# [Unofficial Markscheme] OCR MEI AS LEVEL MATHS (C1, C2, S1)

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Hey guys, theres quite a few threads for all the exams and it takes a while to post my answers on each and everyone so Ive decided to make a thread. I'm going to be uploading all the answers I recieved for C1 C2 and S1 here and Ill probably make a thread for AQA BIO/CHEM AS but that isnt certain, if I do Ill be posting the link to that here.

C1:

C2:

S1:

~L

edit: better s1 mark scheme here just didn't see it before x.x: https://docs.google.com/document/d/1...kfpLx2sO0/edit

pictures of c2 paper at: https://twitter.com/freddyfincham/st...98315039666176

C1:

Spoiler:

1. Sketched y= -2x + 1 (0,1) and gradient was -2 (2 marks)

2.

i) (1 7/9)^-1/2 = 3/4 (3 marks)

ii) (6x^5y^2)^3/18y^10 = 12x^15/y^4 (2 marks)

3. 6-x > 5(x-3) = 6 - x > 5x - 15 = -6x > -21 = x< 21/6 (3 marks)

4. Intersection of 2x+5y=5 and x-2y = (10/3, -1/3) (4 marjs)

5. (x+2)^2 + (y-3)^2 = 5

i) radius is root 5, centre is (-2,3) (2 marks)

ii) Line parallel to 5x + y = 4 and touches centre is y= -5x-7 (2 marks)

6. r = root (v/a+b) - Make B the subject : Two possible answers are b = (v-ar^2)/r^2 or b = v/r^2 - a (4 marks)

7.

i) (5-2root7)/3+root7 = (29-11root7)/2 (3 marks)

ii) 12/root2 + root 98 = 13root2 (2 marks)

8. (a+bx)^5 x^3 coefficient is -1080, constant is 32. a^5 = 32, so a = 2. b= -3 (5 marks)

9. Three consecutive integers and n is the smallest (therefore n, n+1 and n+2) show difference between squares of smallest and largest is 4 * middle. (n+2)^2 - n^2 = 4n+4. 4n+4 = 4(n+1). (4 marks.

Section B

10. A(3,3) B (-2,-2) C (5,-1)

i) Show AB is BC. AB = ROOT 50, BC = ROOT 50 (2 marks)

ii)Line perpendicular to AC which passes by B

Grad of AC is -2, perp gradient is 1/2. Line is y= 1/2x-1 (4 marks)

iii) D is (10,4) because difference between B and C is 7, 1. Add 7 to x of A and 1 to y of A to get 10,4 (2 marks)

iv) 3.8 < 26/7. Find equation of lines and prove 8, 3.8 does not belong in them therefore it is outside rhombus (4 marks)

11. f(x) = (x-2)(2x-3)(x+5)

i) Sketch graph (3 marks)

ii) Show when (-3 0) occurs that g(x) is 2x^3 + 21x^2 + 43x + 24. You use f(x+3) and from that x+3 = 2, 3/2, -5

x = -1, -3/2, -8 are new roots

factors are (x+1)(2x+3)(x+8), multiply out to get the answer (3 marks)

iii)x = -2 is root of g(x)=6, prove this and find other roots. Other roots = (-17+-root217)/4 - Not sure of this (6 marks)

12.

i) Show y= x^2 + x + 3 as (x+a)^2 + b and it doesnt touch the x axis. Completed square is (x+1/2)^2 + 11/4 So minimum point is -1/2, 11/4 which is above x axis (4 marks)

ii) Intersection of x^2+x+3 and 2x^2-3x-9 are (6,45) and (-2,5) (4 marks)

iii) For what values of k is there no intersection of x^2 + x + k and 2x^2 - 3x- 9

Use b^2 - 4ac<0

k< -13

Show

1. Sketched y= -2x + 1 (0,1) and gradient was -2 (2 marks)

2.

i) (1 7/9)^-1/2 = 3/4 (3 marks)

ii) (6x^5y^2)^3/18y^10 = 12x^15/y^4 (2 marks)

3. 6-x > 5(x-3) = 6 - x > 5x - 15 = -6x > -21 = x< 21/6 (3 marks)

4. Intersection of 2x+5y=5 and x-2y = (10/3, -1/3) (4 marjs)

5. (x+2)^2 + (y-3)^2 = 5

i) radius is root 5, centre is (-2,3) (2 marks)

ii) Line parallel to 5x + y = 4 and touches centre is y= -5x-7 (2 marks)

6. r = root (v/a+b) - Make B the subject : Two possible answers are b = (v-ar^2)/r^2 or b = v/r^2 - a (4 marks)

7.

i) (5-2root7)/3+root7 = (29-11root7)/2 (3 marks)

ii) 12/root2 + root 98 = 13root2 (2 marks)

8. (a+bx)^5 x^3 coefficient is -1080, constant is 32. a^5 = 32, so a = 2. b= -3 (5 marks)

9. Three consecutive integers and n is the smallest (therefore n, n+1 and n+2) show difference between squares of smallest and largest is 4 * middle. (n+2)^2 - n^2 = 4n+4. 4n+4 = 4(n+1). (4 marks.

Section B

10. A(3,3) B (-2,-2) C (5,-1)

i) Show AB is BC. AB = ROOT 50, BC = ROOT 50 (2 marks)

ii)Line perpendicular to AC which passes by B

Grad of AC is -2, perp gradient is 1/2. Line is y= 1/2x-1 (4 marks)

iii) D is (10,4) because difference between B and C is 7, 1. Add 7 to x of A and 1 to y of A to get 10,4 (2 marks)

iv) 3.8 < 26/7. Find equation of lines and prove 8, 3.8 does not belong in them therefore it is outside rhombus (4 marks)

11. f(x) = (x-2)(2x-3)(x+5)

i) Sketch graph (3 marks)

ii) Show when (-3 0) occurs that g(x) is 2x^3 + 21x^2 + 43x + 24. You use f(x+3) and from that x+3 = 2, 3/2, -5

x = -1, -3/2, -8 are new roots

factors are (x+1)(2x+3)(x+8), multiply out to get the answer (3 marks)

iii)x = -2 is root of g(x)=6, prove this and find other roots. Other roots = (-17+-root217)/4 - Not sure of this (6 marks)

12.

i) Show y= x^2 + x + 3 as (x+a)^2 + b and it doesnt touch the x axis. Completed square is (x+1/2)^2 + 11/4 So minimum point is -1/2, 11/4 which is above x axis (4 marks)

ii) Intersection of x^2+x+3 and 2x^2-3x-9 are (6,45) and (-2,5) (4 marks)

iii) For what values of k is there no intersection of x^2 + x + k and 2x^2 - 3x- 9

Use b^2 - 4ac<0

k< -13

C2:

Spoiler:

Q1: Sequence

i)Find the sum from 1 to 5 of the sequence 3a+2 (SIGMA Notation) (2 marks)

5+8+11+14+17 = 55

ii) An arithmetic progression has first term 4.2 and sixth term 1.2, find common difference. (2 marks)

A = 4.2 6th term = 4.2 + 5d

4.2 + 5d = 1.2

d = -0.48

Q2: Integration

i) Use definite integration from 5 to 1 for 4x (3 marks)

4x integrated = 2x^2 2(5)^2 - 2(1)^2 = 48

ii) Use indefinite integration for 6x^1/2 (2 marks)

Integrated = (6x^1/2+1) / (1/2 + 1) + c

= 4x^3/2 + c

Q3: Given the graph of y= logx

i) Find the gradient between the points A and B where x = 0.1 and x = 0.2 (2 marks)

(-0.699 - -1)/(0.2-0.1) = 3.01 (3 s.f)

ii) Show in graph, where you could plot a point to get a better estimate of the gradient at A (1 mark)

Plot in between the two points

Q4. You are given the graph y = 2x^3. Find the equation of the normal at x = 2 and put it in the form ax + by = c (5 marks)

dy/dx = 6x^2, put x in as 2.

dy/dx = 24. 24 is the gradient of the tangent, therefore the gradient of the normal = -1/24.

At the point x = 2, y = 16. Put into formula y - y1 = m(x-x1)

y-16 = -1/24(x-2), rearrange to get y = -1/24x + 386/24.

Multiply all sides by 24 and bring x over:

x + 24y = 386.

Q5. ~Transformations

i) Describe the single transformation that maps the graph y= x^2 + 3 to y = 2x^2 + 6 (2 marks)

An enlargement/stretch of scale factor 2 parallel to the y axis

ii) Describe the single transformation that maps the graph of y = 2x^2 to y = 2(x-3)^2 blabla (2 marks)

A translation of (3 0)

Q6. You are given that a curve goes through (2,10) and dy/dx = 12x^3 - 7. Find the equation of the curve (5 marks)

Integrate 12x^3 - 7 to get 3x^4 - 7x + c

3(2)^4 - 7(4) + c = 10

34 + c = 10 so c = -25

Equation = 3x^4 - 7x - 24

Q7.

i) Sketch y= 2^x (2 marks)

Show that it passes through (0,1)

ii)Simplify logaW = 3 + logaX^5 - logaX + loga6 in its simplest form and express w in terms of a and x (3 marks)

Ignore logaW for now, and focus on right hand side:

3 becomes logaA^3

logaX^5 - loga2X becomes logaX^5 /2X

logaX^4 /2

logaA^3 + logaX^4 /2 + loga6 = loga6*A^3*X^4 /2 logaW = loga3*A^3*X^4

w = 3a^3x^4

Q8. x is given to you in radians. You are given that 6cos^2x = 5 - sinx, Show that 6sinx^2 - sinx - 1 = 0 and find the values of x for 0 < x < 2pi (5 marks)

6cosx^2 = 5 - sinx

cos^2x + sin^2x = 1

6(1-sin^2x) = 5 - sinx

Rearrange to get 6sin^2x - sinx- 1 = 0

Factorise to get (3sinx + 1)(2sinx - 1)

sinx = -1/3 or 1/2

Values of x= 3.48, 5.94 (3s.f), pi/6, 5pi/6

Q9. You are given a cylinder blabla with radius r and height h. The volume of the cylinder = pi*r^2*h Area of the cylinder = 2*pi*r*h + 2*pi*r^2. You are also given that V = 400

i) Show that the area of the cylinder is 2*pi*r^2 + 800/r (2 marks)

Rearrange 400 = pi*r^2*h to 400/pi*r^2 = h

Put that into the formula of area to get A = 2*pi*r^2 + 2*pi*r * (400/pi*r^2)

A = 2*pi*r^2 + 2*pi*r*400 / pi*r^2 ~ pi and r cancel out to give you

A = 2*pi*r^2 + 800/r

ii) Find dA/dr and d^2A/dr^2 (4 marks)

dA/dr = 4*pi*r - 800r^-2

d^2A/dr^2 = 4*pi + 1600r^-3

iii) Hence, find the minimum value for r in this cylinder and find the corresponding surface area of cylinder. (4 marks)

Minimum value found when dA/dr > 0

4*pi*r = 800/r^2

r^3 = 800/4pi

cube root the answer to get r.

R= 3.99 (3 s.f).

Prove it’s a minimum point using d^2A/dr^2 as it is > 0.

Then put r into the equation to get A = 301 (3 s.f)

Q10. You are given a trapezium with AD and BC being parallel. AD = 32m, AB = 80m, and DE = 15m. Consists of a triangle shaped meadow, a sector shaped pond with radius 10 (inside the triangle) and a car park.

i) Find the distance AE. (2 marks)

AE^2 = 32^2 + 15^2 - 2(32)(15)cos116

AE = root answer

AE = 40.9 (3 s.f)

ii) Find the perpendicular distance from AE to D and hence show that the pond lies completely within the triangle (3 marks)

Formula for area of triangle: 1/2bh or 1/2abSinC

1/2abSinC = 1/2 * 32 * 15 * sin116 = 215.7

1/2 * b * h = 215.7, rearrange to get h = (215.7 * 2)/40.9 h = 10.55.

10.55 > 10 which is the radius of the pond so the pond lies completely within the triangle

iii) Find the area of the meadow, which consists of everything in the triangle except the pond (4 marks)

Area of sector = angle/360 * pi * r^2

Area of meadow = triangle - sector

215.7 - (116/360 * pi * 100)

Area of meadow = 114.5m^2

iv) Show that the area of the car park takes up more than 90% of the shape (4 marks)

You use the angles to find the base of the trapezium and u can use 1/2(a+b)*h or split the trapezium into two more triangles and find the area blabla it’s really long and if someone has the working out for it comment below and I’ll add it but in the end trapezium takes up like 94% of the whole thing and that proves it’s more than 90%.

Q12. Arif earns 30000, gets annual increase of 1000 (therefore arithmetic equation with formula = 30000 + (n-1)10000)

Bettina earns 25000, gets annual increase of 5% from each year (therefore geometric sequence with formula = 250000 * 1.05^n-1)

i) Show that on year 10, Arif earns more than Bettina, but in year 11, Bettina earns more than Arif. (4 marks)

For year 10: Arif is 39000, Bettina is 38783 so Asif has more

For year 11: Arif has 40000, Bettina has 40722, so Bettina has more.

ii) Find the total earned after 17 years for both Arif and Bettina, round it off to the nearest £100 and show they are both equal (4 marks)

Both Asif and Bettina earned 646000 in total rounded off to nearest 100 (Asifs was exact, Bettinas had to be rounded off) therefore you've shown they're the same.

iii) Prove that when the total Bettina has earned is greater than M; n > [log(M+500000) - log(500000)] / log(1.05), hence find the value of N when Bettina has earned more than £1.2 million (5 marks)

Proof using the sum formula from formula booklet:

25000(1.05^n – 1)/1.05-1 > M

25000(1.05^n – 1)/0.05 > M

25000(1.05^n – 1) > 0.05M

500000(1.05^n – 1) > M

(500000 x 1.05^n) – 500000 > M

500000 x 1.05^n > M + 500000

1.05^n > (M + 500000)/500000

log1.05^n > log (M + 500000) – log500000

nlog1.05 > log (M + 500000) – log500000

n > (log (M + 500000) – log500000)/log1.05

>Substitute M in as 1,200,000 n > 25.08… n must equal 26 as you can’t round down, and if you insert 25 as the value, sum is less than £1.2 million.

I may have gotten some of the order of the questions wrong and if there’s anything someone knows for sure or if anyone knows the marks for certain questions please let me know and I’ll add them in.

Find paper at: https://twitter.com/freddyfincham/st...98315039666176

Show

Q1: Sequence

i)Find the sum from 1 to 5 of the sequence 3a+2 (SIGMA Notation) (2 marks)

5+8+11+14+17 = 55

ii) An arithmetic progression has first term 4.2 and sixth term 1.2, find common difference. (2 marks)

A = 4.2 6th term = 4.2 + 5d

4.2 + 5d = 1.2

d = -0.48

Q2: Integration

i) Use definite integration from 5 to 1 for 4x (3 marks)

4x integrated = 2x^2 2(5)^2 - 2(1)^2 = 48

ii) Use indefinite integration for 6x^1/2 (2 marks)

Integrated = (6x^1/2+1) / (1/2 + 1) + c

= 4x^3/2 + c

Q3: Given the graph of y= logx

i) Find the gradient between the points A and B where x = 0.1 and x = 0.2 (2 marks)

(-0.699 - -1)/(0.2-0.1) = 3.01 (3 s.f)

ii) Show in graph, where you could plot a point to get a better estimate of the gradient at A (1 mark)

Plot in between the two points

Q4. You are given the graph y = 2x^3. Find the equation of the normal at x = 2 and put it in the form ax + by = c (5 marks)

dy/dx = 6x^2, put x in as 2.

dy/dx = 24. 24 is the gradient of the tangent, therefore the gradient of the normal = -1/24.

At the point x = 2, y = 16. Put into formula y - y1 = m(x-x1)

y-16 = -1/24(x-2), rearrange to get y = -1/24x + 386/24.

Multiply all sides by 24 and bring x over:

x + 24y = 386.

Q5. ~Transformations

i) Describe the single transformation that maps the graph y= x^2 + 3 to y = 2x^2 + 6 (2 marks)

An enlargement/stretch of scale factor 2 parallel to the y axis

ii) Describe the single transformation that maps the graph of y = 2x^2 to y = 2(x-3)^2 blabla (2 marks)

A translation of (3 0)

Q6. You are given that a curve goes through (2,10) and dy/dx = 12x^3 - 7. Find the equation of the curve (5 marks)

Integrate 12x^3 - 7 to get 3x^4 - 7x + c

3(2)^4 - 7(4) + c = 10

34 + c = 10 so c = -25

Equation = 3x^4 - 7x - 24

Q7.

i) Sketch y= 2^x (2 marks)

Show that it passes through (0,1)

ii)Simplify logaW = 3 + logaX^5 - logaX + loga6 in its simplest form and express w in terms of a and x (3 marks)

Ignore logaW for now, and focus on right hand side:

3 becomes logaA^3

logaX^5 - loga2X becomes logaX^5 /2X

logaX^4 /2

logaA^3 + logaX^4 /2 + loga6 = loga6*A^3*X^4 /2 logaW = loga3*A^3*X^4

w = 3a^3x^4

Q8. x is given to you in radians. You are given that 6cos^2x = 5 - sinx, Show that 6sinx^2 - sinx - 1 = 0 and find the values of x for 0 < x < 2pi (5 marks)

6cosx^2 = 5 - sinx

cos^2x + sin^2x = 1

6(1-sin^2x) = 5 - sinx

Rearrange to get 6sin^2x - sinx- 1 = 0

Factorise to get (3sinx + 1)(2sinx - 1)

sinx = -1/3 or 1/2

Values of x= 3.48, 5.94 (3s.f), pi/6, 5pi/6

Q9. You are given a cylinder blabla with radius r and height h. The volume of the cylinder = pi*r^2*h Area of the cylinder = 2*pi*r*h + 2*pi*r^2. You are also given that V = 400

i) Show that the area of the cylinder is 2*pi*r^2 + 800/r (2 marks)

Rearrange 400 = pi*r^2*h to 400/pi*r^2 = h

Put that into the formula of area to get A = 2*pi*r^2 + 2*pi*r * (400/pi*r^2)

A = 2*pi*r^2 + 2*pi*r*400 / pi*r^2 ~ pi and r cancel out to give you

A = 2*pi*r^2 + 800/r

ii) Find dA/dr and d^2A/dr^2 (4 marks)

dA/dr = 4*pi*r - 800r^-2

d^2A/dr^2 = 4*pi + 1600r^-3

iii) Hence, find the minimum value for r in this cylinder and find the corresponding surface area of cylinder. (4 marks)

Minimum value found when dA/dr > 0

4*pi*r = 800/r^2

r^3 = 800/4pi

cube root the answer to get r.

R= 3.99 (3 s.f).

Prove it’s a minimum point using d^2A/dr^2 as it is > 0.

Then put r into the equation to get A = 301 (3 s.f)

Q10. You are given a trapezium with AD and BC being parallel. AD = 32m, AB = 80m, and DE = 15m. Consists of a triangle shaped meadow, a sector shaped pond with radius 10 (inside the triangle) and a car park.

i) Find the distance AE. (2 marks)

AE^2 = 32^2 + 15^2 - 2(32)(15)cos116

AE = root answer

AE = 40.9 (3 s.f)

ii) Find the perpendicular distance from AE to D and hence show that the pond lies completely within the triangle (3 marks)

Formula for area of triangle: 1/2bh or 1/2abSinC

1/2abSinC = 1/2 * 32 * 15 * sin116 = 215.7

1/2 * b * h = 215.7, rearrange to get h = (215.7 * 2)/40.9 h = 10.55.

10.55 > 10 which is the radius of the pond so the pond lies completely within the triangle

iii) Find the area of the meadow, which consists of everything in the triangle except the pond (4 marks)

Area of sector = angle/360 * pi * r^2

Area of meadow = triangle - sector

215.7 - (116/360 * pi * 100)

Area of meadow = 114.5m^2

iv) Show that the area of the car park takes up more than 90% of the shape (4 marks)

You use the angles to find the base of the trapezium and u can use 1/2(a+b)*h or split the trapezium into two more triangles and find the area blabla it’s really long and if someone has the working out for it comment below and I’ll add it but in the end trapezium takes up like 94% of the whole thing and that proves it’s more than 90%.

Q12. Arif earns 30000, gets annual increase of 1000 (therefore arithmetic equation with formula = 30000 + (n-1)10000)

Bettina earns 25000, gets annual increase of 5% from each year (therefore geometric sequence with formula = 250000 * 1.05^n-1)

i) Show that on year 10, Arif earns more than Bettina, but in year 11, Bettina earns more than Arif. (4 marks)

For year 10: Arif is 39000, Bettina is 38783 so Asif has more

For year 11: Arif has 40000, Bettina has 40722, so Bettina has more.

ii) Find the total earned after 17 years for both Arif and Bettina, round it off to the nearest £100 and show they are both equal (4 marks)

Both Asif and Bettina earned 646000 in total rounded off to nearest 100 (Asifs was exact, Bettinas had to be rounded off) therefore you've shown they're the same.

iii) Prove that when the total Bettina has earned is greater than M; n > [log(M+500000) - log(500000)] / log(1.05), hence find the value of N when Bettina has earned more than £1.2 million (5 marks)

Proof using the sum formula from formula booklet:

25000(1.05^n – 1)/1.05-1 > M

25000(1.05^n – 1)/0.05 > M

25000(1.05^n – 1) > 0.05M

500000(1.05^n – 1) > M

(500000 x 1.05^n) – 500000 > M

500000 x 1.05^n > M + 500000

1.05^n > (M + 500000)/500000

log1.05^n > log (M + 500000) – log500000

nlog1.05 > log (M + 500000) – log500000

n > (log (M + 500000) – log500000)/log1.05

>Substitute M in as 1,200,000 n > 25.08… n must equal 26 as you can’t round down, and if you insert 25 as the value, sum is less than £1.2 million.

I may have gotten some of the order of the questions wrong and if there’s anything someone knows for sure or if anyone knows the marks for certain questions please let me know and I’ll add them in.

Find paper at: https://twitter.com/freddyfincham/st...98315039666176

S1:

Spoiler:

Q1. Sum of x = 17100, n= 50, Sum of x^2= something like 5998112 I forgot though.

i) Find mean and standard deviation (?? marks)

Standard deviation I found was 73.8 and mean was 342

ii) lalalal new data: y= 0.108x + ?? (?? marks) To find new mean, you put the mean into the equation and write out the answer, to find standard deviation you just multiply it by 0.108.

Q2. Kids walk, cycle, drive, bus to school. Data taken of 50 children who do this to school and numbers of each are listed in a table. Four children are chosen at random.

i) Number of combinations for choosing 4 (?? marks)

50C4

ii) Probability that all 4 children chosen walk (?? marks)

17C4/ 50C4

ii) Probability that at least 2 of any given situation (walking, driving, cycling, bus) are chosen. (4? marks)

Probability at least 2 = 1 - probability all use different: 1 - (17C1 * 9C1 * 13C1 * 11C1)/(50C4)

Q3: Getting a seat going to work = 0.4, not getting it = 0.6. Getting a seat going home = 0.8, not getting it = 0.2

i) Whats the probability of getting a seat going home but not going to work (?? marks)

0.8 * 0.6 = 0.48

ii)Probability of getting atleast 1 seat (?? marks)

Probability of getting at least one seat = 1 - probability of getting no seats. 1 - (0.6*0.2) = 0.88

iii) No idea but I know it was a question using P(A|B) and B was given that he had gotten at least one seat. So x/0.88 ?

Q4. Question had discrete random variables, throwing two dice(going up to 4) and variable X was difference between two scores. r 0 1 2 3, probabilities are 1/4, 3/8, 1/4, 1/8

i) Show that P(X=1) is 3/8. (?? marks)

You can use a two way table or list out the possible combinations of getting 1: (1 2, 2 1, 2 3, 3 2, 3 4, 4 3) and add all the probabilities [6 * (1/4 * 1/4)]

ii) Find E(X) and Var(X) (?? marks)

E(X) = 1.25. Var(X) = 2.5 - 1.25^2 = 0.938 (3 s.f)

Q5. Emily winning was 0.45, S winning was 0.55. Needed to win 3 games to win the table tennis match.

i) Probability S wins all her games (2-4 marks)

0.55^3 = 0.166 to 3s.f

ii) Probability that E wins the match (5 marks)

Got this wrong because I added an extra probability but answer was 6 x (0.55^2 x 0.45^3) + (0.45^3 + 3 x (0.45^3 * 0.55). This is because theres a maximum of 5 games that can occur, Emily can lose twice, then win three times. LLWWW, This can happen 6 times. Emily can lose once, then win three times, LWWW. This can happen 3 times. Emily can win all three games at once, WWW. This can happen once. Answer was 0.407 to 3 s.f

Q6. Pollution levels judged in 358 days on Marylebourne Road. On seven of the days, pollution levels weren’t calculated or bla-bla-bla. (19 marks in total)

i)Draw the cumulative frequency graph on graph paper provided (5 marks)

~Start curve from 40, plot end points, add up value of frequency each time you go up. Probably wrong in some parts but the values I remember are: 40-80 80-120 120-160 160-200, 200-220, 220-300 With frequencies 29, 74, 53, 129, 63, 10

ii) Using graph, find median and IQR. (?? marks)

As far as I can remember, I got LQ as 112, M as 146, and 176 as UQ. IQR was 64, hopefully they accept a range of values or accept it as long as the graph was used to calculate it.

iii) Check for any outliers, discussing whether on either end there is definitely no outlier, possibly an outlier, or definitely an outlier (?? marks)

Lower end definitely had no outliers, upper end possibly had outliers as outliers were >273 or smith and the values were from 220-300 so could possibly contain outliers.

iv) Draw box plot for your data (2? marks)

Box plot started at 40, ended at 300, with line for LQ, median and UQ.

v) Compare skewness of data in marylebourne road with data from a road in tower hamlet (2 marks)

Marylebourne road and Tower Hamlet both has positive skews, but positive skew in Tower hamlet was stronger/more positively skewed.

Q7. About dogs and shampoo which relieves their symptoms for this disease, forgot the name. Probability of the symptoms being relieved using the shampoo was 0.75 (17 marks in total)

i)) Forgot the n value they gave us but

(A) Work out the probability of exactly 9 dogs being treated. (3 marks)

P(X=9) = ???

(B) Work out probability of at leasteast 9 dogs being treated (3 marks)

P(X>=9) = 1 - P(X<=8), use data tables for answer.

ii) New shampoo introduced, and it is believed that it relieves higher proportion of symptoms

(A?) Write down null and alternative hypothesis, giving reasons for your alternative hypothesis (2 marks)

Let p be the probability that the shampoo relieves the symptoms of a randomly chosen dog H0: P = 0.75 H1: P > 0.75 Believed that shampoo relieved greater proportion of dogs symptoms so p> 0.75

(B?) Check effectiveness of new shampoo at 10%, with n = 18 and 16 dogs found relieved of symptoms. (5 marks)

P(X>= 16) = 1 - P(X<= 15) Answer is >0.10 Accept H0, reject H1 as there is insufficient evidence to suggest greater proportion of dogs had relieved symptoms.

(C?) In another bla-bla-bla, 50 dogs were taken as a sample and 42 had their symptoms relieved. Conduct a hypothesis test for this at 10% significance level for X~B(50, 0.75) (4 marks)

Gave values for P(X= 41, 42) and P(X<= 41, 42) I believe. You had to do P(X>= 42) so 1 - P(X<= 41) and the answer was < 0.10 therefore accept H1, reject H0. There is sufficient evidence to suggest greater proportion of dogs had relieved symptoms.

Show

Q1. Sum of x = 17100, n= 50, Sum of x^2= something like 5998112 I forgot though.

i) Find mean and standard deviation (?? marks)

Standard deviation I found was 73.8 and mean was 342

ii) lalalal new data: y= 0.108x + ?? (?? marks) To find new mean, you put the mean into the equation and write out the answer, to find standard deviation you just multiply it by 0.108.

Q2. Kids walk, cycle, drive, bus to school. Data taken of 50 children who do this to school and numbers of each are listed in a table. Four children are chosen at random.

i) Number of combinations for choosing 4 (?? marks)

50C4

ii) Probability that all 4 children chosen walk (?? marks)

17C4/ 50C4

ii) Probability that at least 2 of any given situation (walking, driving, cycling, bus) are chosen. (4? marks)

Probability at least 2 = 1 - probability all use different: 1 - (17C1 * 9C1 * 13C1 * 11C1)/(50C4)

Q3: Getting a seat going to work = 0.4, not getting it = 0.6. Getting a seat going home = 0.8, not getting it = 0.2

i) Whats the probability of getting a seat going home but not going to work (?? marks)

0.8 * 0.6 = 0.48

ii)Probability of getting atleast 1 seat (?? marks)

Probability of getting at least one seat = 1 - probability of getting no seats. 1 - (0.6*0.2) = 0.88

iii) No idea but I know it was a question using P(A|B) and B was given that he had gotten at least one seat. So x/0.88 ?

Q4. Question had discrete random variables, throwing two dice(going up to 4) and variable X was difference between two scores. r 0 1 2 3, probabilities are 1/4, 3/8, 1/4, 1/8

i) Show that P(X=1) is 3/8. (?? marks)

You can use a two way table or list out the possible combinations of getting 1: (1 2, 2 1, 2 3, 3 2, 3 4, 4 3) and add all the probabilities [6 * (1/4 * 1/4)]

ii) Find E(X) and Var(X) (?? marks)

E(X) = 1.25. Var(X) = 2.5 - 1.25^2 = 0.938 (3 s.f)

Q5. Emily winning was 0.45, S winning was 0.55. Needed to win 3 games to win the table tennis match.

i) Probability S wins all her games (2-4 marks)

0.55^3 = 0.166 to 3s.f

ii) Probability that E wins the match (5 marks)

Got this wrong because I added an extra probability but answer was 6 x (0.55^2 x 0.45^3) + (0.45^3 + 3 x (0.45^3 * 0.55). This is because theres a maximum of 5 games that can occur, Emily can lose twice, then win three times. LLWWW, This can happen 6 times. Emily can lose once, then win three times, LWWW. This can happen 3 times. Emily can win all three games at once, WWW. This can happen once. Answer was 0.407 to 3 s.f

Q6. Pollution levels judged in 358 days on Marylebourne Road. On seven of the days, pollution levels weren’t calculated or bla-bla-bla. (19 marks in total)

i)Draw the cumulative frequency graph on graph paper provided (5 marks)

~Start curve from 40, plot end points, add up value of frequency each time you go up. Probably wrong in some parts but the values I remember are: 40-80 80-120 120-160 160-200, 200-220, 220-300 With frequencies 29, 74, 53, 129, 63, 10

ii) Using graph, find median and IQR. (?? marks)

As far as I can remember, I got LQ as 112, M as 146, and 176 as UQ. IQR was 64, hopefully they accept a range of values or accept it as long as the graph was used to calculate it.

iii) Check for any outliers, discussing whether on either end there is definitely no outlier, possibly an outlier, or definitely an outlier (?? marks)

Lower end definitely had no outliers, upper end possibly had outliers as outliers were >273 or smith and the values were from 220-300 so could possibly contain outliers.

iv) Draw box plot for your data (2? marks)

Box plot started at 40, ended at 300, with line for LQ, median and UQ.

v) Compare skewness of data in marylebourne road with data from a road in tower hamlet (2 marks)

Marylebourne road and Tower Hamlet both has positive skews, but positive skew in Tower hamlet was stronger/more positively skewed.

Q7. About dogs and shampoo which relieves their symptoms for this disease, forgot the name. Probability of the symptoms being relieved using the shampoo was 0.75 (17 marks in total)

i)) Forgot the n value they gave us but

(A) Work out the probability of exactly 9 dogs being treated. (3 marks)

P(X=9) = ???

(B) Work out probability of at leasteast 9 dogs being treated (3 marks)

P(X>=9) = 1 - P(X<=8), use data tables for answer.

ii) New shampoo introduced, and it is believed that it relieves higher proportion of symptoms

(A?) Write down null and alternative hypothesis, giving reasons for your alternative hypothesis (2 marks)

Let p be the probability that the shampoo relieves the symptoms of a randomly chosen dog H0: P = 0.75 H1: P > 0.75 Believed that shampoo relieved greater proportion of dogs symptoms so p> 0.75

(B?) Check effectiveness of new shampoo at 10%, with n = 18 and 16 dogs found relieved of symptoms. (5 marks)

P(X>= 16) = 1 - P(X<= 15) Answer is >0.10 Accept H0, reject H1 as there is insufficient evidence to suggest greater proportion of dogs had relieved symptoms.

(C?) In another bla-bla-bla, 50 dogs were taken as a sample and 42 had their symptoms relieved. Conduct a hypothesis test for this at 10% significance level for X~B(50, 0.75) (4 marks)

Gave values for P(X= 41, 42) and P(X<= 41, 42) I believe. You had to do P(X>= 42) so 1 - P(X<= 41) and the answer was < 0.10 therefore accept H1, reject H0. There is sufficient evidence to suggest greater proportion of dogs had relieved symptoms.

~L

edit: better s1 mark scheme here just didn't see it before x.x: https://docs.google.com/document/d/1...kfpLx2sO0/edit

pictures of c2 paper at: https://twitter.com/freddyfincham/st...98315039666176

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#3

How did People find C1?

It wasn't too bad IMO but I think, looking at past papers and past grade boundaries that an A should be around 57/58 for an A.

What do you guys think?

The Rhombus question was weird, (C1 never has suprises lol), the 4 mark proof was fine, but I'm they always split proofs into parts to make it easier, the biomonial expansion was fine , but again it was different to previous years. They really changed up the style of questions this year (a bit.)

It wasn't too bad IMO but I think, looking at past papers and past grade boundaries that an A should be around 57/58 for an A.

What do you guys think?

The Rhombus question was weird, (C1 never has suprises lol), the 4 mark proof was fine, but I'm they always split proofs into parts to make it easier, the biomonial expansion was fine , but again it was different to previous years. They really changed up the style of questions this year (a bit.)

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(Original post by

How did People find C1?

It wasn't too bad IMO but I think, looking at past papers and past grade boundaries that an A should be around 57/58 for an A.

What do you guys think?

The Rhombus question was weird, (C1 never has suprises lol), the 4 mark proof was fine, but I'm they always split proofs into parts to make it easier, the biomonial expansion was fine , but again it was different to previous years. They really changed up the style of questions this year (a bit.)

**Kira Yagami**)How did People find C1?

It wasn't too bad IMO but I think, looking at past papers and past grade boundaries that an A should be around 57/58 for an A.

What do you guys think?

The Rhombus question was weird, (C1 never has suprises lol), the 4 mark proof was fine, but I'm they always split proofs into parts to make it easier, the biomonial expansion was fine , but again it was different to previous years. They really changed up the style of questions this year (a bit.)

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#5

(Original post by

idk recently they've been gradually increasing and 2015 and 2016 were both 63 for an A and this paper was harder compared to those so I think it'll be like 59 or 60

**LokiiR**)idk recently they've been gradually increasing and 2015 and 2016 were both 63 for an A and this paper was harder compared to those so I think it'll be like 59 or 60

If you like at June 11 which was 55 for an A, I think the grade boundaries should be similar because that was a decent paper. But I doubt they'd be that low (but even people on TSR found the C1 a bit weird, when it's always just VERY easy, so I dunno).

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#6

these were my answers for c2:

1) 55

2) -0.48

3) 48

4) 4x^3/2 + c

5) i) 3.01

ii) draw nearer to A

6)-1/24x + 193/12

7) i) Enlargement SF 2

ii)Translation vector (3 0)

8)3a-7x(not sure on this one)

9) x= 0.52,2.62,3.48,5.94

10) i) divide by r, times by 2 to get 800/r, and substitute

ii) minimum stationary point

r=3.99

SA= 301

iii) distance AE=40.9

iv) prove pond is inside. perpendicular distance=10.56 and radius was 10 therefore inside

v) pond area= 101

meadow area=115

vi) car park/total trapezeium= 0.947.., therefore greater than 90%

11) i) prove benetta has lower in year 10 than arif, but higher in year 11

arif = 30000 +(n-1)1000

benetta= 25000x1.05^n-1

ii) sum of 17 years

iii) 646000 and 646009.16, therefore to the nearest 100 same

iv) a(1.05^n-1)/0.05 > M

rearrange for 1.05^n, then take logs and simplify

v) n=25.082 but sum of 25 years is 1.19million, therefore first year to be >1.2million is year 26

therefore n=26

1) 55

2) -0.48

3) 48

4) 4x^3/2 + c

5) i) 3.01

ii) draw nearer to A

6)-1/24x + 193/12

7) i) Enlargement SF 2

ii)Translation vector (3 0)

8)3a-7x(not sure on this one)

9) x= 0.52,2.62,3.48,5.94

10) i) divide by r, times by 2 to get 800/r, and substitute

ii) minimum stationary point

r=3.99

SA= 301

iii) distance AE=40.9

iv) prove pond is inside. perpendicular distance=10.56 and radius was 10 therefore inside

v) pond area= 101

meadow area=115

vi) car park/total trapezeium= 0.947.., therefore greater than 90%

11) i) prove benetta has lower in year 10 than arif, but higher in year 11

arif = 30000 +(n-1)1000

benetta= 25000x1.05^n-1

ii) sum of 17 years

iii) 646000 and 646009.16, therefore to the nearest 100 same

iv) a(1.05^n-1)/0.05 > M

rearrange for 1.05^n, then take logs and simplify

v) n=25.082 but sum of 25 years is 1.19million, therefore first year to be >1.2million is year 26

therefore n=26

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Added unofficial markscheme for c2 if i've made any mistakes please correct them and let me know

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(Original post by

Would you happen to know how many marks each question in C2 was worth?

**Wizard80**)Would you happen to know how many marks each question in C2 was worth?

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(Original post by

Here is the C2 paper:

https://twitter.com/freddyfincham/st...98315039666176

**gemma.louise**)Here is the C2 paper:

https://twitter.com/freddyfincham/st...98315039666176

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#12

(Original post by

Hey guys, theres quite a few threads for all the exams and it takes a while to post my answers on each and everyone so Ive decided to make a thread. I'm going to be uploading all the answers I recieved for C1 C2 and S1 here and Ill probably make a thread for AQA BIO/CHEM AS but that isnt certain, if I do Ill be posting the link to that here.

C1:

C2:

S1:

~L

edit: better s1 mark scheme here just didn't see it before x.x: https://docs.google.com/document/d/1...kfpLx2sO0/edit

pictures of c2 paper at: https://twitter.com/freddyfincham/st...98315039666176

**LokiiR**)Hey guys, theres quite a few threads for all the exams and it takes a while to post my answers on each and everyone so Ive decided to make a thread. I'm going to be uploading all the answers I recieved for C1 C2 and S1 here and Ill probably make a thread for AQA BIO/CHEM AS but that isnt certain, if I do Ill be posting the link to that here.

C1:

Spoiler:

1. Sketched y= -2x + 1 (0,1) and gradient was -2 (2 marks)

2.

i) (1 7/9)^-1/2 = 3/4 (3 marks)

ii) (6x^5y^2)^3/18y^10 = 12x^15/y^4 (2 marks)

3. 6-x > 5(x-3) = 6 - x > 5x - 15 = -6x > -21 = x< 21/6 (3 marks)

4. Intersection of 2x+5y=5 and x-2y = (10/3, -1/3) (4 marjs)

5. (x+2)^2 + (y-3)^2 = 5

i) radius is root 5, centre is (-2,3) (2 marks)

ii) Line parallel to 5x + y = 4 and touches centre is y= -5x-7 (2 marks)

6. r = root (v/a+b) - Make B the subject : Two possible answers are b = (v-ar^2)/r^2 or b = v/r^2 - a (4 marks)

7.

i) (5-2root7)/3+root7 = (29-11root7)/2 (3 marks)

ii) 12/root2 + root 98 = 13root2 (2 marks)

8. (a+bx)^5 x^3 coefficient is -1080, constant is 32. a^5 = 32, so a = 2. b= -3 (5 marks)

9. Three consecutive integers and n is the smallest (therefore n, n+1 and n+2) show difference between squares of smallest and largest is 4 * middle. (n+2)^2 - n^2 = 4n+4. 4n+4 = 4(n+1). (4 marks.

Section B

10. A(3,3) B (-2,-2) C (5,-1)

i) Show AB is BC. AB = ROOT 50, BC = ROOT 50 (2 marks)

ii)Line perpendicular to AC which passes by B

Grad of AC is -2, perp gradient is 1/2. Line is y= 1/2x-1 (4 marks)

iii) D is (10,4) because difference between B and C is 7, 1. Add 7 to x of A and 1 to y of A to get 10,4 (2 marks)

iv) 3.8 < 26/7. Find equation of lines and prove 8, 3.8 does not belong in them therefore it is outside rhombus (4 marks)

11. f(x) = (x-2)(2x-3)(x+5)

i) Sketch graph (3 marks)

ii) Show when (-3 0) occurs that g(x) is 2x^3 + 21x^2 + 43x + 24. You use f(x+3) and from that x+3 = 2, 3/2, -5

x = -1, -3/2, -8 are new roots

factors are (x+1)(2x+3)(x+8), multiply out to get the answer (3 marks)

iii)x = -2 is root of g(x)=6, prove this and find other roots. Other roots = (-17+-root217)/4 - Not sure of this (6 marks)

12.

i) Show y= x^2 + x + 3 as (x+a)^2 + b and it doesnt touch the x axis. Completed square is (x+1/2)^2 + 11/4 So minimum point is -1/2, 11/4 which is above x axis (4 marks)

ii) Intersection of x^2+x+3 and 2x^2-3x-9 are (6,45) and (-2,5) (4 marks)

iii) For what values of k is there no intersection of x^2 + x + k and 2x^2 - 3x- 9

Use b^2 - 4ac<0

k< -13

Show

1. Sketched y= -2x + 1 (0,1) and gradient was -2 (2 marks)

2.

i) (1 7/9)^-1/2 = 3/4 (3 marks)

ii) (6x^5y^2)^3/18y^10 = 12x^15/y^4 (2 marks)

3. 6-x > 5(x-3) = 6 - x > 5x - 15 = -6x > -21 = x< 21/6 (3 marks)

4. Intersection of 2x+5y=5 and x-2y = (10/3, -1/3) (4 marjs)

5. (x+2)^2 + (y-3)^2 = 5

i) radius is root 5, centre is (-2,3) (2 marks)

ii) Line parallel to 5x + y = 4 and touches centre is y= -5x-7 (2 marks)

6. r = root (v/a+b) - Make B the subject : Two possible answers are b = (v-ar^2)/r^2 or b = v/r^2 - a (4 marks)

7.

i) (5-2root7)/3+root7 = (29-11root7)/2 (3 marks)

ii) 12/root2 + root 98 = 13root2 (2 marks)

8. (a+bx)^5 x^3 coefficient is -1080, constant is 32. a^5 = 32, so a = 2. b= -3 (5 marks)

9. Three consecutive integers and n is the smallest (therefore n, n+1 and n+2) show difference between squares of smallest and largest is 4 * middle. (n+2)^2 - n^2 = 4n+4. 4n+4 = 4(n+1). (4 marks.

Section B

10. A(3,3) B (-2,-2) C (5,-1)

i) Show AB is BC. AB = ROOT 50, BC = ROOT 50 (2 marks)

ii)Line perpendicular to AC which passes by B

Grad of AC is -2, perp gradient is 1/2. Line is y= 1/2x-1 (4 marks)

iii) D is (10,4) because difference between B and C is 7, 1. Add 7 to x of A and 1 to y of A to get 10,4 (2 marks)

iv) 3.8 < 26/7. Find equation of lines and prove 8, 3.8 does not belong in them therefore it is outside rhombus (4 marks)

11. f(x) = (x-2)(2x-3)(x+5)

i) Sketch graph (3 marks)

ii) Show when (-3 0) occurs that g(x) is 2x^3 + 21x^2 + 43x + 24. You use f(x+3) and from that x+3 = 2, 3/2, -5

x = -1, -3/2, -8 are new roots

factors are (x+1)(2x+3)(x+8), multiply out to get the answer (3 marks)

iii)x = -2 is root of g(x)=6, prove this and find other roots. Other roots = (-17+-root217)/4 - Not sure of this (6 marks)

12.

i) Show y= x^2 + x + 3 as (x+a)^2 + b and it doesnt touch the x axis. Completed square is (x+1/2)^2 + 11/4 So minimum point is -1/2, 11/4 which is above x axis (4 marks)

ii) Intersection of x^2+x+3 and 2x^2-3x-9 are (6,45) and (-2,5) (4 marks)

iii) For what values of k is there no intersection of x^2 + x + k and 2x^2 - 3x- 9

Use b^2 - 4ac<0

k< -13

C2:

Spoiler:

Q1: Sequence

i)Find the sum from 1 to 5 of the sequence 3a+2 (SIGMA Notation) (2 marks)

5+8+11+14+17 = 55

ii) An arithmetic progression has first term 4.2 and sixth term 1.2, find common difference. (2 marks)

A = 4.2 6th term = 4.2 + 5d

4.2 + 5d = 1.2

d = -0.48

Q2: Integration

i) Use definite integration from 5 to 1 for 4x (3 marks)

4x integrated = 2x^2 2(5)^2 - 2(1)^2 = 48

ii) Use indefinite integration for 6x^1/2 (2 marks)

Integrated = (6x^1/2+1) / (1/2 + 1) + c

= 4x^3/2 + c

Q3: Given the graph of y= logx

i) Find the gradient between the points A and B where x = 0.1 and x = 0.2 (2 marks)

(-0.699 - -1)/(0.2-0.1) = 3.01 (3 s.f)

ii) Show in graph, where you could plot a point to get a better estimate of the gradient at A (1 mark)

Plot in between the two points

Q4. You are given the graph y = 2x^3. Find the equation of the normal at x = 2 and put it in the form ax + by = c (5 marks)

dy/dx = 6x^2, put x in as 2.

dy/dx = 24. 24 is the gradient of the tangent, therefore the gradient of the normal = -1/24.

At the point x = 2, y = 16. Put into formula y - y1 = m(x-x1)

y-16 = -1/24(x-2), rearrange to get y = -1/24x + 386/24.

Multiply all sides by 24 and bring x over: 24x + y = 386.

Q5. ~Transformations

i) Describe the single transformation that maps the graph y= x^2 + 3 to y = 2x^2 + 6 (2 marks)

An enlargement/stretch of scale factor 2 parallel to the y axis

ii) Describe the single transformation that maps the graph of y = 2x^2 to y = 2(x-3)^2 blabla (2 marks)

A translation of (3 0)

Q6. You are given that a curve goes through (2,10) and dy/dx = 12x^3 - 7. Find the equation of the curve (5 marks)

Integrate 12x^3 - 7 to get 3x^4 - 7x + c

3(2)^4 - 7(4) + c = 10

34 + c = 10 so c = -25

Equation = 3x^4 - 7x - 24

Q7.

i) Sketch y= 2^x (2 marks)

Show that it passes through (0,1)

ii)Simplify logaW = 3 + logaX^5 - logaX + loga6 in its simplest form and express w in terms of a and x (3 marks)

Ignore logaW for now, and focus on right hand side:

3 becomes logaA^3

logaX^5 - loga2X becomes logaX^5 /2X

logaX^4 /2

logaA^3 + logaX^4 /2 + loga6 = loga6*A^3*X^4 /2 logaW = loga3*A^3*X^4

w = 3a^3x^4

Q8. x is given to you in radians. You are given that 6cos^2x = 5 - sinx, Show that 6sinx^2 - sinx - 1 = 0 and find the values of x for 0 < x < 2pi (5 marks)

6cosx^2 = 5 - sinx

cos^2x + sin^2x = 1

6(1-sin^2x) = 5 - sinx

Rearrange to get 6sin^2x - sinx- 1 = 0

Factorise to get (3sinx + 1)(2sinx - 1)

sinx = -1/3 or 1/2

Values of x= 3.48, 5.94 (3s.f), pi/6, 5pi/6

Q9. You are given a cylinder blabla with radius r and height h. The volume of the cylinder = pi*r^2*h Area of the cylinder = 2*pi*r*h + 2*pi*r^2. You are also given that V = 400

i) Show that the area of the cylinder is 2*pi*r^2 + 800/r (2 marks)

Rearrange 400 = pi*r^2*h to 400/pi*r^2 = h

Put that into the formula of area to get A = 2*pi*r^2 + 2*pi*r * (400/pi*r^2)

A = 2*pi*r^2 + 2*pi*r*400 / pi*r^2 ~ pi and r cancel out to give you

A = 2*pi*r^2 + 800/r

ii) Find dA/dr and d^2A/dr^2 (4 marks)

dA/dr = 4*pi*r - 800r^-2

d^2A/dr^2 = 4*pi + 1600r^-3

iii) Hence, find the minimum value for r in this cylinder and find the corresponding surface area of cylinder. (4 marks)

Minimum value found when dA/dr > 0

4*pi*r = 800/r^2

r^3 = 800/4pi

cube root the answer to get r.

R= 3.99 (3 s.f).

Prove it’s a minimum point using d^2A/dr^2 as it is > 0.

Then put r into the equation to get A = 301 (3 s.f)

Q10. You are given a trapezium with AD and BC being parallel. AD = 32m, AB = 80m, and DE = 15m. Consists of a triangle shaped meadow, a sector shaped pond with radius 10 (inside the triangle) and a car park.

i) Find the distance AE. (2 marks)

AE^2 = 32^2 + 15^2 - 2(32)(15)cos116

AE = root answer

AE = 40.9 (3 s.f)

ii) Find the perpendicular distance from AE to D and hence show that the pond lies completely within the triangle (3 marks)

Formula for area of triangle: 1/2bh or 1/2abSinC

1/2abSinC = 1/2 * 32 * 15 * sin116 = 215.7

1/2 * b * h = 215.7, rearrange to get h = (215.7 * 2)/40.9 h = 10.55.

10.55 > 10 which is the radius of the pond so the pond lies completely within the triangle

iii) Find the area of the meadow, which consists of everything in the triangle except the pond (4 marks)

Area of sector = angle/360 * pi * r^2

Area of meadow = triangle - sector

215.7 - (116/360 * pi * 100)

Area of meadow = 114.5m^2

iv) Show that the area of the car park takes up more than 90% of the shape (4 marks)

You use the angles to find the base of the trapezium and u can use 1/2(a+b)*h or split the trapezium into two more triangles and find the area blabla it’s really long and if someone has the working out for it comment below and I’ll add it but in the end trapezium takes up like 94% of the whole thing and that proves it’s more than 90%.

Q12. Arif earns 30000, gets annual increase of 1000 (therefore arithmetic equation with formula = 30000 + (n-1)10000)

Bettina earns 25000, gets annual increase of 5% from each year (therefore geometric sequence with formula = 250000 * 1.05^n-1)

i) Show that on year 10, Arif earns more than Bettina, but in year 11, Bettina earns more than Arif. (4 marks)

For year 10: Arif is 39000, Bettina is 38783 so Asif has more

For year 11: Arif has 40000, Bettina has 40722, so Bettina has more.

ii) Find the total earned after 17 years for both Arif and Bettina, round it off to the nearest £100 and show they are both equal (4 marks)

Both Asif and Bettina earned 646000 in total rounded off to nearest 100 (Asifs was exact, Bettinas had to be rounded off) therefore you've shown they're the same.

iii) Prove that when the total Bettina has earned is greater than M; n > [log(M+500000) - log(500000)] / log(1.05), hence find the value of N when Bettina has earned more than £1.2 million (5 marks)

Proof using the sum formula from formula booklet:

25000(1.05^n – 1)/1.05-1 > M

25000(1.05^n – 1)/0.05 > M

25000(1.05^n – 1) > 0.05M

500000(1.05^n – 1) > M

(500000 x 1.05^n) – 500000 > M

500000 x 1.05^n > M + 500000

1.05^n > (M + 500000)/500000

log1.05^n > log (M + 500000) – log500000

nlog1.05 > log (M + 500000) – log500000

n > (log (M + 500000) – log500000)/log1.05

>Substitute M in as 1,200,000 n > 25.08… n must equal 26 as you can’t round down, and if you insert 25 as the value, sum is less than £1.2 million.

I may have gotten some of the order of the questions wrong and if there’s anything someone knows for sure or if anyone knows the marks for certain questions please let me know and I’ll add them in.

Find paper at: https://twitter.com/freddyfincham/st...98315039666176

Show

Q1: Sequence

i)Find the sum from 1 to 5 of the sequence 3a+2 (SIGMA Notation) (2 marks)

5+8+11+14+17 = 55

ii) An arithmetic progression has first term 4.2 and sixth term 1.2, find common difference. (2 marks)

A = 4.2 6th term = 4.2 + 5d

4.2 + 5d = 1.2

d = -0.48

Q2: Integration

i) Use definite integration from 5 to 1 for 4x (3 marks)

4x integrated = 2x^2 2(5)^2 - 2(1)^2 = 48

ii) Use indefinite integration for 6x^1/2 (2 marks)

Integrated = (6x^1/2+1) / (1/2 + 1) + c

= 4x^3/2 + c

Q3: Given the graph of y= logx

i) Find the gradient between the points A and B where x = 0.1 and x = 0.2 (2 marks)

(-0.699 - -1)/(0.2-0.1) = 3.01 (3 s.f)

ii) Show in graph, where you could plot a point to get a better estimate of the gradient at A (1 mark)

Plot in between the two points

Q4. You are given the graph y = 2x^3. Find the equation of the normal at x = 2 and put it in the form ax + by = c (5 marks)

dy/dx = 6x^2, put x in as 2.

dy/dx = 24. 24 is the gradient of the tangent, therefore the gradient of the normal = -1/24.

At the point x = 2, y = 16. Put into formula y - y1 = m(x-x1)

y-16 = -1/24(x-2), rearrange to get y = -1/24x + 386/24.

Multiply all sides by 24 and bring x over: 24x + y = 386.

Q5. ~Transformations

i) Describe the single transformation that maps the graph y= x^2 + 3 to y = 2x^2 + 6 (2 marks)

An enlargement/stretch of scale factor 2 parallel to the y axis

ii) Describe the single transformation that maps the graph of y = 2x^2 to y = 2(x-3)^2 blabla (2 marks)

A translation of (3 0)

Q6. You are given that a curve goes through (2,10) and dy/dx = 12x^3 - 7. Find the equation of the curve (5 marks)

Integrate 12x^3 - 7 to get 3x^4 - 7x + c

3(2)^4 - 7(4) + c = 10

34 + c = 10 so c = -25

Equation = 3x^4 - 7x - 24

Q7.

i) Sketch y= 2^x (2 marks)

Show that it passes through (0,1)

ii)Simplify logaW = 3 + logaX^5 - logaX + loga6 in its simplest form and express w in terms of a and x (3 marks)

Ignore logaW for now, and focus on right hand side:

3 becomes logaA^3

logaX^5 - loga2X becomes logaX^5 /2X

logaX^4 /2

logaA^3 + logaX^4 /2 + loga6 = loga6*A^3*X^4 /2 logaW = loga3*A^3*X^4

w = 3a^3x^4

Q8. x is given to you in radians. You are given that 6cos^2x = 5 - sinx, Show that 6sinx^2 - sinx - 1 = 0 and find the values of x for 0 < x < 2pi (5 marks)

6cosx^2 = 5 - sinx

cos^2x + sin^2x = 1

6(1-sin^2x) = 5 - sinx

Rearrange to get 6sin^2x - sinx- 1 = 0

Factorise to get (3sinx + 1)(2sinx - 1)

sinx = -1/3 or 1/2

Values of x= 3.48, 5.94 (3s.f), pi/6, 5pi/6

Q9. You are given a cylinder blabla with radius r and height h. The volume of the cylinder = pi*r^2*h Area of the cylinder = 2*pi*r*h + 2*pi*r^2. You are also given that V = 400

i) Show that the area of the cylinder is 2*pi*r^2 + 800/r (2 marks)

Rearrange 400 = pi*r^2*h to 400/pi*r^2 = h

Put that into the formula of area to get A = 2*pi*r^2 + 2*pi*r * (400/pi*r^2)

A = 2*pi*r^2 + 2*pi*r*400 / pi*r^2 ~ pi and r cancel out to give you

A = 2*pi*r^2 + 800/r

ii) Find dA/dr and d^2A/dr^2 (4 marks)

dA/dr = 4*pi*r - 800r^-2

d^2A/dr^2 = 4*pi + 1600r^-3

iii) Hence, find the minimum value for r in this cylinder and find the corresponding surface area of cylinder. (4 marks)

Minimum value found when dA/dr > 0

4*pi*r = 800/r^2

r^3 = 800/4pi

cube root the answer to get r.

R= 3.99 (3 s.f).

Prove it’s a minimum point using d^2A/dr^2 as it is > 0.

Then put r into the equation to get A = 301 (3 s.f)

Q10. You are given a trapezium with AD and BC being parallel. AD = 32m, AB = 80m, and DE = 15m. Consists of a triangle shaped meadow, a sector shaped pond with radius 10 (inside the triangle) and a car park.

i) Find the distance AE. (2 marks)

AE^2 = 32^2 + 15^2 - 2(32)(15)cos116

AE = root answer

AE = 40.9 (3 s.f)

ii) Find the perpendicular distance from AE to D and hence show that the pond lies completely within the triangle (3 marks)

Formula for area of triangle: 1/2bh or 1/2abSinC

1/2abSinC = 1/2 * 32 * 15 * sin116 = 215.7

1/2 * b * h = 215.7, rearrange to get h = (215.7 * 2)/40.9 h = 10.55.

10.55 > 10 which is the radius of the pond so the pond lies completely within the triangle

iii) Find the area of the meadow, which consists of everything in the triangle except the pond (4 marks)

Area of sector = angle/360 * pi * r^2

Area of meadow = triangle - sector

215.7 - (116/360 * pi * 100)

Area of meadow = 114.5m^2

iv) Show that the area of the car park takes up more than 90% of the shape (4 marks)

You use the angles to find the base of the trapezium and u can use 1/2(a+b)*h or split the trapezium into two more triangles and find the area blabla it’s really long and if someone has the working out for it comment below and I’ll add it but in the end trapezium takes up like 94% of the whole thing and that proves it’s more than 90%.

Q12. Arif earns 30000, gets annual increase of 1000 (therefore arithmetic equation with formula = 30000 + (n-1)10000)

Bettina earns 25000, gets annual increase of 5% from each year (therefore geometric sequence with formula = 250000 * 1.05^n-1)

i) Show that on year 10, Arif earns more than Bettina, but in year 11, Bettina earns more than Arif. (4 marks)

For year 10: Arif is 39000, Bettina is 38783 so Asif has more

For year 11: Arif has 40000, Bettina has 40722, so Bettina has more.

ii) Find the total earned after 17 years for both Arif and Bettina, round it off to the nearest £100 and show they are both equal (4 marks)

Both Asif and Bettina earned 646000 in total rounded off to nearest 100 (Asifs was exact, Bettinas had to be rounded off) therefore you've shown they're the same.

iii) Prove that when the total Bettina has earned is greater than M; n > [log(M+500000) - log(500000)] / log(1.05), hence find the value of N when Bettina has earned more than £1.2 million (5 marks)

Proof using the sum formula from formula booklet:

25000(1.05^n – 1)/1.05-1 > M

25000(1.05^n – 1)/0.05 > M

25000(1.05^n – 1) > 0.05M

500000(1.05^n – 1) > M

(500000 x 1.05^n) – 500000 > M

500000 x 1.05^n > M + 500000

1.05^n > (M + 500000)/500000

log1.05^n > log (M + 500000) – log500000

nlog1.05 > log (M + 500000) – log500000

n > (log (M + 500000) – log500000)/log1.05

>Substitute M in as 1,200,000 n > 25.08… n must equal 26 as you can’t round down, and if you insert 25 as the value, sum is less than £1.2 million.

I may have gotten some of the order of the questions wrong and if there’s anything someone knows for sure or if anyone knows the marks for certain questions please let me know and I’ll add them in.

Find paper at: https://twitter.com/freddyfincham/st...98315039666176

S1:

Spoiler:

Q1. Sum of x = 17100, n= 50, Sum of x^2= something like 5998112 I forgot though.

i) Find mean and standard deviation (?? marks)

Standard deviation I found was 73.8 and mean was 342

ii) lalalal new data: y= 0.108x + ?? (?? marks) To find new mean, you put the mean into the equation and write out the answer, to find standard deviation you just multiply it by 0.108.

Q2. Kids walk, cycle, drive, bus to school. Data taken of 50 children who do this to school and numbers of each are listed in a table. Four children are chosen at random.

i) Number of combinations for choosing 4 (?? marks)

50C4

ii) Probability that all 4 children chosen walk (?? marks)

17C4/ 50C4

ii) Probability that at least 2 of any given situation (walking, driving, cycling, bus) are chosen. (4? marks)

Probability at least 2 = 1 - probability all use different: 1 - (17C1 * 9C1 * 13C1 * 11C1)/(50C4)

Q3: Getting a seat going to work = 0.4, not getting it = 0.6. Getting a seat going home = 0.8, not getting it = 0.2

i) Whats the probability of getting a seat going home but not going to work (?? marks)

0.8 * 0.6 = 0.48

ii)Probability of getting atleast 1 seat (?? marks)

Probability of getting at least one seat = 1 - probability of getting no seats. 1 - (0.6*0.2) = 0.88

iii) No idea but I know it was a question using P(A|B) and B was given that he had gotten at least one seat. So x/0.88 ?

Q4. Question had discrete random variables, throwing two dice(going up to 4) and variable X was difference between two scores. r 0 1 2 3, probabilities are 1/4, 3/8, 1/4, 1/8

i) Show that P(X=1) is 3/8. (?? marks)

You can use a two way table or list out the possible combinations of getting 1: (1 2, 2 1, 2 3, 3 2, 3 4, 4 3) and add all the probabilities [6 * (1/4 * 1/4)]

ii) Find E(X) and Var(X) (?? marks)

E(X) = 1.25. Var(X) = 2.5 - 1.25^2 = 0.938 (3 s.f)

Q5. Emily winning was 0.45, S winning was 0.55. Needed to win 3 games to win the table tennis match.

i) Probability S wins all her games (2-4 marks)

0.55^3 = 0.166 to 3s.f

ii) Probability that E wins the match (5 marks)

Got this wrong because I added an extra probability but answer was 6 x (0.55^2 x 0.45^3) + (0.45^3 + 3 x (0.45^3 * 0.55). This is because theres a maximum of 5 games that can occur, Emily can lose twice, then win three times. LLWWW, This can happen 6 times. Emily can lose once, then win three times, LWWW. This can happen 3 times. Emily can win all three games at once, WWW. This can happen once. Answer was 0.407 to 3 s.f

Q6. Pollution levels judged in 358 days on Marylebourne Road. On seven of the days, pollution levels weren’t calculated or bla-bla-bla. (19 marks in total)

i)Draw the cumulative frequency graph on graph paper provided (5 marks)

~Start curve from 40, plot end points, add up value of frequency each time you go up. Probably wrong in some parts but the values I remember are: 40-80 80-120 120-160 160-200, 200-220, 220-300 With frequencies 29, 74, 53, 129, 63, 10

ii) Using graph, find median and IQR. (?? marks)

As far as I can remember, I got LQ as 112, M as 146, and 176 as UQ. IQR was 64, hopefully they accept a range of values or accept it as long as the graph was used to calculate it.

iii) Check for any outliers, discussing whether on either end there is definitely no outlier, possibly an outlier, or definitely an outlier (?? marks)

Lower end definitely had no outliers, upper end possibly had outliers as outliers were >273 or smith and the values were from 220-300 so could possibly contain outliers.

iv) Draw box plot for your data (2? marks)

Box plot started at 40, ended at 300, with line for LQ, median and UQ.

v) Compare skewness of data in marylebourne road with data from a road in tower hamlet (2 marks)

Marylebourne road and Tower Hamlet both has positive skews, but positive skew in Tower hamlet was stronger/more positively skewed.

Q7. About dogs and shampoo which relieves their symptoms for this disease, forgot the name. Probability of the symptoms being relieved using the shampoo was 0.75 (17 marks in total)

i)) Forgot the n value they gave us but

(A) Work out the probability of exactly 9 dogs being treated. (3 marks)

P(X=9) = ???

(B) Work out probability of at leasteast 9 dogs being treated (3 marks)

P(X>=9) = 1 - P(X<=8), use data tables for answer.

ii) New shampoo introduced, and it is believed that it relieves higher proportion of symptoms

(A?) Write down null and alternative hypothesis, giving reasons for your alternative hypothesis (2 marks)

Let p be the probability that the shampoo relieves the symptoms of a randomly chosen dog H0: P = 0.75 H1: P > 0.75 Believed that shampoo relieved greater proportion of dogs symptoms so p> 0.75

(B?) Check effectiveness of new shampoo at 10%, with n = 18 and 16 dogs found relieved of symptoms. (5 marks)

P(X>= 16) = 1 - P(X<= 15) Answer is >0.10 Accept H0, reject H1 as there is insufficient evidence to suggest greater proportion of dogs had relieved symptoms.

(C?) In another bla-bla-bla, 50 dogs were taken as a sample and 42 had their symptoms relieved. Conduct a hypothesis test for this at 10% significance level for X~B(50, 0.75) (4 marks)

Gave values for P(X= 41, 42) and P(X<= 41, 42) I believe. You had to do P(X>= 42) so 1 - P(X<= 41) and the answer was < 0.10 therefore accept H1, reject H0. There is sufficient evidence to suggest greater proportion of dogs had relieved symptoms.

Show

Q1. Sum of x = 17100, n= 50, Sum of x^2= something like 5998112 I forgot though.

i) Find mean and standard deviation (?? marks)

Standard deviation I found was 73.8 and mean was 342

ii) lalalal new data: y= 0.108x + ?? (?? marks) To find new mean, you put the mean into the equation and write out the answer, to find standard deviation you just multiply it by 0.108.

Q2. Kids walk, cycle, drive, bus to school. Data taken of 50 children who do this to school and numbers of each are listed in a table. Four children are chosen at random.

i) Number of combinations for choosing 4 (?? marks)

50C4

ii) Probability that all 4 children chosen walk (?? marks)

17C4/ 50C4

ii) Probability that at least 2 of any given situation (walking, driving, cycling, bus) are chosen. (4? marks)

Probability at least 2 = 1 - probability all use different: 1 - (17C1 * 9C1 * 13C1 * 11C1)/(50C4)

Q3: Getting a seat going to work = 0.4, not getting it = 0.6. Getting a seat going home = 0.8, not getting it = 0.2

i) Whats the probability of getting a seat going home but not going to work (?? marks)

0.8 * 0.6 = 0.48

ii)Probability of getting atleast 1 seat (?? marks)

Probability of getting at least one seat = 1 - probability of getting no seats. 1 - (0.6*0.2) = 0.88

iii) No idea but I know it was a question using P(A|B) and B was given that he had gotten at least one seat. So x/0.88 ?

Q4. Question had discrete random variables, throwing two dice(going up to 4) and variable X was difference between two scores. r 0 1 2 3, probabilities are 1/4, 3/8, 1/4, 1/8

i) Show that P(X=1) is 3/8. (?? marks)

You can use a two way table or list out the possible combinations of getting 1: (1 2, 2 1, 2 3, 3 2, 3 4, 4 3) and add all the probabilities [6 * (1/4 * 1/4)]

ii) Find E(X) and Var(X) (?? marks)

E(X) = 1.25. Var(X) = 2.5 - 1.25^2 = 0.938 (3 s.f)

Q5. Emily winning was 0.45, S winning was 0.55. Needed to win 3 games to win the table tennis match.

i) Probability S wins all her games (2-4 marks)

0.55^3 = 0.166 to 3s.f

ii) Probability that E wins the match (5 marks)

Got this wrong because I added an extra probability but answer was 6 x (0.55^2 x 0.45^3) + (0.45^3 + 3 x (0.45^3 * 0.55). This is because theres a maximum of 5 games that can occur, Emily can lose twice, then win three times. LLWWW, This can happen 6 times. Emily can lose once, then win three times, LWWW. This can happen 3 times. Emily can win all three games at once, WWW. This can happen once. Answer was 0.407 to 3 s.f

Q6. Pollution levels judged in 358 days on Marylebourne Road. On seven of the days, pollution levels weren’t calculated or bla-bla-bla. (19 marks in total)

i)Draw the cumulative frequency graph on graph paper provided (5 marks)

~Start curve from 40, plot end points, add up value of frequency each time you go up. Probably wrong in some parts but the values I remember are: 40-80 80-120 120-160 160-200, 200-220, 220-300 With frequencies 29, 74, 53, 129, 63, 10

ii) Using graph, find median and IQR. (?? marks)

As far as I can remember, I got LQ as 112, M as 146, and 176 as UQ. IQR was 64, hopefully they accept a range of values or accept it as long as the graph was used to calculate it.

iii) Check for any outliers, discussing whether on either end there is definitely no outlier, possibly an outlier, or definitely an outlier (?? marks)

Lower end definitely had no outliers, upper end possibly had outliers as outliers were >273 or smith and the values were from 220-300 so could possibly contain outliers.

iv) Draw box plot for your data (2? marks)

Box plot started at 40, ended at 300, with line for LQ, median and UQ.

v) Compare skewness of data in marylebourne road with data from a road in tower hamlet (2 marks)

Marylebourne road and Tower Hamlet both has positive skews, but positive skew in Tower hamlet was stronger/more positively skewed.

Q7. About dogs and shampoo which relieves their symptoms for this disease, forgot the name. Probability of the symptoms being relieved using the shampoo was 0.75 (17 marks in total)

i)) Forgot the n value they gave us but

(A) Work out the probability of exactly 9 dogs being treated. (3 marks)

P(X=9) = ???

(B) Work out probability of at leasteast 9 dogs being treated (3 marks)

P(X>=9) = 1 - P(X<=8), use data tables for answer.

ii) New shampoo introduced, and it is believed that it relieves higher proportion of symptoms

(A?) Write down null and alternative hypothesis, giving reasons for your alternative hypothesis (2 marks)

Let p be the probability that the shampoo relieves the symptoms of a randomly chosen dog H0: P = 0.75 H1: P > 0.75 Believed that shampoo relieved greater proportion of dogs symptoms so p> 0.75

(B?) Check effectiveness of new shampoo at 10%, with n = 18 and 16 dogs found relieved of symptoms. (5 marks)

P(X>= 16) = 1 - P(X<= 15) Answer is >0.10 Accept H0, reject H1 as there is insufficient evidence to suggest greater proportion of dogs had relieved symptoms.

(C?) In another bla-bla-bla, 50 dogs were taken as a sample and 42 had their symptoms relieved. Conduct a hypothesis test for this at 10% significance level for X~B(50, 0.75) (4 marks)

Gave values for P(X= 41, 42) and P(X<= 41, 42) I believe. You had to do P(X>= 42) so 1 - P(X<= 41) and the answer was < 0.10 therefore accept H1, reject H0. There is sufficient evidence to suggest greater proportion of dogs had relieved symptoms.

~L

edit: better s1 mark scheme here just didn't see it before x.x: https://docs.google.com/document/d/1...kfpLx2sO0/edit

pictures of c2 paper at: https://twitter.com/freddyfincham/st...98315039666176

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(Original post by

Heads up on question 4 on the c2 mark scheme, I think you need to have 24y not 24x as you're multiplying up by 24?

**mdmamazing----**)Heads up on question 4 on the c2 mark scheme, I think you need to have 24y not 24x as you're multiplying up by 24?

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#14

(Original post by

i have like estimates but im not sure of most of them, if you know any let me know and I'll add them to the op

**LokiiR**)i have like estimates but im not sure of most of them, if you know any let me know and I'll add them to the op

ii)Finding common difference of AP [2]

2i) Definite integration [3]

ii) Indefinite integration [2]

3i) Calculate gradient of chord [2]

ii) Suggest a more accurate point to gain a closer estimate of gradient of curve [1]

4) Find equation of normal [5]

5i) Describe transformation [2]

ii) Describe transformation [2]

6) Find equation from gradient [5]

7i) Sketch curve [2]

ii) Log question - find expression for w [3]

8) Trig solving question [5]

9i) Show that surface area equal... [2]

ii) Differentiate and again [4]

iii) Find value of r that gives min A, find A [4]

10i) calculate length AE [2]

ii) Find perp distance, prove pond lies in triangle ADE [3]

iii) Calculate area of pond and meadow [4]

iv) Show car park uses over 90% of field [4]

11i) Show Arif earns more in year 10 and less in year 11 [4]

ii) Show total amounts earned in 17 years are equal [4]

iii) Show n>etc, find in which year total earnings exceed 1.2mill [5]

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