## Anonymous asked:

Are the questions in your PWN the Math SAT supposed to be harder than the actual SAT math tests?Some of them are a bit harder, but most of them are in the difficulty 4-5 range.

Anonymous

Some of them are a bit harder, but most of them are in the difficulty 4-5 range.

Anonymous

Most faulty pronoun mistakes occur because the writer tries to use a clause as an antecedent. For example:

Sally ran 10 miles yesterday, and it felt good.

That’s a faulty “it,” because “Sally ran 10 miles” can’t be an antecedent. I know people *talk* like that all the time, but on the SAT, if you can’t point to a noun or gerund (which is really just a noun) as an antecedent, then you shouldn’t be using a pronoun.

Anonymous

Can we all just take a minute to be amazed at how awesome I am for answering questions from other dudes’ books?

…

OK, cool.

Anyway, the key insight here is that the number of students must be a multiple of 7 (4/7 of students are boys) and also a multiple of 5 (the ratio of 2 old to 3 young can be converted easily to 2 old to 5 total). So let’s see what happens if we start by assuming that there are 35 students in the class, and working back from there.

4/7 of the students are boys, which means 3/7 are girls. So if there are 35 students, there are **20 boys** and **15 girls**.

If 2/3 of the girls are less than ten years old, then 10 girls are less than ten years old, which means **5 girls are ten years old or older**.

So far so good?

OK, now back to the proportions of old and young. I noted before that 2/5 of the students are older than or equal to ten years old. That means **14 out of the 35 students are ten years old or older**. 5 of those older students are girls, which means **9 of them are boys**.

The question asks what fraction of boys are ten years old or older. That’s 9/20.

Anonymous

I’m not finding *AB* there, I’m finding *AO*, which is also the radius of the circle. Once you have *AO*, there’s nothing to do but square it and multiply it by π to get the circle area.

For that to be negative, you need one factor to be negative and one to be positive. So you might start by plugging in some numbers that might work:

- When
*n*= 0, you’ll have (1)(–1) = –1 - When
*n*= 1, you get (3)(2) = 6 - When
*n*= –1, you get (–1)(–2) = 2

Therefore, the only integer* n* for which (2*n* + 1)(3*n* – 1) is negative is 0.

You can also see this easily by graphing if you have a graphing calculator:

3(*n *– 1) = 2(*n* + 14)

3*n* – 3 = 2*n* + 28*n* = 31

Anonymous

It should take no longer than 20 seconds to just type them into your calculator, so even if you don’t know a shortcut there’s really no reason to get this question wrong. (Also, not a real SAT question, right?)

1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 = 144

Because a question like this won’t appear on your SAT, I’m ambivalent about saying the following because it feels like Kitchen Sink Prep to even mention it, but the shortcut is that you can find the sum of a set of evenly spaced integers by multiplying their average by how many numbers there are in the set.

In this case, there are 12 positive odd integers less than 24, and their average is 12. 12 × 12 = 144

Seriously, though, by the time you’ve counted the number of terms and figured out their average, you’ve probably already figured out their sum, or could have if you’d just started by doing that instead.

Anonymous

Sounds like what you’re doing is working pretty well—a 150 point improvement in a month is exceptional. Just stay on your grind.

Anonymous

You can’t easily do this with algebra because of the divisibility restrictions. Algebra will just give you a line (*m* = 7*k*/9) that ends abruptly when the sum of *m* and *k* reaches 500. There are an infinite number of solutions on that line. Besides, why would you want to overcomplicate a problem like this with algebra?

70 is divisible by 2 and 90 is divisible by 5. So one easy possible value of *m* + *k* is 70 + 90 = 160.

Anonymous

Expectations aren’t a sure thing, so you need a conditional tense.

Anonymous

I think you’re asking how you can tell, just from looking, whether a number is divisible by 4. There are some neat divisibility rules, but honestly for SAT purposes you can just use a calculator to test a number’s divisibility.

Anonymous

A system of two linear equations has no solution when the lines are parallel.

2*x* – 5*y* = 18

–5*y* = –2*x* + 18*y* = (2/5)*x* – 18/5

4*x* + *ky* = 17*ky* = –4*x* + 17*y* = (–4/*k*) + 17/*k*

Parallel lines have the same slope, so you need –4/*k* to equal 2/5. That happens when *k* = –10.

Anonymous

For some people, yes. But most people benefit from a longer runway.

Anonymous

Yeah, it can sometimes, but not here. I can’t put my finger exactly on why right now—maybe a commenter can help me out. I think it’s parallelism (“to trap… is to greatly affect…” would work), but I’m not sure that’s a complete explanation.