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Retired Moderator
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Status: 2000 posts! I don't know whether I should feel great or sad about it! LOL
Joined: 04 Oct 2009
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Location: Peru
Schools: Harvard, Stanford, Wharton, MIT & HKS (Government)
WE 1: Economic research
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WE 3: Government: Foreign Trade and SMEs
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Packages [#permalink] New post 11 Nov 2009, 21:22
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

100% (02:46) correct 0% (00:00) wrong based on 1 sessions
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Please your help with the following question:

The weights of four packages are 1, 3, 5, and 7 pounds, respectively. Which of the following CANNOT be the total weight, in pounds, of any combination of the packages?
a) 9
b) 10
c) 12
d) 13
e) 14
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Manager
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Joined: 11 Sep 2009
Posts: 129
Followers: 5

Kudos [?]: 244 [0], given: 6

Re: Packages [#permalink] New post 11 Nov 2009, 21:43
I think the easiest thing to do is simply eliminate choices:

a) 9 = 5 + 3 + 1 (OK)
b) 10 = 7 + 3 (OK)
c) 12 = 7 + 5 (OK)
d) 13 = 7 + 5 + 1 (OK)
e) 14 = ???

The answer is E.

I think any "mathematical" way of solving this would just take longer and is unnecessary (unless someone can prove me wrong?)
Manager
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Joined: 10 Sep 2009
Posts: 119
Followers: 2

Kudos [?]: 34 [0], given: 10

Re: Packages [#permalink] New post 12 Nov 2009, 01:19
Picking number is the best technique for this example

It took me 20/25 seconds to choose E.
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Joined: 13 Aug 2009
Posts: 203
Schools: Sloan '14 (S)
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Kudos [?]: 83 [0], given: 16

Re: Packages [#permalink] New post 12 Nov 2009, 02:25
The only way to solve this is to run quickly through the numbers.

I have read suggestions that have said for questions where you are backsolving, start from E and work backwards.

E. 12:
No combination of those numbers 1,3,5,7 can add up to 12
1+3+5+7 = 16
16-1 = 15
16-3 = 13
16-5 = 11

ANSWER: E. 12
Re: Packages   [#permalink] 12 Nov 2009, 02:25
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