<p>On SAT Practice TEST #4
Section 4
can some one explain to me how they answered number 7, and 8</p>
<h1>7</h1>
<p>Area of a triangle = (1/2)<em>b</em>h
it says that the base (XZ) is (6/7)h
fill in (6/7)h for b in the triangle area formula</p>
<p>this gives you (1/2)*(6/7)h * h
then you get (6/14)h^2
then reduce fraction to (3/7)h^2</p>
<p>(this is on page 596)</p>
<p>#7
The area of the triangle is half of base*height, and the base is given at 6/7 of h, or (6h/7). The height is h. (1/2) * (6h/7) * (h) = (6h^2)/14, or (3h^2)/7, which is choice B.</p>
<p>#8
By the law of exponents, you know that the expression on the left side of the equation sign (complicated enough that it's useless reproducing in plain text) is equal to the 6th power of each individual term, and that a power raised to another power yields an exponent number that is the product of the two. </p>
<p>Therefore, a^(1/2) raised to the 6th power would be a^3;
and b^(1/3) raised to the 6th power would be b^2. </p>
<p>From there I just guessed and got a = 3 and b = 4, so ab = 12, (B). I'm sure there's a more sound mathematical way to do it though.</p>
<h1>8</h1>
<p>factor 432 give you 2,2,2,2,3,3,3</p>
<p>So you know 4^2 *3^3</p>
<p>Lol, I just took that section.</p>
<h1>8:</h1>
<p>(a^(1/2)) * a^(1/3))^6 = a^3*b^2</p>
<p>432 factored = 2^4*3^3</p>
<p>So by looking at that, we can match up the 3^3 with a^3 easily.</p>
<p>so a = 3</p>
<p>Then, we know 2^4 ALSO EQUALS (2^2)^2 = 4^2. At this point we can match up 4^2 with b^2.</p>
<p>so b = 4</p>
<p>a<em>b = 3</em>4 = 12</p>
<p>Honestly, that's one of the hardest math SAT questions I've ever seen.</p>
<p>thanks guys, u made that look really easy, im kicking myself in
the head for not thinking about factoring that one.</p>
<p>On SAT Practice TEST #4
Section 8
did anyone get 15 and 16</p>
<h1>15</h1>
<p>just pick a number for n, let's pick 100.<br>
So we know the supplementary angle of n must be 80. </p>
<p>That means the two arrows in that triangle must be equal to 100 (since triangle degrees add up to 180).<br>
The other arrows in the other triangle are the same, so they equal 100. </p>
<p>So the total is 200 degrees.<br>
Now just fill in n for the choices to see which one gives you 200. </p>
<p>It turns out that choice B does, since 2n = 2*100 = 200. Choice B is the answer.</p>
<h1>16</h1>
<p>Pick a number that goes into 1/3 evenly. I chose 3.</p>
<p>So 1/3 of 3 is 1, plus 3 is 4.<br>
So the first term is three, the second term is 4. </p>
<p>It says t is the first term (3 in our case), just fill in 3 for t in the answer choices to see what will yield the ratio of (4/3).</p>
<p>It turns out that choice C is the answer because 12/9 = 4/3</p>
<p>thanks 4 quick post</p>
<p>Tweek, when I did those problems I used symbols instead. On #15 I think your method worked a lot faster, but on #16 it took me a split second because I punched in what I got using symbols (which took me no time at all) and punched it in my ti-89 which auto-simplified it to the correct answer choice. My question to you is do you always try to pick a number to solve problems, and if so, how does it work for you? I'm pretty conceptual when it comes to things, so I always tend to use symbols.</p>
<p>I fill in numbers for variables on a ton of the questions for the SAT. Works for me. The only time it won't work is if you have two variables and the problem doesn't state how the second variable is related to the first variable. Got a 780 on the math section, only got one wrong, and it was due to a stupid mistake. I generally end up with 5 extra minutes at the end of every math section to check for mistakes.</p>
<p>I think I'm going to try this strategy tweek and thank you for your input. To comment, you got a minus one and got a 780? It was under my impression a minus one was forgiving enough to get an 800 in math (but nothing more).</p>
<p>George, I got minus 1, and I wasn't even deducted a quarter of a point (it was in the grid-in section) and STILL got a 780! Yeah I was so angry, I really wanted an 800. My version of the test had a tough math curve.</p>
<p>On SAT Practice TEST #4
Section 2
number 11 , 15 , 19</p>
<p>thanks in advance , this is better than a tutor</p>
<p>anyone know how to do those ?</p>
<p>bump bump bump</p>
<h1>11 The answer is 3. First, reverse the entire list: X Y W Z</h1>
<p>Then switch W and Y: X W Y Z
Then switch X and W: W X Y Z</p>
<h1>15. If the plane took off at noon and landed at 4, it must have taken 7 hours to make the flight (the 4 hours indicated by the time plus the 3 hour time difference).</h1>
<p>Going west to east, the plane still takes 7 hours, but you have to add the three hour time difference, so if it takes off at noon, it arrives at 10pm.</p>
<h1>19: You could just cross-multiply to get:</h1>
<p>A) ac = bf
B) af = bc
C) af = bc
D) af = bc
E) af = bc</p>
<p>A is the odd man out.</p>
<p>Another way to solve the problem is to make up some numbers for the answer choices and try them:</p>
<p>For answer choice A, try a=6, f=3, b=4, c=2. So 6/3 = 4/2. Now try those numbers in the other answer choices:</p>
<p>B: 3/2 = 4/6 But that's not true! So, you know that the correct answer is either A or B. Let's try the same numbers for C:</p>
<p>2/6 = 3/4 That's not true either! So A must be the answer.</p>
<p>You could try making up numbers that work with B, and then try the others. THe numbers would work with C, D and E, but not A.</p>
<p>Well for #11 I think the best thing to do is just fiddle around... thats how I got it.
First reverse the order of the letters (that gives XYWZ). Then swap W and Y (XWYZ). Then swap W and X (WXYZ). All done in three steps, which is choice B.</p>
<h1>15 is pretty hard. First note that the time difference is 3hrs: PST is 3hrs behind EST. So if the plane takes off noon EST, it is taking off at 9am PST. That plane lands at 4pm PST, so it flew for 7hrs.</h1>
<p>Now, the second plane takes off at noon PST, which is 3pm EST. It flies for the same length of time as the 1st plane (7hrs) and so must land at 10pm EST. Choice A.</p>
<p>In #19 you just have to cross multiply all the choices and see which is the odd one out - A. All the others reduce to af=bc, but A is ac=bf.</p>
<p>QED</p>