<p>Stupid mistakes on easy ones?</p>
<p>mexcian food, italian food-the one with over 100%. I said that was the only possibility. Everyone who picked Italian didn't have to pick Mexican.</p>
<p>37500 radios?</p>
<p>9 for the number line with the even spaces.</p>
<p>Raisins, chocolate bits, nuts: 7.5 or something with a .5</p>
<p>Any more easy ones...</p>
<p>the numberline was 6. they asked for y, not 2x</p>
<p>7.5 - right
37500 - right
9- wrong - 9 would be the last number on the line. Y would have to be 6 because then it would read 3,4.5,6,7.5,9
The food one - two of the possibilities made sense. The one with over 100% and the obvious one (more than half liked italian or something)</p>
<p>what about the combination one with 360 in it. and the one with 4 or 5 as the tens digit and 7 8 9 for units digit i think. i got 6 for that. and whats the 7.5 one? do u remember the full problem?</p>
<p>Yeah that was 6. 7.5 was a chocolate bits ratio one. I don't really remember the 360 question - perhaps you can describe it in more detail.</p>
<p>it was a combination question, but i dont remember exactly. it seemed easy cause all i did was multiply the combination numbers they gave so i wasnt sure.</p>
<p>RCMan- I still don't understand the pail question! "The pail one was definitely 'c' - the little piece on top because the first pail's height is the constant at the end and the 2n is that c times the number of top pieces."</p>
<p>n represented the total number of pails. In figure II, there were three pails, so n=3. C makes no sense to me- because you'd be multiplying it by n, the total number of pails (3), not just the number of top pieces (2)...</p>
<p>That is not what's important.</p>
<p>In my explanation, the constant 8 could very well represent the height of the bottom pail OTHER THAN its top part. In that case, you would multiply c by the number of pails.</p>
<p>So basically, if the bottom pail has height 10 - then the constant would be 8, and 2n would be perfectly fine.</p>
<p>Hopefully my explanation makes sense to you.</p>
<p>does anyone remember the answer to the question with the equilateral triangle within an equilateral triangle of perimeter 18 and you had to find the perimeter of the smaller triangle?</p>
<p>^it was 6root(3)</p>
<p>what was the answer to the section with the retiree inequality and the question where they asked what an average would be if a value (I think 30 or 70) was taken away?</p>
<p>Did anyone have a last question like this?...</p>
<p>`"you have a high school club. there are twice as many boys than girls. for every 2 juniors there are 3 seniors. if x is the number of junior girls and y is the number of junior boys, what is the total number of people at the club in terms of (x+y)?"</p>
<p>after futilely trying to solve it with algebra (with 2 minutes left) i just let 360 be the total. that left 144 juniors and 216 seniors... and so x+y = 144.</p>
<p>then let a be the fraction coefficient... a(x+y) = 360
substituting (x+y), you have... a(144) = 360</p>
<p>and then a = (5/3)</p>
<p>was the answer to that one (5/3) * (x+y)??</p>
<p>This is really random - but a few pages back the google adds switched from college app stuff to plastic beach pails.</p>
<p>Anyways - I got 360 for the combo one as well, but it seems too easy. The problem was something like 'How many four digit numbers can be created from the digits 1 through 6 if no digits are repeated within a number'</p>
<p>For the Mexican food one, I put two statements as right - the voting twice one and something about more than 1/2 liking Italian.</p>
<p>Does anyone remeber the question with a 30-60-90 triangle with an side of 4 where you had to find the area? How about the one where t^2=ab and they wanted some expression in terms of t?</p>
<p>and second question... i know some people have talked about it but i'm 95% sure i have it right. its this question:</p>
<p>If xy + x^2 = an odd number, then which of the following MUST be true?</p>
<p>I. x is odd
II. y is odd
III. x + y is odd</p>
<p>I said only III... here's why</p>
<p>if x and y are both even, then the expression is even. (doesn't work)
if x is even and y is odd, then the expession is odd. (works)
if x is odd and y is even, then the expression is odd. (works again)
if x and y are both odd, then the expression is even. (doesn't work)</p>
<p>so, looking at the two situations that work (the middle two), in one case y is even and, in the other, it's odd. so, option I can be eliminated.
II can be eliminated for the same reason, x is odd in one, even in the other.
III must be true. in order for xy + x^2 to be odd, the variables cannot both be odd or even. one has to be odd, and the other even. ergo, x + y must be odd.</p>
<p>My Answer: III only</p>
<p>ever_after, i remember the 30-60-90 problem... the hypotenuse was 4, so the legs flanking the 90-degree angle had to be 2 and 2*root(3).</p>
<p>i just turned the triangle around so that the base was 2 and the height 2root(3).</p>
<p>base * height * (1/2) =
2 * 2root(3) * (1/2) =
4root(3) * (1/2) =
2root(3)</p>
<p>wat was the combo problem?</p>
<p>Alot of people keep referencing the last grid-in as really hard. Does anyone remember what the actual problem is or some keywords from it b/c i can't remember remotely what the question was? was it the what is f(10)</p>
<p>"if x is even and y is odd, then the expession is odd. (works)"</p>
<p>That doesn't work. Take for example we choose x=4 and y=3
When we plug that in it yields the following:
(4)(3) + (4)^2 = 28 (even)</p>
<p>Therefore, the answer must be I and III.</p>
<p>i jsut factored the x out and got x(x+y) and just looked at it from there...tricky but not too hard of a problem if you just plug in stuff</p>
<p>you have a high school club. there are twice as many boys than girls. for every 2 juniors there are 3 seniors. if x is the number of junior girls and y is the number of junior boys, what is the total number of people at the club in terms of (x+y)?</p>
<p>i got (5/3)(x+y) too</p>