EO Practice Questions

First, let us simplify the original expression: \(p + q + p = 2p + q\)

Since the product of an even number and any other integer will always be even, the value of \(2p\) must be even. If \(q\) were even, \(2p + q\) would be the sum of two even integers and would thus have to be even. But the problem stem tells us that \(2p + q\) is odd. Therefore, \(q\) cannot be even, and must be odd.

Alternatively, we can reach this same conclusion by testing numbers. We simply test even and odd values of \(p\) and \(q\) to see whether they meet our condition that \(p + q + p\) must be odd.

1) even + even + even = even (for example, \(4 + 2 + 4 = 10\)). The combination (\(p\) even, \(q\) even) does not meet our condition.

2) odd + odd + odd = odd (for example, \(5 + 3 + 5 = 13\)). The combination (\(p\) odd, \(q\) odd) does meet our condition.

3) even + odd + even = odd (for example, \(4 + 3 + 4 = 11\)). The combination (\(p\) even, \(q\) odd) does meet our condition.

4) odd + even + odd = even (for example, \(3 + 4 + 3 = 10\)). The combination (\(p\) odd, \(q\) even) does not meet our condition.

If we examine our results, we see that \(q\) has to be odd, while \(p\) can be either odd or even. Our question asks us which answer must be odd; since \(q\) is an answer choice, we don't have to test the more complicated answer choices.

The correct answer is B.

If \( ab^2 \) were odd, the quotient would never be divisible by 2, regardless of what c is. To prove this try to divide an odd number by any integer to come up with an even number; you cannot. If \( ab^2 \) is even, either a or b is even.
(I) TRUE: Since a or b is even, the product ab must be even.
(II) NOT NECESSARILY: For the quotient to be positive, a and c must have the same sign since \( b^2 \) is definitely positive. We know nothing about the sign of b. The product of ab could be negative or positive.
(III) NOT NECESSARILY: For the quotient to be even, \( ab^2 \) must be even but c could be even or odd. An even number divided by an odd number could be even (ex: 18/3), as could an even number divided by an even number (ex: 16/4).
The correct answer is A.

(A) UNCERTAIN: k could be odd or even.
(B) UNCERTAIN: k could be odd or even.
(C) TRUE: If the sum of two integers is odd, one of them must be even and one of them must be odd. Whether k is odd or even, 10k is going to be even; therefore, y must be odd.
(D) FALSE: If the sum of two integers is odd, one of them must be even and one of them must be odd. Whether k is odd or even, 10k is going to be even; therefore, y must be odd.
(E) UNCERTAIN: k could be odd or even.
The correct answer is C.