Q48. Let P, Q, R, S and T be five statements such that:
I. If P is true, then both Q and S are true.
II. If R and S are true, then T is false.

Which of the following can be concluded? 1. If T is true, then at least one of P and R must be false.
2. If Q is true, then P is true.

Select the correct answer using the code given below:

Let's translate the given statements into logical notation: I. P → (Q ∧ S) (If P is true, then Q and S are true) II. (R ∧ S) → ¬T (If R and S are true, then T is false) Now, let's evaluate the conclusions: Conclusion 1: 'If T is true, then at least one of P and R must be false.' This is equivalent to T → (¬P ∨ ¬R).

From statement II, the contrapositive is T → ¬(R ∧ S), which simplifies to T → (¬R ∨ ¬S). So, if T is true, then R is false OR S is false. From statement I, the contrapositive is ¬(Q ∧ S) → ¬P, which simplifies to (¬Q ∨ ¬S) → ¬P.

So, if Q is false OR S is false, then P is false. If T is true, then (¬R ∨ ¬S). If ¬S is true, then from the contrapositive of I, ¬P is true. So, if T is true, then (¬R ∨ ¬P) is true, meaning at least one of P and R must be false.

Thus, Conclusion 1 is correct. Conclusion 2: 'If Q is true, then P is true.' (Q → P) Statement I is P → (Q ∧ S). This means P implies Q, but Q does not necessarily imply P. If Q is true, P could still be false (e.g., Q is true, S is true, but P is false).

Thus, Conclusion 2 is incorrect. Therefore, only Conclusion 1 follows. This question tests a deep understanding of conditional logic and contrapositives, a challenging aspect of CSAT.