Yr 6 maths sats help - problem solving?

[FedupofTurkey]

Ds has homework, ie, 3 parcels cost £22. Parcel 1 and 2 cost £9, parcel 1 and 3 cost, £7, what is the price of each parcel. How is that worked out? I can't find what we need to look at on bbc ks2 bitesize. Is it algebra or problem solving?

1 + 2 + 3 = £22
1 + 2 = £9
You can therefore work out what 3 =

Similarly, you know that 1 + 3 = £7

You can work out what 2 = from that.

Working out what 1 costs is all that's left.

I think it's problem solving.

I think you would work it something like

1. Are two of the parcels P1 and P2? They can't be, because that would mean that the last parcel cost £13, which is impossible

1. So are two of the parcelts P1 and P3? Also can't be as this would make the last parcel £15 (impossible)

1. We also can't have 3 parcel 1s as this would mean parcel 1 cost £7.33 which is impossible.

So we have worked out parcel 1 is not one of our 3 parcels.

So that means the combinations of parcels we can have are:
222
333 - not possible as parcel 3 can't be £7.33
223
332

We know that parcel 2 costs £2 more than parcel 3

So we could have

• 3 lots of parcel 2, each costing £7.33

• 2 x parcel 2 and 1x parcel 3 - where parcel 2 costs £8 and parcel 3 costs £6

• 2 x parcel 3 and 1 x parcel 2 - where parcel 2 costs £6.67 and parcel 3 costs £4.67

I guess you could argue it couldn't be the 1st or last because it doesn't quite add up due to rounding.

So it must be parcel 2 costs £8, parcel 3 costs £6 and parcel 1 costs £1.

You could also have written down all the combinations to start with and then eliminated them 1 by 1.

It has to be worked out by trial and error until you arrive at the right answer.
A lot of algebra uses some trial and error.

Oops, we haven't got the total. This is the problem;

3 parcels, all different weights
1 and 2 weigh 7kg
2 and 3 weigh 8kg
1 and 3 weigh 11kg
What is the individual weight? Help!

Parcel 3 = 6kg
Parcel 1=5kg
Parcel 2 = 2kg

Probably the easiest way to do that is to add up all your values

This means that 2 lots of (P1+P2+P3) weigh 26kg
So 1 lot of P1+P2+P3 weighs 13kg

You can now easily work out the individual values by "taking away" the original sums.

^^

this is correct for the second problem you posted.

So are you saying to ignore the original post?

With algebra I calculated this way

P1+p2= 7
p2+p3=8
p1+p3=11. P3=11-p1

In the second equation a put the value of p3

P2+p3= 8. P2+11-p1= 8. P2= 8-11+p1

Now I use equation number one.

p1+p2=7. P1+8-11+p1=7. 2p1=7-8+11. P1=5

P2=8-11+5= 2

P3= 11-p1. P3=11-5=6

P1= 5.

P2= 2. P3=6

Eh?

Dont worry about the complicated algebra methods. In year 6 he will just be expected to find the answer and that can be done using simple trial and error (which is how I worked it out above).
Looking at the weights it looks quite obvious that parcel 3 is likely to be the heaviest so you guess what that weighs and then work out what the others will weigh and see if it fits; if it doesn't fit then you guess another weight for parcel 3 and so on until you get the right answer that works for all three parcels.

1 and 2 weigh 7kg
2 and 3 weigh 8kg
1 and 3 weigh 11kg

Each parcel is mentioned twice.

So 26kg (7 + 8 + 11kg) must be double the total weight of the 3 parcels.

So the total weight of the parcels = half 26 = 13kg

Now you know that, use the statements you've been given:

1 and 2 weigh 7kg, so parcel 3 must equal 13 - 7 = 6kg

2 and 3 weigh 8kg, so parcel 1 must equal 13 - 8 = 5kg

Parcel 1 must equal 2kg

Soz still don't get it, surely there must be an easier way for a 10 year old to work it out??

Sorry, final line should say

Parcel 2 must equal 2kg

I have just asked DS12 and DD10 - He would use my method above, she said she'd use trial and error - make parcel 1 3kg and see if that worked, then adjust higher or lower.

I think for KS2 level 3-5 SATs they are probably aiming for trial and error.

I looked at the first two pieces of info:
Parcel 1 and 2 weigh 7kg
Parcel 2 and 3 weigh 8kg

Parcel 2 weighs the same in each so from that you can deduce that parcel 3 is 1kg heavier than parcel 1.

I then used the 3rd piece of info and knowledge of number bonds for 11 to work out that parcels 1 and 3 could only be 5kg and 6kg. If you substitute one of those values into either of the first two pieces of information you can then work out parcel 2.

It's sort of a half way between trial and error and the 'proper' algebraic method but I found it much quicker given the size of the numbers. If the numbers had been a lot bigger, I would have used the method redsky and others used.

I just asked my 9 year old what he would do to solve this problem, and he would also have used trial and error. Although he first said that those sort of problems were really hard and he would avoid it if possible - he does lots of similar at school, so surely OP's DS must have some idea from what he's done in class?

My 9 year old solved the problem quite quickly and he also used trial and error.