GO Classes CS 2025 | Weekly Quiz 1 | Propositional Logic | Question: 14

Answer 1

Here is how I have solved,

The statements provided by Mrs. Baker can be represented as follows and these are “True statements”

1. ~ Bill is youngest → Alice is youngest
2. ~ Carl is youngest → ~ Alice is youngest

Now lets consider each option

Option A : Alice is youngest. If we consider this true then

LHS of statement 1 is true hence RHS must be true for statement 1 to be true so Alice is youngest is true. This favors our assumption

Now if we see 2nd statement the LHS is true hence RHS must also be true for statement 2 to be true which means Alice is not youngest is true. This doesn't favors our assumption

Hence option A cannot be true

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Option B : Bill is youngest. If we consider this true then

LHS of statement 1 is false hence RHS can be anything for statement 1 to be true but in this case RHS will be false (because only one of them is youngest and we have assumed Bill is youngest) so Alice is youngest is false i.e. Alice is not youngest

Now if we see 2nd statement the LHS is true hence RHS must also be true for statement 2 to be true which means Alice is not youngest

Hence there is no contradiction the conclusions of both the statements is same so B is a possible answer.

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Option C : Carl is youngest. If we consider this true then

LHS of statement 1 is true hence RHS must also be true for statement 1 to be true so Alice is youngest is true. This contradicts our assumption hence Carl cannot be youngest so option C cannot be answer.

Hence B is the only possible answer. @GO Classes please let me now if this argument is correct or not?

 

 

Comments

@swapnilbanerjee It's a good answer. Have selected as the "best answer".

Detailed Video Solution: https://youtu.be/nclBhBmtz2g?t=1841

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Good Question and Good Analysis. What I've observed is sometimes making cases by options is more worth it than making cases from Question directly. It would've have save some time in exam and quick analysis through options as well.

Answer 2

Detailed Video Solution: https://youtu.be/nclBhBmtz2g?t=1841

Let the propositions $A,\; B$ and $C$ denote that Mrs. Baker’s youngest child is Alice, Bill and Carl, respectively.
The following clauses represent the information from Mrs. Baker:
1. $B\; \vee \; A$ (Alice is her youngest child if Bill is not her youngest child. That is, $\neg B \Rightarrow A$.)
2. $C\; \vee \; \neg A$ (Alice is not her youngest child if Carl is not her youngest child. That is, $\neg C \Rightarrow \neg A.$)

We have the following knowledge:
Exactly one of $A,\;B,\;C$ is True. (Only One child has to be the youngest.)
So, we know that only three possibilities we have for $\text{(ABC)}$, that are $\text{(TFF)}$ or $\text{(FTF)}$ or $\text{(FFT)}$.

Only for $\text{(ABC)} = \text{(FTF)}$, both $C ∨ ¬A$ and $B\; ∨\; A$ are true.
Hence, $B$ is true, $A,\;C$ are false.
So, Bill is her youngest child.

Answer 3

option B is correct . 

Answer 4

This puzzle type of questions can be easily solved by truth table.

Assume 3 propositional variables – A, B, C

A: “Alice is youngest”

B: “Bill is youngest”

C: “Carl is youngest”

Given that only 1 of the three child is youngest child and the rest 2 are not youngest. So out of 8 possible combinations with 3 variables, only 3 combinations need to be considered as per the given information.

Let’s assume that the statement made by Mr Baker is represented by wff, S.

S: “Alice is her youngest child if Bill is not her youngest child, and that Alice is not her youngest child if Carl is not her youngest child.” 

Therefore, $S=(\sim B\rightarrow A)\wedge (\sim C \rightarrow \sim A)$

A	B	C	S
TRUE	FALSE	FALSE	FALSE
FALSE	TRUE	FALSE	TRUE
FALSE	FALSE	TRUE	FALSE

 

From the truth table, it’s clear that Mr Baker’s statement is true only in 2^nd case.

Therefore, the answer would be Option B (Bill is youngest since propositional variable B is true here)