### A-AA including marksheet

```Q1.
Diagram NOT
accurately drawn
CEAY and BDAX are straight lines.
XY, ED and CB are parallel.
AE = 5 cm.
AX = 9 cm.
BC = 4 cm.
BD = 2 cm.
CE = x cm.
XY = y cm.
Find the value of x and the value of y .
x = .....................................
y = .....................................
(Total 4 marks)
Page 1 of 23
Q2.
Diagram NOT accurately drawn
A cylinder has base radius x cm and height 2x cm.
A cone has base radius x cm and height h cm.
The volume of the cylinder and the volume of the cone are equal.
Find h in terms of x .
h = .............................
(Total 3 marks)
Page 2 of 23
Q3.
Diagram NOT accurately drawn
The sketch shows a curve with equation
y = kax
where k and a are constants, and a > 0
The curve passes through the points (1, 7) and (3, 175).
Calculate the value of k and the value of a .
k = ................................
a = ................................
(Total 3 marks)
Page 3 of 23
Q4.
(a)
Show that the equation
can be rearranged to give 3x 2 + 7x – 13 = 0
(3)
(b)
Solve 3x 2 + 7x – 13 = 0
Give your solutions correct to 2 decimal places.
x = ............................... or x = ................................
(3)
(Total 6 marks)
Q5.
Katy drove for 238 miles, correct to the nearest mile.
She used 27.3 litres of petrol, to the nearest tenth of a litre.
Petrol consumption =
Page 4 of 23
Work out the upper bound for the petrol consumption for Katy’s journey.
...................................... miles per litre
(Total 3 marks)
Q6.
Simplify fully
................................................
(Total 3 marks)
Page 5 of 23
Q7.
Solve the simultaneous equations
x2 + y2 = 5
y = 3x + 1
x = .................... y = .....................
or x = .................... y = .....................
(Total 6 marks)
Page 6 of 23
The diagram shows a sketch of the curve y = sin x ° for 0 ≤ x ≤ 360
Q8.
The exact value of sin 60° =
(a)
Write down the exact value of
(i)
sin 120°,
....................
(ii)
sin 240°.
....................
(2)
(b)
On the grid below, sketch the graph of y = 4 sin 2x ° for 0 ≤ x ≤ 360
(2)
(Total 4 marks)
Page 7 of 23
Q9.
Express the recurring decimal
as a fraction.
....................
(Total 3 marks)
Q10.
Prove that the difference between the squares of consecutive odd numbers is a multiple of
8
(Total 6 marks)
Page 8 of 23
Q11.
Diagram NOT accurately drawn
OAB is a triangle.
=a
=b
(a)
Find the vector
in terms of a and b.
= ........................................
(1)
P is the point on AB such that AP : PB = 3 : 2
(b)
Show that
(2a + 3b)
(3)
(Total 4 marks)
Page 9 of 23
Prove that (3n + 1)2 – (3n –1)2 is a multiple of 4, for all positive integer values of n .
Q12.
(Total 3 marks)
Q13.
Fred and Jim pay Malcolm to do some gardening.
Fred has £x .
Jim has ten pounds less than Fred.
Fred pays one third of his money to Malcolm.
Jim pays half of his money to Malcolm.
(a)
Show that the amount that Malcolm is paid is
.
(1)
Page 10 of 23
Malcolm is paid a total of £170.
(b)
Use algebra to show how much money Fred has left.
.....................................
(4)
(Total 5 marks)
Q14.
The time, T seconds, for a hot sphere to cool is proportional to the square root of the
surface area, A m 2, of the sphere.
When A = 100, T = 40.
Find the value of T when A = 60.
Page 11 of 23
..................................... seconds
(Total 4 marks)
Page 12 of 23
Q15.
Here is a rectangle.
Diagram NOT
accurately drawn
a = 8.3 cm correct to 1 decimal place.
b = 3.6 cm correct to 1 decimal place.
(a)
Calculate the upper bound of the area of this rectangle.
Write down all the figures on your calculator.
..................................... cm2
(2)
(b)
Find the area of this rectangle correct to an appropriate number of significant figures.
..................................... cm2
(2)
(Total 4 marks)
Page 13 of 23
M1.
Working
Mark
x = 2.05
4
M1 a correct expression for x involving
ratios of sides, e.g.
oe
A1 cao
y = 11.25
M1
oe
OR
A1 cao
Total for Question: 4 marks
M2.
Working
2
πx (2x) =
Mark
6x
3
M1 for a correct volume formula in terms of x , e.g.
2
π(x ) h
πx2(2x) or
π x 2h
A1 for π(2x ) =
πh or 3Πx2(2x) = πx2h or
x2(2x) = x2h (or better)
A1 for 6x cao
Total for Question: 3 marks
Page 14 of 23
M3.
Working
7 = ka 1; 175 = ka 3
k = , 175 =
, 175 = 7a
a2 = 25, so a = 5, k = 1.4
Or
73 = k 3a 3, 175 = ka 3
k2 =
2
Mark
k = 1.4
a=5
3
M1 either a 2 = 25
or 7 = ka (or 7 = ka 1) and 175 = ka 3
A1 k = 1.4 oe
A1 a = 5
SC Either a = 5 or k = 1.4 oe gets B2
, k = 1.4, a = 5
Total for Question: 3 marks
Page 15 of 23
M4.
Working
(a)
5(x – 1) = (4 – 3x )(x + 2)
Mark
Proof
3
M1 multiply through by (x – 1)(x + 2)
and cancel correctly M1 expand
5(x – 1) and (4 – 3x )(x + 2) correctly,
need not be simplified
A1 rearrange to give required equation
(dep on both Ms and fully correct
algebra)
1.22
–3.55
3
M1 correct substitution in formula of
a = 3, b = 7 and c = ±13
2
5x – 5 = 4x + 8 – 3x – 6x
(= 8 – 2x – 3x 2)
(3x 2 + 6x + 5x – 4x – 5 – 8 = 0)
3x 2 + 7x – 13 = 0
(b)
a = 3, b = 7, c = – 13
x=
M1 reduction to
A1 1.215 to 1.22 and –3.55 to –3.555
=
x = 1.2196… or – 3.55297….
Or
M1
Or
M1
A1 1.215 to 1.22 and –3.55 to –3.555
SC T&I 1 mark for 1 correct root, 3
marks for both correct roots
x = 1.2196… or –3.55297….
Total for Question: 6 marks
Page 16 of 23
M5.
Working
238 has an UB 238.5,
a LB of 237.5
27.3 has an UB of 27.35,
a LB of 27.25
Upper:
Mark
8.75
3
B1 for one of 238.5, 237.5, 27.35, 27.25,
seen
M1 for ‘UB no of miles’÷’LB no of litres’
Where 238 < ‘UB no of miles’≤238.5 and
27.25 ≤ 'LB no of litres’ < 27.3
A1 8.75 or 8.752 or 8.7522 or 8.7523 or better
= 8.75229
SC 238.4 ÷ 27.25 which leads to 8.748…B1
M1 A0
Total for Question: 3 marks
M6.
Working
Mark
3
B1 for (x – 3)(x – 5)or x (x – 5)-3(x – 5)
M1 for (2x ± 3)(x ± 5) or 2x (x + 5) ± 3(x + 5) or
2x (x – 5) ± 3(x – 5)
A1 for
Total for Question: 3 marks
Page 17 of 23
M7.
Working
y2 = (3x + 1)2
x2 + 9x2 + 6x + 1 = 5
10x 2 + 6x + 1 = 5
10x 2 + 6x – 4 = 0
2(5 x 2 + 3x – 2) = 0
2(5x – 2)(x + 1) = 0
Mark
x = 0.4
y = 2.2
6
M1 for (3x + 1)2 seen or implied by sight of 9x 2 + 1
A1 for x 2 + 9x 2 + 6x + 1 = 5 or equivalent expanded
form
M1 (dep) for correct attempt to solve a 3-term
quadratic equation (condone omission of = 0)
A1 for x = 0.4, x = –1
M1 (dep on previous Ms) for sub one value of x into
either equation
A1 for y = 2.2, y = –2 (correctly paired with x values)
[SC: B1 for one correct pair of solutions if M0
scored]
x = –1
y = –2
Total for Question: 6 marks
M8.
(a)(i)
Mark
2
(ii)
(b)
B1 cao
B1 cao
2
B2 cao
[B1 for sine curve,starting from the origin with
amplitude 4, OR
B1 cuts x axis at 90, 180, 270, 360 and starts from
0]
Total for Question: 4 marks
Page 18 of 23
M9.
Working
x = 0.213131313…
Mark
3
M1 for 0.2131313…. or 0.2 + 0.0131313…. (dots
MUST be included)
M1 for two correct recurring decimals that, when
subtracted, leave a terminating decimal
10x = 2.13131313…
1000x = 213.131313…
990x = 211
A1 for
Total for Question: 3 marks
M10.
Working
QWC (2n +1)2 – (2n –1)2 =
ii, iii
4n2 + 4n + 1 – (4n2 – 4n + 1)
= 8n
Mark
Fully
algebraic
argument,
set out in
a logical
and
coherent
manner
6
B2 the nth term for consecutive odd
numbers is 2n – 1 oe
(B1 2n + k, k ≠ –1 or n = 2n – 1 or 2x –
1
B1 use of 2n + 1 and 2n – 1 oe
M1 (2n +1)2 – (2n –1)2
M1 4n2 + 4n + 1 – (4n2 – 4n + 1)
C1 conclusion based on correct
algebra QWC: Conclusion should be
stated, with correct supporting
algebra.
OR
(2n +1)2 – (2n –1)2 =
((2n +1) – (2n –1))(2n +1 + 2n –
1)
=2 × 4n = 8n
OR
B1 use of 2n + 1 and 2n –1 oe
M1 (2n +1)2 – (2n –1)2
M1 ((2n +1) – (2n –1))(2n +1 + 2n –1)
C1 conclusion based on correct
algebra QWC: Conclusion should be
stated, with correct supporting
algebra.
Total for Question: 6 marks
Page 19 of 23
M11.
Working
Mark
(a)
b–a
1
(b)
proof
3
B1 for b – a or – a + b oe
M1 for
oe
M1 for
× “(b – a)” oe or
× “(a – b)” oe
A1 for a +
× (b – a) oe or b +
× (a – b) oe
brackets seen
Total for Question: 4 marks
M12.
Working
(9n 2 + 6n + 1) –
(9n 2 – 6n + 1)
= 12n
12n
correct
comment
Mark
3
M1 for (3n )2 + 3n + 3n + 1 or (3n )2 – 3n – 3n + 1 or
((3n + 1) – (3n – 1))((3n + 1) + (3n – 1))
A1 for 12n from correct expansion of both brackets
A1 for 12n is a multiple of 4 or 12n = 3 × 4n or
12n = 4 × 3n or
= 3n or
= 4n
NB: Trials using different values for n score no marks.
Total for Question: 3 marks
Page 20 of 23
M13.
Working
(a)
Fred pays
and Jim pays
Mark
Clear and
coherent
explanation
1
C1 a clear and coherent
explanation
£140
4
M1 multiply through by 6
and cancels fractions
Malcolm gets £170 for Fred and
Jim, so Malcolm gets
+
= 170
(b)
Fred has
using
left, so solving for x
M1 (dep)expand 3(x − 10)
+
= 170
2x + 3(x – 10) = 170 × 6
5x = 1050
M1 (dep)collect terms on
each side correctly
A1 cao
x = 210
OR
OR
M1 collects terms over 6
M1(dep) expand 3(x − 10)
M1(dep) multiply through by
6 and collect terms
5x = 1050
A1 cao
x = 210
Total for Question: 5 marks
Page 21 of 23
M14.
Working
Mark
31.0
4
k=4
M1
M1
A1
A1 for 30.98... or 31(.0)
OR
M2 for
oe
M1 for
oe
A1 for 30.98... or 31(.0)
Total for Question: 4 marks
M15.
(a)
Working
Mark
UB 8.35 × 3.65 = 30.4775
30.4775
2