Maths homework - help please : )

[littlenicky61]

My son has the following question :

The aquarium has 198 guppies angelfish and catfish altogether.
There are 4 times as many guppies as angelfish. There are 6 more angelfish than catfish.
How many catfish are there ?

I get the answer to be - 28 catfish ( 136 guppies 34 angelfish ) but I did this by trial and error and I am sure there is a set method that I should be able to show him so that he knows how to tackle similar questions - he is 9 by the way .

Thanks : )

Isn't it solving simultaneous equations?

G+A+C=198
G=4C
A=C+6

4C+C+6+C=198

Let a be the number of angelfish
That means there are :
4a guppies, and
(a-6) catfish

Therefore:
a + 4a + a - 6 = 198
6a - 6 = 198
6a = 204
a = 34

So there are
34 angelfish
136 guppies (4 x 34)
28 catfish (34 - 6)

Does that make sense?

Number of guppies G, angelfish A and catfish C

G+A+C=198

G also =4A (4 times as many guppies as angelfish)

and C+6=A

So go from G+A+C to 4A+A+C or 5A+C

and then 5A+ (A-6)=198
6A=198+6
6A=204
A=34
C=34-6=28

It's about substitution then solving the equation

G + A +C = 198
G=4A, so 4A+A+C=198
A= C+6 so 4(c+6) + C+6 + c = 198

4c + 24 + c + 6 + c = 198

30 + 6c = 198
6c = 168
c = 168/6 = 28

Oh yeah, -6

Thanks all- that makes sense to me but they haven't done any equations / algebra yet . Their topic at the moment is long division and remainders and multiplication of 3 digit numbers by 1 digit numbers so this question seems bit out of place within the homework if you see what I mean . All the other questions were things like 239x9 842 x 7 867/ 9 etc and the only other word problem was simple and linked to the topic - A bag of sausages costs £ 9 . Sam has £120 . At the very most, how many sausages can Sam buy

The aquarium question just seemed a bit odd within this context

Will show him the ways you have suggested though - thank you all : )

For all those using simultaneous equations, you are overlooking that the pupil is 9yo (unless OP meant in Y9), so method may not be appropriate, and trial and error might be better - eg if 1 catfish, then 7 angelfish, 28 guppies - total too low; if 50 catfish, .... - total too high.

Cross post. If he hasn't done any algebra, I would NOT show the methods above, unless you think he is very able in Maths and would lap it up. (I'd be worried about just confusing him).

Simultaneous equations in year 4/5?! Thank fuck I'm good at algebra, DD's in year 3!

We had a similar question for my 9yo recently. Easily solved by algebra but as they had not introduced algebra yet I got her to solve by trial and error, which took about 6 iterations. She got a house point for answering it so I assume it must have been approved by the teacher.

The 9yo-friendly approach (which is really a word form of solving simultaneous equations) might be:

"Well, if we added 6 extra c/fish, then there would be 204 fish and we could put them in groups of 6: C A G G G G. As 204 divided by 6 is 34, there would be 34 groups, which means 34 c/fish. But we added 6 c/fish, so there must have been 28 to start with."

But again, this is a bit of a stretch, and trial and error is a good general technique that can be used elsewhere, so always worth practising.

Thanks all - yes he is in year 4 ( aged 9 ) . Think I will advise him to just trial and error it as they haven't touched on algebra etc and they seem to teach things in a completely different way now , even from when my now 13 year old was in year 4 and don't want to confuse the hell out of him .

Thanks again to everyone : )