Solution:

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.

Solution:

12 steps:

Steps 11 Liter jug 6 Liter Jug
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 ==> i found 9 liter in 11liter jug(as required)

Solution:

The five cards must contain at least one card of the same suit.
Let's assume Rhodes placed one of the two cards of the same suit at the end and the second card was given to Daniel.
In that case, Daniel will know the suit of the card.

Now the real question is how Rhodes made sure that Daniel also knows the number of the card.

Daniel got four cards and one of them will determine the suit, so we are left with 3 cards.
3 cards can be arranged in 3! ways i.e. 6 ways but our card numbers can vary from 1 to 13.
Since we have two cards of the same suit we will make sure that we don't pick the King (13). This leaves us with 12 numbers.

We can also distinguish card from upside down position, therefore, we now have 6 * 2 ways to arrange these card.
"TADA.... We cracked the magic trick".

Let us assume that the smallest card is C1U (upwards position) and C1D (downwards position). Similarly, we have C2U, C2D, C3U and C3D

According to that, the 12 possible arrangements are:
C1U C2U C3U => NUMBER 1
C1U C3U C2U => NUMBER 2
C2U C1U C3U => NUMBER 3
C2U C3U C1U => NUMBER 4
C3U C1U C2U => NUMBER 5
C3U C2U C1U => NUMBER 6
C1D C2D C3D => NDMBER 7
C1D C3D C2D => NDMBER 8
C2D C1D C3D => NDMBER 9
C2D C3D C1D => NDMBER 10
C3D C1D C2D => NDMBER 11
C3D C2D C1D => NDMBER 12

Please note that king (number 13) is not possible as discussed already.

Solution:

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.

Solution:

17 min

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 minutes. 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 minutes

Solution:

Just One!

example
If u get Blue rock with label as "blue+red" => this is red as all labels are wrong.
In this way u can logically solve this puzzle.

Solution:

0%

Explanation:
1) why cant be 1/4 : If the answer is 1/4, then as we know two out of four answer choices is '1/4', the answer has be 1/2.
This is a contradiction, so the answer cannot be 1/4.

2) why cant be 1/2 : If the answer is 1/2 then because answer:"1/2" is 1 out of 4 answer choices, the answer must be 1/4. This is also a contradiction. So the answer cannot be 1/2.

3) why cant be 1 : If the answer is 1 then because answer:"1" is 1 out of 4 answer choices, the answer must be 1/4. Again the same contradiction and therefore answer cannot be 1.

Solution:

If we have to measure precisely, it is impossible. Because we are asked to measure odd number of liters whereas all the jars we have can contain only even liters of water.

* Tricky question asked in ias/infosys/infoedge/ibibo interview

Solution:

Let's say the Mice are X and Y and Bottle are B1, B2....B9

Step1:
X drinks B1 B2 & B3
Y drinks B1 B4 & B5

Possible Outcomes
If both mice die, B1 is the poison bottle.
If only mice X dies, Either B2 or B3 is the poison bottle.
If only mice Y dies, Either B4 or B5 is the poison bottle.
If nobody dies, Poison is in B6 B7 B8 or B9


Step2: (If nobody dies)
X drinks B6 B7
Y drinks B6 B8

Possible Outcomes
If both mice die, B6 is the poison bottle.
If only mice X dies, B7 is the poison bottle.
If only mice Y dies, B8 is the poison bottle.
If nobody dies, B9 is the poison bottle.

Step2: (Only X dies - Either B2 or B3 is the poison bottle)
Y drinks B2, if Y dies B2 is the poison bottle else B3

Step2: (Only Y dies - Either B4 or B5 is the poison bottle)
X drinks B4, if X dies B4 is the poison bottle else B5

Solution:

Take 10 number of cards in a new deck and change their face direction. For example- You create a new deck of 10 cards and out of 10, 3 faces up in the new deck. So remaining 7 faces up are in the old deck. But hey! while creating the new deck you reversed the face direction of new cards. So actually the 3 cards which were facing up are actually face down in the new deck while 7 faces up.