Could get 2. using only my head.
3. required some paper.
I would probably have not got 3. when i was an eighth grader.
~199 sqft, or 625/pi sqft
Do I get a prize?
Not quite real life. In your example, the cloth might stretch, the stitching might affect the length somehow, etc.
You were able to set up a concrete example that fits the generalized problem presented. Did visualizing it bore you? If not, what’s the problem?
It seems to me that the easy way to do the second problem is to count the number of possible values for n/3. 100 is the lowest and 111 is the highest so you just count from 100 to 111 inclusively.
I was able to do both of these in my head correctly, although I flubbed badly the last time one of these kinds of threads came up, so I have nothing about which to brag.
I was able to both of these in my head fairly easily. To be fair, I have a lot of practice thinking about these types of problems, as this general class of problem comes up very often when doing graphics programming. Seeing as my job involves lots of that, I have a bit of an unfair advantage.
That this is on a standardized test is especially bad, IMO. With a normal test, the students get their test papers back, and would have the chance to challenge a “wrong” answer - I know that there were a few times in my junior high and high school math where the book’s answer key was clearly wrong. But standardized tests are graded by machine.
I wonder if the test-makers do any kinds of analysis to see if the answers to questions correlate with each other? For example, if you have a question which everybody who scores very highly otherwise gets wrong, it wouldn’t correlate with the rest of the test, so you might be able to find those types of questions. I can’t imagine that they do, but it sounds like an interesting project for someone to do. 
How so? My method gives you the correct answer of
a: 12. The values of X which work are 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111.
How are you supposed to find the answer to question 2? Brute force?
There are more formal ways of proving it, but here’s how I thought of it:
The smallest possible area is obtained when the shorter side of the rectangle is at its minimum, 1, because every single unit square then has at least two exposed sides which use up some of the 50 units allowed for the perimeter. By making the rectangle squarer, more unit squares are in the interior and don’t touch the perimeter, or touch it on fewer sides. More area for the same perimeter.
Any increase in the shorter side leads to an increase in area . And you can’t increase the shorter side beyond 12 because it will then be longer than the other side (50/4 = 12.5). Therefore the largest area is obtained when the sides are 12 and 13.
I’m sure I’m being whooshed, but those are not correct values for X (n).
The lowest possible value is (do we still need spoilers?) 300 and the greatest is 333.
In taking the test, I came up with eleven and since I didn’t see it on the list I tested twelve and realized that I forgot to take 300 into account. 333-300 = 33. Since you only want multiples of three 33/3 = 11. But of course, as I mentioned, this does not count 300 as a valid answer.
So the set should be 300, 303, 306, 309, 312, 315, 318, 321, 324, 327, 330 and 333. This is obviously your set x3.
Right, I was clarifying my first post (#14, I think) where I said I did it by reducing the question to “how many values of X are there where X and 9X are both 3-digit numbers?”
I also missed question 3 in the same way others did…I’d have probably gotten both these right as an 8th grader 
Indeed, as it happens, I am generally inclined to count as a rectangle the one of width 25 and height 0 (giving an answer to the first question of 156).
Similarly, I in some contexts would count 000 through 099 as three-digit numbers as well (giving an answer to the second question of 111), though this is perhaps less worrisome an ambiguity.
Of course, seeing the proffered choices readily resolved, in this case, all the interpretational issues, but only because I knew what to watch out for.
But we’re all agreed that they’re not, right?
Sure, in the sense that the questioner here is reasonably taking them to not be.
I think so inside the box. I like those who don’t, like you! I wouldn’t have visualized a circle!
I got 'em both right, but it took about 5 minutes to get the second one (mainly because I didn’t read the question correctly).
P.S. Ms Osborne is lame.
But that is not a rectangle, it is a line. A rectangle has to have 4 sides.
D
A
Wow, I suck
I got D for 2. and A for 3.
I suck because when I saw the problems, a voice in my head went “Oh goodie! Math!” And I excitedly worked them out.