Anonymous asked:Hi! I'm an intl that wrote a while back. I struggle with time management on both the SAT and the Subject Tests. I tend to spend a huge amount of time on questions that I am not sure about. In my country, exams are very difficult and thus we have a lot of time for solving and checking our work (e.g. the math exam is ~18 questions in 3 hrs). On the other hand, the SAT pretty much requires you to solve questions in about 30 sec. Any tips with respect to making the transition and adapting? Thanks!
This advice may seem counterintuitive at first, but you should stop focusing on the time and instead focus on adapting to the kinds of questions the SAT presents. The reason you have so little time (not quite as little as you say—you have more than a minute per question) is that the questions are not as complex as the ones you’re used to.
Get comfortable with the content, and the timing will follow. As for how to get comfortable will the content, well… ::cough cough::.
Anonymous asked:i don't know why this has never occurred to me to ask you before, but what do you think is the best way to go through mistakes made in the CR and W section? I have been just reading the answers, and never really thinking about them... and I'm taking the SAT for the first time ever on March 8! Your opinion and advice is greatly appreciated! I have been following this blog since 9th grade~
Wow—assuming your’e a junior now, you’ve been following a long time! Thanks for that.
The best way to review your errors in CR and W is to look not only at why the right answer is right (in CR, that means going back to the passage and finding exactly where it’s supported) but also figuring out why your answer (and while you’re at it, why all the other answers) are wrong. Every wrong answer is wrong for a reason, and spending the time to figure out why each one is wrong instead of just why your answer is wrong quadruples the learning value of the question. This will help you eliminate wrong answers going forward as you begin to internalize the patterns wrong answers follow.
Anonymous asked:Ok! So I read your book and you seem really knowledgable about the SAT. So, I am aiming for 2400. Last time, number 20 on Math seemed to be really easy. Is that ever the case? Also, I did every question and thought the Math was easy, but I got 3 wrong. So is it just carelessness? What can you advise? Anything special? Next, in reading, I know to do the following, avoid extremes, avoid assumptions, and be literal. THE ANSWER IS IN THE TEST. Main idea is necessary as well. Any further tips?
#20 is never really easy as in anyone could do it with their eyes closed, but if you’re well prepped, the hardest questions on the test might seem very familiar to you. Make sure you’re not taking any shortcuts, though. Easy answers to hard questions can sometimes trap even the most sedulous students.
General advice for reading: don’t pick an answer until you can either justify it with a line from the text, or you’ve carefully eliminated every other answer due to an obvious flaw (e.g. contradicts the passage, contains information that wasn’t addressed in the passage).
Anonymous asked:414 #3... How is there enough information to answer this question? Thank you
Try backsolving. If (C) is right, then there are six 5-person tables (and thus thirteen 4-person tables). That would mean the restaurant seats (6 × 5) + (13 × 4) = 82 people. But the question tells us that the restaurant seats 84 people, so we need more 5-person tables.
Try (D), which says there are seven 5-person tables (and therefore twelve 4-person tables). That means the restaurant seats (7 × 5) + (12 × 4) = 83 people. Closer, but we’re not there yet.
So the answer must be (E). If there are eight 5-person tables and 11 4-person tables, then the restaurant seats (8 × 5) + (11 × 4) = 84 people, just like the question says.
You could also write two equations and solve. If x is the number of 5-person tables and y is the number of 4-person tables:
x + y = 19
5x + 4y = 84
y = 19 – x
5x + 4(19 – x) = 84
5x + 76 – 4x = 84
x = 8
Anonymous asked:I'm going to be taking the SAT during the transition--do you think I should take the Old or New? The Old will have more things to help me prepare, but the New does seem more appealing, and may be easier for me. What do you think? How will colleges compare applicants of my class?
I think many people in your position will end up taking both as a hedge. It’s too early to know for sure, but I have a hunch that if you prepare assiduously for the current test, much of that hard work will also help you with the new test.
I’d caution you not to buy too much into the hype of the new test. It will be different, but it probably won’t be easier. Colleges still want the SAT to spread students out along a normal curve. It’s still going to be very difficult, for example, to get a top score, or score in the 99th percentile. (It’s statistically impossible for the 99th percentile to get bigger.) If it becomes easier to get top scores, then the test becomes meaningless from an admissions perspective, and the College Board will have ushered itself into irrelevance. I give CB enough credit to assume that it won’t do that.
We’ll know much more on April 16th, when some actual test content will be released. As they say on terrestrial radio stations, keep it locked. I’ll tell you everything I know as I know it.
Anonymous asked:If 75 percent of m is equal to k percent of 25, where k> 0 , what is the value of m/k??
The toughest part about this problem is translating it from words to math. Remember, with percents, “of” means “times,” “is” means “equals,” and “percent” means “/100.” Use those facts to translate the words above into the equation below.
Then just solve for m/k.
Of course, 25/75 simplifies to 1/3, so that’s your answer.
Anonymous asked:In the cube above, diagonals of three faces form AGC. What is the measure of <AGC? So there is a cube shown with 3 diagonals (the left face, the back face, and the bottom face) that form a triangle. Sorry that it's hard to visualize. answer says 60 but isn't it 90 because the diagonals of a square bisect the right angle?
The easiest way to get at this one is to remember that all the faces of the cube are the same, so all the diagonals of those faces are the same. Therefore, you’re dealing with an equilateral triangle, which will have three 60 degree angles.
Anonymous asked:Hey, I'm taking the SAT in May and I wanted to ask you for your help. I haven't studied much for it yet and I want to get around a 2100. I have the Barrons 2400, and the Blue book. Can you suggest a study plan? Im a good math student but on the SAT I struggle. I'm around a 1800 right now. Thanks!
Here’s something I wrote a while back that I like because it’s flexible. It’s not dependent on a timeframe or a desired improvement—it’s just what you need to get done to arrive at your goal. Use it as a broad framework and apply it to your specific situation.
Hi Mike! I was looking at a few Critical Reading strategies and someone mentioned that "playing Devil's Advocate" is the right way to go. Do you believe looking for wrong answers is better than trying to justify why an answer choice is correct?
I am indeed a fan of eliminating answers when the correct answer doesn’t just jump right out at you. I’m also in favor of really trying to eliminate a choice when it does jump out at you to protect yourself from trap answers that have only a word or two wrong with them. If it jumped out at you, and even after you consider every word you can’t eliminate it, then it’s probably the right answer.
Anonymous asked:Blue book page 398 #10 and page 399 #14.
I have answered both of those questions before, here and here.
For Blue Book questions, always check the Blue Book Solutions page before you ask. At this point, it’s more likely than not that your questions have already been answered, and you’ll get the solution you seek immediately!
Anonymous asked:What must be added to x/y to make 2y/x ?
If you want to know what must be added to something to make something else, subtract the something from the something else. Sorry, that was super-confusing. LOL. For example, if you want to know what must be added to 2 to make 10, subtract 2 from 10. 10 – 2 = 8, so 8 must be added to 2 to make 10.
To find what must be added to x/y to make 2y/x, subtract:
Of course, this is a really inefficient way of doing this question in an SAT context. If this question appeared on an SAT, it would need to have answer choices, and it would therefore be a very easy plug in.
Say x = 2 and y = 3. What do you have to add to 2/3 to make 6/2 (AKA 3)? Easy—you have to add 7/3. Find the answer choice that gives you 7/3.
…yup, that worked.
Anonymous asked:What PWN stand for?
It doesn’t stand for anything (just like SAT, incidentally). It’s a word that, in certain circles, means to completely dominate.
Anonymous asked:hi can you please elaborate on this statement: When two polynomials are equal for all values of a variable (x) their corresponding coefficients (which are constants) are equal.... I'm confused on how you would use it and why this can occur... also if you can please provide an example!
Great question, thanks for asking. I’m actually working on the 3rd edition of the math guide now and I’ll be elaborating on that a great deal there. Until that’s ready to go (probably not until mid-late April)…
Let’s use parabolas to talk this through—parabolas are just 2nd degree polynomials. If you have two parabolas that are equal for one value of x, that means they intersect once. Like so:
Contrast that with parabolas that are equal for every value of x. They must be right on top of each other! They must be the same parabola. So even if the equation for one of them has some unknown constants in it, or it is not simplified as much as another one, you know that they’re the same exact parabola, which means the equations must be exactly the same.
That’s what I mean when I say that if ax^2 + bx + c = dx^2 + ex + f for all values of x, you can conclude that a = d, b = e, and c = f. Does that help a little?
Here’s an example of when this fact might be useful:
To figure out n – m, first foil the left hand side:
x^2 – 12x + 35 = x^2 + mx + n
Because that equation is true for all values of x, we can conclude that m = –12, and n = 35. Therefore, n – m = 35 – (–12) = 47. Easy, right?
You can find examples of this being tested in the Blue Book only twice—page 527 #8, and page 969 #17. Both are rated hard, but if you know this concept cold I think they’re pretty easy.