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Grade 9 Maths Paper GCSE (and other difficult GCSE questions) watch

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    This paper is great! How long did it take you to make?
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    (Original post by NineOfDiamonds)
    This paper is great! How long did it take you to make?
    Thank you so much I’d say just shy of one week
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    (Original post by Gent2324)
    ah ok, is that the hardest bit or is the -x bit harder? im saying this because im assuming theres some catch to make it really hard or something that i cant see yet?
    Made any progress?
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    (Original post by Y11_Maths)
    Made any progress?
    kind of, i got that 9root5 -5x + 3 = 0 is that correct so far?
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    (Original post by Gent2324)
    kind of, i got that 9root5 -5x + 3 = 0 is that correct so far?
    No, how have you come to this conclusion?
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    this paper looks so interesting; I can't wait to do it later:woo:
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    (Original post by Toastiekid)
    this paper looks so interesting; I can't wait to do it later:woo:
    Thank youuuuu
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    (Original post by Y11_Maths)
    No, how have you come to this conclusion?
    y^2 = 45-x^2
    so y = root45 - x
    y = 3root5 - x

    since 3y-2x+3 = 0
    3y = 9root5 - 3x
    thefore its 9root5-3x-2x+3=0
    9root5-5x+3 = 0
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    (Original post by Gent2324)
    y^2 = 45-x^2
    so y = root45 - x
    y = 3root5 - x

    since 3y-2x+3 = 0
    3y = 9root5 - 3x
    thefore its 9root5-3x-2x+3=0
    9root5-5x+3 = 0
    Have you been taught how to solve simultaneous equations with circle equations?
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    (Original post by Y11_Maths)
    Have you been taught how to solve simultaneous equations with circle equations?
    no i thought this was just a quadratic one?
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    (Original post by Gent2324)
    no i thought this was just a quadratic one?
    The equation of a circle is x^2 + y^2 = radius^2
    To begin to solve this question you must
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    rearrange your first simultaneous equation in the form y= then use this to begin to solve the next one
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    (Original post by Y11_Maths)
    The equation of a circle is x^2 + y^2 = radius^2
    To begin to solve this question you must
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    rearrange your first simultaneous equation in the form y= then use this to begin to solve the next one

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    so does x^2+(2/3x-1)^2 = 45 ?

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    (Original post by Gent2324)
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    so does x^2+(2/3x-1)^2 = 45 ?

    Not quite but right idea.
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    3y=2x-3 then divide both sides by 3 and leave as 1 fraction don’t try to simplify. Then do what you were going to do before
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    (Original post by Y11_Maths)
    Not quite but right idea.
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    3y=2x-3 then divide both sides by 3 and leave as 1 fraction don’t try to simplify. Then do what you were going to do before
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    ok so i factored that in and got x^2 + (2x-3/3)^2 = 45
    i then got x^2 + (4x^2-12x+9/9) = 45
    that correct? if so do i do 45x9 and then equate to zero to get a quadratic and solve for x?
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    (Original post by Gent2324)
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    ok so i factored that in and got x^2 + (2x-3/3)^2 = 45
    i then got x^2 + (4x^2-12x+9/9) = 45
    that correct? if so do i do 45x9 and then equate to zero to get a quadratic and solve for x?
    Yes this is correct
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    Bloody fantastic effort. It's such a legitimate looking paper. Isn't question 15 a bit harsh for 4 marks though?

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    Let a be an integer. Then we're looking at n^2+60=a^2 \Rightarrow 60 = (a+n)(a-n). Without loss of generality for n let a be a positive integer (can't be zero as n^2+60>0). So a+n > a-n, and considering the prime factorisation of 60, we get a+n=30 and  a-n=2 which leads to a=16 and n=14, or a+n=10 and a-n=6 which lead to a=8 and n=2. We can discount cases like a+n=60 and a-n=1 as if the parities of a+n and a-n are different, then that would imply a and n are not integers. Clearly the only way to split 2^2\times3\times5 into a product of two integers with the same parity is (2\times3\times5)\times2 = 30\times2 or (2\times3)\times(2\times5) = 6\times10, so x^3=(14\times2)^3=21952.
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    (Original post by Y11_Maths)
    Yes this is correct
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    ok i done that now but i got 5x^2 - 12x - 396 = 0 and i need to find 2 numbers that add to make -12 and times to make -1580?
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    (Original post by I hate maths)
    Bloody fantastic effort. It's such a legitimate looking paper. Isn't question 15 a bit harsh for 4 marks though?

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    Let a be an integer. Then we're looking at n^2+60=a^2 \Rightarrow 60 = (a+n)(a-n). Without loss of generality for n let a be a positive integer (can't be zero as n^2+60>0). So a+n > a-n, and considering the prime factorisation of 60, we get a+n=30 and  a-n=2 which leads to a=16 and n=14, or a+n=10 and a-n=6 which lead to a=8 and n=2. We can discount cases like a+n=60 and a-n=1 as if the parities of a+n and a-n are different, then that would imply a and n are not integers. Clearly the only way to split 2^2\times3\times5 into a product of two integers with the same parity is (2\times3\times5)\times2 = 30\times2 or (2\times3)\times(2\times5) = 6\times10, so x^3=(14\times2)^3=21952.
    Haha thank you so much. And yes this is correct. I see what you mean and I’m a harsh individual. They have to work hard for each individual mark
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    (Original post by I hate maths)
    Bloody fantastic effort. It's such a legitimate looking paper. Isn't question 15 a bit harsh for 4 marks though?

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    Let a be an integer. Then we're looking at n^2+60=a^2 \Rightarrow 60 = (a+n)(a-n). Without loss of generality for n let a be a positive integer (can't be zero as n^2+60>0). So a+n > a-n, and considering the prime factorisation of 60, we get a+n=30 and  a-n=2 which leads to a=16 and n=14, or a+n=10 and a-n=6 which lead to a=8 and n=2. We can discount cases like a+n=60 and a-n=1 as if the parities of a+n and a-n are different, then that would imply a and n are not integers. Clearly the only way to split 2^2\times3\times5 into a product of two integers with the same parity is (2\times3\times5)\times2 = 30\times2 or (2\times3)\times(2\times5) = 6\times10, so x^3=(14\times2)^3=21952.
    I glanced at all the questions a few days ago and I felt the same about this one. I thought I might have missed an easier method.

    I haven’t tried all of them but I agree that it looks like a brilliant paper and I bet most A Level students couldn’t come up with a GCSE paper this good.
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    (Original post by Gent2324)
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    ok i done that now but i got 5x^2 - 12x - 396 = 0 and i need to find 2 numbers that add to make -12 and times to make -1580?
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    Your x^2 term is connected by an addition therefore when you multiply both sides by 9 this turns into 9x^2. And yes you are correct with your method
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