PSAT math questions!!!!

<p>and .2/1.2 = .166667</p>

<p>the answer was 18 to the question with the reflection. (bh)/2 or 6x6 / 2</p>

<p>yeah.^
any other questions....
nobody remembers any from section 1?</p>

<p>oiseaurebelle, it was 18</p>

<p>Margaret41757, as you said a>0 and following your question, doens't 0.2/1.2 > 0.2 x 1.2 which is 0.66667 > 0.24 contradict a > 1 and b <1? Also, doesn't 1.5/0.8 > 1.5 x 0.8 which is 1.875 > 1.2 contradict a <1 and b>1? they both have a>0 and the numbers are positive, no? i mean, they don't have to be integers, it only said 'numbers' as i recall. Does anyone want to add on to this?</p>

<p>yeah I agree with everything that you are saying except for first example you read your calculator wrong .2/1.2 = .166667 not .66667 so it is not greater than .24. b must be less than 1.</p>

<p>this thread makes me sad :( after looking through all these I think I got 2-3 wrong, and w/ the PSAT nasty curve Im now doubtful of my NMSF chances....esp. w/ the SAT coming up for me in 2 weeks (nov 3)</p>

<p>You wouldn't happed to remember the exact question and choices to the reflection porb, would you? </p>

<p>Anyone know the curve?</p>

<p>Point P (-2,3) is reflected across the X axis to make point Q. Point Q is reflected across the line X=1. What is the area of the resulting triangle</p>

<p>RootBeerCaesar: I'm with you there, expect I probably got more than 3 wrong...</p>

<p>I'm realizing now that I made many silly mistakes. Ergh.</p>

<p>okay Margaret41757 i agree on what you said but did the question realy say that the numbers should be positive?</p>

<p>student91: Yes I think that the question said that the numbers should be positive, which sucks for me because I put iii</p>

<p>So the answer was only II right? Just want to make sure.</p>

<p>yes. only II</p>

<p>For the x^2 +kx + 4 = (x+c)^2</p>

<p>The c was there so you can solve the problem algebraically
If you substitute 0 for x (since the problem specified it is true for all real numbers of x), then you get 4 = c^2, so c = 2. If you substitute in another number for x, like 1, you get 1 + 1k + 4 = (1+2)^2, and 5 + k = 9
So, k = 4</p>

<p>^ I think I guessed 3, darn >_< (ran out of time)</p>

<p>Here are the answers I remember getting in places and I'm pretty sure they're correct:</p>

<ul>
<li>4rt3 + 8</li>
<li>II Only B (I had a huge fight on this one w/ another kid...I won) Try using .1 for a and .2 for b. a/b=.5 and ab=.02. EASY!</li>
<li>27 deg C</li>
<li>1.5 for last one</li>
<li>20 for permutation</li>
<li>12 for number of green</li>
<li>1/6 for c somewhere</li>
<li>that problem we're all discussing adamantly definitely asked for k which was 4</li>
<li>b=42 for that k, a, b problem</li>
<li>1010 on first gridin</li>
<li>5.5 on some gridin</li>
<li>$250,000 on some percent problem</li>
<li>18 for the area of that triangle</li>
<li>For that one gridin, I remember putting 82 cus rt1 + rt81 = 10. I'm not sure if it asked for an odd combination of numbers but thankfully, this was the first combo that came to mind, so I guess I got lucky.</li>
</ul>

<p>What did the first gridin ask?</p>

<p>what k, a, b problem?</p>

<p>least number bigger than 1000 divisible by 10</p>