<p>Well, this is for the math set for the second SAT version, crate and barrel and so on. Also, was the crate and barrel question answer $2000?</p>
<p>Can you rewrite the crate and barrel one? I put something above 2000 - maybe 2750. I’ll check my calc. </p>
<p>Also for the a+1/b+1 = 1/3…
I put a= 4, b= 11. It worked when switched because 4/12 = 1/3, and 3/12 = 1/4.</p>
<p>The crate and barrel one was 2000 kg
The cost of the buses was $2750
The a+1/b+1=1/3 and a/b+1=1/4 was a=3 and b=11. It asked for a/b so the answer was 3/11</p>
<p>For the grid-in factors question:</p>
<p>“Given that 6 has four factors: 1, 2, 3, 6, What is one possible positive 2-digit integer that has exactly 3 different factors?”</p>
<p>Did anyone interpret what it was asking differently? Some people including myself thought it was asking for a number with exactly 3 different factors from those of 6</p>
<p>No, the factors just have to be different from each other, otherwise it wouldn’t work… So it could be 25 (1,5,25) or 49 (1,7,49)</p>
<p>It said 6 has 4 factors (1,2,3,6). Now find an odd two-digit number with 3 factors
meaning
1, x, the odd 2-digit number
so that means it has to be a perfect square
1, 7, 49 or 1, 5, 25</p>
<p>I went about looking it both ways and it just seems really ambiguous:</p>
<p>Interpret as TOTAL factors:
25 (1, 5, 25)
49 (1, 7, 49)</p>
<p>Interpret as DIFFERENT factors:
16 (1, 2, 4, 8, 16) The 1 and 2 are shared factors = 3 different factors
24 (1, 2, 4, 6, 12, 24) The 1, 2, and 6 are shared factors = 3 different factors
35 (1, 5, 7, 35) The 1 is a shared factor = 3 different factors
Possibly a couple others</p>
<p>Honestly if they had just omitted the word ‘different’, then the question would have made sense. I mean it makes less sense to count 4 as a factor of 16 twice, so I don’t see why they would need to use the word ‘different’. All that managed to do was make the phrasing confusing on what exactly it wanted.</p>
<p>Edit: And lol it definitely did not say an “odd” number and it also said the word “different”. I definitely remember that</p>
<p>Oops, I meant a=3, b=11. That one was mult. choice, right/</p>
<p>SAT128, please don’t bring this discussion everywhere. The consensus is that you’re wrong and the answer is 25 or 49. Writing to CB won’t change that. Missing one question most likely won’t affect your goal of 800.</p>
<p>It was pretty clear to me that it was just showing that 6 had 4 different factors as an example.</p>
<p>Alright fine, but if you were one of the people that was genuinely confused then you wouldn’t be so close-minded about it. And ofc it would affect me getting an 800. Missing one would be a 780 as a best case scenario. I thought I would mention it since asking this question clearly makes sense as a topic under a “Dec 1st Math Ver II” thread</p>
<p>@SAT128 I completely agree. I also interpreted the question differently and I really hope I only missed that one question. Fingers crossed for a generous 780.</p>
<p>Ay brutthas what you guys get for that cube problem put into different cubes. You had to calculate how many cubes had two faces.</p>
<p>And what you get for that graph with an absolute value function with a range from -3,2. Yo had to calculate the lowest value.</p>
<p>Okay first off, who is to say that SAT and I are not the ones who interpreted it correctly? Second to the cube problem: I put 12. I even broke out some Legos to figure it out afterwards.</p>
<p>@Shizz Wait - a cube question the ver II math asking for how many had 2 faces? Can you explain that question? I don’t remember seeing it.</p>
<p>I put 16 tho. Idr why. I think I figured out how many 4x4x1 cubes fit into an 8x8x8 cube. Was I wrong?</p>
<p>Also, for the “-3,2” question, I think the answer was -3. I don’t remember.</p>
<p>The question was a 3by3by3 cube. With each cube being 1by1by1. If you paint the outside of the cube, how many will have paint on exactly 2 faces.</p>
<p>@Youknowwho I think there were 2 cube questions, both of which I got 12 to be the answer.
One of them was if you painted a 3 by 3 by 3 cube red, which is made up of 27 smaller 1 by 1 by 1 cubes, how many cubes would have exactly 2 sides painted red.
The other was if you had a 4 by 4 by 1 rectangle and a 8 by 8 by 3 container how many of those rectangles can you fit in the container, or something along those lines.</p>
<p>@Vanilla I don’t remember that question. do you remember if it was at the end of the section? It could have been experimental… Any questions you remember before/after that one? </p>
<p>Also what what was the answer to the -3,2 question? Or, do you remember what the question was at all?</p>
<p>@Mystic - I thought the rectangle was 8<em>8</em>4, but i’m probably wrong.</p>
<p>I recall the problem now. I put 3. the lowest value was at x=2 and f(x) = 3. So it was three
f(x) = 2|x - 3| + 1</p>
<p>For the cube one I forgot the dimensions but, I think it looked like a rubik’s cube where all the two sided mofos are the pieces perpendicular to the center piece. It’s a little messed because a lot of them can be counted twice. I dunno, I put 12</p>