<p>If k is a positive constant and (x+k)(x-2k)=0, what is the sum of the different possible values of x in terms of k?</p>
<p>If 18(x+y)= ax+2by for all x and y where a and b are constants, what is the value of a+b?</p>
<p>These were problems on the 2003 PSAT.</p>
<p>If k is a positive constant and (x+k)(x-2k)=0, what is the sum of the different possible values of x in terms of k?</p>
<p>well, if (x+k)(x-2k) = 0 then either (x+k) or (x-2k) = 0, because the only way the product of two numbers can equal zero is one of the numbers is zero. </p>
<p>So if x+k = 0 then x = -k. (any number plus its opposite is zero)</p>
<p>If x - 2k = 0 then x = 2k. The sum -k + 2k (which are the 2 different values of x in terms of k) = k.</p>
<ol>
<li>If 18(x+y)= ax+2by for all x and y where a and b are constants, what is the value of a+b?</li>
</ol>
<p>Use the distributive property to expand the left side of the equation.
Now you have:
18x + 18y = ax+2by</p>
<p>Visually, the answer comes naturally. a = 18, and b = 1/2 of 18 (since it's multiplied by 2 to equal 18y), so it is therefore 9.</p>
<p>a+b = 18 + 9 = 27.</p>
<p>Hope that helps. Feel free to post more of these types of questions, I love 'em.</p>
<p>Andre</p>
<p>I'm in preparation for the psat too and really want an 80 on the math. Please do post as many difficult PSAT math problems as you can find!</p>
<p>Those were easy. I have been given a boost of confidence! Please keep posting more hard problems for me to study!</p>
<p>ExRunner, how is the prep going? What has been your highest so far? In math, my highest is 640 (since this summer). I'll take more practice tests as the PSAT nears and we'll see if I can finally reach that 750 mark!</p>
<p>I got a 780 on the june SAT math. Careless mistake/nerves. PSAT is really important to me because I plan on going to a school where national merit gets quite a large scholarship. I haven't really done much math prep yet for the PSAT, besides reading this forum of course, as I've been mainly focusing on my weakness, critical reading. I guess I will start soon.</p>
<p>hey thanks for the answer...... but are you sure #2 is right? Because if it is, its really easy.</p>
<p>The Acme Plumbing Complany will send a team of 3 plumbers to work on a certain job. The company has 4 experienced plumbers and 4 trainees. If a team consists of 1 experienced plumber and 2 trainees, how many different such teams are possible?</p>
<p>I had trouble with this one. Please Help.</p>
<p>(4C2)<em>4=6</em>4=24</p>
<p>For the trainees, you have 4 different possibilities. For the experienced plumbers, you need to choose 2 out of 4 which is 4C2 (imagine the plumbers as A,B,C,D....you need to choose 2 so it could be AB, AC, AD, BC, BD, or CD--6 different possibilities).</p>
<p>Sanguine007, you know the answers. Is #2 27? If it is, that's how you solve it. Most of this stuff is really easy once you get over the "gee golly howddya know?" stage.</p>
<p>OK, this next question is the type of question where I do some fast systematic listing. Smarter people know exactly what numbers to multiply as soon as they read the question and get the answer in 2 seconds (and I'd like those people to respond too). It takes me a little longer because I list, this way:</p>
<p>4 leaders --- A, B, C, D</p>
<p>4 trainees, 2 of which will be in each team w/ the leader, call them 1, 2, 3, 4</p>
<p>If a team has A,1 in it, what groupings are possible? 3 others (the 4 plumbers, minus #1 already grouped for all)</p>
<p>If a team has A,2 - what groupings are possible? 2 others (remember we grouped one already with A, 1)</p>
<p>THIS PATTERN CONTINUES...
A,3 = 1 other partner who hasn't been paired yet</p>
<p>So we have 3+2+1 = 6 possibilities groups with leader A. There are 4 leaders, so we multiply 6 x 4 to get 24 total number of different groups. </p>
<p>That should be the answer.</p>
<p>i forgot to mention that I dont have the answer key to that, only the test booklet.</p>