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    (Original post by sivthasan)
    Yeah why not? Please suggest corrections if you find mistakes
    1. -1<x<2
    2. y=1.5x-5.5 (or equivalent)
    3. y=5x-16
    4i) 2root5
    ii) (-1,6)
    5i) x^2+y^2=50
    ii) (7,1) and (5,5)
    6i) Find factors of 6, solve x where f(x) = 0, answer is -3
    ii) x=-3, 2+root2, 2-root2
    7i) 56/3
    ii) Negative area cancels out positive area on graph, so total area is negative, therefore...
    8. 0.132 (or as a fraction)
    9i) Integrate and sub in 0 and 8, to get 0
    ii) 56m
    10i) Prove = (4d^2+a^2-4b^2)/4ad
    ii) (4d^2+a^2-4c^2)/4ad
    iii) Prove = (2b^2+2c^2-a^2)/4
    iv) 2root10
    11i) Sub x=0 (no fertiliser) to get c=24
    ii) a=9/2 (sim equation)
    iii) 3kg per plot - 37.5 tonnes yield
    12i) x+3y≥30
    ii) x≤15, y≤8
    iii) 2x+3y≤48 (after simplifying)
    v) 15 minibuses, 5 coaches = £2250
    vi) 6 minibuses, 8 coaches = £1800
    13ai) 1hr24min
    ii) 2hr30min
    iii) (x/5) + (sqrt(x^2-8x+25))/2
    13b) x=2.7km => 2.1747 hours, which was minimum
    14i) 46.9m
    ii) 1052m
    iii) 8.37°
    13ai) was 2.3 hours
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    (Original post by sivthasan)
    Yeah why not? Please suggest corrections if you find mistakes
    1. -1<x<2
    2. y=1.5x-5.5 (or equivalent)
    3. y=5x-16
    4i) 2root5
    ii) (-1,6)
    5i) x^2+y^2=50
    ii) (7,1) and (5,5)
    6i) Find factors of 6, solve x where f(x) = 0, answer is -3
    ii) x=-3, 2+root2, 2-root2
    7i) 56/3
    ii) Negative area cancels out positive area on graph, so total area is negative, therefore...
    8. 0.132 (or as a fraction)
    9i) Integrate and sub in 0 and 8, to get 0
    ii) 56m
    10i) Prove = (4d^2+a^2-4b^2)/4ad
    ii) (4d^2+a^2-4c^2)/4ad
    iii) Prove = (2b^2+2c^2-a^2)/4
    iv) 2root10
    11i) Sub x=0 (no fertiliser) to get c=24
    ii) a=9/2 (sim equation)
    iii) 3kg per plot - 37.5 tonnes yield
    12i) x+3y≥30
    ii) x≤15, y≤8
    iii) 2x+3y≤48 (after simplifying)
    v) 15 minibuses, 5 coaches = £2250
    vi) 6 minibuses, 8 coaches = £1800
    13ai) 1hr24min
    ii) 2hr30min
    iii) (x/5) + (sqrt(x^2-8x+25))/2
    13b) x=2.7km => 2.1747 hours, which was minimum
    14i) 46.9m
    ii) 1052m
    iii) 8.37°
    13ai was 2.3 hours. You had the time along the path which was 4/5 = 0.8 hours and the time along the marsh which was 3/2 = 1.5 hours. This total 138 minutes
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    Grade boundary predictions??
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    I remember sitting this exam a while ago. If you do well on it, you pretty much got C1/2 sorted apart from logs and a few other stuff.
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    Well I failed this wth pls say I can get a B
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    (Original post by bo.beynon)
    13ai was 2.3 hours. You had the time along the path which was 4/5 = 0.8 hours and the time along the marsh which was 3/2 = 1.5 hours. This total 138 minutes
    Didn't the first part ask for the time taken if the path was taken the whole way? 4km along and 3km up = 7km at 5km/h - 7/5kmh = 1.4hrs = 1hr 24min
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    Grade A : 60 marks
    Grade B : 52 marks
    Grade C : 46 marks
    Grade D : 38 marks
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    (Original post by sivthasan)
    Didn't the first part ask for the time taken if the path was taken the whole way? 4km along and 3km up = 7km at 5km/h - 7/5kmh = 1.4hrs = 1hr 24min
    But for the 3km bit up, that was moorland and so his speed was only 2km/hr for this bit so you had to calculate the time for the path (4km along) separately to the moorland bit (3km up)
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    (Original post by johnadams122212)
    Grade A : 60 marks
    Grade B : 52 marks
    Grade C : 46 marks
    Grade D : 38 marks
    I think they will be higher because this year's paper was easier than last year's one and those boundaries look very similar to last years
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    (Original post by rxmbi)
    But for the 3km bit up, that was moorland and so his speed was only 2km/hr for this bit so you had to calculate the time for the path (4km along) separately to the moorland bit (3km up)
    I thought it was a path, nevermind. I think I got the rest of the question though,
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    Possible predictions
    A = 64
    B = 55
    C = 46
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    Apologies:

    Grade A : 65 marks
    Grade B : 57 marks
    Grade C : 49 marks
    Grade D : 41 marks
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    (Original post by sivthasan)
    Yeah why not? Please suggest corrections if you find mistakes
    1. -1<x<2
    2. y=1.5x-5.5 (or equivalent)
    3. y=5x-16
    4i) 2root5
    ii) (-1,6)
    5i) x^2+y^2=50
    ii) (7,1) and (5,5)
    6i) Find factors of 6, solve x where f(x) = 0, answer is -3
    ii) x=-3, 2+root2, 2-root2
    7i) 56/3
    ii) Negative area cancels out positive area on graph, so total area is negative, therefore...
    8. 0.132 (or as a fraction)
    9i) Integrate and sub in 0 and 8, to get 0
    ii) 56m
    10i) Prove = (4d^2+a^2-4b^2)/4ad
    ii) (4d^2+a^2-4c^2)/4ad
    iii) Prove = (2b^2+2c^2-a^2)/4
    iv) 2root10
    11i) Sub x=0 (no fertiliser) to get c=24
    ii) a=9/2 (sim equation)
    iii) 3kg per plot - 37.5 tonnes yield
    12i) x+3y≥30
    ii) x≤15, y≤8
    iii) 2x+3y≤48 (after simplifying)
    v) 15 minibuses, 5 coaches = £2250
    vi) 6 minibuses, 8 coaches = £1800
    13ai) 2hr18min
    ii) 2hr30min
    iii) (x/5) + (sqrt(x^2-8x+25))/2
    13b) x=2.7km => 2.1747 hours, which was minimum
    14i) 46.9m
    ii) 1052m
    iii) 8.37°
    Wasnt 14 (ii) 1012 m?
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    (Original post by johnadams122212)
    Grade A : 60 marks
    Grade B : 52 marks
    Grade C : 46 marks
    Grade D : 38 marks
    I wish it was that Cause i really messed up the last question, like i definitely lost 10 marks on it.

    I think itll be more like

    Grade A 68 marks. <<<<Hopefully its that
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    (Original post by sivthasan)
    Yeah why not? Please suggest corrections if you find mistakes
    1. -1<x<2
    2. y=1.5x-5.5 (or equivalent)
    3. y=5x-16
    4i) 2root5
    ii) (-1,6)
    5i) x^2+y^2=50
    ii) (7,1) and (5,5)
    6i) Find factors of 6, solve x where f(x) = 0, answer is -3
    ii) x=-3, 2+root2, 2-root2
    7i) 56/3
    ii) Negative area cancels out positive area on graph, so total area is negative, therefore...
    8. 0.132 (or as a fraction)
    9i) Integrate and sub in 0 and 8, to get 0
    ii) 56m
    10i) Prove = (4d^2+a^2-4b^2)/4ad
    ii) (4d^2+a^2-4c^2)/4ad
    iii) Prove = (2b^2+2c^2-a^2)/4
    iv) 2root10
    11i) Sub x=0 (no fertiliser) to get c=24
    ii) a=9/2 (sim equation)
    iii) 3kg per plot - 37.5 tonnes yield
    12i) x+3y≥30
    ii) x≤15, y≤8
    iii) 2x+3y≤48 (after simplifying)
    v) 15 minibuses, 5 coaches = £2250
    vi) 6 minibuses, 8 coaches = £1800
    13ai) 2hr18min
    ii) 2hr30min
    iii) (x/5) + (sqrt(x^2-8x+25))/2
    13b) x=2.7km => 2.1747 hours, which was minimum
    14i) 46.9m
    ii) 1052m
    iii) 8.37°
    For 10iv) how many marks do you think would be lost for writing 6.325, which isn't an exact value?
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    (Original post by sivthasan)
    Didn't the first part ask for the time taken if the path was taken the whole way? 4km along and 3km up = 7km at 5km/h - 7/5kmh = 1.4hrs = 1hr 24min
    The path didn't go up to C, it was a direct route from A to B. The dotted line was there to show that the lines were perpendicular , meaning Pythagoras could be used. Therefore, for the second half of the journey speed was 2km/h not 5km/h.
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    (Original post by georgetbarber)
    For 10iv) how many marks do you think would be lost for writing 6.325, which isn't an exact value?
    Probably like one mark cause i think it was only 2 marks
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    (Original post by rxmbi)
    But for the 3km bit up, that was moorland and so his speed was only 2km/hr for this bit so you had to calculate the time for the path (4km along) separately to the moorland bit (3km up)
    Are you sure that it was the moorland? I swear the moorland was other direct path and not the paths AO and OC

    Posted from TSR Mobile
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    (Original post by y.u.mad.bro?)
    Are you sure that it was the moorland? I swear the moorland was other direct path and not the paths AO and OC

    Posted from TSR Mobile
    Yeah, that's what I thought as well.
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    How many marks do you think you got?
 
 
 
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