The Student Room Group

Scroll to see replies

Reply 20
what do u think u need to get for an A?
Reply 21
Mr M
OCR FSMQ Additional Maths Answers

I hope these are all correct?! I only picked up the paper at 2.55 p.m. so these really are hot off the press! Shout if you disagree and I will check my work.


1. Inequality

(3 marks)


2. Binomial expansion

(3 marks)


3. Factors and remainders

(i) Remainder = 0 (2 marks)

(ii) (3 marks)


4. Binomial probability

(i) (2 marks)

(ii) (4 marks)


5. Maximum and minimum points

(i) Maximum at (-1, 12) (5 marks)

(ii) Positive cubic with vertices at (-1, 12) maximum (3, -20) minimum and crossing y axis at (0, 7). (1 mark)


6. Constant acceleration

(i) (3 marks)

(ii) 42 seconds (2 marks)


7. Trigonometric equation

(i) Show ... (1 mark)

(ii) Show ... (2 marks)

(iii) 75 degrees or 15 degrees (4 marks)


8. Variable acceleration

6.25 km (5 marks)


9. Circle

(i) Centre (8,2) (2 marks)

(ii) Show radius (2 marks)

(iii) (2 marks)


10. Modelling

(i) k = 0.5 (2 marks)

(ii) c = -0.25 (2 marks)

(iii) John's model is best as it gives the correct y value when x = 3. (2 marks)


11. Trigonometry

(i) Show AF=122.7. BF = 113.1 m (4 s.f.) (5 marks)

(ii) Height = 21.63 m (4 s.f.) (2 marks)

(ii) CF = 106.2 m (4 s.f.) and angle of elevation = 11.5 degrees (1 d.p.) (5 marks)


12. Shape made from card

(i) Equation of AB (4 marks)

(ii) Coordinates of A (-4, 6) (3 marks)

(iii) Area of AOB = 21.25 square units (5 marks)


13. Factory components

(i) Ali makes 72/t components per hour and Beth makes 72/(t+2) components per hour (3 marks)

(ii) Derive equation (5 marks)

(iii) Ali takes 6 hours and Beth takes 8 hours (4 marks)


14. Linear programming

(i) (2 marks)

(ii) (1 mark)

(iii) Obtain V shaped graph (3 marks)

(iv) 5 large vans and 5 small vans cost £700 (4 marks)

(v) 7 large vans and 1 small van = £620 (2 marks)


Edit: Correction to answer 1 and answer 11

How do you get the answer to Q4 (ii)
Isn't it 1 - ( P(X=0) + P(X=1)) ?
I get P(X=1) =
Reply 22
I disagree with your 4 part ii. I get: 19144\frac{19}{144}

What I did was add the answer from part i to the probability of there being one six rolled, and deduct from one. Which went something like:

6251296+4531296=11251296\frac{625}{1296} + \frac{4*5^3}{1296} = \frac{1125}{1296}

111251296=1771296191441- \frac{1125}{1296} = \frac{177}{1296} \equiv \frac{19}{144}

Is that not right?

EDIT: Yeah, what mrmanps said.
I was just wondering: would I lose a mark because I did not use the 1cm scale for one of the axes?
Reply 24
grade boundaries rough idea?
Reply 25
Is Q 4) ii) Wrong? If not, then how do we do it? Using the 1- thing.
maznaryk
Isn't the height of the tower 21.64 (4SF) ?

122.7tan(10)

Q11 (ii)


i was going to ask about this and the rest of the question...but i felt abit rude *embarrassed* looll

oh...you know the second question to the binomial expansion i got something like 0.132

i thought you had to do this>>

4C0 x 1/6^0 x 5/6^4 + 4C1 x 1/6^1 x 5/6^3 = .......
1-ANSWER = 0.131944...

could you tell me where i went wrong please? :smile:

EDIT: clearly i should read all the posts before i comment :p:
hearmecry
grade boundaries rough idea?


The grade boundary for an A has been as low as 68 and as high as 79 in the past 4 years. And as most people have said this year's paper was more difficult, they'll probably be in the low 70s.

I think. :yep:
mrmanps
How do you get the answer to Q4 (ii)
Isn't it 1 - ( P(X=0) + P(X=1)) ?
I get P(X=1) =


I did it wrong and thought it was asking for exactly 2 sixes!
Reply 29
can u go through question 8?
u know when u integrated it did u have to find +c?
hearmecry
can u go through question 8?
u know when u integrated it did u have to find +c?


Not if you integrated between limits.

You needed to integrate 60(t410t3+25t2)60 (t^4-10t^3+25t^2) with respect to t between 0 and 5.
Mr M
Not if you integrated between limits.

You needed to integrate 60(t410t3+25t2)60 (t^4-10t^3+25t^2) with respect to t between 0 and 5.

Thanks a lot for the thread.
just for my curiosity are you a teacher? Or how did you have access to the paper?
Rammiejimjams
i was going to ask about this and the rest of the question...but i felt abit rude *embarrassed* looll

oh...you know the second question to the binomial expansion i got something like 0.132

i thought you had to do this>>

4C0 x 1/6^0 x 5/6^4 + 4C1 x 1/6^1 x 5/6^3 = .......
1-ANSWER = 0.131944...

could you tell me where i went wrong please? :smile:

EDIT: clearly i should read all the posts before i comment :p:


Don't be embarrassed. I am happy to be corrected. When I first started posting answers, I used to carefully check them before posting to make sure they were all correct. Now I just knock them out at top speed knowing that people will let me know if they are wrong. The collective wisdom of crowds I think.
hamster33
Thanks a lot for the thread.
just for my curiosity are you a teacher? Or how did you have access to the paper?


yes
hearmecry
what do u think u need to get for an A?


I don't know. This is the first year I have taught this qualification so I don't have a huge amount of experience to draw on.
Reply 35
thankyou for posting these - I actually feel alot better about the exam now! I was really panicking towards the end because I was running out of time :/

Can you run through the first parts (i. and ii.)on question 7 please? Even though I know both the trigonometric equations, I didn't understand/couldn't do it. Also, I didn't rule a line after each question - do you think that matters? Thanks :smile:
Mr M
yes

Thank god.
If you were taking the exam you would seriously damage the grade boundaries.
Reply 37
can u go through 10 and tan question they were so hard!!
Mr M
OCR FSMQ Additional Maths Answers

I hope these are all correct?! I only picked up the paper at 2.55 p.m. so these really are hot off the press! Shout if you disagree and I will check my work.


1. Inequality

x>1.4x > 1.4 (3 marks)


2. Binomial expansion

112x+66x2220x3+...1-12x+66x^2-220x^3 + ... (3 marks)


3. Factors and remainders

(i) Remainder = 0 (2 marks)

(ii) x=2,x=4,x=1x=2, x=4, x=-1 (3 marks)


4. Binomial probability

(i) 6251296\frac{625}{1296} (2 marks)

(ii) 191296\frac{19}{1296} (4 marks)


5. Maximum and minimum points

(i) Maximum at (-1, 12) (5 marks)

(ii) Positive cubic with vertices at (-1, 12) maximum (3, -20) minimum and crossing y axis at (0, 7). (1 mark)


6. Constant acceleration

(i) 2ms22 ms^{-2} (3 marks)

(ii) 42 seconds (2 marks)


7. Trigonometric equation

(i) Show ... (1 mark)

(ii) Show ... (2 marks)

(iii) 75 degrees or 15 degrees (4 marks)


8. Variable acceleration

6.25 km (5 marks)


9. Circle

(i) Centre (8,2) (2 marks)

(ii) Show radius (2 marks)

(iii) x2+y216x4y+18=0x^2 + y^2 -16x -4y + 18=0 (2 marks)


10. Modelling

(i) k = 0.5 (2 marks)

(ii) c = -0.25 (2 marks)

(iii) John's model is best as it gives the correct y value when x = 3. (2 marks)


11. Trigonometry

(i) Show AF=122.7. BF = 113.1 m (4 s.f.) (5 marks)

(ii) Height = 21.63 m (4 s.f.) (2 marks)

(ii) CF = 106.2 m (4 s.f.) and angle of elevation = 11.5 degrees (1 d.p.) (5 marks)


12. Shape made from card

(i) Equation of AB 3y=2x+103y = -2x + 10 (4 marks)

(ii) Coordinates of A (-4, 6) (3 marks)

(iii) Area of AOB = 21.25 square units (5 marks)


13. Factory components

(i) Ali makes 72/t components per hour and Beth makes 72/(t+2) components per hour (3 marks)

(ii) Derive equation (5 marks)

(iii) Ali takes 6 hours and Beth takes 8 hours (4 marks)


14. Linear programming

(i) 2x+y152x + y \geq 15 (2 marks)

(ii) yxy \geq x (1 mark)

(iii) Obtain V shaped graph (3 marks)

(iv) C=80x+60yC=80x+60y 5 large vans and 5 small vans cost £700 (4 marks)

(v) 7 large vans and 1 small van = £620 (2 marks)


Edit: All correct now I believe. The lesson to be learned here is to read the question properly (unlike me).


Isn't your answer for question 9 wrong? Its supposed to be (7,1) because in order for you to get the second part right 8x8 +2x2 must equal r x r which in this is 5 root 2. In order for this to work 7x7 =49 + 1x1= 1. 49+1 =50 root 50 = 5 root 2 according to my calculator.
misskimono
Isn't your answer for question 9 wrong? Its supposed to be (7,1) because in order for you to get the second part right 8x8 +2x2 must equal r x r which in this is 5 root 2. In order for this to work 7x7 =49 + 1x1= 1. 49+1 =50 root 50 = 5 root 2 according to my calculator.


No this one is correct. (8, 2) is the centre of the circle. You do not find the sums of the squares of the coordinates of the centre of the circle in order to find the radius.

Latest

Trending

Trending