OCR FSMQ Additional Maths - 20th June 2017

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

Reply 101

bo.beynon

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
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

Reply 102

username2960722

7

Grade boundary predictions??

Reply 103

TSRUser12345

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.

Reply 104

username3047154

4

Well I failed this wth pls say I can get a B

Reply 105

sivthasan

14

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

Reply 106

johnadams122212

Grade A : 60 marks
Grade B : 52 marks
Grade C : 46 marks
Grade D : 38 marks

Reply 107

moongrape

12

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)

Reply 108

moongrape

12

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

Reply 109

sivthasan

14

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,

Reply 110

username2960722

7

Possible predictions
A = 64
B = 55
C = 46

Reply 111

johnadams122212

Apologies:

Grade A : 65 marks
Grade B : 57 marks
Grade C : 49 marks
Grade D : 41 marks

Reply 112

kish131

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
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?

Reply 113

dx31

15

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 :frown:

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

Reply 114

georgetbarber

12

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?

Reply 115

bo.beynon

2

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.

Reply 116

dx31

15

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

Reply 117

y.u.mad.bro?

21

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)