<p>Post your hardest SAT problems here. This way, we can discuss them and be more prepared for the next SAT (even though they won't be the same, chances are they were all types of problems that use weird words in order to mask the easy problem it really is).</p>
<p>A few that should be posted:</p>
<p>-The weird phone call graphs problems
-The graph of a circle with a line through it that asked for the equation of a line
-The medicine problem (when will they have it at the same time again?)
-Muscle vs Body Fat Make your own Equation Problem (I think I made a = to 2, and k equal to something else to get 20 and found an equation)</p>
<p>A few I didn't get but have read about:
- Something like (2n - 1)(4n +11) (don't know exact equation) - How many negative values of x for something
-"Pail Problem"</p>
<p>So far I know I omitted 2 (ran out of time - one of the phone call graphs and the circle and line). I guessed educated on medicine (I think it may have been experimental though). Muscle vs fat may be wrong, and the token I got careless - but I thought you needed to use both types of tokens, so I put 193</p>
<p>I think it was asking how many 18 wheelers there were, which would be 6. I heard someone say it was asking for the difference though, so I don't remember, since in that case it would be 2. I think I got E.</p>
<p>For the chairs, you find out that every time the first girl chooses a chair, since the other girl can't sit horizontally or verticallly from her, the second girl only has 6 possible seats. Since there are twelve chairs then there are 12 chairs times 6 possible outcomes, or 72 total possible outcomes.</p>
<p>the one with an equilateral triangle within an equilateral triangle of perimeter 18...what is the perimeter of the smaller equilateral triangle...and it was something like side ad to de is a ratio of 3:1</p>
<p>That one was a 30-60-90 problem and the answer was 6 root of 3.</p>
<p>Can anyone explain the phone call graph problems?</p>
<p>Oh, I remember that there was a weighted average problem that not many knew how to do. I think you just multiply each amount of tree times its percentage, or at least that's what I remember from chem.</p>
<p>The problem that gave a rectangle and asked for the area of a shaded region in terms of t gave me 2t I think. At first I had the more complicated one (a term plus a constant) but then I saw that only worked if two values you used were equal, and if you chose different values it was just 2t.</p>
<p>There was one problem with an organism multiplying. I think I got 2 times a value for that one.</p>
<p>For the phone call question the answer is 5.85 if the call was made at 6pm. </p>
<p>You can figure it out by setting up a ratio:</p>
<p>After 6 PM the call cost 3.15 after a 65% discount (meaning the price was 35% of the original price) so the ratio is</p>
<p>35 3.15
---- = ------
100 x</p>
<p>35x=315
x=9</p>
<p>then you want to find how much that call would cost if there was 35% discount instead of 65% and since since 35% off means you're paying 65% of the actual price you just do .65*9= 5.85. </p>
<p>You can do this because the prob stipulates that the call after 6AM was in one rate zone so since that rate zone (65% off) ends at 8AM the call could have been 2 hours at most and a call from 6-8 (the most it could possibly be) falls entirely within the 35% off zone.</p>
<p>For the Trucks problem it was asking how many more 18 wheelers there was so the answer was 2.</p>
<p>how about the one where there is a circle with radius of 5 and a point (13,0) and they ask to find the distance of the shortest line. I ran out of time on that one. Do you need to use distance formula or something?</p>
<p>Sweetdream - you definitely do not need distance formula.</p>
<p>Just think about it. A circle centered at the origin with radius=5 will extend to the point (5,0) on the x-axis. This is the point at which the distance is the shortest. The distance between (5,0) and (13,0) is just horizontal, so the distance is 8.</p>
<p>The little circle had r=sqrt(2)-1. The diagonal in the square was sqrt32because each side was 4. You subtract 4 from because of the large circles. You then divide by four to find the radius because in addition to the small circle in the middle the corners had empty space too; each empty space equalled the radius. I hope that made sense.</p>
<p>The much easier way to do that question....</p>
<p>Draw a right, isoscoles triangle with the hypotenuse from the center of a large circle to the center of the small circle. Since the legs are each 1 (both radii), the hypotenuse is rad(2). Since the hypotenuse includes the radius of the large circle, you subtract 1. Rad(2)-1.</p>