7 questions from 8767
7 questions from 8767
need some questions for the following questions. thanks!
8767
#14 A newscast contained the statement that the total use of electricity in city A had declined in one billing period by 5%. while household use had declined by 4% and all other uses increased by 25%. Which of the following must be true about the billing period?
(A) the statement was in error
(B) the ratio of all other uses to household use was 29/1
(C) the ratio of all other uses to household use was 29/16
(D) the ratio of all other uses to household use was 29/19
(E) None of the above
Answer:A
#38
answer:A, but when I used block matrix multiplication I got C
#58 If f(z) is an analytic function that maps the entire finite complex plane into the real axis, then the imaginary axis must be mapped onto
(A) the entire real axis
(B) a point
(C) a ray
(D) an open finite interval
(E) the empty set
Answer:B
#59
(A)2 (B)7 (C)12 (D)16 (E)21
Answer:C
#60 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
Answer:C
#64 Let S be a compact topological space, let T be a topological space, and let f be a function from S onto T. Of
the following conditions on f, which is the weakest condition sufficient to ensure the compactness of T?
(A) f is a homeomorphism
(B) f is continuous and 11
(C) f is continuous
(D) f is 11
(E) f is bounded
Answer:C
8767
#14 A newscast contained the statement that the total use of electricity in city A had declined in one billing period by 5%. while household use had declined by 4% and all other uses increased by 25%. Which of the following must be true about the billing period?
(A) the statement was in error
(B) the ratio of all other uses to household use was 29/1
(C) the ratio of all other uses to household use was 29/16
(D) the ratio of all other uses to household use was 29/19
(E) None of the above
Answer:A
#38
answer:A, but when I used block matrix multiplication I got C
#58 If f(z) is an analytic function that maps the entire finite complex plane into the real axis, then the imaginary axis must be mapped onto
(A) the entire real axis
(B) a point
(C) a ray
(D) an open finite interval
(E) the empty set
Answer:B
#59
(A)2 (B)7 (C)12 (D)16 (E)21
Answer:C
#60 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
Answer:C
#64 Let S be a compact topological space, let T be a topological space, and let f be a function from S onto T. Of
the following conditions on f, which is the weakest condition sufficient to ensure the compactness of T?
(A) f is a homeomorphism
(B) f is continuous and 11
(C) f is continuous
(D) f is 11
(E) f is bounded
Answer:C
Re: 7 questions from 8767
weilan8 wrote: #60 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
Answer:C
For this one it looks susceptible to transforming it into a standard normal random variable (we have large N here).
Set X to the "the number of times six is observed" and let N[*] be the Normal cdf.
So, we find that Pr[X>=70] =1Pr[X<70] = 1 N[(70mu)/sigma], with mu = (1/6)*360 = 60 and sigma =sqrt((360/6)*(5/6)).
Re: 7 questions from 8767
Hi diogenes,
thank you for your explanation!
Now, I only have two problems in 8767: #58 and #64.
For #64, I know that a continuous function can map a compact space to another compact space and I also know that if a mapping is homomorphism it will remain the topological property. However, I'm not sure if 11 or bounded function can also do that.
thank you for your explanation!
Now, I only have two problems in 8767: #58 and #64.
For #64, I know that a continuous function can map a compact space to another compact space and I also know that if a mapping is homomorphism it will remain the topological property. However, I'm not sure if 11 or bounded function can also do that.
Re: 7 questions from 8767
A continuous map preserves compactness, and this is the weakest property among the first 3 answers. Since the question asks, what is the weakest condition to ensure that T is compact, if you can find counterexamples for answers d and e then you know the answer must be c. Note that at this point you could probably just eliminate the other choices because its hard to say whether 11 or boundedness are "weaker" conditions than continuity. There are 11 maps that arent continuous and bounded maps that arent 11 and so on. They are independent conditions. But here are come counterexamples anyway.
The easiest way to generate counterexamples for this type of problem is to pick a space you know is compact and find a 11 or bounded map from that space to another that you know is not compact. Consider the closed unit disk in R^2, clearly compact, and a bounded map to the open unit disk, not compact, which is the identity everywhere except on the boundary of the closed disk, where it is 0. This is a bounded map but cannot preserve compactness (note it is not continuous). Now since both the closed and open unit disks have the same cardinality, there exists 11 and onto map between them. But that map also cannot preserve compactness since the open disk is not compact.
The easiest way to generate counterexamples for this type of problem is to pick a space you know is compact and find a 11 or bounded map from that space to another that you know is not compact. Consider the closed unit disk in R^2, clearly compact, and a bounded map to the open unit disk, not compact, which is the identity everywhere except on the boundary of the closed disk, where it is 0. This is a bounded map but cannot preserve compactness (note it is not continuous). Now since both the closed and open unit disks have the same cardinality, there exists 11 and onto map between them. But that map also cannot preserve compactness since the open disk is not compact.
Re: 7 questions from 8767
thank you! the explanation is very helpful.

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Re: 7 questions from 8767
There's a very simple way to solve #58. If we write $$f = f_1+if_2$$, with both $$f_1$$ and $$f_2$$ realvalued, then the hypotesis implies that $$f_2\equiv 0$$. By the CauchyRiemann equations, $$\frac{\partial f_1}{\partial x} = \frac{\partial f_2}{\partial y} \equiv 0$$ and $$\frac{\partial f_1}{\partial y} = \frac{\partial f_2}{\partial x} \equiv 0$$, which imply that $$f_1$$ is a constant function. Therefore $$f$$ must be constant, and in particular the imaginary axis must be sent into a single point.
In #64, although the solution given by mk is absolutely correct, I'd like to note that (e) is automatically false since the notion of boundedness doesn't make sense in the context of general topological spaces.
In #64, although the solution given by mk is absolutely correct, I'd like to note that (e) is automatically false since the notion of boundedness doesn't make sense in the context of general topological spaces.
Re: 7 questions from 8767
#58 is in fact even easier, you just have to consider a specific example. f(z) = 0 is certainly analytic, and maps C into (notice the problem does not require onto!) the real axis. And it is simple to see that the image of the imaginary axis is just {0} which is (B) and isn't (A), (C), (D), (E).

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Re: 7 questions from 8767
Prong, you solution is ok (for a GRE test, but not in real life). The only problem I see is that it seems you came to this by knowing that $$f$$ must be constant, and if this is true your solution is no better than mine.
Re: 7 questions from 8767
Oh yes, it is absolutely no use unless someone's given you convenient answer choices to work from.
Re: 7 questions from 8767
For #58
There is a theorem (Open mapping theorem for analytic functions) which states that an analytic function maps open sets to open sets unless it is a constant. Thus the only way that f can map C to R is to be a constant. That is why the image of the imaginary axis has to be mapped to a point.
There is a theorem (Open mapping theorem for analytic functions) which states that an analytic function maps open sets to open sets unless it is a constant. Thus the only way that f can map C to R is to be a constant. That is why the image of the imaginary axis has to be mapped to a point.

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Re: 7 questions from 8767
How can we calculate N[(70mu)/sigma]?just by hand?or we have the distrbution form at the back of our test paper?diogenes wrote:weilan8 wrote: #60 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
Answer:C
For this one it looks susceptible to transforming it into a standard normal random variable (we have large N here).
Set X to the "the number of times six is observed" and let N[*] be the Normal cdf.
So, we find that Pr[X>=70] =1Pr[X<70] = 1 N[(70mu)/sigma], with mu = (1/6)*360 = 60 and sigma =sqrt((360/6)*(5/6)).
Re: 7 questions from 8767
I also wonder this. There is a similar question on one of the other practice tests. I don't really know much probability. I know the general idea behind how one would calculate this, but I have no idea how to do it quick enough for the test and without tables or computers. I have asked this question on another forum, and seen it asked by a couple different people, and have yet to see anyone give a good answer (I have seen several people who actually did know probability say there is no good way to calculate this as a GRE question).breezeintopl wrote:diogenes wrote:How can we calculate N[(70mu)/sigma]?just by hand?or we have the distrbution form at the back of our test paper?weilan8 wrote: #60 A fair die is tossed 360 times. The probability that a six comes up on 70 or more of the tosses is
....
Re: 7 questions from 8767
I agree. This question was outrageous... I knew that I need $$\Phi(\sqrt{2})$$ but well, I do not want to remember useless numbers... So I knew that it was either (C) or (B) and I left this question blank. I may say that I know probability well, and there are certain hints  like the area under $$(\sigma, \sigma)$$ which in means of standard normal distribution is $$(1, 1)$$ is about 66%. It is symmetrical so this helps too. But besides that, the best way I can imagine is to compute the integral with trapezoidal rule which is time consuming and almost impossible in test conditions. Maybe the best way is to memorize some quantiles of the normal distribution like Leduc suggests in Cracking the Gre.mrb wrote:I also wonder this. There is a similar question on one of the other practice tests. I don't really know much probability. I know the general idea behind how one would calculate this, but I have no idea how to do it quick enough for the test and without tables or computers. I have asked this question on another forum, and seen it asked by a couple different people, and have yet to see anyone give a good answer (I have seen several people who actually did know probability say there is no good way to calculate this as a GRE question).
Re: 7 questions from 8767
That seems to be the thing to do. I made flash cards for $$\Phi(1), \Phi(2), \Phi(3).$$ The normal approximation to the binomial distribution appears frequently on old GREs. If there's a classier way to do these problems, I don't know it, and memorizing a few values will work fine.mtey wrote:Maybe the best way is to memorize some quantiles of the normal distribution like Leduc suggests in Cracking the Gre.
Re: 7 questions from 8767
I also wonder this. There is a similar question on one of the other practice tests. I don't really know much probability. I know the general idea behind how one would calculate this, but I have no idea how to do it quick enough for the test and without tables or computers. I have asked this question on another forum, and seen it asked by a couple different people, and have yet to see anyone give a good answer (I have seen several people who actually did know probability say there is no good way to calculate this as a GRE question).[/quote]
there is a easier way to solve this problem, by intuition.
the probability that 6 comes 70 times is exactly less than 1/6
the probability 0,5 coudnt be
after these statements two answer are frown, then by intuition implying that 0,010,02 is too little chose answer C
there is a easier way to solve this problem, by intuition.
the probability that 6 comes 70 times is exactly less than 1/6
the probability 0,5 coudnt be
after these statements two answer are frown, then by intuition implying that 0,010,02 is too little chose answer C

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Re: 7 questions from 8767

Last edited by Adarsh Raj on Sun Sep 18, 2011 10:39 am, edited 1 time in total.

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Re: 7 questions from 8767
Here are two more proofs for #58:
Use Picard's Little Theorem ( http://en.wikipedia.org/wiki/Picard_theorem )
Little Picard says that the image of a nonconstant entire function omits at most one point.
Our function is entire and omits all the good parts of the complex plane. Hence it is constant.
Use Liouville's Theorem ( http://en.wikipedia.org/wiki/Liouville% ... _analysis) )
Liouville says that any bounded entire function is constant. To apply it, compose f(z) with g(z) =1/(zi) .
A composition of analytic functions is analytic, and g is analytic on the Real line. Liouville says that g(f(z)) is constant, so we can deduce f(z) is constant.
Use Picard's Little Theorem ( http://en.wikipedia.org/wiki/Picard_theorem )
Little Picard says that the image of a nonconstant entire function omits at most one point.
Our function is entire and omits all the good parts of the complex plane. Hence it is constant.
Use Liouville's Theorem ( http://en.wikipedia.org/wiki/Liouville% ... _analysis) )
Liouville says that any bounded entire function is constant. To apply it, compose f(z) with g(z) =1/(zi) .
A composition of analytic functions is analytic, and g is analytic on the Real line. Liouville says that g(f(z)) is constant, so we can deduce f(z) is constant.
Re: 7 questions from 8767
Here's something almost feasible:mrb wrote: I also wonder this. There is a similar question on one of the other practice tests. I don't really know much probability. I know the general idea behind how one would calculate this, but I have no idea how to do it quick enough for the test and without tables or computers. I have asked this question on another forum, and seen it asked by a couple different people, and have yet to see anyone give a good answer (I have seen several people who actually did know probability say there is no good way to calculate this as a GRE question).
We use the Central Limit Theorem "backwards". The Binomial(n, 0.5) distribution is approximately N(n/2, n/4), and we can explicitly calculate the cdf for the binomial distribution.
Take e.g. n=4.
The 4th row of Pascal's triangle is
1 4 6 4 1
We're looking for roughly P(Z<1.4), for Z a standard normal random variable. This corresponds to a Bin(4, 0.5) random variable being less than 3.4. Using the cumulative sums above, and the crude discreteness correction of "smearing out" the point mass on k uniformly to [k0.5,k+0.5] we get the approximate answer
(1 + 4 + 6 + 4*0.9) / 16 = 0.9125
which is surprisingly (to me) close to the true value of 0.9214, and certainly good enough to answer the problem.

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Re: 7 questions from 8767
For #38:
M^3=I
So:
M^100=M.I=M
M^3=I
So:
M^100=M.I=M

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Re: 7 questions from 8767
For #60:
We have estimated a binomial distribution with a normal distribution. So why the Chebyshev inequality cannot provide rough estimation on the probability of the required range?
We have estimated a binomial distribution with a normal distribution. So why the Chebyshev inequality cannot provide rough estimation on the probability of the required range?
Re: 7 questions from 8767
hadimotamedi wrote:For #60:
We have estimated a binomial distribution with a normal distribution. So why the Chebyshev inequality cannot provide rough estimation on the probability of the required range?
I tried this question using the Chebyshev inequality, the bound is not strong enough, didn't give you much information given the choices provided.
For this question, I need to make use of the following facts:
P(Z>3) = 0.2%
P(Z>2) = 2.4%
P(Z>1) = 16%
P(Z>0) = 50%
I am NOT 100% sure if I applied the Chebyshev inequality correctly, so pls correct me if I am wrong.
Thanks

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Re: 7 questions from 8767
Hi,
Could you also shed some light on Q 59.
Thanks
Could you also shed some light on Q 59.
Thanks
Re: 7 questions from 8767
Adarsh Raj wrote:Hi,
Could you also shed some light on Q 59.
Thanks
Sure, first of all, the correct answer should be D: 16. It goes as follows:
f(1)  f(0) = 1
f(2)  f(1) = 1
f(5)  f(2) = 5
f(8)  f(5) = 7
f(12)  f(8) = 2
You add all these up, you will get 16.

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Re: 7 questions from 8767
Excuse me, for the Chebyshev inequality , I mean what outlined as :
http://mathworld.wolfram.com/ChebyshevInequality.html
It seems that it must provide fair approximation for the mentioned case.Isn't it?
http://mathworld.wolfram.com/ChebyshevInequality.html
It seems that it must provide fair approximation for the mentioned case.Isn't it?

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Re: 7 questions from 8767
I think you're missing the point. Chebyshev only provides a bound, not an "approximation". It is useless here unless it tells you the probability is less than 0.01
Re: 7 questions from 8767
Pls give it a try, and see if you can solve it using Chebyshev inequality. Pls share with us if you have other ways to solve it.hadimotamedi wrote:Excuse me, for the Chebyshev inequality , I mean what outlined as :
http://mathworld.wolfram.com/ChebyshevInequality.html
It seems that it must provide fair approximation for the mentioned case.Isn't it?
Re: 7 questions from 8767
20second solution to #38 using a random fact: Any element of the symmetric group on n elements can be represented by an nxn matrix with exactly one 1 in each row/column and zeros everywhere else. Knowing this, one can recognize M as a nonidentity element of the symmetric group on 3 elements, and deduce that either $$M^2 = I$$ or $$M^3 = I$$. Computing the first entry of $$M^2$$ reveals that it is not $$I$$, so it must be that $$M^3 = I$$, so $$M^{100} = M^{99}M = IM = M$$.
Re: 7 questions from 8767
I think we can solve it by approximation to normal distribution:weilan8 wrote:need some questions for the following questions. thanks!
#60 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
Answer:C
$$\mu=np=60$$
$$\sigma^2=npq=360\times 1/6 \times 5/6=50$$
$$\sigma=\sqrt{50}\approx 7$$
Hence, 70 is between 1 s.d. and 2s.d. away from the mean (to the right of mean)
Using the "689599.7 rule http://en.wikipedia.org/wiki/689599.7_rule", it is approximately 17/2=8.5%, which gives (C)