Posted: Wed Oct 08, 2003 1:33 am
sheep are cool....
That's what I missed
damn "per"
Cantr II Discussion Forum
A forum for discussion about the PBBRPG Cantr II
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Another math problem
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Posted: Wed Oct 08, 2003 2:04 am
Posted: Wed Oct 08, 2003 2:25 am
kroner wrote:rklenseth wrote:But how do you determine how many sheep there are when they don't tell you how many coins there are? It says the amount of money received is equal to the amount of sheep. But they don't tell you how many coins the brothers earned.
I thought the same thing when I first saw the problem, but it says the number of coins they recieved per sheep is equal to the number of sheep they sold. So the number of coins they got was the number of sheep squared. Limiting the number of coins to a perfect square gives you the added info you need.
But how do you determine the number of coins and sheep. The problem never tells you how many coins there are but neither does it ever tell you how many sheep there are. Is this one of those problems where the number of sheep and coins don't matter and it all depends on a wierd formula? I hate those problems.
Posted: Wed Oct 08, 2003 2:27 am
I think I'm going to stick to history and English. I will Amen to that, Jerry. 
By the way, 'tomatos' are spelt wrong in the game. It should be 'tomatoes'.
By the way, 'tomatos' are spelt wrong in the game. It should be 'tomatoes'.
Posted: Wed Oct 08, 2003 2:54 am
rklenseth wrote:kroner wrote:rklenseth wrote:But how do you determine how many sheep there are when they don't tell you how many coins there are? It says the amount of money received is equal to the amount of sheep. But they don't tell you how many coins the brothers earned.
I thought the same thing when I first saw the problem, but it says the number of coins they recieved per sheep is equal to the number of sheep they sold. So the number of coins they got was the number of sheep squared. Limiting the number of coins to a perfect square gives you the added info you need.
But how do you determine the number of coins and sheep. The problem never tells you how many coins there are but neither does it ever tell you how many sheep there are. Is this one of those problems where the number of sheep and coins don't matter and it all depends on a wierd formula? I hate those problems.
You don't know how many sheep. It could be 6 or 14 or 16 or 24 or 26 or 34 or 36 or..... It doesn't matter for the problem.
Posted: Wed Oct 08, 2003 3:19 am
But how does it not matter when it determines how many coins the brothers earned and thus determine the worth of the knife?
I could easily figure out the problem if I knew how many sheep there would which would allow me to determine how many coins the brothers got which would allow me to determine how much the knife was worth. This is why I never liked math in that they always try to make you figure out the answer through one way and never give you all the information that is relevant to the problem so that can be. They try to teach you problem solving skills (well, at least that was my math teacher's argument as to why we need to learn all of this stuff) so that you can apply them to real life situations but in real life there is always more than one way to a solution. The Boy Scouts do a really good job of teaching problem solving skills to people. One way is to form a group of people then have someone give this group of people a problem they have to solve. Now it is up to that group to pick a leader, if they even bother, and then go about solving the problem. I learned though this experience, which always seemed to better represent real life situations then any math problem could ever, that there is always more than one way to solve a problem. That in fact there are an infinite number of ways to solve a problem, each with their own advantages and disadvantages as well as factors that determine the outcome.
I believe in the Laws of Physics and all but I also believe that they can be broken.
I could easily figure out the problem if I knew how many sheep there would which would allow me to determine how many coins the brothers got which would allow me to determine how much the knife was worth. This is why I never liked math in that they always try to make you figure out the answer through one way and never give you all the information that is relevant to the problem so that can be. They try to teach you problem solving skills (well, at least that was my math teacher's argument as to why we need to learn all of this stuff) so that you can apply them to real life situations but in real life there is always more than one way to a solution. The Boy Scouts do a really good job of teaching problem solving skills to people. One way is to form a group of people then have someone give this group of people a problem they have to solve. Now it is up to that group to pick a leader, if they even bother, and then go about solving the problem. I learned though this experience, which always seemed to better represent real life situations then any math problem could ever, that there is always more than one way to solve a problem. That in fact there are an infinite number of ways to solve a problem, each with their own advantages and disadvantages as well as factors that determine the outcome.
I believe in the Laws of Physics and all but I also believe that they can be broken.
Posted: Wed Oct 08, 2003 3:36 am
Since you know that after the the older brother takes his last 10 there are some number of coins between 1 and 9 left, it forces the tens digit of the total number of coins to be odd. As we showed before, this forces the number of sheep to end in 4 or 6. As watermelonnose showed, this forces the number of coins to end in 6. This mean the younger brother gets 4 less than his older brother. The knife must then be worth 2 coins. So the information forces a certain set of possible numbers of sheep, all of which yield the same answer. Finding the possible numbers of sheep is the whole problem. From there, the answer is easy to get.
Posted: Wed Oct 08, 2003 12:30 pm
OK lets see you do this one -3=4+m or maybe this one 7(a+5)3(8-a)=3
or I love this one. The temp. out side is 13C what is the speed of sound?
Lets see you get those.
or I love this one. The temp. out side is 13C what is the speed of sound?
Lets see you get those.
Posted: Thu Oct 09, 2003 4:58 am
I think...i hate you all now....
::goes into anti-math rage::
muuuuch better
Posted: Thu Oct 09, 2003 8:06 pm
Heh, today I had my first math class since highschool (I started an evening program in computer science) ... and the lecturer started the first lecture with giving a whole list of math puzzles and false proofs 
... (and then, after half an hour, brought us all to the pub - it is Ireland, after all ...) So now my head is full of annoying puzzles
... (and then, after half an hour, brought us all to the pub - it is Ireland, after all ...) So now my head is full of annoying puzzles Posted: Thu Oct 09, 2003 11:24 pm
-3=4+m
-7=m
That was simple Algebra
7(a+5)3(8-a)=3
(7a+35)(-3a+24)=3
-21a(2nd)+168a-105a+840=3
-21a(2nd)+63a+837=0
Using Quadratic Formula
a= (-63(plus or minus)(Square route of)74277)/-42
I wish a keyboard has more math symbols on it, I don't know if I'm right on the 2nd problem as my main problem in math is I make up numbers half way through a problem, but the genral way of going about solving the problem is there. Your making me remember all of this Algebra that was long forgotten.
v=331.4 + .6T m/s
v= Speed of Sound
T = temperature (in celsius)
m = meters
s = seconds
v = 331.4 + .6(13) m/s
v= 331.4 + 7.8 m/s
v = 339.2 m/s
This is an aprroximate speed as If I remember correct humitidy also has a factor. Also don't ask me how I remembered that formula.
-7=m
That was simple Algebra
7(a+5)3(8-a)=3
(7a+35)(-3a+24)=3
-21a(2nd)+168a-105a+840=3
-21a(2nd)+63a+837=0
Using Quadratic Formula
a= (-63(plus or minus)(Square route of)74277)/-42
I wish a keyboard has more math symbols on it, I don't know if I'm right on the 2nd problem as my main problem in math is I make up numbers half way through a problem, but the genral way of going about solving the problem is there. Your making me remember all of this Algebra that was long forgotten.
v=331.4 + .6T m/s
v= Speed of Sound
T = temperature (in celsius)
m = meters
s = seconds
v = 331.4 + .6(13) m/s
v= 331.4 + 7.8 m/s
v = 339.2 m/s
This is an aprroximate speed as If I remember correct humitidy also has a factor. Also don't ask me how I remembered that formula.
Posted: Thu Oct 09, 2003 11:50 pm
Posted: Fri Oct 10, 2003 12:30 am
Ditto.
Posted: Fri Oct 10, 2003 1:43 am
This is grade ten in Canada
7(a+5)3(8-a)=3
7a+35+24-3a=3
then collect trems
4a+59=3
then move +59 over to the right side and change the sign.
4a=-59+3
4a=-56
4a/4=-56/4
a=-14
This is grade 11 physics
And the thingy I got is v=332+0.59(t) but the way you did it is ok
7(a+5)3(8-a)=3
7a+35+24-3a=3
then collect trems
4a+59=3
then move +59 over to the right side and change the sign.
4a=-59+3
4a=-56
4a/4=-56/4
a=-14
This is grade 11 physics
And the thingy I got is v=332+0.59(t) but the way you did it is ok
Posted: Fri Oct 10, 2003 11:15 am
I think you changed a sign somewhere though the problem was
7(a+5)3(8-a)=3
not
7(a+5)+3(8-a)=3
I see how you did the distributive property that much we agreed on, but then you have to multiply not add, but maybe I'm just stupid. Last years Geometry really crewed up my Algebra and now I have to relearn everything for Alg II which I'm taking this year. (My school system is screwed up and puts geometry between Alg I and Alg II,)
7(a+5)3(8-a)=3
not
7(a+5)+3(8-a)=3
I see how you did the distributive property that much we agreed on, but then you have to multiply not add, but maybe I'm just stupid. Last years Geometry really crewed up my Algebra and now I have to relearn everything for Alg II which I'm taking this year. (My school system is screwed up and puts geometry between Alg I and Alg II,)