It is currently 18 Nov 2017, 16:59

### 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

Author Message
TAGS:

### Hide Tags

Manager
Joined: 03 Oct 2009
Posts: 92

Kudos [?]: 54 [3], 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
3
KUDOS
20
This post was
BOOKMARKED
00:00

Difficulty:

85% (hard)

Question Stats:

53% (03:30) correct 47% (02:36) wrong based on 224 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.

Kudos [?]: 54 [3], given: 2

Math Expert
Joined: 02 Sep 2009
Posts: 42249

Kudos [?]: 132618 [3], given: 12326

### Show Tags

27 Oct 2009, 23:39
3
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 ($$\frac{11}{16}$$)
_________________

Kudos [?]: 132618 [3], given: 12326

Manager
Joined: 03 Oct 2009
Posts: 92

Kudos [?]: 54 [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??

Kudos [?]: 54 [0], given: 2

Math Expert
Joined: 02 Sep 2009
Posts: 42249

Kudos [?]: 132618 [1], given: 12326

### 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.
_________________

Kudos [?]: 132618 [1], given: 12326

Manager
Joined: 03 Oct 2009
Posts: 92

Kudos [?]: 54 [0], given: 2

### Show Tags

28 Oct 2009, 00:01
Thanks Buddy..much clear now..

Kudos [?]: 54 [0], given: 2

Senior Manager
Joined: 18 Aug 2009
Posts: 416

Kudos [?]: 145 [0], given: 16

Schools: UT at Austin, Indiana State University, UC at Berkeley
WE 1: 5.5
WE 2: 5.5
WE 3: 6.0

### 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,,,

Kudos [?]: 145 [0], given: 16

Manager
Joined: 25 Dec 2009
Posts: 99

Kudos [?]: 266 [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

Kudos [?]: 266 [0], given: 3

Joined: 20 Aug 2009
Posts: 305

Kudos [?]: 165 [0], given: 69

Location: Tbilisi, Georgia
Schools: Stanford (in), Tuck (WL), Wharton (ding), Cornell (in)

### Show Tags

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

Why

Kudos [?]: 165 [0], given: 69

Math Expert
Joined: 02 Aug 2009
Posts: 5201

Kudos [?]: 5827 [0], given: 117

### 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

Kudos [?]: 5827 [0], given: 117

Intern
Joined: 18 Mar 2009
Posts: 17

Kudos [?]: 11 [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

Kudos [?]: 11 [0], given: 1

Joined: 20 Aug 2009
Posts: 305

Kudos [?]: 165 [0], given: 69

Location: Tbilisi, Georgia
Schools: Stanford (in), Tuck (WL), Wharton (ding), Cornell (in)

### 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

Kudos [?]: 165 [0], given: 69

Non-Human User
Joined: 09 Sep 2013
Posts: 15704

Kudos [?]: 282 [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.
_________________

Kudos [?]: 282 [0], given: 0

Non-Human User
Joined: 09 Sep 2013
Posts: 15704

Kudos [?]: 282 [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.
_________________

Kudos [?]: 282 [0], given: 0

Non-Human User
Joined: 09 Sep 2013
Posts: 15704

Kudos [?]: 282 [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.
_________________

Kudos [?]: 282 [0], given: 0

Manager
Joined: 05 Dec 2016
Posts: 223

Kudos [?]: 24 [0], given: 48

Concentration: Strategy, Finance
GMAT 1: 620 Q46 V29
Re: a, b, and c are integers and a < b < c. S is the set of all [#permalink]

### Show Tags

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.

Kudos [?]: 24 [0], given: 48

Math Expert
Joined: 02 Sep 2009
Posts: 42249

Kudos [?]: 132618 [1], given: 12326

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

### Show Tags

26 Dec 2016, 02:53
1
KUDOS
Expert's post
Alexey1989x wrote:
Bunuel
Hello, can you please clarify how it is clear from basic condition that sets are evenly spaced or consecutive?
Thanks.

We are told that S is the set of all integers from a to b, inclusive. For example, S can be from 1 to 10, inclusive: 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10. In any way S will be set of consecutive integers.
_________________

Kudos [?]: 132618 [1], given: 12326

Manager
Joined: 05 Dec 2016
Posts: 223

Kudos [?]: 24 [0], given: 48

Concentration: Strategy, Finance
GMAT 1: 620 Q46 V29
Re: a, b, and c are integers and a < b < c. S is the set of all [#permalink]

### Show Tags

28 Dec 2016, 23:59
Bunuel wrote:
Alexey1989x wrote:
Bunuel
Hello, can you please clarify how it is clear from basic condition that sets are evenly spaced or consecutive?
Thanks.

We are told that S is the set of all integers from a to b, inclusive. For example, S can be from 1 to 10, inclusive: 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10. In any way S will be set of consecutive integers.

Now it is clear,thanks!

Kudos [?]: 24 [0], given: 48

Senior Manager
Joined: 29 Jun 2017
Posts: 370

Kudos [?]: 70 [0], given: 66

GPA: 4
WE: Engineering (Transportation)
a, b, and c are integers and a < b < c. S is the set of all [#permalink]

### Show Tags

04 Sep 2017, 08:42
Ans is C

S : a,........,b
Q: b,........,c
R: a,.........,c

median of Q= 7/8 c , let us say c=8 ,16,...
median of Q= 7 if c = 8 ___________________________________and median Q = 14 if c = 16 and so on...
if c is 8 median is 7 then b = 6 and ________________________on other hand c=14 means series will be 12,13,14,15,16 , b=12
if b=6 then median 3/4xb = 18/4 =4.5 ____________________if b=12 then median of S = 9
if median =4.5 then series is 3,4,5,6
if a=3 then series is 3,4,5,6,7,8____________________________if median S=9 then series of S = 6,7,8,9,10,11,12 which implies a =6
so its median is 11/2 ______________________________________if a=6 then series R: 6,7,8,9..........,15,16 and median = 11 and 16
and c x fraction=11/2______________________________________if median =11 and c=16 then fraction is 11/16
8X=11/2___________________________________________________fraction x 11 =16
X=$$\frac{11}{16}$$__________________________________________________fraction = $$\frac{11}{16}$$
_________________

Give Kudos for correct answer and/or if you like the solution.

Kudos [?]: 70 [0], given: 66

a, b, and c are integers and a < b < c. S is the set of all   [#permalink] 04 Sep 2017, 08:42
Display posts from previous: Sort by