a, b, and c are integers and a < b < c. S is the set of all : GMAT Problem Solving (PS)
Check GMAT Club App Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

It is currently 07 Dec 2016, 07:03
GMAT Club Tests

Chicago-Booth

is Releasing R1 Admission Decisions | Keep Watch on App Tracker | Join Chat Room2 for Live Updates


Close

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
Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

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

1 KUDOS received
Manager
Manager
avatar
Joined: 02 Oct 2009
Posts: 93
Followers: 1

Kudos [?]: 34 [1] , given: 2

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

Show Tags

New post 27 Oct 2009, 22:18
1
This post received
KUDOS
16
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

55% (04:44) correct 45% (02:57) wrong based on 163 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, 00:50, edited 1 time in total.
Edited the question and added the OA.
Expert Post
2 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 35911
Followers: 6851

Kudos [?]: 90014 [2] , given: 10402

Re: Statistics.. [#permalink]

Show Tags

New post 27 Oct 2009, 22:39
2
This post received
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 (\(\frac{11}{16}\))
_________________

New to the Math Forum?
Please read this: All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Manager
Manager
avatar
Joined: 02 Oct 2009
Posts: 93
Followers: 1

Kudos [?]: 34 [0], given: 2

Re: Statistics.. [#permalink]

Show Tags

New post 27 Oct 2009, 22:46
Set R has three numbers..a , b , c..so why did we take median as
(a+c)/2 shudnt it be only b??
Expert Post
1 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 35911
Followers: 6851

Kudos [?]: 90014 [1] , given: 10402

Re: Statistics.. [#permalink]

Show Tags

New post 27 Oct 2009, 22:59
1
This post received
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.
_________________

New to the Math Forum?
Please read this: All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Manager
Manager
avatar
Joined: 02 Oct 2009
Posts: 93
Followers: 1

Kudos [?]: 34 [0], given: 2

Re: Statistics.. [#permalink]

Show Tags

New post 27 Oct 2009, 23:01
Thanks Buddy..much clear now..
Senior Manager
Senior Manager
avatar
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 [?]: 121 [0], given: 16

Re: Statistics.. [#permalink]

Show Tags

New post 28 Oct 2009, 16:27
got the same C

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

Never give up,,,

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

Kudos [?]: 151 [0], given: 3

Median advanced and tricky [#permalink]

Show Tags

New post 12 Jan 2010, 03:31
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
Tuck Thread Master
User avatar
Joined: 20 Aug 2009
Posts: 311
Location: Tbilisi, Georgia
Schools: Stanford (in), Tuck (WL), Wharton (ding), Cornell (in)
Followers: 17

Kudos [?]: 139 [0], given: 69

Re: Median advanced and tricky [#permalink]

Show Tags

New post 12 Jan 2010, 04:21
chetan2u wrote:
C..


Why :?:
Expert Post
Verbal Forum Moderator
Verbal Forum Moderator
User avatar
Joined: 02 Aug 2009
Posts: 4136
Followers: 306

Kudos [?]: 3245 [0], given: 100

Re: Median advanced and tricky [#permalink]

Show Tags

New post 12 Jan 2010, 04:46
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
Intern
avatar
Joined: 18 Mar 2009
Posts: 17
Followers: 0

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

Re: Median advanced and tricky [#permalink]

Show Tags

New post 16 Jan 2010, 11:54
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
Tuck Thread Master
User avatar
Joined: 20 Aug 2009
Posts: 311
Location: Tbilisi, Georgia
Schools: Stanford (in), Tuck (WL), Wharton (ding), Cornell (in)
Followers: 17

Kudos [?]: 139 [0], given: 69

Re: Median advanced and tricky [#permalink]

Show Tags

New post 16 Jan 2010, 12:43
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
GMAT Club Legend
User avatar
Joined: 09 Sep 2013
Posts: 12885
Followers: 561

Kudos [?]: 158 [0], given: 0

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

Show Tags

New post 23 Feb 2014, 02:51
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 Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

GMAT Club Legend
GMAT Club Legend
User avatar
Joined: 09 Sep 2013
Posts: 12885
Followers: 561

Kudos [?]: 158 [0], given: 0

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

Show Tags

New post 15 Apr 2015, 06:20
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 Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

GMAT Club Legend
GMAT Club Legend
User avatar
Joined: 09 Sep 2013
Posts: 12885
Followers: 561

Kudos [?]: 158 [0], given: 0

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

Show Tags

New post 05 Aug 2016, 22:17
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.
_________________

GMAT Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

Re: a, b, and c are integers and a < b < c. S is the set of all   [#permalink] 05 Aug 2016, 22:17
    Similar topics Author Replies Last post
Similar
Topics:
Experts publish their posts in the topic If a, b, and c are consecutive integers such that a < b < c and a is Bunuel 2 13 Apr 2016, 04:31
12 Experts publish their posts in the topic If a, b, and c are consecutive positive integers and a < b < Walkabout 8 29 Dec 2012, 05:44
42 Experts publish their posts in the topic a, b, and c are integers and a < b < c. S is the set of all nonameee 22 06 Feb 2012, 02:37
31 Experts publish their posts in the topic Positive integers a, b, c, d and e are such that a<b<c<d<e shrive555 13 18 Oct 2010, 09:57
19 Experts publish their posts in the topic a, b, and c are integers and a<b<c. S is the set of all integers from goldgoldandgold 10 19 Aug 2009, 11:04
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  


GMAT Club MBA Forum Home| About| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

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®.