Solution:

As the message contains one truth, the third says that the cash is in the second bag, if it is to be true, then the first bag message will also become true. So cash cannot be in second and third bags. cash is in the first bag.

Solution:

   9567
+ 1085
======
 10652

Solution:

The two numbers are 4 and 13 whose product is 52 and sum is 17.

Solution:

99 matches.

The number of matches will always sum up to one less than the number of players in a knock out tournament. You may calculate it in any manner. Thus 99 matches were played.

Solution:

2 people

Say 2 helpers are Helper1 & Helper2

Helper1 food Helper2 food McDuck's Food
End of Day 1 : 1 4 4
Now 'A' will head back

End of Day 2 : N/A 2 4
Now 'B' will head back

Now Mr. McDuck have 4 day food :-)

Solution:

Four

three placed flat on the table in a triangle(touching each other) and put the fourth one on top of them in the middle.

Solution:

9 - 8 + 76 - 5 + 4 + 3 + 21 = 100


Other ways of making number 100 (withiout using number 32)
98 + 7 + 6 - 5 - 4 - 3 + 2 - 1 = 100
98 - 7 - 6 - 5 - 4 + 3 + 21 = 100
9 - 8 + 76 - 5 + 4 + 3 + 21 = 100
98 - 7 + 6 + 5 + 4 - 3 - 2 - 1 = 100
98 + 7 - 6 + 5 - 4 + 3 - 2 - 1 = 100
98 + 7 - 6 + 5 - 4 - 3 + 2 + 1 = 100
98 - 7 + 6 + 5 - 4 + 3 - 2 + 1 = 100
98 - 7 + 6 - 5 + 4 + 3 + 2 - 1 = 100
98 + 7 - 6 - 5 + 4 + 3 - 2 + 1 = 100
98 - 7 - 6 + 5 + 4 + 3 + 2 + 1 = 100
9 + 8 + 76 + 5 + 4 - 3 + 2 - 1 = 100
9 + 8 + 76 + 5 - 4 + 3 + 2 + 1 = 100

Solution:

I am Spice

Explanation:
Spice is hot if you remove first two alphabets then it will become "ice", which is really cold.

Solution:

If you are thinking to hold one chocolate from each box in hand and then balancing weight in bare hands, you are thinking all wrong.

Let us begin by labelling boxes as 1, 2, 3 and so on till 10.

Now pick one chocolate from box 1, two chocolates from box 2, three from box 3 and so on. In total, you will have 55 chocolates now. (1 + 2 + 3 + ..... + 10)

The ideal weight of the chocolates should be 55 * 20 = 1100. However, somewhere in there are the defected chocolate/s.

You can judge that clearly by noting down the result of 1100 - total weight of chocolates. If the weight is less than 1 gram, the defected box is box 1, if the weight is less than 2 grams, the defected box is box 2 and so on.

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:

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