<p>my highest is 127.5 under the current scoring system. That's 20 questions correct and the rest blank. I am at a point where I am not entirely confident that I won't run into trouble in the first 20 problems and am not completely hopeless at solving one of the last five.</p>
<p>
[quote]
I am at a point where I am not entirely confident that I won't run into trouble in the first 20 problems and am not completely hopeless at solving one of the last five.
[/quote]
wow. my scores RANGE so much. i've done like 5 and my highest is a 138 and my lowest is like a 90</p>
<p>oh wow, which test did you get the 138 on and which did you get the 90 on?</p>
<p>I have about a month and twenty days to refresh number theory and learn how to do counting problems.</p>
<p>wow you guys are good.</p>
<p>i don't remember what i got on practice tests, but 135 on last year's A (which was much better than the B hehe)</p>
<p>
[quote]
I have about a month and twenty days to refresh number theory and learn how to do counting problems.
[/quote]
OH MY GOD</p>
<p>138 - amc12 A 2002
90...like the 1991 AHSME or something but i think they're equivalent??</p>
<p>hmm what are the "concepts" needed to get 130+ and where can i find them?</p>
<p>
[quote]
some schools do one or the other cuz of spring break
[/quote]
</p>
<p>Since when is spring break in February?</p>
<p>oh i meant winter break</p>
<p>oh can u continue posting the test =) can't find it</p>
<p>AMC12B 2007</p>
<p>13.
A traffic light has this cycle: green for 30 seconds, then yellow for 3 seconds, and then red for 30 seconds. Joe picks a random 3-second time interval to watch the light. What is the probability that the color changes while he is watching?</p>
<p>A) 1/63 B) 1/21 C) 1/10 D) 1/7 E) 1/3</p>
<ol>
<li>Point A is inside equilateral triangle XYZ. Points B, C, and D are the feet of the perpendiculars from A to XY, YZ, and Zx, respectively. If AB = 1, AC = 2, and AD = 3, what is XY?</li>
</ol>
<p>A) 4 B) 3<em>sqrt(3) C) 6 D) 4</em>sqrt(3) E) 9</p>
<ol>
<li>Geometric series a + ar + ar^2 + ... = 7 and the terms involving odd powers of r sum to 3. Find a + r.</li>
</ol>
<p>A) 4/3 B) 12/7 C) 3/2 D) 7/3 E) 5/2</p>
<p>Only felt like doing number 15...answer is E</p>
<p>Why?</p>
<p>Subtract one from the other to get </p>
<p>a + ar^2 + ar^4 +.... = 4</p>
<p>Multiply r on each side to get 4r = 3 because ar + ar^3 + ar^5 + ... = 3</p>
<p>so r = 3/4</p>
<p>Summation of a geometric series = a/(1-r) = 3, so a = 7/4 so a + r = 10/4 or 5/2 (very easy for a number 15 IMO)</p>
<p>Also practice tests arent the end all be all in predicting your score...I got a 132 on my last practice test before the AMC 12A last year and ended up with a 100.5 (that question about Yan....)</p>
<p>^ugh that sux</p>
<p>The test already happened this year?</p>
<p>last year...</p>
<p>Hi Guys!!!!!!!!!!!!!!!!!!!!111</p>
<p>where do I get the answers to the AoPS practice AMC12 tests?</p>
<p>the mock AMC12s? they usually have links to discussions on AoPSWiki, google "Mock AMC aopswiki"</p>
<p>or if it's the actual AMCs then the answers are on the AMC website</p>
<p>Good luck tomorrow, everyone!</p>