If the sum of the first four numbers in a list of six consecutive even : GMAT Problem Solving (PS)
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# If the sum of the first four numbers in a list of six consecutive even

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Director
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If the sum of the first four numbers in a list of six consecutive even [#permalink]

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04 Aug 2010, 10:03
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If the sum of the first four numbers in a list of six consecutive even numbers is 908, what is the sum of the last four numbers in the list?

A. 912
B. 914
C. 916
D. 920
E. 924

[Reveal] Spoiler:
I tried it, but got screwed up:
let 2x be one of the numbers;
list: 2x-4, 2x-2, 2x, 2x+2, 2x+4, 2x+6
sum of the first four: (2x-4) + (2x-2) + (2x) + 2x+2)
--> 4x-4=908
x=228

sum of last four: 2x + (2x+2) + (2x+4) + (2x+6)
= 8x+12
=8(228) + 12

Please what is the correct approach?
[Reveal] Spoiler: OA

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Re: If the sum of the first four numbers in a list of six consecutive even [#permalink]

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04 Aug 2010, 10:31
3
KUDOS
Expert's post
gmatbull wrote:
If the sum of the first four numbers in a list of six consecutive even numbers is 908,
what is the sum of the last four numbers in the list?
A. 912
B. 914
C. 916
D. 920
E. 924

I tried it, but got screwed up:
let 2x be one of the numbers;
list: 2x-4, 2x-2, 2x, 2x+2, 2x+4, 2x+6
sum of the first four: (2x-4) + (2x-2) + (2x) + 2x+2)
--> 4x-4=908
x=228

sum of last four: 2x + (2x+2) + (2x+4) + (2x+6)
= 8x+12
=8(228) + 12

Please what is the correct approach?

Let the six consecutive even numbers be $$x$$, $$x+2$$, $$x+4$$, $$x+6$$, $$x+8$$, $$x+10$$.

Given: $$x+(x+2)+(x+4)+(x+6)=4x+12=908$$. Question: $$(x+4)+(x+6)+(x+8)+(x+10)=4x+28=?$$

$$(x+4)+(x+6)+(x+8)+(x+10)=4x+28=(4x+12)+16=908+16=924$$.

The way you are doing is also valid. You've just made an error in calculations, plus no need even number to be $$2x$$ it can be just even $$x$$.

Sum of the first four: $$(2x-4)+(2x-2)+(2x)+(2x+2)=8x-4=908$$;
Sum of the last four: $$2x+(2x+2)+(2x+4)+(2x+6)=8x+12=(8x-4)+16=908+16=924$$.

Hope it's clear.
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Re: If the sum of the first four numbers in a list of six consecutive even [#permalink]

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04 Aug 2010, 11:11
oh, thanks Bunuel for the corrections.
Hope you aren't tired of receiving kudos; you know, people like you have made
kudos a trite, and one simply wonders what else to give.

Methinks, there should be a different category of kudos for genius such as Bunuel.
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Re: If the sum of the first four numbers in a list of six consecutive even [#permalink]

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07 Aug 2010, 03:44
Bunuel wrote:
gmatbull wrote:
If the sum of the first four numbers in a list of six consecutive even numbers is 908,
what is the sum of the last four numbers in the list?
A. 912
B. 914
C. 916
D. 920
E. 924

I tried it, but got screwed up:
let 2x be one of the numbers;
list: 2x-4, 2x-2, 2x, 2x+2, 2x+4, 2x+6
sum of the first four: (2x-4) + (2x-2) + (2x) + 2x+2)
--> 4x-4=908
x=228

sum of last four: 2x + (2x+2) + (2x+4) + (2x+6)
= 8x+12
=8(228) + 12

Please what is the correct approach?

Let the six consecutive even numbers be $$x$$, $$x+2$$, $$x+4$$, $$x+6$$, $$x+8$$, $$x+10$$.

Given: $$x+(x+2)+(x+4)+(x+6)=4x+12=908$$. Question: $$(x+4)+(x+6)+(x+8)+(x+10)=4x+28=?$$

$$(x+4)+(x+6)+(x+8)+(x+10)=4x+28=(4x+12)+16=908+16=924$$.

Hi Bunuel
i did the same way ..
however kindly explain this if I solve
4x+ 12 =908 , I get x = 99 ..which is not an even number . hence I was confuded whether I missed something !!
thanx
Math Expert
Joined: 02 Sep 2009
Posts: 37108
Followers: 7254

Kudos [?]: 96548 [0], given: 10753

Re: If the sum of the first four numbers in a list of six consecutive even [#permalink]

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07 Aug 2010, 04:04
gauravnagpal wrote:
Hi Bunuel
i did the same way ..
however kindly explain this if I solve
4x+ 12 =908 , I get x = 99 ..which is not an even number . hence I was confuded whether I missed something !!
thanx

$$4x+ 12 =908$$ --> $$x=224=even$$.
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Re: If the sum of the first four numbers in a list of six consecutive even [#permalink]

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07 Aug 2010, 05:38
Bunuel wrote:
gmatbull wrote:
If the sum of the first four numbers in a list of six consecutive even numbers is 908,
what is the sum of the last four numbers in the list?
A. 912
B. 914
C. 916
D. 920
E. 924

I tried it, but got screwed up:
let 2x be one of the numbers;
list: 2x-4, 2x-2, 2x, 2x+2, 2x+4, 2x+6
sum of the first four: (2x-4) + (2x-2) + (2x) + 2x+2)
--> 4x-4=908
x=228

sum of last four: 2x + (2x+2) + (2x+4) + (2x+6)
= 8x+12
=8(228) + 12

Please what is the correct approach?

Let the six consecutive even numbers be $$x$$, $$x+2$$, $$x+4$$, $$x+6$$, $$x+8$$, $$x+10$$.

Given: $$x+(x+2)+(x+4)+(x+6)=4x+12=908$$. Question: $$(x+4)+(x+6)+(x+8)+(x+10)=4x+28=?$$

$$(x+4)+(x+6)+(x+8)+(x+10)=4x+28=(4x+12)+16=908+16=924$$.

The way you are doing is also valid. You've just made an error in calculations, plus no need even number to be $$2x$$ it can be just even $$x$$.

Sum of the first four: $$(2x-4)+(2x-2)+(2x)+(2x+2)=8x-4=908$$;
Sum of the last four: $$2x+(2x+2)+(2x+4)+(2x+6)=8x+12=(8x-4)+16=908+16=924$$.

Hope it's clear.

Well done!!!!

Thx and kudos
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Re: If the sum of the first four numbers in a list of six consecutive even [#permalink]

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08 Aug 2010, 00:49
Bunuel wrote:
gauravnagpal wrote:
Hi Bunuel
i did the same way ..
however kindly explain this if I solve
4x+ 12 =908 , I get x = 99 ..which is not an even number . hence I was confuded whether I missed something !!
thanx

$$4x+ 12 =908$$ --> $$x=224=even$$.

i am so sorry ...i reckon sleep of mind ...i dont know how could I write this ..thanx anyways for spending time on this
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Re: If the sum of the first four numbers in a list of six consecutive even [#permalink]

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10 Oct 2015, 10:02
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Re: If the sum of the first four numbers in a list of six consecutive even [#permalink]

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10 Oct 2015, 10:42
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gmatbull wrote:
If the sum of the first four numbers in a list of six consecutive even numbers is 908, what is the sum of the last four numbers in the list?

A. 912
B. 914
C. 916
D. 920
E. 924

[Reveal] Spoiler:
I tried it, but got screwed up:
let 2x be one of the numbers;
list: 2x-4, 2x-2, 2x, 2x+2, 2x+4, 2x+6
sum of the first four: (2x-4) + (2x-2) + (2x) + 2x+2)
--> 4x-4=908
x=228

sum of last four: 2x + (2x+2) + (2x+4) + (2x+6)
= 8x+12
=8(228) + 12

Please what is the correct approach?

Since this is an evenly spaced set, the median is equal to the average of the set. The average and the median of the first four consecutive even integers are $$\frac{908}{4}=227.$$This implies that the first 4 consecutive even integers are 224,226,228, and 230. So the last two consecutive integers are 232 and 234. To find the sum of 228,230,232, and 234, take the median 231 and multiply by 4. (231*4) = 924

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Re: If the sum of the first four numbers in a list of six consecutive even [#permalink]

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10 Oct 2015, 16:55
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Hi All,

If you recognize the 'comparison' taking place in this question, you can avoid a 'calculation-heavy' approach and use the patterns to your advantage.

We're told that the first 4 CONSECUTIVE EVEN numbers (in a group of 6) has a sum of 908.

We can call those 4 terms...

X + (X+2) + (X+4) + (X+6) = 908

We're asked for the sum of the LAST 4 terms in this sequence... We can call those terms...

(X+4) + (X+6) + (X+8) + (X+10)

Notice how each of these four terms is EXACTLY 4 MORE than each of the 4 terms in the original sequence? Those 'differences' lead to an increase of 4(4) = 16 over the original sum.

Thus, the sum of the last 4 terms is 908 + 16 = 924

[Reveal] Spoiler:
E

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Re: If the sum of the first four numbers in a list of six consecutive even   [#permalink] 10 Oct 2015, 16:55
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