# TIFR2011-A-12

16 votes
736 views

The action for this problem takes place in an island of Knights and Knaves, where Knights always make true statements and Knaves always make false statements and everybody is either a Knight or a Knave. Two friends A and B lives in a house. The census taker (an outsider) knocks on the door and it is opened by A. The census taker says ''I need information about you and your friend. Which if either is a Knight and which if either is a Knave?". "We are both Knaves" says A angrily and slams the door. What, if any thing can the census taker conclude?

1. A is a Knight and B is a Knave.
2. A is a Knave and B is a Knight.
3. Both are Knaves.
4. Both are Knights.
5. No conclusion can be drawn.

edited
0
I can't unerstand this. Which if either is a Knight and which if either is a Knave? Please someone clarify
0
Here A cant be knight since the statement " Both are knaves " he is telling is false.If A is knight , then both cant be knaves.
0

The question that outsider ask is as-

Which, if either of you, is a Knight and which, if either of you, is a Knave?"

....

## 2 Answers

26 votes

Best answer

Option (B) should be the correct answer, that is A is a Knave & B is a Knight.

A must be either a Knight or a Knave.

Suppose A is a Knight, it means that the statement "We are both Knaves." must be true.

This is contradicting our assumption.

So the assumption that "A is a Knight" is not logically satisfiable simultaneously with the statement he made, which implies that A must be a Knave.

Now since A is a Knave, the statement made by him : "We are both Knaves." must be false.

The statement "We are both Knaves." will be false in any one of the following 3 conditions :

1. A is a Knight, B is a Knave.
2. A is a Knave, B is a Knight.
3. A is a Knight, B is a Knight.

Bus since we have already deduced that A is a Knave so in order to make the statement "We are both Knaves." false, we are only left with condition 2.

So B must be a Knight.

edited by
0
Why cant both be Knaves?
0
1

Another knight and knave prob https://gateoverflow.in/43225/iiith-pgee-2016

0

Both cannot be knaves because then the statement by A will be true. This means A speaks truth. This implies A is a Knight, but this is a contradiction. Thus, A has to be knave.

Since A is a Knave, it always lies. This means B has to a Knight in order ot make the statement issued by A to be false.

0

The Beauty of this Puzzle is also Identifying that

Understanding the fact that while we assume A is a knave and B is knight and as A says " We both are Knave is ALSO A LIE NOT TRUE

Bcoz it partial truth as only A is knave (while solving 1st time i thought this is true), but this is partially true which makes it Lie ( and since Knave always lie ) the answer is (B).

0

OK, but here it is written a/q "What if anything can the census taker conclude?". And since the census taker is an outsider, he thinks what A tells is right.

Its Logical Reasoning means we need to be precise with words, they are asking the census taker's conclusion and not our conclusion. And acc to this I think C is right

–3 votes

## Related questions

15 votes
4 answers
1
1.1k views
If either wages or prices are raised, there will be inflation. If there is inflation, then either the government must regulate it or the people will suffer. If the people suffer, the government will be unpopular. Government will not be unpopular. Which of the ... wages are not raised Prices are not raised If the inflation is not regulated, then the prices are not raised Wages are not raised
22 votes
6 answers
2
1.2k views
Three dice are rolled independently. What is the probability that the highest and the lowest value differ by $4$? $\left(\dfrac{1}{3}\right)$ $\left(\dfrac{1}{6}\right)$ $\left(\dfrac{1}{9}\right)$ $\left(\dfrac{5}{18}\right)$ $\left(\dfrac{2}{9}\right)$
5 votes
2 answers
3
424 views
Let $n> 1$ be an odd integer. The number of zeros at the end of the number $99^{n}+1$ is. $1$ $2$ $3$ $4$ None of the above.
8 votes
1 answer
4
449 views
The equation of the tangent to the unit circle at point ($\cos \alpha, \sin \alpha$) is $x\cos \alpha-y \sin\alpha=1$ $x\sin \alpha-y \cos\alpha =1$ $x\cos \alpha+ y\sin\alpha=1$ $x\sin \alpha-y \cos\alpha=1$ None of the above.