<p>Maybe you all already figured this out, but #16 in the grid-ins can’t be 20!!</p>
<p>That’s because 3 and 20 cross paths at then number 60, so, the multiples of 20 would only remove 4 numbers, not 5…right?</p>
<p>@rayankees94
missed three MC and missed two grid ins. I just counted the gridins as omits. So, maybe 700?</p>
<p>well you do need to remove 4 numbers</p>
<p>Hard to do. -4 Raw points a while back was a 720 -6 is around 680/700 if you are super lucky</p>
<p>@141421356 - don’t you need to remove 5? because there are 33 multiples of 3 in 1-100 inclusive. so, you have to remove 5 more? (101-33 = 68, to get to 63, you need 5 less…)?</p>
<p>1-100 inclusive is 100 numbers…</p>
<p>so you have to remove 4</p>
<p>and for the g(-1) question, did you guys get 5?</p>
<p>$^loca</p>
<p>you have the correct approach to the problem, however</p>
<p>1-100 inclusive has 100 integers, not 101. 0-100 inclusive = 101.</p>
<p>rayankees94, idr the exact question, but i remember that all i did was found f(-1) and multiplied it by (-1)… the problem implies g(x) = f(x) * -1 because g(x) is f(x) reflected over the y-axis.</p>
<p>So the section with the g(p)= p-1 problem and the number of integers divisible by 53 with remainder of 3 problem was an experimental section right?</p>
<p>oh geez, that’s the luckies thing that every happened. i made 2 mistakes, which (i guess) canceled to get the right answer!!</p>
<p>What would
-3 wrong of them being grid in be
or
-4 wrong 2 of them being grid in</p>
<p>Im asking because I put 3.3 for the one that was 10/3 :</p>
<p>lacamotif: This one took me a while to figure out as well. With numbers 1-100, multiples of 3 are crossed off. This leaves 66 numbers. You have to find a way to remove 3 more numbers that are multiples of n. With 25, you cross out 25,50,75, and 100. That’s four numbers, but keep in mind that 75 was a multiple of 3, and was already crossed off. The other three are gone, leaving 63 numbers. It’s a tricky one.</p>
<p>People speculate that numbers other than 25 could be the answer. I heard 19 multiple times.</p>
<p>25 cannot be an answer…
once you remove multiples of 3 you are left with 67 numbers and need to remove 4 more thus a number such as 19 or 20 would be a correct answer</p>
<p>Brolex: Unfortunately, 3.3 will be considered wrong, as it’s not accurate enough an answer. 10/3 is acceptable, and I think 3.33 would be acceptable as it’s as accurate as you can be. I’m 99% sure on the 3.33 situation, but to be safe I put 10/3.</p>
<p>Shoot, I miscounted. there were 67 numbers. Darn.</p>
<p>So was the grid in math section with the g(p)= p-1 problem and the number of integers divisible by 53 with remainder of 3 problem an experimental section?</p>
<p>Mario Kart, First I put 3.33 and then I was like lets be safe 10/3. I sure hope the scantrons are good because I erased it, but it looks like im in more danger now. I’m hoping its not going to read 1./3 or someth</p>
<p>and I put 16 :)</p>
<p>@Mario: 3.33 is considered correct. Decimal answers as accurate as possible in the given space, with proper rounding, are acceptable along with their fraction equivalents. If you read the instructions before the grid-ins (which nobody has time for), it tells you that.
Personally, I don’t think it’s fair to include reading the directions as part of the time limit. People who have already taken the test and can safely skip those have an advantage. I think they should be printed on the back cover or something, so you can read them before you begin.</p>
<p>@wowowow: I believe so because I didn’t have either of those questions.</p>
<p>Dont think that one was exp. </p>
<p>also can someone elaborate on the 6 one. the one with g(x) and the graph?</p>
<p>can someone please state the land lease question again. id greatly appreciate it if someone remeberd the question and posted it. Thanks!</p>