Math Guru's please break down for this SoU Riddle to an algebraic solution.(Spoiler)

[magnetism]

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I saw a battle between griffons and hippogryphs. Each
race had male and female members involved in the battle. Each member
of both races attacked every member of the other race once. When a male griffon attacked a male hippogryph, they would swipe with
their claws. When a male and a female fought, or two females fought
each other, they would bite. I witnessed 21 pairs swipe, while 34
pairs bit each other. How many male creatures and how many female
creatures were fighting?
Answer: 10 males and 6 females

Try as I might I just can't seem to recall formula developement and equation building for a word problem like this.

I would be quite greatfull thanks

[LordKathen]

What are you talking about exactly? ... [img]graemlins/saywhat.gif[/img]

[magnetism]

I am asking someone to solve this problem for example.

Bites = 21
Swipes = 34


males = x
females = y

(x^2 + 1)/2 = bites

((x^2 + 1) + (y^2 + 1))/2 = swipes

where x and y are solved for.. Obviously this is not the solution.

[Gabrielles blades]

MxM = swipe
MxF = bite
FXF = bite
21 swipe 34 bite

now, it says all attack each other once, you simply need to find out for the males what numbers are multiples of 21
ie 1, 3, 7, 21
so its either 10 M or 22 M

(y1) + 21(y2) + (y1)(y2) = 34
now y2 cant be greater than 1 for this one, so
y1 + 21 + y1 = 34
2(y1) = 13
so y1 = 13/2 which is an impossible number for this question (cant have halves...)

Ok, so we know that it has to be 10 males, so how many females?
3(y1) + 7(y2) + (y1)(y2) = 34
i forget how to solve this type of problem (so good old plugging and chugging worked...) and you get that y1 = 4 and y2 = 2 to make that equation true or a total of 6 F

[LordKathen]

STOP IT PLEASE! You're both giving me a headache!

[Juddy[SCGW]]

wat the!! huh??!! Wat the hell i didnt understand a word of that!! :S

[LennonCook]

Originally posted by Gabrielles blades: y2 cant be greater than 1 for this one

Why?

[Davros]

Originally posted by LennonCook: quote: Originally posted by Gabrielles blades: y2 cant be greater than 1 for this one

Why? [/QUOTE]Beacuse if Y2 is greater than one then the 2nd term (21 x Y2) is greater than 42 and that just can't be less than 34.

PS - Maybe you missed that Y2 is "Y2" the variable, and not Y squared.

PPS - the other corollary I guess is that Y2 could be greater than 1 if Y2 was a negative number, but this fails to give a soluble algorithm. Oh, and just how do you have a negative number of females . I can see how it might be possible to have a number of negative females ..... - Oh - that just goes to show if one keeps on talking it is possible to put one's both feet in one's mouth.

Time to practice the ole "duck and cover" move .

[ 02-05-2004, 07:17 AM: Message edited by: Davros ]