Solution:

Yes We can do it.

9^(9-9)

Solution:

* First, the fireman and the athlete will go spending two minutes.
* The fireman will come back taking one minute.
* The drunk guy and the old lady will go spending five minutes.
* The athlete will come back taking two minutes.
* The athlete and fireman will go spending two minutes.

Two + One + Five + Two + Two = Twelve.

Thus all of them will be out of the house before it collapses.

Solution:

It has been clearly stated that none of the two students gave same number of correct answers which leads us to two different possibilities of correct answers:
Either (1 + 2 +3 + 4 + 5 = 15) or (0 + 1 + 2 + 3 + 4 = 10) stands true.

After analyzing the question, the maximum number of correct answers possible with the answers Gerry, Billy, Clark, Peeta and Jonathan gave:
Question 1 = 2 (b or c)
Question 2 = 2 (b or c)
Question 3= 4 (True)
Question 4 = 4 (True)
Question 5 = 3 (True)

Let us first assume with the maximum number of correct answers possible according to our data which are 15 (2+2+4+4+3). According to it, Billy must have given all correct answers as he was the only one who answered True for questions 3, 4 and 5. But in that case, Gerry and Clark must have given exactly three correct answers. Furthermore, Peeta and Jonathan must have given two correct answers. Therefore none of them got all the five correct answers which is wrong according to what we have analyzed.

Thus we move on to our next assumption according to which, the total number of correct answers are 10 (0+1+2+3+4). Now we must acknowledge that Questions 3 as well as Question 4 cannot be False as it will mean that the number of correct answers would not be 10. So the students giving wrong answers cannot be Gerry, Billy and Peeta.

Assume that Jonathan got all wrong, it will suggest that Gerry, Clark and Peeta each would have at least two correct answers. Then, Billy would have to be the person with only one correct answer and in that situation, the correct answers for questions 1 and 2 would be a and a respectively. Since that is not possible as then the total number of correct answers would be 1 (a) + 1 (a) + 1 (False) + 4 (True) + 2 (Flase) = 9.

Thus it must be Clark who has given all wrong answers. The correct answers are then b, a, True, False and False. Also, the scores are Clark (0), Billy (1), Gerry (2), Jonathan (3) and Peeta (4).

Solution:

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

Solution:

I am DAVID.

At the core I have 5
If we talk in roman numeric system, 5 = V

Similarly, 500 stands in the beginning and end
500 = D

I also include the first letter i.e. A and the first number i.e. I (in roman numeric system)

All of them combine to form DAVID.

Solution:

54

Explanation:
7^2 + 2^2 + 1^1
6^2 + 3^2 + 3^2
2^2 + 5^2 + 5^2

Solution:

There is only one possible shape i.e. a cone.

Solution:

The girl was born on 29 February. Thus her birthday came once in four years only while her sister was born on a normal day and celebrated her birthday every year. Thus the girl had a chance of depositing money only 10 times in 4 years through which she collected $2500 while her sister took $50 from the piggy bank every year making the total amount to be $2000. Thus when the piggy bank was opened, it had just $500.

Solution:

1:1 (approx)

Explanation:
We know that the probability of having a boy or having a girl is same and thus, half of the couples will stop after having a boy child. Half of the others will have a girl and will have a baby again. Out of those half of the couples, half will have a boy and will stop and half will have a girl again. This will keep on going like this.

Now, if there are X number of couples, there will be X boys.

1/2 have a boy and stop: 0 girls
1/4 have a girl, then a boy: X/4 girls
1/8 have 2 girls, then a boy: 2*X/8 girls
1/16 have 3 girls, then a boy: 3*X/16 girls
1/32 have 4 girls, then a boy: 4*X/32 girls
…
Total: X boys and
1X 2X 3X 4X
– + – + – + — +… = ~X

Therefore, the proportion of boys to girls will be extremely close to 1:1

Solution:

(0! + 0! + 0!)! = 6
(1 + 1 + 1)! = 6
2 + 2 + 2 = 6
3 x 3 - 3 = 6
√4 + √4 + √4 = 6
5 + 5/5 = 6
6 + 6 - 6 = 6
7 - 7/7 = 6
8 - √√(8 + 8) = 6
√(9 x 9) - √9 = 6