Solution:

Costa

Explanation:
c + 1 characters-> d
o + 2 characters-> q
s + 3 characters-> v
t + 4 characters-> x
a + 5 characters-> f

=> dqvxf = costa

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

Solution:

A) Fill 5 ml gallon ( 5mlGallon - 5, 3mlGallon - 0)
B) Transfer to 3 ml gallon (5mlGallon - 2, 3mlGallon - 3)
C) Empty 3 ml gallon ( 5mlGallon - 2, 3mlGallon - 0)
D) Transfer 2 ml from 5 ml gallon to 3 ml gallon (5mlGallon - 0, 3mlGallon - 2)
E) Fill 5 ml gallon(5mlGallon - 5, 3mlGallon - 2)
F) Transfer 1 ml from 5 ml gallon to 3 ml gallons(5mlGallon - 4, 3mlGallon - 3)

Solution:

Difference 1
Toy plane direction is different.

Difference 2
On the clock, number 1 & 2 is missing.

Difference 3
The colors of lamp's light is changed.

Difference 4
The ribbon on the gift is missing.

Difference 5
Duck is added in the gift section.

Solution:

A Spider

Solution:

The last marble will be yellow.

Since marbles can only be taken out in pairs and you started off with an odd number of yellows there is always going to be one yellow left over that you'll keep putting back in the box until it's left on it's own.

Solution:

The competition was held on 11 November 2013. The logic is simple. In America, the dates are mentioned as Month/Day/Year and in Britain, dates are written as Day/Month/Year. But on 11 November 2013, the date and month will be the same. And thus both will write in the same manner i.e. 11/11/2013. If it would have been any other date, the evaluators would have been able to identify it.

Solution:

8/(1−1/5)

Solution:

Everyone will done one day or other.

Abraham Lincoln era is too old and we expect most of them(if not all) to be dead by now.

Solution:

30,450 miles.

Explanation:
For each 5 miles driven using this rotation, each tire will lose (2/29000)+(2/21000) of its tread.So, suppose we drive 5*N miles like this. Then, each tire will lose 2*N*((1/29000)+(1/21000)) of its tread. We wish to determine N so that the latter expression gets to be 1, representing all of each tire's tread. That happens when N=6090, which is when 5*N (the total miles driven) is 30,450.

Solution:

So that the following plan can be followed, let us number the coins from 1 to 12. For the first weighing let us put on the left pan coins 1,2,3,4 and on the right pan coins 5,6,7,8.

There are two possibilities. Either they balance, or they don't. If they balance, then the different coin is in the group 9,10,11,12. So for our second one possibility is to weigh 9,10,11 against 1,2,3

(1) They balance, in which case you know 12 is the different coin, and you just weigh it against any other to determine whether it is heavy or light.
(2) 9,10,11 is heavy. In this case, you know that the different coin is 9, 10, or 11, and that that coin is heavy. Simply weigh 9 against 10; if they balance, 11 is the heavy coin. If not, the heavier one is the heavy coin.
(3) 9,10,11 is light. Proceed as in the step above, but the coin you're looking for is the light one.

That was the easy part.

What if the first weighing 1,2,3,4 vs 5,6,7,8 does not balance? Then any one of these coins could be the different coin. Now, in order to proceed, we must keep track of which side is heavy for each of the following weighings.

Suppose that 5,6,7,8 is the heavy side. We now weigh 1,5,6 against 2,7,8. If they balance, then the different coin is either 3 or 4. Weigh 4 against 9, a known good coin. If they balance then the different coin is 3, otherwise it is 4. The direction of the tilts can tell us whwther the offending coin is heavier or lighter.

Now, if 1,5,6 vs 2,7,8 does not balance, and 2,7,8 is the heavy side, then either 7 or 8 is a different, heavy coin, or 1 is a different, light coin.

For the third weighing, weigh 7 against 8. Whichever side is heavy is the different coin. If they balance, then 1 is the different coin. Should the weighing of 1,5, 6 vs 2,7,8 show 1,5,6 to be the heavy side, then either 5 or 6 is a different heavy coin or 2 is a light different coin. Weigh 5 against 6. The heavier one is the different coin. If they balance, then 2 is a different light coin.