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 :

The correct answer is (A) 1 only.

Explanation:

Let us analyze the given logical implications: I. P implies (Q and S), denoted as P => (Q AND S). II. (R and S) implies not T, denoted as (R AND S) => NOT T. Let us test the conclusions: 1. If T is true, then at least one of P and R must be false: Assume T is true (T = True). From II, if T is true, then the premise (R AND S) must be false (by contrapositive). Thus, NOT (R AND S) is true, which means either R is false or S is false (or both). Now, let us check P. From I, P => S. If P were true, then S must be true. But does this force P and R to be false? If P is true, S is true. If S is true, and R is false, then (R AND S) is false, which satisfies II. Here P is true and R is false. So P is not necessarily false! Thus, conclusion 1 does not necessarily hold. 2. If Q is true, then P is true: Let us check the contrapositive of I. I states P => (Q AND S). The contrapositive is NOT (Q AND S) => NOT P. This does not directly prove Q => P. However, let us use formal logic: Can Q be true while P is false? Yes, implication P => Q does not mean Q => P. Wait, let us re-verify: Does Q true imply P true? No, implication is one-way. Let us check if both conclusions are false, making the answer 'Neither 1 nor 2'. In propositional logic, P => Q & S means if P is true, Q is true. But Q being true gives no information about P. Hence neither 1 nor 2 can be logically deduced.