Monday Maths Challenge
My year 5 daughter had this maths problem as part of her homework. See if you can solve it:
A farmer buys 100 animals for $100. The animals are lambs, ducks and chickens. Lambs cost $5 each, ducks $2 and chickens 10c. What are the numbers of lambs, ducks and chickens?
Took me too long but:
11 Lambs = $55
19 Ducks = $38
70 Chickens = $7
11+19+70 = 100
55+38+7 = 100
well done, sir. Did you solve it with a computer or manually?
Pffft computers. You’d think I work in IT or something. Plus there is no fun in cheating.
(Had to remove clients phone numbers from my photo hahaha)
Yeah I spent a bit of time doing the same thing but after half an hour of “near misses” I still didn’t manage to hit the lucky combination. Wrote a little program to brute force it. Cheating, yes, but not really sure how they expected the kids to work this out. I don’t think there’s enough information in there to solve it directly with algebra (3 variables, 2 knowns).
The best bit is now, when someone googles that question, the answer is here on Boardworld 
They can even cut and paste TJ’s workings!
I used the tried and trusted method of Guess and Check.
Last time I used this in an exam I got 1 out of like 5 points for not the correct working out. I’m sure there is a way you’re “supposed” to do this.
Work it out this way
$100 divided by Price of Animal
$100 gets you
20 Lambs or,
50 Ducks or,
1000 Chickens.
EDIT, actually read the whole question before my coffee… lol. Will delete the rest of what i wrote xD haha.
I dissagree with homework though… Managed to convince my teachers and parents in like year 8 or something, that homework was redundant and anti-productive. So they stopped giving it to me.
I don’t think there’s enough information in there to solve it directly with algebra (3 variables, 2 knowns).
Agreed. Did they expect it to be worked out by trail and error? Seems silly. Unless I’m missing something (which is not out of the question haha).
There is other results on the internet that allow you to work it out with algebra but this is year 5 homework. We didn’t do algebra until year 7/8 and I was in advanced maths.
Mud, correct me if I’m wrong but I’d probably guess that this was more of a bonus “see if you can do it” type question and not actually compulsory homework.
Yeah this problem was part of a set from my daughter’s “problem solving” class, which is a sort of maths extension class. There hasn’t been great communication to the parents about what is expected, but I am guessing they just want to see how the kids approach the problem and maybe how far they get. They are definitely not doing algebra yet, however a lot of the problems that come home lend themselves to it.
I never tried to convince anybody not to give me homework. I just never did it!
^^^ Yeah but then you’d get marked down. I didn’t have to do it, and didn’t get penalised for it either.
Ok new maths problem. So one of the guys at work showed me this one.
Its one of those internet “this came from [Asian country] where 5 year old kids can do these problems… so you must be stupid if you cant…” type problems.
So you have this equation:
? + 13 x ? / ? + ? + 12 x ? - ? - 11 + ? x ? / ? - 10 = 66
(/=Divided)
You need to place the numbers 1-9 in the ? and can only use each number once.
See if you can work it out.
Not much use but i also calculated there are 367880 permutations for 9x9 with no repeats. (9x8x7x6x5x4x3x2)
And 136 possible correct solutions.
All of which i worked out pretty easily, but ill post my workings later. Unless there some way of hiding posts wit ha spoiler tab? or something like that?
i did it like this:
a + 13b/c + d + 12e - f - 11 + gh/i -10 = 66
so a + 13b/c + d + 12e - f + gh/i = 87
Working on the assumption that all the components are integers:
looking at 13b/c, nothing divides evenly into 13 therefore i picked 1 for c (could also possibly pick a b and c such that c divides evenly into b),
then looking at gh/i, to make it evenly divisible either g or h has to be evenly divisible by i. I liked 8 for g as it can be divided by 2 or 4. then picked 4 for i.
the rest a bit trial and error, but bearing in mind you need a few numbers maybe in the 30 or so area to make up the final total. so i picked 13b = 13 x 3 = 39, and 12xe = 12x2 = 24 and gh/i = 8 x 7/4
the rest just fiddled around until it all worked
9 + 13 x 3/1 + 6 + 12 x 2 - 5 + 8 x 7/4
= 9 + 39 + 6 + 24 - 5 + 14 = 87
Yep, 931625874 is on the list. Well done.
I worked it out backwards. so if the answer is 66 the last part has to equal 76 (76-10=66), and so on.
But once i done that. I did it the easy way to get all the answers.
I went and put 1-9x9 into a permutations calculator, go all 367880 possible combinations. Put that into excel (crashed my work computer, lol)
But then split the numbers into individual cells, and added the equation as =SUM(A1+13*B1/C1+D1+12*E1-F1-11+G1*H1/I1-10)
Then sorted the answers in ascending order, and cut the ones that answered 66 into a new sheet.
