Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 27 Oct 2011
Posts: 151
Location: United States
Concentration: Finance, Strategy
GPA: 3.7
WE: Account Management (Consumer Products)

The median of the list of positive integers above is 5.5
[#permalink]
Show Tags
25 Mar 2012, 18:46
Question Stats:
66% (01:23) correct 34% (00:57) wrong based on 517 sessions
HideShow timer Statistics
8, 5, x, 6 The median of the list of positive integers above is 5.5. Which of the following could be the average (arithmetic mean) of the list? A. 3 B. 5.5 C. 6.25 D. 7 E. 7.5
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
DETERMINED TO BREAK 700!!!




Math Expert
Joined: 02 Sep 2009
Posts: 49300

The median of the list of positive integers above is 5.5
[#permalink]
Show Tags
26 Mar 2012, 01:05
piyatiwari wrote: Given "The median of the list of positive integers above is 5.5"
So arranging the numbers from lowest to highest, avg of 2 middle terms needs to be 5.5
so the sequence will be x 5 6 8
Lets say x = 4, which gives us mean = sum/4 = 5.5
B is correct 8, 5, x, 6
The median of the list of positive integers above is 5.5. Which of the following could be the average (arithmetic mean) of the list?A. 3 B. 5.5 C. 6.25 D. 7 E. 7.5 Notice that we are told that x is a positive integer.The median of a set with even number of terms is the average of two middle terms when arranged in ascending/descending order. Since given that the median is 5.5 (the average of 5 and 6) then integer x must be less than or equal to 5, in order 5 and 6 to be two middle numbers. So our list in ascending order is: {x, 5, 6, 8}. Now, since \(x\leq{5}\) then the maximum sum of the numbers is 5+5+6+8=24, so the maximum average is 24/4=6 (options C, D, and E are out). 3 cannot be the average because in this case the sum of the numbers must be 4*3=12, which is less than the sum of just two numbers 6 and 8 (remember here that since x is a positive integer it cannot decrease the sum to 12). So we are left only with option B (for x=3 the average is (3+5+6+8)/4=22/4=5.5). Answer: B.
_________________
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? Extrahard Quant Tests with Brilliant Analytics




Senior Manager
Joined: 28 Jun 2009
Posts: 399
Location: United States (MA)

Re: Mediating the mean
[#permalink]
Show Tags
25 Mar 2012, 18:57
Given "The median of the list of positive integers above is 5.5"
So arranging the numbers from lowest to highest, avg of 2 middle terms needs to be 5.5
so the sequence will be x 5 6 8
Lets say x = 4, which gives us mean = sum/4 = 5.5
B is correct



Senior Manager
Status: Gonna rock this time!!!
Joined: 22 Jul 2012
Posts: 468
Location: India
GMAT 1: 640 Q43 V34 GMAT 2: 630 Q47 V29
WE: Information Technology (Computer Software)

Re: The median of the list of positive integers above is 5.5
[#permalink]
Show Tags
20 Oct 2012, 20:58
This is amazing soln I solved by plugging in.. (8+5+x+6)/4 . . plug nos from options 1 by 1.. when ans 5.5 is plugged in,you get x=3. . so you have 3,5,6,8 as the order which confirms with the given data
_________________
hope is a good thing, maybe the best of things. And no good thing ever dies.
Who says you need a 700 ?Check this out : http://gmatclub.com/forum/whosaysyouneeda149706.html#p1201595
My GMAT Journey : http://gmatclub.com/forum/endofmygmatjourney149328.html#p1197992



VP
Status: Far, far away!
Joined: 02 Sep 2012
Posts: 1086
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8

Re: 8, 5, x, 6 The median of the list of positive integers
[#permalink]
Show Tags
27 Mar 2013, 03:40
8, 5, x, 6 The median of the list of positive integers above is 5.5. Which of the following could be the average (arithmetic mean) of the list? If the median is 5.5, then x cannot be >8 ( 5,6,8,9 median 7>5,5). It cannot be > 6 (5,6,7,8 median 6.5>5.5). So we have found that \(x<=5\) \(\frac{x+5+6+8}{4}=mean\) \(x=mean*419\) x is an integer so 4*mean  19 must be an integer, option (C)6.25 is out. (D)7 (E)7.5 these options return values of x > 9, so they cannot be right. Option (A)3 returns a negative integer, but x must be positive. Option (B)5.5 is correct, and the value of x is 3, which is possible given the initial condition of "The median of the list of positive integers above is 5.5"
_________________
It is beyond a doubt that all our knowledge that begins with experience.
Kant , Critique of Pure Reason Tips and tricks: Inequalities , Mixture  Review: MGMAT workshop Strategy: SmartGMAT v1.0  Questions: Verbal challenge SC III CR New SC set out !! , My QuantRules for Posting in the Verbal Forum  Rules for Posting in the Quant Forum[/size][/color][/b]



Intern
Joined: 22 Jan 2010
Posts: 24
Location: India
Concentration: Finance, Technology
GPA: 3.5
WE: Programming (Telecommunications)

Re: The median of the list of positive integers above is 5.5
[#permalink]
Show Tags
27 Mar 2013, 04:26
The median will be the average of the two elements in the middle as the list contains even number of elements. So,the sum of the two numbers in the middle will be 5.5 * 2 = 11. With the above data,we can write the list as follows : x 5 6 8 where 1 < x < 6. Avg = (x+5+6+8)/4 = (19+x)/4. So, 5< Avg <6 Hence, B will be the answer. ** Alternatively, we can work out this problem taking options one by one.
_________________
Please press +1 KUDOS if you like my post.



Manager
Joined: 24 Nov 2012
Posts: 166
Concentration: Sustainability, Entrepreneurship
WE: Business Development (Internet and New Media)

Re: 8, 5, x, 6 The median of the list of positive integers
[#permalink]
Show Tags
03 Aug 2013, 05:43
Zarrolou wrote: n  19 must be an integer, option (C)6.25 is out.
The reason c is out is because if the average was 6.25, x = 6 and the media would become 6 (6.25 x 4) = 25 x = 25  19
_________________
You've been walking the ocean's edge, holding up your robes to keep them dry. You must dive naked under, and deeper under, a thousand times deeper!  Rumi
http://www.manhattangmat.com/blog/index.php/author/cbermanmanhattanprepcom/  This is worth its weight in gold
Economist GMAT Test  730, Q50, V41 Aug 9th, 2013 Manhattan GMAT Test  670, Q45, V36 Aug 11th, 2013 Manhattan GMAT Test  680, Q47, V36 Aug 17th, 2013 GmatPrep CAT 1  770, Q50, V44 Aug 24th, 2013 Manhattan GMAT Test  690, Q45, V39 Aug 30th, 2013 Manhattan GMAT Test  710, Q48, V39 Sep 13th, 2013 GmatPrep CAT 2  740, Q49, V41 Oct 6th, 2013
GMAT  770, Q50, V44, Oct 7th, 2013 My Debrief  http://gmatclub.com/forum/fromtheashesthoushallrise770q50v44awa5ir162299.html#p1284542



Manager
Joined: 03 May 2013
Posts: 72

Re: The median of the list of positive integers above is 5.5
[#permalink]
Show Tags
20 Aug 2015, 22:46
8, 5, x, 6
The median of the list of positive integers above is 5.5. Which of the following could be the average (arithmetic mean) of the list? A. 3 B. 5.5 C. 6.25 D. 7 E. 7.5
Notice that we are told that x is a positive integer.
Experts your view please??
In my opinion this q is flawed unless it should explicitly mentioned "list of different positive no"
list could be 5,6,6,8 and in that case we have mean 6.25



Board of Directors
Joined: 17 Jul 2014
Posts: 2683
Location: United States (IL)
Concentration: Finance, Economics
GPA: 3.92
WE: General Management (Transportation)

The median of the list of positive integers above is 5.5
[#permalink]
Show Tags
26 Dec 2015, 11:17
the question directly says that the median is 5.5. this can be only if 5 and 6 are the middle numbers. thus, x must be <=5. now, let's take x=5 = the sum of the numbers is 24, and the average is 6. since the maximum value of x can be 5, it must be that the maximum value of the mean can't be greater than 6. thus, we can eliminate C, D, and E. now, we have the sum = 24. we have in the answer choices 3 and 5.5 since these numbers should be the mean, it means that 12 or 22 is the sum of the integers. Now, we have 5,6, and 8. the sum of these numbers is way above 12. We are told that we have only positive numbers. thus, 12 can't be the sum and 3 can't be the average. A is out, and B must be the possible average. to test, let's say that x=3. now, we have 3+5+6+8=22. 22/4 = 5.5. all good. looks like I got to the right answer the way bunuel did



Current Student
Joined: 12 Aug 2015
Posts: 2648

Re: The median of the list of positive integers above is 5.5
[#permalink]
Show Tags
15 Dec 2016, 16:26
Excellent Question this one. Here is my solution >
Notice that the number of terms here are even => The median of a set with even number of terms is the average of two middle terms when arranged in either increasing or decreasing order.
Median =5.5
Hence the sum of two middle terms must be 11
For that allowed values of positive integer x => 1,2,3,4,5
Hence the mean => 19+x/4 = Value x=4*Value 19
This value is nothing but the option.
Lets put in each value and see which on of them gives us x as {1,2,3,4,5) A)Value=3 => x=1219=7 => Rejected B)Value=>5.5=> x=2219=3=> Allowed C)Value=6.25=>x=2519=6=> Rejected
D)Value=7=>x=2819=9=> Rejected E)Rejected
Hence B
Method 2=> Back solving
Let Mean = 3 Hence x=7 => Rejected Mean = 5.5 => x=3 => Median of set = 5.5 => Accepted.
Hence B
_________________
MBA Financing: INDIAN PUBLIC BANKS vs PRODIGY FINANCE! Getting into HOLLYWOOD with an MBA! The MOST AFFORDABLE MBA programs!STONECOLD's BRUTAL Mock Tests for GMATQuant(700+)AVERAGE GRE Scores At The Top Business Schools!



SVP
Joined: 08 Jul 2010
Posts: 2334
Location: India
GMAT: INSIGHT
WE: Education (Education)

Re: The median of the list of positive integers above is 5.5
[#permalink]
Show Tags
05 Jan 2018, 06:42
calreg11 wrote: 8, 5, x, 6
The median of the list of positive integers above is 5.5. Which of the following could be the average (arithmetic mean) of the list?
A. 3 B. 5.5 C. 6.25 D. 7 E. 7.5 Median = 5.5 = Average of two middle numbers in case the number of terms in teh set is even. i.e. 5 and 6 are the two middle terms of the set i.e. x must be smaller than 5 for it to be the second term when terms are arranged in the ascending order so the sum of the terms in the set = x+5+6+8 = x+19 average = (x+19)/4 = (x/4)+4.75 and since x is smaller than or equal to 5 average ≤ (5/4)+4.75 i.e. average ≤ (1.25)+4.75 average ≤ 6 since x is positive hence average > 4.75 ANswer: Option B
_________________
Prosper!!! GMATinsight Bhoopendra Singh and Dr.Sushma Jha email: info@GMATinsight.com I Call us : +919999687183 / 9891333772 Online OneonOne Skype based classes and Classroom Coaching in South and West Delhi http://www.GMATinsight.com/testimonials.html
22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION



Intern
Joined: 06 Dec 2017
Posts: 3

Re: The median of the list of positive integers above is 5.5
[#permalink]
Show Tags
05 Jan 2018, 09:13
The median is above 5.5, shouldn't we consider options like x,5,6,8 & 5,6,8,x & 5,6,x,8 ?
What Am I missing?



Math Expert
Joined: 02 Sep 2009
Posts: 49300

Re: The median of the list of positive integers above is 5.5
[#permalink]
Show Tags
05 Jan 2018, 09:22
siddharthselvamohan wrote: The median is above 5.5, shouldn't we consider options like x,5,6,8 & 5,6,8,x & 5,6,x,8 ?
What Am I missing? The median of a set with even number of terms is the average of two middle terms when arranged in ascending/descending order. So, if the set in ascending order is {5, 6, x, 8} (meaning if \(6\leq x \leq 8\)), then the median would be the average of 6 and x, and it cannot be 5.5, it could be at least 6, for x = 6. Similarly if if the set in ascending order is {5, 6, 8, x} (meaning if \(x \geq 8\)), then the median would be the average of 6 and 8, so 7, not 5.5. Thus, since given that the median is 5.5 (the average of 5 and 6) then integer x must be less than or equal to 5, in order 5 and 6 to be two middle numbers. So our list in ascending order is: {x, 5, 6, 8}. Check complete solution here: https://gmatclub.com/forum/themediano ... l#p1064955Hope it helps.
_________________
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? Extrahard Quant Tests with Brilliant Analytics




Re: The median of the list of positive integers above is 5.5 &nbs
[#permalink]
05 Jan 2018, 09:22






