GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 17 Jan 2019, 18:07

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

Confirm Cancel
Events & Promotions in January
PrevNext
SuMoTuWeThFrSa
303112345
6789101112
13141516171819
20212223242526
272829303112
Open Detailed Calendar
  • The winning strategy for a high GRE score

     January 17, 2019

     January 17, 2019

     08:00 AM PST

     09:00 AM PST

    Learn the winning strategy for a high GRE score — what do people who reach a high score do differently? We're going to share insights, tips and strategies from data we've collected from over 50,000 students who used examPAL.
  • Free GMAT Strategy Webinar

     January 19, 2019

     January 19, 2019

     07:00 AM PST

     09:00 AM PST

    Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT.

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

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

Hide Tags

 
Manager
Manager
avatar
Status: Berkeley Haas 2013
Joined: 23 Jul 2009
Posts: 184
a, b, and c are integers and a<b<c. S is the set of all integers from  [#permalink]

Show Tags

New post 19 Aug 2009, 11:04
3
24
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

57% (03:11) correct 43% (02:38) wrong based on 191 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
Most Helpful Expert Reply
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 52231
Re: a, b, and c are integers and a<b<c. S is the set of all integers from  [#permalink]

Show Tags

New post 15 Dec 2010, 03:19
2
4
rishabh26m wrote:
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 \(R=\frac{a+c}{2}=\frac{c*\frac{3}{8}+c}{2}=c*\frac{11}{16}\)

Answer: C (\(\frac{11}{16}\)).
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | 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

Most Helpful Community Reply
Manager
Manager
avatar
Joined: 11 Sep 2009
Posts: 129
Re: a, b, and c are integers and a<b<c. S is the set of all integers from  [#permalink]

Show Tags

New post 18 Sep 2009, 00:15
18
5
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
General Discussion
Current Student
User avatar
Status: What's your raashee?
Joined: 12 Jun 2009
Posts: 1765
Location: United States (NC)
Concentration: Strategy, Finance
Schools: UNC (Kenan-Flagler) - Class of 2013
GMAT 1: 720 Q49 V39
WE: Programming (Computer Software)
Reviews Badge
Re: a, b, and c are integers and a<b<c. S is the set of all integers from  [#permalink]

Show Tags

New post 19 Aug 2009, 11:25
1
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...

Kudos for the question!
_________________

If you like my answers please +1 kudos!

Manager
Manager
User avatar
Joined: 10 Jul 2009
Posts: 104
Location: Ukraine, Kyiv
Re: a, b, and c are integers and a<b<c. S is the set of all integers from  [#permalink]

Show Tags

New post 18 Sep 2009, 05:08
1
very good question, goldgoldandgold!
Clear and easy explanation, AKProdigy87!
Thank you too, guys. Kudos.
_________________

Never, never, never give up

Intern
Intern
avatar
Joined: 11 Oct 2010
Posts: 9
Reviews Badge
Re: a, b, and c are integers and a<b<c. S is the set of all integers from  [#permalink]

Show Tags

New post 05 Nov 2010, 00:31
tough question to solve in real-time... looks easy when u check soln
Intern
Intern
avatar
Joined: 22 Nov 2010
Posts: 1
Re: a, b, and c are integers and a<b<c. S is the set of all integers from  [#permalink]

Show Tags

New post 15 Dec 2010, 02:57
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?
Veritas Prep GMAT Instructor
User avatar
D
Joined: 16 Oct 2010
Posts: 8789
Location: Pune, India
Re: a, b, and c are integers and a<b<c. S is the set of all integers from  [#permalink]

Show Tags

New post 07 Mar 2016, 22:00
1
1
goldgoldandgold 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



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)

Answer (C)
_________________

Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >

Intern
Intern
avatar
Joined: 06 Jun 2014
Posts: 45
Re: a, b, and c are integers and a<b<c. S is the set of all integers from  [#permalink]

Show Tags

New post 26 Mar 2016, 04:08
1
Hi Bunuel, VeritasPrepKarishma


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
Veritas Prep GMAT Instructor
User avatar
D
Joined: 16 Oct 2010
Posts: 8789
Location: Pune, India
Re: a, b, and c are integers and a<b<c. S is the set of all integers from  [#permalink]

Show Tags

New post 28 Mar 2016, 23:05
kzivrev wrote:
Hi Bunuel, VeritasPrepKarishma


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.
_________________

Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >

Intern
Intern
avatar
B
Joined: 10 Nov 2013
Posts: 18
Re: a, b, and c are integers and a<b<c. S is the set of all integers from  [#permalink]

Show Tags

New post 04 Jul 2017, 17:29
VeritasPrepKarishma wrote:
goldgoldandgold 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



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)

Answer (C)


Where does 28 come from?
Intern
Intern
avatar
B
Joined: 01 May 2017
Posts: 30
Re: a, b, and c are integers and a [#permalink]

Show Tags

New post 04 Jul 2017, 19:40
Consider set S =(a.. 3/4b...b) because 3/4b is the median of the set S
Since the integers in a given set would be consecutive, median, (a+b) /2=3/4b

Now consider set Q=(b... 7/8c...c) because 7/8c is the median of the set Q
Since the integers in a given set would be consecutive, median, (b+c)/2=7/8c

set R=(a..... c).
Thus median =(a+c) /2

Adding the median equations of set S and set Q and simplifying we get,
7/8c + 1/4b = (a+c) /2
But b = 3/4c.

Thus (a+c) /2 = 11/16c.

Answer is Option C.

Regards,
Shreya

Sent from my ONEPLUS A3003 using GMAT Club Forum mobile app
Veritas Prep GMAT Instructor
User avatar
D
Joined: 16 Oct 2010
Posts: 8789
Location: Pune, India
Re: a, b, and c are integers and a<b<c. S is the set of all integers from  [#permalink]

Show Tags

New post 05 Jul 2017, 08:50
azelastine wrote:
VeritasPrepKarishma wrote:
goldgoldandgold 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



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)

Answer (C)


Where does 28 come from?


We assumed c = 32
Median of Q = (7/8)*c = (7/8)*32 = 28

Since all data is given in fractions and we are asked to give the answer in fraction too, assuming any suitable value of c will help get the answer.
_________________

Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >

Intern
Intern
avatar
B
Joined: 26 Sep 2016
Posts: 7
Re: a, b, and c are integers and a<b<c. S is the set of all integers from  [#permalink]

Show Tags

New post 05 Jul 2017, 09:29
When I was doing the task I tried just to make a common numbers.
3/4 and 7/8
3/4 = 21/28 (x7)
7/8 = 28/32 (x4)
Hence, 21/32 = 10,5/16 or 11/16
Tell me please is that coincidence???
Veritas Prep GMAT Instructor
User avatar
D
Joined: 16 Oct 2010
Posts: 8789
Location: Pune, India
Re: a, b, and c are integers and a<b<c. S is the set of all integers from  [#permalink]

Show Tags

New post 06 Jul 2017, 03:34
Jahfors wrote:
When I was doing the task I tried just to make a common numbers.
3/4 and 7/8
3/4 = 21/28 (x7)
7/8 = 28/32 (x4)
Hence, 21/32 = 10,5/16 or 11/16
Tell me please is that coincidence???


What logic are you using here?
And 21/32 is not the same as 11/16.
_________________

Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >

Intern
Intern
avatar
B
Joined: 10 Nov 2013
Posts: 18
a, b, and c are integers and a<b<c. S is the set of all integers from  [#permalink]

Show Tags

New post 06 Jul 2017, 09:49
shaselai wrote:
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...

Kudos for the question!


Hi VeritasPrepKarishma

Is the way above where the person does not rely on the medians given but simply assumes numbers that fit the set correct?

Also, as long as the values are integers, the median does not have to be a whole number, right? I can have set S consist of [3,4,5,6] which would meet the requirements but the median in this case would be 4.5, i.e. not an integer.

Thanks.
Veritas Prep GMAT Instructor
User avatar
D
Joined: 16 Oct 2010
Posts: 8789
Location: Pune, India
Re: a, b, and c are integers and a<b<c. S is the set of all integers from  [#permalink]

Show Tags

New post 07 Jul 2017, 01:57
1
azelastine wrote:
shaselai wrote:
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...

Kudos for the question!


Hi VeritasPrepKarishma

Is the way above where the person does not rely on the medians given but simply assumes numbers that fit the set correct?

Also, as long as the values are integers, the median does not have to be a whole number, right? I can have set S consist of [3,4,5,6] which would meet the requirements but the median in this case would be 4.5, i.e. not an integer.

Thanks.


Yes, it is correct. The person does rely on the medians to get the elements of the set.
You start with c = 8, get median as (7/8)*c so median is 7. This would mean that the set Q would be {6, 7, 8}. This means b = 6.
So median of (3/4)b = 4.5. Set S would need median of 4.5 which means its elements will be {3, 4, 5, 6}. Median of even number of consecutive integers will be a non-integer.
Set R = {3, 4, 5, 6, 7, 8}
Its median is 5.5.
Fraction = 5.5/8 = 11/16

Yes, median of a set of integers can certainly be a non integer.
_________________

Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >

Intern
Intern
avatar
B
Joined: 31 May 2018
Posts: 11
a, b, and c are integers and a<b<c. S is the set of all integers from  [#permalink]

Show Tags

New post 21 Nov 2018, 01:11
VeritasKarishma wrote:
goldgoldandgold 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



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)

Answer (C)


Hi Karishma,

loved your explanation, got every word of it.

but why are we considering only consecutive integers?

thank you
GMAT Club Bot
a, b, and c are integers and a<b<c. S is the set of all integers from &nbs [#permalink] 21 Nov 2018, 01:11
Display posts from previous: Sort by

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

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


Copyright

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

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

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®.