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Author Topic: To All You BFM Geniuses  (Read 23580 times)

Offline Racerrr

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To All You BFM Geniuses
« on: July 22, 2009, 12:45:43 AM »
okay this isnt really homework but then again i hope this is the best section to ask if not then feel free 2 move it.. okay so i applied to an engineering school and the department gave me a test that i took.  i still have my scratch people along with notes so i will reconstruct some of the questions asked. some of them are like obvious and stuff but some are not and some are tough that i had no idea. there were 50 questions so here are a few of them.

1) How many unique 10-character password exist, if valid characters consist of only lowercase letters and numbers, and at least one character in the password must be a number?

2) Suppose you are gambling with a friend of yours with a fair game, where the chance of winning and losing are 50/50. If you play 100 rounds, what's the probability you will win 50 times, no more & no less?

3) Write 2.3417417417417... as a fraction. For example, 0.9375 can be written as (30/32).

4) What is the solution to this differential equation: (2x-5y)dx + (-5x+3y^2)dy=0

5) Suppose a dealer for a casino rolls two standard dice. You saw one of the dice; it was a 2. What is the chance that the sum of the face value of the two dice adds up to 6?

6) Find a polynomial function f(x)=ax^2 + bx + c, where the graph passes though points (-1,14), (1,4), and (3,10).

7) Suppose your friend invites you to play a game. He shows you three cups, and exactly one of the cups has the prize. You don't know which one, and you have to tell him which cup you think conceals the prize. He will always reveal one of the empty cups which does not have the prize. So there are exactly two cups left: one with the prize, and one without the prize. He persuades you to switch cup. Should you switch, stay with your cup, or it doesn't really matter at all?

8 ) Suppose the profit a company makes can be model by the function: (X/10)*(1+Y/100)*(2+Z/80) subjected to the constraint that X+Y+Z=4000. What value of X,Y,Z will maximize the company's profit?

9) How many zeros are at the end of 200! (200 factorial)?
« Last Edit: July 22, 2009, 01:03:36 AM by Racerrr »

Offline BFM_NavyJHk

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Re: To All You BFM Geniuses
« Reply #1 on: July 22, 2009, 01:30:46 AM »
Quote
7) Suppose your friend invites you to play a game. He shows you three cups, and exactly one of the cups has the prize. You don't know which one, and you have to tell him which cup you think conceals the prize. He will always reveal one of the empty cups which does not have the prize. So there are exactly two cups left: one with the prize, and one without the prize. He persuades you to switch cup. Should you switch, stay with your cup, or it doesn't really matter at all?

Stay with the same one you originally chose.  If you switch cups your probability goes down to 1/3.  If you stay the same, your probability is still 2/3.  (Or something like that).

I feel stupid now... I had it backwards  ;D ;D
« Last Edit: July 22, 2009, 02:47:54 PM by BFM_NavyJHk »
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Offline jim360

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Re: To All You BFM Geniuses
« Reply #2 on: July 22, 2009, 04:34:30 AM »
Give me some time and I'll try to answer as many of these as possible. Here's the answer to number 9:

Quote
9) How many zeros are at the end of 200! (200 factorial)?

Simply go through the numbers 1 - 200 and find all the ones that divide by ten. these are 10, 20, 30... 100 ... 190, 200. there are 22 tens in these numbers so there are 22 zeros at the end of 200!.

and to number 2:

Quote
2) Suppose you are gambling with a friend of yours with a fair game, where the chance of winning and losing are 50/50. If you play 100 rounds, what's the probability you will win 50 times, no more & no less?

Using the combination formula gives the answer as 50!*50!/100! which is probably too high for most calculators to give the exact answer, but is close to 9.9*10-30.

Also, navy's answer to number 7 is exactly wrong. You should always switch which gives probability 2/3 of winning (so why your friend's asking you to do it is beyond me). This is because you had a 2./3 chance of picking the wrong cup to start with. Google "The Monty Hall Problem" for full details.
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Offline ·WídgêT·

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Re: To All You BFM Geniuses
« Reply #3 on: July 22, 2009, 05:16:44 AM »
Also, navy's answer to number 7 is exactly wrong. You should always switch which gives probability 2/3 of winning (so why your friend's asking you to do it is beyond me). This is because you had a 2./3 chance of picking the wrong cup to start with. Google "The Monty Hall Problem" for full details.

That's not what the movie 21 tells us :P

I could probably do 1 or 2 of these but I need time to work them out myself, and since I don't have time I won't try any just yet


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Offline BFM_Edison

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Re: To All You BFM Geniuses
« Reply #4 on: July 22, 2009, 05:30:58 AM »
1. 10(10)(26^9) This is wrong
2. Already done.
3. 23394/9990
4. -5xy + x2 + y3
5. 1/6
6. 2x2 - 5x + 7 (used matrix row reduction)
7. Already done.
8. This would take longer than the others, so I don't feel like doing it at 6:30 am.
9. You didn't quite do this correctly, 360. You forget that the 2 and 5 would make another zero, and there are other pairs like that as well. Would take longer as well. I'll just do it quickly even though it may be wrong by assuming each ending with 5 will have a corresponding multiple of 2, which would give 20 more I believe, giving 42 zeros.

Edit:
1. I'll just do this in summation notation to the best of my abilities.
Summation from n = 1 to n = 10 (n will essentially be number of numbers) of (10 C n)(10n)(2610 - n)
« Last Edit: July 22, 2009, 05:38:17 AM by BFM_Edison »
52.87   60.07   46.40   72.73   68.23   55.10   98.27   84.73

Offline Racerrr

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Re: To All You BFM Geniuses
« Reply #5 on: July 22, 2009, 09:36:04 AM »
wow thank you for the quick reply

i suspected #7 was tricky so i guess the correct answer lol.

edson... how u compute the ans for #3 and #4?


and if any1 can work out #8 also, and is there a formula to solve it?

Offline BFM_Edison

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Re: To All You BFM Geniuses
« Reply #6 on: July 22, 2009, 02:20:02 PM »
Number three is basically done by utilizing the repeating aspect of the .0417417417. For repeating stuff like that, it's just over something like 999 or 9 or 99, etc. So take 417/999 for .417417, then divide by 10 for .0417417417 and 417/9990. Then you just multiply 2.3 by 9990 and add it to 417.

Number four was a pretty simple differential equation, so I really just did it by inspection.

As for number 8, which I didn't do, I feel tempted to use Lagrangian Multipliers.
52.87   60.07   46.40   72.73   68.23   55.10   98.27   84.73

Offline Goalie

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Re: To All You BFM Geniuses
« Reply #7 on: July 22, 2009, 02:39:06 PM »
8. Yeah, having just taken Calc 3, that's a definite use of LaGrange Multipliers.  Though I can't remember how you do that.

What school was this for, may I ask?  I haven't heard of entrance exams like that around here.
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Offline Racerrr

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Re: To All You BFM Geniuses
« Reply #8 on: July 22, 2009, 05:30:14 PM »
Number three is basically done by utilizing the repeating aspect of the .0417417417. For repeating stuff like that, it's just over something like 999 or 9 or 99, etc. So take 417/999 for .417417, then divide by 10 for .0417417417 and 417/9990. Then you just multiply 2.3 by 9990 and add it to 417.


As for number 8, which I didn't do, I feel tempted to use Lagrangian Multipliers.

ahaa, i was like trying to figure out how to make something repeating... wow can't believe it was that simple. ty for the explanation. Can you calculate and tell me the answer for #8, also please?


Goalie, its some nonofficial benchmarking kind of thing to see where u stand and see if u need any leveling work. hopefully i did okay but i dont think that is the case but u never no.

Offline BFM_Edison

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Re: To All You BFM Geniuses
« Reply #9 on: July 22, 2009, 06:13:43 PM »
This should explain enough for you to try it out.
52.87   60.07   46.40   72.73   68.23   55.10   98.27   84.73

Offline Racerrr

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Re: To All You BFM Geniuses
« Reply #10 on: July 23, 2009, 09:49:33 AM »
sry im dyslexic when it comes to reading somethin with bunch of math terminology, symbols, etc. cant figure out what that wiki thing is sayin. so how many steps does it take to do a problem like that and how much time for an avg. person? and how long u think it will take u to do it?

Offline BFM_Edison

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Re: To All You BFM Geniuses
« Reply #11 on: July 23, 2009, 05:39:19 PM »
I'd have to review it really quickly, but I wouldn't think that it would take terribly long.
52.87   60.07   46.40   72.73   68.23   55.10   98.27   84.73

Offline jim360

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Re: To All You BFM Geniuses
« Reply #12 on: July 24, 2009, 12:28:32 AM »
Whoops, Edison, I didn't pay enough attention there. Oo-er, all the 2*5 pairs... Also there  are more fives buried in 25, 125, 200, etc., which means a bit more counting.

Well, 22 zeros so far from all the tens, then let's count all the fives (it's fairly safe to assume that we will have more than enough 2's for all these extra 5's I'd have thought since 64 and 128 give 13 2's alone). Exhaustive counting gives us 27 fives since 25, 75, 125, and 175 give 5 more fives over what you counted, while also we need to look at 150 and 50 which have an extra factor of 5 in each. So that allows for 27 more zeros or 49 in total.
« Last Edit: July 24, 2009, 12:34:12 AM by BFM_three60 »
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Offline Titanium»X«

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Re: To All You BFM Geniuses
« Reply #13 on: July 24, 2009, 01:57:32 AM »
I'm graduating with a Bachelor of Science in Mathematics this year, so hopefully I can help you with your questions.


Quote
1) How many unique 10-character password exist, if valid characters consist of only lowercase letters and numbers, and at least one character in the password must be a number?

1. It's a combinatorics (counting) problem. You basically calculate all the possible passwords that can be generated with 36 (26 letters + 10 digits) characters and then subtract the total number of passwords that consist of only letters. So the answer is 36^10-26^10. It's a lot easier than using summations and combinations to add up the sum of all possible cases like Edison did. Edison, you make something easy.... so complicated  :LOL: I have my friends that are like that.

2. Going to disagree with the answer. It should be about 7.959%, not almost 0%.  Each round, you have a 50-50 chance, so if you play a total of 100 rounds, then there are 2^100 unique possible outcomes, each equally likely with a probability of 1/(2^100). Since the person wins exactly 50 times, you have the multiply 1/(2^100) by the number of ways you can get exactly 50 wins out of 100 rounds, which is isomorphic to a combination problem. So the answer is 1/(2^100) * 100C50

Answer in LaTeX generated by mathurl.com:


3. Done by Edison

4. Done by Edison.

5. Going to disagree with Edison. It should be higher than 1/6. It's a Game Theory question where knowing a piece of information will dynamically change the probability.  It also has basis in probability theory. I computed it to be 18.18%.

6. Done by Edison

7. Done by BFM_NavyJHk... backward.

8. @Racerrr.... there's not really a "formula" but it has only one simple constraint so it's easier to just set Fx=Fy=Fz=0 and solve for x,y,z instead of using Lagrange Multipliers. There are other easier methods if you ever take a course in Optimization Theory.  If you still need a step-by-step guidance for this problem or any other, feel free to let me know.

9. I computed it to be 49 zeroes.... Number Theory certainly helps.


I'm still a n00b in math though, so if anyone thinks any of my answers are incorrect, then let me know.



Offline BFM_Edison

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Re: To All You BFM Geniuses
« Reply #14 on: July 24, 2009, 02:43:43 AM »
On number 5, if you have two die, and one is a 2, there's only one possible scenario where you could get a total value of 6, which is with a 4 on the other die. The die are mutually exclusive so the first die has no impact on the second, making the chance 1/6, or the chance that it was a 4.

On the first one, I mainly just had misread it first off, thinking only one number, so I ended up just doing for two numbers and so on >.>

I knew something was missing in 360's for number two >.> It is 100 C 50 over 2100.

Figured there was a number theory way for 9.
52.87   60.07   46.40   72.73   68.23   55.10   98.27   84.73

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