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 [#permalink]

Show Tags

06 Feb 2012, 03:37

4

This post received KUDOS

10

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

47% (03:12) correct
53% (02:08) wrong based on 192 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, 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?

Bunuel or someone else, where am I going wrong with this one?

Median of a combined interval will be in the middle between the median of Q and the median of S:

(\(3/4\) b + \(7/8\) c) * \(1/2\) (1)

From the formula for median of Q we get:

(b+c)/2 = \(7/8\) c ==> b = \(3/4\) c (2)

Substituting b from (2) into (1) we get:

(\(3/4\) *\(3/4\)c + \(7/8\) c) * 1/2 ==> \(23/32\) c

Please help.

Thank you.

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

Another quick question, so for this question we're assuming that the medium is equal to mean. I thought the only way for that to happen is if there is no skewness in the set, but it doesn't say that anywhere. Is there any sort of general rule to tell if medium = mean?

Another quick question, so for this question we're assuming that the medium is equal to mean. I thought the only way for that to happen is if there is no skewness in the set, but it doesn't say that anywhere. Is there any sort of general rule to tell if medium = mean?

Thanks so much Bunuel

For any evenly spaced set (aka AP) the arithmetic mean (average) is equal to the median (consecutive integers are evenly spaced set). _________________

Bunuel, so in order to determine a median of two intervals of integers (a,b) and (b,c) (where a<b<c), you should always use the formula: (a+c)/2?

The median (mean) of the integers from a to c, inclusive is always (a+c)/2 (if you have some additional info you can obtain this value in another way but this way is ALWAYS true).

Consider two sets: {1, 2, 3} and {3, 4, 5, 6, 7, 8, 9} --> combined set {1, 2, 3, 4, 5, 6, 7 8, 9}

As you've written the median (mean) of combined set should be (2+6)/2=4, which is wrong as median of combined set is 5.

Another quick question, so for this question we're assuming that the medium is equal to mean. I thought the only way for that to happen is if there is no skewness in the set, but it doesn't say that anywhere. Is there any sort of general rule to tell if medium = mean?

Thanks so much Bunuel

For any evenly spaced set (aka AP) the arithmetic mean (average) is equal to the median (consecutive integers are evenly spaced set).

Bunuel,

how do we know that they are evenly spaced. The a<b< c can be 1<2<3 or random 4<78<125 (not evenly spaced). Am i missing something?

Another quick question, so for this question we're assuming that the medium is equal to mean. I thought the only way for that to happen is if there is no skewness in the set, but it doesn't say that anywhere. Is there any sort of general rule to tell if medium = mean?

Thanks so much Bunuel

For any evenly spaced set (aka AP) the arithmetic mean (average) is equal to the median (consecutive integers are evenly spaced set).

Bunuel,

how do we know that they are evenly spaced. The a<b< c can be 1<2<3 or random 4<78<125 (not evenly spaced). Am i missing something?

Given that "S is the set of all integers from a to b, inclusive" and "Q is the set of all integers from b to c, inclusive", which means that both S and Q are sets of consecutive integers, thus evenly spaced sets.

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

Show Tags

10 Jul 2013, 02:58

Bunnel, if i take set S as 3,6,8 and set Q as 8,14,16 whats wrong with it? satisfy questions requirement and are not in AP. We cant apply consecutive integers formula then.

Bunnel, if i take set S as 3,6,8 and set Q as 8,14,16 whats wrong with it? satisfy questions requirement and are not in AP. We cant apply consecutive integers formula then.

S is the set of all integers from a to b, inclusive. Say a=3 and b=8. What is set S then? S={3, 4, 5, 6, 7, 8} not {3, 6, 8}, where did 4, 5 and 7 go? Aren't they integers in the range from 3 to 8?

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

Show Tags

08 Oct 2014, 23:02

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]

Show Tags

12 Nov 2015, 07:47

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 [#permalink]

Show Tags

06 Dec 2015, 06:52

nonameee 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?

Bunuel or someone else, where am I going wrong with this one?

Median of a combined interval will be in the middle between the median of Q and the median of S:

(\(3/4\) b + \(7/8\) c) * \(1/2\) (1)

From the formula for median of Q we get:

(b+c)/2 = \(7/8\) c ==> b = \(3/4\) c (2)

Substituting b from (2) into (1) we get:

(\(3/4\) *\(3/4\)c + \(7/8\) c) * 1/2 ==> \(23/32\) c

Please help.

Thank you.

Bunuels solution looks easier.. but may be my approach will help some of you

1.\(\frac{(b+c)}{2} = \frac{7}{8}c\) after some calculations we get\(\frac{c}{b}=\frac{4}{3}\) and using this ratio let's say c=8, b=6 2. Let's make the same for a&b: \(\frac{(a+b)}{2} = \frac{3}{4}b\) after some calculations we get b=2a, and from (1) we know that b=6 than a=3 Our numbers in set R look like this 3,4,5,6,7,8. The Median of the set is equal to \(\frac{(8+3)}{2}\) = \(\frac{11}{2}\) What fraction of c is the median of set R --> \(\frac{11}{2}=x*8\) (which is C) x\(=11/16\) Answer C _________________

When you’re up, your friends know who you are. When you’re down, you know who your friends are.

Share some Kudos, if my posts help you. Thank you !

Hi.. So for this question only way to solve is by knowing the fact that median will lie between the 2 and eliminating the options... right?

Another method is "assuming values".

a < b < c

Median of Q is 7/8 of c so c must be a multiple of 8. Say c = 8. Median of c = 7. Set Q = {6, 7, 8} Then b = 6 Median of S = (3/4) of b which is (3/4)*6 = 4.5. Set S = {3, 4, 5, 6}

So the entire set R = {3, 4, 5, 6, 7, 8} Median = 5.5

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...

“Oh! Looks like your passport expires soon” – these were the first words at the airport in London I remember last Friday. Shocked that I might not be...