What the ?!?!
by wantunn on Nov.23, 2006, under Rants
My wife asked me the following question today. Apparently it was given to one of her collegue’s kid (a primary one student) in a math test for a measly 4 marks.
The question:
A Hen costs $6
A Rooster costs $3
Chicks are $0.1 a piece.
How many Hens, Roosters and Chicks do you need if you have exactly $100 and the sum total of Hens, Roosters and Chicks must also be 100?
Yeah, I couldn’t solve it via algebra or any other mathematical method I knew. In the end, 2 minutes with perl and a text editor solved the equation in 2 seconds but bearing in mind the average P1 kid is not going to have a computer during the test (heck, not even a calculator), I’m baffled at why this would be a math question.
Anyone who knows how to solve this mathematically, drop me a note.
FYI, if you have to know the answer, click on the more link…
Answer, 1 hens 29 roosters and 70 chicks
Derived via brute force calculations and verified that there is only 1 answer possible (assuming we don’t consider parts of chickens, i.e. whole numbers only).
December 1st, 2006 on 9:44 am
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?
December 6th, 2006 on 9:54 am
100 chicks x 0.1 = 10
10 hen x 3 = 30
10 rooster x 6 = 60
Isn’t it just that?
December 6th, 2006 on 7:28 pm
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).
December 26th, 2006 on 3:31 am
#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!!!
January 10th, 2007 on 1:50 pm
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 ????