Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 46297

List S consists of 10 consecutive odd integers, and list T c [#permalink]
Show Tags
30 Jan 2014, 01:57
Question Stats:
73% (02:04) correct 27% (01:54) wrong based on 1001 sessions
HideShow timer Statistics
The Official Guide For GMAT® Quantitative Review, 2ND EditionList S consists of 10 consecutive odd integers, and list T consists of 5 consecutive even integers. If the least integer in S is 7 more than the least integer in T, how much greater is the average (arithmetic mean) of the integers in S than the average of the integers in T? (A) 2 (B) 7 (C) 8 (D) 12 (E) 22 Problem Solving Question: 70 Category: Arithmetic Statistics Page: 70 Difficulty: 600 GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition  Quantitative Questions ProjectEach week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution. We'll be glad if you participate in development of this project: 1. Please provide your solutions to the questions; 2. Please vote for the best solutions by pressing Kudos button; 3. Please vote for the questions themselves by pressing Kudos button; 4. Please share your views on difficulty level of the questions, so that we have most precise evaluation. Thank you!
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
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



Math Expert
Joined: 02 Sep 2009
Posts: 46297

Re: List S consists of 10 consecutive odd integers, and list T c [#permalink]
Show Tags
30 Jan 2014, 01:57



Director
Joined: 25 Apr 2012
Posts: 702
Location: India
GPA: 3.21
WE: Business Development (Other)

Re: List S consists of 10 consecutive odd integers, and list T c [#permalink]
Show Tags
30 Jan 2014, 02:54
List S consists of 10 consecutive odd integers, and list T consists of 5 consecutive even integers. If the least integer in Sis 7 more than the least integer in T, how much greater is the average (arithmetic mean) of the integers in S than the average of the integers in T? (A) 2 (B) 7 (C) 8 (D) 12 (E) 22 Sol: Let List T has the following members : 2,4,6,8 and 10 Then S has : 9,11,13,15,17,19,21,23,25,27 Now If we find the average of List T is 6 and average of List S is (19+17)/2 =18 So Ans is 12. Suppose if we S also had 5 members and all the other condition remains same then Average of S would have been 13 and diferecne between the 2 would be 7 cause when the same number is added/subtracted from a given set then the average of the new set increases or decreases by the same number So ans is D. Average difficulty level of 650 is okay
_________________
“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”



Senior Manager
Joined: 20 Dec 2013
Posts: 250
Location: India

Re: List S consists of 10 consecutive odd integers, and list T c [#permalink]
Show Tags
30 Jan 2014, 03:18
We could do this by taking value for the lists List T=4,2,0,2,4.Mean=0 List S=3,5,7,...21=>Mean=12;(21+3)/2 (S has started from 3 as 4+7=3) Difference=12 Ans.D



Manager
Joined: 18 Oct 2013
Posts: 77
Location: India
Concentration: Technology, Finance
Schools: Duke '16, Johnson '16, Kelley '16, Tepper '16, Marshall '16, McDonough '16, Insead '14, HKUST '16, HSG '15, Schulich '15, Erasmus '16, IE April'15, Neeley '15
GMAT 1: 580 Q48 V21 GMAT 2: 530 Q49 V13 GMAT 3: 590 Q49 V21
WE: Information Technology (Computer Software)

Re: List S consists of 10 consecutive odd integers, and list T c [#permalink]
Show Tags
30 Jan 2014, 10:17
Easy one. Let us consider two set S and T. 1) T is the even consecutive set and S is odd consecutive set . 2) Least value of T +7=Least value of S
So if least value of T is 2 then least value of S is 9.
Its a series of even and odd consecutive integer. So T 5th term= 2+4*2=10 ... Mean=(10+2)/2=6 Similarly S 10 th term = 9+2*9=27.... Mean=(27+9)/2=18
Difference is 186=12
answer is D



Math Expert
Joined: 02 Sep 2009
Posts: 46297

Re: List S consists of 10 consecutive odd integers, and list T c [#permalink]
Show Tags
01 Feb 2014, 08:59



Retired Moderator
Joined: 29 Oct 2013
Posts: 272
Concentration: Finance
GPA: 3.7
WE: Corporate Finance (Retail Banking)

Re: List S consists of 10 consecutive odd integers, and list T c [#permalink]
Show Tags
28 May 2014, 02:47
Since the least no. in S is 7 greater than the least no. in T, lets assume S starts at 7 so T will start at 0. For S mean will be the average of 5th and 6th no.: {7, 9, 11, 13, 15, 17....} = (15+17)/2 = 16 For T mean will be the 3rd no. {0, 2, 4...} = 4 Answer=164=12 D!
_________________
Please contact me for super inexpensive quality private tutoring
My journey V46 and 750 > http://gmatclub.com/forum/myjourneyto46onverbal750overall171722.html#p1367876



Intern
Joined: 12 Mar 2015
Posts: 4

Re: List S consists of 10 consecutive odd integers, and list T c [#permalink]
Show Tags
24 Jan 2016, 23:49
Bunuel wrote: SOLUTION
List S consists of 10 consecutive odd integers, and list T consists of 5 consecutive even integers. If the least integer in S is 7 more than the least integer in T, how much greater is the average (arithmetic mean) of the integers in S than the average of the integers in T?
(A) 2 (B) 7 (C) 8 (D) 12 (E) 22
For any evenly spaced set median = mean = the average of the first and the last terms.
So the mean of S will be the average of the first and the last terms: mean = (x + x + 9*2)/2 = x+9, where x is the first term;
The mean of T will simply be the median or the third term: mean = (x  7) + 2*2 = x  3;
The difference will be (x + 9)  (x  3) = 12.
Answer: D. Hi Bunel, I could not understand how x+9*2 is the final term and similarly "the mean of T will simply be the median or the third term: mean = (x  7) + 2*2 = x  3;"



Math Expert
Joined: 02 Aug 2009
Posts: 5938

Re: List S consists of 10 consecutive odd integers, and list T c [#permalink]
Show Tags
25 Jan 2016, 00:32
amanlalwani wrote: Bunuel wrote: SOLUTION
List S consists of 10 consecutive odd integers, and list T consists of 5 consecutive even integers. If the least integer in S is 7 more than the least integer in T, how much greater is the average (arithmetic mean) of the integers in S than the average of the integers in T?
(A) 2 (B) 7 (C) 8 (D) 12 (E) 22
For any evenly spaced set median = mean = the average of the first and the last terms.
So the mean of S will be the average of the first and the last terms: mean = (x + x + 9*2)/2 = x+9, where x is the first term;
The mean of T will simply be the median or the third term: mean = (x  7) + 2*2 = x  3;
The difference will be (x + 9)  (x  3) = 12.
Answer: D. Hi Bunel, I could not understand how x+9*2 is the final term and similarly "the mean of T will simply be the median or the third term: mean = (x  7) + 2*2 = x  3;" Hi, there are 10 consecutive odd numbers , means each number is 2 more than the previous number... if the least number here is x, the next number will be x+2, third will be x+2*2... and so on till 10th term= x+9*2.. also we can find this through arithmetic progression.. Nth term = first term + (N1)d, d is the constant difference between two consecutive numbers.. 2ND part.. "the mean of T will simply be the median or the third term: mean = (x  7) + 2*2 = x  3in the second set, there are only five consecutive numbers so the median=mean=the central number, which is third number here.. the least integer in s is 7 less than T, so it will become x7... the third term here will be (x7) + 2*2.. same as nthterm above
_________________
Absolute modulus :http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
GMAT online Tutor



Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 5600
GPA: 3.82

Re: List S consists of 10 consecutive odd integers, and list T c [#permalink]
Show Tags
25 Jan 2016, 02:20
Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer. List S consists of 10 consecutive odd integers, and list T consists of 5 consecutive even integers. If the least integer in S is 7 more than the least integer in T, how much greater is the average (arithmetic mean) of the integers in S than the average of the integers in T? (A) 2 (B) 7 (C) 8 (D) 12 (E) 22 Since S is the list consisting of 10 consecutive odd integers we can put S={s, s + 2, s + 4, ...., s + 18}, where s is the least odd integer of S. So the average of S is (10*s + 2+4+....+18)/10=(10*s + 90)/10= s+9. Similarly we may put T={t, t+2, ..., t+8}, where t is the least even integer of T. So the average of T is (5*t + 2+ 4+ ....+8)/5 = t+4. s+9(t+4)=st+5=7+5=12. So the answer is 12. > (D).
_________________
MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The oneandonly World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only $99 for 3 month Online Course" "Free Resources30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons  try it yourself"



Board of Directors
Status: QA & VA Forum Moderator
Joined: 11 Jun 2011
Posts: 3511
Location: India
GPA: 3.5
WE: Business Development (Commercial Banking)

List S consists of 10 consecutive odd integers, and list T c [#permalink]
Show Tags
22 Apr 2016, 12:27
Bunuel wrote: The Official Guide For GMAT® Quantitative Review, 2ND EditionList S consists of 10 consecutive odd integers, and list T consists of 5 consecutive even integers. If the least integer in S is 7 more than the least integer in T, how much greater is the average (arithmetic mean) of the integers in S than the average of the integers in T? (A) 2 (B) 7 (C) 8 (D) 12 (E) 22 Let the set of numbers be S = { 9 , 11 , 13 , 15 , 17 , 19 , 21 , 23 , 25 , 27 } T = { 2 , 4 , 6 , 8 , 10 } Sum of the set S = 180 Mean of set T = 18 Sum of the set T = 30 Mean of set T = 6 So, The arithmetic mean of set S is 12 more than the mean of set T
_________________
Thanks and Regards
Abhishek....
PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS
How to use Search Function in GMAT Club  Rules for Posting in QA forum  Writing Mathematical Formulas Rules for Posting in VA forum  Request Expert's Reply ( VA Forum Only )



Manager
Joined: 09 Aug 2016
Posts: 68

Re: List S consists of 10 consecutive odd integers, and list T c [#permalink]
Show Tags
17 Nov 2016, 15:16
General formula for odd numbers is 2n + 1 and even 2n
Assume that 2 is the least in the even set then 2+7 = 9 has to be the first in the odd set.
So 2n + 1 = 9 gives n = 4 so the index n of the 10th value for the odd set is (4 + 10)  1 = 13 AND THE magical 1 occurs because the formula has "zero based" indexing. Hence value n for the 10th is 13 AND NOT 14.
Therefore 2(13) + 1 = 27. This means that for the ODD set min = 9 and max = 27 so mean = 18
The mean of the EVEN set is the median which is equal to 6 so the difference is 12 and Correct answer D



Target Test Prep Representative
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2570

Re: List S consists of 10 consecutive odd integers, and list T c [#permalink]
Show Tags
29 Nov 2016, 16:43
Bunuel wrote: List S consists of 10 consecutive odd integers, and list T consists of 5 consecutive even integers. If the least integer in S is 7 more than the least integer in T, how much greater is the average (arithmetic mean) of the integers in S than the average of the integers in T?
(A) 2 (B) 7 (C) 8 (D) 12 (E) 22
We can let x = the least integer in T. Thus, T contains the following integers: x, x + 2, x + 4, x + 6, and x + 8. Since the least integer in S is 7 more than the least integer in T, x + 7 = the least integer in S, and so S has the following integers: x + 7, x + 9, x + 11, x + 13, x + 15, x + 17, x + 19, x + 21, x + 23, and x + 25. Since each list is an evenly spaced set, the average of each list is the respective median. Since the median of the integers in T is x + 4, and the median of integers in S is [(x +15) + (x + 17)]/2 = (2x + 32)/2 = x + 16, the averages of the integers in T and S are x + 4 and x +16, respectively. Therefore, the average of list S is (x + 16)  (x + 4) = 12 more than the average of list T. Answer: D
_________________
Jeffery Miller
Head of GMAT Instruction
GMAT Quant SelfStudy Course
500+ lessons 3000+ practice problems 800+ HD solutions



Intern
Joined: 06 Dec 2016
Posts: 3

Re: List S consists of 10 consecutive odd integers, and list T c [#permalink]
Show Tags
11 Dec 2016, 11:48
S = 10 consecutive interegers T = 5 c. integers
S>7t
If the least integer of T is 1, than the least integer of S is 8.
The largest number of T is thus: 1+2x4=9, 9+1=10/2 = 5, so mean is 5 And for S is 8+2x9= 26, 26+8, 34/2=17, so mean is 17
The difference is 175, which is 12. Hence answer D



Manager
Joined: 26 Mar 2017
Posts: 145

Re: List S consists of 10 consecutive odd integers, and list T c [#permalink]
Show Tags
07 May 2017, 03:24
Abhishek009 wrote: Bunuel wrote: The Official Guide For GMAT® Quantitative Review, 2ND EditionList S consists of 10 consecutive odd integers, and list T consists of 5 consecutive even integers. If the least integer in S is 7 more than the least integer in T, how much greater is the average (arithmetic mean) of the integers in S than the average of the integers in T? (A) 2 (B) 7 (C) 8 (D) 12 (E) 22 Let the set of numbers be S = { 9 , 11 , 13 , 15 , 17 , 19 , 21 , 23 , 25 , 27 } T = { 2 , 4 , 6 , 8 , 10 } Sum of the set S = 180 Mean of set T = 18 Sum of the set T = 30 Mean of set T = 6 So, The arithmetic mean of set S is 12 more than the mean of set T guess thats the easiest approach. However, we could also start with 0 and recognise that we are dealing with an evenly spaced set, hence median = mean
_________________
I hate long and complicated explanations!



Study Buddy Forum Moderator
Joined: 04 Sep 2016
Posts: 1016
Location: India
WE: Engineering (Other)

Re: List S consists of 10 consecutive odd integers, and list T c [#permalink]
Show Tags
28 Sep 2017, 21:10
JeffTargetTestPrep VeritasPrepKarishma Bunuel Engr2012 Quote: We can let x = the least integer in T. Thus, T contains the following integers: x, x + 2, x + 4, x + 6, and x + 8. Is not a set of even consecutive no represented by 2x, 2x+2, 2x+4 .. ? Did we took 2 common to reach above step ? I messed up taking first T as 2n, 2n+2 , 2n+4 ... and S as 2n+7, 2n+9... which was far more calculation intensive. Please let me know flaw in approach ?



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8102
Location: Pune, India

Re: List S consists of 10 consecutive odd integers, and list T c [#permalink]
Show Tags
11 Oct 2017, 05:29
Bunuel wrote: The Official Guide For GMAT® Quantitative Review, 2ND EditionList S consists of 10 consecutive odd integers, and list T consists of 5 consecutive even integers. If the least integer in S is 7 more than the least integer in T, how much greater is the average (arithmetic mean) of the integers in S than the average of the integers in T? (A) 2 (B) 7 (C) 8 (D) 12 (E) 22 Problem Solving Question: 70 Category: Arithmetic Statistics Page: 70 Difficulty: 600 GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition  Quantitative Questions ProjectEach week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution. We'll be glad if you participate in development of this project: 1. Please provide your solutions to the questions; 2. Please vote for the best solutions by pressing Kudos button; 3. Please vote for the questions themselves by pressing Kudos button; 4. Please share your views on difficulty level of the questions, so that we have most precise evaluation. Thank you! Responding to a pm: I would simply take an example since constraints are few. "If the least integer in S is 7 more than the least integer in T" S has odd integers so say it starts from 11. 11, 13, 15, 17, 19, 21 .... (10 numbers) Average = 20 (middle of 19 and 21) T will start from 117 = 4 4, 6, 8, 10, 12 Average = 8 So average of S is 12 greater than average of T. Answer (D)
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8102
Location: Pune, India

Re: List S consists of 10 consecutive odd integers, and list T c [#permalink]
Show Tags
12 Oct 2017, 23:33
adkikani wrote: JeffTargetTestPrep VeritasPrepKarishma Bunuel Engr2012 Quote: We can let x = the least integer in T. Thus, T contains the following integers: x, x + 2, x + 4, x + 6, and x + 8. Is not a set of even consecutive no represented by 2x, 2x+2, 2x+4 .. ? Did we took 2 common to reach above step ? I messed up taking first T as 2n, 2n+2 , 2n+4 ... and S as 2n+7, 2n+9... which was far more calculation intensive. Please let me know flaw in approach ? There isn't a flaw in your approach. You can consider the first term of T as 2n and first term of S as 2n+7. Of course the more complicated your terms, more calculation intensive it will become. Since all you need is the difference between the averages, no matter how you take your integers, the answer will always be the same. So in such cases, I wouldn't take a variable at all.
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Intern
Joined: 24 Jan 2017
Posts: 7

Re: List S consists of 10 consecutive odd integers, and list T c [#permalink]
Show Tags
19 Oct 2017, 10:41
daviddaviddavid wrote: Abhishek009 wrote: Bunuel wrote: The Official Guide For GMAT® Quantitative Review, 2ND EditionList S consists of 10 consecutive odd integers, and list T consists of 5 consecutive even integers. If the least integer in S is 7 more than the least integer in T, how much greater is the average (arithmetic mean) of the integers in S than the average of the integers in T? (A) 2 (B) 7 (C) 8 (D) 12 (E) 22 Let the set of numbers be S = { 9 , 11 , 13 , 15 , 17 , 19 , 21 , 23 , 25 , 27 } T = { 2 , 4 , 6 , 8 , 10 } Sum of the set S = 180 Mean of set T = 18 Sum of the set T = 30 Mean of set T = 6 So, The arithmetic mean of set S is 12 more than the mean of set T guess thats the easiest approach. However, we could also start with 0 and recognise that we are dealing with an evenly spaced set, hence median = mean In this approach how did you find the sum of these sets? I do not think there will be time in exam to calculate the numbers. there has to be some logic to get sums of consecutive even/odd integers.



VP
Joined: 07 Dec 2014
Posts: 1020

List S consists of 10 consecutive odd integers, and list T c [#permalink]
Show Tags
19 Oct 2017, 15:11
Bunuel wrote: The Official Guide For GMAT® Quantitative Review, 2ND EditionList S consists of 10 consecutive odd integers, and list T consists of 5 consecutive even integers. If the least integer in S is 7 more than the least integer in T, how much greater is the average (arithmetic mean) of the integers in S than the average of the integers in T? (A) 2 (B) 7 (C) 8 (D) 12 (E) 2 let 2=least term of T 2+2*2=6=third term=mean of T let 9=least term of S 9+4*2+1=18=fifth term+1=mean of S 186=12 D




List S consists of 10 consecutive odd integers, and list T c
[#permalink]
19 Oct 2017, 15:11






