Math questions

Test Preparation SAT Preparation
1

<p>Please explain these:</p>

<p>x f(x)
0 a
1 24
2 b</p>

<li>The table above shows some values for the function f. If f is a linear function, what is the value of a + b?</li>
</ol>

<p>a. 24
b. 36
c. 48
d. 72
e. Cannot be determined with information given.</p>

<p>The answer is C</p>

<li>The first term in the sequences of numbers 3, 5, -5 is 3. Each even-numbered term is 2 more than the previous term and each odd numbered term, after the first, is -1 times the previous term. For example, the second term is 3+2, and the third term is (-1)x5. What is the 55th term of the sequence?</li>
</ol>

<p>a. -5
b. -3
c. -1
d. 3
e. 5</p>

<p>The answer is A</p>

<li>(grid in) Ester drove to work in the morning at average speed of 45 mph. Returned home later at 30 mph. If she spent a total of one hour commuting to and from work, how many miles did she drive to work in the morning?</li>
</ol>

<p>The answer is 18</p>

<li>If (a+b)^1/2 = (a-b)^-1/2 , which of the following must be true?</li>
</ol>

<p>a. b=0
b. a+b=1
c. a-b=1
d. a^2 + b^2 = 1
e. a^2 - b^2 = 1</p>

<p>The answer is E</p>

<p>Also I need help with number 5 from the Blue Book, Test 1, Section 6 but it has a picture with it which I cannot draw here. I would appreciate if someone took the time to look that up and explain.</p>

<p>Thanks all</p>

2

<ol>
<li>Write out the terms. You'll see that it's recurring...3, 5, -5, -3, 3, 5, -5, -3...</li>
</ol>

<p>5th term: 3
10th term: 5
15th term: -5
20th term: -3
25th term: 3</p>

<p>Get the pattern?</p>

<p>This way, you'll find that the 55th term is -5.</p>

<ol>
<li><p>Use the classic xy/(x+y) formula --> 30*40/75. Answer is 18, no pains.</p></li>
<li><p>Set up the equation. root a+b= 1/root a-b.
Square both sides.
a+b=1/(a-b)
Take a-b to the left side.
a^2-b^2=1. Voila!</p></li>
</ol>

3

<p>Oops didn't answer 12.</p>

<p>From f(1), you can see that f(x)=24x. So f(0)=0 (a), and f(2)=48 (b).
Thus a+b=0+48=48. Tada!</p>

4

<p>Tetrisfan: the 2xy/(x+y) formula, you made it something a little different. You had xy/(x+y), and the one that I know is 2xy/(x+y). That gives you the average speed for both trips, which is also the distance(one hour) for both trips, so divide it by 2 to get 18. I never knew that xy/(x+y) got you just one, that's cool though. </p>

<p>Also for #12, is there a way to find the slope and use that for the other values if you are given more than just 1 and 24?</p>

5

<p>Half the total distance covered gives you the one way distance, so it becomes xy/(x+y) :)</p>

6

<p>Don't forget that there is this valuable resource on CC:
<a href="http://talk.collegeconfidential.com/sat-preparation/339734-consolidated-list-blue-book-math-solutions-3rd-ed.html%5B/url%5D"&gt;http://talk.collegeconfidential.com/sat-preparation/339734-consolidated-list-blue-book-math-solutions-3rd-ed.html&lt;/a&gt;&lt;/p>

7

<p>thanks tetris!</p>

<p>But I still don't really see how you did 12...how do you just assume that f(0)=0 and that f(x)=24x</p>

<p>It could also be something like f(x)=36x-12 so f(1) sill gives you 24, but doesn't give 0 and 48 for f(0) and f(2).</p>

<p>I don't see how its possible to solve this problem.</p>

<p>Any help is appreciated :)</p>

8

<p>^^Once again, Shubham92 - why don't you look for these questions on the stickie given above?</p>

9

<p>oh sorry, gfc, I forgot CC had a stickied thread that provided solutions.
I didn't see your link until after I had posted the previous post.</p>

10

<p>Well this is fairly simple to solve, you the line is linear so the slopes have to be equal, so the slope between point 1 and 2 have to be equal to the slope between points 2 and 3</p>

<p>therefore </p>

<p>(24-a)/(1-0) = (b -24)/(2-1) (simple rise over run)</p>

<p>24-a = b-24
24=a+b-24
48=a+b QED :-)</p>

11

<p>wow, that seriously was amazingly done and explained vasudevank
thanks</p>

12

<p>Guys --</p>

<p>I don't get it. Here we have an incredibly valuable thread with tons of solutions to the most of the difficult questions from the Blue Book, and you are reinventing the wheel.</p>

<p>Posted on 11-07-2005: #3 in <a href="http://talk.collegeconfidential.com/sat-act-tests-test-preparation/112708-math-sat-i-question-help.html%5B/url%5D"&gt;http://talk.collegeconfidential.com/sat-act-tests-test-preparation/112708-math-sat-i-question-help.html&lt;/a&gt;&lt;/p>

<p>(see 398 / 12 / in <a href="http://talk.collegeconfidential.com/sat-preparation/339734-consolidated-list-blue-book-math-solutions-3rd-ed.html%5B/url%5D"&gt;http://talk.collegeconfidential.com/sat-preparation/339734-consolidated-list-blue-book-math-solutions-3rd-ed.html&lt;/a&gt;)&lt;/p>

<p>


</p>

<p>There are a bunch of other cool solution on that same thread. One of them is mine, but I am pointing you in the direction of <a href="http://talk.collegeconfidential.com/sat-preparation/339734-consolidated-list-blue-book-math-solutions-3rd-ed.html%5B/url%5D"&gt;http://talk.collegeconfidential.com/sat-preparation/339734-consolidated-list-blue-book-math-solutions-3rd-ed.html&lt;/a> not to satisfy my vanity - I don't have too much of it. I sound vain, don't I. :D</p>

13

<p>Speaking of Esther. I start picturing her as a horse galloping back and forth unending number of times foaming at the mouth until the poor thing is ridden to death. CCers are still beating this dead horse way past her passing without mercy, let alone compassion. Look at all the discussions on 412 / 18 in <a href="http://talk.collegeconfidential.com/sat-preparation/339734-consolidated-list-blue-book-math-solutions-3rd-ed.html%5B/url%5D"&gt;http://talk.collegeconfidential.com/sat-preparation/339734-consolidated-list-blue-book-math-solutions-3rd-ed.html&lt;/a>
The saga goes on.</p>