Solution:

110 kms

72927+110 = 73037 , a palindrome

Solution:

All Digits [0-9] follow the following pattern: There is one occurrence of all digits in every 10 numbers Example. Digit-9 occurred one time between 10 and 19 in 9, between 20 and 29 in 29
However, for tens(91,92,93...99) and hundreds (900,901.... 999), the occurrence is much more. Since we have included 1000, Therefor 1 is the most occurred digit.
Note: Despite three '0s' in 1000, 0 is not most occurred digit as they don't have tens and hundreds like other digits.

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:

Sansa has the siamese (conjoined) twin sister.

Solution:

What is the correct time ?

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:

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.

Solution:

31

Explanation:
Ramanujan lighting the candles in binary.
11110 =30
11111 =31

Solution:

Birbal drew another line that was larger than the pre-drawn line. Now the pre drawn line was smaller.

Solution:

192 021

If you look at it as two-digit numbers you'll see:10 11 12 13 14 15 16 17 18 19 20 21

Solution:

This puzzle has no answer if you are considering this figure in a 2d domain. However if you follow it on a three-dimensional plane, you can do it like shown in the figure given.