Summer is Coming! Join the Game of Timers Competition to Win Epic Prizes. Registration is Open. Game starts Mon July 1st.

 It is currently 19 Jul 2019, 16:02

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

M18-14

Author Message
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 56300

Show Tags

16 Sep 2014, 01:03
00:00

Difficulty:

65% (hard)

Question Stats:

58% (01:52) correct 42% (01:57) wrong based on 214 sessions

HideShow timer Statistics

What is the mean of four consecutive even integers $$a$$, $$b$$, $$c$$ and $$d$$, where $$a \lt b \lt c \lt d$$?

(1) $$a + d = b + c$$

(2) $$b + c = d - a$$

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 56300

Show Tags

16 Sep 2014, 01:03
1
1
Official Solution:

(1) $$a + d = b + c$$. That statement is true for ANY four consecutive even integers. Not sufficient.

(2) $$b + c = d - a$$. Let $$a=2k$$, for some integer $$k$$. Now, since $$a$$, $$b$$, $$c$$ and $$d$$ are consecutive even integers then $$b=2k+2$$, $$c=2k+4$$ and $$d=2k+6$$. So, we have that $$(2k+2)+(2k+4)=(2k+6)-2k$$. Solving gives $$k=0$$, so $$a=0$$, $$b=2$$, $$c=4$$ and $$d=6$$. Hence, $$mean=3$$. Sufficient.

_________________
Retired Moderator
Joined: 29 Apr 2015
Posts: 830
Location: Switzerland
Concentration: Economics, Finance
Schools: LBS MIF '19
WE: Asset Management (Investment Banking)

Show Tags

17 Jun 2015, 11:32
Bunuel wrote:
Official Solution:

(1) $$a + d = b + c$$. That statement is true for ANY four consecutive even integers. Not sufficient.

(2) $$b + c = d - a$$. Let $$a=2k$$, for some integer $$k$$. Now, since $$a$$, $$b$$, $$c$$ and $$d$$ are consecutive even integers then $$b=2k+2$$, $$c=2k+4$$ and $$d=2k+6$$. So, we have that $$(2k+2)+(2k+4)=(2k+6)-2k$$. Solving gives $$k=0$$, so $$a=0$$, $$b=2$$, $$c=4$$ and $$d=6$$. Hence, $$mean=3$$. Sufficient.

Wow, this approach is just brilliant. Once you see it, it's so easy. The big question is, how do you ever come to this idea to plug in 2k for a? Of course 2k, 2x whatever stands for an even integer. What's really bothering me is the logic behind. I stare at this task and never in my life would have come up with this genious approach ...

Please direct me to this thinking... Any help is very very very appreciated.

Thanks
_________________
Saving was yesterday, heat up the gmatclub.forum's sentiment by spending KUDOS!

PS Please send me PM if I do not respond to your question within 24 hours.
Senior Manager
Joined: 10 Mar 2013
Posts: 487
Location: Germany
Concentration: Finance, Entrepreneurship
Schools: WHU MBA"20 (A)
GMAT 1: 580 Q46 V24
GPA: 3.88
WE: Information Technology (Consulting)

Show Tags

04 Oct 2015, 05:19
2
We can solve this one with number picking
1) Doesn't give us any additional information - Not Suff
2) Just pick 4 even integers (it doesn't matter which .. the second is always 2 more than the previous) 2 4 6 8 and check this expression b+c=d−a
4+6 is not equal to 8-2, Why ? Because an even integer we need here is Zero. 0,2,4,6 2+4=6-0 -> 2+4/2=3 Sufficient
_________________
When you’re up, your friends know who you are. When you’re down, you know who your friends are.

800Score ONLY QUANT CAT1 51, CAT2 50, CAT3 50
GMAT PREP 670
MGMAT CAT 630
KAPLAN CAT 660
Director
Joined: 02 Sep 2016
Posts: 657

Show Tags

09 Jan 2017, 22:48
Bunuel wrote:
Official Solution:

(1) $$a + d = b + c$$. That statement is true for ANY four consecutive even integers. Not sufficient.

(2) $$b + c = d - a$$. Let $$a=2k$$, for some integer $$k$$. Now, since $$a$$, $$b$$, $$c$$ and $$d$$ are consecutive even integers then $$b=2k+2$$, $$c=2k+4$$ and $$d=2k+6$$. So, we have that $$(2k+2)+(2k+4)=(2k+6)-2k$$. Solving gives $$k=0$$, so $$a=0$$, $$b=2$$, $$c=4$$ and $$d=6$$. Hence, $$mean=3$$. Sufficient.

Hi Bunuel

I solved statement 1 like this:

The 4 nos. are 2n, 2n+2, 2n+4, and 2n+6

a+d=b+c
2n+2n+6= 2n+2+2n+4

4n+6= 4n+6

We get no value from this as both the sides become zero.

Therefore insufficient.

The reason you explained for this result makes sense. Thanks.
Intern
Joined: 16 Jan 2017
Posts: 6

Show Tags

29 Apr 2017, 10:24

What if set would be -2, 0, 2, 4?

Or does the question stem somehow rule negative numbers out?
Math Expert
Joined: 02 Sep 2009
Posts: 56300

Show Tags

29 Apr 2017, 11:41
Kontaxis wrote:

What if set would be -2, 0, 2, 4?

Or does the question stem somehow rule negative numbers out?

These numbers does not satisfy b+c=d−a.
_________________
SVP
Joined: 26 Mar 2013
Posts: 2284

Show Tags

30 Apr 2017, 03:52
What is the mean of four consecutive even integers $$a$$, $$b$$, $$c$$ and $$d$$, where $$a \lt b \lt c \lt d$$?

As number are Even Consecutive numbers

a, a+2, a+4, a+6

(1) $$a + d = b + c$$

a+a+6 = a+2+a+4

2a+6 = 2a +6

0=0..does not lead to get value of a

Insufficient

(2) $$b + c = d - a$$

a+2+a+4 = a+6-a

2a +6 =6........2a = 0.......a=0...........Hence we can get the mean of the four numbers.

Sufficient

Senior Manager
Joined: 08 Jun 2015
Posts: 421
Location: India
GMAT 1: 640 Q48 V29
GMAT 2: 700 Q48 V38
GPA: 3.33

Show Tags

20 Jun 2017, 05:01
1
The answer has to be option B. My take :

The questions asks us to find the mean of four consecutive even numbers. If a,b,c,& d are four consecutive even numbers , a+d=b+c. We need to find the mean of the four numbers, the mean will be (b+c)/2.

Statement 1 - Does not give us the value of (b+c)/2
Statement 2 - b+c=d-a. This implies d+a=d-a , or a=0. The series is 0,2,4,6. We can find the mean now. Hence statement 2 is sufficient.

The answer has to be option B.
_________________
" The few , the fearless "
Intern
Joined: 31 Mar 2017
Posts: 38
GMAT 1: 710 Q45 V42
GMAT 2: 730 Q50 V39
GMAT 3: 740 Q50 V40
GPA: 3.73

Show Tags

20 Jun 2017, 05:30
1
reto wrote:
Bunuel wrote:
Official Solution:

(1) $$a + d = b + c$$. That statement is true for ANY four consecutive even integers. Not sufficient.

(2) $$b + c = d - a$$. Let $$a=2k$$, for some integer $$k$$. Now, since $$a$$, $$b$$, $$c$$ and $$d$$ are consecutive even integers then $$b=2k+2$$, $$c=2k+4$$ and $$d=2k+6$$. So, we have that $$(2k+2)+(2k+4)=(2k+6)-2k$$. Solving gives $$k=0$$, so $$a=0$$, $$b=2$$, $$c=4$$ and $$d=6$$. Hence, $$mean=3$$. Sufficient.

Wow, this approach is just brilliant. Once you see it, it's so easy. The big question is, how do you ever come to this idea to plug in 2k for a? Of course 2k, 2x whatever stands for an even integer. What's really bothering me is the logic behind. I stare at this task and never in my life would have come up with this genious approach ...

Please direct me to this thinking... Any help is very very very appreciated.

Thanks

I don't know if this will help, but I used to have the same issues.

Anytime you see these problems where each number is related (consecutive ints, ratios etc.) you can often relate all the numbers to one variable

For example, 4 consecutive even intergers are a, a+2, a +4 and a+6
In S2, we know that a +2 + a + 4 = a + 6 - a
solving for this we get: 2a + 6 = 6
2a = 0
Therefore a = 0

I like to double check just to make sure, so if we have 0 , 2, 4 and 6. Then 2 + 4 = 6 - 0
This satisfies the equation.

Hope this helps.
Manager
Joined: 27 Mar 2016
Posts: 96

Show Tags

20 Jun 2017, 10:42
a) a+d=b+c this statement is obvious as if a=n,b =n+2,c=n+4 and d=n+6 then a+d=b+c
b) b+c=a-d
if a =n then d=a+6(as a,b,c,d are consecutive even integers)
b+c=6=a+d
hence B
Intern
Joined: 16 Feb 2010
Posts: 4

Show Tags

21 Jun 2017, 23:16
I think it is easier than it seems to be...
if a+d=b+c and d-a=b+c, then a=0, otherwise equations are not true.
So, if a,b,c,d are consecutive even integers, then a=0, b=2, c=4, d=6
the mean is equal to (0+2+4+6)/4=3
Manager
Joined: 11 May 2018
Posts: 91
Location: India
GMAT 1: 460 Q42 V14

Show Tags

02 Sep 2018, 00:19
Quote:
What is the mean of four consecutive even integers a, b, c and d, where a<b<c<d?

(1) a+d=b+c

(2) b+c=d−a

Quote:
Given a<b<c<d and they are even consecutive integers.
STMT1:
case 1:
2,4,6,8 mean =5
case 2:
-8,-6,-4,-2 mean = -5

2 solutions not sufficient
STMT2:
b+c=d−a
b+a=d-c
now take
0,2,4,6
2=2 mean =3
Try another case : Not possible!!
so B is the winner

_________________
If you want to Thank me Give me a KUDOS
"I’ve spent months preparing for the day I’d face you. I’ve come a long way, GMAT"- SonGoku
Re: M18-14   [#permalink] 02 Sep 2018, 00:19
Display posts from previous: Sort by

M18-14

Moderators: chetan2u, Bunuel