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# OCR FSMQ Additional Maths - 20th June 2017 watch

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1. (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
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
2. (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
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
4. 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.
5. Well I failed this wth pls say I can get a B
6. (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
7. Grade A : 60 marks
8. (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)
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
10. (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,
11. Possible predictions
A = 64
B = 55
C = 46
12. Apologies:

13. (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
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?
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
15. (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
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?
16. (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.
17. (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
18. (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

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Are you sure that it was the moorland? I swear the moorland was other direct path and not the paths AO and OC

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Yeah, that's what I thought as well.
20. How many marks do you think you got?

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