<p>OK. The 2011 Practice PSAT answers are in never never land. Please help explain this answer.</p>
<p>I'll make an attenpt at a number line here:</p>
<pre><code> x y z
</code></pre>
<p><--------.---------.----------.-------></p>
<ol>
<li> The three points shown on the number line above have coordinates x, y and z. If x, y, and z are each integers such that xyz is a negative odd integer, which of the following MUST be true?</li>
</ol>
<p>I. x, y, and z are each odd integers.
II. x < 0
III. yz >0</p>
<p>A. II only
B. I and II only
C. I and III only
D. II and III only
E. I. II. and III only</p>
<p>College Board says the answer is E. Please explain.</p>
<p>I do not understand why all three integers could not be negative numbers, making (III) not always true. </p>
<p>Am I missing a number line rule here somewhere?</p>
<p>Sorry! The number line just didnt turn out as planned. :(</p>
<p>I’m assuming there is no 0 marked on the number line, but that x is to the left of y, is to the left of z.</p>
<p>Let’s go one by one:
I: if you multiply an even number to ANYTHING, the number becomes even. Therefore all three numbers must be odd, so that xyz is odd.</p>
<p>II: to get a negative number with the product of 3 numbers, you can either have one or three numbers be negative. You can’t have only two be negative, because then the product would be positive. Since x is the left most number, it must be negative in both cases (whether 1 or 3 numbers are negative)</p>
<p>III: again, either one or three numbers are negative. If only x is negative, then y and z are positive, so yz is positive. If x, y, and z are negative, then yz is also positive.</p>
<p>Thus answer E!</p>
<p>Thanks for the answer, though I had already (finally!) figured it out. For some reason, I wasn’t mentally processing yz as “y times z”, but instead incorrectly as “y and z are each both greater than zero”. And once I got past that error, it all made sense. </p>
<p>May I ask that where do you find the past psat tests?</p>