Official December SAT Math Thread

<p>Yeah I'm pretty sure it was D.</p>

<p>dammit that was the only math problem i got wrong!!!, i had no idea wut it was asking, is one wrong 790?</p>

<p>-1 could be 800 depending on the curve. I felt it was pretty tough. Especially the inversely proportionaly question.</p>

<p>good thing i did the blue book 8th practice test 3 days before! it had practically the same question.</p>

<p>Does anyone remember the question about a triangle with a perimeter (a+b+c) of 10, a<b, b<c? What is the maximum possible value for b?</p>

<p>it said perimeter was 8. But yeah i remember it.</p>

<p>a is less than or equal to b, b is less than or equal to c</p>

<p>i dont know did they ask for b??, i thought the largest side</p>

<p>It was less than or equal to, and the perimeter was 8, and they had to be integers. My answer was 3. The sides (a,b,c) can be 2,3,3. I don't see how it can be any larger than that.</p>

<p>Oh and they asked for the highest value of B.</p>

<p>yeah they was asking for b.</p>

<p>btw was it 8 or 10? i thought it was 8.</p>

<p>haha i got 3 i think. But by fluke luck i think. god i thought i had another incorrect.</p>

<p>i thought i read that the sides couldnt be equal </p>

<p>so the only possibility would have been 4,3,1 which wouldnt have been a triangle</p>

<p>so 3,3,2 was the answer. phew.</p>

<p>the answer was 3</p>

<p>3-3-2 were the sides.</p>

<p>it was 8..........</p>

<p>yea i put 3</p>

<p>haha sweet. im stupid and lucky! yay! this makes me very happy. only 1 wrong so far! i will be happy with the result. stupid 180-70 = 120. oh well. I was retaking for CR anyway. Satisfied with 730 math, but hoping for better and 1 wrong would be cool.</p>

<p>im just really <em>ed off i got that one rate prob wrong with t= 0 and t=4 i was like *</em> are they asking?</p>

<p>i just plugged in random percentage until i got that in 4 years it would double. which was 19%. then i did something and i got 2. I just kind of guessed somehow</p>

<p>ugh..any pther probs u guyz remember?</p>

<p>the 4 paralleograms finding the area of one of the 4. with knowing the area of the others.</p>

<p>i got 8*6 = 48</p>

<p>also the xyz</p>

<p>which of these must be true</p>

<p>I
II
III</p>

<p>i got II, III</p>

<p>i dunt remember the parallelogram question....but for xyz i got all 3</p>

<p>If the avg of a list X is x and of a list Z is z, and x<y<z, for a no. y</p>

<p>if the two lists are combined, which of the following can be the avg of the two lists combined</p>

<p>I. x
II. y
III z</p>

<p>i chose only II...what did you choose</p>

<p>i didnt have that questions</p>

<p>oh well, even though you didn't had this question but you could probably answer it</p>