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Expert replies
by rahul.s » Tue Feb 23, 2010 1:21 am
If twelve consecutive even integers are listed in increasing order, what is the value of the largest of the twelve integers?

(1) The sum of the first six integers is 18.
(2) The sum of the first three integers is 54 less than the sum of the last three.

OA: A
Source: Knewton
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Source: — Data Sufficiency |

by ajith » Tue Feb 23, 2010 1:37 am
rahul.s wrote:If twelve consecutive even integers are listed in increasing order, what is the value of the largest of the twelve integers?

(1) The sum of the first six integers is 18.
(2) The sum of the first three integers is 54 less than the sum of the last three.

OA: A
Source: Knewton
k, k+2 , ....., k+22 be the numbers

1)k+ k+2 +k+4 +k+6 +k+8 +k+10 =18

6k +30 = 18
k = -12/6
k = -2

the largest k+22 = 20
2) 3k+6 +54 = 3k +22+20+18

3k +60 = 3k +60

doesn't yield an equation. Hence insufficient
[spoiler]
A[/spoiler]
Always borrow money from a pessimist, he doesn't expect to be paid back.
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by rahul.s » Tue Feb 23, 2010 2:55 am
i fell for the trap by opting for D.

Knewton: In any list of 12 consecutive even integers, the sum of the first 3 will be 54 less than the sum of the last three.
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by sumanr84 » Tue Feb 23, 2010 4:47 am
rahul.s wrote:i fell for the trap by opting for D.

Knewton: In any list of 12 consecutive even integers, the sum of the first 3 will be 54 less than the sum of the last three.
I too opted for D ( within less than 15 secs)..I have to be careful hereon.
I am on a break !!
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