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

 It is currently 24 Apr 2019, 11:52

### 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

# Consider a set S = {2, 4, 6, 8, x, y} with distinct elements. If x and

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 54496
Consider a set S = {2, 4, 6, 8, x, y} with distinct elements. If x and  [#permalink]

### Show Tags

30 Jan 2016, 02:55
00:00

Difficulty:

95% (hard)

Question Stats:

44% (02:46) correct 56% (02:38) wrong based on 190 sessions

### HideShow timer Statistics

Consider a set S = {2, 4, 6, 8, x, y} with distinct elements. If x and y are both prime numbers and 0 < x < 40 and 0 < y < 40, which of the following MUST be true?

I. The maximum possible range of the set is greater than 33.
II. The median can never be an even number.
III. If y = 37, the average of the set will be greater than the median.

A. I only
B. I and II only
C. I and III only
D. III only
E. I, II, and III

_________________
Verbal Forum Moderator
Status: Greatness begins beyond your comfort zone
Joined: 08 Dec 2013
Posts: 2258
Location: India
Concentration: General Management, Strategy
Schools: Kelley '20, ISB '19
GPA: 3.2
WE: Information Technology (Consulting)
Re: Consider a set S = {2, 4, 6, 8, x, y} with distinct elements. If x and  [#permalink]

### Show Tags

30 Jan 2016, 05:55
1
1
S = {2, 4, 6, 8, x, y}
x and y are both prime numbers and 0 < x < 40 and 0 < y < 40
I. The maximum possible range of the set is greater than 33.
The biggest prime number < 40 is 37 . To obtain the max possible range either x or y has to be 37
Range = 37-2 = 35
True
II. The median can never be an even number.
S = {2, 4, 6, 8, x, y}
If x and y are prime numbers greater than 8 , then median will be (6+8)/2 = 7
If x and y are 3 and 5 , then set S = {2,3,4,5,6,8} , then median will be = 9/2
If x = 3 , then set S = { 2,3,4,6,8, y} , then median will be = 5
If x = 5 , then set S = { 2,4,5,6,8,y} , then median will be = 11/2
If x = 7 , then set S ={ 2,4,6,7,8,y} , then median will be = 13/2
Median will either be an odd number or a decimal .
True
III. If y = 37, the average of the set will be greater than the median.
S = {2, 4, 6, 8, x, 37}
The average will be least when x is smallest
S = {2, 3, 4, 6, 8, 37}
If x=3 , then average will be 60/6 = 10 and median will be 5 .
The median will be highest when x>8 , median = 7 , now the average will be more than 10 .
True
_________________
When everything seems to be going against you, remember that the airplane takes off against the wind, not with it. - Henry Ford
The Moment You Think About Giving Up, Think Of The Reason Why You Held On So Long
+1 Kudos if you find this post helpful
Manager
Joined: 15 Dec 2016
Posts: 102
Re: Consider a set S = {2, 4, 6, 8, x, y} with distinct elements. If x and  [#permalink]

### Show Tags

29 Jun 2017, 13:17
how to solve this within 2 mins---- seems hard to do this within 2 mins ?

any suggestions on cutting above steps perhaps ?
Intern
Joined: 19 Jul 2016
Posts: 2
Re: Consider a set S = {2, 4, 6, 8, x, y} with distinct elements. If x and  [#permalink]

### Show Tags

02 Jul 2017, 11:33
I have a question, the first line say "with distinct elements", and for me means that x and y are totally different from the numbers of the set. For this reason the statement III does not fix.

Could you clarify it please or why do you consider x=3?
Intern
Joined: 27 Mar 2017
Posts: 10
Re: Consider a set S = {2, 4, 6, 8, x, y} with distinct elements. If x and  [#permalink]

### Show Tags

03 Jul 2017, 09:21
leysbals wrote:
I have a question, the first line say "with distinct elements", and for me means that x and y are totally different from the numbers of the set. For this reason the statement III does not fix.

Could you clarify it please or why do you consider x=3?

in the given set they have mentioned all the numbers are distinct and x and y are prime number, now considering that, 3 is the number not included in the set given and is the smallest prime number so x taken as 3 to get least possible mean
Re: Consider a set S = {2, 4, 6, 8, x, y} with distinct elements. If x and   [#permalink] 03 Jul 2017, 09:21
Display posts from previous: Sort by