Always use scratch paper and write down the givens, etc. It is really the only way to become proficient.

I was working on that today at my job.

Oh and Uber, not sure if you have seen this or not.

http://www.ets.org/portal/site/ets/menuitem.1488512ecfd...CM10000022f95190RCRD

I am looking through them just to make sure I am comfortable with the topics

My GRE's = Greatly Reduced Expectations.

OK I have been trying to wrap my head around this for two days.

see the attached pic.

a 20 foot ladder leaning against a vertical wall with the base of the ladder 10 feet from the wall is pulled 2 feet further out from the wall, causing the top of the ladder to drop x feet.

which is larger x or 2 or are they the same?

the answer is 2 is larger, but I can't figure out how without changing the length of the hypotenuese the perimeter of the triangle can change.

untitled.JPG (13 Kb, 10 downloads)

Fuck off!

But when you do the math with the pythagoreum theorem they are right.

The hypotenuse is the only fixed thing dimension in the problem. In the extreme case where 1 side approaches 0, the perimeter approaches 40.

I guess also the extreme case could be viewed as the perimeter having a value of 20 too.

I understand that algebraically it changes the perimeter but I just can't figure out how in geometry this would actually decrease it.

I just need to stop thinking about it

quote:

Originally posted by editengine:
Oh and Uber, not sure if you have seen this or not.

http://www.ets.org/portal/site/ets/menuitem.1488512ecfd...CM10000022f95190RCRD

I am looking through them just to make sure I am comfortable with the topics

I had not, thanks much.

You should not worry about why the answer is right, the GRE does not care. the GRE is spock but with a standardization racket going on.

Bastards.

So, what is the consensus on the writing sections? Take the topic, come up with three arguments for or against, lead with the best, summarize at the end?

How fun.

quote:

Originally posted by editengine:
OK I have been trying to wrap my head around this for two days.

see the attached pic.

a 20 foot ladder leaning against a vertical wall with the base of the ladder 10 feet from the wall is pulled 2 feet further out from the wall, causing the top of the ladder to drop x feet.

which is larger x or 2 or are they the same?

the answer is 2 is larger, but I can't figure out how without changing the length of the hypotenuese the perimeter of the triangle can change.

Equations for the two triangles. y is the height of the ladder on the wall after being pulled out 2 feet.

10^2 + (x + y)^2 = 20^2
12^2 + y^2 = 20^2

Thus:
100 + x^2 + 2xy + y^2 = 144 + y^2
x^2 + 2xy = 44

Now, if x = 2 then 4 + 4y = 44 -> y = 10

So if x < 2, y > 10 and if x > 2, y < 10

Using the second equation y^2 = 400 - 144 = 256 -> y = 16

So x < 2.

For the question of how the perimeter of a triangle can change without changing the hypotenuse, consider the 20 foot ladder laid flat on the ground: perimeter 40 (0 feet off the ground on the wall, 20 feet out from the wall at the bottom). Now, put it at exactly a 45 degree angle. The perimeter is obviously larger than 40, isn't it? When traveling around the triangle you now have to make that huge detour to the corner where the wall meets the floor, where, when it was laid flat, you could just to travel along the other side of the ladder.

quote:

Originally posted by colin:

quote:

Originally posted by editengine:
OK I have been trying to wrap my head around this for two days.

see the attached pic.

a 20 foot ladder leaning against a vertical wall with the base of the ladder 10 feet from the wall is pulled 2 feet further out from the wall, causing the top of the ladder to drop x feet.

which is larger x or 2 or are they the same?

the answer is 2 is larger, but I can't figure out how without changing the length of the hypotenuese the perimeter of the triangle can change.

Equations for the two triangles. y is the height of the ladder on the wall after being pulled out 2 feet.

10^2 + (x + y)^2 = 20^2
12^2 + y^2 = 20^2

Thus:
100 + x^2 + 2xy + y^2 = 144 + y^2
x^2 + 2xy = 44

Now, if x = 2 then 4 + 4y = 44 -> y = 10

So if x < 2, y > 10 and if x > 2, y < 10

Using the second equation y^2 = 400 - 144 = 256 -> y = 16

So x < 2.

Ah, you, too, have the fucking-shit-boring-pointless-math-you-desperately-do-not-need-to-function-in-greater-society questions, too.

I pretty much flunked my QCS...mostly out of laziness (I spent the study week playing UT) but also for utterly huge erection I got from the idea that were my score low enough, it would lower the scores of the over-achievers around me. All hail the bell curve!

Didn't do too shabbily on the writing bits.

'Course, QCS is out highschool final exam, not some entry to a postgrad course. You spend five years preparing for 'em, then realise that, come next February, the whole thing is about as useful as priest's cock, and that holding down a job for five minutes gives you greater qualifications than any school mark.

I really don't think the world works that way, Lith. More's the pity, but what we might want to count in life and what largely does (at least within the confines of 1st world civilization participation)are not often the same.

Well, I wouldn't restrict it to math. Pretty much everything you learn in high school is useless.

quote:

Originally posted by colin:
Well, I wouldn't restrict it to math. Pretty much everything you learn in high school is useless.

You still have to jump through hoops after though, and they all start back then. Earlier even, in the warehouses they store us in from K-8.

Sort of true for college. Graduating though at least shows you had enough discipline to do that. Or at least it did. Grade inflation may have killed that.