Question

If d and g are integers, is d + g an even integer?
(1) \(\frac{d}{g}\) is an even integer.
(2) d = g

• A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
• B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
• C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
• D. EACH statement ALONE is sufficient.
• E. Statements (1) and (2) TOGETHER are NOT sufficient.

Hint:

When dealing with number properties like even/odd, testing cases with simple numbers is a very effective strategy. For statement (1), picking an odd 'g' and an even 'g' quickly reveals that the outcome isn't fixed. For statement (2), algebraic simplification often provides a definitive answer.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The question asks about the parity of the sum of two integers, d and g. The sum of two integers is even if both integers are even or if both integers are odd. The sum is odd if one integer is even and the other is odd.
(Even + Even = Even), (Odd + Odd = Even), (Even + Odd = Odd).
Step 2: Detailed Explanation:
Analyzing Statement (1):
This statement says that \(\frac{d}{g}\) is an even integer. Let's call this integer \(k\).
\[ \frac{d}{g} = k, \text{ where } k \text{ is an even integer.} \]
This implies \(d = g \times k\).
Let's test cases for the parity of g:
Case 1: If g is odd. Then \(d = (\text{odd}) \times (\text{even}) = \text{even}\).
In this case, \(d+g = \text{even} + \text{odd} = \text{odd}\).
For example, let g = 3 and k = 2. Then d = 6. \(d+g = 9\) (odd).
Case 2: If g is even. Then \(d = (\text{even}) \times (\text{even}) = \text{even}\).
In this case, \(d+g = \text{even} + \text{even} = \text{even}\).
For example, let g = 4 and k = 2. Then d = 8. \(d+g = 12\) (even).
Since we can get both an odd and an even result for \(d+g\), statement (1) alone is not sufficient.
Analyzing Statement (2):
This statement says that \(d = g\).
We need to find if \(d+g\) is even. We can substitute \(d\) with \(g\).
\[ d + g = g + g = 2g \]
Since g is an integer, \(2g\) is by definition a multiple of 2, which means it is always an even integer, regardless of whether g is even or odd.
Therefore, statement (2) alone is sufficient.
Step 3: Final Answer:
Statement (2) alone is sufficient to determine that \(d+g\) is even, while statement (1) alone is not. Therefore, the correct option is (B).