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

 It is currently 12 Dec 2018, 20:08

### 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

## Events & Promotions

###### Events & Promotions in December
PrevNext
SuMoTuWeThFrSa
2526272829301
2345678
9101112131415
16171819202122
23242526272829
303112345
Open Detailed Calendar
• ### The winning strategy for 700+ on the GMAT

December 13, 2018

December 13, 2018

08:00 AM PST

09:00 AM PST

What people who reach the high 700's do differently? We're going to share insights, tips and strategies from data we collected on over 50,000 students who used examPAL.
• ### GMATbuster's Weekly GMAT Quant Quiz, Tomorrow, Saturday at 9 AM PST

December 14, 2018

December 14, 2018

09:00 AM PST

10:00 AM PST

10 Questions will be posted on the forum and we will post a reply in this Topic with a link to each question. There are prizes for the winners.

# List S consists of 10 consecutive odd integers, and list T

Author Message
TAGS:

### Hide Tags

Director
Status: No dream is too large, no dreamer is too small
Joined: 14 Jul 2010
Posts: 521
List S consists of 10 consecutive odd integers, and list T  [#permalink]

### Show Tags

16 Feb 2011, 08:42
3
53
00:00

Difficulty:

55% (hard)

Question Stats:

72% (01:51) correct 28% (02:12) wrong based on 959 sessions

### HideShow timer Statistics

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

_________________

Collections:-
PSof OG solved by GC members: http://gmatclub.com/forum/collection-ps-with-solution-from-gmatclub-110005.html
DS of OG solved by GC members: http://gmatclub.com/forum/collection-ds-with-solution-from-gmatclub-110004.html
100 GMAT PREP Quantitative collection http://gmatclub.com/forum/gmat-prep-problem-collections-114358.html
Collections of work/rate problems with solutions http://gmatclub.com/forum/collections-of-work-rate-problem-with-solutions-118919.html
Mixture problems in a file with best solutions: http://gmatclub.com/forum/mixture-problems-with-best-and-easy-solutions-all-together-124644.html

Math Expert
Joined: 02 Sep 2009
Posts: 51121

### Show Tags

16 Feb 2011, 08:59
12
20
Baten80 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

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.

_________________
SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1825
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re: List S consists of 10 consecutive odd integers, and list T  [#permalink]

### Show Tags

23 Jun 2014, 01:53
12
1
Picked up numbers:

T = 2 , 4 , 6, 8 , 10 (Mean = 6)

S = 9, 11, 13, 15, 17, 19, 21, 23, 25, 27 (Mean $$= \frac{17+19}{2} = 18$$)

Difference = 18-6 = 12

_________________

Kindly press "+1 Kudos" to appreciate

##### General Discussion
Retired Moderator
Joined: 20 Dec 2010
Posts: 1820

### Show Tags

16 Feb 2011, 08:56
3
5
Sum of evenly spaced progression:
$$\frac{n}{2}(2a+(n-1)d)$$

$$Average = \frac{Sum}{n}$$
or
$$Average =\frac{1}{2}(2a+(n-1)d)$$

$$\frac{1}{2}(2a+(10-1)*2) - \frac{1}{2}(2(a-7)+(5-1)*2)$$

$$\frac{1}{2}(2a+(10-1)*2) - \frac{1}{2}(2(a-7)+(5-1)*2)$$

$$a+9-(a-7+4)$$

$$a+9-a+7-4$$

$$12$$

Ans: "D"
_________________
Senior Manager
Status: Up again.
Joined: 31 Oct 2010
Posts: 493
Concentration: Strategy, Operations
GMAT 1: 710 Q48 V40
GMAT 2: 740 Q49 V42

### Show Tags

17 Feb 2011, 04:51
5
4
If you didnt know the formula for evenly spaced sets, you can pick numbers and solve this question very easily.

Pick first number of set T=2. Since number of elements = 5, the mean will be the middle term ie the 3rd term, which will be 6.

From the question, you can infer that first number of the set T, will be 2+7= 9. Since this set consists of even number of terms,ie 10, the mean will be the average of the middle 2 terms, ie 5th and 6th term, which are 17 and 19 respectively and their average will be 18= mean of the set.

Therefore, the difference between the mean of two sets= 18-6= 12.

_________________

My GMAT debrief: http://gmatclub.com/forum/from-620-to-710-my-gmat-journey-114437.html

Senior Manager
Joined: 08 Nov 2010
Posts: 337

### Show Tags

17 Feb 2011, 12:01
Fluke - when u say 2a - what do u mean?
shouldnt it be A1 (the first number in the series)?

thanks.
_________________
Manager
Joined: 17 Feb 2011
Posts: 147
Concentration: Real Estate, Finance
Schools: MIT (Sloan) - Class of 2014
GMAT 1: 760 Q50 V44

### Show Tags

17 Feb 2011, 17:24
Easy by picking numbers, but I didn't know that formula. Thanks.
Retired Moderator
Joined: 20 Dec 2010
Posts: 1820

### Show Tags

17 Feb 2011, 23:46
1
144144 wrote:
Fluke - when u say 2a - what do u mean?
shouldnt it be A1 (the first number in the series)?

thanks.

You are half right;

It should be $$2*a_1$$ i.e. twice of the first term. "a" is another way of depicting $$a_1$$

The formula for Sum of n numbers in an evenly spaced progression is denoted by:
$$\frac{n}{2}(2*a_1+(n-1)d)$$

If the progression is: 2,4,6. Sum=12

Using formula:
$$a_1=2$$. First term of the series
$$n=3$$ Number of elements in the progression(series)
$$d=a_2-a_1=4-2=2$$

$$Sum=\frac{n}{2}(2a_1+(n-1)d)$$
$$Sum=\frac{3}{2}(2*2+(3-1)2)$$
$$Sum=\frac{3}{2}*8$$
$$Sum=12$$
_________________
Senior Manager
Joined: 08 Nov 2010
Posts: 337

### Show Tags

18 Feb 2011, 05:55
ye, i mixed with N formula. thanks.
_________________
VP
Status: There is always something new !!
Affiliations: PMI,QAI Global,eXampleCG
Joined: 08 May 2009
Posts: 1033

### Show Tags

18 May 2011, 22:43
(2a + 2a + 8) /2

(2a+7 + 2a + 25)/2

leaves 12 after subtracting.
Manager
Joined: 12 Oct 2009
Posts: 151

### Show Tags

19 May 2011, 18:57
I got 12 by solving but i think Bunuel's method is brilliant ( as always )
Intern
Joined: 26 Jan 2014
Posts: 2
Re: List S consists of 10 consecutive odd integers, and list T  [#permalink]

### Show Tags

29 Jun 2014, 21:14
PareshGmat wrote:
Picked up numbers:

T = 2 , 4 , 6, 8 , 10 (Mean = 6)

S = 9, 11, 13, 15, 17, 19, 21, 23, 25, 27 (Mean $$= \frac{17+19}{2} = 18$$)

Difference = 18-6 = 12

PareshGmat's solution is the quickest way. Anyway, here is mine:
T=(t1+t2+t3+t4+t5)/5
=[t1+(t1+1*2)+(t1+2*2)+(t1+3*2)+(t1+4*2)]/5
=[5t1+2*(1+2+3+4)]/5
=[5t1+2*(4*(4+1)/2)]/5
=[5t1+20]/5=t1+4
S=(s1+...+s10)/10
=[s1+(s1+1*2)+...+(s1+9*2)]/10
=[10s1+2*(1+...+9)]/10
=[10s1+2*(9*(9+1)/2)]/10
=[10s1+90]/10=s1+9
We have s1=t1+7
S-T=s1+9-t1-4=t1+7+9-t1-4=12 =>D
P/S: Sum of n consecutive integers:
n(n+1)/2
EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 13074
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Re: List S consists of 10 consecutive odd integers, and list T  [#permalink]

### Show Tags

26 Aug 2015, 10:25
2
Hi All,

This question can be solved by using Number Property Rules or by TESTing VALUES:

Here's how TESTing VALUES works:

List S: 10 consecutive ODD integers
List T: 5 consecutive EVEN integers
The least integer is S is 7 more than the least integer in T.

Let's say that….
T = {2, 4, 6, 8, 10}
Since the least integers in S is 7 MORE than the least integer in T…
S = {9, 11, 13, 15, 17, 19, 21, 23, 25, 27}

The average of T = 6
The average of S = 18

So, the average of S is 18 - 6 = 12 more than the average of T.

GMAT assassins aren't born, they're made,
Rich
_________________

760+: Learn What GMAT Assassins Do to Score at the Highest Levels
Contact Rich at: Rich.C@empowergmat.com

# Rich Cohen

Co-Founder & GMAT Assassin

Special Offer: Save \$75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee
www.empowergmat.com/

*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*****

Intern
Joined: 12 Apr 2014
Posts: 8
Re: List S consists of 10 consecutive odd integers, and list T  [#permalink]

### Show Tags

18 Jun 2016, 00:21
1
the idea is to remember the rule for consecutive numbers whether even or odd that it will have median=mean=average(1st term+last term)....then for S and T series have a difference of 7 of first digit of series when creating hypothetical digits for series....this will result in exact solution
Current Student
Joined: 27 Mar 2016
Posts: 330
Location: United States (CO)
GMAT 1: 770 Q51 V44
GPA: 3.69
WE: Analyst (Consulting)
Re: List S consists of 10 consecutive odd integers, and list T  [#permalink]

### Show Tags

19 Jun 2016, 01:21
Mean of 10 consecutive odd integers is 9 more than the least.
Mean of 5 consecutive even integers is 4 more than the least.

9 - 4 + 7 = 12.
Intern
Joined: 12 Apr 2014
Posts: 8
Re: List S consists of 10 consecutive odd integers, and list T  [#permalink]

### Show Tags

20 Jun 2016, 01:02
when we read consecutive odd/even integers...that implies mean(average)=median=average(1st term+last term).....make series S and T with hypothetical numbers according to the condition given....on setting up the numbers.....find A.M according to the method described above leading to the right solution
VP
Joined: 07 Dec 2014
Posts: 1128
List S consists of 10 consecutive odd integers, and list T  [#permalink]

### Show Tags

23 Jun 2017, 08:52
Baten80 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

let s=least integer of S
t=least integer of T
mean of S=s+range/2=s+18/2=s+9
mean of T=t+range/2=t+8/2=t+4
substituting s-7 for t,
(s+9)-(s-3)=12
D
Target Test Prep Representative
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2830
Re: List S consists of 10 consecutive odd integers, and list T  [#permalink]

### Show Tags

26 Jun 2017, 16:53
3
Baten80 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.

_________________

Jeffery Miller

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

Intern
Joined: 28 Apr 2017
Posts: 40
Re: List S consists of 10 consecutive odd integers, and list T  [#permalink]

### Show Tags

28 Jun 2017, 00:05
This is can also by using the concept of evenly spaced numbers.
Let Set S be s-10, s-8, s-6.....s+6-- 10 numbers.
Total =(10s-10)=> Avg=s-1

Set T be e-4,e-2,e,e+2,e+4--5 numbers
Total=5e => Avg =e

diff of avg is s-1-e

Other statement for least numbers gives us (s-10)=7+(e-4)
>s-e=13

Putting it back to s-e-1= 12
Manager
Joined: 29 Jul 2018
Posts: 106
Concentration: Finance, Statistics
Re: List S consists of 10 consecutive odd integers, and list T  [#permalink]

### Show Tags

02 Oct 2018, 03:18
since numbers are evenly spaced
so mean = median
7,9,11,13,15.........
median = 13+15/2 = 16
0,2,4,6,8
median = 4
16-4=12
Re: List S consists of 10 consecutive odd integers, and list T &nbs [#permalink] 02 Oct 2018, 03:18
Display posts from previous: Sort by