1. Crossing the Bridge Puzzle
Four people need to cross a rickety bridge at night. Unfortunately, they have only one torch and the bridge is too dangerous to cross without one. The bridge is only strong enough to support two people at a time. Not all people take the same time to cross the bridge. Times for each person: 1 min, 2 mins, 7 mins and 10 mins. What is the shortest time needed for all four of them to cross the bridge?
Solution: 17 mins
1 and 2 go first, then 1 comes back. Then 7 and 10 go and 2 comes back. Then 1 and 2 go again, it makes a total of 17 minutes.
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2. Ant and triangle Problem
Three ants are sitting at the three corners of an equilateral triangle. Each ant starts randomly picks a direction and starts to move along the edge of the triangle. What is the probability that none of the ants collide?
Solution: 0.25
Every ants can move in two direction so total possible move is 2x2x2=8. If ants avoid collision then there are only two moves (clockwise and anticlockwise).So by the Simple probablity 2/8 = 0.25 is the answer.
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3. Burning Rope Timer Puzzle
A man has two ropes of varying thickness (Those two ropes are not identical, they are not the same d ensity nor the same length nor the same width). Each rope burns in 60 minutes. He actually wants to measure 45 mins. How can he measure 45 mins using only these two ropes. He can not cut the one rope in half because the ropes are non-homogeneous and he can not be sure how long it will burn.
Solution:
He will burn one of the rope at both the ends and the second rope at one end. After half an hour, the first one burns completely and at this point of time, he will burn the other end of the second rope so now it will take 15 mins more to completely burn. so total time is 30+15 i.e. 45mins.
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4. Heaven or Hell Puzzle
You are standing before two doors. One of the path leads to heaven and the other one leads to hell. There are two guardians, one by each door. You know one of them always tells the truth and the other always lies, but you don not know who is the honest one and who is the liar. You can only ask one question to one of them in order to find the way to heaven. What is the question?
Solution:
The question you should ask is "If I ask the other guard about which side leads to heaven, what would he answer?". It should be fairly easy to see that irrespective of whom do you ask this question, you will always get an answer which leads to hell. So you can chose the other path to continue your journey to heaven.This idea was famously used in the 1986 film Labyrinth.
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5. 10 Coins Puzzle
You are blindfolded and 10 coins are place in front of you on table. You are allowed to touch the coins, but can’t tell which way up they are by feel. You are told that there are 5 coins head up, and 5 coins tails up but not which ones are which. How do you make two piles of coins each with the same number of heads up? You can flip the coins any number of times.This puzzle was asked in Yahoo Interview.
Solution:
Make 2 piles with equal number of coins. Now, flip all the coins in one of the pile.How this will work? lets take an example.
So initially there are 5 heads, so suppose you divide it in 2 piles.
Case:
P1 : H H T T T
P2 : H H H T T
Now when P1 will be flipped
P1 : T T H H H
P1(Heads) = P2(Heads)
Another case:
P1 : H T T T T
P2 : H H H H T
Now when P1 will be flipped
P1 : H H H H T
P1(Heads) = P2(Heads)
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6. 3 Mislabeled Jars
This problem is also called Jelly Beans problem. This is the most commonly asked interview puzzle. You have 3 jars that are all mislabeled. One jar contains Apple, another contains Oranges and the third jar contains a mixture of both Apple and Oranges. You are allowed to pick as many fruits as you want from each jar to fix the labels on the jars. What is the minimum number of fruits that you have to pick and from which jars to correctly label them?
Solution: one
This is very simple puzzle.First take a fruit from second jar there may be only two possibilities:case 1:
The fruit taken out is Apple.As we already know that every jar must be mislabeled.So the second jar will be Apple.First will be for orange and third for mixed.
case 2:
The fruit taken out is Orange.As we already know that every jar must be mislabeled.So the second jar will be Orange.First will be for mixed and third for Apple.
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7. Red and Blue Marbles
You have two jars, 50 red marbles and 50 blue marbles. You need to place all the marbles into the jars such that when you blindly pick one marble out of one jar, you maximize the chances that it will be red. When picking, you will first randomly pick a jar, and then randomly pick a marble out of that jar. You can arrange the marbles however you like, but each marble must be in a jar.
Solution:
Suppose there two jars A and B:
Put one red marbel in A and rest of them in B. Now analyze the Total probality.
jar A : (1/2)*1 = 1/2 (selecting the jar A = 1/2, red marble from jar A = 1/1)
jar B : (1/2)*(49/99) = 0 (selecting the jar B = 1/2, red marble from jar B = 49/99)
Total probability = 74/99 (~3/4)
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8. 100 Doors Puzzle
You have 100 doors in a row that are all initially closed. you make 100 passes by the doors starting with the first door every time. the first time through you visit every door and toggle the door (if the door is closed, you open it, if its open, you close it). the second time you only visit every 2nd door (door #2, #4, #6). the third time, every 3rd door (door #3, #6, #9), ec, until you only visit the 100th door.What state are the doors in after the last pass? Which are open which are closed?
Puzzle asked in: Google/Adobe/Amazon/Oracle
Solution:
You can figure out that for any given door, say door #38, you will visit it for every divisor it has. so has 1 & 38, 2 & 19. so on pass 1 i will open the door, pass 2 i will close it, pass 19 open, pass 38 close. For every pair of divisors the door will just end up back in its initial state. so you might think that every door will end up closed? well what about door #9. 9 has the divisors 1 & 9, 3 & 3. but 3 is repeated because 9 is a perfect square, so you will only visit door #9, on pass 1, 3, and 9… leaving it open at the end. only perfect square doors will be open at the end.
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9. Gold Bar Puzzle
You have got someone working for you for seven days and a gold bar to pay him. The gold bar is segmented into seven connected pieces. You must give them a piece of gold at the end of every day. What and where are the fewest number of cuts to the bar of gold that will allow you to pay him 1/7th each day?
Solution:
Lets split the chain as,
Day 2: Get back A, give B (-1, +2)
Day 3: Give A (+1)
Day 4:Get back A and B, give C (-2,-1,+4)
Day 5:Give A (+1)
Day 6:Get back A, give B (-1,+2)
Day 7:Give A (+1)
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10. King and Wine Bottles
A bad king has a cellar of 1000 bottles of delightful and very expensive wine. A neighboring queen plots to kill the bad king and sends a servant to poison the wine. Fortunately (or say unfortunately) the bad king's guards catch the servant after he has only poisoned one bottle. Alas, the guards do not know which bottle but know that the poison is so strong that even if diluted 100,000 times it would still kill the king. Furthermore, it takes one month to have an effect. The bad king decides he will get some of the prisoners in his vast dungeons to drink the wine. Being a clever bad king he knows he needs to murder no more than 10 prisoners believing he can fob off such a low death rate and will still be able to drink the rest of the wine (999 bottles) at his anniversary party in 5 weeks time. Explain what is in mind of the king, how will he be able to do so ? (of course he has less then 1000 prisoners in his prisons)
Solution:
Number the bottles 1 to 1000 and write the number in binary format.
bottle 1 = 0000000001 (10 digit binary)
bottle 2 = 0000000010
bottle 500 = 0111110100
bottle 1000 = 1111101000
Now take 10 prisoners and number them 1 to 10, now let prisoner 1 take a sip from every bottle that has a 1 in its least significant bit. Let prisoner 10 take a sip from every bottle with a 1 in its most significant bit. etc.
prisoner = 10 9 8 7 6 5 4 3 2 1
bottle 924 = 1 1 1 0 0 1 1 1 0 0
For instance, bottle no. 924 would be sipped by 10,9,8,5,4 and 3. That way if bottle no. 924 was the poisoned one, only those prisoners would die.
After four weeks, line the prisoners up in their bit order and read each living prisoner as a 0 bit and each dead prisoner as a 1 bit. The number that you get is the bottle of wine that was poisoned.
1000 is less than 1024 (2^10). If there were 1024 or more bottles of wine it would take more than 10 prisoners.
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