Blue Book Test Question

Test Preparation
1

<p>Test #5
Section 7 (Math)
Number 19:</p>

<p>I wasted like 3 minutes to get this one right doing wacky stuff on my ti-89 (solved for individual variables, plugged into expression, and then discriminated the ti-89 non-simplified answer into choice A). I read the college board explanation on how they "officially" got it, but am still a bit confused. I understand how the algebra checks out, but the way they "looked" at it is seems idiosyncratic to me.</p>

<p>My question to CC is how would you all look at it? When reviewing this question I plugged in numbers and it seemed to go a lot faster, but is it better to discern the problem algebraically or what? I appreciate the responses.</p>

<p>PS - God Bless the freedom loving city of London.</p>

2

<p>well what's the question? we(those of us who don't have the blue book in front of us) can't solve it unless we know what it is...</p>

3

<p>If k, n, x, and y are positive numbers satisfying x^(-4/3) = k^(-2) and y^(4/3) = n^2, what is (x*y)^(-2/3) in terms of n and k?</p>

<p>A) 1 / (n*k)</p>

<p>B) n / k</p>

<p>C) k / n</p>

<p>D) n*k</p>

<p>E) 1</p>

<p>Correct Answer = A</p>

<p>BTW, I assumed everyone on CC had the blue book =P</p>

4

<p>okay well we know that x^(4/3)=k^2 and y^(4/3) = n^2 and the question asks for (x<em>y)^(-2/3) which means 1/(x</em>y)^(2/3). going back to k^2 and n^2, we see that if we square root it we get x^(2/3) and y^(2/3) respectively. i'll let you do the rest</p>

5

<p>oh ya and that's why you learn to do math instead of using a ti-89. also if you want to learn more about problem solving try reading how to solve it by polya</p>

6

<p>
[quote]
oh ya and that's why you learn to do math instead of using a ti-89. also if you want to learn more about problem solving try reading how to solve it by polya

[/quote]
</p>

<p>What's "polya"?</p>

7

<p>polynomials I assume ..</p>

8

<p><a href="http://www.amazon.com/exec/obidos/ASIN/069111966X/qid=1120791909/sr=2-1/ref=pd_bbs_b_2_1/103-1202762-9819009%5B/url%5D"&gt;http://www.amazon.com/exec/obidos/ASIN/069111966X/qid=1120791909/sr=2-1/ref=pd_bbs_b_2_1/103-1202762-9819009&lt;/a&gt;&lt;/p>

<p>This is another excellent book:
<a href="http://www.amazon.com/exec/obidos/tg/detail/-/0471135712/ref=pd_sim_b_5/103-1202762-9819009?%5Fencoding=UTF8&v=glance%5B/url%5D"&gt;http://www.amazon.com/exec/obidos/tg/detail/-/0471135712/ref=pd_sim_b_5/103-1202762-9819009?%5Fencoding=UTF8&v=glance&lt;/a&gt;&lt;/p>

9

<p>Why be sarcastic?</p>

10

<p>Of course both books are totally unnecessary for the SAT.</p>

<p>As a matter of fact, finishing them does not guarantee a high score on the SAT: you can do all the hard questions and make a bunch of stupid mistakes on the rest.</p>

<p>Only if you are seriously in love with math, you'll find them indispensable.</p>

11

<p>the fastest way would be to plug in...unless ur really good at algebra</p>

12

<p>What numbers would you plug?</p>

13

<p>If k, n, x, and y are positive numbers satisfying x^(-4/3) = k^(-2) and y^(4/3) = n^2, what is (x*y)^(-2/3) in terms of n and k?</p>

<p>k=4<br>
x=8</p>

<p>8^(-4/3)=4^(-2)</p>

<p>n=4<br>
y=8</p>

<p>8^(4/3) = 4^2</p>

<p>so then you know (x<em>y)^(-2/3) is (8</em>8)^(-2/3)=1/16</p>

<p>so then you plug 'n' and 'k' into the answer choice until you get 1/16</p>

<p>the only one that works is </p>

<p>A) 1 / (n<em>k)=1/(4</em>4)=1/16</p>

14

<h1>harry, you did use algebra in your solution, you just did that implicitly.</h1>

<p>x^(-4/3) = k^(-2)</p>

<h2>y^(4/3) = n^2</h2>

<p>x^(-4/3) = k^(-2)
y^(-4/3) = n^(-2)</p>

<p>Multiplying left and right sides of equations respectively</p>

<p>(xy)^(-4/3) = (kn)^(-2)</p>

<p>( (xy)^(-2/3) )^2 = ( (kn)^(-1) )^2</p>

<p>(xy)^(-2/3) = (kn)^(-1) = 1/(n*k)</p>

15

<p>however, wont it be square root of 3 in (xy)^(-2/3) ?</p>

16

<p>guys its reallly simple</p>

<p>solve for x by raising it to the -3/4
so... x=k^(6/4)
and same with y
so.. y=n^(6/4)
then merely substitute
((k^(6/4))(n^(6/4)))^(-2/3)
and look they cancel with a -1 as power
so k^-1*n^-1 and that's the equivalent to 1/kn
i know its hard to see on computer jargon so copy it down and it will be clear.
and bobobabob is right.. i love math and therefore i never ever use a calculator.. i see no use for it.</p>

17

<p>solve(x^(-4/3)=k^(-2) and y^(4/3)=n^2 and z=(x*y)^(-2/3),{x,y,z})|x>0 and y>0 and k>0 and n>0</p>

18

<p>sorry but i dont really get it. I get it all the way up to ((k^(6/4))(n^(6/4)))^(-2/3)... Then how do you simplify to get -1. Sorry if this sounds like really stupid question to you guys that are really talented in math and all.. but i really dont get it.. thanks for helping!</p>

19

<p>6/4=3/2</p>

<p>k^t * n^t = (kn)^t</p>

<p>(k^t)^u = k^(tu)</p>

<p>==================</p>

<p>((k^(6/4))(n^(6/4)))^(-2/3) =</p>

<p>( (kn)^(3/2) )^(-2/3) =</p>

<p>(kn)^( (3/2) * (-2/3) ) =</p>

<p>(kn)^(-1)</p>

20

<p>Some of you are making this look more complicated than it has to be. Why raise it to 6/4 or whatever? Why did you bring in this new variable t? Ugh, stop!</p>

<p>Also, that plug in technique won't work. What are the chances that you would choose the right numbers that would work unless you can do the math in your head, you do not have time.</p>