#1 - Hard Measuring Water Riddle

You have two buckets of 11liter and 6liter.
How can you measure exactly 8liter ?

Hard Measuring Water Riddle

Steps 11-Liter 6Liter
1. 11 -
2. 5 6
3. 5 0
4. 0 5
5. 11 5
6. 10 6
7. 10 0
8. 4 6
9. 4 0
10. 0 4
11. 11 4
12. 9 6
13. 9 0
14. 3 6
15. 3 0
16. 0 3
17. 11 3
18. 8 6 ==> got it

#2 - Hardest Balance Logic Puzzle

You are given a set of scales and 12 marbles. The scales are of the old balance variety. That is, a small dish hangs from each end of a rod that is balanced in the middle. The device enables you to conclude either that the contents of the dishes weigh the same or that the dish that falls lower has heavier contents than the other.
The 12 marbles appear to be identical. In fact, 11 of them are identical, and one is of a different weight. Your task is to identify the unusual marble and discard it. You are allowed to use the scales three times if you wish, but no more.

Note that the unusual marble may be heavier or lighter than the others. You are asked to both identify it and determine whether it is heavy or light.

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.

#3 - Tough Probability Interview Question

There is a country where everyone wants a boy. Every family continue to have babies till a boy is born. If the probability of having a girl or a boy is the same, what is the proportion of boys to girls after some time in that country?

Tough Probability Interview Question

Since the probability of having a girl or a boy is same, half of the families will have a boy first and stop. The other half of the families will have a girl and from half of those families, the second born will be a boy and they will stop while the others will again have a girl. This process will continue.

Suppose the number of couples are N, the number of boys will be N.

1/2 have a boy and stop: 0 girls
1/4 have a girl, then a boy: N/4 girls
1/8 have 2 girls, then a boy: 2*N/8 girls
1/16 have 3 girls, then a boy: 3*N/16 girls
1/32 have 4 girls, then a boy: 4*N/32 girls

Total: N boys and
1N 2N 3N 4N
– + – + – + — +… = ~N

Therefore the proportion of boys to girl will be quite close to 1:1.

#4 - Famous Probability puzzle SHOOT

Mr. Black, Mr. Gray, and Mr. White are fighting in a truel. They each get a gun and take turns shooting at each other until only one person is left. Mr. Black, who hits his shot 1/3 of the time, gets to shoot first. Mr. Gray, who hits his shot 2/3 of the time, gets to shoot next, assuming he is still alive. Mr. White, who hits his shot all the time, shoots next, assuming he is also alive. The cycle repeats. If you are Mr. Black, where should you shoot first for the highest chance of survival?

Famous Probability puzzle SHOOT

He should shoot at the ground.

If Mr. Black shoots the ground, it is Mr. Gray's turn. Mr. Gray would rather shoot at Mr. White than Mr. Black, because he is better. If Mr. Gray kills Mr. White, it is just Mr. Black and Mr. Gray left, giving Mr. Black a fair chance of winning. If Mr. Gray does not kill Mr. White, it is Mr. White's turn. He would rather shoot at Mr. Gray and will definitely kill him. Even though it is now Mr. Black against Mr. White, Mr. Black has a better chance of winning than before.


You have two strings whose only known property is that when you light one end of either string it takes exactly one hour to burn. The rate at which the strings will burn is completely random and each string is different.

How do you measure 45 minutes?

Light both the ends of the first string and one end of the second string. 30 minutes will have passed when the first string is fully burned, which means 30 minutes have burned off the second string. Light the end of the second string and when it is fully burned, 45 minutes will have passed.

#6 - Challenging Mind puzzles

You are the ruler of a medieval empire and you are about to have a celebration tomorrow. The celebration is the most important party you have ever hosted. You've got 1000 bottles of wine you were planning to open for the celebration, but you find out that one of them is poisoned.

The poison exhibits no symptoms until death. Death occurs within ten to twenty hours after consuming even the minutest amount of poison.

You have over a thousand slaves at your disposal and just under 24 hours to determine which single bottle is poisoned.

You have a handful of prisoners about to be executed, and it would mar your celebration to have anyone else killed.

What is the smallest number of prisoners you must have to drink from the bottles to be absolutely sure to find the poisoned bottle within 24 hours?

10 prisoners must sample the wine. Bonus points if you worked out a way to ensure than no more than 8 prisoners die.

Number all bottles using binary digits. Assign each prisoner to one of the binary flags. Prisoners must take a sip from each bottle where their binary flag is set.

Here is how you would find one poisoned bottle out of eight total bottles of wine.

Bottle 1 Bottle 2 Bottle 3 Bottle 4 Bottle 5 Bottle 6 Bottle 7 Bottle 8
Prisoner A X X X X
Prisoner B X X X X
Prisoner C X X X X
In the above example, if all prisoners die, bottle 8 is bad. If none die, bottle 1 is bad. If A & B dies, bottle 4 is bad.

With ten people there are 1024 unique combinations so you could test up to 1024 bottles of wine.

Each of the ten prisoners will take a small sip from about 500 bottles. Each sip should take no longer than 30 seconds and should be a very small amount. Small sips not only leave more wine for guests. Small sips also avoid death by alcohol poisoning. As long as each prisoner is administered about a millilitre from each bottle, they will only consume the equivalent of about one bottle of wine each.

Each prisoner will have at least a fifty percent chance of living. There is only one binary combination where all prisoners must sip from the wine. If there are ten prisoners then there are ten more combinations where all but one prisoner must sip from the wine. By avoiding these two types of combinations you can ensure no more than 8 prisoners die.

One viewer felt that this solution was in flagrant contempt of restaurant etiquette. The emperor paid for this wine, so there should be no need to prove to the guests that wine is the same as the label. I am not even sure if ancient wine even came with labels affixed. However, it is true that after leaving the wine open for a day, that this medieval wine will taste more like vinegar than it ever did. C'est la vie.

#7 - Logical Interview Puzzle

Can you divide numbers from 1 to 9 (1 2 3 4 5 6 7 8 9) into two groups so that sum of number of each group is equal.

Note : 9 cannot be turned over to make it 6.

Logical Interview Puzzle

1 2 3 4 5 9 and 7 8 9

sum of numbers 1 2 3 4 5 6 7 8 9 is 45 , so numbers cant be split into 2 equal parts.
I know i cant turn 9 upside down and make it 6 but i can turn 6 to make it 9.

#8 - Maths Logical Problem

Accidentally, two trains are running in the opposite direction and enter a tunnel that is 200 miles long. A supersonic bird that has fled the lab and taken shelter in the tunnel starts flying from one train towards the other at a speed of 1000 mph. As soon as it reaches the second train, he starts flying back to avoid collision and meets the first train again at the other end. The bird keeps flying to and fro till the trains collide with each other.

What is the total distance that the supersonic bird has traveled till the trains collided?

Maths Logical Problem

Let us consider the length of the tunnel first; which is 200 miles. Now, the trains are running on the same speed which means that they will collide at the center of the tunnel and will take an hour to reach the center. Now the bird is travelling at a speed of 1000 mph and it is flying for an hour (since the trains will take an hour to collide). Thus the bird will travel 1000 miles in the process.

#9 - Tricky Maths Problem

I drive at an average speed of 30 miles per hour to the railroad station each morning and just catch my train. On a particular morning there was a lot of traffic and at the halfway point I found I had averaged only 15 miles per hour. How fast must I drive for the rest of the way to catch my train?

Tricky Maths Problem

The train is just about to leave the station and there is no way I will be able to catch it this time.

#10 - Water Jugs Problem

Six jugs are in a row.
The first three are filled with coke, and the last three are empty.
By moving only one glass, can you arrange them so that the full and the empty glasses alternate?

Move and then pour all coke from second glass to fifth glass.