What recommendations can be used to speed up solving a logic puzzle problem involving placing in order a set of elements using a logic grid? The problem is as follows:

A group of archaeologists found an artifact to which they believe
  tells an account of the results of a race in an ancient Hippodrome.
  After days of studying the document they were able to decode the
  message which is summarized below.
The race was comprised of six horses each one wearing an insignia with
  a number on it. The judges declare that there was no tie at the end
  and this was reassured by the spectators. We also noted:
I. The order of arrival does not match with the numbers used by the
  horses. 
II. The sum of the numbers of those who went in the last two places is
  7. 
III. The horse which wears the badge with the number 3 on it arrived
  on the second place. 
IV. The horse which arrived in third place has an odd number written
  in its identification badge.
Find the sum of the numbers on horses badges which arrived in 4th and
  5th places.

The alternatives given in my book are:


*

*7

*6

*5

*9

*8


In my attempt to solve this puzzle and going in circles for several minutes I concluded the following using a grid as I will explain in the following lines.
The first clue that it was clear to me is III.
This was a tricky thing since the number on the badges can be confused with the number that states the order of arrival from the horses so to avoid that confusion I put a circle outside the number to represent the badges used by the horses. Since it was mentioned that the one which arrived in second place wears number 3 then this cancels the other choices as seen below:

Now for the second clue that I used was IV which states that the horse who went 3rd had an odd number insignia so this meant that it cannot be neither 2, 4 or 6. This is seen below.

Now that was kind of easy, here is where it all became into guessing. And for this part I used the second clue which states that the sum of the horses who went in the last two places is 7. From those numbers the only choices that can make 7 are 3+4, 1+6 or 2+5. 3 and 4 can't be as it is known 3 is in second place and by extension four cannot be in the last two places. Seen below:

So the remaining choices can be 1 and 6 or 2 and 5 for the places of the remaining horses. To add up with the first clue which meant that no horse arrived with the same number written on its badge I tried with 1 being the 6th place so that 6 being the one which ended 5th place. This is shown below:

By this time, all cascaded into plugging checks with remaining choices so that they do not share the same number:

So finally that's what I've found and since what it is being asked is the sum of the numbers on the badges from the horses which arrived in 4th and 5th place, from the table can be concluded:
$$2+6=8$$
And this appears within the alternatives. To which I hope that is correct. But this whole process took me a while to complete and needless to say time that is kind of limited in an exam. Moreover what If i had started filling the table using 2 and 5?. It doesn't seem that it may had got me tangled with trial and error. 
So, Is there a better method or alternative that can be used to speed up this process? 
 A: 

Now that was kind of easy, here is where it all became into guessing.
...
To add up with the first clue which meant that no horse arrived with
  the same number written on its badge ...

It was not at all obvious to me that that's what the first clue means, but assuming that you've interpreted it correctly we don't need to guess yet. We cross out the diagonal:
    1 2 3 4 5 6

(1) x x . . . .
(2) . x x . . .
(3) x o x x x x
(4) . x x x . .
(5) . x . . x .
(6) . x x . . x

Applying the same reasoning that prevents (4) from being in position 5 or 6 because that would require (3) to be in the other position, we have
    1 2 3 4 5 6

(1) x x . . . .
(2) . x x . . .
(3) x o x x x x
(4) . x x x x x
(5) . x . . x .
(6) . x x . . x

and then (4) is forced to position 1, ruling out all other possibilities for position 1:
    1 2 3 4 5 6

(1) x x . . . .
(2) x x x . . .
(3) x o x x x x
(4) o x x x x x
(5) x x . . x .
(6) x x x . . x

At this point (6) is definitely one of the numbers in the sum. If (6) is in 5th place then (1) is in 6th place and (2) can only go in 4th place, giving result 8. Alternatively (6) is in 4th place, meaning that (2) and (5) take 5th and 6th places, so (5) is in 6th place and (2) in 5th place, again giving result 8. QED

Alternatively, from the last table, we apply clue II to say that (2) cannot be in 6th place since that would put (5) in 5th place: then we have
    1 2 3 4 5 6

(1) x x . . . .
(2) x x x . . x
(3) x o x x x x
(4) o x x x x x
(5) x x . . x .
(6) x x x . . x

at which point it's obvious that since (2) and (6) are both restricted to 4th and 5th place1, the sum of 4th and 5th place is 8.
This is about as much as we can say for certain: if we update the table to show this last point we get 
    1 2 3 4 5 6

(1) x x . x x .
(2) x x x . . x
(3) x o x x x x
(4) o x x x x x
(5) x x . x x .
(6) x x x . . x

but each of the remaining .s can be o in one valid solution and x in another:
    1 2 3 4 5 6             1 2 3 4 5 6

(1) x x o x x x            (1)  x x x x x o
(2) x x x x o x            (2)  x x x o x x
(3) x o x x x x            (3)  x o x x x x
(4) o x x x x x            (4)  o x x x x x
(5) x x x x x o            (5)  x x o x x x
(6) x x x o x x            (6)  x x x x o x

1 In sudoku circles this is called a naked pair.
