And no one in real life would put fringe on a quilt.
I like the completely unapplied versions of the questions (today’s math books are too touchy-feely for me) but the first was easier to me than the second.
I needed less than two minutes to get both answers correct. No sweat. The first was trivial geometry and arithmentic, and the second was make up a list and count the number of entries.
That’s one way to look at things. But another way to look at things is by seeing what I described as having 4 sides: only, two of them have length 0, and two of them happen to be coextensional.
Areas…
Rectangle, 1x24; 24 square units; 13x12, 156 square units. 156-24=132
300, 303, 306, 309, 312, 315, 318, 321, 324, 327, , 330, 333
Boy, I haven’t done calculus in 30 years…
If the Area is A, and the Sides are 2x and 50-2x, then A = 100x - 4x^2. This is an upside down parabola, with an X intercept at x=0, and -25.
DA/DX = 100 - 8X. Maximum area is where DA/DX=0; so 0 = 100 - 8X. Solving for X, you get 12.5. Since you can’t have a decimal, the next nearest values would be 12 and 13 for the sides of a rectangle with maximum area.
For minimum area, you want A to be zero for some value of x. With the formula for the parabola of A = 100x = 4x^2, the quadratic formula gives two answers for x of 0, and 25. Zero can’t be the size of one side, or it isn’t a rectangle. So we take 25. However, 24 is the maximum length for a side, so the smallest area rectangle is sides of 1 and 24.
I agree with Indistinguishable that 25-by-0 is a perfectly good rectangle, giving an answer of 156*. Since that answer isn’t listed, the test-maker must want 24-by-1.
I didn’t do the long multiply that chrisk used above. 1213 = 144 + 12 = 156.
For the other question, I saw we have three values varying by a factor of 3 at each of two steps, so a factor of 9 in all. The smallest 3-digit number is 100, and the largest we can have is 999. Dividing 999 by 9 gives 111, so our result is the inclusive range 100-111, or 12 values.
It took me a while, but I did both in my head, as so many have done above.
d=132
a=12, there are 34 overlapping sets, but only one third can be integers
Try this:
Simplify the following expression:
(x-a)(x-b)(x-c)…(x-z)
Does too.
Sure, but how did you get 100 and 111? Show your work, young man!
I just went: 999/3 - 99*3= 333- 297= 36. Only every three count, so 36/3=12=A.
I’m surprised by the number (heh) of you that tossed out 300 (or 100), only to bring it back later and count it. Why not subtract what doesn’t work? That is, 297 or 99 or what have you…
Pretty sure they do. They see how many missed a certain question and if it’s an outlier, they go from there.
No, Sublight’s assertion does not explicitly say that the numbers must be positive. You’re referring to the OP, I was referring to RickJay’s questioning of Sublight’s statement.
An engineer, a physicist, and a mathematician are asked to come up with the shortest fence that will enclose a certain amount of drug crop. The engineer calculates the amount of area needed for the crop and, knowing that a circle minimizes the circumference for a given area, designs a circular fence. The physicist starts by speculating a fence of infinite size and makes it smaller until there’s no extra space. The mathematician takes the limit as the fence length approaches zero and defines where the crop is as inside.
Just dusting off and modifying an old joke from college.
That semipenultimate term is a killer.
A rectangle is defined as a polygon. A polygon is defined as having sides that are line segments. A line segment is defined as having two endpoints, and all the points in between. Therefore, your visualization of a rectangle with sides of “0” is definitionally incorrect. 
If you go with those definitions, and in particular define a line segment as having two distinct endpoints, then sure. If you don’t, then you don’t. There are, after all, a multitude of contexts in which it’s perfectly natural not to.
Is one of those contexts the eighth grade?