Three best logicians are brought into a room. Each of them has been painted a number on their forehead with a number that is unique and greater than zero. You are one of them and can see other two foreheads with the number 20 and 30 painted on them. The game begins and the host circles around the logicians asking each one of them to guess the number written on their forehead. Each of the logician is unable to answer and says that he can't guess it. Then the question is asked again.
This particular answer depends on what the third person says.
You know that the other two logicians have 20 and 30 painted on their foreheads. If the first logician is 10, the second will either say he has either 40 or 20 painted on his forehead. The second logician will say "I'm either 30 or 10 but I can't be 10 because every number is unique as explained earlier. If the third logician is able to deduce that he has 30 painted on his forehead, but the third logician is not able to deduce the number. Thus the first logician is not numbered 10 and he must be painted 50 on his forehead.
As shown in picture, there are four gears with the following specification
* Gear A has 60 teeth
* Gear B has 40 teeth
* Gear C has 20 teeth
* Gear D has 60 teeth
* Every minute, Gear B makes 15 complete turns.
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?
There can be two possibilities for the given situation.
One is if I run on the perimeter. Then, the lion will eventually catch me as the lion can follow my radius and then his trajectory will be half of the circle and not a spiral. Therefore, the lion will subsequently catch me in a finite amount of time.
Secondly if I don’t run on the perimeter, then clearly, I have an infinite amount of time before the lion catches me.
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.
* They have only one torch and the river is too risky to cross without the torch.
* If all people cross simultaneously then torch light wont be sufficient.
* Speed of each person of crossing the river is different.cross time for each person is 1 min, 2 minutes, 7 minutes and 10 minutes.
What is the shortest time needed for all four of them to cross the river ?
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)
In a recreational activity, you are given four different jars of 2 liters, 4 liters, 6 liters and 8 liters respectively with an unlimited water supply. Then you are asked to measure exactly 5 liters of water using them.
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.
There are a hundred statements.
1st person says: At least one of the statements is incorrect.
2nd person says: At least two of the statements is incorrect.
3rd person says: At least three of the statements are incorrect.
4th person says: At least four of the statements are incorrect.
100th person says: At least a hundred of the statements are incorrect.
Now analyze all the statements and find out how many of them are incorrect and how many are true?
The 100th statement for sure is incorrect because it says that at least 100 of the statements are incorrect.
Suppose if that is correct, then 100 statements cannot be true.
This suggests that the 100th statement is incorrect and that the first statement is true.
Similarly 99 statements cannot be true because if they were true, then two statements would become correct i.e. the 1st and the 99th.
But the 99th statement says that at least 99 are incorrect.
This suggests that the 99th statement is incorrect and that 2ndone is true.
If we keep analyzing is the same way till the end, we will find out that only the first fifty statements are true and all the remaining ones are incorrect.
There are two glasses in front of you. One of the glasses is full of coke and the other glass is full of lemonade. You take a spoonful of coke and mix it into the glass of lemonade. Now the lemonade glass has a mixture of coke and lemonade. You take a spoonful of that mixture and mix it inside the coke glass.
Now what do you think? - The glass with coke has more quantity of lemonade or the glass with lemonade have more quantity of coke mixed with it?
This problem can be solved with algebra as well as logical reasoning. We are going to tell you how it is logically possible.
Be it any quantity that was present in the beginning and any liquid as well. We know that we have taken a spoonful from one glass and put it into another. Then, we have taken a spoonful from the other glass and put it into the first glass. So, at the end of it, the quantity of liquid in either glass remains same as it was in the beginning.
Therefore, we can conclude the fact that any amount of coke that is missing in the glass with coke will be present in the lemonade glass. Also, the same quantity of lemonade will be missing from the lemonade glass and will be present in the coke glass.
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