Some *difficult* math questions - need help

Test Preparation SAT Preparation
1

<p>Hey everyone,</p>

<p>I've been working with a few PSATS and other tests and have come across these few problems which have completely stumped me. Does anyone know how to solve them?</p>

<p>

<a href="http://i39.tinypic.com/2uh8ldj.png%5B/img%5D"&gt;http://i39.tinypic.com/2uh8ldj.png

</a>
*edit: forgot one.. <a href="http://i43.tinypic.com/29p5c36.png%5B/url%5D"&gt;http://i43.tinypic.com/29p5c36.png&lt;/a&gt;&lt;/p>

<p>Thanks! :)</p>

2

<p>I’d be glad to help but could you please host it on imageshack or photobucket? ■■■■■■■ never loads for me.</p>

3

<p>ummm.
34 could be x = 49, y = 9… so x+y = 56</p>

4

<p>yeah thanks… but is there any algebraic method of solving this question? or is it just guess and check?</p>

<p>reup:
<a href=“ImageShack - Best place for all of your image hosting and image sharing needs”>ImageShack - Best place for all of your image hosting and image sharing needs;
and
<a href=“ImageShack - Best place for all of your image hosting and image sharing needs”>ImageShack - Best place for all of your image hosting and image sharing needs;

5

<ol>
<li><p>We know that a square number is only odd if its square root is also odd; that is, x^2 is if odd if and only if x is also odd. So we want to make both x and y odd, so that sqrt(x) and sqrt(y) will both be odd. We can pick x = 81 and y = 1, so that sqrt(x) = 9 and sqrt(y) = 1. sqrt(x)+sqrt(y) = 10, so that checks out. Thus, x+y=82. Other possible values are 58 and 50. </p></li>
<li><p>Add all three equations. Then, 3z = (x-y+4) + (y-w-3) + (w-x+5). All the w, x, and y cancel. This leaves us with 3z = 4 - 3 + 5. Thus, z = 2. </p></li>
<li><p>arclength = (radius)(angle). In the smaller circle, 6 = (r)(angle). Solve for angle to get angle=6/r. Now, for the larger circle, x = (r+3)(6/r) because the angle in both arcs is the same. The answer is thus E.</p></li>
<li><p>The answer is D, but I’m not sure how to explain it.</p></li>
<li><p>If a circle has exactly one point in common with the x-axis, that point must be immediately above or below the center. You can confirm this for yourself by trying to draw a picture that disagrees with this statement. In any case, that point must then be at (6, 0). The radius of this circle is hence 2.5. So the point (3.5, t) must also be 2.5 units away from the center (6, 2.5). Apply the distance formula or something, and you get t = 2.5.</p></li>
</ol>