Aptitude Questions :
Find the Truth Teller Interview Riddle
Just to test the brilliance of Sherlock's mind, Dr. Watson makes five people stand in front of him. Out of the five, only one is the truth teller and the other four are togglers which means they may tell the truth or lie on being asked. But on being asked again, they will switch which means if they told a lie the first time, they will tell the truth on second question and vice versa.
Sherlock is challenged to ask only two questions to determine who the truth teller is. He can ask both the questions to the same person or ask two different people. How will he determine who is the truth teller?
Sherlock will ask any one of them "Are you the truth teller?" There can be two responses to this question - Yes or No.
If the person says yes, then he is the truth teller or a lying toggle.
He will ask the second question to the same person "Who is the truth teller?"
If he is the truth teller, he will admit it. If he is a lying toggler, he will speak truth on the second question if he lied on the first, telling you who the truth teller is.
If the person says no, Sherlock will be confirmed of the fact that he is not the truth teller and thus he is a toggle who has told truth and will say a lie on the second question.
The next question he will ask to the same person "Who is not the truth teller?"
Since the person told truth in the first question, he will speak a lie this time and in order to tell a lie, he will tell Sherlock exactly who the truth teller is.
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
We all know that New Year occurs after a week from Christmas and thus it falls on the same day as of Christmas. But this will not happen in 2050. In 2050, Christmas will appear on Sunday while New Year will appear on Saturday.
Read the question carefully again. New Year do falls after Christmas but that happens if two different years. The question is put up against the year 2050 and thus there will be 51 weeks and 2 days in between them as New Year will appear on 1 January 2050 and Christmas will happen on 25 December 2050.
You have been given three jars of 3 liters, 5 liters and 8 liters capacity out of which the 8 liters jar is filled completely with water. Now you have to use these three jars to divide the water into two parts of 4 liters each.
How can you do it making the least amount of transfers?
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.
There are 100 doors. 100 strangers have been gathered in the adjacent room. The first one goes and opens every door. The second one goes and shuts down all the even numbered doors – second, fourth, sixth... and so on. The third one goes and reverses the current position of every third door (third, sixth, ninth… and so on.) i.e. if the door is open, he shuts it and if the door is shut, he switches opens it. All the 100 strangers progresses in the similar fashion.
After the last person has done what he wanted, which doors will be left open and which ones will be shut at the end?
Think deeply about the door number 56, people will visit it for every divisor it has. So 56 has 1 & 56, 2 & 28, 4 & 14, 7 & 8. So on pass 1, the 1st person will open the door; pass 2, 2nd one will close it; pass 4, open; pass 7, close; pass 8, open; pass 14, close; pass 28, open; pass 56, close.
Thus we can say that the door will just end up back in its original state for each pair of divisor. But what about the cases in which the pair of divisor has analogous number for example door number 16? 16 has the divisors 1 & 16, 2 & 8, 4&4. But 4 is recurrent because 16 is a perfect square, so you will only visit door number 16, on pass 1, 2, 4, 8 and 16… leaving it open at the end. So only perfect square doors will remain open at the end.
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.
Three college toppers are summoned by the inspecting faculty. To identify the best from them, the faculty takes them into a room and places one hat on each of their heads. Now all of them can see the hats on other’s heads but can’t see his own. There are two colored hats – green and red.
Now the faculty announces that he had made sure that the competition is extremely fair to all three of them. He also gives them a hint that at least one of them is wearing a red hat. Now the first one who is able to deduce his own hat color will be awarded the most intelligent student of all award. After a few minutes, one of them raises his hand and is able to deduce the color correctly.
There are two things to keep in mind:
Firstly there is at least one red hat. (There can be two or three as well).
Secondly the competition is fair for everyone.
Thus if there is only one red hat, that person will see two green hats on other heads and will be able to deduce his own color as red. However the other students will see one red and one green hat and can never be sure. In such manner, the competition will prove to be partial for one student.
Suppose if there are two red hats. Then the students who are wearing red hats will see one red and one green hat on others. Now they must have deduced that there can’t be just one red hat. Thus they will know that they are also wearing a red hat. But the one who is wearing a green hat will see two red hats and can never be sure of his own color. In this case as well, the competition will not be fair.
Thus the only possible and fair means is if all of them are wearing a red hat. The one who is able to deduce the situation first, will raise his hand and will tell the correct answer.
Jonathan has three boxes containing milk chocolates and dark chocolates. The problem is that all of them have been labeled incorrectly as follows.
Box1: Dark Chocolates
Box2: Milk Chocolates
Box3: Dark Chocolates and Milk Chocolates
How will he label all the boxes correctly by just opening one box?
It has been clearly mentioned that all the boxes are labeled incorrectly. If he opens the Box3, then he will get either Dark Chocolates or Milk Chocolates as it is labeled incorrectly. Let us suppose he finds Dark Chocolates in there. Now since all are labeled incorrectly, Box B A must contain Milk Chocolates and Box B must contain Milk Chocolates and Dark Chocolates.
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