Can someone do this Y4 maths question for me please?

[cantdomaths]

Brain is muddled...

There are 170 people in a hall.
There are 6 times as many girls as boys.
There are 6 more boys than teachers.
How many teachers are there?

I have Maths A level...but have also been drinking wine....

Wait divided 176 wrong it is 16, long time since I've done any maths.

176/8 is 22...

Yeah did that wrong so when subtracted 6 got 15 instead of 16.

It was the last question on a worksheet. DC has done a bit on equations but think it was a bit too hard. I have other older children so can't remember what they're expected to know at this stage!

Where did you get this from @cantdomaths ?
I teach Year 4 and I can’t think how this would be taught in Y4. Was it a challenge for your DC and therefore not actually Y4 level?

Bought back lots of work last week and I was just looking through it tonight ( as you do...). Was the last question on a revision worksheet. To be fair DC had answered all the other questions. Year 4- state school- DC good at maths though so not sure if all got the same worksheet...

As in bought home all their school workbooks from this term.

The easiest way is to use algebra but this would not be expected in year 4 (or at all in primary). Your ds should try to solve it using trial and error. Start with multiples of 6.

Older child definitely had to do simultaneous equations when they did the 11 plus though so she was taught it at some point in y5/ early y6.

Have to say, my y6 has done algebra and (to be clear I haven't asked him but am fairly sure) he would need quite a bit of help with this, and he is working above expected for maths. I think this is y7 sort of level! But maybe my bar is low.....

170 = 6g + b + b -6
176=6g + 2b
88= 3G + b
G=b

1b or g =22
Teacher =22-6

Are the teachers boys or girls? :o

To be fair I think he would have would be able to do it but by substituting numbers and then adjusting up or down according to answers. Tricky for y4!

Never noticed that. I assume teachers have no sex or are inanimate objects?

I teach Year 4...There's no way we would ever give this to a Year 4 child

This is Year 7/8 maths..

The easiest way is to use algebra but this would not be expected in year 4 (or at all in primary). Your ds should try to solve it using trial and error. Start with multiples of 6.

This. Simultaneous equations are not in the KS2 maths curriculum, but reasoning about multiples is. Even so, it's tricky for Y4.

It’s algebra

Boys x
Girls 6x
Teacher x-6

X + 6x + x - 6 = 170
8x = 176
X=22

I solved it using bar modelling! I teach Y6 and think it's hard for Y4!!

176 divided by 8 is 22

You don't need algebra or simultaneous equations to do this, but you do need to know the technique.
You also need to draw the problem to understand it (something I cannot do satisfactorily here).

You have the whole which is 170. This whole has been divided in some number of equal parts.
The girls have 6 shares, the boys one share and the teachers part of one share. To solve it you could, for example, add 6 to the number of teachers, so they would have one whole share as well. What you have is:

X X X X X X (girls, six shares)
X (boys, one share)
X (teachers + 6, one share)

In total you have 8 equal shares. The whole is now 170+6=176, so each share is worth 176:8=22
Hence, the number of teachers is 22-6=16.

Just to make sure I have done the maths right:
226=132 (girls)
221=22 (boys)
22-6=16 (teachers)
The total number of people is
132+22+16=170.

The problem is easy if you know the technique (there are much more difficult ones as well), but it is pretty much undoable if you don't.

@keepsmiling1, if an adult, let alone a child, has not been exposed to bar modelling, they won't be able to solve that problem.
Instead even a young(er) child, who knows bar modelling, will solve that question in 3 seconds. It is not a matter of age, but of knowledge.
You could use algebra, but clearly it is like using a cannon to shoot a fly ;)

That's very much a mastery style question though, and they do quite often seem to be "keep 'em busy" style so maybe trial and error is expected. DS is currently competing with his friend to see who can find all 232 different alternative solutions to one of their "extending" Y4 questions . It seems like time-filling to me.

Anyway, I absolutely love Arkadia's solution, but I'm not sure DS would have made the leap to adding 6 to both sides. I think he'd systemise a bit and use trial and error a bit. If you think of a group of 6 girls and 1 boy as a "unit" of 7, that helps. So the max sets of [6 girls, 1 boy] is 24, giving 168 children, remainder 2. Not enough teachers. But the number of sets of children is the same as the number of boys. If he wrote out:
24 x 7 = 168 (remainder to 170 = 2)
23 x 7 = 161 (remainder to 170 = 9)
22 x 7 = 154 (remainder to 170 = 16)

Then the 24..22 are the number of boys (and sets of [6 girls, one boy]) and the remainders are the number of teachers. It won't take many trials to get to 22 boys, 16 teachers.

It's certainly not the neatest solution on the thread, but I do think it would be doable this way for my very mathematical Y4 child. I might ask him later and see!

I had DD (she knows the bar method) try it. She found it difficult to draw the "6 more than" bit, but after that she solved it differently from me.
In short she did (you need to see the drawing to follow it, really...):

(170 - (6*7))/8 = 16

Which is neither better nor worse than my version, just different. Once you know how to sketch these kind of problems you can solve them in many different ways, all pretty much easy.

There are some fiendishly difficult problems at this level :D this one is medium difficulty.

I suspect that in this case who put the work sheet together didn't give it much thought and probably didn't know how to do it either :D

Arkadia - would you mind postings pictures of your and your daughters bar method pictures please? I hadn't heard of the bar method before, but have just googled, and can see that it looks really useful for the sort of "Bob spends 3/5 of his money and saves 1/3 of the rest" type problems. But I can't see how I would do the 6 more than bit either. (I default to algebra for anything like this, but I think DS2 tends to do something akin to the bar method in his head - and I'm sure he would find it easier if he had a technique for putting it on paper.)

stealth
noeuf that’s how I do it! But I’ve always assumed my strategies are long winded and slow compared to what DDs will be expected to be using at school. Also, nice picture, did you use your toes to count on?