Solution:

In GP, a/r+a+ar....=38
a (1+r+r^2) = 38r --- eq1
Product a^3 = 1728
a = 12
Putting the value in eq1
12 (1+r+r^2) = 38r
r=2/3 and 3/2

Thus the smallest required number is 8

Solution:

David to play second

Explanation:
Let's suppose that David plays first and he picks 9. Then Albert will definitely pick 8. Now, David will have to pick 7 or Albert will pick 7 in his turn. But if David picks up 7, then he will score 16 that is beyond 15 and will lose. So one thing is for sure, no one will be willing to start with the highest digits.

Suppose David plays first and picks up 1, Albert will pick 2. Then David will pick 3 and Albert will pick 4. Now David will be forced to pick 9. The score is 6 to 13 and thus David will have no chance of winning.

If David Picks 9 after Albert has picked up 2, then Albert will pick 8 and the score will become 10 to 10. Thus David will pick 3 as picking 7 will send him past 15. Now Albert will pick 4 and David has nothing to pick for winning. Thus Albert wins.

Therefore, you should suggest David to play second.

Solution:

10,15,9,20,12

Explanation:
The score obtained on the first throw is 10.
The score obtained on the second throw is 15.
The score obtained on the third throw is 9.
The score obtained on the fourth throw is 20.
The score obtained on the fifth throw is 12.

Solution:

There are 3 fruits (1 grapes, 1 apples, and 1 oranges).

Solution:

Put one red marble in one box and all other marbles in the 2nd box.

Probability :
50 + 24/49

Solution:

Read the question carefully again. New Year do falls after Christmas but that happens if two different years. The question is put up against the year 2050 and thus there will be 51 weeks and 2 days in between them as New Year will appear on 1 January 2050 and Christmas will happen on 25 December 2050.

Solution:

There are two things to keep in mind:
Firstly there is at least one red hat. (There can be two or three as well).
Secondly the competition is fair for everyone.

Thus if there is only one red hat, that person will see two green hats on other heads and will be able to deduce his own color as red. However the other students will see one red and one green hat and can never be sure. In such manner, the competition will prove to be partial for one student.

Suppose if there are two red hats. Then the students who are wearing red hats will see one red and one green hat on others. Now they must have deduced that there can’t be just one red hat. Thus they will know that they are also wearing a red hat. But the one who is wearing a green hat will see two red hats and can never be sure of his own color. In this case as well, the competition will not be fair.

Thus the only possible and fair means is if all of them are wearing a red hat. The one who is able to deduce the situation first, will raise his hand and will tell the correct answer.

Solution:

Think deeply about the door number 56, people will visit it for every divisor it has. So 56 has 1 & 56, 2 & 28, 4 & 14, 7 & 8. So on pass 1, the 1st person will open the door; pass 2, 2nd one will close it; pass 4, open; pass 7, close; pass 8, open; pass 14, close; pass 28, open; pass 56, close.
Thus we can say that the door will just end up back in its original state for each pair of divisor. But what about the cases in which the pair of divisor has analogous number for example door number 16? 16 has the divisors 1 & 16, 2 & 8, 4&4. But 4 is recurrent because 16 is a perfect square, so you will only visit door number 16, on pass 1, 2, 4, 8 and 16… leaving it open at the end. So only perfect square doors will remain open at the end.

Solution:

It can be down as shown below.

Solution:

45 minutes

Explanation :
All you have to do is burn the first rope from both the ends and the second rope from one end only simultaneously. The first rope will burn in 30 minutes (half of an hour since we burned from both sides) while the other rope would have burnt half. At this moment, light the second rope from the other end as well. Where, the second rope would have taken half an hour more to burn completely, it will take just 15 minutes as we have lit it from the other end too.

Thus you have successfully calculated 30+15 = 45 minutes with the help of the two given ropes.

Solution:

Step 5 is invalid, because we are dividing by (x-y), and since x=y, we are thus dividing by 0. This is an invalid mathematical operation (division by 0), and so by not following basic mathematical rules