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

 It is currently 20 Jun 2019, 12:30 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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  x, y, and z are integers and x<y<z. C is the set of all integers from

Author Message
TAGS:

Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 55732
x, y, and z are integers and x<y<z. C is the set of all integers from  [#permalink]

Show Tags

2
14 00:00

Difficulty:   75% (hard)

Question Stats: 57% (02:32) correct 43% (03:10) wrong based on 130 sessions

HideShow timer Statistics

x, y, and z are integers and x<y<z. C is the set of all integers from x to y, inclusive, and D is the set of all integers from y to z, inclusive. The median of set C is $$\frac{5}{8}y$$ and the median of set D is $$\frac{3}{4}z$$. If E is the set of all integers from x to z, inclusive, then the median of set E is what fraction of z?

A. 1/2
B. 9/16
C. 5/8
D. 11/16
E. 3/4

_________________
CEO  D
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2943
Location: India
GMAT: INSIGHT
Schools: Darden '21
WE: Education (Education)
Re: x, y, and z are integers and x<y<z. C is the set of all integers from  [#permalink]

Show Tags

5
1
Bunuel wrote:
x, y, and z are integers and x<y<z. C is the set of all integers from x to y, inclusive, and D is the set of all integers from y to z, inclusive. The median of set C is $$\frac{5}{8}y$$ and the median of set D is $$\frac{3}{4}z$$. If E is the set of all integers from x to z, inclusive, then the median of set E is what fraction of z?

A. 1/2
B. 9/16
C. 5/8
D. 11/16
E. 3/4

C is the set of all integers from x to y

CONCEPT: For terms in Arithmatic progression (in which adjacent terms are equidistant) the MEAN = MEDIAN = Average of First and Last Term

i.e. Median of C = (x+y)/2 = (5/8)y (Given)
i.e. 4x + 4y = 5y
i.e. y = 4x

D is the set of all integers from y to z
i.e. Median of D = (y+z)/2 = (3/4)z (Given)
i.e. 2y + 2z = 3z
i.e. z = 2y

Median of Set E = (x+z)/2 = [(y/4)+2y]/2 = 9y/8 = (9/8)*(z/2) = (9/16)z

_________________
Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

ACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION
General Discussion
Intern  S
Joined: 29 Jan 2017
Posts: 44
Re: x, y, and z are integers and x<y<z. C is the set of all integers from  [#permalink]

Show Tags

1
Set C: make y= 8 then the median is 5 and x=2
Set Z: make z=4 then the median is 3 and y=2

x<y<z
C: 2<8
Z: 2<4 (x4)
E: 2<8<16 median is 9

9/16
Retired Moderator P
Joined: 22 Aug 2013
Posts: 1437
Location: India
Re: x, y, and z are integers and x<y<z. C is the set of all integers from  [#permalink]

Show Tags

2
Lets take y=8, so take median of C = 5/8 * 8 = 5.
So set C is a set of consecutive integers which goes upto 8 and its median is 5.
This means C must be: C = {2, 3, 4, 5, 6, 7, 8}

Now set D should start from 8 and go upto z such that median = 3z/4
lets try for z=16, so median becomes 3*16/4 = 12
So set D goes from 8 to 16 and median is 12 (which fits in properly, no mistake)
D = {8, 9, 10, 11, 12, 13, 14, 15, 16}

now Set E = {2, 3, 4, ....., 16}
15 terms, median will be the middle term, which is (15+1)/2 = 8th term
8th term of set E is 9.

We have to say 9 is what fraction of z.
9/16
Intern  B
Joined: 12 Sep 2017
Posts: 28
Re: x, y, and z are integers and x<y<z. C is the set of all integers from  [#permalink]

Show Tags

How do we know that the numbers are consecutive?
Math Expert V
Joined: 02 Sep 2009
Posts: 55732
Re: x, y, and z are integers and x<y<z. C is the set of all integers from  [#permalink]

Show Tags

@s wrote:
How do we know that the numbers are consecutive?

We are told that C is the set of all integers from x to y, inclusive.
_________________
Non-Human User Joined: 09 Sep 2013
Posts: 11398
Re: x, y, and z are integers and x<y<z. C is the set of all integers from  [#permalink]

Show Tags

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: x, y, and z are integers and x<y<z. C is the set of all integers from   [#permalink] 25 Dec 2018, 20:56
Display posts from previous: Sort by

x, y, and z are integers and x<y<z. C is the set of all integers from  