<p>post some answers, tests, and ask questions ^_^</p>
<p>(AMC12 - 07B)
1. Bob's house has 3 bedrooms. Each bedroom is 12 feet long, 10 feet wide, and 8 feet high. Bob must paint the walls of all the bedrooms. Doorways and windows, which will not be painted, occupy 60 square feet in each bedroom. How many square feet of walls must be painted?</p>
<p>A) 678 B) 768 C) 786 D) 867 E) 876</p>
<ol>
<li>Bob drove his compact car 120 miles home for the weekend and averaged 30 miles per gallon. On the return trip Bob drove his parent's SUV and averaged only 20 miles per gallon. What was the average gas mileage, in miles per gallon, for the round trip?</li>
</ol>
<p>A) 22 B) 24 C) 25 D) 26 E) 28</p>
<ol>
<li>(Problem has a diagram, but see if you can draw it based on info)
The point O is the center of the circle circumscribed about triangle(ABC) with angle(BOC) = 120 deg and angle(AOB) = 140 deg. What is the degree measure of angle(ABC) ?</li>
</ol>
<p>A) 35 B) 40 C) 45 D) 50 E)60</p>
<p>to be continued...</p>
<p>hmm this is from last year's? 2007?
the AIMEs are all online but only some of the AMC's so if anyone has any old AMC tests can u put them online...</p>
<p>note that in problem three, the circle is is of course the circumcircle so that the center is the intersection (point of concurrency) of the perpendicular bisectors of the triangle. thus the center O will be equidistant from the three vertices.</p>
<p>if i had a photocopier i would, only the A version of last year was on AoPS so... :-</p>
<p>is this the B version?</p>
<p>yep, the header should be AMC12B 2007</p>
<p>wow thanks i need that ^__^</p>
<p>Bad bad test. I liked the A much better last year. :)</p>
<p>lol no one's even doing the problem.</p>
<p>where do we find past contest questions?</p>
<ol>
<li>Bob's house has 3 bedrooms. Each bedroom is 12 feet long, 10 feet wide, and 8 feet high. Bob must paint the walls of all the bedrooms. Doorways and windows, which will not be painted, occupy 60 square feet in each bedroom. How many square feet of walls must be painted?</li>
</ol>
<p>A) 678 B) 768 C) 786 D) 867 E) 876</p>
<p>2 * 3 * (10 * 8 + 12 * 8) - 60 * 3 = 6 * (80 + 96) - 180 = 6 * 176 - 180
= 876</p>
<p>did it in my HEAD, that's why it's so long. dunno if its right</p>
<p>oh and username: AOPS!</p>
<p>That was in last year's AMC 12 B test...i remember...I got it right :D ... go AIME!</p>
<p>lol yay u got the first one right!</p>
<p>jenkster, I think the problems are rather easy ;) so to keep this thread alive, I'll post what I thought were questions that transitioned from the basics to intermediate level. I've modified the wording so I don't avoid copyright issues :).</p>
<p>AMC12B 2007
8. Bob's age is B years, which is the same as the sum of the ages of his 3 children. His age Y years ago was twice the sum of their ages then. Find B/Y</p>
<p>a) 2 B) 3 C) 4 D) 5 E) 6</p>
<ol>
<li>Let g be a function such that g(3a-1) = a^2 + a + 1 for all real numbers a. Find f(5).</li>
</ol>
<p>A) 7 B) 13 C) 31 D) 111 E) 211</p>
<ol>
<li>A group of boys and girls formed a facebook group in their combined hatred of facebook applications. Initially, 40% of the group are girls. After two girls leave and two boys arrive, 30% of the group are girls. Find how many girls were initially in the group.</li>
</ol>
<p>A) 4 B) 6 C) 8 D) 10 E) 12</p>
<ol>
<li>In quadrilateral ABCD we have that A = 2B = 3C = 4D (angles). Find the degree measure of angle A to the nearest whole number.</li>
</ol>
<p>A) 125 B) 144 C) 153 D) 173 E) 180</p>
<ol>
<li>A proficiency exam is given to a class in which 10% of the kids are juniors and 90% are seniors. The average score on the exam was 84. The juniors all received the same score while the average score of the seniors was an 83. What score did each of the juniors receive on the test?</li>
</ol>
<p>A) 85 B) 88 C) 93 D) 94 E) 98</p>
<p>Enjoy :)</p>
<p>EDIT oh and for something similar to questions 9...</p>
<p>AHSME 1991
21. For all real numbers x, except x = 0 and x = 1, the function g is defined by g[x/(x-1)] = 1/x. Let 0 < theta < pi/2. What is g[(sec(theta))^2]?</p>
<p>what's the difference btw. A test and B test? should i take a or b? for AMC10 and 12</p>
<p>same
some schools do one or the other cuz of spring break</p>
<p>I spent the time to do this problem so why not post a solution haha...</p>
<ol>
<li>x/(x-1) = 1/cos^2(x)
cross multiply to xcos^2(x) = x - 1
(1 - cos^2(x))*x = 1
x = 1/sin^2(x)</li>
</ol>
<p>So g(sec^2(x)) = sin^2(x)</p>
<p>nice solution, i decided to stick with sec^2 and use the identity tan^2(x) + 1 = sec^2(x).</p>
<p>if you want to avoid trig substitutions altogether, substitute y = x/(x-1), solve for x and then plug back into the equation on the right hand side. this change of variable allows you to plug sec^2(theta) right into an explicit formula.</p>
<p>what was ur highest practice AMC12 score?</p>