A puzzle on quiz?

Question

here is a puzzle and Analysis is given below

http://i218.photobucket.com/albums/cc298/curseofgoldendragon/save-1.png

please note the green shaded area . i made that green .......because i had some problems there . i don't know how to read those and how to utilize them . whats the systemetic way to utilize these facts ? which one to use first ? there are 5 facts and i am in trouble what to take into.......how to decide that ?

Here is the Analysis solution

part1:
http://i218.photobucket.com/albums/cc298/curseofgoldendragon/t1.png

part2
http://i218.photobucket.com/albums/cc298/curseofgoldendragon/t3.png

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I don't understand how they started and where from to start.

I need a process / technique .
How do i start ?

I don't understand the solution.

Answer 1

The best place to start is the point iii). Enter 2 into the table in row C column C. Row C must sum to 4 since C was given 4 direct questions, so fill in 0 for the other entries.

Then go to point ii) the wording of which is confusing. But essentially the direct questions are represented by the main diagonal. The questions passed once are represented by the diagonal next to it. And so on.

In the end though there is not enough info to completely fill in the table. The number of questions passed once is 2. And the numbers in the main diagonal range from 1 to 3. So there is no unique solution beyond this.

Answer 2

Note that clue (v) just tells you that all the questions asked are in the grid somewhere. Since the preamble says 4 questions were directed at each candidate, each row must add to 4. (Clue (v) also says the questions were answered correctly, which is only relevant if you have to calculate point scores later). I'll use cell XY to refer to the cell in row X and column Y.

Start with the simplest parts first: clue (iii) says you can fill in cell CC (the box in row C, column C) with the value 2. The filled values in row C now add to 4, so enter 0 for the remaining values in row C. The grid now looks like this, where x means unknown:
x x 2 x x x
x x x x 1 x
1 0 2 0 0 1
x 0 x 2 x x
2 x x x x x
x 1 0 x 1 x

Clues (i) and (iv) aren't helpful in themselves, but can be used in combination with clue (ii). But we can also get some information from clue (ii) directly. Note that 2+5+3+1+1 = 12, so 12 questions were passed and 12 must have been answered directly.

Now the number of questions passed once is AB + BC + CD + DE + EF + FA = 2. This doesn't help much at this stage. For questions passed twice, though, we have AC + BD + CE + DF + EA + FB = 5; but AC, EA and FB already add up to 5, so BD and DF must be 0 (we already know CE is 0).

For questions passed three times, we have AD + BE + CF + DA + EB + FC = 3. Putting in known values we get AD + DA + EB = 1. But clue (iv) says that AD = EB, so they must be 0 and DA must be 1.

At this stage the grid is
x x 2 0 x x
x x x 0 1 x
1 0 2 0 0 1
1 0 x 2 x 0
2 0 x x x x
x 1 0 x 1 x

For questions passed four times, we have AE + BF + CA + DB + EC + FD = 1. But CA is 1, so all the others must be 0. Similarly, for questions passed five times, we have AF + BA + CB + DC + ED + FE = 1, and we know FE is 1, so the others must be 0.

Now the grid is
x x 2 0 0 0
0 x x 0 1 0
1 0 2 0 0 1
1 0 0 2 x 0
2 0 0 0 x x
x 1 0 0 1 x

Note that we now know all cells in row D except DE. Since each row must add to 4, we can conclude that DE = 1. So we have the final grid
x x 2 0 0 0
0 x x 0 1 0
1 0 2 0 0 1
1 0 0 2 1 0
2 0 0 0 x x
x 1 0 0 1 x

Now, we know there were 12 direct questions, so there must be an average of 2 for each candidate. But since each row must sum to 4, you can see that only B can answer more than 2. So there are two cases:
(a) B answers 2 direct questions, hence everyone else does too and it's easy to see that the final grid is
2 0 2 0 0 0
0 2 1 0 1 0
1 0 2 0 0 1
1 0 0 2 1 0
2 0 0 0 2 0
0 1 0 0 1 2
(b) B answers three direct questions:
x x 2 0 0 0
0 3 0 0 1 0
1 0 2 0 0 1
1 0 0 2 1 0
2 0 0 0 x x
x 1 0 0 1 x
In this case each of rows A, E and F has two consecutive unknowns summing to 2. One of them must be 1 1 and the other two must be 2 0 (remembering that we need 5 direct questions out of the 6 remaining to be placed). However, there is no way of telling which row has the indirect question.

Note that you can also consider these two cases as a single case by looking at the four possibilities of where to place the remaining indirect question, but I think it's a little easier to understand this way. In any event, there are four possible solutions:

1 1 2 0 0 0
0 3 0 0 1 0
1 0 2 0 0 1
1 0 0 2 1 0
2 0 0 0 2 0
0 1 0 0 1 2

2 0 2 0 0 0
0 2 1 0 1 0
1 0 2 0 0 1
1 0 0 2 1 0
2 0 0 0 2 0
0 1 0 0 1 2

2 0 2 0 0 0
0 3 0 0 1 0
1 0 2 0 0 1
1 0 0 2 1 0
2 0 0 0 1 1
0 1 0 0 1 2

2 0 2 0 0 0
0 3 0 0 1 0
1 0 2 0 0 1
1 0 0 2 1 0
2 0 0 0 2 0
1 1 0 0 1 1

Answer 3

(For convenience, I'll denote the row X column Y entry as XY.)

In general I would say you want to look for things that you almost know, or things where some sort of "extreme" information is given.

For example, the "extreme" information here is that a whopping 5 questions were passed by two people (which is where the analysis starts). This might be particularly useful.
Notice that the number of questions passed by n people corresponds to a diagonal in the table (for example, the diagonal DA, EB, FC, AD, BE, CF corresponds to the number of questions passed by 3 people; here, AD corresponds to the number of questions posed to A and passed by A, B, and C).

So let's look at the first bit of analysis: using the fact that 5 questions were passed by two people, the sum of EA, FB, AC, BD, CE, DF must be 5. But the first three of these alone add to 5, so the latter three must be 0's.

Without studying the rest of the analysis, it would appear that they use this diagonal trick along with fact (ii) a few more times. They also note that since 4 questions are posed to each person, and fact (v) tells us that every question is answered, each row must sum to 4.

Fact (i) tells us that XX (the main diagonal) is always between 1 and 3 (inclusive).

Fact (iii) tells us that CC is 2. (I think this would be the first thing I would do, as it's just so simple to use this information.)

Fact (iv) says EB=AD.

I have to get going away from this computer; if I've helped a little but you have a couple of more specific questions, post up additional details, and if no one else gets around to it by around 8PM, I'll be back to help some more.