Pricing

PP1 (Shorter) Quant Section 2 (Easy) Q14

Solving the Problem Method 1 (Conceptual) $j -k$ means the distance between $j$ and $k$ on the number line. If two integers have an even distance between them, we know that they are either both odd or both even. Since the question has the word must, the answer choice must be guaranteed to be even, regardless of whether $j$ and $k$ are both even or both odd. If $j$ and $k$ were both odd: <ul> <li>A: $k = odd$, so we can eliminate this.</li> <li>B: $jk = odd*odd = odd$, so we can eliminate this.</li> <li>C: $j + 2k = odd + even*odd = odd + even = odd$, so we can eliminate this.</li> <li>D:  $jk + j = odd*odd + odd = odd + odd = even$, so we can keep this.</li> <li>E: $jk - 2j = odd*odd - even*odd = odd - even = odd$, so we can eliminate this.</li> </ul> If $j$ and $k$ were both odd (which they can be), only $jk+j$ can be guaranteed to be even. Method 2 (Choosing Numbers) We can experiment with odd and even values for j and k: <table border="1" cellpadding="1" cellspacing="1" style="width:500px;"> <thead> <tr> <th scope="col">j</th> <th scope="col">k</th> <th scope="col">A: k</th> <th scope="col">B: jk</th> <th scope="col">C: j+2k</th> <th scope="col">D: jk+j</th> <th scope="col">E: jk-2j</th> </tr> </thead> <tbody> <tr> <td>0</td> <td>2</td> <td>2</td> <td>0</td> <td>4</td> <td>0</td> <td>0</td> </tr> <tr> <td>1</td> <td>3</td> <td>3</td> <td>3</td> <td>7</td> <td>4</td> <td>1</td> </tr> </tbody> </table> We can see that only $jk+j$ gives us an even value in all instances. So, the answer must be D.