A bad king has a cellar of 1000 bottles of delightful and very expensive wine. A neighboring queen plots to kill the bad king and sends a servant to poison the wine.
Fortunately (or say unfortunately) the bad king's guards catch the servant after he has only poisoned one bottle.
Alas, the guards don't know which bottle but know that the poison is so strong that even if diluted 100,000 times it would still kill the king. Furthermore, it takes one month to have an effect.
The bad king decides he will get some of the prisoners in his vast dungeons to drink the wine. Being a clever bad king he knows he needs to murder no more than 10 prisoners – believing he can fob off such a low death rate – and will still be able to drink the rest of the wine (999 bottles) at his anniversary party in 5 weeks time.
Explain what is in mind of the king, how will he be able to do so?

Think in terms of binary numbers. (now don’t read the solution, give a try).
Number the bottles 1 to 1000 and write the number in binary format.
bottle 1 = 0000000001 (10 digit binary)
bottle 2 = 0000000010
bottle 500 = 0111110100
bottle 1000 = 1111101000
Now take 10 prisoners and number them 1 to 10, now let prisoner 1 take a sip from every bottle that has a 1 in its least significant bit. Let prisoner 10 take a sip from every bottle with a 1 in its most significant bit. etc.
prisoner = 10 9 8 7 6 5 4 3 2 1
bottle 924 = 1 1 1 0 0 1 1 1 0 0
For instance, bottle no. 924 would be sipped by 10,9,8,5,4 and 3. That way if bottle no. 924 was the poisoned one, only those prisoners would die.
After four weeks, line the prisoners up in their bit order and read each living prisoner as a 0 bit and each dead prisoner as a 1 bit. The number that you get is the bottle of wine that was poisoned.
1000 is less than 1024 (2^10). If there were 1024 or more bottles of wine it would take more than 10 prisoners.

Mick and John were in a 100 meter race. When Mick crossed the finish line, John was only at the 90 meter mark. Mick suggested they run another race. This time, Mick would start ten meters behind the starting line. All other things being equal, will John win, lose, or will it be a tie in the second race?

John will lose again. In the second race, Mick started ten meters back. By the time John reaches the 90 meter mark, Mick will have caught up him. Therefore, the final ten meters will belong to the faster of the two. Since Mick is faster than John, he will win the final 10 meters and of course the race.

There are n coins in a line. (Assume n is even). Two players take turns to take a coin from one of the ends of the line until there are no more coins left. The player with the larger amount of money wins.
Would you rather go first or second? Does it matter?
Assume that you go first, describe an algorithm to compute the maximum amount of money you can win.
Note that the strategy to pick maximum of two corners may not work. In the following example, first player looses the game when he/she uses strategy to pick maximum of two corners.
Example 18 20 15 30 10 14
First Player picks 18, now row of coins is
20 15 30 10 14
Second player picks 20, now row of coins is
15 30 10 14
First Player picks 15, now row of coins is
30 10 14
Second player picks 30, now row of coins is
10 14
First Player picks 14, now row of coins is
10
Second player picks 10, game over.
The total value collected by second player is more (20 + 30 + 10) compared to first player (18 + 15 + 14). So the second player wins.

Going first will guarantee that you will not lose. By following the strategy below, you will always win the game (or get a possible tie).
(1) Count the sum of all coins that are odd-numbered. (Call this X)
(2) Count the sum of all coins that are even-numbered. (Call this Y)
(3) If X > Y, take the left-most coin first. Choose all odd-numbered coins in subsequent moves.
(4) If X < Y, take the right-most coin first. Choose all even-numbered coins in subsequent moves.
(5) If X == Y, you will guarantee to get a tie if you stick with taking only even-numbered/odd-numbered coins.
You might be wondering how you can always choose odd-numbered/even-numbered coins. Let me illustrate this using an example where you have 6 coins:
Example
18 20 15 30 10 14
Sum of odd coins = 18 + 15 + 10 = 43
Sum of even coins = 20 + 30 + 14 = 64.
Since the sum of even coins is more, the first player decides to collect all even coins. He first picks 14, now the other player can only pick a coin (10 or 18). Whichever is picked the other player, the first player again gets an opportunity to pick an even coin and block all even coins.

The owner of a banana plantation has a camel. He wants to transport his 3000 bananas to the market, which is located after the desert. The distance between his banana plantation and the market is about 1000 kilometer. So he decided to take his camel to carry the bananas. The camel can carry at the maximum of 1000 bananas at a time, and it eats one banana for every kilometer it travels.
What is the most bananas you can bring over to your destination?

First of all, the brute-force approach does not work. If the Camel starts by picking up the 1000 bananas and try to reach point B, then he will eat up all the 1000 bananas on the way and there will be no bananas left for him to return to point A.
So we have to take an approach that the Camel drops the bananas in between and then returns to point A to pick up bananas again.
Since there are 3000 bananas and the Camel can only carry 1000 bananas, he will have to make 3 trips to carry them all to any point in between.
When bananas are reduced to 2000 then the Camel can shift them to another point in 2 trips and when the number of bananas left are <= 1000, then he should not return and only move forward.
In the first part, P1, to shift the bananas by 1Km, the Camel will have to
Move forward with 1000 bananas – Will eat up 1 banana in the way forward
Leave 998 banana after 1 km and return with 1 banana – will eat up 1 banana in the way back
Pick up the next 1000 bananas and move forward – Will eat up 1 banana in the way forward
Leave 998 banana after 1 km and return with 1 banana – will eat up 1 banana in the way back
Will carry the last 1000 bananas from point a and move forward – will eat up 1 banana
Note: After point 5 the Camel does not need to return to point A again.
So to shift 3000 bananas by 1km, the Camel will eat up 5 bananas.
After moving to 200 km the Camel would have eaten up 1000 bananas and is now left with 2000 bananas.
Now in the Part P2, the Camel needs to do the following to shift the Bananas by 1km.
Move forward with 1000 bananas – Will eat up 1 banana in the way forward
Leave 998 banana after 1 km and return with 1 banana – will eat up this 1 banana in the way back
Pick up the next 1000 bananas and move forward – Will eat up 1 banana in the way forward
Note: After point 3 the Camel does not need to return to the starting point of P2.
So to shift 2000 bananas by 1km, the Camel will eat up 3 bananas.
After moving to 333 km the camel would have eaten up 1000 bananas and is now left with the last 1000 bananas.
The Camel will actually be able to cover 333.33 km, I have ignored the decimal part because it will not make a difference in this example.
Hence the length of part P2 is 333 Km.
Now, for the last part, P3, the Camel only has to move forward. He has already covered 533 (200+333) out of 1000 km in Parts P1 & P2. Now he has to cover only 467 km and he has 1000 bananas.
He will eat up 467 bananas on the way forward, and at point B the Camel will be left with only 533 Bananas.

As I was going to the mall I met a man with seven wives. Each wive held two bags, each bag held a mother cat, each mother cat had six babies,
How many people were going to the mall?

A 400 yard long train, travelling at 30 mph, enters a 4.5 mile long tunnel.
How long will elapse between the moment the front of the train enters the tunnel and the moment the end of the train clears the tunnel?