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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

This post received KUDOS

4

This post was BOOKMARKED

00:00

A

B

C

D

E

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?

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...

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

This post received KUDOS

2

This post was BOOKMARKED

The answer is C: 11/16.

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

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?

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 \(R=\frac{a+c}{2}=\frac{c*\frac{3}{8}+c}{2}=c*\frac{11}{16}\)

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. _________________

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)

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.

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. _________________

This is the kickoff for my 2016-2017 application season. After a summer of introspect and debate I have decided to relaunch my b-school application journey. Why would anyone want...

Check out this awesome article about Anderson on Poets Quants, http://poetsandquants.com/2015/01/02/uclas-anderson-school-morphs-into-a-friendly-tech-hub/ . Anderson is a great place! Sorry for the lack of updates recently. I...

Time is a weird concept. It can stretch for seemingly forever (like when you are watching the “Time to destination” clock mid-flight) and it can compress and...