Math

Ugh...so, I figure I am going to start documenting the errors that I make on the math portion of the GRE.

One thing that I notice is that I tend to panic when I don't recognize what is being asked in a particular question.  For instance, I know absolutely nothing about 3-D figures, or solids.  I don't know what a diagonal is, and I don't know how to find an area.  Although, I think it may be s^3.  Whatever it is, I panic when I am confronted by a question like that...

I am able to draw 3D cubes and rectangles; and so I do.  And then I label the sides.  I look at it, and I wonder, what is a diagonal on this image.  It's a definition I just don't know the meaning of...and so I panic...I make something up and pick a answer.  What am I supposed to do?  I read the answer on the practices I'm doing and walk away...if I am confronted by that exact same question on the test, I'd get it correct...but, that's highly unlikely, and it seems like a very poor way to study.

In the back of my mind, I am thinking...well, if only I could get more organized.

Things seem simply when they are axiomatically stated.  For instance, the sum of the lengths of any two sides of a triangle is greater than the length of the other side.  Ok, so, if I have a triangle with sides a, b, c and I want to compare the quantity a + b + c to 2b, I could reason that if the sum of the length of any two sides of a triangle is greater than the length of the other side, then when two sides of a triangle are added together, and compared to a third side, the added two sides will be larger, but if I double the third side, what follows?  The added two sides will be a + c > b and the third side would be 2b.  Now, distribute a 2 to all sides of the equation 2a + 2c > 2b, what follows?  Not much of help.  If the equation said that a + c > 2b, I'd be money.  But, it doesn't.  I suppose I could try to -2b from one side, and get 2a + 2c - 2b > 0 and then I could factor out the 2, leaving me with 2(a + c - b) > 0, but that doesn't do me any good either.

So, here is another one of my problems when it comes to the GRE.  I can manipulate the elements of an equation and hope to hit upon a solution, but what sort of thing am I doing if I am not driven by a purpose?  The purpose should be to get either 2b > (a  + b + c) OR  (a + b + c) > 2b OR  (a + b + c) = 2b OR come up with a proof that shows that the two quantities are ambiguous.

So, list what the end result should look like...
2b > (a  + b + c) v (a + b + c) > 2b v (a + b + c) = 2b v ambiguous
Assume that...
a + c > b since we're dealing with a rule that says that the sum of two sides of a triangle is greater than the third side alone.  
Ok, then, what?  How can I make (a + c) > b look like one of the end results....
I could add b to both sides:  a + c + b > b + b; thus, (a + b + c) > 2b

So, if I know the rule that is being tested in the GRE, then I could work with this sort of strategy.  I can list out the possible solutions, and the rule.  And then, I could ask myself, how can I may the rule look like one of the outcomes.  Really, rules seem to be the wrong way to think about these simple statements, statements like the sum of two sides of a triangle is greater than the third side alone; instead, let's call them facts.  As facts, they serve as my true premise.  Insofar as I use valid reasoning, I ought to be able start with a true premise and reason to a true conclusion.  Since the GRE supplies only four possible true conclusions, all I need to know is the fact that they want me begin with...

Take another example,
x/y = z/4; and all the variables are positive.
Compare 6x and 2yz
Well, I know that positives divided by positives are positive.  I know that an integer divided by fraction always equals a larger integer.  For instance, 4/1/2 = 8 and so on.  But, we're getting ahead of ourselves.  What is the true conclusion that I am supposed to reason to find?  There are four possibilities.  Those possibilities are...
A) 6x > 2yz or  6x < 2yz or 6x = 2yz or ambiguous
So, what do I know...
I know that an integer divided by a fraction equals a bigger integer; and I know that a fraction divided by an integer equals a smaller fraction...
But, that's not exactly helpful information for me.
So, what is being tested here?
How can I make x/y = z/4 look like one of my true conclusions?
I could try to make it look like the one with the equal sign.  How?

Multiple both sides by 6...
6(x/y) = 12yz

Divide each side by 6
x/y = 2yz

Divide x/y, a fraction, by 2, an integer...
x/y/2=x/y*1/2=x/2y
x/2y=yz....

But, wait....
When I had...x/y = 2yz, I can substitute in x/y in my conclusions...

So, either 6x > x/y or  6x < x/y
And we know that x/y = z/4
So, 6x > z/4 or 6x < z/4
So, 6x > x/y or  6x < x/y
So, 6x > 2yz or 6x < 2yz

I need these...x/y = z/4 = 2yz...to look like something above....

The problem is that I have an integer on one side and fractions on both sides....

So, the question is asking me to compare 6x to quantities that I know little about....
x/y=z/4= (4x=yz)
8x=2yz
6x<2yz

So, the problem I have in this question doesn't appear to be capable of being solved in the way that I solved earlier.  Or at last, there is something that I missed in trying to solve it.  I have the vague idea that I have true conclusions, and I need to validly reason to only one of them, but I'm not sure of how to go about doing that....something is missing...I know not what.....

Another....
1/x > 1/x^2;  x ~= 0
x or 1/4
x>1/4 v x<1/4 v x=1/4 v ambiguous
What's the easy way to do this?
Well, for starters, I feel like I am dealing with three different things.
I could pick numbers.
That's about the only thing I feel like i know how to do here...
4>8 is false....so x can't be 1/4
But, we know if x was a fraction, that a fraction squared will always be less than the original, and an integer divided by fraction will always produce a larger integer, therefore, an integer divided by a fraction squared will result in a larger integer had the fraction not be squared.
Proof.....2/1/4=8 and 2/1/4^2=2/1/16=32
So, if x were a fraction, then it would turn out that the original statement is false; but the GRE is not giving us false information; therefore, we can assume that x is not a fraction.  And if x is not a fraction, and x ~=0, then it would have to be the case that x was bigger than 1/4.  Therefore, A would be correct.

Ok, here's something ratios....
I don't know shit about ratios....
If for every 1 freshman I have 3 of sophomores, we say this is a ratio of 1F to 3S.
Apparently, I can set up a fraction with variables.  f/s=1/3
Simp. 3f=s
Now, if for every 3 sophomores there are 5 juniors....we say this is a ratio of 3S to 5J
s/j=3/5 = 5s=3j
And, 5(3f)=3j =15f=3j =5 freshmen to 1junior
What is the ratio of freshmen to juniors?
F
SSS
JJJJJ

Another....
55 student
4 boys to 7 girls
How many girls do I need to add to make the ratio become 1 to 2?
First, how many boys and how many girls are there?
For every 11 students, there will be 4 boys and 7 girls.  If the group is 55 students, than there will be
5 x 11 = 55, 5 groups of students.  And if there are 7 girls in each group there will be 35 girls and 20 boys, which equals out to 55 students.  If I want to ratio of boys to girls to be 1 to 2, then if I add 5 more girls to the class, I will get 20 boys to 40 girls, which reduces to 1 to 2....

Fuck man....I'm totally and utterly overwhelmed right now...