Three problems from GRE

Forum for the GRE subject test in mathematics.
volodja
Posts: 5
Joined: Mon Jan 18, 2010 5:18 pm

Three problems from GRE

Postby volodja » Wed Apr 07, 2010 3:35 am

1. Let the bottom edge of a rectangular mirror on a vertical wall be parallel to and h feet above the level floor. If a person with eyes t feet above the floor is standing erect at a distance d feet from the mirror, what is the relationship among h,d, and t if the person can just see his own feet in the mirror?

(A) t=2h and d does not matter.
(B) t=4d and h does not matter.
(С) h^2+d^2=\frac {t^2}4.
(D) t-h=d.
(E) (t-h)^2=4d.

2. Let A and B be subsets of a set M and let S_0=\{A,B\}. For i\ge0, define S_{i+1} inductively to be the collection of subsets X of M that are of the form C\cup D, C\cap D, or M-C (the complement of C in M), where C,D\in S_i. Let S=\bigcup_{i=0}^\infty S_i. What is the largest possible number of elements of S?

3. A fair die is tossed 360 times. The probability that a six comes up on 70 or more of the tosses is

(A) greater than 0.50
(B) between 0.16 and 0.50
(C) between 0.02 and 0.16
(D) between 0.01 and 0.02
(E) less than 0.01

EugeneKudashev
Posts: 27
Joined: Tue Apr 06, 2010 8:22 am

Re: Three problems from GRE

Postby EugeneKudashev » Wed Apr 07, 2010 4:04 am

second:

very boring, and I dont see any chances of solving it on the real exam.

you have to just follow the definition to obtain every possible set.
S_1 = \{ A\cup B, A\cap B, M-A, M-B\}
S_2 = \{ A,B,M, A\cup B, A\cap B, M-A, M-B, M-\{A\cup B\}, M - \{A\cap B\}, A\cup B - A, A\cup B - B, M - \{A\cup B - A\}, M - \{A\cup B - B\}, A\cup B - A\cap B, M - \{A\cupB - A\cap B\}, \emptyset \}

this gives you exactly 16 elements, and constructing further S_i does not increase the number. this is plain from a diagram (plot a "big" set M, and two intersecting proper subsets of it, namely, A and B. S_2 already gives you every possible intersection. diagram also could be useful to build the S_2 itself. however, this solution is extremely time-consuming and anxiety on the test will almost certainly prevent you from succeeding. also it is easy to forget including empty set and ending up with answer '15', which is among the choices.

I'd be happy to look at other solutions.
Last edited by EugeneKudashev on Wed Apr 07, 2010 4:05 am, edited 1 time in total.

EugeneKudashev
Posts: 27
Joined: Tue Apr 06, 2010 8:22 am

Re: Three problems from GRE

Postby EugeneKudashev » Wed Apr 07, 2010 4:05 am

third is easy and has already been discussed here: viewtopic.php?f=1&t=256

speedychaos4
Posts: 47
Joined: Mon Mar 22, 2010 2:42 am

Re: Three problems from GRE

Postby speedychaos4 » Wed Apr 07, 2010 4:39 am

EugeneKudashev wrote:second:

very boring, and I dont see any chances of solving it on the real exam.

you have to just follow the definition to obtain every possible set.
S_1 = \{ A\cup B, A\cap B, M-A, M-B\}
S_2 = \{ A,B,M, A\cup B, A\cap B, M-A, M-B, M-\{A\cup B\}, M - \{A\cap B\}, A\cup B - A, A\cup B - B, M - \{A\cup B - A\}, M - \{A\cup B - B\}, A\cup B - A\cap B, M - \{A\cupB - A\cap B\}, \emptyset \}

this gives you exactly 16 elements, and constructing further S_i does not increase the number. this is plain from a diagram (plot a "big" set M, and two intersecting proper subsets of it, namely, A and B. S_2 already gives you every possible intersection. diagram also could be useful to build the S_2 itself. however, this solution is extremely time-consuming and anxiety on the test will almost certainly prevent you from succeeding. also it is easy to forget including empty set and ending up with answer '15', which is among the choices.

I'd be happy to look at other solutions.


Hey man, let's consider this is this way,
Sets A and B separate M into four distinct parts, namely, A-B, B-A, AnB and M-(AuB), therefore, no matter how you manipulate those sets, the result will be the combination of these four distinct parts, then, it is easy to see the ans is 2^4=16.

EugeneKudashev
Posts: 27
Joined: Tue Apr 06, 2010 8:22 am

Re: Three problems from GRE

Postby EugeneKudashev » Wed Apr 07, 2010 5:01 am

power set (set of all possible subsets) is good here when you already know the answer. unfortunately, on step 1 (having S_1) you do not even have these four distinct parts. so you can't say for sure what you will get. and you can't say for sure that the collection will be finite.

speedychaos4
Posts: 47
Joined: Mon Mar 22, 2010 2:42 am

Re: Three problems from GRE

Postby speedychaos4 » Wed Apr 07, 2010 5:13 am

EugeneKudashev wrote:power set (set of all possible subsets) is good here when you already know the answer. unfortunately, on step 1 (having S_1) you do not even have these four distinct parts. so you can't say for sure what you will get. and you can't say for sure that the collection will be finite.


But don't you think the question could be interpreted as "How many different sets could be generated by A B and M" ?

volodja
Posts: 5
Joined: Mon Jan 18, 2010 5:18 pm

Re: Three problems from GRE

Postby volodja » Wed Apr 07, 2010 5:14 am

Yeah, problem 3 isn't hard if there is a computer at hand... But how to figure out the answer in a fast way? As to problem 1, I think it is very easy but I really don't understand it; perhaps some well-known properties of a mirror are expected to be applied or something like that.
Last edited by volodja on Wed Apr 07, 2010 5:22 am, edited 1 time in total.

EugeneKudashev
Posts: 27
Joined: Tue Apr 06, 2010 8:22 am

Re: Three problems from GRE

Postby EugeneKudashev » Wed Apr 07, 2010 5:16 am

intuitive solution to #2 is perfectly described in the last post in that topic. (D), (E) are too small, (A), (B) are too big. thus, (C).

anyway it's useful to remember brief table for standard normal distribution values.

speedychaos4
Posts: 47
Joined: Mon Mar 22, 2010 2:42 am

Re: Three problems from GRE

Postby speedychaos4 » Wed Apr 07, 2010 5:21 am

volodja wrote:Yeah, problem 2 isn't hard if there is a computer at hand... But how to figure out the answer in a fast way? As to problem 1, I think it is very easy but I really don't understand it; perhaps some well-known properties of a mirror are expected to be applied or something like that.


It is easy if you sketch out the figure, it is just about symmetric.

enork
Posts: 33
Joined: Fri Sep 18, 2009 3:16 am

Re: Three problems from GRE

Postby enork » Wed Apr 07, 2010 11:15 am

yeah, the well known property of a mirror is that it reflects. Reflect your diagram against the plane of the mirror.

volodja
Posts: 5
Joined: Mon Jan 18, 2010 5:18 pm

Re: Three problems from GRE

Postby volodja » Wed Apr 07, 2010 1:40 pm

Thanks for suggestions. :)

ArcTan
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Joined: Sat Feb 27, 2010 10:43 am

Re: Three problems from GRE

Postby ArcTan » Thu Apr 08, 2010 1:22 pm

For question 3, this is a sequence of Bernoulli Trial, meaning a sequence of independent events where, at each step, there a porobablilty p of a success, and a proibability q of a failure. In this case, the probablity of getting a 6, p, is 1/6; and the probability of not getting a 6, q, is 5/6.

There is a theorem that says the mean number if successes (X) in a sequence of Bernoulli trials is np. Here, np=360(1/6)=60. It also says that the standard deviation sd=(npq)^(1/2). Here sd=[360*(1/6)*(5/6)]^(1/2)=(50)^1/2. This is just slightly more than 7.

Technically, this is a discrete random variable, but when n is very large (like 360), we can approximate the distribution with a normal distribution.

So, 70 sixes, which is 70 successes according to this setup, is roughly 1.5 standard deviations above the mean of 60 successes (mean+1.5sd). If you happen to have memorized a table/diagram of the normal distribution (as I suppose we should) then you will know the following...

probability X < mean = 0.5
probability X < mean + 1sd = 0.84
probability X < mean + 2sd = 0.97

This lists the probability P(X) that X is LESS than these amounts. The probability that X is MORE than these amounts is the compliment 1-P(X).

So the correct answer is somewhere between .03 and .16. Hence the answer must be C.

astathis
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Joined: Fri Apr 09, 2010 5:23 am

Re: Three problems from GRE

Postby astathis » Fri Apr 09, 2010 5:43 am

Problem 1 is pretty straight forward. You just need to use a bit of similar triangles.
You can draw a triangle that can be split into two smaller right triangles. It has a shape similar to |>.
The whole left side has length t. The "point" of the arrow is distance d from the left side and h from the bottom. Split the triangle in the middle (connect the point to the left side with a straight line). The line will have length d. The length of the left side above the split is t-h, and the length of the side below the split is h. We now have two similar triangles, the upper right triangle, and the lower right triangle. From these we know that the ratio of the sides are equal, and we get:
(t-h)/d = h/d
since the lengths of the long non-hypotenuse sides of the right triangles are t-h and h respectively, and the d is the short side that they both share.
The d's cancel, and you get:
t=2h, and d doesn't matter. (A).

Hopefully you can follow that. Ideally a picture would help a bunch, but I don't have the ability to make one here...




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