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GMAT Odds and Evens Practice Questions

Practice Odds and Evens questions with worked explanations and timing guidance for Quantitative Reasoning.

Five-question preview. Answer 3 now without an account.
Question 1 of 3 free Easy

If \( k \) and \( y \) are integers, and \( 10k + y \) is odd, which of the following must be true?

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Correct answer: C

(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.
Question 2 of 3 free Easy

If \( a \) is an even integer and \( b \) is an odd integer, which of the following cannot be an even integer?

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Correct answer: C

Let us analyze each answer choice:

(A) CAN BE EVEN: Since \( a \) is even, the product \( ab \) will always be even. For example, \( 2 \times 7 = 14 \).

(B) CAN BE EVEN: An even number divided by an odd number might be even if the odd number's prime factors are also present in the even number's prime factorization. For example, \( \frac{6}{3} = 2 \), which is even.

(C) CANNOT BE EVEN: An odd number divided by an even number will never yield an integer. By definition, an odd number is not divisible by 2, and an even number contains a factor of 2. The quotient \( \frac{b}{a} \) will not be an integer, let alone an even integer. For example, \( \frac{15}{4} = 3.75 \).

(D) CAN BE EVEN: An even number raised to any positive integer power will always be even. For example, \( 2^1 = 2 \).

(E) CAN BE EVEN: An even number raised to any positive integer power will always be even. For example, \( 2^3 = 8 \).

The correct answer is C.
Question 3 of 3 free Medium

If p and q are integers and p + q + p is odd, which of the following must be odd?

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Correct answer: B

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.
Question 4 of 5 preview Medium

If \( a \), \( b \), and \( c \) are integers and \( \frac{ab^2}{c} \) is a positive even integer, which of the following must be true?
I. \( ab \) is even
II. \( ab > 0 \)
III. \( c \) is even

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Question 5 of 5 preview Hard

If \( q \), \( r \), and \( s \) are consecutive even integers and \( q < r < s \), which of the following CANNOT be the value of \( s^2 - r^2 - q^2 \)?

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