Download Gold 1 - Maths Tallis

Gold: 1 of 4
Practice Paper –Gold 1
Instructions





Use black ink or ball-point pen.
Fill in the boxes at the top of this page with your name,
centre number and candidate number.
Answer all questions.
Answer the questions in the spaces provided
– there may be more space than you need.
Calculators must not be used.
Information



The total mark for this paper is 99
The marks for each question are shown in brackets
– use this as a guide as to how much time to spend on each question.
Questions labelled with an asterisk (*) are ones where the quality of your
written communication will be assessed.
Advice
 Read each question carefully before you start to answer it.
 Keep an eye on the time.
 Try to answer every question.
 Check your answers if you have time at the end.

Suggested Grade Boundaries (for guidance only)
A*
A
B
C
D
91
73
53
37
24
Practice Paper – Gold 1 (1 of 4)
This publication may only be reproduced in accordance with Pearson Education Limited copyright policy.
©2015 Pearson Education Limited.
GCSE Mathematics 1MA0
Formulae: Higher Tier
You must not write on this formulae page.
Anything you write on this formulae page will gain NO credit.
Volume of prism = area of cross section × length
4 3
πr
3
Surface area of sphere = 4πr2
Area of trapezium =
1
(a + b)h
2
1 2
πr h
3
Curved surface area of cone = πrl
Volume of sphere
Volume of cone
In any triangle ABC
The Quadratic Equation
The solutions of ax2+ bx + c = 0
where a ≠ 0, are given by
x=
Sine Rule
a
b
c


sin A sin B sin C
Cosine Rule a2 = b2+ c2– 2bc cos A
Area of triangle =
Practice Paper: Gold 1 of 4
1
2
ab sin C
2
 b  (b 2  4ac)
2a
Answer ALL questions.
Write your answers in the spaces provided.
You must write down all stages in your working.
You must NOT use a calculator.
1.
A box contains milk chocolates and dark chocolates only.
The number of milk chocolates to the number of dark chocolates is in the ratio 2 : 1
There are 24 milk chocolates.
Work out the total number of chocolates.
.....................................
(2 marks)
___________________________________________________________________________
*2.
Karen got 32 out of 80 in a maths test.
She got 38% in an English test.
Karen wants to know if she got a higher percentage in maths or in English.
Did Karen get a higher percentage in maths or in English?
(Total 2 marks)
___________________________________________________________________________
Practice Paper: Gold 1 of 4
3
3.
50 people each did one activity at a sports centre.
Some of the people went swimming.
Some of the people played squash.
The rest of the people used the gym.
21 of the people were female.
6 of the 8 people who played squash were male.
18 of the people used the gym.
9 males went swimming.
Work out the number of females who used the gym.
..........................................
(Total for Question 3 is 4 marks)
___________________________________________________________________________
Practice Paper: Gold 1 of 4
4
4.
Here are the first five terms of an arithmetic sequence.
2
6
10
14
18
(a) Find, in terms of n, an expression for the nth term of this sequence.
.....................................
(2)
(b) An expression for the nth term of another sequence is 10 − n2
(i) Find the third term of this sequence.
.....................................
(ii) Find the fifth term of this sequence.
.....................................
(2)
(Total 4 marks)
___________________________________________________________________________
Practice Paper: Gold 1 of 4
5
5.
(a) Solve the inequality 6y + 5 > 8
..........................................
(2)
(b) Here is an inequality, in x, shown on a number line.
Write down the inequality.
..........................................
(2)
(Total 4 marks)
___________________________________________________________________________
Practice Paper: Gold 1 of 4
6
6.
 3 
(a) Translate shape A by the vector   .
2
(1)
Practice Paper: Gold 1 of 4
7
(b) Describe fully the single transformation that maps shape Q onto shape R.
......................................................................................................................................................
......................................................................................................................................................
......................................................................................................................................................
......................................................................................................................................................
(3)
(Total 4 marks)
___________________________________________________________________________
Practice Paper: Gold 1 of 4
8
7.
(a) Solve
3(2t – 4) = 2t + 12
t = ................................
(3)
(b) Expand and simplify
2(x – y) – 3(x – 2y)
.....................................
(2)
(c) Expand and simplify
(x – 5)(x + 7)
.....................................
(2)
(Total 7 marks)
___________________________________________________________________________
Practice Paper: Gold 1 of 4
9
8.
Sasha carried out a survey of 60 students.
She asked them how many CDs they each have.
This table shows information about the numbers of CDs these students have.
Number of CDs
Frequency
0–4
5–9
10 – 14
15 – 19
20 – 24
8
11
9
14
18
(a) Write down the class interval containing the median.
.....................................
(1)
(b) On the grid, draw a frequency polygon to show the information given in the table.
(2)
(Total 3 marks)
___________________________________________________________________________
Practice Paper: Gold 1 of 4
10
9.
Two shops both sell the same type of suit.
In both shops the price of the suit was £180
One shop increases the price of the suit by 17 12 %.
The other shop increases the price of the suit by 22 12 %.
Calculate the difference between the new prices of the suits in the two shops.
£ ..................................
(Total 3 marks)
___________________________________________________________________________
10.
Jim has only 5p coins and 10p coins.
The ratio of the number of 5p coins to the number of 10p coins is 2 : 3
Work out the ratio of
the total value of the 5p coins : the total value of the 10p coins.
Give your answer in its simplest form.
..................................
(Total 2 marks)
___________________________________________________________________________
Practice Paper: Gold 1 of 4
11
11.
Suzy did an experiment to study the times, in minutes, it took 1 cm ice cubes to melt at
different temperatures.
Some information about her results is given in the scatter graph.
The table shows information from two more experiments.
Temperature (C)
15
55
Time (Minutes)
22
15
(a) On the scatter graph, plot the information from the table.
(1)
(b) Describe the relationship between the temperature and the time it takes a 1 cm ice cube
to melt.
......................................................................................................................................................
(1)
(c) Find an estimate for the time it takes a 1 cm ice cube to melt when the temperature
is 25 C.
....................... minutes
(2)
Suzy’s data cannot be used to predict how long it will take a 1 cm ice cube to melt when the
temperature is 100 C.
(d) Explain why.
......................................................................................................................................................
(1)
(Total 5 marks)
___________________________________________________________________________
Practice Paper: Gold 1 of 4
12
12.
Batteries are sold in packets and boxes.
Each packet contains 4 batteries.
Each box contains 20 batteries.
Bill buys p packets of batteries
and b boxes of batteries.
Bill buys a total of N batteries.
Write down a formula for N in
terms of p and b.
...............................................
(Total 3 marks)
Practice Paper: Gold 1 of 4
13
13.
(a) Work out the value of (6 × 108) × (4 × 107)
Give your answer in standard form.
.....................................
(2)
(b) Work out the value of (6 × 108) + (4 × 107)
Give your answer in standard form.
.....................................
(2)
(Total 4 marks)
___________________________________________________________________________
Practice Paper: Gold 1 of 4
14
14. (a) Write 6.43 × 105 as an ordinary number.
.....................................
(1)
(b) Work out the value of 2 ×107 × 8 ×10−12
Give your answer in standard form.
.....................................
(2)
(Total 3 marks)
___________________________________________________________________________
15.
(a) Factorise fully
2x2 − 4xy
.....................................
(2)
(b) Factorise
p2 – 6p + 8
.....................................
(2)
(c) Simplify
( x  2) 2
x2
.....................................
(1)
(d) Simplify
2a2b × 3a3b
.....................................
(2)
(Total 7 marks)
___________________________________________________________________________
Practice Paper: Gold 1 of 4
15
16.
Work out
3
1
2
×2
4
3
Give your answer in its simplest form.
.....................................
(Total 3 marks)
___________________________________________________________________________
Practice Paper: Gold 1 of 4
16
17.
y = p – 2qx2
p = –10
q=3
x = –5
(a) Work out the value of y.
.....................................
(2)
(b) Rearrange y = p – 2qx2
to make x the subject of the formula.
.....................................
(3)
(Total 5 marks)
___________________________________________________________________________
Practice Paper: Gold 1 of 4
17
18.
(a) Write down the value of 20
.....................................
(1)
2y =
1
4
(b) Write down the value of y.
y = ...............................
(1)
(c) Work out the value of 9

3
2
.....................................
(2)
(Total 4 marks)
___________________________________________________________________________
Practice Paper: Gold 1 of 4
18
19.
The diagram shows a straight line, L1, drawn on a grid.
A straight line, L2, is parallel to the straight line L1 and passes through the point (0, –5).
Find an equation of the straight line L2.
....................................................................................
(Total for Question 19 is 3 marks)
___________________________________________________________________________
Practice Paper: Gold 1 of 4
19
20.
(a) Here are some expressions.
a3b
a2( c+ b)
ab + c3
4abc
4πc2
The letters a, b, and c represent lengths.
π and 4 are numbers that have no dimension.
Two of the expressions could represent volumes.
Tick the boxes underneath these two expressions.
(2)
The volume of this cube is 8 m3.
(b) Change 8 m3 into cm3.
............................. cm3
(2)
(Total 4 marks)
___________________________________________________________________________
Practice Paper: Gold 1 of 4
20
*21
B, C and D are points on the circumference of a circle, centre O.
AB and AD are tangents to the circle.
Angle DAB = 50°
Work out the size of angle BCD.
Give a reason for each stage in your working.
(Total for Question 21 is 4 marks)
___________________________________________________________________________
Practice Paper: Gold 1 of 4
21
22.
Enlarge the shaded shape by scale factor 
1
with centre (–1, –2).
2
(Total 3 marks)
___________________________________________________________________________
Practice Paper: Gold 1 of 4
22
23.
ABCDEF is a regular hexagon, with centre O.
OA = a , OB = b.
(a) Write the vector AB in terms of a and b.
.....................................
(1)
The line AB is extended to the point K so that AB : BK = 1 : 2
(b) Write the vector CK in terms of a and b.
Give your answer in its simplest form.
.....................................
(3)
(Total 4 marks)
___________________________________________________________________________
Practice Paper: Gold 1 of 4
23
24.
Make t the subject of the formula
p=
3  2t
4t
..............................................................................................
(Total for Question 24 is 4 marks)
___________________________________________________________________________
Practice Paper: Gold 1 of 4
24
25.
The diagram shows two similar solids, A and B.
Solid A has a volume of 80 cm3.
(a) Work out the volume of solid B.
....................................cm3
(2)
Solid B has a total surface area of 160 cm2.
(b) Work out the total surface area of solid A.
....................................cm2
(2)
(Total 4 marks)
___________________________________________________________________________
Practice Paper: Gold 1 of 4
25
26.
P is inversely proportional to V.
When V = 8, P = 5
(a) Find a formula for P in terms of V.
P = ...............................
(3)
(b) Calculate the value of P when V = 2
.....................................
(1)
(Total 4 marks)
TOTAL FOR PAPER IS 99 MARKS
Practice Paper: Gold 1 of 4
26
1
24 ÷ 2
36
2
M1 for 24 ÷ 2 or
3
 24 oe or 12
2
A1 cao
*2
Maths with correct
comparative
figure(s)
2
M1 for correct method to find figure(s) to compare,
eg
32
 100 (=40) oe or 0.38×80 oe (=30.4)
80
C1 for maths with 40% or 30.4 or
3
4
Sq
F
2
M 15
6
9
Tot 8
24
G S Tot
21
4
29
14
18
50
Practice Paper – Gold 1 (1 of 4)
This publication may only be reproduced in accordance with Pearson Education Limited copyright policy.
©2015 Pearson Education Limited.
4
and
oe
M1 for a correct first step which results in a value that could be in the
table: eg. 50 – 18 – 8 (= 24) or 50 – 21 (= 29) or 8 – 6 (= 2)
M1 for a correct method to find a second value that could be in the
table using their first value eg “29” – 9 – 6 (=14) or “24” – 9 (=15)
M1 for a fully correct and complete method.
A1 cao
4
5
6
(a)
4n − 2
2
B2 for 4n − 2 (oe including unsimplified)
(B1 for 4n or 4n + k , k ≠ -2 or 4n – k, k ≠ 2
or n = 4n – 2 )
2
B1 cao
(b)(i)
10 − 3²
1
(ii)
10 − 5²
− 15
B1 cao
(a)
y > 0.5
2
M1 for clear intention to subtract 5 from both sides of
inequality or equation or divide all terms of the inequality or
equation by 6 or 6y > 3 or 0.5 oe seen
A1 for y > 0.5 oe as final answer
(b)
−3 < x ≤ 4
2
B2 for −3 < x ≤ 4 oe
(B1 for one correct inequality, eg −3 < x or x > −3 or
x ≤ 4 or 4 ≥ x or −3 ≤ x < 4)
(a)
Shape with vertices
at
(–1, 3), (0, 6),
(2, 6), (1, 3)
1
B1 for correct shape in correct position
(b)
Rotation
centre (0,0)
90˚ anticlockwise
3
B1 rotation
B1 (centre) (0,0)
B1 90˚ anticlockwise or 270˚ clockwise
Note: award no marks if more than one transformation is
given
Practice Paper: Gold 1 of 4
28
7
3(2t – 4)
6t – 12
6t – 2t
4t
(a)
=
=
=
=
2t +12
2t +12
12 +12
24
2(x − y) – 3(x − 2y)
= 2x – 2y – 3x + 6y
(b)
6
3
B1
M1
terms
–x + 4y
2
2t
12
+
3
3
correctly isolating their terms in t or their constant
6t – 12
or
A1
in an equation
cao
M1
A1
2x – 2y
–x + 4y
or
or
3x – 6y
4y – x
or
–3x + 6y
[SC: B1 for – x – 8y or x + 4y with or without working
if M0 scored]
(c)
( x  5)( x  7)
= x2 – 5x + 7x – 35
x2 + 2x – 35
2
M1
A1
Practice Paper: Gold 1 of 4
29
3 out of 4 terms correct with correct signs
or all 4 terms correct ignoring any sign errors
cao
Practice Paper: Gold 1 of 4
30
9
22.5% – 17.5% = 5%
5
180 ×
100
9
3
22.5 – 17.5
5
M1 180 × ‘
’ oe
100
A1 cao
M1
OR
180 ×
180 ×
10
1
2
100
1
17
2
100
22
OR
= 40.50
M1
= 31.50
180 ×
1
2
100
22
oe
or
180 ×
1
2
100
17
oe
40.50 – 31.50
M1 (dep) ‘40.50’ – ‘31.50’
A1
cao
OR
OR
1.225 × 180 = 220.5
1.175 × 180 = 211.5
220.5 – 211.5
M1
1.225 × 180 or 1.175 × 180
M1 (dep) ‘220.5’ – ‘211.5’
A1 cao
[ SC: Award M2 A0 for an answer of 9 with 1 arithmetic
error]
2 × 5 : 3 × 10 = 10 : 30 = 1 : 3
Practice Paper: Gold 1 of 4
1:3
2
31
M1 2 × 5 : 3 × 10 or 2 × 1 : 3 × 2 or sight of 10 and 30 or
10p and 30p
A1 for 1 : 3 cao
(SC B1 for 3 : 1 or 1p : 3p or 10 : 30 or 5 : 15 or 10p : 30p)
11
(a)
Plot (15, 22) and (55, 15)
Points plotted
1
B1
cao
(b)
Describe
relationship
1
B1
If the temperature increases so the time taken
decreases oe
(accept negative correlation)
(c)
18 – 20
2
M1 draw LOBF between (20,18) and (20, 22) to (70,3) and
(70,8)
A1 18 – 20 (if M0 allow B2 for an answer in the range
18–20)
(d)
Reason
1
B1
Practice Paper: Gold 1 of 4
32
± ½ square
reason
e.g LOBF would give negative time,
you should not use the LOBF beyond your data
13
(a)
(6 × 108) × (4 × 107)
= 24 × 10
2.4 × 10 16
M1
24 × 10 8+7oe or
× 10n
2
8+7
24 × 10 15
(b)
(6 × 108) + (4 × 107)
= 6 × 10 8 + 0.4 × 108
6.4 × 10 8
2
24 000 000 000 000 000
cao
M1
6  108  0.4  108 or 60  10 7  4  10 7
or 600 000 000 + 40 000 000
or 640 000 000
A1
(a)
(b)
2 107  8 1012  16 10712  16 105  1.6 104
Practice Paper: Gold 1 of 4
or 6.4 × 10n
cao
643000
1
B1 cao
1.6 104
2
M1 for 16 107 12 or 16 105 or 0.00016 or
16
1.6 × 10n where n is an integer or
oe or
100000
16
simplified correctly
100000
A1 cao
33
2.4
A1
oe
14
or
15
(a)
(b)
p2  6 p  8
(c)
( x  2) 2 ( x  2)
=
x2
1
(d)
Practice Paper: Gold 1 of 4
2x(x – 2y)
2
B2 cao
(B1 2x(linear expression) or
x(2x – 4y) or 2(x2 – 2xy) or nx(x – 2y) where n is
an integer)
(p – 4)(p – 2)
2
M1 for (p ± 4)(p ± 2) or
(p + a)(p + b) with a,b ≠0 , a + b = –6 or ab = 8
or
p(p – 2) – 4(p – 2) or p(p – 4) – 2(p – 4)
A1 for (p – 4)(p – 2)
(accept others letters)
x+2
1
6a5b2
2
34
B1 x + 2 or
( x  2)
1
B2 cao
(B1 exactly 2 out of 3 terms correct in a product
or a5b2 or 6a2+3b1+1)
17
(a)
–10 – 2 × 3 × (– 5)2 = – 10 – 150
(b)
y = p – 2qx2
2qx2 = p – y
p y
x2 = 2 q
–160
x
p y
2q
2
M1 –10 – 2 × 3 × (–5)2
seen
A1
cao
3
M1
35
or –150
at least one correct process from isolate 2qx2,
divide by q, or by 2 or by 2q
M1 (dep on M1) attempt to square root both sides of x2 =
p y
‘
’
2q
A1
Practice Paper: Gold 1 of 4
or 75 or 150
x=±
p y
oe
2q
condone omission of ±
18
(a)
1
1
B1
cao
(b)
–2
1
B1
cao
1
27
2
M1
use of reciprocal eg 1/ 9 3/2 or
(c)
9 –3/2 = 1/ 9 3/2 = 1/33
y=
1
x–5
2
3
or
cao
729 seen
or
27 seen
M1 for method to find gradient of L1 e.g
or
– 27 seen
63  1 
 
60  2
1
”x + c or y = mx – 5 (c, m do not have to be
2
1
numerical, or correct numerical values) or for ( L ) x  5
2
1
A1 y =
x – 5 oe
2
M1 for y = “
Practice Paper: Gold 1 of 4
eg 3- 3,
1
33
A1
19
square root
36
21*
Practice Paper: Gold 1 of 4
ABO = ADO = 90°
(Angle between tangent and radius
is 90°)
DOB = 360 – 90 – 90 – 50
(Angles in a quadrilateral add up
to 360°)
BCD = 130 ÷ 2
(Angle at centre is twice angle at
circumference)
65o
4
B1 for ABO = 90 or ADO = 90 (may be on diagram)
B1 for BCD = 65 (may be on diagram)
C2 for BCD = 65o stated or DCB = 65o stated or angle C = 65o
stated with all reasons:
angle between tangent and radius is 90o;
angles in a quadrilateral sum to 360o;
angle at centre is twice angle at circumference
1
(accept angle at circumference is half (or
) the angle at the
2
centre)
(C1 for one correct and appropriate circle theorem reason)
QWC: Working clearly laid out and reasons given using correct
language
37
22
23
Vertices at
(–2, –4), (–4, –4),
(–4, –6), (–2, –5)
Correct
diagram
3
(a)
(b)


BK = 2 × AB = 2 × (b – a)


M1 for a similar shape in the correct orientation in the
third quadrant
M1 for an image in the correct orientation of the correct
size
A1 cao
b–a
1
B1 b – a or – a + b
2b – a
3
M1 for a correct vector statement for CK




eg. CK  CA  AK or CK  CB  BK
CK  CB  BK = a + 2 × (b – a)
M1 for BK  2 AB or BK = 2(‘b – a’) or
AK  3AB or AK = 3(‘b – a’)
(may be seen as part of a vector equation BUT
2(b – a) or ‘2(b – a)’ or 3(b – a) or ‘3(b – a)’ by
itself does not score M1)
A1 2b – a or – a + 2b
24
Practice Paper: Gold 1 of 4
t
3 4p
p2
4
M1
for intention to multiply both sides by 4+t
eg p  4  t  3  2t
M1 for intention to correctly move their t terms to one side,
and correctly move their other terms to the other side
eg p  4  t  4 p  2t  3  2t  2t  4 p
M1
for intention to factorise eg t  p  2
A1
for t 
38
3 4p
oe
p2
25
26
(a)
640
2
8
4
M1 for 80    or 80   
4
8
A1 cao
(b)
40
2
8
4
M1 for 160    or 160    or ft their scale factor
4
8
from (a)
A1 cao
(a)
P=
k
k
: 5 = ; k = 40
V
8
P=
3
3
2
3
40
V
2
M1 for P 
1
k
or P = , k algebraic
V
V
M1 for subs P = 5 and V = 8 into P =
A1 for P =
(b)
P=
Practice Paper: Gold 1 of 4
40
2
20
1
39
40
V
B1 ft on k for P =
'k'
V
k
V
Examiner report: Gold 1
Question 1
This, the first question on the paper, was disappointingly done. The majority of errors arose
from the failure to read the question carefully. Many students worked out the correct number
of dark chocolates but then gave this as their answer rather than the total number of
chocolates. The other very frequently seen error was to divide 24 in the ratio 2 : 1 which was
not the technique required.
Question 2
This question was not done well. A significant number of students were unable to express
32
 80 . Many of those
100
32
students who were able to write down the correct calculation for the percentage, i.e.
 100 ,
80
32 out of 80 as a percentage. A common incorrect answer here was
were then unable to work this out correctly. A surprising number of students attempted to
change
32
to a percentage by adding 20 to both the numerator and the denominator. Some
80
students attempted to work out 32% of 80, usually by employing a multiplication grid or a
method of decomposition. Most students adopting a method of decomposition were
unsuccessful in their attempts.
Question 3
There were many correct answers to this question. Candidates who designed and completed a
two-way table were generally successful in gaining all 4 marks. Others were less so as they
often lacked the organising principle already built in to the table. They generally started off
confidently by finding the number of males (29) or the number of females who play squash
(2). Subsequent calculations were then often confused as candidates could not keep track of
what it was they were actually working out. In particular, they wrote down their calculations
without making it clear (e.g. 2GSq would have done) what they were actually finding. There
were too many cases of 50 – 21 = 19 seen.
Question 4
The two most popular answer given were 4n – 2 (correct) and n + 4 (incorrect) Many students
realised that 4 and 2 had to figure in their answer somehow but often combined these
incorrectly with 2n – 4 being the most common incorrect variation. Answers of 4n and 4n + 2
were seen reasonably frequently; both of these were awarded one mark.
Question 5
Many students were able to score at least 1 mark in each part of this question. In part (a),
many students were able to rearrange the inequality to the form 6y > 3 but a surprising
number of these were then unable to solve this for y. A common incorrect answer here was
y > 2.
In part (b), many students were able to write down the correct inequality for one side of the
solution, e.g. x > –3 or x ≤ 4, but relatively few were able to do this for both sides of the
solution. Common incorrect answers included, eg –3 > x ≥ 4, –3 > x ≥ 4, –3 < x ≥ 4 and
–2, –1, 0, 1, 2, 3, 4. Some students gave their answer in the form –3 < n ≤ 4, i.e. using an n
Practice Paper – Gold 1 (1 of 4)
This publication may only be reproduced in accordance with Pearson Education Limited copyright policy.
©2015 Pearson Education Limited.
(or some other letter) instead of x, which was accepted. Students should be advised to use the
letter of the variable provided in the question rather than make up their own variable.
Question 6
In part (a) the vast majority of candidates understood what was meant by translation and
many were able to translate the shape correctly. Incorrect answers were often the result of
translations in the wrong direction or of the x and y movements being interchanged.
Part (b) was generally answered well with the majority of candidates being able to identify
the transformation as a rotation. Common errors were to give the direction of rotation as 90º
clockwise or to give no direction at all and some candidates failed to give a centre of rotation.
A significant number of candidates failed to score any marks as they applied two
transformations, usually a rotation combined with a translation, despite the question asking
for a single transformation.
Question 7
Part (a) was a seemingly innocent algebra question which if attempted in logical steps yielded
the correct value of x. Unfortunately many missed the correct expansion of the left hand side
of the equation and floundered in further simplification with only 48% of the candidates
scoring all 3 marks. 18% of the candidates did manage to score 1 mark for either expanding
the bracket correctly or rearranging their equation with at least the terms in x or the constant
terms isolated correctly. However a significant number made mistakes with signs when
rearranging ending up with 8t or 0 rather than 4t and 24. 25% failed to score.
In part (b) 57% scored one mark as large numbers of students failed to correctly expand the
2nd term of the 2nd bracket with –3x – 6y seen rather than –x + 6y. Most managed to expand
the first bracket correctly. A significant minority treated the question as expanding double
brackets to obtain a quadratic equation. Only 21% of the candidates expanded and simplified
correctly.
Part (c) was generally well done with 43% getting it fully correct and a further 24% scoring
one mark for either writing down 4 correct terms with incorrect signs or 3 correct terms out
of 4 with the correct signs. Many students used the grid method which generally resulted in at
least one mark. There are still many candidates who do not realise that the expansion should
contain a term in x2 and there were many who combined the constant terms to get +2 or –2
rather than –35.
Question 8
In part (a), the majority of candidates gained the mark for identifying the correct modal class
interval, although answers of 5 – 9 and 10 – 14 were not uncommon. In part (b), many
candidates failed to use mid interval values for their plotted points choosing the upper and
sometimes the lower interval values. Many candidates drew extra lines in an attempt to
complete a closed polygon or joined points with curves and consequently failed to gain full
credit. Candidates who drew a frequency diagram with bars first and then identified the
midpoint of the top of their bars to construct the polygon were usually most successful with
their accuracy. A significant number of candidates drew a bar chart only.
Practice Paper: Gold 1 of 4
41
Question 9
Not many candidates took the easiest route of using 22 12 % – 17 12 % and then finding 5% of
£180. Numerous different attempts were seen, some of which were productive. Most
candidates made a good attempt at this question and encouragingly lots of working was
shown with 43% scoring all 3 marks.
However a high proportion of candidates made arithmetic errors in their calculations. The
most common method used to find the percentages was to break them down to 10%, 5%, 2%,
1% and 12 %. Mistakes came from an error in this or an error when adding them. 26% of
candidates failed to score.
Workings were too often slanted on the page and scattered everywhere. A more organised
approach should be encouraged.
Question 10
The ability to combine both bits of information was the key to success in this question. Those
candidates who used either the true value of the coins or the fact that the 10p coin was worth
two times as much as the 5p coin with the given ratio in the correct way generally scored at
least one mark. It was disappointing to see so many candidates who had overcome this hurdle
then ignore the instruction to give their ratio in its simplest form. Answers of 10 : 30 or 5 : 15
were seen as often as 1 : 3 as the final answer. The most common incorrect method of
solution was to attempt to divide an amount in the ratio 2 : 3 rather than use this in the correct
way.
Question 11
In part (a) many candidates plotted the two points correctly and then went on to provide an
acceptable description of the relationship between the 2 variables. 64% of candidates got both
part (a) and part (b) correct with only 6% failing to score.
Many lost a mark in (b) by only writing ‘negative’ rather than ‘negative correlation’.
In part (c) many found the approximate value without drawing a line of best fit on the
diagram. Whilst this was not penalised, candidates risked losing all the marks if they wrote
down an incorrect value.
In part (d) many statements referred to the physical attributes of the situation. There was not a
great appreciation that correct comments had to refer to the data stopping at 70° or a
reference to the line of best fit extending to negative time, not the laws of physics. Some
candidates gave irrelevant information about the graph needing to be in seconds or
information about the boiling point of water rather than answer the actual question which
concerned why Suzy’s data cannot be used. There were many interesting, unacceptable
comments which referred to ice cubes melting and water boiling at 100°. Only a quarter of
the candidates were successful in both parts (c) and (d) with around half the candidates not
scoring in part (d).
Question 12
This question was a fairly straightforward way of requiring candidates to write a formula
using a familiar situation. The responses were hardly electrifying with many which were
shocking. N  p  b or even N  p  b were often seen indicating no insight into the
meaning of the symbols.
Practice Paper: Gold 1 of 4
42
Question 13
The vast majority of the candidature gained at least one mark for drawing a rectangle in part
(a). Drawing a rectangle of area 20 cm2 was less successful, many producing a 5 by 5 square
and many drawing a rectangle of perimeter of 20 cm.
In part (b), an isosceles triangle was often seen but rarely of area 12 cm 2. Far too often the
product of the base and the perpendicular height was equal to 12. Some candidates did not
make good use of the cm. square grid, instead drew triangles using fractions of the sides of
the squares.
Question 14
Common errors in part (a) were to have the wrong number of zeros or to write the answer as
0.0000643.
In part (b) those candidates that worked with the numbers in standard form were more likely
to be awarded marks than those who attempted to first take both numbers out of standard
form. It was common to be able to award a mark for 16 × 10–5 but then candidates were more
likely to write their final answer incorrectly as 1.6 × 10–6.
Question 15
Success throughout this question was varied. Candidates who understood the instruction
‘factorise’ were generally able to score at least one mark by taking out one common factor
correctly but not all recognised that 2x was a common factor.
It should be noted that part (b) of this question also appeared on the calculator paper 4H.
There were many good attempts at the factorisation and it was only a lack of confidence with
signs which prevented a large number of candidates from scoring two marks for a fully
correct factorisation.
Common errors in part (d) included writing the answer as 5a5b2, 6a5b, or 6a5 + b2.
Question 16
This question was poorly answered with many candidates either making simple arithmetic
errors or demonstrating an inability to multiply two fractions together. Many correctly
13 8
39 32
 often became

converted the given fractions to improper fractions but then
4 3
12 12
followed by attempts to add the two fractions or multiply only the denominators or ‘cross
104
multiply’. Some candidates correctly found
then failed to simplify accurately. The
12
ability to “cancel” seems to be lacking. The most common mistake in efforts to find the
product of the two given mixed numbers was to multiply the whole numbers and the fractions
1
separately giving an incorrect answer of 6 .
6
Question 17
This was one of the most poorly answered questions on the paper, largely because a great
many students just could not interpret a mileage chart correctly. In finding a distance between
two towns, many calculated the differences between the numbers beneath each town. This
was evident immediately in part (a).
Practice Paper: Gold 1 of 4
43
In part (b) many, who could read a mileage chart correctly, often correctly selected just two
of the 3 required distances.
In part (c), many candidates demonstrated a complete inability to communicate a structured,
organised and clear solution to the problem. Calculations dotted about the working space
without explanation were rife. A small number of candidates were able to correctly solve the
problem fully, many stumbling by incorrect distances from the table but more often the
inability to deal with any distance, speed, time calculation. The most common mistake was to
assume that 1 mph equated to 1 minute so distances of say 95 miles or 105 miles instantly
became times of ‘1 hour 45 minutes’ and ‘1 hour 55 minutes’ respectively. When a total time
of travel was found by a reasonable method (usually incorrect however), many candidates
were then able to gain credit for correctly attempting to work out the time for the end of the
journey.
Question 18
Very few candidates used any correct algebra to solve this problem. The value of x, Jim’s
share of the £23 was often correctly found in part (b) by trial and improvement methods.
It was very rare, in part (a), to see a correct equation formed. Some were able to quote x + 4
and x – 2 as the shares of Gemma and Jo, but could go no further.
In (a), there were many statements like x + 4 – 2 = 23 followed often in (b) by answers of
£17.
Question 19
Candidates were expected to show how they could find the gradient of the given line by using
a variant of rise ÷ run. Many candidates were unable to do this and had no idea of what a
gradient is. Some candidates were able to give the correct gradient for the given line L1, but
then gave a different coefficient of x for L2. Candidates were much more confident in
assigning the value of −5 to c in y = mx + c.
Question 20
[NB: Part (a) of this question would not be tested on the 1MA0 specification]
In part (a), the expression 4abc was usually identified as representing a volume. The
expressions a3b and ab + c3 were often chosen in error. Part (b) was poorly answered with an
answer of 800 cm3 being the modal mistake. Some candidates wrote 200 × 200 × 200 and
then were unable to complete the calculation.
Question 21
When asked to give reasons in a geometry questions, reasons must be correct and must use
correct mathematical language. Reasons given in responses seen to this question were often
incomplete or not completely correct. ‘Angle between tangent and circle is 90’ and ‘angle at
origin is twice the angle at the edge of the circle’ are both examples where a communication
mark was not awarded as the statements were not accurate enough. It is also important to
ensure that the final answer is communicated properly. In this case the value of the angle had
to be linked with the angle itself so sight of Angle BCD = 65 (or similar) was expected
rather than just to see a 65 somewhere amongst the candidate's working. Very few
candidates used the alternate segment theorem as part of their explanation.
Practice Paper: Gold 1 of 4
44
Question 22
This question on transformation geometry was not very well answered with a small
percentage of candidates giving a fully correct answer. More than three quarters of candidates
scored no marks but 1 mark was awarded for showing a similar-sized shape in the correct
orientation in the third quadrant or for a shape of the correct size in the correct orientation. If
they showed both of these, they scored 2 marks.
The negative scale factor of this transformation proved a major stumbling block with many
candidates instead using a scale factor of + 1 .
2
Question 23
In part (b) candidates do need to show their method of solution. When the answer was
incorrect it was very often difficult to follow through working. Vector equations should be
used to show how the required vector is being calculated.
Question 24
This question was poorly completed, with few candidates managing to gain more than one
mark for an intention to multiply through by 4 + t. Often the bracket was missing and p(4 + t)
became 4p + t. Candidates did appear to realise that they needed to find ‘t = something’ but
lacked the ability to achieve this. Of those who did successfully isolate the term sin t, only the
most able went on to factorise correctly.
Question 25
Correct answers were rare in this question, with most candidates incorrectly assuming a scale
factor of 2 and giving the answers 160 and 80. Attempts to work out the surface area or the
volume frequently led nowhere.
Question 26
Those students who were able to write down the correct proportionality equation generally
went on to score full marks for the question. The majority of students were unable to gain any
credit. The most common errors seen were to give the incorrect equation of P = V – 3 or to use
direct rather than inverse proportion.
Practice Paper: Gold 1 of 4
45
Practice paper: Gold 1
Mean score for students achieving Grade:
Spec
1380
1MA0
1MA0
1380
1MA0
1MA0
1380
1380
1380
1380
1380
1380
1380
1380
1380
1380
1380
1380
1MA0
1380
1MA0
1MA0
1380
1MA0
1MA0
1380
Paper
1H
1H
1H
1H
1H
1H
1H
1H
1H
1H
1H
1H
1H
1H
1H
1H
1H
1H
1H
1H
1H
1H
1H
1H
1H
1H
Session
YYMM
1011
1411
1406
1011
1411
1311
1111
1006
1111
1203
1111
911
1111
1203
1203
1006
1111
1111
1406
1006
1206
1303
1203
1211
1211
1011
Question
Q01
Q02
Q03
Q04
Q05
Q06
Q07
Q08
Q09
Q10
Q11
Q12
Q13
Q14
Q15
Q16
Q17
Q18
Q19
Q20
Q21
Q22
Q23
Q24
Q25
Q26
Mean
score
1.30
0.74
3.00
1.73
1.02
2.37
3.91
1.53
1.84
0.93
3.42
1.50
1.25
1.19
3.47
1.35
0.96
1.03
0.62
1.60
0.89
0.41
0.54
0.32
0.36
0.58
37.86
Max
score
2
2
4
4
4
4
7
3
3
2
5
3
4
3
7
3
5
4
3
4
4
3
4
4
4
4
99
Practice Paper – Gold 1 (1 of 4)
This publication may only be reproduced in accordance with Pearson Education Limited copyright policy.
©2015 Pearson Education Limited.
Mean
%
65
37
75
43
26
75
56
51
61
47
68
50
31
40
50
45
19
26
21
40
22
14
14
8
9
15
38
ALL
1.30
0.74
3.00
1.73
1.02
2.37
3.91
1.53
1.84
0.93
3.42
1.50
1.25
1.19
3.47
1.35
0.96
1.03
0.62
1.60
0.89
0.41
0.54
0.32
0.36
0.58
37.86
A*
1.77
1.87
3.82
3.83
3.66
3.59
6.61
2.63
2.78
1.63
4.39
2.77
3.53
2.40
6.61
2.72
4.30
3.56
2.65
3.54
3.00
2.48
2.63
2.95
2.97
3.33
86.02
A
1.64
1.76
3.56
3.26
3.28
3.31
6.18
2.13
2.58
1.37
4.08
2.30
2.71
1.99
5.86
2.07
2.87
2.43
1.76
2.39
1.94
1.46
1.42
1.43
1.15
1.52
66.45
%A
82.0
88.0
89.0
81.5
82.0
82.8
88.3
71.0
86.0
68.5
81.6
76.7
67.8
66.3
83.7
69.0
57.4
60.8
58.7
59.8
48.5
48.7
35.5
35.8
28.8
38.0
67.12
B
1.47
1.47
3.30
2.28
2.32
2.95
5.42
1.49
2.35
1.08
3.83
1.65
1.86
1.54
4.40
1.32
1.35
1.32
0.56
1.34
0.91
0.54
0.51
0.38
0.35
0.35
46.34
C
1.22
0.86
2.91
1.06
1.09
2.27
3.65
0.96
1.87
0.77
3.39
1.17
0.90
0.87
2.54
0.64
0.40
0.66
0.08
0.83
0.23
0.13
0.11
0.05
0.17
0.05
28.88
%C
61.0
43.0
72.8
26.5
27.3
56.8
52.1
32.0
62.3
38.5
67.8
39.0
22.5
29.0
36.3
21.3
8.0
16.5
2.7
20.8
5.8
4.3
2.8
1.3
4.3
1.3
29.17
D
0.92
0.41
2.12
0.44
0.37
1.34
2.10
0.56
1.20
0.50
2.95
0.84
0.34
0.37
1.23
0.23
0.10
0.39
0.01
0.60
0.04
0.04
0.02
0.01
0.12
0.01
17.26
E
0.70
0.11
1.17
0.23
0.11
0.62
1.36
0.34
0.70
0.37
2.63
0.60
0.19
0.16
0.63
0.10
0.05
0.27
0.00
0.51
0.01
0.03
0.01
0.00
0.09
0.01
11.00