Points to Remember:

• The sum or difference of even numbers is always even. (Note: Zero is an even number.)
• The sum or difference of an odd number and an even number is always odd.

Solution of the question:

We are given that three of the five positive integers 𝑝, 𝑞, 𝑟, 𝑠, 𝑡 are even, and two of them are odd.

Let's consider the possibilities:

Statement 1: p + q + r – s - t is definitely even.

•   The sum of p, q, r, s, and t always results in an even number, regardless of the arrangement.

•   If we have three even numbers and two odd numbers, the sum of three even numbers will always be even.

•   Subtracting two odd numbers from an even number will also result in an even number.

•   So, statement 1 is always correct.

statement 2: 2p + q + 2r – 2s + t is definitely odd.

Evaluating the statement:

•   We can write 2p + q + 2r – 2s + t  = (2𝑝 + 2𝑟 – 2𝑠) + (q + t)

•   Say, A + B = (2𝑝 + 2𝑟 – 2𝑠) + (q + t)

• 2𝑝, 2𝑟 and 2𝑠 are always even since they are multiples of 2. Hence, their sum or difference will always be even. Hence, A is even.
• 𝑞 and 𝑡 could be either odd or even. Hence, B or (q + t) can be even or odd.
• Hence, A + B can be both even or odd.

•   So, statement 2 is incorrect.

•   Hence, option (a) is correct.

Alternate Approach for Statement 2: 2p + q + 2r -2s + t is definitely odd.

•   We can evaluate the expression for some specific values to determine if it results in an even number.

•   Let’s put some values in this expression and check.

•   (2×5) + 6 + (2×3) – (2×2) + 4 = 10 + 6 + 6 – 4 + 4 = 22 (an even number)

•   The result of substituting values into the expression was 22, which is an even number.