These are the most asked interview puzzles with answers or solutions of 2020. Collected from logically, analytics and geeks for geeks.

**1. Using two wires measure 45minutes of time**

**Puzzle question:**

**If a wire takes one hour to burn completely, and the condition is the wire will burn non uniformly. (For example first half of wire may burn in 10min and second half may burn in 50min). Then how will you measure the 45 minutes using the given wires?**

Puzzle solution:

Here the idea is to take a wire and burn from both the ends, which will get burn in half of the time that is in 30 minutes.

To get the answer to the asked question first take the wire 1 and burn from both ends and burn wire 2 only from one end. After 30min when the first wire will get burnt completely start burning the other end of the wire 2. By the time wire, 2 get burned completely the 45minutes will be measured.

**2. Total distance traveled by a honey bee**

**Puzzle question:**

**There are two trains A and B which are running opposite to each other on the same track. The two trains have their speeds 50 km per hour and 70 km per hour respectively. A bee with a speed of 80km per hour starts flying between the trains when the trains were 100km apart. The bee starts flying from first train A to the second train B to and fro continuously. Until the trains, A and B collide with each other. What will be the total distance traveled by the bee?**

Puzzle solution:

The total time taken by the trains A and B which were at 100km apart will be d/(u+v) = 100/(50+70) = 0.833hrs.

Now the distance traveled by the bee in that time with the speed 80km per hour will be 80 * 0.833 = 66.67km. So the answer will be 66.67km.

**3. Payment using 7units of the Gold bar.**

**Puzzle question:**

**An employee works under a boss for about 7 days. As a payment to the employee, the boss had a 7 unit gold bar. The employer earns a unit daily. The boss needs to pay him as a unit of gold each day as a daily wage. The boss cannot cut the gold bar more than twice. How the boss will pay unit of gold to an employee?**

Puzzle solution:

The answer is the boss will me two cuts on the gold bar which divides as 1unit, 2unit, and 4unit.

1. So the first-day boss will pay 1unit gold.

2. The second-day boss will give a 2unit piece and get back the 1unit bar piece.

3. On the third day boss will give the 1 unit gold bar he has to the employee.

4. Next, the fourth-day boss will give the 4unit bar and take back the 1unit and 2unit gold bar from the employee.

5. The fifth-day boss will give 1unit.

6. And the sixth-day boss takes back previously given one unit and give 2unit of the gold bar from his hand.

7. The last seventh-day boss will give him the remaining one unit of gold bar.

**4. Monty hall puzzle**

**Puzzle question:**

**There are three doors and there is a car behind one of the doors and there is a goat behind each of the other two doors. A person will be there behind the door who will assist you in some way. you need to pick a door number, say door number 1. The person who knows what is behind the door he will reply that No. Then you say No. 3 and guess there is a goat. The person will ask you do you want to go with option door number 2? Which is better switching the choice of staying with the previous choice.?**

Puzzle solution:

Since the probability of goat is two by three 2/3, the switching is a good choice always. Because only 1/3 probability will be a car.

**5. Three bulbs and there three switches**

**Puzzle question:**

**A room with a single door has three bulbs inside the room. The three switches of the room are outside the door and the room is closed right now. Also currently, all of the three bulbs are currently off. You have only one choice to open the door. you need to choose the right switch for the right bulb. you can manipulate the switches several times if you wish but opening the door only once is allowed. How do you find?**

Puzzle solution:

Let us assume the switch names as A, B, and C. The idea is turning ON any one switch say A for about 5 to 10 minutes and do this OFF. And turn On one say switch B of the switches among two and enter the door. The bulb which is glowing will be of second switch B you changed. And touch and sense the bulb among remaining two, the bulb which is still hot that will be of switch A you operated first and the cool one will be of switch C that is unoperated.

**6. 1000 coins and 10 bags**

**Puzzle question:**

**A dealer having a thousand coins and ten bags has to divide coins among ten bags. Which has to make him easy to few give bags with coins and decide a number of coins. How the dealer must have to divide his coins of money into ten bags?**

Puzzle solution:

Using the factorization concept we can solve this puzzle. We have to use a bag with one number of coins, a bag with 3 numbers of coins and, a bag with 9 numbers of coins such that we can form the patterns.

1 = 1,

2 = 1+1

3 = 1+3-1

4 = 1+3

5 = 9-3-1

6 = 9-3

7 = 9+1-3

8 = 9 -1

9 = 9

10 = 9+1 etc.

**7. Maximum white ball probability**

**Puzzle question:**

**You are given 50 white balls and 50 Black balls and also there two empty boxes in the room. Place the balls in the boxes in such a way that, when you pick a random ball from them the white ball probability should be maximum.**

Puzzle solution:

Case1: If we place 50 white balls in the first box and 50 black balls in the second box the white ball probability will be equaled to

(1/2)*(0/50)+(1/2)*(50/50) = 1/2.

Case2: Let us place 49 white balls and 50 black balls in the first box and the remaining one white ball in the second box. Then the white ball probability will be equaled to

(1/2)*(49/99)+(1/2)*(1/1) = 0.747.

So in the second case, the white ball probability will be maximum.

**8. Transport 3000 Bananas with 1 camel for 1000kms**

**Puzzle question:**

**A person has three thousand bananas and one camel, and he wants to carry the maximum number of bananas to a place which is 1000 km far. The mode of transport is through only one camel and no other mode of transport is allowed. The camel will eat a banana after traveling 1 km every time and carry a load of the maximum of 1000 bananas. How many maximum numbers of bananas can be carried to the destination?**

Puzzle solution:

The person has to make 3 trips as there are 3000 bananas and to a distance of 1000 km.

1. The camel will have to move forward with the first 1000 bananas. It will eat one banana while moving forward and one while returning. so it will drop 998 bananas at 1km distance. 2. Do the same for the next thousand bananas so 998+998 bananas at 1km away point.

3. Third time take 1000 and it will eat one and drop 999(since no need to return), so 2995 bananas at 1km away point.

Let us move like this 200 times to pass 200km, so the camel will eat around 5*200=1000 bananas to take bananas to a distance of 200km, leaving behind 2000 bananas.

Now repeating the same as the above, the camel will eat around 3 bananas to move forward 1km by taking 2000 bananas (In the first case eats 5 bananas for carrying 3000 bananas to travel 1km). If it moves 333km like this it will eat 3*333=999 and leaving behind 2000-999=1001 at a distance of 200+333=533kms. After the 467kms need to be traveled and 1001 bananas are left.

Let the camel eat 1 banana after every km so the camel will eat 467 bananas and leaving behind 1001-467=534 bananas after traveling a total of 1000kms.

**9. Toggling 100 doors**

**Puzzle question:**

**There is a big building with 100 doors in a row and initially, all doors are closed. A person walks through all the doors and changes the door condition if open he will close it and if it is closed he will open it. The person will toggle the door in certain way as listed below.**

**1. First time when he walks he toggles every door.**

**2. The second time when he walks he toggles only even-numbered doors such as 2,4,6,8,10, etc.**

**3. The third time when he walks, he toggles only third doors such as 3,6,9,12,15, etc. continues to do like this…**

**…..**

**…..**

**100. On the 100th time when he walks, toggles the 100th door.**

**So, What is the condition of the door? open/close.**

Puzzle solution:

The idea is the, when you realize with the first 10 trials you will come to the perfect squares will always be open. for example 1,4,9,16,25,36,49,64,81,100. So, the 100th door will be open.

**10. Poison and Rat**

**Puzzle question:**

**There are 1000 alcohol bottles out of the one bottle contains poison. By sacrificing the number of Rats you can identify the poisoned bottle. A rat drinking the poison will die after 1hr. How many minimum numbers of Rats required to find out the poisoned bottle?**

Puzzle solution:

Let us give numbers to all bottles from 1 to 100. The idea here is making use of binary number systems. Since there are 1000 alcohol bottles the minimum binary numbers required to represent 1000 will be 10. So assign 10 Rats to 10 binary bit positions will help to find the poisoned bottle.

Let’s see how for Bottle1: 0000000001 is the binary number (Rat10,Rat9,Rat8,Rat7,Rat6,Rat5,Rat4,Rat3,Rat2,Rat1) for first time Rat1 will drink bottle one. Similarly, do the same for all other bottles.

for example, 0101010000. If Rat5, Rat7, and Rat9 die the Bottle 42 will be a poisoned one.