More SAT Level 2 Question help

Archive SAT Subject Test Preparation
1

<p>Can someone please assist in solving this?</p>

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<p>Is there anywhere I can find trig questions, and probability similar to these? Because I’ve found that in the prep books I’ve used (Barron’s, Princeton), it’s only in the practice tests that lots of concepts from different topics are incorporated into a question. It’s so hard to find resources for more questions. :confused: </p>

<p>Any tips for Math Level 2? The concepts are easy, I just find that some questions are really tricky.</p>

<p>Thank youuu.</p>

2

<p>My answer is1,C 2,D. I think the only way to get a good score in math is having enough practice. As you say， basic concepts are really easy but the combinational use of different concepts is a little bit tricky. However, even the most tricky question can be tackled down if you dissect the question into several basic conceptual ones. Take question 1 for example, it’s a quite complicated one at first glance. But if you dissect the questions into smaller ones like sinA, cosB and tanA and express them in terms of a, b and c, you are able to tackle them down easily. I think the practice tests in Barron are really enough. You just need to go over all the questions in practice tests and diagnostic test and look at the explanation at the back of the book. Make sure you know the answer for every question and most importantly, the way that figures it out. The actual test is easier than those in Barron so if you can tackle questions in Barron, you will get a pretty high score in real test. By the way, I studied Barron by myself and got a 800 in actual test. Good luck.</p>

3

<p>these are my answers (though i’m not completely confident about the second one)

  1. tan^2a/sinacosb=(sina/cosa)^2/sinacosb
    sina = cosb in an acute angled right triangle (u can verify by substituting in the values of sin a and cos b)
    so, (sina/cosa)^2/sin^2a
    = 1/cos^2a
    = sec^2a
    by defn of secant, we know, sec x = hypotenuse/base
    so sec^2 a =( c/a)^2
    that is optn C</p>

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<li> there are 2 black marbles and only one white marble, and the second marble taken out is white. this means that the first marble taken out must’ve been black.
P(taking out a black marble)= 2/3
P(taking out a white marble after first trial)= 1/2
P(taking out a white marble after having taken out a black one) = 2/3 * 1/2 = 1/3
that is optn B</li>
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<p>hope that helped</p>

4

<p>Sorry, I got black and white inverted in the second question. It should be B. Don’t make such stupid mistake on the exam.</p>

5

<p>Thank you for the advice, ZZK1120 :smiley: I do prefer Barrons and use that, but for some reason found Princeton’s questions more difficult.</p>

<p>xtanyax, could you please explain “sina = cosb in an acute angled right triangle”? My trig can be so bad sometimes. I get very confused with statements like that. Thank you.</p>

6

<p>try to look at it this way-
sin a = opposite side/hypotenuse
= b/c
cos b = adjacent side/hypotenuse
= b/c</p>

<p>another way is
using the trig identity cos(A-B)= cosAcosB+sinAsinB<br>
cos(90-a)=cos90cosa+sin90sina= sina ------------------------- (1)
since a+b=90
b=90-a
and so you can replace (90-a) in eq (1) with b</p>