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The Math Game :D
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| ODST |
Posted on 09/09/2013 07:44:04
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Is it, idk, 60mph? lol your posts got hidden under me! lol
�RG� (Renegade Gamers) Senior Admin
[C�R] Veteren
^Don't mess with ODST 
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| ninja |
Posted on 09/09/2013 08:08:53
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Posts: 4174
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ODST wrote:
Is it, idk, 60mph? lol your posts got hidden under me! lol
Nope, it's not 60mph! I'm asking for the average speed over the journey, not just the (mean) average of the two numbers.
My brain is open.
- Paul Erdős |
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| BuhBiggieback |
Posted on 09/09/2013 10:43:23
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Wanker
Posts: 582
Joined: 18.07.13
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How can we ever expect to win over a soon to be math major 
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| ninja |
Posted on 09/09/2013 12:56:45
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Posts: 4174
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BuhBiggieback wrote:
How can we ever expect to win over a soon to be math major
It's the taking part that counts!
If noone gets it by tomorrow, I'll post a different question ...
My brain is open.
- Paul Erdős |
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| BuhBiggieback |
Posted on 09/09/2013 13:19:18
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Wanker
Posts: 582
Joined: 18.07.13
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180mph??
2x60x70/70+60
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| ninja |
Posted on 09/10/2013 11:25:49
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Posts: 4174
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BuhBiggieback wrote:
180mph??
2x60x70/70+60
No, but close. Your formula looks like one correct way of working out the answer, but it has the wrong numbers in it (making me think you copy and pasted it), and you evaluated it wrong.
My brain is open.
- Paul Erdős |
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| ninja |
Posted on 09/12/2013 06:47:34
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Super Admin
Posts: 4174
Joined: 15.11.10
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Looks like noone's going to get this one ...
ninja wrote:
Yesterday I drove from home to a shop/store, and then back again. My average speed on the way there was 50mph. My average speed on the way back was 70mph. What was my average speed over the whole journey?
Hint: Average Speed = (Total Distance) / (Total Time Taken)
(One possible) Solution:
To find the overall average speed, I need to know the total distance and the total time of the journey.
Let the distance from home to the shop be d.
Therefore, the time taken on the first part of the journey is d/50.
Similarly, the time taken on the second part of the journey is d/70.
So, the overall time taken is d/50 + d/70 while the total distance is d + d = 2d.
Therefore the average speed over the whole journey is 2d/(d/50 + d/70) = (2*50*70)/(70+50) = 58.333... mph.
In fact, it turns out that the answer is the harmonic mean of the two speeds - this is because they're rates.
----
New question ...
What is the sum of the first 5 prime numbers?
My brain is open.
- Paul Erdős |
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| ODST |
Posted on 09/12/2013 08:09:43
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Super Admin
Posts: 2047
Joined: 11.09.12
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2+3+5+7+11 = 28
Edited by ODST on 09/12/2013 08:27:30
�RG� (Renegade Gamers) Senior Admin
[C�R] Veteren
^Don't mess with ODST
Quis custodiet ipsos custodes? |
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| ninja |
Posted on 09/12/2013 08:17:53
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Super Admin
Posts: 4174
Joined: 15.11.10
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ODST wrote:
1+3+5+7+11 = 27
1 isn't a prime number, and you missed out 2.
My brain is open.
- Paul Erdős |
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| SkyDiv3r17 |
Posted on 09/12/2013 12:53:00
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Senior Member
Posts: 437
Joined: 25.10.12
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ninja wrote:
Looks like noone's going to get this one ...
ninja wrote:
Yesterday I drove from home to a shop/store, and then back again. My average speed on the way there was 50mph. My average speed on the way back was 70mph. What was my average speed over the whole journey?
Hint: Average Speed = (Total Distance) / (Total Time Taken)
(One possible) Solution:
To find the overall average speed, I need to know the total distance and the total time of the journey.
Let the distance from home to the shop be d.
Therefore, the time taken on the first part of the journey is d/50.
Similarly, the time taken on the second part of the journey is d/70.
So, the overall time taken is d/50 + d/70 while the total distance is d + d = 2d.
Therefore the average speed over the whole journey is 2d/(d/50 + d/70) = (2*50*70)/(70+50) = 58.333... mph.
In fact, it turns out that the answer is the harmonic mean of the two speeds - this is because they're rates.
----
New question ...
What is the sum of the first 5 prime numbers?
This doesn't make sense to me. If you drove 50mph one way and 70mph back. The average of that is 60 mph. It all depends on t, time. If you drove 50mph for 1 hour and then 70mph for one hour. Your average speed is still 60mph.
More in depth: If you drove 50mph for 10 minutes and 70mph for a 2nd 10 minutes, you would have (minutes increments) 50,50,50,50,50,50,50,50,50,50,70,70,70,70,70,70,70,70,70,70. (That's /minute for the 20 minutes). And the average of that is still 60..
You are leaving out rate of change and initial velocity if I'm correct so I don't figure how you get 58mph.
Edited by SkyDiv3r17 on 09/12/2013 12:53:41
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| ninja |
Posted on 09/12/2013 13:20:58
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Super Admin
Posts: 4174
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SkyDiv3r17 wrote:
This doesn't make sense to me. If you drove 50mph one way and 70mph back. The average of that is 60 mph. It all depends on t, time. If you drove 50mph for 1 hour and then 70mph for one hour. Your average speed is still 60mph.
More in depth: If you drove 50mph for 10 minutes and 70mph for a 2nd 10 minutes, you would have (minutes increments) 50,50,50,50,50,50,50,50,50,50,70,70,70,70,70,70,70,70,70,70. (That's /minute for the 20 minutes). And the average of that is still 60..
You are leaving out rate of change and initial velocity if I'm correct so I don't figure how you get 58mph.
The key here is that it's not for the same time. The journey is split into 2 parts: the distance is the same (home to shop), but the average speed over each part of the journey is different (and hence the time taken for each part is different). On the return part of the journey, I'm going faster so it's quicker - this means that I spend less time travelling at 70mph than at 50mph, so you'd expect the overall average speed to be closer to 50 than 70, which it is!
You're right that 60 is the average of 50 and 70. You're also correct that in your example, the average of 50mph and 70mph is 60mph, but this is only true if you're travelling at each of those speeds for the same proportion of time!
My brain is open.
- Paul Erdős |
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| SkyDiv3r17 |
Posted on 09/12/2013 13:58:10
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Senior Member
Posts: 437
Joined: 25.10.12
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ninja wrote:
The key here is that it's not for the same time. The journey is split into 2 parts: the distance is the same (home to shop), but the average speed over each part of the journey is different (and hence the time taken for each part is different). On the return part of the journey, I'm going faster so it's quicker - this means that I spend less time travelling at 70mph than at 50mph, so you'd expect the overall average speed to be closer to 50 than 70, which it is!
You're right that 60 is the average of 50 and 70. You're also correct that in your example, the average of 50mph and 70mph is 60mph, but this is only true if you're travelling at each of those speeds for the same proportion of time!
Ahhhhhhhhhhhh. The ninja math master, you are. Quicker pace = lesser amount of time = an average closer to 50... right on. Next!!
Edited by SkyDiv3r17 on 09/12/2013 14:00:52
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| ninja |
Posted on 09/12/2013 14:13:00
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Super Admin
Posts: 4174
Joined: 15.11.10
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SkyDiv3r17 wrote:
Ahhhhhhhhhhhh. The ninja math master, you are. Quicker pace = lesser amount of time = an average closer to 50... right on. Next!!
 Exactly.
Noone's correctly answered 'what is the sum of the first 5 primes' yet. If you answer it, you get to ask a question!
My brain is open.
- Paul Erdős |
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| SkyDiv3r17 |
Posted on 09/18/2013 13:16:17
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Senior Member
Posts: 437
Joined: 25.10.12
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ninja wrote:
SkyDiv3r17 wrote:
Ahhhhhhhhhhhh. The ninja math master, you are. Quicker pace = lesser amount of time = an average closer to 50... right on. Next!!
Exactly.
Noone's correctly answered 'what is the sum of the first 5 primes' yet. If you answer it, you get to ask a question!
28
Find the derivative of
[(x+5)/(x^2 +2)]^2 <---- the homework problem im working on right now hahaha
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| ninja |
Posted on 09/18/2013 13:28:08
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Super Admin
Posts: 4174
Joined: 15.11.10
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SkyDiv3r17 wrote:
28
Find the derivative of
[(x+5)/(x^2 +2)]^2 <---- the homework problem im working on right now hahaha
Good.
d/dx ([(x+5)/(x^2 +2)]^2) = 2 * (d/dx ((x+5)/(x^2 +2))) * ((x+5)/(x^2 +2)) = 2((x^2+2)-2x(x+5))/(x^2+2)^2 * ((x+5)/(x^2 +2))
= -2(x^2+10x-2)(x+5)/(x^2+2)^3
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Which is larger: two to the power of three (2^3) or three to the power of two (3^2)?
(Bonus question, if you liked that one ... which is larger: 2^333 or 3^222?)
Edited by ninja on 09/18/2013 13:29:20
My brain is open.
- Paul Erdős |
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| SkyDiv3r17 |
Posted on 09/18/2013 13:41:51
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Senior Member
Posts: 437
Joined: 25.10.12
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ninja wrote:
Good.
d/dx ([(x+5)/(x^2 +2)]^2) = 2 * (d/dx ((x+5)/(x^2 +2))) * ((x+5)/(x^2 +2)) = 2((x^2+2)-2x(x+5))/(x^2+2)^2 * ((x+5)/(x^2 +2))
= -2(x^2+10x-2)(x+5)/(x^2+2)^3
---
Which is larger: two to the power of three (2^3) or three to the power of two (3^2)?
(Bonus question, if you liked that one ... which is larger: 2^333 or 3^222?)
I hope you did that on paper.. otherwise O.O haha nice!
Since I know calc im going to leave that to someone else 
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| ninja |
Posted on 09/18/2013 14:02:38
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Super Admin
Posts: 4174
Joined: 15.11.10
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SkyDiv3r17 wrote:
ninja wrote:
Good.
d/dx ([(x+5)/(x^2 +2)]^2) = 2 * (d/dx ((x+5)/(x^2 +2))) * ((x+5)/(x^2 +2)) = 2((x^2+2)-2x(x+5))/(x^2+2)^2 * ((x+5)/(x^2 +2))
= -2(x^2+10x-2)(x+5)/(x^2+2)^3
---
Which is larger: two to the power of three (2^3) or three to the power of two (3^2)?
(Bonus question, if you liked that one ... which is larger: 2^333 or 3^222?)
I hope you did that on paper.. otherwise O.O haha nice!
Since I know calc im going to leave that to someone else
I scribbled a few workings on paper, then typed it up, yes.
Leave what to someone else - my question? If you have a good way of answering it, go ahead!
(You shouldn't need calculus, at least not for the main question!)
My brain is open.
- Paul Erdős |
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| SkyDiv3r17 |
Posted on 09/18/2013 14:15:16
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Senior Member
Posts: 437
Joined: 25.10.12
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ninja wrote:
Which is larger: two to the power of three (2^3) or three to the power of two (3^2)?
(Bonus question, if you liked that one ... which is larger: 2^333 or 3^222?)
2^3 = 8. 3^2 = 9. 9>8.
I think 2^333 is greater. But I have no idea how to do it, my calculator says "Error, Overflow" LOL
---- What is 6! ?
Edited by SkyDiv3r17 on 09/18/2013 14:17:16
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| BuhBiggieback |
Posted on 09/18/2013 17:04:29
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Wanker
Posts: 582
Joined: 18.07.13
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six is equivalent to the product of two and three; one more than five, or four less than ten.
also an arithmetical value, expressed by a word, symbol, or figure, representing a particular quantity and used in counting and making calculations and for showing order in a series or for identification.
tell me everything that isn't two plus two.
Edited by BuhBiggieback on 09/18/2013 17:05:39
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| ninja |
Posted on 09/18/2013 18:32:19
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Super Admin
Posts: 4174
Joined: 15.11.10
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SkyDiv3r17 wrote:
ninja wrote:
Which is larger: two to the power of three (2^3) or three to the power of two (3^2)?
(Bonus question, if you liked that one ... which is larger: 2^333 or 3^222?)
2^3 = 8. 3^2 = 9. 9>8.
I think 2^333 is greater. But I have no idea how to do it, my calculator says "Error, Overflow" LOL
---- What is 6! ?
I thought the same thing at first, but actually 2^333 > 3^222. I'm not surprised that your calculator throws an error: 2^333 has 101 digits, and 3^222 has 106 digits!
I got asked that question in my university interview - one way to answer it without a powerful calculator or computer is to use logarithms (and the fact that log(a^n)=n*log(a)).
I assume that you mean 6 factorial.
6! = 6 * 5 * 4 * 3 * 2 * 1 = 720
tell me everything that isn't two plus two.
I can't get it to display properly on this forum, but this is how I'd answer your question:
http://www.forkos...eq+4%5C%7D
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What is the probability of rolling 3 sixes in a row on a fair six-sided die?
Edited by ninja on 09/18/2013 18:39:35
My brain is open.
- Paul Erdős |
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