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GR0568 Questions 4, 11, 13

Posted: Sun Oct 04, 2009 12:07 pm
by Quakerbrat
Question #4

Which of the following circles has the greatest number of points of intersection with the parabola x^2 = y + 4?

I thought about this intuitively, thinking that a circle with radius 4 would hit the parabola at its x intercept as well as crossing the parabola at 2 other points, but clearly I was wrong.

Is there a concise mathematical way to solve this?

Question #11

Of the following, which is the best approximation of sqrt(1.5)*(266)^(3/2)?

a) 1,000 b) 2,700 c) 3,200 d) 4,100 e) 5,300

I thought it would simple to use linear approximation on this, setting f(x) = sqrt(x) * 266 ^ (x), but I am having trouble and cannot arrive at a good answer. Is the trick to make 266 a y variable and make the equation f(x,y) = sqrt(x) * (y)^x?

Question #13

A total of x feet of fecing is to form three sides of a level rectangular yard. What is the maximum possible area of the yard, in terms of x?

I thought about it this way:

Area = W x L, from the constraints I know that 2W + L = x or 2L + W = x. Unfortunately, that's about as far as I get.

Re: GR0568 Questions 4, 11, 13

Posted: Sun Oct 04, 2009 12:39 pm
by isih
for #11

$$\sqrt{\frac{2}{3}}266^{\frac{3}{2}}=\sqrt{\frac{3}{2}266 (266)^{2}}=\sqrt{399 (266)^{2}}\simeq 20(266)\simeq5300$$

Re: GR0568 Questions 4, 11, 13

Posted: Sun Oct 04, 2009 12:58 pm
by diogenes
Quakerbrat wrote:Question #4

Which of the following circles has the greatest number of points of intersection with the parabola x^2 = y + 4?

I thought about this intuitively, thinking that a circle with radius 4 would hit the parabola at its x intercept as well as crossing the parabola at 2 other points, but clearly I was wrong.

Is there a concise mathematical way to solve this?


Question #13

A total of x feet of fencing is to form three sides of a level rectangular yard. What is the maximum possible area of the yard, in terms of x?

I thought about it this way:

Area = W x L, from the constraints I know that 2W + L = x or 2L + W = x. Unfortunately, that's about as far as I get.
For #4: Just draw the circles and count the number of "hits." Edit: For the circle with radius 4 the trap is thinking that one hit is actually two. So, take a look at radius three, too.

For #13: You are really close and just need to write down the equation to take the derivative, set to zero, and then solve. Edit: Set A= w*l, and x = 2w+l, as you have. Now l = x-2*w, so A = w*(x-2*w). So, dA/dw = wx-2*w^2.
Set dA/dw = 0 and solve for w, then l to get the optimal points. Plugging back into A will give you the solution.

Re: GR0568 Questions 4, 11, 13

Posted: Mon Oct 05, 2009 8:01 am
by Quakerbrat
Thanks so much you guys!

Re: GR0568 Questions 4, 11, 13

Posted: Mon Oct 05, 2009 12:38 pm
by zuluyankee
Everytime we have a "which is the best approximation of" question, how do we judge it is best?

To approximate an expression as in the question, we can take various approximations and get different values. A quadratic approximation is generally seen as being better than a linear one, and a qubic is better than a quadratic, etc.

Unless we have finitely many choices as in a GRE Exam, best approximation simply doesn't make sense, generally speaking.

So, in the question given here, there are five choices, and question is, which is best? Unless the exact value is known, 4100 could be "the" best approximation by, say Klix method (I made this up). The method as delineated above happens to hit a given option--5300--perhaps?

How do we decide, then?

Re: GR0568 Questions 4, 11, 13

Posted: Mon Oct 05, 2009 8:48 pm
by diogenes
zuluyankee wrote:Everytime we have a "which is the best approximation of" question, how do we judge it is best?
My understanding of the questions on the GRE Subject are that they tempt you into straight calculations, but are generally questions on if you are able to "feel" what the problem is about, e.g., here it seems to be a simple definition and pattern recognition question. Another good example of this is the series question from this exam (iirc), that has a fraction always less than one cluttering up the "real" series.

So, my metric is if the pattern in the question is clear, then go with the quick calculation.