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    I'm struggling with question 3 on the new gcse maths paper. I got the answer 125/4, could anyone please check to see if it is right and if it isn't could you explain how to work it out. This type of question is a level 9(A**) in the new gcse.
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    (Original post by theBranicAc)
    I'm struggling with question 3 on the new gcse maths paper. I got the answer 125/4, could anyone please check to see if it is right and if it isn't could you explain how to work it out. This type of question is a level 9(A**) in the new gcse.
    I got 6.25 instead of 125/4.

    To get point D, substitute 0 in place of x. This makes y = 5, so point D is (0,5). To get point A, make substitute 0 in place of y. This leaves you with 0 = -2x + 5 , which gives x = 2.5. So the coordinates of A are (2.5,0).

    Line (P)AB is perpendicular to line AD, so the gradient of AB must be -1/-2 , which is 1/2. The difference between the x coordinates point P (0,y) and point A (2.5,0) is 2.5. The change in y can be found by multiplying the change in x (2.5) by the gradient (0.5), which gives 1.25. Subtract this from the y coordinate of A (0), to get the the y value of P, -1.25. So the coordinates of P are (0,-1.25).

    From this point, just work out the difference between the y coordinates (as the x coordinates just give 0), which is 6.25.
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    (Original post by leafcannon)
    I got 6.25 instead of 125/4.

    To get point D, substitute 0 in place of x. This makes y = 5, so point D is (0,5). To get point A, make substitute 0 in place of y. This leaves you with 0 = -2x + 5 , which gives x = 2.5. So the coordinates of A are (2.5,0).

    Line (P)AB is perpendicular to line AD, so the gradient of AB must be -1/-2 , which is 1/2. The difference between the x coordinates point P (0,y) and point A (2.5,0) is 2.5. The change in y can be found by multiplying the change in x (2.5) by the gradient (0.5), which gives 1.25. Subtract this from the y coordinate of A (0), to get the the y value of P, -1.25. So the coordinates of P are (0,-1.25).

    From this point, just work out the difference between the y coordinates (as the x coordinates just give 0), which is 6.25.
    ohh okay i seee cheers!
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    (Original post by theBranicAc)
    ohh okay i seee cheers!
    I actually did this before and I think I might've got 6.25 as well, but looking back on it, while it looks pretty solid I'm still not entirely sure.
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    (Original post by leafcannon)
    I actually did this before and I think I might've got 6.25 as well, but looking back on it, while it looks pretty solid I'm still not entirely sure.
    how would you express 0.213 with the repeating digits of 13 as fraction, question 10
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    (Original post by theBranicAc)
    I'm struggling with question 3 on the new gcse maths paper. I got the answer 125/4, could anyone please check to see if it is right and if it isn't could you explain how to work it out. This type of question is a level 9(A**) in the new gcse.
    Wait, these are the new hard questions? I thought they were making GCSEs harder, it seems these are just normal questions youd find at the end of a paper, it makes no sense to change the grading systems aif the standard is the same, the only difference is that you guys get a free double A* for doing the same Questions as us.
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    (Original post by 1jonam16)
    Wait, these are the new hard questions? I thought they were making GCSEs harder, it seems these are just normal questions youd find at the end of a paper, it makes no sense to change the grading systems aif the standard is the same, the only difference is that you guys get a free double A* for doing the same Questions as us.
    you may not find it hard, but theres harder questions than that if you think it was easy

    what did you get for question 16?
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    (Original post by theBranicAc)
    you may not find it hard, but theres harder questions than that if you think it was easy

    what did you get for question 16?
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    Question 16 is 152 Degrees
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    (Original post by ituneserror)
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    Question 16 is 152 Degrees
    how did you work it out may i ask?
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    (Original post by theBranicAc)
    how did you work it out may i ask?
    38 Degrees = Angle BAE = Angle DBA because alternate angles are equal ('z' angles).

    Angles in a triangle add up to 180 and the base angles of an isosceles triangle are equal, so Angle DAB plus Angle CBA plus Angle ADB is 180 Degrees. And we know that Angle DBA is 38, so 38x2= 76. 180-76=104 Degrees.

    We also know that Triangles ADB and CDB are equal becuase they share two equal sides and the same angle so they must be equal (SAS). This means that Angle ADB = Angle CDB = 104 Degrees. And we know that an angle by itself adds up to 360 Degrees. This means that Angle ADB plus Angle CDB plus Angle ADC is 360. 360-(104x2)= 152 Degrees. So x = 152 Degrees.
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    (Original post by theBranicAc)
    how would you express 0.213 with the repeating digits of 13 as fraction, question 10
    Take 0.2 (2/10) away from 0.213... , leaving 0.013... . Label this x. Times that by 1000 to get 13.13... or 1000x. Subtract 10x from this to get exactly 13, or 990x.

    This leaves x = 13/990. Add the 2 tenths (198/990) you originally took away to yield the answer, 211/990.
 
 
 
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