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Paper test 42- table question

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Senior Manager
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Paper test 42- table question [#permalink] New post 18 Jan 2009, 05:59
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The table above shows the cost, in dollars, of traveling to and from cities A, B, C, D, E and F. A sales representative wants to leave from A, travel to C, E and F, and return to A. If the first city that the sales representative travels to must be E, what is the minimum possible cost for the entire trip?


A- 13
B- 14
C- 16
D- 18
E- 20
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Senior Manager
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Re: Paper test 42- table question [#permalink] New post 18 Jan 2009, 06:07
I know a way to solve this, but it's not very effective:

Total cost of the trip:A-E-B-C-D-F-A
Total cost of the trip:A-E-B-D-C-F-A
Total cost of the trip:A-E-B-D-F-C-A
Total cost of the trip:A-E-B-C-D-F-A
...
and so on until you get all the possibilities.

There must be a better solution that than. Does anyone know?
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Re: Paper test 42- table question [#permalink] New post 18 Jan 2009, 08:11
lumone wrote:
I know a way to solve this, but it's not very effective:

Total cost of the trip:A-E-B-C-D-F-A
Total cost of the trip:A-E-B-D-C-F-A
Total cost of the trip:A-E-B-D-F-C-A
Total cost of the trip:A-E-B-C-D-F-A
...
and so on until you get all the possibilities.

There must be a better solution that than. Does anyone know?



Do only for the cities (from A to E to either C or F to A) that the sales person goes. There are only two possibilities:

1: Total cost of the trip: A-E-C-F-A = 7+2+4+3 = 16
2: Total cost of the trip: A-E-F-C-A = 7+6+4+3 = 20

select the lower.

Editied for typo: C-F = 4 not 6.
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Re: Paper test 42- table question [#permalink] New post 18 Jan 2009, 11:57
how about:

A->E = 7
E->C = 2
C->F = 4
F->A = 3

Total is 16
Senior Manager
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Re: Paper test 42- table question [#permalink] New post 18 Jan 2009, 13:09
16 is the OA

All I had to do is properly read the question. :|
Re: Paper test 42- table question   [#permalink] 18 Jan 2009, 13:09
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