Detective Hoffman wakes up in a room to find himself in a trap set by Jigsaw. The room has two exits i.e. two doors. One of the door leads to a chamber made by magnifying glass. If Hoffman exits through that door, the burning sun will fry him up within moments. The second door exit leads to a fire breathing dragon.
What is the possible way in which Detective Hoffman can set himself free from the Jigsaw trap?
This is quite a famous interview puzzle and is usually asked as a warm up question. While you can figure out endless possibilities, the answer is quite simple. Detective Hoffman will just have to wait till the night. When there is no sun, he can simply walk out of the door that leads to the chamber built with magnifying glass.
This one is a bit of tricky river crossing puzzle than you might have solved till now. We have a whole family out on a picnic on one side of the river. The family includes Mother and Father, two sons, two daughters, a maid and a dog. The bridge broke down and all they have is a boat that can take them towards the other side of the river. But there is a condition with the boat. It can hold just two persons at one time (count the dog as one person).
No it does not limit to that and there are other complications. The dog can’t be left without the maid or it will bite the family members. The father can’t be left with daughters without the mother and in the same manner, the mother can’t be left alone with the sons without the father. Also an adult is needed to drive the boat and it can’t drive by itself.
How will all of them reach the other side of the river?
A couple had to take shelter in a hotel for they could not proceed their journey in the rain. Having nothing to do at all, they started playing cards. Suddenly there was a short circuit and the lights went off. The husband inverted the position of 15 cards in the deck (52 cards normal deck) and shuffled the deck.
Now he asked his wife to divide the deck into two different piles which may not be equal but both of them should have equal number of cards facing up. There was no source of light in the room and the wife was unable to see the cards.
For a certain amount of time, she thought and then divided the cards in two piles. To the husband’s astonishment, both of the piles had equal number of cards facing up.
The answer is very simple. All she had to do is take the fifteen cards from the top and reverse them. This would make another pile out of that and there will be two piles - one of 15 cards and one of 37 cards. Also both of them will have the same number of inverted cards.
Just think about it and if the mathematical explanation will help you understand better, here it is.
Assume that there were p inverted cards initially in the top 15 cards. Then the remaining 37 cards will hold 15-p inverted cards.
Now when she reverses the 15 cards on the top, the number of inverted cards will become 15-p and thus the number of inverted cards in both of the piles will become same.
A great meeting is held by a great logician where all the other logicians are called upon. The master logician takes them in a room and makes them sit in circle. A hat is placed on each of their heads. Now all of them can see the color of hats others are wearing but can’t see his own. They are told that there different colors of hats.
The master logician explains that a bell will be rung at regular intervals and the moment when a logician knows the color of his hat, he will leave on the next bell. If anyone leaves at the wrong bell, he will be disqualified and sent home.
All of them are assured of one thing that the puzzle will not be impossible for anyone of them. How will they manage the situation?
The first step that they will take will be a leap of logic. What it means is that they will deduce that every color must appear twice at least. Why? Because the master logician has assured them that the puzzle will not be impossible for anyone of them. And if a color appears only once in the circle, the person wearing it will have no clue about that color which will not be fair for him.
Then the logicians will follow the same and look for all the colors of hats in the circle. If one of them sees a color just once, he can safely assume that he is also wearing the hat of the same color as by leap of logic, no color can appear just once. Thus when the bell is rung, he will leave.
In the similar fashion, if anyone sees another color just once, he can determine that he is wearing the hat of the same color and will leave when the bell rings or will be disqualified and sent home. Unvaryingly, if a color is seen twice, they will be eliminated after the first bell. Hence, there must be at least three hats of any of the remaining color.
Assume that you are sitting in the circle and you don’t see a color once but see it twice. Then if they were the only two hats of the same color, the logicians must have left at the first bell already. But they did not. Which means that there are three hats of that color and you are wearing one. Thus you will leave after the second bell.
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.
In front of you, there are 9 coins. They all look absolutely identical, but one of the coins is fake. However, you know that the fake coin is lighter than the rest, and in front of you is a balance scale. What is the least number of weightings you can use to find the counterfeit coin?
The answer is 2. First, divide the coins into 3 equal piles. Place a pile on each side of the scale, leaving the remaining pile of 3 coins off the scale. If the scale does not tip, you know that the 6 coins on the scale are legitimate, and the counterfeit is in the pile in front of you. If the scale does tip, you know the counterfeit is in the pile on the side of the scale that raised up. Either way, put the 6 legitimate coins aside. Having only 3 coins left, put a coin on each side of the scale, leaving the third in front of you. The same process of elimination will find the counterfeit coin.
In a jar, there are some orange candies and some strawberry candies. You pick up two candies at a time randomly. If the two candies are of same flavor, you throw them away and put a strawberry candy inside. If they are of opposite flavors, you throw them away and put an orange candy inside.
In such manner, you will be reducing the candies in the jar one at a time and will eventually be left with only one candy in the jar.
If you are told about the respective number of orange and strawberry candies at the outset, will it be feasible for you to predict the flavor of the final remaining candy ?
At each draw, the number of strawberry candies are either decreasing by 2 or not decreasing at all. In the case of orange candies, at each draw, they are either increasing by 1 or decreasing by 1.
Thus on an assumed outset with at least one candy in the jar to begin with, if the number of strawberry candies are 0 or are even in numbers, they will finish off leaving an orange candy at the end. If otherwise, the remaining candy will be a strawberry one.
Step1. Lets say y = x
Step2. Multiply through by x xy = x2
Step3. Subtract y2 from each side xy - y2 = x2 - y2
Step4. Factor each side y(x-y) = (x+y)(x-y)
Step5. Divide both sides by (x-y) y = x+y
Step6. Divide both sides by y y/y = x/y + y/y
Step7. And so... 1 = x/y + 1
Step8. Since x=y, x/y = 1 1 = 1 + 1
Step9. And so... 1 = 2
Step 5 is invalid, because we are dividing by (x-y), and since x=y, we are thus dividing by 0. This is an invalid mathematical operation (division by 0), and so by not following basic mathematical rules
In a closed jar, there are three strawberry candies, two mango candies and five pineapple candies. You can't see inside the jar. Now, how many toffees you must take out from the jar to make sure that you have one of each flavor?
2 < 3 < 5
To find out the required number of candies, take one in place of the least number (i.e. take one mango candy) and then add all the greater numbers (i.e. three strawberry and five pineapple candies) to it.
Required number of balls = 1 + 3 + 5 = 9.
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