Cool Interview Puzzles With Answers

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.

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

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.

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

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.

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%

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.

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.

(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.

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