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    (Original post by student004)
    do past papers forget about everything else
    Past papers are not as useful as they were last year but are still good for practicing questions. 9-1 Practice papers are more useful than past papers.
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    Hey I have this rather confusing question which needs a bit of clarification.
    Name:  Copy of 1MA1 - specimen papers set 2 - 7-9 practice paper gold.doc [Compatibility Mode] - Word 1.png
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    So I worked the equation of L to be y=\frac{20}{3}-\frac{1}{3}x which crosses the y axsis at (0,\frac{20}{3}) However, point O hasn't be defined, so assuming it means the center (0, 0) would the area of the triangle be \frac{20 \sqrt 10}{9} ?
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    (Original post by CraigBackner)
    Hey I have this rather confusing question which needs a bit of clarification.
    Name:  Copy of 1MA1 - specimen papers set 2 - 7-9 practice paper gold.doc [Compatibility Mode] - Word 1.png
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    So I worked the equation of L to be y= \frac{20}{3}- \frac{1}{3}x which crossed the x axsis at (0,\frac{20}{3}) However, point O hasn't be defined, so assuming it means the center (0, 0) would the area of the triangle be \frac{20 \sqrt 10}{9} ?
    Yes O is (0, 0) and that would be made clearer in a real exam question.

    It crosses the y-axis at (0, 20/3), not the x-axis.
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    (Original post by notnek)
    Yes O is (0, 0) and that would be made clearer in a real exam question.

    It crosses the y-axis at (0, 20/3), not the x-axis.
    Ok yeah my bad.
    So knowing this, the length of the triangle are \sqrt40, (given by the radius), \frac{20}{3} (given by the difference of the two points ,(0,\frac{20}{3}), (0, 0) and \frac{40}{9} (by Pythagoras theorem).
    Therefore the area is \frac{b*h}{2} which should give \frac{40\sqrt10}{9}. However the mark scheme answer is 60 Is this correct and if so how?

    Name:  Copy of 1MA1 - specimen papers set 2 - 7-9 practice paper gold.doc [Compatibility Mode] - Word 1.png
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    (Original post by CraigBackner)
    Ok yeah my bad.
    So knowing this, the length of the triangle are \sqrt40, (given by the radius), \frac{20}{3} (given by the difference of the two points ,(0,\frac{20}{3}), (0, 0) and \frac{40}{9} (by Pythagoras theorem).
    Therefore the area is \frac{b*h}{2} which should give \frac{20\sqrt10}{9}. However the mark scheme answer is 60 Is this correct and if so how?

    Name:  Copy of 1MA1 - specimen papers set 2 - 7-9 practice paper gold.doc [Compatibility Mode] - Word 1.png
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    It is referencing the x-intercept of the tangent, not the circle, if you have got those confused. (the tangent intercepts the x-axis at (20,0))
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    (Original post by CraigBackner)
    Yes,yes I know that but you need to work out the area of OAP, which includes a point on the circle
    Try sketching it out if you're still unsure.
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    (Original post by _gcx)
    It is referencing the x-intercept of the tangent, not the circle, if you have got those confused. (the tangent intercepts the x-axis at (20,0))
    Ohh i see now,thanks
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    (Original post by Misaki24)
    My teacher said the grade boundary has already been set for this year and any changes would only apply to the year 10s im confused now :P
    But how can the grade boundaries already be set if the whole point of grade boundaries is to see how the whole nation did as a whole and then setting the ranges of the marks to achieve certain grades.
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    do this one:
    nice and easy

    make "q" the subject of the formula 5(q+p)=4+8p
    give your answer in it simplest form

    (Original post by NothingButWaleed)
    But how can the grade boundaries already be set if the whole point of grade boundaries is to see how the whole nation did as a whole and then setting the ranges of the marks to achieve certain grades.
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    (Original post by unillama)
    do this one:
    nice and easy

    make "q" the subject of the formula 5(q+p)=4+8p
    give your answer in it simplest form
    No Problem....

    Expand: 5q+5p = 4+8p
    Rearrange: 5q = 4+8p-5p

    Final Touches: q = (4+8p-5p)/5
    :aetsch::aetsch::aetsch::aetsch::aetsch::aetsch:
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    (Original post by NothingButWaleed)
    No Problem....

    Expand: 5q+5p = 4+8p
    Rearrange: 5q = 4+8p-5p

    Final Touches: q = (4+8p-5p)/5
    YOU WOULD ONLY GET 2 MARKS FOR THAT
    READ THE QUES AGAIN :aetsch::aetsch::aetsch::aetsch::aetsch::aetsch:
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    (Original post by unillama)
    YOU WOULD ONLY GET 2 MARKS FOR THAT
    READ THE QUES AGAIN :aetsch::aetsch::aetsch::aetsch::aetsch::aetsch:
    Oh shame "Simplest form"
    q = (4+3p)/5

    Lol, see this is the kind of stupid mistakes I make in the exam....
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    its ok this is the only place you should make the mistakes not in the exam!!
    you learn from mistakes so make them but dont repeat them!!
    xx
    (Original post by NothingButWaleed)
    Oh shame "Simplest form"
    q = (4+3p)/5

    Lol, see this is the kind of stupid mistakes I make in the exam....
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    (Original post by unillama)
    its ok this is the only place you should make the mistakes not in the exam!!
    you learn from mistakes so make them but dont repeat them!!
    xx
    True....watch me still make the same mistake in the exam *facepalm*. Lol..retard alert!
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    lol!!
    (Original post by NothingButWaleed)
    True....watch me still make the same mistake in the exam *facepalm*. Lol..retard alert!
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    Hey can someone explain the understanding to this question?

    A gardener wants to estimate the number of snails in his garden.
    One day he catches 95 snails. He puts a tag on each snails then releases them.
    Then next day the gardener catches 164 snails.
    68 of these snails have a tag on them.
    Work out an estimate for the total number of snails in his garden.
    Write down any assumptions you have made.

    Then answer is \frac{(95) × (164)}{68}=229 snails because 95 of the snails make \frac{68}{164} of the population. But whys that?
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    (Original post by CraigBackner)
    Hey can someone explain the understanding to this question?

    A gardener wants to estimate the number of snails in his garden.
    One day he catches 95 snails. He puts a tag on each snails then releases them.
    Then next day the gardener catches 164 snails.
    68 of these snails have a tag on them.
    Work out an estimate for the total number of snails in his garden.
    Write down any assumptions you have made.

    Then answer is \frac{(95) × (164)}{68}=229 snails because 95 of the snails make \frac{68}{164} of the population. But whys that?
    We assume that the sample that has been taken is representative of the population. Thus, as \frac{68}{164} of the snails collected have been tagged, we can assume that we have tagged \frac{68}{164} of the population (again, making the assumption that it is representative), and hence \frac{68}{164}x=95, from which you can find the population.
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    (Original post by CraigBackner)
    Hey can someone explain the understanding to this question?

    A gardener wants to estimate the number of snails in his garden.
    One day he catches 95 snails. He puts a tag on each snails then releases them.
    Then next day the gardener catches 164 snails.
    68 of these snails have a tag on them.
    Work out an estimate for the total number of snails in his garden.
    Write down any assumptions you have made.

    Then answer is \frac{(95) × (164)}{68}=229 snails because 95 of the snails make \frac{68}{164} of the population. But whys that?
    I've seen questions similar to this in lots of the 9-1 practice papers. I'm not sure what the obsession is
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    Hey I'm really stuck on this question because the answer I'm getting is not the same as the markscheme.
    Name:  Edexcel GCSE (9-1) Mathematics_ Higher Student Book (Edexcel GCSE Maths 2015) - Google Chrome 26.png
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    I worked out the two equal lengths of the isosceles triangle to be 13 because of Pythagorean triples, 5,12  and 13
    The using \frac{1}{2}absinx=42. I rearranged for Sin x to give  = sin^-1(\frac{84}{a^2}) then subbing in 13 for a and solving I get 29.8.. however the markscheme answer is 32.5
    Anyone know why?
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    (Original post by CraigBackner)
    Hey I'm really stuck on this question because the answer I'm getting is not the same as the markscheme.

    I worked out the two equal lengths of the isosceles triangle to be 13 because of Pythagorean triples, 5,12  and 13
    Where did you get 5 from?

    The first step is to use the fact that the area is 42 to find the length of the base. Have you done this?
 
 
 
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