Solution:

25 Fruits

Remember we told you that Birbal is a witty trader. So his sole motive is to get rid of the sacks as fast as he can.

For the first sack:
He must be able to fill fruits from one sack to other two sacks. Assume that he is able to do that after M check points. Now to find M,

(Space in first sack) M + (Space in second sack) M = (Remaining fruits in Third Sack) 30 – M
M = 10

Thus after 10 checkpoints, Birbal will be left with only 2 sacks containing 30 fruits each.

Now he must get rid of the second sack.
For that, he must fill the fruits from second sack to the first sack. Assume that he manages to do that after N checkpoints.

(Space in First Sack) N = (Remaining fruits in second sack) 30 – N
N = 15

Thus after he has crossed 25 checkpoints, he will be left be one sack with 30 fruits in it. He has to pass five more checkpoints where he will have to give five fruits and he will be left with twenty five fruits once he has crossed all thirty check points.

Solution:

Shikhar hit a sixer and then got retired hurt.
Suresh Raina came but he got run out after he and Kohli had crossed halfway.
On the next ball, Kohli scored a six and was not out at 100.

Solution:

"World War I" was named after after "World War II"

Solution:

The node can be connected as shown.

Solution:

This statement is true but only in a specific case i.e. if both of the siblings continue to live till the day the younger is as old as the elder one was during the time of their birth.

Solution:

There can be two possibilities for the given situation.

One is if I run on the perimeter. Then, the lion will eventually catch me as the lion can follow my radius and then his trajectory will be half of the circle and not a spiral. Therefore, the lion will subsequently catch me in a finite amount of time.

Secondly if I don’t run on the perimeter, then clearly, I have an infinite amount of time before the lion catches me.

Solution:

The Stars

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:

2, 3, 10 and 10
To maintain the difference of eight and keeping in mind the third and fourth hint, the first and third cards can be 2 and 10 or 10 and 2. But keeping in mind the first hint, we know that the first card is 2 and the third is 10. Similarly, if we see the fifth hint, the difference between the second and fourth card should be 7. Thus we know that the second card is 3 and the fourth card is 10.

Solution:

0%

Explanation:
1) why cant be 1/4 : If the answer is 1/4, then as we know two out of four answer choices is '1/4', the answer has be 1/2.
This is a contradiction, so the answer cannot be 1/4.

2) why cant be 1/2 : If the answer is 1/2 then because answer:"1/2" is 1 out of 4 answer choices, the answer must be 1/4. This is also a contradiction. So the answer cannot be 1/2.

3) why cant be 1 : If the answer is 1 then because answer:"1" is 1 out of 4 answer choices, the answer must be 1/4. Again the same contradiction and therefore answer cannot be 1.

Solution:

NONE, holes are empty.