Solution:

They are numbers for houses and it's Rs.100 per digit.

Solution:

For this answer is 3^0, 3^1, 3^2... That is 1,3,9,27,81,243 and 729.

Solution:

The answer is 2. First, divide the coins into 3 equal piles. Place a pile on each side of the scale, leaving the remaining pile of 3 coins off the scale. If the scale does not tip, you know that the 6 coins on the scale are legitimate, and the counterfeit is in the pile in front of you. If the scale does tip, you know the counterfeit is in the pile on the side of the scale that raised up. Either way, put the 6 legitimate coins aside. Having only 3 coins left, put a coin on each side of the scale, leaving the third in front of you. The same process of elimination will find the counterfeit coin.

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:

17 mins

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 mins. 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 mins

Solution:

89

Solution:

2nd

When the cylinder is spun for the first and last time, the following six outcomes are possible:

First:
First chamber is fired first. The first player loses.

Second:
Second chamber is fired first. The first player loses.

Third:
Third chamber is fired first. The first player loses.

Fourth:
Fourth chamber is fired first. The second player loses. (First shot, player 1, chamber 4 empty. Second shot player 2, chamber 5, empty. Third shot player 1, chamber 6 empty. Fourth shot player 2, chamber 1 not empty.)

Fifth:
Fifth chamber is fired first. The first player loses. (First shot, player 1, chamber 5 empty. Second shot player 2, chamber 6, empty. Third shot player 1, chamber 1 not empty.)

Sixth:
Sixth chamber is fired first. The second player loses. (First shot, player 1, chamber 6 empty. Second shot, player 2, chamber 1, not empty)

Thus, if you analyze, you will find that the second player has a 4/6 chance of winning. Therefore if you want to win the game, going second gives you better shot at it.

Solution:

First Person: 3 full bottles, 1 half empty bottle and 3 empty bottles.

Second Person: 3 full bottles, 1 half empty bottle and 3 empty bottles.

Third Person: 1 full bottle, 5 half empty bottles and 1 empty bottle.

Solution:

7 minutes hourglass = p
11 minutes hourglass = q

Turn p and q.
P will finish at 7 minutes and then turn it again.
When q finishes (11 minutes), turn p again.
It will take 4 minutes for p to become empty.

The total time if you count is 15 minutes when p empties.

Solution:

The first step will be to cut the three tablets in equal halves and then put each half in two piles. Now, each of the two piles contains half of pill A and two halves of pill B (or one pill B). Now, take out one pill A from the bottle and cut it into half. Put the two halves in two piles.

In this way, each pile now contains two halves of pill A and two halves of pill B or 1 pill A and 1 pill B. You can now take 1 pile one day and the second pile on the next day.