SAT Blue Book Math Help

<p>I don't know why, but SAT Math is really difficult for me, especially under pressure. That's why I have so many questions. If you guys could explain why these answers are correct, and strategies I could use to do these types of questions faster, that'd be fantastic! Also, if posting SAT Blue Book questions isn't allowed, I'll edit my post and remove them right away.</p>

<p>Answers are in bold.</p>

<p>
[quote]
If x is an integer and 2<x<7, how many different triangles are there with sides of lengths 2, 7, and x?</p>

<p>A. One
B. Two
C. Three
D. Four
E. Five

[/quote]
</p>

<p>
[quote]
If 2|x+3|= 4 and |y+1|/3 = 2, then |x+y| could equal each of the following EXCEPT
A. 0
B. 4
C. 8
D. 10
E. 12

[/quote]
</p>

<p>
[quote]
If a>b and a(b-a)=0, which of the following must be true?
I. a = 0
II. b < 0
III. a-b > 0</p>

<p>A. I only.
B. II only.
C. III only.
D. I and II only.
E. I, II, and III

[/quote]
</p>

<p>Thanks!</p>

<p>For the first question you want to use the triangle rule. This says that the third side of a triangle is between the difference and sum of the other two sides.</p>

<p>In this case, the difference is 7-2=5, and the sum is 7+2=9. So the possible integer values are 6,7,8 and 9.</p>

<p>But the restriction 2<x<7, means that only 6 is acceptable. So the answer is ONE.</p>

<p>
[quote]

If x is an integer and 2<x<7, how many different triangles are there with sides of lengths 2, 7, and x?
A. One
B. Two
C. Three
D. Four
E. Five

[/quote]
</p>

<p>From "2<x<7" and "x is an integer," you can conclude that the only numbers x can be are 3, 4, 5, and 6.</p>

<p>A triangle with lengths 2, 7, and 3 cannot exist because 2 + 3 < 7. (Do you understand what I mean by this? If not, get a ruler and try to draw a triangle using 2 cm, 3 cm, and 7 cm.) The same way, 4 and 5 cannot be possible sides either. (If you don't understand why 5 can't be a possible side either, try drawing that too.) You can also look up the Triangle Inequality Theorem for further clarification. </p>

<p>A triangle can have lengths 2, 7, and 6, therefore, there is only one triangle with sides of lengths 2, 7, and x.</p>

<p>
[quote]

If 2|x+3|= 4 and |y+1|/3 = 2, then |x+y| could equal each of the following EXCEPT
A. 0
B. 4
C. 8
D. 10
E. 12

[/quote]
</p>

<p>There might be an easier way to do this, but I would solve for x and y first.
2 |x + 3| = 4
|x + 3| = 2
x + 3 = 2 or x + 3 = -2
x = -1 or x = -5 </p>

<p>|y + 1| / 3 = 2
|y + 1| = 6
y + 1 = 6 or y + 1 = -6
y = 5 or y = -7</p>

<p>(A) No. |-5 + 5| = 0
(B) No |-1 + 5| = 4
(C) No |-1 + -7| = 8
(D) Correct Answer
(E) No |-5 + -7| = 12</p>

<p>
[quote]

If a>b and a(b-a)=0, which of the following must be true?
I. a = 0
II. b < 0
III. a-b > 0</p>

<p>A. I only.
B. II only.
C. III only.
D. I and II only.
E. I, II, and III

[/quote]
</p>

<p>(I) Correct. Divide "a (b - a) = 0" by "(b - a)" to get "a = 0"
(II) Correct. You know "a = 0," so plug that into "a > b" to get ""0 > b" or b < 0"
(III) Correct. Take "a > b" and subtract b from both sides to get "a - b > 0"</p>

<p>Be careful with the third one, roman numeral I. Silver's suggestion of "dividing by b-a" in general cannot be done. For what if b-a=0? Well in this particular problem that can't happen because a>b. So what he says is in fact ok, but this does need to be checked carefully (or you may get the wrong answer).</p>

<p>Here's a better way to state it:</p>

<p>In general, when you have a product equal to 0, one of the factors must be 0. In this case we have a(b-a)=0. So a=0 OR b-a=0. If b-a=0, then b=a which contradicts the assumption a>b. Thus, a=0.</p>

<p>IMPORTANT: Always be careful about dividing by expressions with variables. Make sure these expressions can't be 0 before performing the division!</p>

<p>This issue will only come up in Level 4 or 5 SAT Math problems. In easier problems you won't get the wrong answer by being a little sloppy.</p>

<p>At the first and last question, I'm slapping myself in the face. I can't believe I couldn't figure those out. The second question though, is a bit more confusing. I seem to have forgotten the triangle rule, but I suppose it's good that I know now. </p>

<p>But I'm still confused about something, combining what both DrSteve and Silver said about it being somewhere between 2-7, how do you know VERY QUICKLY that a triangle couldn't be five? I mean, without measuring it out.</p>

<p>practice :)
Also, imagine a triangle like that. Sides of length 2 and 5 + Another side 7 = a line!</p>

<p>
[quote]
At the first and last question, I'm slapping myself in the face.

[/quote]
</p>

<p>That is what happens to almost everyone!</p>

<p>Ah, yeah you're right Silver. Also, xiggi, I sent you a pm. :)</p>

<p>One more question: how do you guys know this? I'm in Calculus, and I'm struggling with SAT math. I mean, I'm almost sure if there wasn't a time limit I could get them all no problem, but it's that added pressure and the thought of, "Hmm, should I skip this one and risk forgetting all of the thinking I've already done for this, or stay on it and risk losing time on the other probably harder questions?!"</p>

<p>How did you guys study so well that you were able to look at SAt problems and instantly jump into it? I mean, I can do that for the first 70 percent of the test, but when you get to the last parts with level 4 and level 5 questions, HOW DO YOU TRAIN YOURSELF!? Like, what are your study habits!?</p>

<p>I just finished geometry, so a lot of the concepts are still fresh. :)</p>

<p>Take it slowly. Spend the time to work through one section at a time, and make sure you review ALL your answers until you are fuly comfortable. That means to review your correct answers also! </p>

<p>After a while, everything will seem easier and you will start to identify the patterns faster and faster. There will always be something new on the SAT, but if you can answer 90 percent of the problems with enough confidence, you will not be flustered. Your mind WILL work better on certain problems. Blazing through the "easy ones" will leave you a bit more time for the challenging ones. </p>

<p>Calculus is not really relevant to the SAT.</p>

<p>Still, what are your study habits like, Silver?</p>

<p>And yeah, xiggi, that's what I've been trying to do so far. It's just that, I want to try and get better before I get to the 10 tests at the back of the book.</p>

<p>I started studying for SAT math in the middle of Geometry, and I didn't know a lot of the concepts, so I got bogged up on questions about similarity and circles. Needless to say, I did pretty bad on my first practice test.</p>

<p>I also got this book from my dad's friend, whose son went off to college, for SAT Math called Amsco's Preparing for the Sat Mathematics. It was pretty good. I ended up learning about similarity/circles/triangles before my geometry class even started them! I pretty much understood all the concepts by then, and school reinforced the geometry ones. </p>

<p>The only thing I wasn't too good at was functions and combination/permutations. In April and May, I had to study Algebra II and Trig on my own for a placement test, so I got a much better sense at the functions and probability, and some other things too.</p>

<p>From there, all I had to do was do the questions faster to finish them within the time limit. This wasn't too hard; all I had to do was practice. And any time I encountered a seemingly impossible question on a test (usually the last question), I asked my parents for help to fully understand it. Those questions always ended up to be something I should have understood, so I tried my best to make sure I wouldn't make such a mistake again.</p>

<p>Pretty much, that was my journey to a 800. All I have to watch out for is careless mistakes. I am able to finish each section with 3/4 of the time, so I go back right to the beginning of the section to check my work. In a three section test, I can usually find 1 careless mistake. :)</p>

<p>Thanks for the advice guys. I don't know why, but I went to a party last night, woke up today, ate some breakfast, went to practice SAT Maths, and my mind just shut down. I have no idea why, but I just cannot think at the moment.</p>

<p>Here are a few of the problems I need help with. Again, if this isn't in compliance with any rules or Terms of Agreement with College Confidential or the College Board, please let me know!</p>

<p>

<a href="http://img717.imageshack.us/img717/3703/p1010112p.jpg%5B/img%5D"&gt;http://img717.imageshack.us/img717/3703/p1010112p.jpg

</a>
Answer: E</p>

<p>

<a href="http://img215.imageshack.us/img215/9595/p1010113qi.jpg%5B/img%5D"&gt;http://img215.imageshack.us/img215/9595/p1010113qi.jpg

</a>
Answer: C</p>

<p>

<a href="http://img580.imageshack.us/img580/3900/p1010114c.jpg%5B/img%5D"&gt;http://img580.imageshack.us/img580/3900/p1010114c.jpg

</a>
Answer: 5</p>

<p>

<a href="http://img847.imageshack.us/img847/8366/p1010115x.jpg%5B/img%5D"&gt;http://img847.imageshack.us/img847/8366/p1010115x.jpg

</a>
Answer: 5.6</p>

<p>I definitely need more practice with lines and graphs. Anyone have any online resources that I could use to catch up on them?</p>

<p>Guys, I really need help! :(</p>

<p>1) k = f(3)
k = 4 (from chart you can see that if x = 3, then f(x) = 4)
g(4) = 5 (from chart you can see that if x = 4, then g(x) = 5)
(E) 5</p>

<p>2) The steps are congruent triangles.
t = the hypotenuse of all the steps added together.
What to do: calculate the hypotenuse of one triangle and multiply it by 7 (for 7 steps)
Calculate the hypotenuse of one triangle: The hypotenuse of the bottom step is sāˆš2, you can do it by applying the Pythagorean Theorem or by using what you know about 45-45-90 triangles.
Multiply it by 7: 7sāˆš2
(C)</p>

<p>3) Use the slope formula
m = (y2 - y1) / (x2 - x1)
(1/2) = (-2 - 0) / (1 - a)
Simplify.
(1) / (2) = (-2) / (1 - a)
1 - a = -4
a = 5</p>

<p>4) If you translate the line 2 units to the left, you already have two points for the new line: (-2 , 4) and (-7 , 0).
Plug in one of the coordinates into y = (4/5)x + k
0 = (4 * -7)/5 + k
0 = -28/5 + k
k = 28/5 or 5.6</p>

<p>I heard regents prep is a good website. You can check it out [url=<a href="http://www.regentsprep.org/Regents/math/geometry/math-GEOMETRY.htm%5Dhere%5B/url"&gt;http://www.regentsprep.org/Regents/math/geometry/math-GEOMETRY.htm]here[/url&lt;/a&gt;]. I hope I'm allowed to post this link. :) </p>

<p>Happy studying!</p>

<p>Thank you so much Silver! You have pretty much answered every single SAT Math question I've asked so far, and I am so thankful! Keep watching this thread though, because I'll have more questions. Don't you worry. :) </p>

<p>Also, one thing I've noticed while doing SAT practice tests: never think after you finish a test that it was easy. Odds are you messed up one some of the simplest of things. :(</p>

<p>Ok guys, here's another one.
<a href="http://img199.imageshack.us/img199/9868/p1010117e.jpg%5B/url%5D"&gt;http://img199.imageshack.us/img199/9868/p1010117e.jpg&lt;/a&gt;&lt;/p>

<p>No problem! I love to solve math problems. :D</p>

<p>Is the answer (A)?
Here is my work. <a href="http://i51.tinypic.com/25kuc6x.png%5B/url%5D"&gt;http://i51.tinypic.com/25kuc6x.png&lt;/a>
I hope I didn't do anything wrong :)</p>

<p>Darn you...the answer IS A. You genius you.
One thing I'm confused about is where did you get the m(sqrt 2) from? How did you know its value?</p>

<p>The pyramid has a square base, so a diagonal separates the square into to 45-45-90 triangles. If one side of the square is x, then the diagonal would be xāˆš2. :)</p>