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Re: Jorge travels to meet once daily with exactly six clients, Clients A t [#permalink]
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So here we can try the permutations and combinations on our own to see how the path is formed

We have to always follow the logic :

John schedules each subsequent meeting according to which client is the least distance from the meeting immediately prior to it.

So here first John goes to E

Now from E ,then from E the closest client is A with a distance of 2,so John will go to A,next from the closest point is F with a distance of 1,so John will go to client F,

Now from F,again the closest is A,but he already met A,so the next closest in the list is B with a distance of 4 ,so from F John will go to B

Now from B,the closest is C with a distance of 2,so John will go to C

Now from C,the closest is B,but John has already met B,so he will go to D(at a distance of 3 from C),so John will go to D.

Now from D,John will not go to A,since he already met A.

So the answers are Yes,Yes,No

Originally posted by Apt0810 on 19 May 2020, 09:30.
Last edited by Apt0810 on 21 May 2020, 08:00, edited 2 times in total.
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Re: Jorge travels to meet once daily with exactly six clients, Clients A t [#permalink]
Hi SajjadAhmad, could you share the official explanation, I am not able to come to Yes for the second one.

thanks.
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Jorge travels to meet once daily with exactly six clients, Clients A t [#permalink]
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First E, so we have E-2-3-4-5-6

Check E's row (or column). A is closest.
So we have E-A-3-4-5-6

Check A's row. Ignore E-A.
F is closest. So we have E-A-F-4-5-6

Check F's row. Exclude E-A-F.
B is closest. So we have E-A-F-B-5-6

Check B's row. Exclude E-A-F-B.
C Is closest. So Jorge's full schedule is E-A-F-B-C-D.

Now check the answer choices
Client B then Client C - Yes
Client C then Client D - Yes
Client D then Client A - No

Posted from my mobile device
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Re: Jorge travels to meet once daily with exactly six clients, Clients A t [#permalink]
­For part 2, 18 miles is farther away than 12 miles? Therefore it should be D then C, no??
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Re: Jorge travels to meet once daily with exactly six clients, Clients A t [#permalink]
Expert Reply
Fish181

You are going wrong in measuring distances. The distances has to be measured from the last client visited.

So we start with E.
1. Scan for all the distances to the remaining clients from E. Closest is A, 2 km away.
2. From A, scan the distances to the remaining clients less E from A. Closest is F, 1 km away.
3. Now, he is with F and the closest is B, 2 km away.
4. From B, C is 3 km away
5. Finally, D is 3 km from C.

Thus, the clients visited are E-2-A-1-F-4-B-2-C-3-D.
# Client B then Client C - Yes
# Client C then Client D - Yes
# Client D then Client A - No

On a lighter and logical note, B to D is 15 km while B-C-D is just 5 km. Why would someone call B-D as direct route when a route one-third the length is available.
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Re: Jorge travels to meet once daily with exactly six clients, Clients A t [#permalink]
­Most urgent: E

1. Sort by E => A is closest

2. Sort by A => F is closest

3. Sort by F => B is closest (since already met A)

4. Sort by B => C is closest

5. Last is D


1. Client B then Client C => Yes

2. Client C then Client D => Yes

3. Client D then Client A => No
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Re: Jorge travels to meet once daily with exactly six clients, Clients A t [#permalink]
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