from a to b, inclusive", and in set Q there are "all integers from b to c, inclusive"

So these sets are sets of consecutive integers. In this case Mean=Median=(a+b)/2
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 12212
Followers: 542

Kudos [?]: 151 [0], given: 0

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

### Show Tags

23 Feb 2014, 03:51
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.
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 12212
Followers: 542

Kudos [?]: 151 [0], given: 0

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

### Show Tags

15 Apr 2015, 07:20
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.
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 12212
Followers: 542

Kudos [?]: 151 [0], given: 0

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

### Show Tags

05 Aug 2016, 23:17
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.
_________________
Re: a, b, and c are integers and a < b < c. S is the set of all   [#permalink] 05 Aug 2016, 23:17
Similar topics Replies Last post
Similar
Topics:
If a, b, and c are consecutive integers such that a < b < c and a is 2 13 Apr 2016, 05:31
3 a, b, c, and d are positive consecutive integers and a < b < 3 17 Aug 2013, 10:34
42 a, b, and c are integers and a < b < c. S is the set of all 22 06 Feb 2012, 03:37
31 Positive integers a, b, c, d and e are such that a<b<c<d<e 13 18 Oct 2010, 10:57
18 a, b, and c are integers and a<b<c. S is the set of all integers from 10 19 Aug 2009, 12:04
Display posts from previous: Sort by

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

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.