Solution:

As per the puzzle given above The fewest number of aircraft is 3! Imagine 3 aircraft (A, B and C). A is going to fly round the world. All three aircraft start at the same time in the same direction. After 1/6 of the circumference, B passes 1/3 of its fuel to C and returns home, where it is refueled and starts immediately again to follow A and C.

C continues to fly alongside A until they are 1/4 of the distance around the world. At this point C completely fills the tank of A which is now able to fly to a point 3/4 of the way around the world. C has now only 1/3 of its full fuel capacity left, not enough to get back to the home base. But the first 'auxiliary' aircraft reaches it in time in order to refuel it, and both 'auxiliary' aircraft are the able to return safely to the home base.

Now in the same manner as before both B and C fully refueled fly towards A. Again B refuels C and returns home to be refueled. C reaches A at the point where it has flown 3/4 around the world. All 3 aircraft can safely return to the home base, if the refueled process is applied analogously as for the first phase of the flight.

Solution:

1)The first bit is easy, as the first letter of each plate spells WHODUNNIT IX(A challenge to our Detective.)
2)The second bit is a little trickier, but I gave you the solution.If you read the last three letters in each plate from the bottom up and right to left you get ADVANCE EACH LETTER BY THE NUMBER SHOWN, so advance W by 5 to get B, H by 13 to get U and so on until you spell BUTLER DID IT.

Solution:

False- if he had instantly died, how would anybody know his dream

Solution:

A B C
E F G H
I J K L
M N O P
R S T
5 horizontal rows
4 vertical columns
4 diagonals with slope 1 (ROL, NKH, MJG, IFC)
4 diagonals with slope -1 (SNI, TOJE, PKFA, LGB)
2 knight's move diagonals with slope -2 (AJS, BKT)

Solution:

Let us first calculate the total number of matches that will be placed:
9C2 * 2
= 9! / (2!7!) * 2
= 8 * 9
= 72
Therefore, every team will play 16 matches.

Let us now glance over the cases that are possible.

Case 1:
Let us say that Team A, Team B and Team C wins 14 matches each (they lose 1 match each from the other two teams), then the Team D can win only (16 - 3 * 2) = 10 matches.

Case 2:
Suppose that, the top 4 teams win 13 matches each.
Team A will win 13 matches and will lose 3 matches from Team B, Team C and Team D.
Team B will win 13 matches and will lose 3 matches from Team A, Team C and Team D.
Team C will win 13 matches and will lose 3 matches from Team A, Team B and Team D.
Team D will win 13 matches and will lose 3 matches from Team A, Team B and Team C.

If this case actually happens, then the Team E will be able to win 8 matches at maximum as Team A, Team B, Team C and Team D will win all the matches played by E.

Case 3:
Suppose that, the top 5 teams win 12 matches each.
Team A will win 12 matches and will lose 3 matches from Team B, Team C, Team D and Team E.
Team B will win 12 matches and will lose 3 matches from Team A, Team C, Team D and Team E.
Team C will win 12 matches and will lose 3 matches from Team A, Team B, Team D and Team E.
Team D will win 12 matches and will lose 3 matches from Team A, Team B, Team C and Team E.
Team E will win 12 matches and will lose 3 matches from Team A, Team B, Team C and Team D.

If this case actually happens, Team E will win 12 matches but it still won't be able to get into the playoffs assuming that it has the lowest run rate among all.

This implies that from case 2, it is evident that a team must win 13 matches at the least to confirm its position in the playoffs.

Solution:

89

Solution:

It can be done as shown below.

Solution:

3,3,8

Lets break it down. The product of their ages is 72. So what are the possible choices?

2, 2, 18 sum(2, 2, 18) = 22
2, 4, 9 sum(2, 4, 9) = 15
2, 6, 6 sum(2, 6, 6) = 14
2, 3, 12 sum(2, 3, 12) = 17
3, 4, 6 sum(3, 4, 6) = 13
3, 3, 8 sum(3, 3, 8 ) = 14
1, 8, 9 sum(1,8,9) = 18
1, 3, 24 sum(1, 3, 24) = 28
1, 4, 18 sum(1, 4, 18) = 23
1, 2, 36 sum(1, 2, 36) = 39
1, 6, 12 sum(1, 6, 12) = 19

The sum of their ages is the same as your birth date. That could be anything from 1 to 31 but the fact that Torres was unable to find out the ages, it means there are two or more combinations with the same sum. From the choices above, only two of them are possible now.

2, 6, 6 sum(2, 6, 6) = 14
3, 3, 8 sum(3, 3, 8 ) = 14

Since the eldest kid is taking piano lessons, we can eliminate combination 1 since there are two eldest ones. The answer is 3, 3 and 8.

Solution:

888 + 88 + 8 + 8 + 8 = 1000

Solution:

17 min

The initial solution most people will think of is to use the fastest person as an usher to guide everyone across. How long would that take? 10 + 1 + 7 + 1 + 2 = 21 minutes. Is that it? No. That would make this question too simple even as a warm up question.

Let's brainstorm a little further. To reduce the amount of time, we should find a way for 10 and 7 to go together. If they cross together, then we need one of them to come back to get the others. That would not be ideal. How do we get around that? Maybe we can have 1 waiting on the other side to bring the torch back. Ahaa, we are getting closer. The fastest way to get 1 across and be back is to use 2 to usher 1 across. So let's put all this together.

1 and 2 go cross
2 comes back
7 and 10 go across
1 comes back
1 and 2 go across (done)

Total time = 2 + 2 + 10 + 1 + 2 = 17 minutes