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Math Expert V
Joined: 02 Sep 2009
Posts: 56300

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

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Math Expert V
Joined: 02 Sep 2009
Posts: 56300

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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.

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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
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Joined: 10 Mar 2013
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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
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Director  G
Joined: 02 Sep 2016
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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  B
Joined: 16 Jan 2017
Posts: 6

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

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

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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.
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SVP  V
Joined: 26 Mar 2013
Posts: 2284

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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  S
Joined: 08 Jun 2015
Posts: 421
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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.
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Intern  S
Joined: 31 Mar 2017
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GMAT 1: 710 Q45 V42 GMAT 2: 730 Q50 V39 GMAT 3: 740 Q50 V40 GPA: 3.73

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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  B
Joined: 27 Mar 2016
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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

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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  B
Joined: 11 May 2018
Posts: 91
Location: India
GMAT 1: 460 Q42 V14 ### Show Tags

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

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# M18-14

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