<p>Hi,</p>
<p>I learned how to solve these problems in the past but forgot. I'd really appreciate if someone could tell me how to solve them again.</p>
<p>Thanks.</p>
<h1>43 (Page 31)</h1>
<h1>51 (Page 32)</h1>
<h1>63 (Page 33)</h1>
<p><a href="http://www.act.org/aap/pdf/preparing.pdf%5B/url%5D">http://www.act.org/aap/pdf/preparing.pdf</a></p>
<h1>43:</h1>
<p>the triangle formed in the isosceles trapazoid is an isosceles triangle, meaning angle BDC = ACD.
Therefore, angle BDC = 25 degrees = angle ACD = 50
since the three angles of a triangle must = 180, you know that the middle angle in the lower triangle (point X) is equal to 130 degrees.</p>
<p>Since BD is a line, and lines have a 180 degree angle, you know that the angle to the right of X must equal 180 - 130, or 50 degrees.</p>
<p>the last triangle has angle measurements of 35 and 50, and since you know it must be 180 degrees total, you do 180 - 85 = 95 degrees or answer B</p>
<h1>51</h1>
<p>this is an advanced counting problem:
100-109 = 10 numbers
10, 20, 30, 40, 50, 60, 70, 80, 90 = 9 numbers
19 numbers * 9 number sets (100, 200, 300, 400…) = 171 or answer D</p>
<h1>63: this one is easy! there is no #63 so there is no problem!!!</h1>
<p>however, if you meant number 53 or number 60 i can help :)</p>
<p>Thanks!</p>
<p>The first problem - you explained it perfectly! </p>
<p>Is there not an easier way for the second problem? What if the problem was how many numbers between 1-1000 have the digit 3 at least once? It seems like a long process and I don’t even know if it fully registers in my head except if I went through all of them to make sure. By that I mean like, especially with 0, it occurs every 10 numbers, plus 9 times for x0x, x being 1-9, etc. </p>
<p>And I actually meant 60, no idea why I said 63. :)</p>
<p>Okay I think I got it. </p>
<p>Basically you need to know the formula</p>
<p>Sum of a geometric series = (a*(1-r^n))/(1-r) where a is the first term of the series and r is the common ratio. Considering r is 0.15 and n is to the infinity, r^n is going to approach 0 (if you knew calculus you probably already know that). Therefore in an infinite series, the formula is really just = a/(1-r). </p>
<p>So if you set 200 to equal (a/(1-0.15)), you realize that a equals 170. If a equals 170, then the next number in the series has to be 15% of that number which is 25.5 or F.</p>
<p>EDIT: That’s obviously for question 60. Surprised they gave you the formula but hopefully that is standard so we don’t have to remember it.</p>
<p>As for the counting problem, I honestly dont think they’d give you a serious “i have to sit down and count numbers for 5 mins” situation. If anything, they’d give you two or three patterns you’d have to identify and multiply together. the idea is to find that pattern quickly. there probably is a formula - but i dont believe it’d be worth stressing over for one or two problems like that.</p>
<p>As for 60, it is a basic problem hidden within a complex paragraph - it is testing your ability to synthesize information and pick out important forumlas and information. You aren’t expected to know what a geometric series is, just solve for A.
the above poster explained the actual problem much better than i would have done - so i wont bother double posting :P</p>
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<p>Could you please explain why it is 15% and not 85%? :S</p>
<p>Because the common ratio is 0.15 which means the next number in the sequence is 0.15 of what the last number was. </p>
<p>If the common ratio was 5, it means the next number is 5 times what its predecessor was.</p>
<p>Oh okay, thanks so much for clearing that up! :)</p>