Three of the five positive integers p, q, r, s, t are even and two of them are odd (not necessarily in order). Consider the following: 1. p + q + r – s – t is definitely even. 2. 2p + q + 2r – 2s + t is definitely odd. Which of the above statements is/are correct?

1. A. 1 Only
2. B. 2 Only
3. C. Both 1 and 2
4. D. Neither 1 nor 2

Answer: A

Explanation

Let E represent an even number and O represent an odd number. Among p, q, r, s, t, there are 3 Even and 2 Odd numbers.

Statement 1: p + q + r – s – t
The sum of all five numbers (p+q+r+s+t) will have the parity of (3 Even numbers + 2 Odd numbers) = E + E = Even.
We can rewrite the expression as (p+q+r+s+t) – 2s – 2t. Since 2s and 2t are always even, subtracting them from an even sum (p+q+r+s+t) will result in an even number. Thus, p + q + r – s – t is definitely even. Statement 1 is correct.

Statement 2: 2p + q + 2r – 2s + t
Any number multiplied by 2 is even. So, 2p, 2r, and 2s are all even. The expression simplifies to (Even) + q + (Even) – (Even) + t, which has the same parity as q + t.
Since there are 2 odd and 3 even numbers among the five, q and t could be:
– Both Odd (O+O = Even)
– One Odd and one Even (O+E = Odd)
Since q+t can be either even or odd depending on the specific assignment of p,q,r,s,t, the expression is NOT definitely odd. Statement 2 is incorrect.
Therefore, only statement 1 is correct.