Solution:

117(1+3+5,3+4)

Solution:

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

Solution:

To do it, the marbles must be arranged as the vertices of an octahedron. If you are unsure, check the image below.

Solution:

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

Probability :
50 + 24/49

Solution:

16-days

In 15-days, he will climb 15-inch and when he climbs 5 inch more, he will be out of the well :-)

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:

1

Explanation:
4 + 9 = 1
The trick to solve is to assume the first number as the time in AM and the second number is the number of hours you have to add to it. You'll find the time in PM when you do that.

For example
7 AM + 7 hours = 14:00 or 2 PM

4 AM + 9 hours = 13:00 or 1 PM.

Solution:

To do this, he must be pushing the cork inside the bottle and then taking out the coin.

Solution:

28 =>1+2+4+7+14

cannot be 1 (as number cannot be included)

Solution:

Either your read from start or from ends, it will be same

Solution:

1653

Explanation:
For an ideal case, the batsman will hit a six on each ball. But if he hits six on the last ball of the over, the strike will change in the next over. Thus, the best he can do in the last ball is run 3 runs so that he retains the strike even in the next over. Thus the total runs that he can score in each over:
6 * 5 + 3 = 33
But this will have to go like it only till the 49th over. In the last over, he can hit a six in the last ball as well as that will be the last ball of the match.
Thus runs for the last over will be 6 * 6 = 36.
Hence the maximum runs = 33 * 49 + 36 = 1653