# Year 7 homework problem

Discussion in 'Mathematics' started by DeborahCarol, Mar 19, 2012.

1. ### DeborahCarolNew commenter

Found this on another forum. Maths undergraduate drawing a blank on his niece's 'Year 7' homework problem. Ashamed to say I'm also drawing a blank, although I'm meant to be able to teach Year 7! Perhaps we're both missing something obvious and trying to make it more complicated than it is.
'In a class, one-third of the pupils bring a teddy to school. During the term, each boy took 12 books out of the library, each girl 17 and each teddy 9. In total, 305 books were taken out. How many girls are in the class?'
Need a 'Year 7-friendly' way of solving it.

2. ### DeborahCarolNew commenter

Found this on another forum. Maths undergraduate drawing a blank on his niece's 'Year 7' homework problem. Ashamed to say I'm also drawing a blank, although I'm meant to be able to teach Year 7! Perhaps we're both missing something obvious and trying to make it more complicated than it is.
'In a class, one-third of the pupils bring a teddy to school. During the term, each boy took 12 books out of the library, each girl 17 and each teddy 9. In total, 305 books were taken out. How many girls are in the class?'
Need a 'Year 7-friendly' way of solving it.

3. ### bobboots

Maybe for a very bright y7:
boys books + girls books + teddy's books = 305
12b + 17g + 1/3 * (b + g) * 9 = 305
simplifies to 15b + 17g = 305
By inspection the number of girls is a multiple of 5 otherwise you'd never end in a 5 when adding the boys books so g = 5, 10 or 15.
Try each and you only end up with an integer number of boys when g = 10.

You'll need to flesh out the explanation. Hope this helps

4. ### PaulDGOccasional commenter

They could simply try all the options for class sizes of up to, say, 35.

As a whole class or grouped activity, they'd need to divide the problem up into work packages for individual members, ("you try it with 25 in the class. I'll try 26.. Angie can try 27...")

Perhaps split the class into 3.. Give a little bit of scaffolding with the "formula".. Make it a race..?

Maybe direct the brightest towards the method above?

5. ### DeborahCarolNew commenter

Many thanks, bobboots and pauldg. Relieved that I wasn't missing an 'easy way' of doing this! (Unless someone else comes up with one!)
This problem, and another equally tricky one involving parallelograms, was posted (on an OU Maths forum by uncle struggling to help niece!). These certainly aren't standard Year 7 homeworks - have asked the poster to find out if they're UKMT Maths Challenge questions!

6. ### KarvolOccasional commenter

12b + 17g + 1/3 * (b + g) * 9 = 305
simplifies to 15b + 20g = 305

This then gives you 3b + 4g = 61, which has a number of different natural number solutions. However, only one of these solutions - 11 boys and 7 girls - would allow you to have a third of the number of students bringing teddies into the class, as required.

7. ### PaulDGOccasional commenter

They could be if that school is trying to encourage "functional skills".

8. ### bobboots

Ooops! I shouldn't post on here when I'm half asleep.

9. ### cackhanded

Edit - 3/4 asleep!

10. ### PiranhaStar commenter

JMC 2009, Q25; the possible answers given are 4, 7, 10, 13, 16. As the last question on the paper, this is designed to identify the very best.

And once you have the possible answers, there's another way to tackle it, as you can work from each of those.

12. ### DeborahCarolNew commenter

Piranha, many thanks for that information. Quite comforting actually, as I have no problem with some eleven year-olds in the UK being brighter than I am!

13. ### PiranhaStar commenter

I remember discussing this question with my Year 7s when preparing for the JMC last year - it was such a striking question that is stuck in my mind. As frustum points out, the sample answers make it easier.

14. ### hassan2008

I agree with Karvo's solution.
In fact if you draw the graph of 3x+4y=61, x=boys, y= girls,
On the straight line, you will find only 3 possible pairs of integer values
(11, 7), (7, 10) and (3, 13). The one that you want is (11, 7), 11+7 = 18
You could try making up a table of values of 3b, 4g, until you get (3b+4g) = 61.
You could do this easily in Excel.