Comments on: What the ?!?!

By: James Lee

James Lee — Wed, 10 Jan 2007 17:50:46 +0000

Urrm .. that doesn’t look like 3 steps. And they don’t teach algebra until pri 6. My wife’s niece’s just started learning that. So how’s a pri 1 student going to do that???
BTW goo, nice job.
Any Primary school teachers out there ????

By: qoo

qoo — Tue, 26 Dec 2006 07:31:41 +0000

#hens = x, #roosters = y, #chicks = z.
x + y + z = 100
$6x + $3y + $0.1z = $100
z has to be multiples of 10.
and price of 1h = 2r
logically, z > x + y, if not it can never be within $100.
So then it’s working backwards.
if z = 90
x + y = 10
h + r = 100-9 = 91
2r + r = 91 (this amt has to be divisible by 3 to get whole chickens and roosters), so we’re ditching z=90.
if z = 80
3r = 92 (not divisible by 3)
if z = 70
3r = 93 (divisible! r = 21)
so
hx + ry = $93
6x + 3y = $93
3(2x + y) = 93
2x + y = 31
x + y = 30
y = 30 – x
2x + 30 – x = 31
x + 30 = 31
x = 31-30 = 1
y = 30 – 1 = 29
x = 1, y = 29, z = 70
ergh!!!! i took three hours to do this!!!

By: wantunn

wantunn — Wed, 06 Dec 2006 23:28:06 +0000

Rage:
100 chicks x 0.1 = 10
10 hen x 3 = 30
10 rooster x 6 = 60
gives you 120 animals in total. You need 100 animals @ $100 total
James:
Apparently they teach something like check and guess (?) in school now to solve problems like this. Sort of like logical elimination? Not too sure since I haven’t been to primary school in a while.
It’s supposed to be solveable in 3 steps though (yah.. beats me).

By: Rage

Rage — Wed, 06 Dec 2006 13:54:11 +0000

100 chicks x 0.1 = 10
10 hen x 3 = 30
10 rooster x 6 = 60
Isn’t it just that?

By: James

James — Fri, 01 Dec 2006 13:44:59 +0000

Ooi …. how to solve. 2 equations with 3 variables?????
Other than brute force, I don’t think there is any other method.
Any maths teacher out there wants to enlighten us? Urrm …. Miss Tan, you reading this?