### #1 - 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.

### #2 - Hard Measuring Water Riddle

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

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

### #3 - Challenging Puzzle

Outside a room there are three light switches. One of switch is connected to a light bulb inside the room.
Each of the three switches can be either 'ON' or 'OFF'.

You are allowed to set each switch the way you want it and then enter the room(note: you can enter the room only once)

Your task is to then determine which switch controls the bulb ??

Set the first switches on for abt 10min, and then switch on the second switch and then enter the room.
Three cases are possible
1.Bulb is on => second switch is the ans
2.Bulb is off and on touching bulb , you will find bulb to be warm
=>1st switch is the ans.
3.Bulb is off and on touching second bulb , you will find bulb to be normal(not warm)
=>3rd bulb is the ans.

### #4 - River Riddle

Four people need to cross a dark river at night.They have only one torch and the river is too risky to cross without the tourch. if all people cross simultanoesly then torch light wont be sufficient.Speed of each person of crossing the river is different.cross time for each person is 1 min, 2 mins, 7 mins and 10 mins. What is the shortest time needed for all four of them to cross the river ?

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

### #5 - Logic Interview Question

The first box has two white balls. The second box has two black balls. The third box has a white and a black ball.

Boxes are labeled but all labels are wrong!

You are allowed to open one box, pick one ball at random, see its color and put it back into the box, without seeing the color of the other ball.

How many such operations are necessary to correctly label the boxes?

Just One!

Because we know all labels are wrong.
So the BW box must be either BB or WW. Selecting one ball from BW will let you know which.
And the other two boxes can then be worked out logically.

### #6 - Oracle Probability Interview Question

What is the probability of choosing the correct answer at random from the options below.

a) 1/4
b) 1/2
c) 1
d) 1/4

If the answer is 1/4, then because 2 out of 4 answer choices are '1/4', the answer must actually be 1/2. This is a contradiction. So the answer cannot be 1/4.
If the answer is 1/2 (or 1), then because 1/2 (or 1) is 1 out of 4 answer choices, the answer must be 1/4. This is again a contradiction. So the answer cannot be 1/2 (nor 1).

So none of the provided answer choices are correct. Therefore the probability of choosing the correct answer is 0%

### #7 - Classic Logical Interview Puzzle

I have two rectangular wires.
Both of them have property that when i light the fire from one end , it will take 60 minutes to get completely burn.
However they do not burn at consistent speed (i.e it might be possible 1st 20% burn in 50 minutes and 80% can burn in 10 minutes).

So how could i measure 45 minutes ?

Steps:
A) burn 1st wire from both end and 2nd wire from one end
B) After 30 minutes(when 1st wire gets completely burned out) , burn the second wire from 2nd end as well
C) when 2nd run completely gets burned , you know its 45 minutes.

### #8 - Qualcomm Logic Interview Puzzle

A convention is held where all the big logicians are summoned. The master places a band on everyone’s forehead. Now all of them can see others’ bands but can’t see his own. Then they are told that there are different colors of bands. All the logicians sit in circle and they are further explained that a bell will ring at regular intervals. The moment when a logician knew the color of band on his forehead, he will leave at the next bell. If anyone leaves at the wrong bell, he will be disqualified.

The master assures the logicians that the puzzle will not be impossible for anyone of them. How will the logicians manage ?

First the logicians will have to take a leap of logic. The master has told them that it won’t be impossible for any logician to solve the puzzle. Thus it is assure that any of the color can’t exist only once. If it did, the one wearing it will have no clue about that color which would be unfair for him.

Now every logician will look around in the circle and count the number of times they see a particular color. If a color is seen only once, then the logician will know that the color on his band must be of the same color (as per the leap of logic). And then the logician will leave on the first bell.

In the similar fashion, any logician who see any color just once will be able to identify his own color and they will leave when the bell rings or they will be disqualified and asked to leave. Equivalently, any color for which there are two bands, will be eliminated after the first bell has rung. Thus there must be three bands of any remaining color at least.

Assume that a logician don’t see any color once, but sees a color twice. If they were the only bands of this color then the two logicians must have left at the first bell. But they did not. Thus it means that his band color is the same and he will leave on the second bell.

### #9 - Knockout Tournament Matches Puzzle

A chess tournament is taking place on knock-out terms (the one who loses the match is out of the game).
(a) If 15 matches are played in total, how many players participated?
(b) If 50 players took part in the tournament, how many matches were played?

(a) If you look at the question closely, it is a knockout tournament and the winner will be just one. Thus no. of players is equal to 15+1=16.
(b) The number of matches will always be one less than the number of players (in knockout tournaments). Thus the answer is 49.

### #10 - Marbles Puzzle

There is a box full of marbles,
all but two are blue,
all but two are green,
and all but two are red.

How many marbles are in the box ?

There are 3 marbles (1 blue, 1 green, and 1 red).