Helllllllllp

[friskybivalves]

DD has come home with a Y6 maths paper and I can't do the last question 🙈

Can anyone help - such a ritzy way to be spending a Saturday night!

'There are 5 times as many pens in box A than box B.
Tom moves 76 pens from box A to box B.
Both boxes now have the same number of pens.
How many pens are in box A now?'

Bully for Tom. But is it algebra? Ratios?something to do with a bar chart? I'm totally stumped. As is DD. It's going to be something really obvious - I'm braced for scorn...

Box A:Box B
5:1(ratio)
When you put 76 pens into Box B, they become equal (3:3)
Which means that 2 parts of the whole is 76, so 76 x 6 = total in both boxes. Halve that to find out what number are in Box A now.

I’ve had wine and gin. (It’s been a long day.) Lots of fucking pens.

To begin:
Box A = 5x pens
Box B = x pens

After:
A = 5x - 76
B = x + 76

5x - 76 = x + 76
5x - x = 76 + 76
4x = 152
x = 152 / 4
x = 38

So Box A has 5(38) - 76 = 190 - 78 = 114 pens (as does Box B).

Think you’re correct SallyGardens (and very nearly missed my train stop working it out your way!)

Sally is right, not really sure what wary was doing...

You are all way impressive.

But does waryandbored technique come to 228 and sallygardens to 114? ( I'm so bad at numbers and so grateful for all this help).

I feel that for a 10year old, wary's technique seems the simpler and therefore more likely but not if I cant get it to come up with the right number!

Sally's method just seems more like GCSE than yr 6 in a state primary though....

It’s akgebra. And at 10, we did that kind of thing. Sally is right.

Wary said 76 was 2 parts, but then multiplied by 6 instead of 3 to get 6 parts. That's why the answers are different.

Sorry, should’ve been 76 x 3 then halve that total to find what is in Box A now.

Wary is right up to here 2 parts of the whole is 76. That means 1 part of the while is 76/2 = 38 so 3 parts (or half) of the whole is 38 x 3 which is 114.

This can be solved by algebra.

Let A = number of pens now in box A
Let B = number of pens now in box B

Then what we have been told translates into 2 equations
A = 5B [A equals 5 times B] and
A-76 = B+76 [moving 76 pens will take that many away from box A AND add that many to box B, and leave each box containing the same number of pens]

Combining the two ,
5B - 76 = B +76
4B = 152
B = 38
A = 5x 38 = 190

Whole not "while". And I missed that it was for a 10yo - I'd do stuff like this with 13 and 14 year olds.

Sally's method just seems more like GCSE than yr 6 in a state primary though

Quadratic equations are just around the corner! Doing those in yr9. Time to brush up on the maths! It will be algebra. I agree with Sally too.

The method I used is how I would solve this with Y6 pupils. We do use algebra but not usually for this kind of problem, unless supported with bar model... really depends what your daughter has been taught.

Ok I think I've got it using ratios though.

Box A:Box B
5:1(ratio)
When you put 76 pens into Box B, they become equal (3:3)
Which means that 2 parts of the whole is 76, so 76 x 6 = total in both boxes. Halve that to find out what number are in Box A now.

Ratio was 5:1
Then it went to 3:3 by moving 76 pens.
One side went down by two and the other up by two.
So two parts of the whole is 76.
Two parts of the whole is a third, so 76 is a third of the total.
76 x 3 is the overall number of pencils = 228
If there are now equal numbers in box a and box b you need to divide that in half.

So 114 in each.

We know that

Right ... ... I mean wrong ...
A = 190 = number of pens originally in box A

A-76 = 114 = number of pens now in box A, which is what was asked

so, to summarise, what Sally said.

Sorry - it took so long for my mental cogs to turn that everyone else kindly posted before me.

I think she has done bar charts too but said she missed a key moment of explanation while she stuck her worksheet in her book with pritt stick. Sigh.

I'm still slightly hazy about getting to the 'so two parts of the whole is 76'. Was I right that you can just work out the difference between each side of the ratio to reach that conclusion.

Yes. Because we are told that Box a is 5x Box b, we can create the ration of 5:1. So 6 parts altogether. Then we are told that they become equal when 76 pens have moved. So that tells us it must be 3:3 because the number of parts doesn’t change. Does that make sense?

And I also realize that I have added to the conclusion by having to copy and paste wAry's first answer, without identifying it as such, and then hav8g a random 'we know that' adrift st them bottom.

It's too late. But I'm so grateful to all of you. I have a semblance of an explanation to go through with her in the morning. Maybe the light bulb will go on as we begin to go through it. You never know!

It's a white rose maths paper by the way. I discovered that by googling the whole question out of desperation. But no answer sheet available!

Confusion. Not conclusion. I give up.

It does make sense to me, Wary. Whether it will to DD, who knows? It does seem suddenly far more advanced than what she has done before.

You could start with simpler numbers, say 5 pens and one pen or ten pens and 2 pens and actually show your dd by putting the pens in 2 piles how ratios work. How many pens does she have to move to make the piles equal? Then ask her to imagine what would happen if you started with more pens in B. How many pens would you need in pile A for the ratios to be the same? What if she started with 38 pens in B?

She may not get it all at once, but by looking at the patterns, it with help build her understanding.

White Rose are obsessed with using bar models to solve this sort of question. I’ve attached a worked answer to illustrate.

Box A has 5 times. 5/6
Box B has 1times.. 1/6
So six lots in total...

SO YOU MOVE 2/6...

38. DIVIDED BY 2. IS 19.. WHICH IS 1/

TIMES THIS BY 3.TO GET 3/6.

114....