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# 3 non calc gcse maths problems watch

1. So I've got my Paper 1 mock tomorrow, and I was doing a Maths Practice Paper. I don't have the mark scheme and these 3 questions are bugging me. I'm completely lost with 1 and 3, but on 2 I just got stuck on how to explain

1) Mark has 18 bags of counters and 12 boxes of counters. The mean number of counters in all 30 bags and boxes is 14. The mean number of counters in the 18 bags is 10. Mark says, "The mean number of counters per box is 4". Is Mark right? You must show how you get your answer.

I am COMPLETELY lost

2) Prove that the sum of the squares of any three consecutive odd numbers is always 11 more than a multiple of 12.

I used general formulae 2n-1, 2n+1, 2n+3 and got as far as:
(2n-1)^2 + (2n+1)^2 + (2n+3)^2 = 16n^2 + 12n + 11. I know it needs to = 11+12n, but how do I show that?

3) The diagram shows a quadrilateral XBYA. The diagonals AB and XY intersect at the point M. Given that the area of AXB is equal to the area of triangle AYB, prove that XY is bisected by AB.

I guess you have to prove that XBYA is a rhombus because the diagonals bisect each other at right angles. That means you have to prove that all sides are equal and opposite sides are parallel, or something, but how?

Does anyone have any help?
2. (Original post by KawaiiCupcake)

2) Prove that the sum of the squares of any three consecutive odd numbers is always 11 more than a multiple of 12.

I used general formulae 2n-1, 2n+1, 2n+3 and got as far as:
(2n-1)^2 + (2n+1)^2 + (2n+3)^2 = 16n^2 + 12n + 11. I know it needs to = 11+12n, but how do I show that?
It should be - check it. Then the finishing touch is obvious.
3. (Original post by KawaiiCupcake)
So I've got my Paper 1 mock tomorrow, and I was doing a Maths Practice Paper. I don't have the mark scheme and these 3 questions are bugging me. I'm completely lost with 1 and 3, but on 2 I just got stuck on how to explain

1) Mark has 18 bags of counters and 12 boxes of counters. The mean number of counters in all 30 bags and boxes is 14. The mean number of counters in the 18 bags is 10. Mark says, "The mean number of counters per box is 4". Is Mark right? You must show how you get your answer.

I am COMPLETELY lost

2) Prove that the sum of the squares of any three consecutive odd numbers is always 11 more than a multiple of 12.

I used general formulae 2n-1, 2n+1, 2n+3 and got as far as:
(2n-1)^2 + (2n+1)^2 + (2n+3)^2 = 16n^2 + 12n + 11. I know it needs to = 11+12n, but how do I show that?

3) The diagram shows a quadrilateral XBYA. The diagonals AB and XY intersect at the point M. Given that the area of AXB is equal to the area of triangle AYB, prove that XY is bisected by AB.

I guess you have to prove that XBYA is a rhombus because the diagonals bisect each other at right angles. That means you have to prove that all sides are equal and opposite sides are parallel, or something, but how?

Does anyone have any help?
1) You know that the mean = total/(number of numbers) so if you know the mean and the number of numbers you can work out the total of a set of numbers.

2) You have made a mistake in you expanding or simplifying. Check that and it should work out OK. It will be 12(........) + 11.

3) Think about two triangles with equal area and the same base. It doesn't have to be a rhombus.
4. (Original post by RDKGames)
It should be - check it. Then the finishing touch is obvious.
Thanks! I just noticed, that's really helpful
5. (Original post by BuryMathsTutor)
1) You know that the mean = total/(number of numbers) so if you know the mean and the number of numbers you can work out the total of a set of numbers.

2) You have made a mistake in you expanding or simplifying. Check that and it should work out OK. It will be 12(........) + 11.

3) Think about two triangles with equal area and the same base. It doesn't have to be a rhombus.
Thanks so much for the help!
6. (Original post by BuryMathsTutor)
1) You know that the mean = total/(number of numbers) so if you know the mean and the number of numbers you can work out the total of a set of numbers.

2) You have made a mistake in you expanding or simplifying. Check that and it should work out OK. It will be 12(........) + 11.

3) Think about two triangles with equal area and the same base. It doesn't have to be a rhombus.
Question 2 is the last question on my paper (I added it because I liked this question ). It can be written like this but it should really be illustrated as 12(....)-1
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