Maths help please!

[Land0r]

I have worked this out using trial and error (I think!) so I know the answer (I think!) but how should it be written out to 'show working' please? I know it's algebra but can't remember how to do it!

There are 72 children in the library and classroom altogether. After 12 pupils leave the library and enter the classroom, there are 3 times as many children in the classroom as the library. How many children were there in each room to start with?

This is part of DC2's homework - strong at maths but they haven't done algebra yet! The sheet says to show working not use trial and error (like me!).

You don't really need algebra here, and the 12 people moving classrooms isn't needed until the end. There are 72 students in total, and you know there are 3 times as many in the classroom. If you split the 72 into 4 equal parts (18) then 3 groups are in the classroom (54) and 1 group (18) in the library. You can then take the 12 students out of the classroom total and put them into the library - 42 classroom and 30 library originally.

Thank you for your very clear answer. The numbers 42 and 30 are what I got with my trial and error and a whole page of scribbling! DD hasn't even got this homework yet, it's just been emailed from the maths teacher for tonight, but I thought I'd have a look at it first. She may well be able to do this without any help, but if she does need help then I'll be able to explain using your words. Very grateful to you, thank you.

If you do need algebra, it's a bloody hard one, or I'm a lot rustier than I thought.

I got as far as L+C = 72 and then was trying to get an equation for L -12 = 3c or something and just couldn't do it.

If it is algebra for beginners then it seems like quite a tough start.

BarbaraofSeville I tried the same algebra (and lots of other things!) but couldn't do it either! DD is 9 but good at maths so this is extension work.

If you want to use algebra:
x= number of kids in library (after the departure of the 12), so:

x+3x=72
4x=72
x=18

So to answer the question:
There were originally
x+12 kids in library:
30 in library
42 in class

Thank you, I think I was getting confused with taking 12 off in the beginning when I was attempting algebra. (Or just confused all round really!)

There's another question on the sheet which I can't do either!

I wrote down all the info given as equations as follows:

Lb = number in library before
La = number in library after
Cb = number in classroom before
Ca= number in classroom after

1. Lb + Cb = La + Ca = 72

2a) La = Lb - 12
2b) Ca =Cb + 12

3. Ca = 3La

Use 3) in 1) to get La, then 2a) to get Lb, then 1) to get Cb.

I think I was putting too much thought into expressing the movement of the 12 mathematically too.

claras post shows the algebraic working that looks correct.

Would anyone like to see the question?!

The OTHER question, sorry!

Yes please!

HotInWinter Here you go:

Mary has a piece of ribbon that is 90cm long. She cuts it into two parts. The first part is 2cm shorter than 3 times the length of the second part. How long is each part?

Oh god this is giving me flash backs to school. I couldn't do them then and still can't apparently.

23 and 67. But these are hideous questions as they wrap up simple concepts in complex language. There is some bullshit about “real world problems” but these are not real world problems.

Me neither! DD2 is only 9 as well. She's just doing the first part of the homework, she hasn't got to the extension questions yet so don't know what she'll make of them. She's good at maths but dislikes wordy problems usually!

Ribbon via algebra
x +y = 90 and x=3y-2
substitute for x in formula 1 3y-2+y=90 4y-2=90 4y=92 y=23
therefore x=67
check 3 x 23=69 2 more than 67 ie x

So, it's cut into two parts, x and 90-x. The first part is 2 less than 3 times the second part, so x = 3(90-x)-2

And the solution is indeed x = 67 as others have shown in their cross posts!

ReflectentMonatomism Are you able to explain how you got your answers please? I tried doing what CoffeeandChocolateplease* did for the first question, dividing it into 4 equal parts then taking 2cm off the total of 3, but I didn't get the same answer as you. Actually I have so many scribbles I don't know what my answers were! Not whole numbers though, they ended in .5 I think.

Length of ribbon =a
So
3times the length of ribbon -2 =3a-2

(3a-2) + a=90
4a-2=90
4a=90+2
A=92÷4=23

3a -2= (3×23)-2=69-2=67
Pita question to explain simply tho

Sorry, I've just seen some more answers have come in while I was posting!

If they haven't been taught algebra then they will not be expected to use it.
Have they been doing bar modelling? because this looks very much like questions that would be used to practice this.

Just asked and she said they 'drew number sentences' in class. Is that the same as bar modelling?

whole piece is 90cm
you want 2 bits
one bit is three times the size of the otther (give or take 2cm!)
so you're dealing with 4 bits - 1 for one piece of ribbon, 3 for the other
90 / 4 = 22.5cm
so piece 2 = 22.5cm + 2cm = 24.5cm
piece 2 - 22.5 x 3 = 67.5 - 2cm
so final answer = 65.5cm

answer must add up to 90cm as that is what you started with

Not really, bar modelling looks like this
thirdspacelearning.com/blog/bar-modelling-techniques-maths-word-problems-sats-multi-step/

If she is yr 4 in a state school she will not be using algebra!