Q1.
A trapezium has parallel sides of length (x + 1) cm and (x + 2) cm.
The perpendicular distance between the parallel sides is x cm.
The area of the trapezium is 10 cm2.
Not drawn accurately
Find the value of x.
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Answer x = .............................................. cm
(Total 5 marks)
Page 1 of 64
Q2.
Here are the equations of four straight lines.
(a)
Line 1:
y=x+4
Line 2:
y = 3x
Line 3:
y = 3x + 5
Line 4:
y = –x + 5
Which two lines are parallel?
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Answer .............................. and ..............................
(1)
(b)
Which two lines intersect the y axis at the same point?
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Answer .............................. and ..............................
(1)
(Total 2 marks)
Q3.
(a)
Simplify fully
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Answer .................................................
(2)
(b)
Given that
work out the value of
Write your answer in its simplest form.
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Answer .................................................
(2)
(Total 4 marks)
Page 2 of 64
Q4.
Solve the equation
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Answer x = .................................................
(Total 4 marks)
Q5.
(a)
Simplify fully
You must show your working.
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Answer .................................................
(2)
(b)
Rationalise the denominator and simplify
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Answer .................................................
(2)
(Total 4 marks)
Page 3 of 64
Q6.
On Friday the ratio of the time Priya is sleeping to the time she is awake is 3 : 5.
She is sleeping for less time than she is awake.
(a)
Work out the number of hours that she is sleeping on Friday.
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Answer ................................................. hours
(2)
(b)
On Saturday she sleeps for one hour more than she did on Friday.
Show that the ratio of the time she is sleeping to the time she is awake on Saturday
is 5 : 7
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(3)
(Total 5 marks)
Q7.
Show that
is an integer.
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(Total 2 marks)
Page 4 of 64
Q8.
Multiply out and simplify
(2p – 5q)(3p + q)
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Answer .................................................
(Total 3 marks)
Q9.
The line PQ is shown on the grid.
(a)
Find the gradient of a line which is perpendicular to PQ.
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Answer .................................................
(3)
Page 5 of 64
(b)
Hence find the equation of the perpendicular bisector of the line PQ.
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Answer .................................................
(2)
(Total 5 marks)
Q10.
(a)
Find the values of a and b such that
x 2 + 6x – 3 = (x + a)2 + b
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Answer a = ........................., b = .........................
(2)
Page 6 of 64
(b)
Hence, or otherwise, solve the equation
x 2 + 6x – 3 = 0
giving your answers in surd form.
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Answer .................................................
(3)
(Total 5 marks)
Q11.
Each term of a Fibonacci sequence is formed by adding the previous two terms.
1, 1, 2, 3, 5, 8, 13, 21, ……
A Fibonacci sequence starts a, b, a + b, …
(a)
Use algebra to show that the 6th term of this Fibonacci sequence is 3a + 5b
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(2)
Page 7 of 64
(b)
Use algebra to prove that the difference between the 9th term and 3rd term of this
sequence is four times the 6th term.
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(3)
(Total 5 marks)
Q12.
The table gives the diameter, in metres, of planets in the solar system.
The diameters are given to an accuracy of 3 significant figures.
Planet
Diameter (metres)
Mercury
4.88 × 106
Venus
1.21 × 107
Earth
1.28 × 107
Mars
6.79 × 106
Jupiter
1.43 × 108
Saturn
1.21 × 108
Uranus
5.11 × 107
Neptune
4.95 × 107
Pluto
2.39 × 106
Page 8 of 64
(a)
Which planet has the largest diameter?
Answer .................................................
(1)
(b)
Which planet has the smallest diameter?
Answer .................................................
(1)
(c)
Which planet has a diameter approximately 10 times that of Venus?
Answer .................................................
(1)
(d)
Write
as an ordinary number.
Answer .................................................
(1)
(e)
What is the diameter of Pluto in kilometres?
Give your answer in standard form.
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Answer ................................................. km
(2)
(Total 6 marks)
Q13.
(a)
Simplify
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Answer .................................................
(2)
Page 9 of 64
(b)
Simplify
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Answer .................................................
(3)
(Total 5 marks)
Q14.
(a) Write down whether each of the following is an expression (X), an identity (I), an
equation (E) or a formula (F).
X, I, E or F
v = u + at
3n + 2n ≡ 5n
3x + 2 = 7
+ 2x – 3
(3)
(b)
Show clearly that
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(2)
(Total 5 marks)
Page 10 of 64
Q15.
Rearrange
to make x the subject.
Simplify your answer as much as possible.
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Answer .................................................
(Total 4 marks)
Q16.
Write each of these in the form p
, where p is an integer.
(a)
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Answer .................................................
(2)
(b)
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Answer .................................................
(2)
Page 11 of 64
(c)
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Answer .................................................
(2)
(Total 6 marks)
Q17.
(a)
Show clearly that
(p + q)2 ≡ p 2 + 2pq + q2
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(1)
(b)
Hence, or otherwise, write the expression below in the form ax2 + bx + c
(2x + 3)2 + 2(2x + 3)(x – 1) + (x – 1)2
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Answer .................................................
(3)
(Total 4 marks)
Page 12 of 64
Q18.
A is the point (2, 9)
B is the point (8, 7)
M is the midpoint of AB
C is the point (8, 18)
Not drawn accurately
Is MC perpendicular to AB?
You must justify your answer.
Do not use graph paper to answer this question.
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(Total 4 marks)
Q19.
(a)
Factorise
5x 2 + 20x
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Answer .................................................
(1)
(b)
Factorise
x 2 – 49
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Answer .................................................
(1)
Page 13 of 64
(c)
Factorise fully
(3x + 4)2 – (2x + 1)2
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Answer .................................................
(3)
(Total 5 marks)
Q20.
Evaluate
(a)
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Answer .................................................
(3)
(b)
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Answer .................................................
(2)
(Total 5 marks)
Page 14 of 64
Q21.
A is the point (1, –2).
B is the point (5, 4).
Find the equation of the line perpendicular to AB, passing through the mid-point of AB.
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Answer .................................................
(Total 4 marks)
Page 15 of 64
Q22.
Find the values of a and b such that
x 2 – 10x + 18 = (x – a)2 + b
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Answer a = ..........................., b = ...........................
(Total 2 marks)
Q23.
Find the equation of the line through (0, –2) and (4, 18).
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Answer .................................................
(Total 3 marks)
Page 16 of 64
Q24.
Solve the simultaneous equations
y=x+2
y = 3x 2
You must show your working.
Do not use trial and improvement.
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Answer .....................................................................................................................
(Total 5 marks)
Q25.
A shape is made from two trapezia.
Not drawn accurately
The area of this shape is given by
A=
(a + b) +
(a + h)
Page 17 of 64
Rearrange the formula to make a the subject.
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Answer a = ................................................
(Total 4 marks)
Page 18 of 64
Q26.
Match each of the shaded regions to one of these inequalities.
A
y ≤ –
B
y ≤ C
y ≥ – 2x + 4
+2
+2
D
y ≥ 2x – 4
E
y ≤ 2x – 4
Region 1 ................................................
Region 2 ................................................
Region 3 ................................................
Region 4 ................................................
(Total 4 marks)
Page 19 of 64
Q27.
Julie has a bag containing x blue marbles and y red marbles.
The ratio of blue marbles to red marbles is 2:3
She adds z blue marbles.
The ratio of blue marbles to red marbles is now 2:1
What is the ratio between x and z?
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Answer .................................................
(Total 3 marks)
Q28.
Make x the subject of the formula
a(x – b) = a2 + bx
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Answer ....................................................
(Total 4 marks)
Page 20 of 64
Q29.
The triangle number sequence is
1, 3, 6, 10, 15, 21, ...
The nth term of this sequence is given by
n(n + 1)
(a)
Write down an algebraic expression for the (n – 1)th term of the sequence.
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Answer .................................................
(1)
(b)
Prove that the sum of any two consecutive triangle numbers is a square number.
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(3)
(Total 4 marks)
Page 21 of 64
Q30.
(a)
This is a page from Zoe’s exercise book.
Give a counter example to show that Zoe is wrong.
Justify your answer.
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(2)
(b)
Prove that
(n + 5)2 – (n + 3)2 = 4(n + 4)
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(3)
(Total 5 marks)
Page 22 of 64
Q31.
Solve these simultaneous equations
x + 3.6y = 2
x – 2.4y = 5
You must show all your working.
Do not use trial and improvement.
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Answer x = ..............................................
y = ..............................................
(Total 3 marks)
Q32.
(a)
Find the value of
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Answer .................................................
(1)
(b)
Find the value of 8x 0
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Answer .................................................
(1)
(Total 2 marks)
Page 23 of 64
The diagram shows the graph of an equation of the form y = x 2+ bx + c
Q33.
Find the values of b and c.
You must show your method.
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Answer b = ................. , c = ...................
(Total 3 marks)
Q34.
Some large numbers are written below.
1 million = 106
1 billion = 109
1 trillion = 1012
(a)
How many millions are there in one trillion?
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Answer ....................................................
(1)
Page 24 of 64
(b)
Write 8 billion in standard form.
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Answer ....................................................
(1)
(c)
Work out 8 billion multiplied by 3 trillion.
Give your answer in standard form.
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Answer ....................................................
(2)
(Total 4 marks)
Q35.
Annie, Bert and Charu are investigating the number sequence
21, 40, 65, 96, 133, ...
(a)
Annie has found the following pattern.
1st term
1 × 2 + 32 + 2 × 5 =
21
2nd term
2 × 3 + 42 + 3 × 6 =
40
3rd term
3 × 4 + 52 + 4 × 7 =
65
4th term
4 × 5 + 62 + 5 × 8 =
96
5th term
5 × 6 + 72 + 6 × 9 = 133
Complete the nth term for Annie’s pattern.
nth term
n × (n + 1) + ........................ + ........................ × ........................
(2)
Page 25 of 64
(b)
Bert has found this formula for the nth term
(3n + 1)(n + 3) + 5
Charu has found this formula for the nth term
(2n + 3)2 – (n + 1)2
Prove that these two formulae are equivalent.
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(3)
(Total 5 marks)
Page 26 of 64
Q36.
(a)
Find the equation of the line AB.
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Answer .................................................
(3)
(b)
Give the y-coordinate of the point on the line with an x-coordinate of 6.
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Answer .................................................
(2)
(c)
Write down the gradient of a line perpendicular to AB.
Answer .................................................
(1)
(Total 6 marks)
Page 27 of 64
Q37.
(a)
Factorise 2n2+ 5n + 3
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Answer .................................................
(2)
(b)
Hence, or otherwise, write 253 as the product of two prime factors.
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Answer .................................................
(1)
(Total 3 marks)
Q38.
(a)
n is a positive integer.
(i)
Explain why n(n + 1) must be an even number.
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(1)
(ii)
Explain why 2n + 1 must be an odd number.
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(1)
(b)
Expand and simplify (2n + 1)2
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Answer .................................................
(2)
Page 28 of 64
(c)
Prove that the square of any odd number is always 1 more than a multiple of 8.
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(3)
(Total 7 marks)
Q39.
(a)
(i)
Factorise x 2 – 10x + 25
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Answer .................................................
(2)
(ii)
Hence, or otherwise, solve the equation
(y – 3)2 – 10(y – 3) + 25 = 0
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Answer y = .................................................
(2)
Page 29 of 64
(b)
Simplify
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Answer .................................................
(3)
(Total 7 marks)
Q40.
Make x the subject of the formula
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Answer x = .......................................
(Total 4 marks)
Page 30 of 64
Q41.
Find the equation of the straight line passing through the point (0, 5) which is perpendicular
to the line
y=
x+3
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Answer .................................................
(Total 2 marks)
Q42.
Make x the subject of the formula
w = x2 + y
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Answer x = .................................................
(Total 2 marks)
Page 31 of 64
Q43.
Solve the simultaneous equations
4x + 3y = 14
2x + y = 5
You must show your working.
Do not use trial and improvement.
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Answer x = .................... , y = ......................
(Total 3 marks)
Q44.
A special packet of breakfast cereal contains 20% more than a normal packet. The special
packet contains 600 g of cereal. How much cereal does the normal packet contain?
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Answer ................................................. g
(Total 3 marks)
Q45.
Two gas supply companies have different ways of charging for the gas they supply.
Alpha gasCO
Fixed Charge
Price per kilowatt hour of gas
£9.60
First 5 kilowatt hours free then
£1.30 for every kilowatt hour over 5.
Beta gasCO
Fixed Charge
Price per kilowatt hour of gas
No fixed charge
£1.50 for every kilowatt hour.
Page 32 of 64
Find the number of kilowatt hours after which Alpha gasCo becomes cheaper than Beta gasCo.
You might want to use some graph paper.
You must show your method clearly.
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Answer ................................... kilowatt hours
(Total 4 marks)
Q46.
The region R is shown shaded below.
Page 33 of 64
Write down three inequalities which together describe the shaded region.
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Answer .................................................
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(Total 3 marks)
Q47.
Solve the equation
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(Total 5 marks)
Page 34 of 64
Q48.
Simplify fully
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Answer ............................................
(Total 4 marks)
Q49.
(a)
(i)
Evaluate
13z0
Answer ............................................
(1)
(ii)
Evaluate
(13z)0
Answer ............................................
(1)
(b)
If 3x =
, find the value of x.
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Answer x = ............................................
(2)
(c)
If 4y =
, find the value of y.
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Answer y = ............................................
(2)
(Total 6 marks)
Page 35 of 64
Q50.
On the grid below, indicate clearly the region defined by the three inequalities
x ≥ 1
y ≥ x – 1
x + y ≤ 7
Mark the region with an R.
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(Total 3 marks)
Page 36 of 64
Q51.
Solve the equation
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Answer .................................................
(Total 5 marks)
Q52.
Simplify
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Answer .................................................
(Total 4 marks)
Page 37 of 64
M1.
(Area =)
x (x + 1 + x + 2)
oe
(x + 1) +
× x × (1)
M1
2x 2 + 3x – 20 = 0
oe
eg
x 2 + 1.5x – 10 = 0
A1
(2x – 5)(x + 4) = 0
M1 for an attempt at using an algebraic method such as
factorising, formula (allow one error) or completing the square
(allow one error) to solve the quadratic
eg for (2x + a)(x + b) where ab = ± 20
A1 for a completely correct method
M1dep
A1
x = 2.5
Do not award last A1 if a negative value given as possible answer
eg if –4 given
2.5 seen with no or incomplete work SC2
2.5 after first M1, A1 give 5/5
A1
[5]
M2.
(a)
2 and 3
oe
B1
(b)
3 and 4
oe
B1
[2]
Page 38 of 64
(a)
M3.
√16 – √4 (= 4 – 2)
or
or
√2(2√2 – √2)
= √2(√2)
both steps needed
or
Both steps needed
M1
2
A1
(b)
or
or
or
Do not allow for
B1
oe
B1
[4]
M4.
3(3x + 1) –2 (2x + 5)
Could have 6 as denominator here
Condone lack of brackets
M1
9x + 3 – 4x – 10
A1
(their 5x – 7) = 6
M1 dep
x = 2.6
or
A1
[4]
Page 39 of 64
(a)
M5.
5√3 or 3√3
M1
8√3 A1
(b)
M1
3√7 A1
[4]
M6.
(a)
Condone
3 unsupported is M0
M1
9
Do not allow
(of a day)
SC1 Answer 15 or 9 and 15
A1
(b)
(their 9 + 1) : 24 – (their 9 + 1)
10 and 14 seen
M1
10:14
Must be integers
A1 ft
5:7
Must have seen previous ratio
A1
[5]
Page 40 of 64
5√2 (– √2 = 4√2)
If attempts to square the bracket
√2500 ± √50√2 ± √50√2 ± √4 M1
M7.
B1
32
32 A1
B1
[2]
M8.
6p2 + 2pq – 15pq – 5q2
For 3 correct terms
M1
6p2 + 2pq – 15pq – 5q2
Fully correct
A1
6p2 – 13pq – 5q2
From 4 terms
Do not ignore fw
B1 ft
[3]
M9.
(a)
Gradient of PQ =
M1
Perp. grad. =
(=
oe)
Drawing method:
Perpendicular line drawn and attempt at finding its gradient M2
M1 dep
oe
A1
Page 41 of 64
(b)
y = (their –
)x+c
M1
y=–
x+
oe
Accept 1.4 to 1.6 for
from graph
A1
[5]
M10.
(a)
(a =) 3
B1
(b =) –12
Allow 12 if –12 given in working
B1
(b)
(x + 3)2 = 12
or (x =)
Using their values from (a)
Substitution into formula (allow 1 error)
M1
x + 3 = √12
or (x =)
Using their values from (a)
M1 dep
(x =)
±√12 – 3
or
A1
[5]
Page 42 of 64
M11.
(a)
4th term = a + 2b
or (a = 1 and b = 1 and)
3(1) + 5(1)
oe
Accept 5th term = 2a + 3b (oe) for M1 if 4th
term not seen.
M1
6th term
Must see 4th and 5th terms
A1
(b)
Continuing sequence to 9th term
= 3a + 5b, 5a + 8b, 8a + 13b, 13a + 21b
Must come from continuing sequence and not
from 4 × 6th – 3rd
M1
Allow subtraction to be ‘assumed’. Condone
missing bracket if answer correct
A1
Either way round, expansion or factorisation
A1
[5]
M12.
(a)
Jupiter
B1
(b)
Pluto
B1
(c)
Saturn
B1
(d)
4 880 000
B1
(e)
or 2 390 oe
B1
A1
[6]
Page 43 of 64
M13.
(a)
3(x + 5) or 3x + 15
B1 for 3
B1 for x + 5
B1 for
B2
(b)
(x – 3)(x + 3)
M1
x(x + 3)
M1
Do not ignore further working
A1
[5]
M14.
Allow embedded solutions, but if contradicted M marks only
(a)
F, I, E, X
–1eeoo
B3
(b)
Must have 4 terms
Condone 1 sign error
M1
Must show cancellation, either by ‘crossing out’
or stating ab – ab = 0
A1
[5]
Page 44 of 64
M15.
y(3x – 4) = xy + 2
y × 3x – 4 = xy + 2 is M0 unless recovered
M1
3xy – 4y = xy + 2
A1
2xy = 4y + 2
dep 3xy – xy = 4y + 2 Allow one ‘sign’ error
M1
oe Do not award if x = not written
SC x =
B2
A1
Alt.
y(3x – 4) = xy + 2
y × 3x – 4 = xy + 2 is M0 unless recovered
M1
3x – 4 = x +
3x – 4 =
A1
2x = 4 +
3x – x = 4 +
Allow one ‘sign’ error
M1 dep
x=2+
oe Deduct mark if x = not written
SC x =
B2
A1
[4]
(a)
M16.
√300
oe eg, √(2 × 3) × √(2 × 25) or √(2 × 2 × 3 × 25)
√ (correct product of factors which includes ‘3’)
M1
10√3 SC1 for 5√12 or 2√75
A1
(b)
4√3 or 5√3 seen
M1
9√3 A1
Page 45 of 64
(c)
Attempt to rationalise
ie, Multiply num. and denom. By √3
oe eg,
scores M1
M1
6√3 A1
[6]
M17.
(a)
Convincing algebra
Must see
or
box method and
B1
(b)
Allow one sign or coefficient error
For middle term accept
or
M1
A1
ft if M1 awarded and no further errors
A1 ft
[4]
Page 46 of 64
M18.
Mid point (5, 8)
B1
Gradient AB
Accept any indication
eg, 6 across, 2 down
B1
Attempt to find gradient MC or ‘stepping’ from M to C
M1 for using ‘Their gradient’
M1
Valid conclusion with justification.
eg, No because gradient MC not 3
Accept any indication
eg, (5, 8) plus (3, 9) = (8, 17), mm’ ≠ – 1
A1
Alt
Mid point (5, 8)
B1
Use of Pythagoras
M1
Three correct lengths
A1
Correct conclusion at least 2 correct values
A1
[4]
M19.
(a)
5x (x + 4)
B1
(b)
(x + 7)(x – 7)
B1
Page 47 of 64
(c)
M1 for expanding and collecting to general quad form,
allow one error but expansions must have x 2 term,
x term and constant term.
Allow misuse of minus.
eg 9x2 + 24x + 16 – 4x2 + 4x + 1
Difference of two squares
((3x + 4) – (2x + 1)) × ((3x + 4) + (2x + 1))
M1
5x 2 + 20x + 15
A1 for either (x + 3) or (5x + 5) if difference of 2 squares used.
A1
5 (x + 3)(x + 1)
Accept
(x + 3)(5x + 5) or (5x + 15)(x + 1)
A1
[5]
M20.
(a)
(±)6
B1
B1
(±)1.5
oe
B1
(b)
oe eg, 0.01
B1 for 100 or
or
B2
[5]
Page 48 of 64
M21.
Attempt to find gradient of perpendicular line
Must be negative reciprocal of their gradient for AB
M1
(Gradient =) –
oe
eg –0.66, –0.67
A1
Use of midpoint (3, 1)
Must be used either on the diagram with an attempt
at a perpendicular or in y = mx + c to find c.
M1
y=–
x +3
ft their gradient if first M1 awarded
Accept equivalents eg 3y + 2x = 9
A1 ft
[4]
M22.
a=5
from expansion x2 – 2ax + a2 and comparing coeffs.
or simply spotting that a = 5
B1
b=–7
ft. from their a using a2 + b = 18
ie. b = 18 – a2
or by inspection
B1 ft
[2]
M23.
Identifying –2 as constant term in equation y = mx + c
B1
Gradient =
Attempt to find gradient
M1
y = 5x – 2
oe
A1
[3]
Page 49 of 64
3x 2 = x + 2
M24.
y = 3(y – 2)2
M1
3x 2 – x – 2 = 0
3y2 – 13y + 12 = 0
A1
(3x + 2)(x – 1) = 0
or (x –
)2 = ±√(
) or ±
x=
(3y – 4)(y – 3) = 0 (Reverse A1 s below)
Must be for factorising a quadratic.
x (or y) terms must have product equal to square term and
number terms must have a product equal to ± constant term.
If completing the square or formula used must be to at least the
stage shown for Method mark. or (y –
)2 = ±√(
) or ±
y=
M1
x = 1 and –
Need both
A1
y = 3 and
Must match appropriate values of y with x
Must use y = x + 2, or x = y – 2. Answers without any working is
B1, otherwise answers must be supported by an algebraic method.
Graphical method is M0.
Special case: x = 1, y = 3 without working B1. (Can be guessed).
NB only award this if no other marks awarded.
A1 ft
[5]
Page 50 of 64
M25.
2A = ah + bh + ab + bh
Accept A= ah/2 + bh/2 + ab/2 + bh/2
Allow one error
NB 4A = ah + bh + ab + bh is one error.
M1
2A – 2bh = ah + ab
A – bh = ah/2 + ab/2
A1
2A – 2bh = a(h + b)
For factorising
DM1
or equivalent ft if both Ms awarded.
oe e.g.
A1 ft
[4]
M26.
1 → D
1 → y ≥ 2x – 4
B1
2 → C
2 → y ≥ –2x + 4
B1
3 → E
3 → y ≤ 2x – 4
B1
4 → A
4 → y ≤ –
x+2
B1
[4]
M27.
6:3 or numerical values in the
ratios 2:3 and 6:3
(x + z) : y = 2: 1
3x = 2y
M1
Page 51 of 64
Finding ‘z’ e.g. 4 or appropriate numerical value
x + z = 2y
If both correct. Accept x + z = 2y
A1
1: 2
oe Accept words e.g. z is twice x.
A1
[3]
M28.
ax – ab = a2 + bx
Allow ax + ab =
M1
ax – bx = a2 + ab
A1
x(a – b) = a2 + ab
For factorising
NB sc x(a – b) = a2 + b Allow Ml and Al if
DM1
oe, e.g.
Follow through on factorisation if DM1 awarded.
Do not award if x = not shown,
fw such as cancelling a’s do not award last Al.
A1 ft
[4]
M29.
(a)
oe e.g.
B1
Page 52 of 64
(b)
oe e.g
ft their a
M1
n2 + 2n + 1
A1
n2
(n + 1)2
A1
[4]
M30.
(a)
Continuation at least once more
e.g. 53 – 43 = 61,
63 – 53 = 91 (allow this to be prime if stated)
Correctly evaluated.
M1
Justification that the answer is not prime.
e.g. 91 = 7 × 13.
83 – 73 = 169 = 13 × 13
Must show the factors.
NB 13 – 03 = 1 (1 not prime) Ml, A1
A1
(b)
n2 + 5n + 5n + 25 – (n2 + 3n + 3n + 9)
Ml for expanding and subtracting (allow 1
arithmetical error). Condone ‘invisible bracket’
M1
n2 + 10n + 25 – n2 – 6n – 9
Al for all terms collected and correct signs
or clear evidence of subtraction.
A1
4n + 16 = 4(n + 4)
Factorisation must be shown. Expanding is AO.
A1
[5]
Page 53 of 64
M31.
trial and improvement is 0
1st-2nd
6y = – 3 allow 1 error eg, 12y = – 3 6y = 3
2 – 3.6y = 5 + 2.4y allow 1 error or
2.4equation(l) + 3.6equation(2)
M1
y = – 0.5 or x = 3.8
A1
y = – 0.5 and x = 3.8
Must have both.
Allow reversed if both seen correct in working
ft if Ml awarded
A1 ft
[3]
M32.
(a)
4
B1
(b)
8
B1
[2]
Page 54 of 64
M33.
Either,
(x + 3)(x – 5) = x 2 – 2x – 15
Expansion not necessary for M1
M1
b = –2
A1
c = –15
A1
Note starting with (x – 3)(x + 5) could give c = –15 and will score
M1, A0, A1
OR, substituting coordinates (–3,0) and (5,0) into equation to get:
0 = 9 –3b + c and
0 = 5 + 5b + c
correct substitution which might be unsimplified
eg. 0 = (–3)2 – 3b + c and 0 = 52 + 5b + c
M1
Solving to give b = –2
A1
c = –15
A1
[3]
M34.
(a)
106
oe
B1
(b)
8 × 109
B1
(c)
2.4 × 1022
B1 2.4 or 22 as power or
24000000000000000000000
oe
e.g. 24 × 1021
B2
[4]
Page 55 of 64
M35.
(a)
(n + 2)2, (n + 1)(n + 4)
–1 eeoo
B2
(b)
Expand Bert 3n2 + 10n + 8
Allow one error but not 3n2 + 10n + 3
M1
Expand Charu 4n2 + 12n + 9 – (n2 + 2n + 1)
4n2 + 6n + 6n + 9 – n2 + n + n + 1
M1
Convincing algebra that these are equivalent.
Allow dealing correctly with – (n2 + 2n + 1) as minimum.
e.g. 4n2 + 12n + 9 – n2 – 2n – 1 is M1 A1
A1
[5]
M36.
(a)
Intercept = 9
i.e. identifying that 9 is the constant
term in the equation.
B1
Gradient = –
=–3
Any attempt at gradient for M1.
i.e ±9/ ±3
M1
y = – 3x + 9
Accept equivalent forms.
NB y = 3x + 9 is B1, Ml, A0
A1
(b)
Substitute x = 6 into their equation
Or recognise that y-step from 0 to 3
is the same as 3 to 6. eg sight of 9.
M1 can be implied by answer only.
M1
–9
A1
Page 56 of 64
(c)
ft on their gradient in (a), Allow an
'embedded' answer in an equation,
e.g. y =
x+9
B1 ft
[6]
M37.
(a)
(2n ± 3)(n ± 1) or (2n ± 1)(n ± 3)
M1
(2n + 3)(n + 1)
A1
(b)
23 × 11
Must see both factors
B1
[3]
M38.
(a)
(i)
Even × odd, so even product
or equivalent
B1
(ii)
2 × n always even, so 2n + 1 is odd
or equivalent
B1
(b)
4n2 + 2n + 2n + 1
3 or 4 terms correct
M1
4n2 + 4n + 1
Must simplify
A1
Page 57 of 64
(c)
Odd2 – 1 = (2n + 1)2 –1
= 4n2 + 4n
= 4n(n + 1)
Must factorise
B1
= 4 × even
Deduce ‘even’ connection
B1
= multiple of 8
Concluding statement
B1
[7]
M39.
(a)
(i)
(x – 5)(x – 5) or (x – 5)2
B1 for incorrect signs
B2
(ii)
Indicating replacement of x by y – 3
Might just see (y – 3 – 5)2 or (y – 8)2
Re-starting ?... must get as far as
y 2 –16y + 64 or (y – 8)2 to score M1
M1
y=8
A1
(b)
(x – 3)(x + 3)
M1
x(x + 3)
M1
A1
[7]
Page 58 of 64
M40.
y(x – 3) = 3x + 4
M1 for cross-multiplying and expanding bracket
M1
yx – 3y = 3x + 4
A1 correct expansion
A1
yx – 3x = 3y + 4
M1
x(y – 3) = 3y + 4
M1 for clollecting terms and factorising
A1
x = (3y + 4)/(y – 3)
A1 correct factorisation and division
[4]
M41.
Sight of –1
or –1.5 or –3/2
accept –1 / (
) or –1 / 0.66 ... for M1 only
M1
y=–
x+5
oe eg. 2y = –3x + 10
A1
[2]
M42.
x 2 = w – y.
Or equivalent -x 2 = y – w
B1
x = √(w – y)
Accept ± √(w – y) and – √(w – y)
B1
[2]
Page 59 of 64
M43.
4x + 3y = 14 4x + 3y = 14
4x + 2y = 10 6x + 3y = 15
allow error in one term
M1
y=4
2x = 1
correct elimination from their equations
M1
x=
and y = 4
oe
SC correct answers with no working or using T & I
A1
[3]
120% → 600
1.2
M44.
B1
600 ÷ 120 × 100
600 ÷ 1.2
M1
500
A1
[3]
M45.
9.60 + (x – 5) × 1.30
Alt: M1 for graph of Alpha parcels
M1
= 1.50x
M1 for graph of Beta
M1
3.10 = 0.20x
A1 accuracy
A1
x = 15.5
A1 answer. Accept 16 but not 15.
T&I gets M1 iff taken as far as 15.
A1 for both schemes at 15
A1 for both schemes at 16
A1 conclusion
A1
[4]
Page 60 of 64
M46.
y ≥ 0
Accept y > 0, or 0 ≤ y ≤ 3,
B1
x ≤ 6
Accept x < 6 or 0 ≤ x ≤ 6,
B1
y ≤ ×
Accept y <
y=<
x or x ≥ 2y or equivalent.
x
Any order.
B1
Special case: All three equations given (no inequalities) B1
Special case: All three inequalities the wrong way around B2.
[3]
M47.
(x – 2) + 5x(x + 1) = 3(x + 1)(x – 2)
Allow 1 error
M1
5x 2 + 6x – 2 = 3x 2 – 3x – 6
A1
2x 2 + 9x + 4 = 0
M1
(2x + 1)(x + 4) = 0
A1
x = –1/2, –4
A1
[5]
Page 61 of 64
M48.
Numerator = (x + 4)(x – 4)
B1
Denominator = (3x 2)(x 4)
M1
= (3x – 2)(x + 4)
A1
2
or 3x + 12x – 2x – 8
or 3x2 – 2x + 12x – 8
Answer = (x – 4)/(3x – 2)
A1
[4]
M49.
(a)
(i)
13
B1
(ii)
1
B1
(b)
3x = 3–3
M1
x = –3
M1 for writing 1/27 as a power of 3, correctly
allow embedded answer
A1
(c)
4y = 41½
M1
y = 1½
M1 for 4y = 8
allow embedded answer
A1
[6]
M50.
Correct region indicated
Award marks dependent upon number of lines drawn correctly and
extent of shading
B3
[3]
Page 62 of 64
M51.
LHS x(x – 1) – 2(x + 1)
Give M1 for x2 – 3x + 2 if first line seen
Allow invisible bracket if recovered.
M1
LHS = x 2 – 3x – 2
Terms need not be collected. e.g.x2 – x – 2x – 2
A1
(x – 1)(x + 1)(= x 2 – 1)
On RHS or as denominator.
x 2 – 1 can be written as x2 – x + x – 1
M1
Their (x 2 – 3x – 2) = their (x 2 – 1)
Dependent on first 2 M1’s
DM1
–
(= 0.33(3...))
Do not follow through.
NB ‘cancelling’ x2 on top and bottom of
Gives correct answer. Give M1, A1, M1. M0, A0.
A1
[5]
M52.
(5x ± a)(x ± b)
M1 for attempt to factorise. Must have (5x ± a)(x ± b)
where ab = ± 3, a, b must be integers.
M1
(5x – 1)(x + 3)
A1
(x – 3)(x + 3)
B1
(5x – 1)(x – 3)
Answer seen and further work then deduct last B1.
B1
[4]
Page 63 of 64
Page 64 of 64