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:

Four

As the window was closed it is obvious that the six candles that were not extinguished were lit till they melted away entirely. Thus only four candles that were extinguished remain at the end.

Solution:

The 100th statement for sure is incorrect because it says that at least 100 of the statements are incorrect.

Suppose if that is correct, then 100 statements cannot be true.
This suggests that the 100th statement is incorrect and that the first statement is true.

Similarly 99 statements cannot be true because if they were true, then two statements would become correct i.e. the 1st and the 99th.
But the 99th statement says that at least 99 are incorrect.
This suggests that the 99th statement is incorrect and that 2ndone is true.

If we keep analyzing is the same way till the end, we will find out that only the first fifty statements are true and all the remaining ones are incorrect.

Solution:

This problem can be solved with algebra as well as logical reasoning. We are going to tell you how it is logically possible.

Be it any quantity that was present in the beginning and any liquid as well. We know that we have taken a spoonful from one glass and put it into another. Then, we have taken a spoonful from the other glass and put it into the first glass. So, at the end of it, the quantity of liquid in either glass remains same as it was in the beginning.

Therefore, we can conclude the fact that any amount of coke that is missing in the glass with coke will be present in the lemonade glass. Also, the same quantity of lemonade will be missing from the lemonade glass and will be present in the coke glass.

Solution:

1. Fill 5 liters jar ( 5l-jar:5, 3l-jar:0)
2. Transfer to 3 liters pail (5l-jar:2, 3l-jar:3)
3. Empty 3 liters jar ( 5l-jar:2, 3l-jar:0)
4. Transfer 2q from 5 pail to 3 pail (5l-jar:0, 3l-jar:2)
5. Fill 5 liters pail(5l-jar:5, 3l-jar:2)
6. Transfer 1q from 5 pail to 3 pail(5l-jar:4, 3l-jar:3)

Solution:

Both will rotate at Same speed.

Explanation:
Gears of the same size will always rotate at constant speeds despite any smaller or larger gears in between them.

Solution:

Difference 1
Toy plane direction is different.

Difference 2
On the clock, number 1 & 2 is missing.

Difference 3
The colors of lamp's light is changed.

Difference 4
The ribbon on the gift is missing.

Difference 5
Duck is added in the gift section.

Solution:

If we have to measure precisely, it is impossible. Because we are asked to measure odd number of liters whereas all the jars we have can contain only even liters of water.

* Tricky question asked in ias/infosys/infoedge/ibibo interview

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:

It is best if the person sits on the back of the train. After the stop, the train will be accelerating as it goes into the tunnel. Thus the train will be much master when the back of the train enters the tunnel. In this way, he will have to spend much less time in the tunnel.