1H
1
Mark Scheme
Specimen Paper
Pearson Edexcel International GCSE
In Mathematics A (4MA1) Paper 1H
3
2
1
50, 100, 150, 200, 250, 300, 350, 400 and
80, 160, 240, 320, 400 OR
2 × 5 × 5 and 2 × 2 × 2 × 2 × 5
2 × 2 × 2 × 2 × 5 × 5 or 400
b
245 = 5 × 7 × 7
140 = 2 × 2 × 5 × 7;
120 ÷ 1002 (=0.012) or 810 ÷ 120 (=6.75)
810 ÷ “0.012” or “6.75” × 1002
Working
5400 ÷ (5 + 3 + 4) (=450)
“450” × 5 or “450” × 3 or “450” × 4
a
Q
A1
M1
M1
A1
2
3
35
16 40
M1
3
3
2250, 1350, 1800
67 500
Mark
Answer
Notes
Total 3 marks
Total 3 marks
LCM found
or 4 40 pm
Total 5 marks
or lists at least 3 factors of each number
(other than 1 and the number)
(1, 2, 4, 10, 14, 35, 70, 140)
(1, 5, 7, 35, 49, 245)
M1
M1
A1
M1
M1
A1
Apart from Questions 4d, 10 (where the mark scheme states otherwise), the correct answer, unless clearly obtained by an incorrect
method, should be taken to imply a correct method.
International GCSE Maths 1H
2
5
4
d
c
a
b
s 1 2
2s 2
 t or
t
a 2
a
5 3
1     or 100% − 80% (=20%) or
8 8
1 3  23 
  
5 8  40 
4 3
5 1
23
 " " or  " " or 1 - " "
5 8
8 5
40
6x – 5 = 2(x + 1) or 6x – 5 = 2x + 2
6x – 2x = 2 + 5
2s = at2 or
m + 9m – 5m − 45
2
1.75
17
40
3
M1
M1
A1
A1
2
2s
a
A1
M1
2
m2 + 4m − 45
t = (±)
B2
M1
2
3y(2y + 5)
3
oe eg.
Total 3 marks
may see decimal or percentage
equivalents
M1
A1
may see decimal equivalents
M1
7
dep on at least M1 scored
4
Total 9 marks
for a correct first step
B1 for 3(2y2 + 5y) or y(6y + 15)
M1 for 3 terms correct
or 4 terms correct ignoring signs or
m2 + 4m + ….. or
…. + 4m – 45
3
8
7
6
a
b
c
b
a
6630
( 78)
85
55 000 000
1.382 × 109 − 1.327 × 109 oe or
42 × 7 (=294) or 8 × 50 (=400)
8 × 50 − 42 × 7
6630 ÷ 85 × 100 or “78” × 100
6630 = 85% oe or
0.03 × 180 000 (=5400)
“5400” + 180 000
5.5 × 107
93 000 000
Singapore
106
7800
185 400
B1
B1
M1
A1
2
M1
M1
A1
M1
A1
1
1
3
3
3
M1
M1
A1
M1
Total 3 marks
Total 6 marks
M2 for 6630 ÷ 0.85
M2 for 1.03 × 180 000
Total 4 marks
or for 5.5 × 10n n ≠ 7
dep
dep
4
10
9
y = −1.5
−26y = 39
21x + 35y = 21
21x + 9y = 60
e.g. (h = )
or
or
26x = 91
x = 3.5
9x + 15y = 9
10
14.5
35x + 15y = 100
14.52  102 (=10.5) or
 10 
(x = ) cos 1 
 (=46.3…)
 14.5 
1
e.g.  20  "10.5" or
2
1
 20 14.5  sin("46.3...")
2
e.g. (h2 = ) 14.52  102 or cos x =
x = 3.5, y = −1.5
105
4
4
A1 one variable correct dep on M1
M1 (dep on M1) for method to find second variable
A1 both variables dep on at least M1
Total 4 marks
M1 for method to eliminate one variable
(condone one arithmetic error)
Total 4 marks
(dep on M1) method to find area
M1
cao
complete method to find height or angle
M1
A1
start to find height or angle
M1
5
median = 26 or LQ = 20 or UQ = 29
iqr = 9 and median = 26
(1 + 4)2 (= 25)
“5”2 × 8 – 8 oe
11
12
192
median = 26; iqr = 9
and two comparisons
4
B1
M1
A1
B1
3
M1
M1
A1
complete method
Total 3 marks
ft comparison of IQR
eg. English results were more spread out
ft comparison of median
eg. Maths results were higher
NB. In order to award both marks, at least
one of the comparisons must be in context
Total 4 marks
6
7  1  8 
 
4 1  3 
8
8
e.g. 7 = “ ”× 4 + c or y – 7 = “ ”(x – 4)
3
3
11
c=−
3
8
11
y =” ”x – “ ”
3
3
Angle DBA = 43o or Angle DAB = 90o
180 – 90 – 43 (=47)
13
14
47 with reasons
5
8x – 3y = 11oe
M1
M1
A1
B2
Total 4 marks
dep on M2
M1
A1
for a method to find c
M1
for full reasons
Angles in the same segment are equal;
Angle in a semi-circle is a right angle
Angles in a triangle add up to 180o
(B1 for a correct and relevant reason using a
circle theorem)
Total 5 marks
4
for a method to find gradient
M1
7
16
15
b
a
3 × (3 + 8.5) = 5 × PR or 3 × (3 + 8.5) = 5 × (5 + PQ)
(3 × (3 + 8.5)) ÷ 5 − 5
2
k
or k = 0.8
0.0064
k
q
 0.8 


 20 
10 
P
0.8
q
1.9
0.0016oe
P
3
2
3
M1
M1
A1
A1
M1
A1
M1
M1
Total 3 marks
for a complete method for PQ
Total 5 marks
oe with P as the subject
implies first M1
8
18
17
b
a
-4
-3
-2
-1
0
1 4
    63 or   62  t or   62 10
2 3
1 4
    63 +   62 10
2 3
1 4
  62  t =     63 +   62 10
2 3
-5
(x – 5)(x + 1)
x2 + 2x – 6x – 5 > 0 or x2– 4x – 5 > 0
1
2
3
4
7
x < −1, x > 5
5
x
4
3
1
Total 4 marks
for forming a correct equation
M1
A1
for total volume
M1
Total 4 marks
for one expression for an
appropriate volume
ft from (a)
M1
M1
M1
A1
B1
9
20
19
4
for simplified answer, may be factorised
for completion of proof
Total 4 marks
proof
M1
A1
8n + 8
(4n2 + 6n + 6n + 9) – (4n2 + 2n + 2n + 1)
M1
Total 4 marks
(2n + 3)2 – (2n + 1)2 =
A1
for algebraic representation of two
consecutive odd numbers
for correct expansion of at least one bracket
4
M1
22
M1 for finding area of 50 – 55 bar
M1 for method to find total area
(condone two errors)
M1 (dep on M2) for complete method
eg. 2n + 1, 2n + 3
"44"
150
"300"
e.g. 8.8 × 5 (=44)
e.g. 3.4 × 10 (=34) + 7.2 × 20 (=144) + 8.8 × 5
(=44) + 4 × 15 (=60) + 1.8 × 10 (=18) (=300)
10
21
x  3 5
x = 3  32  4 or (x – 3)2 = 5
  6  (6)2  4 1 4
x
or
2 1
Correct
graph
6
A1
M1
(x − 3)2 − 5 or x = 3
eg.
B2
A1
M1
dy
= 2x – 6 or (x − 3)2 − 32 + 4 or
dx
'3  5 ' '3  5 '
2
e.g.
for fully correct labelled graph
(see end of mark scheme)
(B1 for parabola intercepting y-axis at (0, 4))
Total 6 marks
or exact equivalents
(ft providing previous M1 scored) for (3, −5)
for correct method to find roots
11
23
22
  31  (31)  4  210
2
AB =
180
 sin(180  55  "25.6")
sin 55
CAB = 25.6(1…)
sin CAB sin 55

95
180
 95sin 55 
CAB = sin 1 
 or
 180 
eg. diagram drawn showing relative positions of A, B and
C
can be implied by angle ABC = 55o
e.g. (x – 21)(x – 10) = 0 or
2
e.g. x2 – 31x + 210 = 0 or 7x2 – 217x + 1470 = 0
7
x7
or
x
x 1
7 x7 7
2 

x x  1 15
217
21
5
5
21 must be selected as final answer
Total 5 marks
A1
Total 5 marks
1802  952  2 180  95  cos(180  55  "25.6"
M1 dep or for
M1 dep
M1
M1 interprets information
A1
correct quadratic ready to solve
method to solve quadratic equation
for equation formed
M1
M1
M1
for a correct expression for one probability
M1
12
13
May 18
Mark Scheme (Results)
Summer 2018
Pearson Edexcel International GCSE
In Mathematics A (4MA1) Paper 1H
b
“56” ÷ 40
0.5 × 19 + 1.5 × 12 + 2.5 × 5 + 3.5 × 2 + 4.5 × 2 (=56)
or
9.5 + 18 + 12.5 + 7 + 9 (=56)
1.4
4
A1
for 1.4 or 1
2
5

M1 dep on at least M1
f provided
Allow division by their
addition or total under column seen
M2 for at least 4 correct products added
(need not be evaluated)
If not M2 then award
M1 for consistent use of value within
interval (including end points) for at least
4 products which must be added
OR
correct mid-points used for at least 4
products and not added
International GCSE Maths 4MA1 1H
Apart from questions 3c, 11b and 20 (where the mark scheme states otherwise) the correct answer, unless clearly obtained from an incorrect
method, should be taken to imply a correct method.
Question
Working
Answer
Mark
Notes
a
0<p≤1
1
B1
1
14
2
Question
2
M1
“1275” ÷ 5 × 2 or 3 × 170
"
(girls = )
A1
M1
M1 dep on M2
5
15
"170 (=1275) or G : A = 2 : 6 oe
2
510
M1 award of this mark implies the first M1
2
(of children)
9
2 3 2 
    (of total)
9 5  15 
2 3 3 2 2

or G : C : A =  : :   : 3: 2 
9 5 5 5 3

(girls = )
A1
M1
“255” × 2 or “1275” – “765” or “1275” ÷ 5 × 2
Alternative scheme
M1 dep on M2
Notes
“765” ÷ 3 (=255) or "765" ÷ 3 × 5 (=1275)
Mark
5
M1
M1 award of this mark implies the first M1
510
Answer
7 × “85” + 170 (=765) or 9 × “85” (=765) or
“595” + 170 (=765) or 170 × “3.5” + 170 (=765)
Working
170 ÷ 2 (=85) or 170 ÷ 2 × 7 (=595) or 7 ÷ 2 (=3.5)
15
4
19
oe
2
Yes with reason
10
oe
12
c
1, 3, 5, 7, 9, 10, 11
b
(ii)
1, 2, 3, 4, 6, 12
4, − 6
(ii)
a (i)
(x – 4)(x + 6)

5x + 15 = 3x – 4 or
x + 3 = 3x  4
5 5
e.g. 5x – 3x = −4 – 15
y14
16m12
Answer
Working
d (i)
c
a
b
Question
3
2
1
1
1
1
2
3
Mark
1
2
A1
M1
A1
M1
B1
B1
B1
B1
a
with a < 12 or
12
for
10
oe or 0.83(3…) or 83(.3..)%
12
10 and 12 used with incorrect notation E.g. 10 : 12
for 12 – 2 (=10) or
e.g. no numbers in both A and C or A and C do not
intersect or A and C do not overlap or A and C are
mutually exclusive
cao
cao or ft from any (x + p)(x + q)
cao
dep on at least M1
for (x + a)(x + b) where either ab = −24 or a + b = +2
e.g (x – 6)(x + 4)
M1
A1
ft from ax + b = cx + d for correctly isolating terms in x on
one side of equation and constant terms on the other side
M1
Notes
if not B2 then
B1 for am12 or 16mb or 24m12 b ≠ 0, 12 a ≠ 1, 16
for removing bracket in a correct equation or dividing all
terms by 5 in a correct equation
B1
B2
16
6
b
Question
a
5
"106.34" (=10.3…)
2
π × “10.3…” or 2 × π × "10.3..."
9.72  3.52 or
9.72 + 3.52 (=106.34)
0.5 × 105 – 8 or 0.0005 or 5 × 10n or
5.0 × 10n
Working
32.4
5 × 10-4
Answer
80 000
4
2
Mark
1
A1
M1
M1
M1
A1
M1
B1
3.5
(=10.3…)
cos 70.2
for answer in range 32.3 – 32.41
dep on M2
eg MN=
M1 for a complete method to find MN
eg cos70.2=
3.5
MN
1
1
or
2000
2 103
M1 for the use of MN and a correct angle
(70.1… or 70.2, 19.8…) in a correct trig
statement
SC : B1 for
for 5 × 10-4 or 5.0 × 10-4
Notes
17
b
Question
a
7
E.g. 252 000 ÷ 1.05
160 000 – “6400” – “6144” − “5898.24”
4
× (160 000 – “6400”) (= 6144)
100
4
× (160 000 – “6400” – “6144”) (= 5898.24)
100
4
× 160 000 oe (=6400)
100
Working
240 000
Answer
141 558
3
3
Mark
A1
M2
A1
M1
M1
If not M2 then award
M1 for 160 000 ×0 .96
(=153 600)
or 160 000 × 0.962
(=147 456)
M2 for 160 000 × 0.963
or 160 000 × 0.964
(=135 895.44..))
NB: An answer of 239 400 scores M0 M0 A0
If not M2
then M1 for
x × 1.05 = 252 000 or 252 000 ÷ 105 oe
SC If no other marks gained, award
B1 for 160 000 × 0.12 oe (=19 200)
or 160 000 × 0.88 oe (=140 800)
or an answer of 140 800
or an answer of in the range 179 978 – 179 978.24
accept (1 – 0.04) in place of 0.96 throughout
for 141 557.76 - 141 558
for a complete
method (condone 4
years rather than 3)
Notes
18
b
(ii)
Question
a (i)
8
Working
2
1
23 × 35 × 5 × 74
4, 2, 1
Mark
1
Answer
3 × 73
A1
M1
B1
B1
for r = 1
or for p = 4 and q = 2
or correct representation of C in terms of prime
factors on a Venn diagram
for 23 × 35 × 5 × 74 oe or 23 337 720
Notes
for 3 × 7 oe or 1029
3
19
Question
9
12.8
or
sin 72
12.8
or
cos(90  72)
"13.4..."2  12.82
12.8
or 12.8tan(90 – 72)
tan 72
12.8
or
cos(90  72)
or 12.8tan(90 – 72) or
5 × (“13.4(58…)” – “4.15(89…)” ) + 5 × 12.8 or
5 × (“13.4…” + “4.15…” + 12.8) – 10 × “4.15…”
12.82  "4.15..."2
13.4(58…) or 13.5 or
One of (hypotenuse = )
or 4.15(89…) or 4.16 or
AND
One of (shortest side = )
13.4(58…) or 13.5
sin 72
(hypotenuse = ) 12.8 or
4.15(89…) or 4.16 or
tan 72
E.g.(shortest side) = 12.8
Working
E.g. tan 72  12.8 or tan(90  72)  o or
a
12.8
12.8
12.8
or cos(90  72) 
sin 72 
h
h
Answer
110
Mark
5
for method to use found lengths to find
perimeter
for answer in range 110 - 111
M1
A1
for a complete method to find both missing
sides of triangle
NB Could use Pythagoras’s theorem with
side found – must be a complete correct
method
for a complete method to find one side of
the triangle
M1
M1
Notes
substitutes correctly into a trig ratio
(including the Sine rule)
M1
20
b
Question
a
10
80
Reading from graph from cf = 52 (=118) or
Reading from graph from time = 120 (=55)
Alternative scheme
0.65 × 80 (=52)
"55"
100 oe (= 68(.75))
80
or
0.35 × 80 (=28) or e.g. 80  "55" 100 oe (=31(.25))
Reading from graph from time = 120 (=55)
or 80 – 55 (=25)
Working
Readings from graph at cf 20 and cf 60
eg. readings of 103 and 123
No with
correct figures
No with
correct figures
Answer
20.5
3
3
Mark
2
eg. No with 28 and 25
or
No with 31.25%
(accept value in range 30% – 31.25%)
or
No with 68.75% and 65%
(accept value in range 68% – 70%)
A1
A1
M1
eg. No with 118 (minutes)
or
No with 52 and 55
accept reading in range 55 – 56
accept a value in the range 30 – 31.25
or a value in the range 68 – 70 for
this mark unless clearly from
incorrect working
M1
M1
accept reading in range 55 – 56
for answer in range 19 – 21
M1
A1
M1
Notes
21
Question
a
11
Alternative scheme
2x3 – 10x2 – x2 + 5x + 6x2 – 30x – 3x + 15
eg.
2x3 + 5x2 – 3x – 10x2 – 25x + 15 or
2x3 – 4x2 – 30x –x2 + 2x + 15 or
2x3 – 11x2 + 5x + 6x2 – 33x + 15
Working
2x – x + 6x – 3 or 2x2 + 5x – 3 or
x2 + 3x – 5x – 15 or x2 – 2x – 15 or
2x2 – 10x – x + 5 or 2x2 – 11x + 5
2
2x3 – 5x2 – 28x + 15
Answer
2x – 5x2 – 28x + 15
3
3
Mark
3
A1
M2
A1
M1
M1
for a complete expansion with 8
terms present, at least 4 of which
must be correct
(dep) ft for at least half of their
terms correct in second expansion
(the correct number of terms must
be present)
Notes
for expansion of any 2 of the 3
brackets (at least 3 of 4 terms
correct)
22
Question
b
11
6  6  60
6
96
(x = ) 1 
5
 1 oe
3
5
( x  1)2  1  ( 0)
3
Alternative scheme
e.g 3((x + 1)2 – 1) – 5 (= 0) or
NB: denominator must be 2 × 3 or 6 and there must be
evidence for correct order of operations in the
numerator
Accept 9.79 – 9.8(0) in place of
6  96
or
6
2
Working
0.633, −2.63
Answer
0.633, −2.63
3
Mark
3
for correct method to isolate x
dep on M1 for answer in range
0.63 to 0.633 , −2.63 to −2.633
Award M2A1 for correct answer
with correct working that would
gain at least M1
A1
for completing the square
dep on M1 for answers in range
0.63 to 0.633 , −2.63 to −2.633
Award M2A1 for correct answer
with correct working that would
gain at least M1
6  62  4  3  5
23
condone one sign error in
substitution;
allow evaluation of individual
terms e.g 36 in place of 62
M1
M1
A1
M2
Notes
If not M2 then award M1 for
23
13
alternate segment theorem
b (i)
(ii)
54
Answer
3, 4
see graph at end of mark
scheme
angle at centre is twice
angle at circumference
27
Working
(ii)
a (i)
Question
(a)
12
(b)
1
1
1
1
Mark
1
3
B1
B1
B1
B1
B1
B3
"54"
2
dep on B1 in (b)(i) accept alternative reason
angle between tangent and radius is 90o
If answer for (b)(i) is ft from (a)(i) then reason must be
angle between tangent and radius is 90o
ft from (a)(i) for
dep on B1 in (a)(i) accept alternative reasons
eg. angle at circumference is half the angle at the centre
NB. May shade wanted or unwanted regions; lines may be
solid or dashed
cao
If not B2 then award
B1 for line x + y = 4 drawn
If not B3 then award
B2 for x + y = 4 drawn (with no additional lines drawn) and a
region identified that satisfies at least 3 of the 5 given
inequalities
for correct region identified
Notes
24
d
c
b
Question
a
14
or
3 19  x  5
oe
4
19  3 oe or f(4) or
3 19  3  5
4
4y = 3x – 5 or 4x = 3y – 5
Working
x > 19
1.75 oe
4x  5
oe
3
Answer
−6.5 oe
2
2
2
Mark
1
B2
A1
A1
M1
M1
B1
for (x) > 19 or an equivalent
statement in words
If not B2 then award
B1 for (x) ≥ 19
for 1.75oe (and no other solution)
Notes
25
b
Question
a
15
2
1(3x  5)

or
2(3x  5)(3x  5) 2(3x  5)(3x  5)
Alternative scheme
6 x  10
9 x 2  25

(9 x 2  25)(6 x  10) (9 x 2  25)(6 x  10)
6 x  10
9 x 2  25

(9 x 2  25)(6 x  10) (9 x 2  25)(6 x  10)
E.g.
1
1

(3x  5)(3x  5) 2(3x  5)
1
4
1 5
x
5 1
5


4x
x
 y

4
E.g. 
or  2  or
or 2
20 
y
4 y 2
 y 
 256 x 
ya
ky a
1
k
or
oe , a = 2, b = 5
b or
b with 2 of k =
x
x
4
ya
or
with 2 of m = 4, a = 2, b = 5
mxb
8
Working
7  3x
2(3x  5)(3x  5)
7  3x
2(3x  5)(3x  5)
y
4 x5
2
Answer
3
3
Mark
2
accept equivalents eg.
7  3x
18 x 2  50
M1 for two correct fractions with a common
denominator
A1
M1 for two correct fractions with a common
denominator
if there is any expansion at this stage
then it must be correct
M1 indep for (3x + 5)(3x – 5)
2
1 2
y
y
y2
A1
4
2 -5
0.25
for 5 or 5 or
5 or 0.25y x
x
x
4x
Notes
M1 for a correct first step leading to a
correct partially simplified expression
26
(7  3x)(3x  5)
(9 x 2  25)(6 x  10)
A1
accept equivalents
M1 Numerator expanded and then factorised
correctly
27
Question
16
"
27  3 
"    or
125  5 
3
"
125  5 
"  
27  3 
98  27 

 or 0.216 or 125 – 98 (=27)
125  125 
3
3
1  " " or h − " " h oe
5
5
5
k= 3
1 2
1
98 1 2
 r h   (kr )2 kh =
  r h oe
3
3
125 3
Alternative scheme
3
3
1  " " or h − " " h oe
5
5
3
1
Working
2
h oe
5
2
h oe
5
Answer
4
Mark
4
A1
for
2
h oe (may not be simplified)
5
for the length scale factor
M1
M1
sets up an equation using scale factor
5
for 2 h oe (may not be simplified)
for the length scale factor
may be seen as a ratio E.g. 3 : 5
Notes
M1
A1
M1
M1
M1
28
b
Question
a
17
8 

10 

with
e.g. “60” ÷ 2 (=30) and “210” ÷ 7 (=30)
e.g. 
 63   3    60  
     

 211  1    210  
with
e.g. “58” ÷ 2 (=29) and “203” ÷ 7 (=29)
OR
 63   5    58  
     

 211  8    203  
e.g. 
 5  8 
10   3 
   "   " or    "   "
8  4
 11   1 
 BC    7    11    4  
2
Working
Proof
Answer
(13, 12)
2
Mark
3
A1
 2
proof with justification eg. BE  29   (or
7
 2
AE  30   ) with ABE is a straight line or
7
210 ÷ 60 = 3.5 and 7 ÷ 2 = 3.5 so ABE is a straight line
may work with A and E, in which case may need to ft
for method mark from (a)
for coordinates (5 – 2 + 10, 8 – 7 + 11) assigned to C
M1
A1
M1
Notes
or coordinates (5 – 2, 8 – 7) (= (3, 1)) assigned to A
(may be seen in vector form) or
(13, y) or (x, 12) given as coordinates for C
M1
29
b
(ii)
Question
a (i)
18
Working
3
1
(−2, −0.5) oe
e.g. 2, 90, 1
Mark
1
Answer
(3, −1)
B3
B1
B1
If not B2 then B1 for any 1 correct value or the graph of
y = sinxo for 0 ≤ x ≤ 360
If not B3 then B2 for any 2 correct values
NB. 2 values from 2, 90, 1 OR 2 values from −2, 270, 1
NB: accept a value of (90 + 360n) in place of 90
or (270 + 360n) in place of 270 where n is an integer
(could be negative)
for all 3 correct values
e.g. 2, 90, 1 or −2, 270, 1
Notes
30
Question
19
9 
 2
11
11
" " " " or  " " " " 
20 
20
20
 20
2
1 2
3 3  11 
 1 3 3 2  11 
      or 1        
4 5
4 5  20 
 4 5 4 5  20 
1 2 2 
3 3 9 
    or    
4 5  20 
4 5  20 
1 3 3 
3 2 6 
  
or     or
4 5  20 
4 5  20 
Working
121
oe
400
Answer
A1 for
M1
121
oe or 0.3025 or 30.25%
400
M1 for a complete method
Mark
Notes
4
M1 for any one correct probability
31
20
2  12  or y  2 x  12 or gradient = 2
x 
3
3
3  3
3
x  43
2
2 × 37 = − 3 × 4 + c
Alternative scheme
2y = −3x + c oe
y
3
37  " " 4  c or
2
or 1 oe
2
" "
3
c = 43
2
(gradient of perpendicular line =)  3 oe
y
3x + 2y = 86
3x + 2y = 86
5
5
A1 for
correct equation
(equation in any form)
for 3x + 2y = 86 oe for a simplified equation with
integer coefficients e.g. 3x = 86 – 2y
A1
A1
A2
M1
M2
3
y  37  " "( x  4)
2
for 3x + 2y = 86 oe for a simplified equation with
integer coefficients e.g. 3x = 86 – 2y
3
y  37   ( x  4)
2
M1 for
M1 (dep on previous M1)
and ft from their
gradient
M1 ft from their gradient
M1
32
Question 12
-2
-1
-2
-1
0
1
2
3
4
5
6
7
8
9
10
y
1
2
3
4
5
6
7
8
9
10
x
33
49
Jan 19
Mark Scheme (Results)
January 2019
Pearson Edexcel International GCSE
In Mathematics A (4MA1) Higher Tier
Paper 1H
2y 1

5 5
E.g. 5y – 2y = 1 or 3 y  1 or 3 y  1  0 or
3y 1

5 5
(c)
5 y  2 y  1 or y 
e 2  3e  5e  15
(b)
1
oe
3
e 2  2e  15
3
2
e.g. 0.3 , 0.3333…

dep on at least M1 for
1
oe
3
for collecting terms in y in a
correct equation
M1
A1
for a correct first step
for 3 correct terms or
for 4 correct terms ignoring signs
or e 2  2e  k for non-zero k
or ...  2e  15
M1
A1
M1
Apart from Questions 1(c), 5, 6(c), 20 and 21 (where the mark scheme states otherwise), the correct answer, unless clearly obtained by
an incorrect method, should be taken to imply a correct method.
Question
Working
Answer
Mark
Notes
2 p (2  3q )
(a)
2
B2
If not B2 then award B1 for
1
2(2 p  3 pq ) or p (4  6q )
or 2p(a two term expression)
or x(2 + 3q) where x ≠ 2p
50
Triangle at (−2, 1), (−2, 3), (−1, 3)
(c)
Answer
Rotation, 90° clockwise, centre ( 2, 3)
Triangle at (−2, 2), (−2, 4), (−1, 4)
Working
(b)
Question
(a)
2
2
1
Mark
3
B2
B1
B1
B1
B1
If not B2 then award
B1 for a triangle of the correct size and
orientation or the wrong size but enlarged
correctly from (-4, 2) with a sf other than 0.5
e.g. a triangle at (4, −2), (4, 6), (8, 6)
for rotation
90° clockwise or −90o (or 270o anticlockwise)
(centre) ( 2, 3)
−2
Note: Do not accept (
) for centre
3
Award no marks is more than one
transformation explicitly stated (the sight of a
vector is not a second transformation)
eg. moved and then rotated; rotation and
translation
cao
Notes
51
Question
3
150  "0.1"
(P(yellow)=)
"0.26" 0.06
or 0.1
2
Working
1  (0.15  0.26  0.33) or 1  0.74
(=0.26)
15
Answer
4
Mark
15
scores M3 A0
150
independent mark
Award for 150 × p where 0 < p < 1
M1
NB: An answer of
for a complete method to find P(yellow)
M1
A1
Notes
can be implied by two values where
P(brown) + P(yellow) = 0.26
(may be seen in table)
M1
52
(b)
Question
4 (a)
1126.5  1.19 oe
or 0.0976(475...)
1236.5  1126.5
"110"
or
1126.5
1126.5
1236.5
1236.5


 1  or 1.09  764...  1 or
 100  100
or 
1126.5
 1126.5 
Working
1236.5
1236.5 − 1126.5 or 110 or
or 1.09(7647...)
1126.5
1236.5
 100 or 109(.7647...)
or
1126.5
1341
9.76
3
3
Answer Mark
A1
M2
A1
M1
M1
for 1340 – 1342
19
 1126.5 oe or 214(.035)
100
if not M2
then award M1 for
for 9.76 - 9.765
for method that would result in
9.76… or 0.0976…
Notes
53
Question
5
A1
B1
For full reasons:
Alternate angles are equal and
angles in a straight line add to 180º OR
Allied angles (or co-interior) add to
180o and
angles in a straight line add to 180º
dep on previous M1
NB : It must be clear which angles are
being found
A1
for substituting their value for x into
the expression NOT used to form the
equation solved
dep on previous M1
M1
M1
Notes
for forming an appropriate equation
E.g. AFC = 145 and FCD = 145 OR
AFC = 145 and BCF = 35
OR
x = 5 from the solution of two equations
5
Mark
OR
forms a second equation in x
Shown
correctly with
reasons
Answer
OR
E.g. 4 x  15  30 x  5  180 AND
30x – 5 = 20x + 45
E.g. 20 × “5” + 45 (=145) or
4 × “5” + 15 (=35) or
30 × “5” – 5 (=145)
Working
E.g. 4 x  15  30 x  5  180 OR
20x + 45 + 4x + 15 = 180 OR
4x + 15 + 20x + 45 = 180 OR
30x – 5 = 20x + 45
x=5
54
7
(c)
(b)
Eg 1.88  10 7 + 3.10  108 + 2.64  108 + 7.18  10 7
or
18 800 000 + 310 000 000 + 264 000 000 + 71 800
000 with at least 3 numbers correct
x2  5x  6  x 1
(c)
(a)
(0, 6), (1, 2), (2, 0), (3, 0), (4, 2), (5, 6)
Working
(b)
Question
(a)
6
9.88  10 6
6.646  108 oe
71 800 000
1.6 and 4.4
Correct curve
1
2
1
3
2
Answer
Mark
(6),2,(0),(0),(2),6
1
B1
A1
M1
B1
for 6.646  108 oe
eg 664 600 000
for a complete method
or for digits 6646
dep on M2
ft from their graph in (b) if at
least 1 mark scored in (b)
for y  x  1 drawn
M1
A1
or for y  x  1
for at least 5 points plotted
correctly (ft their table)
for a correct curve
Notes
For both entries correct
M1
A1
M1
B1
55
Question
8
15.8
4
A1
for 15.8 – 15.83
dep on previous M1
M1
2  "5.41" 5
M1
Notes
NB: 4.8 may be seen on the diagram
ft the candidate’s value for height for
this mark (award of this mark does not
depend on award of previous mark)
Mark
M1
Answer
( x  ) 2.5 2  " 4.8 2 " or ( x ) "29.29 "
or 5.41(202…)
Working
1
1
 5  h  12 oe or  2.5  h  6 oe
2
2
h

4.8
or
56
(c)
(b)
Question
(a)
9
15 and 45 indicated on the cumulative
frequency axis and readings taken from
speed axis
Working
13  15
correct cf graph
2
2
Answer
Mark
3, 19, 43, 53, 58, 60
1
A1
M1
A1
M1
B1
accept 13 – 15
ft from a cf graph
ft from a cf graph
for a correct method to find LQ and UQ
and intention to subtract
Eg for a correct reading from 45/45.75
and 15/15.25 from vertical axis to find
LQ and UQ and an intention to subtract
accept curve or line segments
accept curve which is not joined to (0,0)
ft from (a) if only one addition error
for at least 4 points plotted correctly at
end of interval
or for all 6 points plotted consistently
within each interval in the frequency
table at the correct height
(Eg. using values of 5, 15, 25 etc on x
axis)
Notes
57
10
Question
13
𝐶𝐷
13
8 +"4.73"
or tan(BAD) = 0.97(93...)
E.g. (BAD =) tan–1 ("0.979") or 44.4(024...)
E.g. tan(BAD) =
E.g. (CD =) 13tan20 or 4.7(316...)
E.g. tan20 =
Working
Working with CD and then triangle ABD
24.4
Answer
5
Mark
Award M1A1M1M1A0 for an
answer in the range 44.3 – 44.41
for 24.3 - 24.41
for a correct method to find angle
BAD
M1
A1
for a correct statement or equation
including angle BAD as the only
variable
E.g.
 13 
2

  13
 cos 20 
2
for a correct method to find CD
2
 13 
2
E.g. CD = 
 − 13
 cos 20 
2
for a correct statement or equation
including CD as the only variable
M1
M1
M1
Notes
58
10
Question
A1
for a correct statement or equation
including angle BAC as the only
variable
M1
Award M4A0 for an answer in the
range 44.3 – 44.41
for ans in range 24.3 - 24.41
E.g. 132  (13 tan 20) 2
sin BAC sin110
or

8
"18.1"
82 = “13.8”2 + “18.1”2 – 2 × “13.8”ד18.1”×cos BAC
E.g.
E.g. (AC =)
for a correct method to find AB
for a correct statement or equation
including AC as the only variable
E.g. AC2 = 132 + (13tan20)2
for a correct method to find AC
M1
M1
Notes
E.g. (AB = ) "13.8"2  82  2  13.8  8  cos(110)
(=18.1(9..) or 18.2
5
Mark
M1
24.4
Answer
13
or 13.8(3…)
cos 20
Working
Alternative mark scheme – working with AC and
then triangle ABC
13
E.g. cos20 =
AC
59
Question
11
7 1
10 x  3 x  6

or
6x x
6x
Working
10 x 3( x  2)
10 x  3( x  2)

E.g.
or
6x
6x
6x
7x  6
6x
Answer
3
7x  6
as the final answer
6x
SC: If no marks awarded then award B1
7x  6
for an answer of
6x
for
for a correct single fraction with brackets
expanded
M1
A1
Notes
for two correct fractions with common
denominator or a single correct fraction
Mark
M1
60
(b)
Question
(a)
12
1
3× x2 − 9
3
Working
3
2
x 2  9 oe
3  x  3 oe
Mark
Answer
B3
A1
M1
SC: If no marks awarded and M1 awarded
in (a) then award B1 for “quadratic” < 0
if not B2 then award B1 for
x 2  9  0 or x 2  9 oe
or for ( x  3)( x  3)
or for ( x )  3 (values maybe seen in
incorrect inequalities)
if not B3 then award B2 for
x3
or x  3
or −3 ≤ x ≤ 3
may be seen as two separate inequalities
or for 1x2 – 9 oe
1
for 3× x2 oe or −9 oe
3
Notes
61
(b)
Question
(a)
13
Working
2
3
Correct Venn
diagram
3
oe
18
Mark
Answer
A1
M1
A1
M2
ft from Venn diagram
a
for
where a is an integer
"18"
and 1  a  "18" or
"3"
for
where b is an integer and
b
b  "3"
ft from Venn diagram
Accept omission of 0 for the
award of full marks
NB: For the award of the method
marks do not accept a blank
outside the circles as 0
Notes
for at least 4 correct entries
If not M2 then
M1 for 2 or 3 correct entries
62
(b)
Question
(a)
14
Working
21.76  k  43 oe or k  0.34
T  kr
3
73.44
T  0.34 r 3 oe
Answer
1
3
Mark
B1ft
A1
M1
M1
for their value of k if T  kr 3
Only award if T is the subject
Award M2A1 if T  kr 3 on answer line
and k given as 0.34oe in working
space.
Condone use of proportional sign in
place of equals sign
Award M2 if k  0.34 stated
unambiguously (m = 2.94)
for correct substitution into a correct
equation; implies first M1
Do not allow T  r 3
Notes
Allow r  mT
3
63
Question
15
2
1 4
 4 
     " h "
2 3
 3 
3
4
and   (2  " h ") 2  h
3
A1
dep on award of first M1
ft for candidate’s expression for r or h
for correct expressions for volume of
hemisphere and volume of cylinder ;
both in terms of either r or h
M1
1 4
3
    r 3 and   (2 r ) 2  " r "
2 3
4
Eg
OR
for a correct expression for either r or h
Notes
for use of, for example, r and 2r in an
equation condone omission of flat
surface area
A1
M1
3
4
r or r  h
4
3
4
Mark
h
4.5 oe
Answer
4 r
2
  r  = 2 (2r )h oe
Working
Eg
2
64
(b)
Question
(a)
16
a  4b
a  4b

a  4b
a  4b
Eg
( a  4b )( a  4b )
a 2  4b
2
a  2 b 
 a  2 b  a  2 b 
Eg
or
a2 b
a2 b
Working

a2 b
a2 b
or
2.5 oe
a 2  4 a b  4b
a 2  4b
Answer
1
3
Mark
B1
A1
M1
M1
a 2  4b
a 2  4b
a  2 b 
a  4 a b  4b
for
or
2
denominator
dep on M1 for correctly simplified
2
denominator by a  4b or a  2 b
Notes
For multiplying the numerator and
65
Question
17
sin x sin110

5.3
"7.7"
or
5.3
"7.7"
or

sin x sin110
cosx =
2  4.1  "7.7 "
Eg sin x 
(= 0.764(83…)
sin110
 5.3 (= 0.644(2…)) or
"7.7"
4.12  "7.7 "2  5.32
5.32 = 4.12 + “7.7”2 – 2×4.1×”7.7”×cosx oe
Eg
59.7  641... or 7.7(3073) or AC 2 = 59.7…
(AC = ) 16.81  28.09  14.8(641...) or
2
Working
(AC² = ) 4.1  5.3  2  4.1 5.3  cos(110)
2
40.1
Answer
5
Mark
for 40.1 – 40.11
for isolating sinx or cosx
M1
A1
dep on first M1
for correct use of sine rule or cosine
rule ft for their value of AC or AC2
M1
M1
M1
Notes
for the correct use of Cosine rule to
find AC
NB: there must be evidence of correct
order of operations for this mark to be
awarded
66
(b)
Question
18 (a)
Working
3
( 4, 5), ( 2, 0), (0, 3), (2, 4), (4, 3) (6, 0), (8, 5)
Answer
Parabola through
1
2
Mark
B1
B2
( 4, 5), (8, 5)
If not B2 then B1
For u-shaped parabola with
minimum (2, 4) or
For u-shaped parabola through
( 2, 0), (6, 0) or
For u-shaped parabola through
( 4, 5), ( 2, 0), (0, 3), (4, 3)(6, 0), (8, 5)
Notes
For a parabola with minimum (2, 4)
through at least 5 of
67
(b)
Question
(a)
19
y  9  x  3 or
x9  y3
y  9  ( x  3) 2 or x  9  ( y  3) 2
( y  3) 2  9
( x  3) 2  32 or ( x  3) 2  9 or ( y  3) 2  32 or
Working
3  x  9
y  3
Answer
4
Mark
1
A1
M1
M1
M1
B1
and for 3  x  9
oe
M3A0 for 3  y  9
for completing the
square
Notes
Accept g 1 ( x )  3
68
Question
20
2
Eg ( n  10)( n  3)  0 or
2 1
  13  ( 13) 2  4  1  30
Eg 2n 2  26n  60  0 or n 2  13n  30  0
3n  27 n  60  n  n
2
n4 n5 1


n
n 1 3
Eg 3(n 2  9n  20)  n(n  1) or
n4
n5
or
n
n 1
Working
10
Answer
6
Mark
for a correct quadratic equation
with fractions removed
for a correct quadratic equation
equal to 0
dep on M2 ft
for method to solve 3 term
quadratic
M1
M1
M1
M1
Award M0A0 for an answer of
10 with no working and no
justificastion
NB. Award M5A1 for an
8answer of 10 with justification
6 5 1
e.g.  
10 9 3
for correct answer from correct
working
for the correct equation
M1
A1
Notes
n4
n5
or
n
n 1
69
21
Question
A1
A1
Eg 2 × 5.5 + 23 (=34) and 8 × 5.5 + 2 (=46)
OR
8 × 5.5 + 2 (=46) and 20 × 5.5 − 52 (=58)
M1
x  5.5
4
Mark
M1
shown
Answer
(8x + 2) – (2x + 23) = (20x – 52) – (8x + 2) oe or
(2x + 23) – (8x + 2) = (8x + 2) – (20x – 52) oe
Working
Mark scheme 1 (see next page for alternative mark scheme)
(8x + 2) – (2x + 23) (= 6x – 21) or (2x + 23) – (8x + 2) (= −6x +
21)
or
(20x – 52) – (8x + 2) (= 12x – 54) or (8x + 2) – (20x – 52) (= −12x
+ 54)
for 12 from correct working
for a correct equation
for a correct expression for the
common difference in terms of x
brackets must be present or
removed correctly
Notes
70
21
Question
OR
2x + 23) + 12 = (8x + 2) and (8x + 2) + 12 = (20x – 52)
and gets x = 5.5 both times
x  5.5
or x = 1.5 from (2x + 23) − 12 = (8x + 2)
or x = 3.5 from (8x + 2) − 12 = (20x – 52)
2 × 5.5 + 23 (=34) and 8 × 5.5 + 2 (=46)
and 20 × 5.5 − 52 (=58)
Alternative method – starts by assuming d = 12
E.g.
(2x + 23) + 12 = (8x + 2) or (8x + 2) + 12 = (20x – 52) or
(2x + 23) − 12 = (8x + 2) or (8x + 2) − 12 = (20x – 52)
or
3
(2x + 23) + (8x + 2) + (20x – 52) =  2(2 x  23)  2  12 
2
Working
shown
Answer
4
Mark
A1
A1
M2
OR
solves both
(2x + 23) + 12 = (8x + 2) and
(8x + 2) + 12 = (20x – 52)
and gets x = 5.5 both times
for explicitly showing both
common differences are 12
for a correct equation
If not M2 then award M1 for a
correct expression for the
common difference in terms of x
brackets must be present or
removed correctly e.g
(8x + 2) – (2x + 23) (= 6x – 21)
or
(20x – 52) – (8x + 2) (= 12x – 54)
Notes
71
89
May 19
Mark Scheme (Results)
Summer 2019
Pearson Edexcel International GCSE
In Mathematics A (4MA1)
Paper 1H
14
9
×
3
10
e.g.
14 9 126 21
1
 

4
3 10 30
5
5
14 9 126
6
1
 
4 4
or
3 10 30
30
5
7
3
14
9
21
1
or
 5
4
1
3
10
5
5
126 30 126 21
1
¸
=
=
= 4
or
27 27 30
5
5
e.g.
Shown
A1
M1
126 30
¸
27 27
126
must be
30
1
5
21
5
NB: use of decimals scores no
marks
Total 3 marks
to the multiplication to
seen or correct cancelling prior
multiplication e.g.
from correct working – either
sight of the result of the
Dep on M2 for conclusion to 4
42 10
÷
or
9
9
or for both fractions expressed
as equivalent fractions with
denominators that are a
common multiple of 3 and 9 eg.
International GCSE Maths
Apart from questions 1, 11, 12b, 15 (where the mark scheme states otherwise) the correct answer, unless clearly obtained from an
incorrect method, should be taken to imply a correct method.
Question
Working
Answer
Mark
Notes
14
10
1
3
M1 Both fractions expressed as
e.g.
and
improper fractions
3
9
90
2
(c)
(b)
(a)
10
9
m/sec or 0.2 km/min or
oe
3
3
12 km/h or
1h 45m + 1h 30m or 1 + 0.75 + 1.5 or
3h 15m or 3.25h or 195m oe
(24 × 2) ÷ “3.25” oe eg (48 ÷ 195) × 60
oe
15
25
m/sec or 0.25 km/min or
4
6
15 km/h or
14.8
line from (12:00,
24) to (12:45, 24) to
(14:15, 0)
‘before’ with
reason
3
2
1
M1
A1
M1
B2
B1
e.g. before as gradient is steeper
or before as speed before is 15
km/h speed after is 12 km/h or
before as she goes over 11(allow
11-12) km in ¾ hour but only
goes 9 km in ¾ hour after oe
NB: any figures used for the
reason must be accurate if they
haven’t used ‘gradient is steeper’
oe
If not B2 then B1 for a line from
(12:00, 24) to (12:45, 24) or for a
line from (t, 24) to (t + 1.5, 0) or
for a time of 1.5 hours (oe) seen
ft from their graph for total time
when cycling
ft dep on M1 for full method
awrt 14.8
Total 6 marks
91
4
3
(b)
(a)
(d)
(a)
(b)
(c)
x² + 9x − 2x − 18
2
1
1
2
correct graph
9, 3, (−1), −3, (−3), −1, (3)
x² + 7x  18
4cp2(4c3 + 5p)
e4
y16
2
2
A1
B2
B1
B1
M1
A1
M1
B2
If not B2 then award
B1 for at least 2 correct values
dep on B1 ft from (a) for at least
5 points plotted correctly
for the correct graph (clear
intention to go through all the
points and which must be
curved at the bottom)
Total 4 marks
if not B2 then award B1 for
any correct factorisation with at least
2 factors outside the bracket eg
4cp(4c³p + 5p²) , cp²(16c³ + 20p),
2p(8pc4 + 10cp²) etc or the correct
common factor and a 2 term
expression with just one error
Total 6 marks
for 3 correct terms or
4 correct terms ignoring signs or
x² + 7x + c or .... + 7x  18
92
6
5
336
92
“480” × 0.7
12 × 8 × 5 (= 480)
3
A1
M1
M1
or
Total 3 marks
Total 4 marks
92
, oe with 92 seen
200
dep on M2 and probabilities
between 0 and 1
Dep on M1
A1
M1
eg (0.18 + 4 ד0.07”) × 200
or 0.46 × 200
or 36 + 42 + 14 oe
M1
M1
4
x = (1 – 0.18 – 0.26) ÷ (2 + 2 + 3 + 1) (=0.07)
2x + 0.18 + 2x + 3x + 0.26 + x = 1 or
1 – (0.18 + 0.26) (= 0.56)
93
8
7
(a)
(b)
(c)
0.08 × 170 000 (=13600) or 0.92 × 170 000
(=156400)
e.g. 0.92 × (0.92 × “156400”)
132377
3
A1
(dep)for a complete
method
or 132376.96
(SCB2 for 170 000 × 0.924 )(=121786.(810))
(SCB1 for 170 000 × 0.24 (=40 800) or
170 000 ×0.76 (=129 200) or
170 000 × 1.08 (= 183 600) or
170 000 × 1.08³ (= 214151) or an answer of
129 200 or an answer of 214 151 – 214151.1(0))
Total 3 marks
M2 for 170 000 ×
0.923
If not B2 then award B1 for
320000 or 3.2 × 105 oe or
5 × 10n oe where n ≠ 6
Total 4 marks
M1
B1
B1
B2
oe eg 170 000 ÷ 12.5
1
1
2
M1
5 700 000
4 × 10-3
5 000 000 or 5 ×
106
oe
94
10
9
46.3
(8 =) 2 × 2 × 2 or 2³ or 23+n
2
2n  3  3  5 m
A1
M1
A1
M1
2
 "8.48.." 
0.5 × π × 
 (= 9π or 28….)
2


M1
M1
M1
AC
6

(sin 90) sin 45
5
62  62 (  72 = 6 2 =8.4(85…)or 8.5) or
6(sin 90)
AC 
= 6 2 =8.4(85...)or 8.5) oe
sin 45
(d 2 =) 62 + 62 (=72) or
0.5 × 6 × 6 (=18)
Total 5 marks
Total 2 marks
For clearly writing 8 as a product
of prime factors or as 2³
for 46.2 – 46.3
1
 6  6 2 sin 45 or
2
1
 6 2  3 2 oe
2
For area of triangle, or may use
95
11
17.5 – 5.5
5.5 or 6.5 or 12.5 or 17.5
12
3
A1
M1
M1
•
•
Accept 6.4 9 for 6.5 and 17.49
for 17.5
for UB – LB where
15 < UB ≤ 17.5 and 5.5 ≤ LB < 6
dep on M2
Total 3 marks
96
12
(b)
(a)
4
m  3  7  2m
3
4
m  2m  7  3 oe
3
Alternative
4m + 6m = 21 – 9 or 10m = 12 or
−21 + 9 = −6m – 4m or −10m = −12
4m + 9 = 21  6m
4m + 9 = 3(7  2m)
12
oe
10
(2x – 3)(x – 2)
4
4
2
awarded
M1 for a correct equation with m
terms isolated on one side
ft their equation if first M1
M1 Division of each term on LHS by 3
(from 4m + 9 = 7 – 6m oe)]
m = − 0.2oe with working shown
(from 4m + 9 = 21 – 2m oe) or
m = 2 with working shown
[SC: B2 for an answer of
M1 for a correct equation with m
terms isolated on one side
ft their equation if first M1
awarded
A1 dep on at least M2
in a correct equation
M1 for correct expansion of bracket
or (3 – 2x)(2 – x)
(B1 for (2x + a)(x + b) where ab = 6
or 2b + a = −7 eg (2x + 3)(x + 2),
(2x – 5)(x – 1)), etc or for
M1 for removing fraction
B2
97
12
(c)
contd
1
1
1
1
y4
or 4 y  y 4 or y 4
y
10m = 3 × 4 oe

3
y 4
12
oe
10
2
A1
M1
A1
or b = 
3
4
dep on at least M2
correct equation
Total 8 marks
M1 For removing fraction in a fully
98
13
42
2
 8 7   7 6
" " " or       
117
5
 14 13   10 9 
7 6  42 14

 

 0.35(897...)  or
13 9  117 39

8 7  56  7 6  42 
 
oe  or   

14 13  182  10 9  90 
(c)
"
8 7

14 10
(b)
(a)
28
oe
195
2
oe
5
3 7 3 7
, , ,
10 10 10 10
6 8
,
14 14
3
2
2
A1
M1
M1
A1
M1
B1oe
B1
6 3 8 4
  ,   in correct
14  7  14  7 
3 7 3 7
, , ,
in correct
10 10 10 10
28
oe, e.g. 0.14(3589…)
195
from accurate working
Total 7 marks
for
ft from (b)
7
(  0.54 to 2dp
13
6
 0.67 to 2dp)
9
ft from (a)
ft from (a)
positions.
for
positions. Allow decimals of 2dp
or better (0.43, 0.57)
for
99
14
7, 8, 9, 10, 11
eg 2, 4, 6
(a)
(b)
1
2
B1
B2
for any 3 of 2, 4, 6, 8, 10
Total 3 marks
completely correct.
(B1 for 4 or 5 correct and no
more than 1 incorrect or for all
terms seen correctly placed in a
Venn diagram or for a correct
description of the numbers in
the set but not listed, eg 7 ≤ x <
12)
100
15
5.4 54  3 
2  99  54 252 14



   and
99 990  55 
990
990 55
5.4 54
3
11  3 14
or


and

99 990 55
55
55
100 x  x  5.4545...  0.05454...  5.4 and
252 14

or
990 55
1000x – 10x = 254.545... – 2.545... = 252 and
25.2 14

or
99 55
eg 100x – x = 25.454... – 0.254...= 25.2 and
1000x = 254.54...
10x = 2.5454...
100x = 25.454…
x = 0.25454…
show
2
A1
M1
for completion to
14
55
with intention to subtract.
For 2 recurring decimals that when
subtracted give a whole number or
terminating decimal eg 25.2 or 252 etc eg
100x = 25.454…and x = 0.25454... or
1000x = 254.54... and 10x = 2.5454....
with intention to subtract.
(if recurring dots not shown then showing
at least the digits 25454, ie 5sf)
or
0.2  0.054 and
eg x  0.05454..., 100 x  5.4545...
101
16
100th term is 3 × 100 + 4 (= 304) and
100 × (7 + “304”) ÷ 2
100th term is 7 + (100 – 1) × 3 (= 304) and
100 × (7 + “304”) ÷ 2 or
100
(2  7  (100  1)  3) or
2
a = 7 and d = 3
15 550
2
A1
M1
Total 2 marks
for a method to find the sum brackets (100 – 1) must be used
correctly
Total 2 marks
102
18
17
17.82 + 26.32 – 2 × 17.8 × 26.3 × cos36
e.g. 1008.5... – 757…. or 251(.06…)
3
 24 
 36 
  or 2³ : 3³ oe or   or 3³ : 2³ oe or
 36 
 24 
8
27
or
oe
27
8
3
(b)
24
36
or 2 : 3 oe or
or 3 : 2 oe
36
24
eg
(a)
8
V oe
27
15.8
(A =)
2160
3
2
2
M1
M1
A1
A1
A1
M1
M1
V
3.375
Total 4 marks
"2160"
960
for correct order of operations
for ans in range 15.8 – 15.9
Total 3 marks
oe eg
or ft from
For correct SF for volume
ft from linear scale factor in (a)
for a correct scale factor
103
19
16 ÷ 10 (=1.6)
21 ÷ 5 (=4.2)
48 ÷ 15 (=3.2)
15 ÷ 20 (=0.75)
correct histogram
3
B3
For a fully correct histogram
[If not B3 then B2 for 3 correct
frequency densities (can be
implied by heights) or 3 correct
bars drawn
If not B2 then B1 for 2 correctly
calculated frequency densities
(can be implied by heights) or 2
correct bars drawn.]
Total 3 marks
104
Students can use other methods to gain the correct answer
20
angle ABD = 71 or
5
angle ACD = 71 or
using O as centre of circle,
angle ADO = 90 – 71 (=19)
angle ADB = 71 or
angle ACB = 71 or
angle BAD = 19 × 2 (=38) or
reflex angle BOD = 2 × 142 (=284)
angle BCD = 142
142
dep clearly labelled or stated
Clearly labelled or stated, from no incorrect working for
their method
dep on A1 for fully correct reasons for each stage of
working, repeated if used more than once.
eg alternate segment theorem,
base angles in an isosceles triangle are equal,
angles in a triangle sum to 180o,
angle between tangent and radius(diameter) is 90°
congruent triangles (equal triangles) oe
opposite angles of a cyclic quadrilateral sum to 180o
angles in the same segment
angle at the centre is 2 × angle at circumference oe
equal chords subtend equal angles at the circumference
If not B2 then award B1 dep on M1 for any one correct
circle theorem reason associated with angle(s) found
Total 5 marks
M1
A1
B2
clearly labelled or stated
M1
105
21
A1
A1ft
(r = ) 6 or (h = ) 18
24
M1
M1
1 4
   r 3 +   r 2  3r = 792π oe
2 3
5
M1
h
3
1 4
   r 3 oe or   r 2  3r oe
2 3
h = 3r or r 
3
2
2
4
correctly
3
evaluated dep on M3
Total 5 marks
their “6” × 4 or "18"
1 4 h
h
        h  792
2 3 3
3
or
3
h
oe stated or
3
1 4 h
h
or     or    h
2 3 3
3
used correctly
for h = 3r or r 
106
22
translation of  
 3
1
2
2
(x – 3)2 + 1
(b)(i)
(b)(ii)
2
correct graph (see
end of mark scheme)
[must go through
(60, 2), (150, 0),
(240, −2), (330, 0)] and
not through (0, 0)
(a)
B1
B1
B2
B2
6
)² + n (where n ≠ 1) or for
2
must be column vector
For   ft from (b)(i)
 3
1
x2 − ax − ax + a2 + b with
2a = 6 or a2 + b = 10)
for translation
Total 6 marks
(x – m)² + 1 (where m ≠ 3) or for
(B1 for (x –
point on graph at (150, y) and a point at
(330, y))
or a clear translation of 
 30 
 (ie a
 0 
if not B2 then award B1 for a graph of
the correct shape going through 2 or 3
of the given points or for a clear stretch
of SF2 (ie a maximum point on graph at
(x1, 2) and a minimum point at (x2, −2))
107
(v =) 3t2 – 6 × 2t + 5 (+ 0)
24
(a =) 3 × 2t – 12
6t – 12 = 3
 10  2 7  19 
,

 or (6, 13)
2 
 2
19  7  12 
   oe or 1.5 oe
10  2  8 
3
2
m×
= −1 oe or m = 
2
3
2
“13” = “  ”× “6” + c or c = 17
3
2
oe or y  "13"  " "( x  "6")
3
23
3y + 2x = 51
2.5 oe
5
4
A1
2
x  17
3
M1
2
“ must come
3
A1
M1
M1
M1
Total 4 marks
for differentiating at least 2
terms correctly
dep ft
dep on at least M1 for equating
their acceleration in terms of t
to 3
from correct working]
for 3y + 2x = 51 or 3y = −2x + 51 etc but
must be integer coefficients
Total 5 marks
[NB: “13”, “6” and “ 
Or for y  
for use of m1m2 = −1
M1
M1
M1
108
Q19
0
1
2
3
4
5
Frequency density
10
20
30
40
50
Height (h metres)
109
q22
-3
-2
-1
0
1
2
3
y
60
120
180
240
300
360 x
110
127
Mark Scheme (Results)
Jan 2020
January 2020
Pearson Edexcel International GCSE
In Mathematics A (4MA1)
Paper 1H
2
or
28 105
or 2, 2, 3, 5, 7 oe
7
28, 56, 84, 112… and 105, 210, 315,
420…
or
2, 2, 7 and 3, 5, 7
or
420
2
A1 cao
or a fully correct Venn diagram
Total 2 marks
or 2, 2, 7 and 3, 5, 7 seen (may be in a factor tree and ignore 1)
for starting to list at least four multiples of each number
M1for any correct valid method e.g.
Question Working
Answer
Mark Notes
Apart from questions (where the mark scheme states otherwise) the correct answer, unless clearly obtained by an incorrect method,
should be taken to imply a correct method
1
(a)
2
M1 for a correct method to find one coordinate or for one
5  13
4  1
or
coordinate correct or for (−1.5, 9)
2
2
(9, −1.5)
A1 oe
(b)
−3
1
B1
(c)
No with
1
B1 No (oe) and e.g. line goes through (100, −298) or (101.3(3..),
reason
 304

−302) or 
, 302  or (3 × 100) – 302 = −2 not (+)2
 3

Total 4 marks
128
3
E.g.
12 × 9 (=108) or (9 – 6) × x (= 3x)
E.g.
129 – ‘108’ (= 21) or
‘108’ + ‘3x’ = 129
E.g.
‘21’ ÷ (9 – 6) or
129  '108'
x=
96
7
4
A1 Accept 7 cm
Total 4 marks
M1 (dep on M1) for 129 used correctly with another area
or
for a correct equation (ft) with bracket(s) expanded
M1 for a complete method
M1 for one correct relevant area
129
4
(c)
(a)
(b)
2
1

40 40
‘144’ ÷ 40
or
[(12 × 2.5) + (16 × 3.5) + (9 × 4.5) +
(2 × 5.5) + (1 × 6.5)] ÷ 40
30 + 56 + 40.5 + 11 + 6.5 (= 144)
or
(12 × 2.5) + (16 × 3.5) + (9 × 4.5) +
(2 × 5.5) + (1 × 6.5)
3
40
3.6
3<w≤4
2
1
4
Total 7 marks
3
a
where 0 < a < 40 or
where b > 3 where a and b are
b
40
integers
A1 0.075 oe
M1 for
A1 oe
Allow division by their Σf provided addition or total under column
seen
correct midpoints used for at least 4 products and not added
M1 (dep on at least M1)
or
M1 for consistent use of value within interval (including end points)
for at least 4 products which must be added
If not M2 then award
B1
M2 for at least 4 correct products added (need not be evaluated) or
130
6
5
(a)
(b)
22 500 000 oe e.g. 22.5 × 10
or
2.25 × 10n n ≠ 7
(‘75’ – ‘48’) + (‘45’ – ‘15’)
6
‘45’ ÷ 3 (= 15) or
‘45’ ÷ 3 × 2 (= 30)
16
9
‘75’ ×
(= 48) or ‘75’ ×
(= 27)
25
25
E.g.
9
(‘45’ ÷ 3 × 2) + (‘75’ ×
) oe or
25
‘27’ + ‘30’ or
120 ÷ (3 + 5) (= 15)
‘15’ × 3 (= 45) or
‘15’ × 5 (= 75)
2.25 × 107
0.000 78
57
1
2
6
A1
B1
M1
A1
M1 for a complete method
M1
M1
M1
M1
M2 for
3
120 (= 45) or
8
5
120 (= 75) oe
8
Total 3 marks
Total 6 marks
131
7
A1 dep on at least M2 awarded
Total 9 marks
M1 ft (dep on 4 terms) for terms in x on one side of equation;
number terms on the other
2(‘6x – 15’) = 9 – x oe or
9 x
3(2x – 5) =  oe
2 2
12x + x = 9 + 30 oe or
x 9
6 x    15 oe
2 2
M1 for any 3 correct terms or for 4 out of 4 correct terms
ignoring signs
for m2 – 3m … or
for …– 3m – 40
A1
B2
If not B2 then award B1 for
5(y + 4y2) or y (5 + 20y) or
5y(a + 4y) where a is an integer and a ≠ 0 or
5y(1 + by) where b is an integer and b ≠ 0
B1
M1 for expansion of a correct bracket
M1 for removal of fraction or separating fraction (RHS) in an
equation
1
4
2
2
2 × 3(2x – 5) = 9 – x oe or
3
1
(c)
(d)
E.g. 6 x  15 or 12 x  30 oe
(b)
m2 – 8m + 5m – 40
m2 – 3m – 40
5y(1 + 4y)
(a)
132
9
8
24.3
or
 PQ 
(PQ =)
24.3
or
cos 63
24.3
(PQ =)
or
sin 27
24.3
 sin 90
 PQ  
sin 27
 PQ   24.3 or sin 90  sin 27 oe
sin 90 sin 27
 PQ  24.3
24.3
sin 27 =
or
 PQ 
cos 63 =
53.5
Trapezium
with
vertices at
(6, 3) (8, 3)
(8, 6) (4, 6)
3
2
A1 Accept 53.5 - 53.53
M1 for a correct rearrangement
for PQ
M1 for a correct trigonometric
ratio
Total 2 marks
Total 3 marks
 PQ   '47.6914 '2  24.32 oe
and
 RQ   24.3  tan 63   47.6914..
M2 for
B1 for shape of correct size and orientation or
3 or 4 points plotted correctly
B2
If not B2 then award
133
11
10
6000 × 1.0152 (= 6181.35) or
6000 + (0.015 × 6000) + (0.015 × (6000 +
‘90’)) (= 6181.35) or
(1.015)2 (= 1.030225) or
6311.16
  1.05186 
6000
6311.16 ÷ ‘6181.35’ (= 1.021) (×100) or
6311.16  '6181.35'
(= 1.021) (×100) or
'6181.35'
‘1.05186’ ÷ ‘1.030225’ (= 1.021) (×100)
y≥
2.1
1
x  2 oe
3
x + y ≤ 4 oe
x ≥ −1 oe
3
3
Total 3 marks
A1 awrt 2.1
Total 3 marks
M1 (dep on M1) for a complete method to find the compound
interest multiplier (×100)
M1 for working out the total amount after two years
or working out the compound interest multiplier after two years
or working out the compound interest multiplier after three years
Accept < for ≤ and > for ≥ throughout
1
x  2 oe)
3
(If no marks gained B1 for understanding of equation x + y = 4 e.g.
y > 4 − x)
(SC B3 for x ≤ −1, x + y ≥ 4 and y ≤
(B2 for two correct inequalities
B1 for one correct inequality)
B3 for all 3 correct inequalities
134
14
13
12
(0.5 ×) 9.3 × 14.7 × sin106 or
(9.3 × cos 16) × 14.7 or
(9.3 × sin 74) × 14.7
(GE2 =) 9.32 + 14.72 – 2 × 9.3 × 14.7 × cos106
377(.9….) or 378 or 86.49 + 216.09 + 75.3… or
302.58 + 75.3….
(b)
19.4
131
70
18
8000
(a)
56
or '400 ' (=20)
0.14
‘20’ × ‘20’ × ‘20’ oe
56
oe or 56 ÷ 0.14 (= 400)
w2
[40.5, 43]
‘42’ ÷ 0.6 oe
(b)
0.14 
E.g. 56 – 38
(a)
3
2
4
3
2
Total 5 marks
A1 for 19.4 – 19.5
Total 5 marks
M1 for applying the area of a triangle formula using correct values
(to find half of the area of the parallelogram) or
for a correct method to find the area of the parallelogram
A1 awrt 131
M1 for the correct use of the cosine rule
M1 (dep on M1)for the correct order of operations
M1 (dep on M2) for a method to find the volume of the cube
A1
Total 4 marks
M1 for a method to find w
M1 for using the given formula correctly
M1 for subtracting readings from 60 and 20 oe
A1 for answer in the range 17 – 19
B1
M1 for complete method to find the number of men
A1
135
15
(a)
[(−2x2 + x + 15)(x + 1) =]
−2x3 − 2x2 + 15x + x2 + x + 15
Shown
A1
for at least 4 terms correct
out of a maximum of 8
terms
or
[(−x2 + 2x + 3)(2x + 5) =]
−2x3 − 5x2 + 10x + 4x2 + 6x + 15 or
M1 for multiplying out two
brackets correctly at least 3
terms correct
M1 for at least 3 terms
correct out of a maximum of
6 terms
3
E.g.
[(2x2 + 7x + 5)(3 – x) =]
−2x3 − 7x2 − 5x + 6x2 + 21x + 15 or
(3 – x) (2x + 5) = −2x2 + 6x − 5x + 15
(= −2x2 + x + 15)
(x + 1)(3 – x) = −x2 + 3x – x + 3
(= −x2 + 2x + 3) or
(2x + 5)(x + 1) = 2x2 + 2x + 5x + 5
(= 2x2 + 7x + 5) or
6x2 −2x3 + 6x − 2x2 + 15x − 5x2 +15 − 5x
M2 for at least 4 terms correct out of a
maximum of 8 terms
136
15
(b)
2

E.g.
2
2

1 1 
6   x        16   0  oe

6   6  

(x = )
2  2  4  6  16
oe
26
(accept + in place of ±) or
E.g.
 dV 
 16  2 x  6 x 2 oe

x
d


'16  2 x  6 x 2 '  0 oe

 dV 
 16  2 x  3  2 x 2 oe

x
d


1.47
5
A1 dep on M1 for answer in range 1.47 – 1.5 from correct working
(Must reject −1.80 to −1.81 if calculated)
Total 8 marks
M1 (dep on M1) for a complete method to solve their 3-term
quadratic equation (allow one sign error and some simplification –
2  4  384
)
allow as far as
12
M1 (dep on M1) for equating their differentiated expression to zero
A1 for a correct differentiated expression
16 or −2x or (3 × −2x2)
M1 for the correct differentiation of at least 2 correct terms from
137
16
2  58.45  19.5
(= 27.4366...)
3.55
58.35 or 58.45 or
19.5 or 20.5 or
3.55 or 3.65
27.44
3

58.4 < aUB ≤ 58.45 and
19.5 ≤ cLB < 20 and
3.55 ≤ dLB < 3.6
A1 from correct working
allow 27.4 – 27.5
where
2  aUB  cLB
d LB
M1 for correct substitution into
3.64 9 for 3.65

20.4 9 for 20.5 or

Accept 58.449 for 58.45 or
B1 for any correct bound
Total 3 marks
138
17
(a)

or
36  24 3  24 3  48
or
36  2  24 3  48
or
36  6  2  2 12  48
84  48 3
or
36  12 12  12 12  4 12 12
or
36  12 12  12 12   4  12 
6  6  6  2 12  6  2 12  2  12

2
Shown
3
A1 for fully correct working leading to given expression
M1 (dep on M1)
or
for showing or stating 12 = 2 3 oe
for the use of (a + b)2 = a2 + 2ab + b2
or
M1 for correct expansion of brackets showing four terms (need not
be simplified)
139
17
(b)
2
2
3
8
1
2
8
1

3
 t 30 
a 8
or

24 
9t 10
 729a 
1
3
1 8
2
a
 9a 
 t5 
3 a
9
 10  or 10 or 10 or  4  or
t
t
 t 
 3a 
E.g.
 t15 
 729a 24 
 3a 4 
or  30 
 5  or 
12 
 27 a 
 t

 t 
E.g.
t10
9a8
3
1
t10 a 8
or t10 a 8
9
9
t10
where k is an integer ≠ 0
ka 8
A1 Allow
or
cube rooting or inverting or squaring
or
ka 8
where k is an integer ≠ 0
t 10
M1 for two of
cube rooting or inverting or squaring
M1 for one of
Total 6 marks
140
18
5 4 11  220
11 
5 11 10  550
55 
  

  

 oe or
 oe
16 15 14  3360 168 
16 15 14  3360 336 
24
84
210
168
'
' 3  '
' '
' 3  '
' oe or
3360
3360 3360
3360
80
140
60
210
140 
 60
1  '
' 3  '
' 3  '
' 3  '
'3  '
'6  '
'  oe or
3360
3360
3360
3360
3360 
 3360
220
550 
 60
1  '
' 3  '
' 3  '
'  oe
3360
3360 
 3360
or
5 4 4  80
1 
5 4 7  140
1 
  

  

 oe or
 oe or
16 15 14  3360 42 
16 15 14  3360 24 
5 4 3  60
1 
5 7 6  210
1 
  
  oe or
  
  oe or
16 15 14  3360 56 
16 15 14  3360 16 
5 7 4  140
1 
  

 oe
16 15 14  3360 24 
or
4 3 2  24
1 
7 6 5  210
1 
  

  
  oe or
 oe or
16 15 14  3360 140 
16 15 14  3360 16 
5 4 3  60
1 
  
  oe
16 15 14  3360 56 
4 7 6  168
1 
4 3 7  84
1 
  

  

 oe or
 oe
16 15 14  3360 20 
16 15 14  3360 40 
990
3360
4
A1 for
11 10 9
  oe
16 15 14
M3 for
990
33
oe e.g.
or 0.29(464...)
3360
112
Total 4 marks
M1 for a complete method
LLX or LXX (X = not L)
or
LLB or LLO or LBB or LOO
or LOB
or
BOO or BBO
M1 for finding the following
in any order
M1 for finding BBB or OOO
or
LLL
141
19
6
6
or tan AHF =
or
' 142 '
' 106 '
52  92 (= 106 )
 ' 106 ' 
or cos 1 
 or
 ' 142 ' 
 ' 106 ' 
6 
1 
90  sin 1 
 or 90  cos 

 ' 142 ' 
 ' 142 ' 
 ' 106 ' 
or 90  tan 1 

 6 
sin FAH =
' 106 '
6
or cos FAH =
or
' 142 '
' 142 '
' 106 '
tan FAH =
6
E.g.
6 
 6 
1 
sin 1 
'
 or tan 
 ' 142 ' 
 ' 106 
' 106 '
cos AHF =
or
' 142 '
sin AHF =
(FH = GE =)
E.g.
(AH =) 62  52  92 (= 142 ) or
30.2
4
A1 for 30.2 – 30.3
 ' 142 '2  ' 106 '2  62 
cos 1 
 oe or
 2  ' 142 ' ' 106 ' 
 sin 90

sin 1 
 6  oe
 ' 142 ' 
Allow
M1 for a complete method
 ' 142 '2  ' 106 '2  62 
cos AHF = 
 oe or
2

'
142
'

'
106
'


sin 90
sin AHF =
 6 oe
' 142 '
Allow
Total 4 marks
M1 for a correct method for finding angle AHF or finding angle
FAH
M1 for working out AH or FH or GE
142
20
x = 1
a
oe or y = 4 – a
4
graph drawn in shape of a quadratic
with a minimum in any quadrant
x = 1 , y = 4 (1 – 1)2– a
Correct graph
4
y-axis intercept marked as (0, 4 – a) oe

Total 4 marks
Note: The 0’s can be ignored (as shown in the diagram)

quadratic shape with minimum in the fourth quadrant and
marked as (1, – a) oe
a
x-axis intercepts marked as (1 
, 0) oe on the positive
2
a
x-axis and (1 
, 0) oe on the negative x-axis
2

M1 for finding the turning point (may be seen marked on the graph
as (1, –a))
M1 for finding one of the intercepts (or award for any one correct
coordinate shown on graph)
a
a
(0, 4 – a) or (1 
, 0) or (1 
, 0)
2
2
Note: The 0’s can be ignored (as shown in the diagram)
A1 for a fully correct graph
M1 for a quadratic with a minimum
143
21
x
or
2
4  16  4  3  y 
or x  2  1  y
(x + 2)2 = y + 1 or (y + 2)2 = x +1
2
x  4 x  (3  y )  0 or
y 2  4 y  (3  x)  0
or
(fg(x) = ) (x + 3)2 – 2(x + 3) oe
(fg(x) =) x2 + 4x + 3
(x + 2)2 – 4 + 3 or (x + 2)2 – 1
2  x  1
5
A1 oe
Total 5 marks
Allow same equations with x and y swapped
or
correctly substituting into the quadratic formula
M1 ft (dep on M2) for a correct rearrangement for their
completed the square quadratic
Correctly setting up an equation
or
M1 for substituting g(x) into f(x)
A1 Allow y2 + 4y + 3
M1 ft (dep on M1) for correctly completing the square
on their 3 term quadratic
144
22
2
23
2
 78  4  3  495
 k  13  169  165   0 
or
78 
23
2
 12   4  3  36
or
2
 j  2   4  12   0 
12 
eg 3k2 – 78k + 495 = 0 oe
or 5j² − 60j – 140 = 0 oe
or 5k2 – 150k + 1045 = 0 oe
or 3j2 – 12j – 36 = 0 oe
or
k  15
1
k  15
2
gradient HM: eg
oe
 2 or k = 2j + 15 or j 
j6
2
4
2
eg (j – 6)(j + 2)( = 0)
eg (k – 15)(k – 11)( = 0)
or
or
(j – 6)2 + (k – 15)2 = 80 oe
or
 j  6 k  15 
,

 oe
2 
 2
or
2
2
 j  4   196  100   k  1 oe
gradient of JK = −0.5 or m × 2 = −1
j  24
k  15
1
  or 2k – j = 24 or j = 2k – 24 or k 
oe
6 j
2
2
j = −2, k = 11
6
M1 (dep on M3) for a complete method to solve their 3term quadratic equation (allow one sign error in the use of
the quadratic formula)
or
a correct method to eliminate either j or k
k  15
j  24
eg 2k – 24 =
oe or
= 2j + 15 oe
2
2
A1
Total 6 marks
or
A correct equation for the gradient of HM in terms of j
and k or a correct equivalent equation
or
for equating length HJ with length HK
M1 (dep on M3) writing a correct quadratic expression in
the form ax2 + bx + c (= 0) (allow ax2 + bx = c)
or
for finding the midpoint of M
M1 for finding equation of JK in terms of j and k
M1 for finding the gradient of JK using m1 × m2 = −1
M1 for expressing the gradient of JK in terms of j and k or
a correct equivalent equation
145
ALT
22
or
2
 j  8  64  28   0 
25
2
 12   4  5  44
j = −2, k = 11
or
2
 k  1  1  143   0 
25
E.g.
(5k – 119)(k – 11)( =
0)
or
2
174   174   4  5 1309
E.g.
(5j – 22)(j + 2)( = 0)
or
12 
E.g.
5k2 – 174k + 1309 = 0
or
3k2 + 6k − 429 = 0 oe
E.g.
5j2 – 12j – 44 = 0
or
3j2 + 48j + 84 = 0 oe
 j  6 k  15 
,

 oe
2 
 2
k  15
1
2
 2 or k − 2j = 15 or k = 2j + 15 or
j6
4
2
k  15
j
oe
2
(j – 6)2 + (k – 15)2 = 80 oe
2
2
or  j  4   196  100   k  1 oe
6
A1
Total 6 marks
M1 (dep on M3) for a complete method to solve their 3-term
quadratic equation (allow one sign error in the use of the quadratic
formula)
allow ax2 + bx = c
M1 (dep on M3) writing the correct quadratic expression in form
ax2 + bx + c (= 0)
M1 for finding the length of JK in terms of j and k
or for equating length HJ with length HK
M1 for expressing the gradient of JK in terms of j and k or a correct
equivalent equation
M1 for finding the midpoint of M
146
2H
1
Mark Scheme
Specimen paper
International GCSE Mathematics A
4MA1/2H
2
1
(d)
(b)
(c)
Q
(a)
Working
Answer
 × 8.5² ÷ 22.5
(area of trapezium =) (20 + 25) ÷ 2 × h oe
(=22.5h)
 × 8.5² (=226.98...)
5q ≥ 31 or 2q + 3q ≥ 31
2
10.1
q ≥ 6.2
−2, −1, 0, 1, 2
12e f
9a8
9
Mark
Notes
4
2
2
2
2
A1
M1
M1
M1
A1
B2
B2
M1
B2
Total 4 marks
A correct method to find h
A correct method to find the area
of the circle
Use of correct formula for
trapezium
B1 for 9or a8
For 5q ≥ 31 or 2q + 3q ≥ 31 or 5q
= 31 or q = 6.2 for q ≤ 6.2 or an
answer of 6.2 following q ≥ 6.2 in
working
oe, (q > 6.2 is M1 only)
B1 for 4 correct and none incorrect
or all correct with one addition.
Total 8 marks
B1 for 2 correct parts
Apart from questions 10, 11, 14, 17 and 21 (where the mark scheme states otherwise) the correct answer, unless clearly obtained from an
incorrect method, should be taken to imply a correct method.
International GCSE Maths
2
4
3
78, 76, 74
(b)
(c)
Correct reason
4n + 3
49
0.22
(a)
(91 + 0.3 × “350”) ÷ 4 [ (91 + “105”) ÷ 4] oe
91 ÷ 0.26 (=350) or (0.3 ÷ 0.26) × 91 (=105))
(b)
“0.44” ÷ 2
1 – (0.26 + 0.3) (=0.44)
(a)
B2
B2
B1
2
1
A1
M1
M1
M1
A1
M1
2
3
3
Total 5 marks
B1 for 4n + x where x is any
integer
B1 for one correct term
The first sequence is only odd
numbers and the second is only
even numbers
Total 6 marks
A correct method to find total
number of bricks or number of
blue bricks
A correct method to find number
of layers
3
5
4
18000 oe or 720
100
= 778.75
4
× (18000+’720’+’748.80’)
100
= 748.80
4
× (18000+’720’)
100
Eg
1.043
18000
OR
2248
3
A1
M1
M1
(M1 for
18000 × 1.04
or 18720
or 18000 × 1.04²
or 19468.8
or18000 × 1.044
or 21057.45)
18000 1.043
OR
M2 for
Accept 1 + 0.04 as equivalent to 1.04
throughout
SC: If no other marks gained, award
M1 for 18000 × 1.12 oe or 20160 OR
or 2160
Answers in range 2247 – 2248
Total 3 marks
oe
or 720
for completing
method
4
eg
18000
100
for
4
8
12
8
or sin x =
or cos x =
12
208
208
(31.50 ÷ 7) × 8 oe (=36)
8
‘36’× 1.2 oe
(x = ) 360 – (90 + 90 + 52)
 8 
 12 
8
x = tan-1   or sin 1 
or cos 1 


 12 
 208 
 208 
tan x =
7
6
43.2(0)
128
Correct reasons
33.7
3
4
3
A1
M1
M1
B1
M1
A1
B1
A1
M1
M1
Total 3 marks
Correct method to find the amount
Behnaz has
Correct method to find the amount
Ahmed has
The angle between a tangent and a
radius is 90° oe
Angles in a quadrilateral add up to
360° oe
Total 4 marks
Accept answers which round to
33.7
Total 3 marks
A complete method to find angle x
A correct trig ratio for angle x
5
9
15 and 45 or 15.25 and 45.75 indicated on
cumulative frequency axis or stated
A vertical line from 48 up to the cf graph
(c)
(d)
(a)
(b)
Approx 6
Approx 19
4, 18, 35, 48, 55, 58, 60
Points correct
Curve or line segments
2
M1
2
M1
A1
A1
M1
B1
B1
1
2
If M1 scored ft from CF graph. If
M1 not scored, ft from correct
curve and, if answer is correct
(±½sq) award M1A1
Total 7 marks
If M1 scored ft from CF graph. If
M1 not scored, ft from correct
curve and, if answer is correct
(±½sq) award M1A1
+ ½ sq
ft from points if 4 or 5 correct or if
points are plotted consistently within
each interval at the correct heights
Accept curve which is not joined to
the origin
Correct cumulative frequencies
6
10
135° − 108° (=27°)
180 – 2 × 27
Alternative
360 ÷ 8 (=45) 180 – 45 (=135)
360 ÷ 5 (=72)_180 – 72 (=108)
72° − 45° (=27°)
180 – 2 × 27
360 ÷ 8 (=45)
360 ÷ 5 (=72)_
126
126
5
5
M1
M1
A1
M1
M1
M1
M1
A1
M1
M1
Method to find interior angle of
octagon or pentagon
Method to find interior angle of
both octagon and pentagon
Method to find CAB or CBA
Fully correct method to find angle y
dep on at least M2
Total 5 marks
Method to find exterior angle of
octagon or pentagon
Method to find exterior angle of
both octagon and pentagon
Method to find CAB or CBA
Fully correct method to find angle y
dep on at least M2
7
11
6 x  4  15  20 x
2
10
26x = 39 or
6x + 20x = 20 + 4 + 15
6x + 20x = 39 oe
6x – 4 – 15 + 20x = 2 × 10 oe
2(3x  2) 5(3  4 x)

 2 or
10
10
2(3x  2)  5(3  4 x)
 2 or
10
2(3x − 2) – 5(3 – 4x) = 2 × 10
Eg
1.5
4
Expanding brackets
For correct rearrangement of a
correct equation with terms in x
isolated
Award full marks for a correct
answer if at least M1 scored
Total 4 marks
M1
M1
A1
for clear intention to multiply all
terms by 10 or a multiple of 10
or to express LHS as a single fraction
with a denominator of 10 or a multiple
of 10
M1
8
13
12
(a)
(b)
3
5
2
3
7
2
8
x = −1 or x = 5
3x² − 12x – 15 = 0
(3x + 3)(x – 5) (=0)
20
(−1, 8) (5, −100)
3x² − 12x − 15
A1
M1
M1
3
Total 3 marks
For at least 3 correct entries into
Venn diagram
30 – (3 + 3 + 5 + 7 + 2 + 2)
B2 B1 for 2 correct terms
M1ft
M1 Correct factorisation or correct use
of quadratic formula
A1 One correct pair
A1 Both correct pairs
Total 6 marks
2
4
9
15
14
(b)
(a)
4(7  5)
49  5
0.3 × 0.9 (=0.27)
0.7 + ‘0.27’
e.g.
0.97
7 1

5
11 11
show
3
2
3
A1
M1
M1
A1
M1
M1
A1
M1
The correct product for fail, pass
A fully correct method to find the
probability that Sophie passes 1st
or 2nd time
oe
Total 3 marks
Total 5 marks
dep on correct working seen
For selecting 10x = 3.2424.... and
1000x = 324.2424... oe
321
990
For multiplying the numerator and
denominator by (7 + 5 )
For a correct single fraction with
brackets expanded in denominator
10
17
16
Greatest number of spheres
= 12.55³ ÷ 135 (=14.641899...)
Largest volume of cube = 12.55³
12.45, 12.55, 135 or 145
14
(270, −1)
(ii)
(b)
(180, 0)
(a)(i)
4
4
Correct intersections of 0°, 180°
and 360° with x axis
A1
A1
Dep on M1
Total 4 marks
Units must be consistent
12.55³
M1
M1
For sight of 12.45, 12.55, 124.5,
125.5, 135 or 145
B1
Total 4 marks
Correct shape curve
M1
B1
B1
11
2
½q − ¼p
MN is parallel to BD
BD  4  MN
(b)
1
1
MB  p or BM   p
4
4
1
1
BN  (p  q)or NB  (p  q)
2
2
1
2
2q − p
(a)(i)
(a)(ii)
(6, −4)
9
19
(7, −4)
(3, −12)
(a) (i)
(ii)
(iii)
(b)
18
1
1
1
1
A1
A1
A1
B1
M1
B1
B1
B1
B1
Total 5 marks
With suitable reasons
With suitable reasons
For correctly giving
MB or BM or BN or NB
Total 4 marks
12
20
'15'2  '7.5'2 
15 5
( 16.7705...)
2
CSA =  15 16.77..
1
 (2h) 2 h (=562.5𝜋) or
3
1
1
   r 2  r (=562.5𝜋)
3
2
1
  4h2  h  562.5 or
3
1
1
   r 2  r  562.5
3
2
3  562.5
h 3
( 7.5) or r = 3 3375
4
790
5
A correct expression for l
M1
Total 5 marks
A correct equation for h or r
M1
(786.5 – 791.7)
A correct equation for the volume of the cone
with (2h)² expanded
M1
A1
A correct expression for the volume of the cone
NB: other letters may be used rather than r and h
M1
13
 2 x  3 
7
(2 x  3)(2 x  5)
5
7

2
4 x  25 2 x  5
5  7(2 x  5)
5  14 x  35
or
2
4 x  25
4 x 2  25
a + 2d = 19
10
(2a + 9d) = 290 oe
2
Eg 10a + 45d = 290
10a + 20d = 190
Or 5(2(19 – 2d) + 9d) = 290, a = 11, d = 4
10th term = 11 + 9 × 4
or 290 – 4.5(2 × 11 + 8 × 4)
21
22
47
40  14 x
4 x 2  25
5
4
A1
Total 5 marks
A correct method to find the 10th
term.
M1
M1
A formula for term 3
A formula for the sum of the first
10 terms
A correct method to find a or d
M1
M1
A1
Total 4 marks
Correct subtraction shown
((4x²−25) can be factorised)
Correct single fraction,
unsimplified ((4x² −25) can be
factorised)
40  14 x
oe e.g.
 2 x  5 2 x  5
M1
M1
For inverting and factorising
M1
14
15
Mark Scheme (Results)
Summer 2018
Pearson Edexcel International GCSE
In Mathematics A (4MA1) Paper 2H
1
eg ef (18ef
eg 6e (3 f
2
2
3
M
bd
a
c
c
2
 12e2 )
 2ef ) , eg 2 f (9e f  6e )
2
(c)
3
5x39  4 oe
ac  M  bd or –ac = −M – bd or
(b)
(a)
6e2 f (3 f 2  2e)
3
5
M  bd
c
x8
a
2
2
2
A1
M1
A1
M1
A1
M1
M bd
 M  bd

, a
c
c
c
43
or x  8.6 or
5
Total 6 marks
[−∞, 8.6)
Any correct partially factorised
expression with at least 2 terms in
the common factor or for the correct
common factor and a 2 term
expression inside the brackets with
just one error
Accept x
[must have been seen with a = to
award accuracy mark]
Accept as equation or with the
wrong inequality sign. Also award
M1 for an answer of 8.6 or 8.6 with
an = sign or the incorrect inequality
sign.
oe, eg a 
For a correct first stage
Apart from Questions 4, 9, 15, 16, 21(a) 21(b) and 22, where the mark scheme states otherwise, the correct answer, unless clearly
obtained by an incorrect method, should be taken to imply a correct method.
Question
Working
Answer
Mark
Notes
International GCSE Maths A June 2018 – Paper 2H Mark scheme
16
3
2
Question
3
"21600"19200
"2400"
(100)
19200
 20000 (=2400)
(100) or
100
8
or “21600” ÷ 19200 (×100) oe
19200
"3450
 20000 (=21600) or
(20 000 – 19200) +
100
8
 20000 (=1600)
20000 
100
8
“1610” – “460” or "
1
(7 – 2) × “230” or 7 × “230” – 2 × “230” or
2
3450
 3450( 460) or
(=230) or
267
267
7
72
 1
 3450( 1610) or
 
267
267  3 
Working
12.5
1150
Answer
4
3
Mark
A1
M1
M1
M1oe
A1
M1
M1
9
8
Total 4 marks
or 112.5%
21600
or for 1.125 or
oe
Total 3 marks
Award M2 for
200001.08 or
Notes
17
4
Question
109
109
with RHS shown as
56
56
32
35
 (1)
56
56
1
53
56
eg
eg 3
32 35
1
56 56
88 35
2 1
56 56
Alternative method

3
56
3
109
or 2 
56
56
eg (3)
Alternative method
Or
eg
8 25 713
200 91


or
56
56
56 56
109
53
= 1
56
56
13
25
and
8
7
Working
correctly shown
correctly shown
correctly shown
Answer
3
3
3
Mark
109
53
and 1
but
56
56
53
109
allow showing that 1
=
on
56
56
A1
dep on M2
Total 3 marks
complete correct method
M1
3
109
or 2 
56
56
two correct fractions with a common
denominator, at least one correct
the subtraction eg
dep on M2 with sight of the result of
two improper fractions, with a
common denominator, at least one
correct
correct subtraction of fractional
parts
RHS in working
the subtraction eg
dep on M2 with sight of the result of
correct improper fractions or two
improper fractions with a common
denominator, at least one correct
two correct fractions with a common
denominator
M1
A1
M1
M1
A1
M1
M1
Notes
18
7
6
5
Working
(b)
(a)
4
70.5
63
3
Total 7 marks
or for 33 – “96” or 33 to “96” oe
M1
A1
for 39 + 57(=96) or 33 + 60(=93)
Correct calculation for mean of
class B
Expression for total of both classes
together or total for class A
Expression for total of class B
Total 4 marks
Allow 21 cm oe if units shown
Correct scale factor (given as a
fraction or ratio) or correct equation
in r or a correct expression for r.
Allow 2.6666... to 1 dp rounded or
truncated
Award even if part of a calculation
including 1 or 2 circles
awrt 5.63
Correct reason for 90° angle
[If used alternate segment theorem]
Total 3 marks
For 90° and 18° correctly identified
in the working or on the diagram or
for 90 – 18 or for other fully correct
method
Notes
M1
A1
M1
(28  32)72.6  2875
(=”2256” ÷ 32)
32
Highest in A = 39 + 57 (= 96)
Highest in B = 33 + 60 (= 93)
(39 + 57) – 33
M1
M1
A1
M1
(28 + 32) × 72.6 – 28 × 75 (=2256)
(28 + 32) × 72.6 (=4356) or 28 × 75 (=2100)
0.21
2

8
0.6
1.6
r
0.6

(=0.375) or
(= = 2.6 ) or
1.6
0.6
0.56 1.6
3

0.560.6
or (r )
or 0.56 ÷ 2.6 oe
1.6
A1
(b)
5.63
M1
A1
M1
2 × 𝜋 × 0.56 × 1.6
2
3
Mark
(a)
72
Answer
B1
OQT  90 and OQP 18 or 90 – 18
o
Angle between tangent and radius(or diameter)
is 90 degrees
Question
19
8
Question
( x )
(=
12.6
) or
0.61566...
12.6
12.6
or sin 38 =
x
x
12.6
12.6
or
sin 38
cos 52
cos 52 
Working
20.5
Answer
3
Mark
A1
M1
M1
"16.12..."
oe
x
Total 3 marks
Allow fully rearranged sine rule
20.4 – 20.5
Or (x =)
12.62  "16.12..."2 or
"16.12..."
(x = )
sin 52
Allow correct first stage of sine rule
Accept decimal correct to at least
3SF
sin 52 
first stage to find x eg
x² = 12.6² + “16.12...”² or
12.6
(=16.12...) and a correct
tan 38
Or use of tan to find horizontal side
12.6 × tan 52 or
Notes
20
11
10
9
(b)
(a)
Question
5x + 5y = 75 +
7x – 5y = 3
or
23  27 or
1
1
or
and16  24
4
2
16
9
4
V   1.53 (= 14.1(37)… or  )
2
3
109.6
D
oe
3
4
3  1.5
1324 135
27
23
“6.5” + y = 15 or x + “8.5” = 15 or
7 × “6.5” – 5y = 3 or 7x – 5 × “8.5” = 3
7(15 – y) – 5y = 3 or 7x – 5(15 – x) = 3 oe
eg 7x + 7y = 105 −
7x – 5y = 3
Working
7.75
19
4
x = 6.5, y = 8.5
Answer
3
2
2
3
Mark
7.75 – 7.78
Total 3 marks
dep
M1
A1
Correct expression for volume.
Total 4 marks
M1
19
or for k = −6 × 4 + 5
Accept 13
A1
24
4
for 13
Accept 2
Total 3 marks
Correct method to eliminate x or y:
coefficients of x or y the same and
correct operation to eliminate
selected variable (condone any
one arithmetic error in
multiplication) or
writing x or y in terms of the other
variable and correctly substituting
dep Correct method to find second
variable using their value from a
correct method to find first variable
or for repeating above method to
find second variable
dep on first M1
M1
A1
M1
A1oe
M1
M1
Notes
21
12
or 42 + 144 + “138” + (50 + 96) + DEP = “540” (where
P is on AB and FE extended)
oe
Eʹ = “720” – “138” – 42 −50 – 96 – 144
(= 720 – 470 = 250)
and E = 360 – “250”
or E = “138” + 42 + 50 + 96 + 144 + 360 – “720”
(= 830 – 720)
or “138” + 42 + 50 + 96 + 144 + (360 – E) = “720”
eg “138” + 42 + 50 + 96 + 144 + Eʹ = “720”
(2 × 6 – 4) × 90 (=720)
EDC = 180 − 42 (=138)
110
5
A1
M1
M1
M1indep
M1
Total 5 marks
from no incorrect working
A completely correct calculation for
the correct angle E
obtuse angle of the hexagon and E  is
the interior (reflex) angle
or for an answer of 250 from correct
working
Method to find sum of interior angles
of hexagon or the correct sums for
the interior angles of shapes used (eg
540° & 180° if the line through FE to
point on AB drawn or 720° and 180° if
line drawn from E parallel to AB or
540° & 180° if line through FE
extended and joined to line through
CB extended) oe
dep on previous M marks
Equation for E or E  where E is the
May be marked on diagram.
NB: splitting the shape incorrectly (FDC and DEA are not straight lines) gains no marks for angles calculated from false information.
However angles calculated that follow the scheme, such as EDC = 138° or interior angles of hexagon = 720° can be awarded. Other
ways of correctly splitting the shape can be awarded full marks, eg FE to a point on AB or adding a parallel line eg from E parallel to AB
NB: some students show lots of lines but actually work with the angles correctly so please check carefully.
Question
Working
Answer
Mark
Notes
22
13
(b)
(a)
Question
5 4 4 5 4 3
5 4 4 8
     or
   oe
10 9 10 9 10 9
10 9 10 9
5 4 5 1 4 1 1
or 1− 
       oe
 10 9 10 9 10 9 10 
Working
90
52
4 4 1 5 3 1 5 4
, , , , , , , ,0
9 9 9 9 9 9 9 9
Answer
3
2
Mark
A1
M1
M1
B2oe
0
or the branch
9
Total 5 marks
oe decimals 0.577… or 57.7...%
rounded or truncated to 2 or more sf
crossed out or left blank
Award M1 for one correct product
(ft tree diagram)
A fully correct method (ft tree
diagram)
shown), 0 can be
Award B1 for any 3 correct.
Decimals must be correct (recurring
Notes
23
15
14
−0.8
0.6
2.2
8.305  0.655
7.65
2
A1
A1
-0.8 or 0.6 or 2.2
M1
B2
B2
M1
4
2
2
Mark
Plot y = −2x + 3
x³ − 2x² − 3x + 4 = −2x + 3
correct curve
(c)
-6, 4, 0, -2, 4
Answer
(b)
Working
(a)
Question
A1
M1
Total 2 marks
seen). Accept 0.6549
dep on correct method shown
For either bound correct (used or
Any one correct x value at intersection of
graphs (or one or more points given as
coordinates)
ft dep on second M1 (Award even if curve in
(a) is incorrect)
Accept −0.9 to −0.7
SC B2 for all correct
Accept 0.4 to 0.7
solutions from
Accept 2.1 to 2.4
graph of
(not coordinates)
y  x3  2 x 2  x  1
ft (±1 square) dep on
second M1 must be 3
values
Total 8 marks
Sufficient to cross curve at least once.
For correct smooth curve.
If B2 not awarded, award B1 for at least 5
points plotted correctly ft from table dep on
B1 or B2 in (a) (plots ±1 sq)
Award B1 for 2, 3 or 4 correct.
Notes
24
16
(b)
(a)
Question
5
" t 2 "
5x 2
8
eg 10 = k × 22 or
R = kt² oe
40 = k × 4²
Working
or
k = 2½
t
x
0.8
5
R t2
2
Answer
2
3
Mark
A1
M1
A1
M1
M1
2
25
16

x
1
proportional to
x ]
2
2
Total 5 marks
[allow other clear arguments that
clearly shows t is inversely
required. eg accept t 
ft dep on answer of the form R  kt
Simplification of constant is not
ft dep on answer of the form R  kt
2
Award for R  kt if the value of k is
shown clearly in (a) or (b).
Equation consistent with R  t
Substitute values at any point on
the graph or find the value of k.
(Implies first M1.) Allow readings
from graph for t ± 0.1 and R ± 1
Notes
25
17
(b)
(a)
Question
5
 ,3
3
23
(4)  (4)2  43(15)
(3x + 5)(x – 3)( < 0) or
3x² − 4x – 15 < 0 (or = 0)
Working

3
5
 x3
3x  4 x 15
2
Answer
4
2
Mark
A1oe
M1
M1
M1
B2
Allow x  
5
, x3
3
Total 6 marks
ft from “their (a)” (=0) for 3 term
quadratic, for correct factorisation or
correct use of quadratic formula to find
the two critical values, allow 1 sign
error. [−(−4) could be 4 and (−4)²
could be 4²](condone missing
brackets)
Both critical values correct
Accept -1.66… rounded or truncated to
3SF.
Inequality signs needed
Award B1 for any 2 or 3 of the 4 terms
differentiated correctly.
ft from (a) ie “their (a)” = 0 (or < 0)
Notes
26
19
18
Question
2810
"740"
10
(= 0.3676 …)
sin DBE 
(=0.9428…)
cos DBE =
(= 0.3952…)
or
tan DBE 
M1
M1
Total 3 marks
21.5 – 21.6
(=0.9299...)
Total 3 marks
2  "640"  "740"
10sin 90
or
"740"
"640" "740" 102
Allow use of sine or cosine rule
Complete method to find BE or BE2
or BD or BD2
oe
101.5 to 101.6
 196  100  64 
1  196  64  100 
180  cos1 
  cos 

280
224




cos DBE 
A1
A1
 102  82 142 
cos 1 
 oe ie cos-1 of the correct angle or
 2  10  8 
M1
a fully correct method to find the largest angle eg
Correct substitution in cosine rule for any angle or for
44.4... or 34.047....(the other 2 angles to 1dp or
better)
M1
Notes
"640"
"740"
3
3
Mark
sin DBE 
21.6
101.5
Answer
"640"
10
(BD = 8 10 =
25.298….)
185 =27.202 …)
(BE = 2
BD2  82  242
(=64 + 576 = 640)
oe
(= 100 + 576 + 64 = 740)
BE 2 102  242 82
cos A 
102  82 142
14² = 10² + 8² − 2 × 10 × 8 × cosA or
Working
27
20
Question
4
1
(= 8) or
(=2) or
0.5
2
1 small square vertically = FD of 2 or
1cm vertically = FD of 10 oe
eg (4 × 5 + 20 × 4 + 25 × 2 + 15 × 4) ÷ 5 or
4 + 40 × 0.4 + 50 × 0.2 + 30 × 0.4 or
4 + 16 + 10 + 12 oe
height of last bar 
height of first bar 
eg 4 × 5 + 1 × 10 = 30 small squares for 6 babies or
30 ÷ 6 or
5 small squares represent 1 baby or
Working
42
Answer
3
Mark
A1
M1
M1
Total 3 marks
Fully correct method, allow one
error in products but must be the
sum of 4 parts
Start working with area being
proportional to frequency or show
the height of the first or last bar or
show a correct scale on the
frequency density scale, with no
inconsistent values.
eg could be awarded by seeing total
of little squares ÷ 5 oe
Notes
28
21
(c)
oe
2
(b)
Working
( x  3 2)2  (3 2)2 1
3 1
2( 3 1) 2 3  2

or
or
3 1 3 1
31
2
95 and 45
(a)
Question
2
( x  3 2)  19
1 3
5 5 shown
Answer
2
2
2
Mark
A1
M1
A1
M1
A1
M1
a =3
2 or b = − 19
or ( x  3
3 1
3 1
Total 6 marks
2)2  18 1 or for
 3 1
 3 1
dep on M1
or
seeing multiplication by
3 5  2 5 but we must see where
these come from
Rationalise denominator – award for
or for 45 = 3 × 3 × 5 and
20 = 2 × 2 × 5
dep on M1 cao with sight of
Notes
29
23
22
Question
0
M1
2
(2a  (30 1)d )
A1
M1
(2a  (36 1)d ) oe
M1
30
2
36
5
M1
(2a  (48 1)d )  4
2
(2a  (36 1)d ) oe
2
96a + 1392d = 0 oe eg 4a + 58d = 0,
2a + 29d = 0 or a = −14.5d etc
2
2
48
36
(2a  (48 1)d )
or
48
21
or
16  16  41(80)
A1
M1
(x – 4)(x + 20) (= 0)
16  576
A1
2
M1
x² + 16x – 80 (= 0)
M1
5 × (5 + 4 + 7) = x × (2 × “8” + x)
6
Mark
A1
4
Answer
r = 8 or d = 16
7 × 4 = 2(2r – 2) or 7 × 4 = 2(d – 2)
Working
Incorrect working giving the radius as 16 cm gains M0A0
Notes
Total 5 marks
Indep Allow substitution of any
‘found’ values of a and d
For a correct equation.
For a correct expression for the first
48 terms or the first 36 terms
Total 6 marks
dep on first 2 method marks
Correct factors or evidence of
correct use of quadratic formula.
Accept 5 × 16 = x(2r + x)
Or a correct equation in r
eg 5.5² − 1.5² = 4r − 4
30
47
Jan 19
Mark Scheme (Results)
January 2019
Pearson Edexcel International GCSE
In Mathematics A (4MA1) Higher Tier
Paper 2H
3
2
Question
1
12  x 1

90  x 3
“6” × 2 (=12) or”6” × 13 (=78) or
3(12 + x) = 90 + x
(“78” ÷ 2 ) – “12” or 2x = 54
or “78” × 3/2 –“78” – “12”oe
90 ÷ (2 + 13 ) (= 6) or
Working
73 ÷ 200 (=0.365) or 73 × 100 (= 7300) or
1 cm = 2 m oe
“0.365” × 100 or “7300” ÷ 200
73 ÷ 2
4
3
2
36.5
4n + 3
27
Mark
Answer
A1
M1
M1
M1
A1
B2oe
M1
M1
200
M2 for
dep on a correct method for “78” and
“12”
2
13
 90(  12) or  90(  78)
15
15
M2 for
If not B2 then award B1 for answer
of
4n + k (k ≠ 3) or n = 4n + 3
e.g. 7 + 4(n – 1) or 4n + (7 – 4) etc
allow Tn = 4n + 3 or
x = 4n + 3 etc
100 ÷
Allow their incorrectly
73
converted 73 m ÷ 200 oe
Notes
48
5
Question
4
5
A1
M1
(67 – 2×”11”) × (123 – 2×”11”)
(45 × 101)
or
123 × 67 – 12 × “28” ×” 11”
(8241 – 3696)
4545
M1
67 – 56 (=11) or 67 – 2×”28” (=11)
or 123 – 4×”28” (=11)
4
M1
Fully correct
Venn diagram
dep on M2
for method to find length or
width
for method to find other
dimension
Notes
B4 fully correct Venn diagram with labels
A and B
(If not B4 then B3 for 3 correct regions,
B2 for 2 correct regions
B1 for 1 correct region)
e.g. “56” ÷ 2 (=28)
2 4
6 12
B
Mark
M1
1
3
Answer
8 10
123 – 67 (=56) or 2x = 123 – 67 or 2x + y =
67 or 4x + y = 123 oe
(x = length of tile, y = width of tile)
A
5 7
9 11
Working
49
6
Question
(b)
(a)
6
4
96 120
16 20
4
5
2 × 2 × 2 × 2 × 2 × 3 or 2 × 2 × 2 × 3 × 5
e.g.
2 96 120
2 48 60
2 24 30
3 12 15
4
5
Working
2
2
646 800
Mark
24
Answer
A1
A1
M1
M1
or 2³ × 3 oe
for 2m × 3n × 5p × 7q × 11r with at least two of
m = 4, n = 1, p = 2, q = 2, r = 1 (or omission of one
with others fully correct) NB: e.g.24 could be 2 ×23
or
prime numbers may be seen in a Venn diagram – if
so must be correctly placed
or 24 × 3 × 52 × 72 × 11 oe
Notes
for one number written as product of prime factors
number may be at the end of factor trees or on
‘ladder’ diagrams
or
Use of table method (allow 1 error), 2 examples
shown but could have 2, 3, 4, 6, 12, 24 along the side
or
at least 2 factors for each (excluding 1, 96, 120)
50
(b)
Question
(a)
7
687 700 ÷ 0.92 (=747 500) or
687 700 ÷ 1.15 (=598 000) or
1.15 × 0.92 (=1.058)
687 7000 ÷ (0.92 × 1.15)
8500 × 0.023 (=195.5) or
8500 × 1.023 (=8695.5)
((8500 + “195.5”) × 1.023) × 1.023
Working
Mark
3
3
Answer
9100
650 000
complete method
for 9100 – 9100.1 (answer for 600(.1) gains M2A0)
a correct first step
Dep on M1 for completely correct method
M1
A1
M1
M1
A1
M1
Notes
M2 for 8500 × 1.0233
(M1 for 8500 × 1.023n)
51
(b)
Question
(a)
8
3.5
V
3.5
(V )
0.65
630  1000
oe
60  60
630 × 1000 (=630 000)
60 × 60 (=3600)
eg
630 ÷ 60 (=10.5)
630 000 ÷ 60 (=10 500)
1000 ÷ 60 (=16.66...)
1000 ÷ (60 × 60) (=0.277...)
1 ÷ (60 × 60) (= 0.000277...)
0.65 
Working
175
5.38
Answer
3
3
Mark
3500
0.65
(M2 for 630 ÷ 3.6)
Fully correct method
M1
A1
for converting 630 km to m
or
1 hour to seconds
or
for correct operation(s) using at least 2 of
the numbers 630, 1000, 60, 60
been converted eg V 
for answer in range 5.38 – 5.385
SCB1 for a “correct” equation involving
V with digits 65 and 35 where units have
M1
A1
M1
M1
Notes
52
10
Question
9
4 1
or c = 1
2( 0)
y = “1.5”x + c or y = mx + 1
or eg y – 4 = m(x – 2)
3 ÷ 2 (=1.5) or eg
y = 2x – 9 and 4x + 5(2x – 9) = 4
4x + 5y = 4
10x – 5y = 45
With the operation of adding
Working
e.g. 4x + 5y = 4
4x – 2y = 18
with the operation of subtraction
y = 1.5x + 1 oe
x = 3.5 oe, y = −2
Answer
3
3
Mark
A1
M1
A1
M1
M1
M1
oe eg y  4 
3
( x  2)
2
(dep) for substitution of found variable into
one equation or correct method to eliminate
second variable
Dep on M1
for correct method to find gradient – may see
this on grid. For c = 1, could be (L =) mx + 1 oe
or for 1.5x + c
for use of y = mx + c with either m or c
or for (L =) 1.5x + 1
Notes
for correct method to eliminate one variable –
multiplying one or both equations so the
coefficient of x or y is the same in both with
the intention to add or subtract to eliminate one
variable(condone one arithmetic error) or
isolating x or y in one equation and substituting
into the other equation
53
Question
11
On the whole students have higher marks in
Science
The spread of results is greater for Science
Results are more consistent for Maths
Comparisons in context: eg
The median is greater for Science/less for Maths
The IQR (or range) is higher for Science/less for
Maths
The median is 2.5 marks higher for Science
The IQR (or range) is 7 marks more for Science
Working
Basic comparisons from information: eg
one for IQR
and one for
median
Answer
Two
comparisons
Mark
2
B2
NB; any numbers used must be correct for the
award of the mark
(B1 for 1 or 2 basic statements or for 1
statement in context)
Notes
For 2 comparisons in context
or
1 basic comparison and 1 comparison in context
54
(d)
(c)
Question
(a)
12
(b)
6a  r
5r
m2 × 5r = 6a + r
5rm2 – r = 6a
m2 
2(e2 – 9) or (2e – 6)(e + 3) or (e – 3)(2e + 6)
Working
r
6a
5m 2  1
2(e – 3)(e + 3)
Answer
1
27x6y15
4
2
Mark
1
2
M1
M1
A1
M1
A1
M1
B1
B2
6 a
oe
1  5m 2
6a
6a
in working if
2
5m  1
5m 2  1
alone is given as answer
r
NB: to award A1 we must see
or for r 
If not B2 then
B1 for any two correct terms in a
product
Notes
55
13
Question
"264  8 x "
"39  x "
or “273” – “264”
"264  8 x "
=7 oe eg“264 + 8x” = “(39 + x)” × 7
"39  x "
(4 + 13 + 16 + 6 ) × 7(=273) oe or
4 × 5 + 13 × 6 + 16 × 7 + 8x + 6 × 9
(20 + 78 + 112 + 8x + 54) or
264 + 8x
(4 + 13 + 16 + 6 + x) × 7 (=7(39 + x) = 273+ 7x) or
Working
9
Answer
4
Mark
for use of mean
M1
M1
A1
M1
Notes
at least 3 products correct with
intention to add
56
(b)
Question
(a)
14
0.35 × 0.35 or 0.35 × 0.65 or 0.65 × 0.35 or
0.65 × 0.65
0.35 × 0.35 + 0.35 × 0.65 + 0.65 × 0.35 or
1 – 0.65 × 0.65
Working
0.5775
Answer
0.65
0.35, 0.65
0.35, 0.65
3
Mark
2
A1
M1
M1
B2oe
oe e.g.
231
, 0.58 or 58% or better
400
ft from (a)
Notes
for all correct
If not B2 then award B1 for 0.65 in
any of the 3 possible positions
NB all values may be given as
fractions
ft from (a)
57
(b)
Question
15 (a)
1
2
1
 (8 x 2  12 x  6 x  9)  133
2
or (4x − 25)(2x + 11) (=0)
  6  36  8800
6  36  8800
6  8836
or
or
28
16
16
or 8x² − 12x + 6x – 9 = 266
eg.
or 0.5(4x + 3)(2x – 3) oe
e.g.  ( x  5  3 x  2)  (2 x  3)
Working
6.25 oe
shown
Answer
3
3
Mark
for completion to given equation
dep on M2
A1
dep on M1 and 6.25 oe alone given as
final answer
ft from an incorrect 3 term quadratic
equation
(if student gains M1 and shows both
answers the 2nd M1 can be awarded)
28
Condone one sign error in substitution;
allow evaluation of individual terms e.g. 36
in place of (−6)² [allow −62 or 62 in place of
(−6)2, throughout allow + rather than ± ]
or
(4x ± 25)(2x ± 11) (=0)
  6  ( 6) 2  4  8  275
M2 If not M2
then award M1 for
A1
M1 for correct equation with brackets expanded
Notes
M1 correct algebraic expression for area
58
18
17
16
Question
(a)
(b)
(c)
2
960  4 
405  3 
   (=1.3...)or 3
   (=0.75)
405  3 
960  4 
4.55
25  7.5
7.5 or 8.5 or 4.65 or 4.55
25 or 15
g(−1.5) = 1 ÷ (1 – 2 × −1.5) (=0.25) or
 1 
fg(x) = 4  3  
 oe
 1  2x 
3
4
   928 or 928    oe
4
3
2
e.g. 3
Working
4
2
3.25 oe
0.26 oe
3
1
1
Mark
522
−11
0.5 oe
Answer
A1
A1
M1
M1
M1
LB1
with 4.55 ≤ LB1
UB  LB 2
< 4.6 and
20 < UB ≤25 and 7.5 ≤ LB2 <
8
for 0.26 from correct working
for
g(−1.5) must be the correct
calculation alone.
for a complete method
M1
A1
B1
B1
M1
Notes
for a correct linear scale factor
M1
59
Question
19
2.5 × 2 + 4 × 3 + 3.4 × 5 + 2.2 × 5 + (1 ×) 15
or
5 + 12 + 17 + 11 + 15 (=60)
or e.g.
100 + 240 + 340 + 220 + 300 (=1200)
Working
At least 2 of:
2.5 × 2 (=5) or 4 × 3 (=12) or 3.4 × 5 (=17)
or 2.2 × 5 (=11) or (1 ×) 15 or
(1 ×) 10 (=10)
or
e.g. at least 2 of
100, 240, 340, 220, 300 or 200
1
oe
6
Answer
3
Mark
A1
M1
M1


1
or 16.6 % or 0.16 or 1 in 6
6
(percentage or decimal rounded or
truncated to 3 or more sig figs)
for
for method to find total number of
people (allow one error)
or
total number of squares/blocks for
method used (allow one error)
Notes
for working with area of at least 2
bars
could be using freq density × mins
or
use of counting squares or blocks
60
Question
20
Alternative method
angle OCB = 90 − x
angle BOC = 180 – 2(90 – x)
(=2x)
angle AOB = 2x and angle CDA
= 2x
Alternative method
angle CDB = x or angle CAB = x
angle ACB = x
angle ACQ = 2x and angle CDA
= 2x
Working
angle CDB = x or angle CAB = x
angle CBA = 180 – 2x
angle CDA = 180 – (180 – 2x) =
2x
proof with
reasons
5
5
5
proof with
reasons
proof with
reasons
Mark
Answer
B1
A1
M1
M1
M1
B1
A1
M1
M1
M1
B1
A1
M1
M1
M1
dep for any one appropriate circle theorem reason
for complete proof with full reasons
angle between tangent and radius is 90o oe, angles in a triangle
sum to 180o, isosceles triangle, angle at centre is twice angle at
circumference oe
dep on M1 for any one appropriate circle theorem reason
for complete proof with full reasons
alternate segment theorem, isosceles triangle
dep on M1 for any one appropriate circle theorem reason
for complete proof with full reasons
alternate segment theorem, angles in a triangle sum to 180o,
isosceles triangle, opposite angles of a cyclic quadrilateral sum
to 180o
Notes
61
21
Q20 contd
6
6
oe
x ( 33) or (gradient = )
4
4
6
2
m×
= −1 or (gradient of M =)  oe
4
3
2
k 6
 " "
3
4  5
y
eg angle ABC = 180 – 2x
Angle CAB = angle ACB =
[180 – (180 – 2x)] ÷ 2 = x
BCQ = CAB = x
12
4
M1
dep
A1
M1
M1
M1
B1
A1
M1
M1
or complete method to find equation of line (3y = −2x + 28)
and then substitution of x = −4
Dep on M1 for any one appropriate circle theorem reason
For complete proof with reasons
e.g. opposite angles of cyclic quadrilateral sum to 180°
angles in triangle sum to 180°
isosceles triangle
alternate segment theorem
Alternative method where students assume CDA = 2x and must work to show that BCQ = x
62
Question
22

3

5

3
(  1.62...)
5
]
r
3
2 × sin 1   oe
5
 AVB  3
 AVB 
e.g. sin 
oe eg sin 


 2  5
 2 
l
(  0.9772...) and l 
r 3
 oe or
l 5
[r 


 3
5
3
 r 2 :  r 2   rl  3 : 8 or
 r 2 :  rl  3 : 5 or  r 2  3 and  rl  5
8 r 2  3( r 2   rl ) or 5 r 2  3 rl or
Working
 r2
3
 or
2
 r   rl 8
73.7
Answer
6
Mark
A1
M1
M1
M1
M1
M1
awrt
3
sin 1 ( )  36.86....
5
Notes
63
Question
23
or
oe
 1   2   3    4  
BC              
 4   3   12    5  
 1   3   4
BC            oe or
 7   12    5  
1
AB   
4
1  3 
DC  3        
 4    12  
Working
e.g. AB  AD  DB
 2   1 
  
 3   7 
41 cao
Answer
5
Mark
A1
M1
M1
A1
M1
No isw
Notes
for a correct vector equation for AB
64
4
A
7
11
5
9
3
1
12
2
6
4
B
10
8
65
6b
A
7
5
3
7
2
C
3
11
5
2
B
66
82
May 19
Mark Scheme (Results)
Summer 2019
Pearson Edexcel International GCSE
In Mathematics A (4MA1)
Paper 2H
2
Two pairs of intersecting arcs with
equal radius centre D and E
‘5400’ ÷ ‘48’
112.5
4
M1 for consistent use of value within interval (including end
points) for at least 4 products which must be added
760 + 1260 + 1725 + 1250 + 405
(= 5400)
M1 for 2 pairs of arcs that intersect within guidelines or correct
perpendicular bisector without arcs.
Allow division by their Σf provided addition or total under
column seen
A1 oe accept 112 or 113 from complete working
Accept 112.5 with no working
Do not accept 112 or 113 with no working
Total 4 marks
M1 dep on at least M1
correct midpoints used for at least 4 products and not added
or
If not M2 then award
or
Question Working
Answer
Mark Notes
Apart from questions 9, 16, 20, 23 (where the mark scheme states otherwise) the correct answer, unless clearly obtained by an
incorrect method, should be taken to imply a correct method
1
95 × 8 + 105 × 12 + 115 × 15 +
M2 for at least 4 correct products added (need not be
125 × 10 + 135 × 3 (= 5400)
evaluated) or
83
3
(b)
(c)
(a)
e.g.
Examples
There are no members that are in
both A and B
No members in common (in A and
B)
No numbers the same (in A and B)
B has even numbers. A has odd
numbers except 2 which is not in B
Nothing in A is in B oe
No overlap
A and B don’t share any numbers
1 and 9
1, 2, 8, 9
Correct
statement
Correct
bisector
with arcs
1
2
1
2
Total 2 marks
B1
B2 for fully correct
(B1 for 3 or 4 correct with no more than one addition or a fully
correct Venn diagram)
B1 for a statement which indicates correct meanings for
intersection and empty set
A1
84
(b)
'732 ' '480 '
100 or
'480 '
e.g. 1 + 0.2 (= 1.2) or
100(%) + 20(%) (= 120(%)) or
 '732 '

100   100 or 152.5 – 100 or

 '480 '

 '732 ' 
 1 100 or 0.525 × 100

 '480 ' 
‘252’ ÷ ‘480’ × 100 or
e.g.
4 × 120 (= 480)
e.g. 120 ÷ 2 × 5 (= 300) or
120 × 0.4 × 7 (= 336) or
(120 – ‘60’ – ‘48’) × 8 (= 96) or
120 × 0.1 × 8 (= 96)
e.g. (120 ÷ 2 × 5) + (120 × 0.4 × 7) +
((120 – ‘60’ – ‘48’) × 8) (= 732) or
(120 ÷ 2 × 5) + (120 × 0.4 × 7) +
(120 × 0.1 × 8) (= 732) or
‘300’ + ‘336’ + ‘96’ (= 732)
5
(a)
π × 72 × 20 (= 3078.76...) or 980π
4
52.5
3080
5
2
Total 2 marks
A1 accept 53
M1
M1 for a complete method to find the percentage profit
M1 for a complete method to find the total income
M1
M1 for a method to find the income for one of the selling prices
M1 for complete method to find volume
A1 for answer in range 3077.2 − 3080
Total 4 marks
85
e.g. 15 ÷ 1.2 or
15 ÷ 120 × 100 or
15 × 100 ÷ 120
15
(= 0.125) oe
120
12.5(0)
3
A1 accept (£)12.5, (£)12.50p, 1250p if the £ sign is crossed out
Total 8 marks
M1 dep
86
5
ALT
(b)
(a)
e.g. 15 ÷ 1.2 or
15 ÷ 120 × 100 or
15 × 100 ÷ 120
15
(= 0.125) oe
120
e.g. 1 + 0.2 (= 1.2) or
100(%) + 20(%) (= 120(%)) or
12.5(0)
52.5
3
5
M1 for a method to find the profit of one of the books
e.g. 120 ÷ 2 × 1 (= 60) or
120 × 0.4 × 3 (= 144) or
(120 – ‘60’ – ‘48’) × 4 (= 48) or
120 × 0.1 × 4 (= 48)
e.g. (120 ÷ 2 × 1) + (120 × 0.4 × 3) +
((120 – ‘60’ – ‘48’) × 4) (= 252) or
(120 ÷ 2 × 1) + (120 × 0.4 × 3) +
(120 × 0.1 × 4) (= 252) or
‘60’ + ‘144’ + ‘48’ (= 252)
‘252’ ÷ ‘480’ × 100 oe
A1 accept (£)12.5, (£)12.50p, 1250p if the £ sign is crossed out
Total 8 marks
M1 dep
A1 accept 53
M1
M1 for a complete method to find the percentage profit
M1 for a complete method to find the total profit
M1
4 × 120 (= 480)
87
7
6
19.5 ÷ 2.5 or 19.5 × 0.4 oe or
(b)
19.5
a 
e.g. 30 × 26.8 (= 804) or
13 × 25 (= 325) or
(26.8 – 25) × 30 or
1.8 × 30
e.g. (30 × 26.8 − 13 × 25) ÷ (30 – 13)
(= 28.1764...) or
(‘804’ – ‘325’) ÷ (30 – 13) (= 28.1764...)
or
(‘804’ – ‘325’ ÷ 17) (= 28.1764...) or
((26.8 – 25) × 30) ÷ 17 + 25
(= 28.1764...) or
‘1.8’ × 30 ÷ 17 + 25 (= 28.1764...)
4.2 
2.5 or 0.4 or 0.7 or 1.4(2857……)
15
6
4.2
6
or
or
or
oe
6
15
6
4.2
(a)
28.2
7.8
10.5
3
2
2
Total 4 marks
A1 accept 28.15 – 28.2 (accept without working)
(Accept 28 from complete working)
Total 3 marks
M1 for a complete method to find the mean mark for the girls
M1 for finding the total marks for the boys or the total test
marks
A1 oe
A1 oe
M1 If using DF ft their answer from part (a)
M1 for a correct scale factor, accept ratio notation eg 6 : 15
88
x
1000
oe
60  60
(x) × 1000 or (x) ÷ 60 or
(x) ÷ 60 ÷ 60 or
(x) × 1000 ÷ 60 oe
Using elimination then substitution
9
e.g.
e.g.
x + 2y = −0.5
3x + 6y = −1.5
+ 6x – 2y = 32
− 3x – y = 16
(7x = 31.5)
(7y = −17.5)
e.g.
e.g.
‘4.5’ + 2y = −0.5 x + 2 × ‘−2.5’ =
or
−0.5
3 × ‘4.5’ – y = 16
or
3x – ‘−2.5’ = 16
8
x = 4.5
y = −2.5
5
x
18
3
3
x
1
x
or
3.6
3.6
5
oe
18
Total 3 marks
A1 (dep on first M1) for both solutions
Total 3 marks
for repeating above method to find second variable
M1 (dep) for substituting their value found of one variable into
one of the equations or
M1 for a correct method to eliminate x or y: coefficients of x or y
the same and correct operation to eliminate selected variable
(condone any one arithmetic error)
A1 accept 0.27x or 0.27 x or
M1 (dep) for a complete correct method
M1 for at least one of × 1000 or ÷ 60 or
89
16  ' 2.5'
x
3
x = −0.5 – 2‘–2.5’
or
e.g.
Using substitution
9
3(−0.5 – 2y) – y = 16
(7y = −17.5)
or
16  y
 2 y  0.5
3
(7y = −17.5)
or
0.5  '4.5'
2
y  3' 4.5' 16
y
(7x = 31.5)
e.g.
x  2 3x 16  0.5
(7x = 31.5)
or
 0.5  x 
3x  
  16
2


x = 4.5
y = −2.5
3
A1 (dep on first M1) for both solutions
Total 3 marks
M1 (dep) for substituting their value found of one variable into
one of the equations
M1 for correctly writing x or y in terms of the other variable and
correctly substituting
90
10
0≤x≤2
1≤y≤3
(b)
x ≥ 0, x ≤ 2, y ≥ 1, y ≤ 3 or
y = 5x – 3
oe
(a)
2
2
Accept <, ≤, > and ≥ throughout
(SC B2 y > 3, y < 1, x < 0, x > 2)
Total 4 marks
(Treat double−ended inequalities as two separate inequalities)
(B1 for 2 or 3 out of 4 inequalities correct)
B2 fully correct oe
y = 5x or y = 5x + a or y = bx – 3 or (L=) 5x – 3
If not B2 then B1 for
B2 fully correct equation eg y = 5x + −3 or y − −3 = 5(x – 0)
91
11
(8.3 × 103) × 50 (= 415 000 or 4.15 ×
105 ) or
(4.2 × 104) ÷ 50 (= 840 or 8.4 × 10² )
or
(4.2 × 104) ÷ (8.3 × 103) (= 5(.060...))
1.15 × 0.92 (= 1.058) oe or 105.8
(b)
(c)
e.g.
200 1.15  0.92
200
number or variable
n 1.15  0.92
where n is a
n
±(7.7 × 104 – 9.5 × 103) or
±(7.7 × 104 – 0.95 × 104) or
±(77 000 – 9 500) or
±67 500 oe
(a)
5.8
No
supported
by correct
comparabl
e figures in
the same
form
6.75 × 104
2
2
2
A1
NB. −5.8 (M1A0)
decrease of 5.8% (M1A0)
condone x × 1.15 × 0.92 oe
NO and 5(.060...)
M1
NO and 840 and 8 300 or
NO and 8.4 × 10²
A1 for NO and 415 000 and 42 000 or
NO and 4.15 × 105
A1 allow −6.75 × 104 allow ±6.8 × 104
M1 for a relevant calculation
M1 for clearly subtracting the correct values
92
12
16.7
(= 24.48686...)
sin 43
16.7
(=17.90855...)
tan 43
16.7
(= 24.48686...)
sin 43
 CD  
 ED  
 CD  
16.7 + 21.2 × 2 + ‘24.5’ + ‘17.9’
(= 101.495...)
and
or
16.7
(=17.90855...)
tan 43
 ED  
101
4
A1 accept 101 – 102
Total 4 marks
NB. Sine rule must be in the correct form to give the answer
M1 (dep on M2) complete method with no extra sides
 ED  '24.48....'2 16.72 (=17.90855...)
CD  16.72  '17.90....'2 (= 24.48686...)
use of Pythagoras theorem
NB. Sine rule may be used
M1 for a correct method to find both CD and ED or
(E is the point on line AD from where a vertical line is drawn
downwards from point C)
M1 for a correct method to find length CD or ED
Total 6 marks
93
13
(c)
(d)
(b)
(a)
e.g. readings of 11−13 and 22−24
indicated on horizontal axis
or 23 − 12
For correct use 20 and 60 (20.25
and 60.75) indicated (horizontal
line or mark) on the cumulative
frequency axis and their readings
taken from time taken axis
9 – 13
Correct cf
graph
17 − 19
7, 17, 29,
48, 66, 80
2
1
2
1
A1 accept 9 – 13
ft from a cumulative frequency graph dep on M1 in (b)
Total 6 marks
A1 accept curve or line segments
accept curve that is not joined to (0,0)
B1 ft from a cumulative frequency graph dep on M1 in (b)
M1 for a complete method to ft from a cumulative frequency
graph dep on M1 in (b)
for all 6 points plotted consistently within each interval in the
frequency table at the correct height
for at least 4 points plotted correctly at end of interval or
M1 ft from (a) if only one addition error
B1 cao
94
15
14
x
2  y2
3  y2
4
A1 accept x 
2  y 2
oe
y2  3
Total 4 marks
M1 for isolating terms in x and factorising the correct
expression of the equation
Total 3 marks
y2 + 2 = x(3 – y2) oe
M1 squaring both sides to get a correct equation
A1
1  6 
 1   6   5 
     or      (=   )
 3   9 
 3   9   12 
or
M1 for multiplying by the denominator and expanding the
bracket
3x  2
x 1
3
52  122
 6  1
 6   1   5 
     or      (= 
 ) or
 9   3 
 9   3   12 
If not M2 then M1 for
2
 5  122 or
xy2 + y2 = 3x – 2 oe
y2 
13
2
52   12 or
M2 for
95
16
 2  1
 2  1
= 4 2  4  4  8 oe
4 2 4 8 2  8
or
2 1
4 2 44 8
or
2 1
4 2  4  16  8
or
2 1
e,g,
4 8

2 1
8 + 6√2
3
A1 (dep on M2) or for stating a = 8 and b = 6
4 2  4  8 2  8
4 2  4  4  8
or
or
2  1
2  1
4 2  4  16  8
2  1
e.g.
Using −√2 − 1
Total 3 marks
condone missing brackets
M1 (dep) for expansion of numerator with at least 3 terms
correct oe
M1 for rationalising the denominator by multiplying numerator
and denominator by √2 + 1 (or −√2 − 1)
96
17
(b)
(a)
3
67.5h  '
20
' oe
h2
20h = k × h3 oe
y = kx3 or ky = x3
20x3
h2
1.5h
y
2
3
20x3
oe
h2
A1 accept
3
3h
h or
2
2
Total 5 marks
(a) or in part (b)
M1 ft, dep on at least M1 in part (a), complete method to find x
20
oe is seen in part
h2
20h = k × h3 oe
M2 for
Award 3 marks if answer is y = kx3 and k 
A1 for y 
M1
(NB. Not for y = x3)
Constant of proportionality
must be a symbol such as k
M1 substitution of x and y into
a correct formula
97
18
‘−10’[(x – ‘2.4’)2 –
‘2.4’2] oe
‘−10’ × − ‘2.4’2 or
‘−10’ × − ‘5.76’
(x = 2.4)
48 × ‘2.4’ – 10 ×
‘2.4’2
57.6
5
A1 accept 58 from correct working
Total 5 marks
finding max value of A from completing the square
or
for isolating x and substituting into A
completing the square
M1 ft if previous M1 awarded
or
M1 for differentiating a correct expression for A (allow 1 error)
and equating to zero
M1 for a complete expression for A (6 sides) with brackets
expanded
x2 + x2 + 48x − 12x2 (= 48x − 10x2)
‘48 – 20x’ = 0
M1 for finding an expression for the area of one face
x2 oe
or
x(12 − 3x) oe
98
19
 BC 
sin 95
'30.56 '
 sin 180  95  47 
sin 95
 BC  
250 + 0.5 × ‘30.56’ × ‘22.43’ ×
sin(180 – 95 – 47) (= 461.03....)
or
250 + 0.5 × ‘30.56’ × ‘18.88’ ×
sin(47) (= 461.03....)
(= 18.8(8524...))
(= 22.4(3407...)) or
'30.56 '
 sin 47
sin 95
'30.56 '
oe
sin 95
 AB  
sin 180  95  47 
or
sin 47

461
6
(AC=) 30.5(5579...) or 30.6
A1 accept 461 - 462
22.39 – 22.47 for AB
18.8 – 18.92 for BC
M1 for a complete method to find total area
sight of values in the ranges
Total 6 marks
M1 (dep on previous M marks) for a correct method to find a
missing length or
M1 dep on M1 for correct substitution into sine rule
If this mark is awarded then ft on the remaining M marks
A1
 AB   '30.56 ' oe
M1 for using the area formula correctly
250 = 0.5 × 26 × AC × sin(39) oe
99
20
y = 4.5 and y = 3
x = 4.5 and x = 6
2×2
e.g.
  15 2  15 2 
2   x        27

4   4  

e.g.
2
2

21   21  
2   x        54

4   4  

2×2
−(−15) ± √(−15)2 − 4×2×27
y =
−(−21) ± √(−21) − 4×2×54
2
allow 2y2 – 15y = −27 oe
e.g. (2y – 9)(y – 3) (= 0)
allow 2x2 – 21x = −54 oe
e.g. (2x – 9)(x – 6) (= 0)
x =
(9 – y)2 – 3y(9 – y) + 2y2 (=
0)
e.g. 6y2 – 45y + 81 (= 0)
or
2y2 – 15y + 27 (= 0)
x2 – 3x(9 – x) + 2(9 – x)2 (=
0)
e.g. 6x2 – 63x + 162 (= 0)
or
2x2 – 21x + 54 (= 0)
(4.5, 4.5)
and
(6, 3)
5
Total 5 marks
A1 (dep on M1) both x-values or both
y-values
A1 (dep on M1) oe
Must be paired correctly
21  441  432
)
4
allow ax2 + bx = c
M1 (dep on M1) for a complete
method to solve their 3-term quadratic
equation (allow one sign error and
some simplification – allow as far as
M1 substitution of linear equation into
quadratic
A1 (dep on M1) writing the correct
quadratic expression in form ax2 + bx +
c (= 0)
100
20
Alt
(9 – y – y)( 9 – y – 2y) (= 0)
(9 – 2y)(9 – 3y) (= 0) oe
y = 4.5 and y = 3
(x – (9 – x))(x – 2(9 – x)) (= 0)
(2x – 9)(3x – 18) (= 0) oe
x = 4.5 and x = 6
(x – y)(x – 2y) (= 0)
(4.5, 4.5)
and
(6, 3)
5
A1 (dep on M1) oe
Must be paired correctly
Total 5 marks
A1 (dep on M1) both x-values or both
y-values
M1 (dep on M1)
A1 (dep M1) substitution of L into their
factorised C
M1 for a method to factorise C
101
21
(= √29) or
3x 
 (= 33.854...) or
 ' 20 x ' 

(CAF =) tan 1 

3 
 (= 33.854...) or
 ' 20 ' 
 ' 20 ' 
(CAF =) cos 1 
 ' 29 '  (= 33.854...) or


 3 
(CAF =) sin 1 
 (= 33.854...)
 ' 29 ' 
(CAF =) tan 1 
e.g.
(AF =) √(√20) + (3)
2
33.9°
3

A1 answers in the range 33.85 – 33.9
(CAF =) tan 1 
 1.5 
 (= 33.854...)
' 5'
AB : BC : CF = 2 : 1 : 1.5
x can represent any number
Total 3 marks
M1 for a complete method to find angle CAF using length AC or
for a complete method to find angle CAF using length AF with or
without x or
AC 2  22  12  5

e.g.
AB : BC : CF = 2 : 1 : 1.5
2
2
(AF =) √(4)2 + (2) + (3) (= √29) or
2
x can represent any number
M1 for a method to find an expression for length AC or length AF
with or without x or
(AC =) √(4)2 + (2)2 (= √20) or
(AC =) √(4x)2 + (2x)2 (= √20x) or
e.g.
102
22
x(2x + 5)(3x − 1) or (2x + 5)(3x2 − x)
and (2x + 5)(2x – 5) oe
x(2x + 5)(3x − 1) or (2x + 5)(3x2 − x)
or (2x + 5)(2x – 5) oe
x  3x  1
2x  5
3
A1 accept
3x 2  x
oe Do not ISW
2x  5
Total 3 marks
denominator into 2 brackets where one of the factors must be
(2x + 5)
and
numerator into 2 or 3 factors where one of the factors must be
(2x + 5)
M1 for a correct factorisation of the
denominator into 2 brackets where one of the factors must be
(2x + 5)
or
numerator into 2 or 3 factors where one of the factors must be
(2x + 5)
M1 for a correct factorisation of the
103
23
3 2
t 3 t 4


or
t t 1
t
t 1
3 2

+
t t 1
t  3 t  4 23

=
t
t  1 35
3 t 3
t 3 3


or
t t 1
t
t 1
3 t 3

+
t t 1
t  3 3 12

=
or
t
t  1 35
3 t  3 12
2 
 oe
t t  1 35
2×2
2
−(−37)±√(−37) − 4×2×105
  37 2  37 2 
e.g. 2   t 
      105

4
  4  

e.g. t =
e.g. (2t – 7)(t – 15) = 0
allow 2t2 – 37t = −105
e.g. 2t2 – 37t + 105 (= 0) or
RR and GG method
RG and GR method
12
5
A1 (dep on A1) cao
7
)
2
Total 5 marks
Can be implied by answers of 15 (and
37  1369  840
) or
4
A1 (dep on M2) writing the correct
quadratic expression in form ax2 + bx + c
(= 0)
allow ax2 + bx = c
M1 (dep on A1) for a complete method to
solve the 3-term quadratic equation
(allow one sign error and some
simplification – allow as far as
M1 dep on M1 for a correct equation
M1 for one correct product
104
23
Alt
3
2

or
x3 x2
x
x 1

x3 x2
3
2

+
x3 x2
x
x  1 23

=
x  3 x  2 35
3
x

or
x3 x2
x
3

x3 x2
3
x

+
x3 x2
x
3
12

=
or
x  3 x  2 35
3
x
12
2


oe
x  3 x  2 35
2×2
2
−(−25)±√(−25) − 4×2×12
2
2

25   25  
e.g. 2   x 
      12

4
  4  

e.g. x =
e.g. (2x – 1)(x – 12) = 0
allow 2x2 – 25x = −12
e.g. 2x2 – 25x + 12 (= 0) or
RR and GG method
RG and GR method
12
5
)
A1 (dep on A1) cao
1
2
Total 5 marks
can be implied by answers of 12 (and
25  625  96
) or
4
A1 (dep on M2) writing the correct
quadratic expression in form ax2 + bx + c
(= 0)
allow ax2 + bx = c
M1 (dep on A1) for a complete method
to solve the 3-term quadratic equation
(allow one sign error and some
simplification – allow as far as
M1 dep on M1 for a correct equation
M1 for one correct product
105
(a)
(b)

2
2

2
 x  5 
y  41
oe
2
2
y  2  x  5   41 oe
9
2

y  2   x  5   52   or
2

y  2  x  5  5  9 or
e.g.
9

y  2  x 2  10 x  
2

y  2  x  10 x   9 or
2
Note: Allow candidates to swap x and y when finding the inverse
24
5
x  41
2
13
4
1
A1 oe
M1
M1 dep
Total 5 marks
B1
M1 for a correct equation for a first
step in order to complete the square
106
(b)
(a)
x  5
x
41  y
oe
2
5
x  41
2
4
A1 oe
M1
Total 5 marks
M1 dep
B1
20  400  8(9  y )
or
4
20  400  8(9  y )
x
4
1
M1 for a correct first step
13
2 x2 - 20 x + (9 - y) = 0
Note: Allow candidates to swap x and y when finding the inverse
24
Alt
107
(b)
(a)


2
 x  5 
2
y  41
oe
2
2  x  5   41  y (= 0)
9
2

2   x  5   52    y (= 0) or
2

2  x  5  52  9  y (= 0) or
2
5
x  41
2
4
A1 oe
M1
Total 5 marks
M1 dep
B1
e.g.
1
M1 for a correct first step
13
2 x2  20 x  (9  y)   0
Note: Allow candidates to swap x and y when finding the inverse
24
Alt
108
120
Mark Scheme (Results)
Jan 2020
January 2020
Pearson Edexcel International GCSE
In Mathematics A (4MA1)
Paper 2H
2
1
(b)
(a)
Q
Working
Answer
32.4 × 1003
eg 78 × 74 = 712 or 78 ÷ 73 = 75 or 75 × 74 or 74 ÷ 73
= 7 or 78 × 7 or 7’12’ ÷ 73 = 7’12’−3
32 400 000
79
x
7
Mark
Notes
2
2
1
M1
A1
A1
M1
B1
for 32.4 × 1003 oe
for 32 400 000 accept 3.24 × 107
Total 2 marks
Total 3 marks
for one correct step – must be
written as a power of 7
for 79
Apart from Questions 3, 7b, 12, 17, 20, 22 the correct answer, unless clearly obtained by an incorrect method, should be taken to
imply a correct method
International GCSE Maths
121
3
eg
14  5  19  3
70 57 70a 57a
 or

or
or
3 5
15 15 15a 15a
10 12
22
4 3
7
oe
15 15
15
70 57 127
7
22
7
 
 8 or 7  8
15 15 15
15
15
15
7 127
or if shows 8 
at the beginning then show that the
15 15
127
addition comes to
15
14
19
10
12
10a
12a
( )
or (4)
()(3)
or (4) ()(3)
3
5
15
15
15a
15a
Shown
3
for correct fractions with a common
denominator of 15 or a multiple of 15
dep on M2 for a correct answer from
fully correct working or shows that
127
and fully correct working
RHS =
15
127
shows LHS =
15
M1
A1
Total 3 marks
for correct improper fractions or
fractional part of numbers written
correctly over a common denominator
M1
122
5
4
5x = ‘120’ or ‘120’ ÷ 5
or 180 – 30 – 10 – 20 (=120)
24
Fully correct angle
bisector with all
relevant arcs
shown
e.g. 30 + 4x + 10 + x + 20 = 180 or 5x + 60 = 180 oe
30 + 4x + 10 + x + 20 (= 5x + 60) or 180 – 30 (=150)
4
2
A1
M1
M1
M1
B2
Fully correct angle bisector with all
arcs shown.
B1 for all arcs and no angle bisector
drawn or for a correct angle bisector
within guidelines but not arcs or
insufficient arcs
Total 2 marks
Allow 5x + 60 = n
M2 for
where n ≠ 180 or for
5x + 30 = 150
subtracting 30 from 180 oe
for setting up the
equation or for
subtracting all
numerical values of
angles from 180
for correctly simplifying to ax = b or for
dividing ‘120’ by 5
for 24
Total 4 marks
123
6
‘0.45’ ÷ 3 (= 0.15)
Or
‘0.45’ × 180 (= 81)
Or
180 − 99 (= 81)
‘0.15’ × 180
Or
‘81’ ÷ 3
1 – (0.24 + 0.31) (= 0.45)
Or
(0.24 + 0.31) × 180 (= 99)
27
4
A1
M1
M1
M1
or for an answer of
Total 4 marks
27
180
or for a correct equation for
missing values eg
x + 0.24 + 2x + 0.31 = 1 oe
(can be implied by 2 probabilities
that total 0.45 in table if not
contradicted in working space)
(or 0.15 correctly placed in table
as long as not contradicted)
124
7
(b)
(x − 8)(x + 5)
(x ± 8)(x ± 5)
2
3  9  160
2
3  169
3  13
or
2
2
or
(3)  (3)  4  1 (40)
2 1
(a) 2x ˃ 4 – 7 or x + 3.5 > 2
8, −5
x ˃ −1.5
3
2
A1
M1
3  169
3  13
or
2
2
dep on at least M1 for correct values
Total 5 marks
M1 For a correct first step allow 2x = 4 – 7 or x + 3.5 = 2
or an answer of x = −1.5 or x < −1.5 or −1.5
A1 for x ˃ −1.5 oe
M1 or (x + a)(x + b) where ab = −40 or a + b = −3
OR correct substitution into quadratic formula
(condone one sign error in a, b or c and missing
brackets)
(if + rather than ± shown then award M1 only unless
recovered with answers)
125
45
47
100( 9) and
100 ( 8.6)
500
545
(a) 545 – 500 ( = 45) or 592 – 545 ( = 47)
45
47
100 ( 9) or
100 ( 8.6)
500
545
0.15 × “1120” or “1120” – 952 oe
(b) 952 ÷ 85 × 100 oe (=1120)
Alternative mark scheme for 8(a)
545
545
100(  109) or
( 1.09) and
500
500
592
592
100 (  108.6) or
  1.086 
545
545
8
168
No, 109(%)
and 108.6(%)
No, 9(%) and
8.6(%)
3
4
4
oe eg no and 1.09 and 1.086
M1 for a method to find price
before discount
M1 for a correct method to find
discount
A1
A1
952
15
85
Total 7 marks
M2 for
(if not M3 then award M2 for one of these
expressions)
M3 for both correct expressions which should lead to
109 or 1.09 and 108.6 or1.086
(allow 108 or 108.7 from correct working for 108.6
or 1.08 or 1.087 from correct working for 1.086
throughout)
M1 for both correct expressions or having found “9%”
finds 109% of 545: 1.09 × 545(=594.05) or 9% of
545 (49.05) or having found “8.6%” finds 108.6% of
500: 1.086 × 500(=543) or 8.6% of 500 (43)
A1 for no oe, 9% and 8.6% seen or
no oe and 9% and 594.05 or 8.6% and 543 or
No, 49.05 > 45 or No 594.05 > 592 oe
M1 may be seen as part of a calculation
M1 for one correct expression (allow 8 or 8.7 from a
correct expression for 8.6 throughout)
126
50 
60  60
5
oe eg 50 ÷
1000
18
1000 ÷ 60 ÷ 60 (= 0.27777…. or
5
)
18
50 × 60 (= 3000) or 50 ÷ 1000 (= 0.05 or
10
or
50 × 60 × 60 (= 180 000) or
or
60  60
( 3.6)
1000
or
19.3 × 150
9
1
)
20
180
2895
3
2
A1
M1
M1
M1
A1
Total 2 marks
for 180
(SCB1 for both conversion factors
correct but applying them wrongly
50 1000
)
eg
60  60
Total 3 marks
(dep) for a complete method
1000 ÷ 60 ÷ 60
60  60
( 3.6)
1000
or
or
for 50 with at least one of ÷ 1000
or × 60
for 19.3 × 150
for 2895
127
2
( 4 12.566...)
‘12.566…’+ 15 + 17
2
  '8'
(AC =) 17  15 ( 64  8)
2
(AC2 =) 172 − 152
A1
M1
M1
“12.566” + 15 + 17
5
Total 5 marks
dep on M2 for
  '8'
or 12.5663… or 4π
2
for “12.566” + 15 + 17 and no
additional values
for awrt 44.6
Total 5 marks
for a correct method to find one of
the angles
additional values
for awrt 44.6
oe or 4π
A1
2
  '8'
for ‘12.566’ + 15 + 17 and no
12.5663…
dep on M2 for
M1
M1
M1
M1
M1
M1
44.6
44.6
5
15 × tan (‘28.0724’)(= 8) or 15 ÷ tan (‘61.9275’)(= 8)
  '8'
(= 4π = 12.566…)
2
Alternative mark scheme for 11
 15 
 15 
cos 1   (= 28.0724) or sin 1   (= 61.9275)
 17 
 17 
11
128
12
Cost per litre then conversion
2500000
  5.952..
4.2 105
770 000
(0.895..)
8.6 105
‘5.952’ ÷ 6.57(=0.9059..) or ‘0.895’ × 6.57(= 5.882..)
8.6  105
8.6  105
( 1.1168) or
( 0.1699..)
770 000
'5058900 '
Conversion then litres per amount of money
2500 000
(380517.5..) or 770 000 × 6.57(= 505 8900)
6.57
4.2  105
4.2  105
( 0.168) or
( 1.103..)
2500 000
'380517.5'
4.2  105
( 0.168) l/k
2500 000
A: l/$ to l/k ‘1.1168’ ÷ 6.57 (= 0.1699..)or
D: l/k to l/$ ‘0.168’ × 6.57 (= 1.103..)
Litres per amount of money and then conversion
8.6  105
( 1.1168) l/$
770 000
Arctic Oil and
relevant figures
Arctic Oil and
relevant figures
Number of litres per Krone for A
l/$ to l/k for A or l/k to l/$ for D
M1
M1
Litres per Krone or litres per $ for D
Litres per Krone or litres per $ for A
M1
M1
Price per litre in $ for A
M1
M1 Conversion of Krone to $ or $ to Krone
Price per litre in Krone for D
M1
A1 for Arctic Oil with 1.1168… and
1.10376… or 0.168 and 0.1699..
Changing Krone to $ or $ to Krone
M1
A1 for Arctic Oil with 1.1168… and
1.10376… or 0.168 and 0.1699..
Number of litres per $ for D
M1
129
4.2 105
21
(
 0.488..)
5
8.6  10
43
‘0.488..’ × 770 000 $
(=376046.511..)$
‘376046.511..’× 6.57
=2470625.58..K or
2500 000÷6.57 = 380517..$
Arctic Oil and
relevant figures
For Arctic Oil with 5.952 and 5.882 or
0.895 and 0.9059
For Arctic Oil with 5.952 and 5.882 or
0.895 and 0.9059
M1 Multiplier for same amount of D as A or
same amount of A as D
A1
M1 Cost per litre in $ or cost per litre in
Krone for A
M1 Cost per litre in Krone or cost per litre in
$ for D
M1 Changing Krone to $ or $ to Krone
A1
M1 Cost of equal amount of D as A or
A as D
M1 Converts so can compare costs – either
K to $ or original A to K or
$ to K or original D to $
Arctic Oil and
A1 Arctic Oil and 779154.. or with
relevant figures
2470625..(figures may be rounded)
Or
Arctic Oil with 5119047… and 5058900
or with 376046.. and 380517
Students may compare other equal amounts – please use the scheme that best fits their method and award marks appropriately.
Total 4 marks
Comparing equal amounts
8.6  105
43
(
 2.047..)
5
4.2 10
21
‘2.047..’×2500 000 K
(=5119047.619..)K
‘5119047.619’÷6.57
= 779154.88…$ or
770 000×6.57=5058900 K
Conversion then cost per litre
2500 000
(380517.5..) or 770 000 × 6.57(= 505 8900)
6.57
2500000
'380517.5..'
( 5.952) or
(0.9059..)
5
4.2 10
4.2 105
770 000
'5058900 '
( 0.895) or
( 5.882..)
5
8.6 10
8.6 105
Arctic Oil and
relevant figures
130
13
Angles in the same segment are equal,
angle in a semicircle is 90° (or angle at centre
is double angle at circumference oe)
angles in a triangle add up to 180°/angles in a
triangle
isosceles triangle
alternate angles
vertically opposite angles (or vertically
opposite)
angles at a point
opposite angles in a cyclic quadrilateral
angle between tangent and radius (diameter)
alternate segment theorem
angles subtended by the same arc(or chord)
at the circumference (or on the circle)
Angle CAD = 28° or angle ACB = 32° or
angle ACD = 90° or angle ABD = 90°
30°
4
A1
B2
M1
Total 4 marks
For a correct answer of 30
Dep on M1 for all correct reasons
for their method used
(if not B2 then award B1(dep on
M1) for a correct circle theorem
reason)
131
15
14
(b)
(a)
7 6
 oe only
20 19
21
190
C, B, E
21
190
Correct probabilities
on the tree diagram
3
2
2
B3
(B2
(B1
A1
M1
B1
B1
Total 3 marks
Total 4 marks
21
oe or 0.11… (at least 2
190
for all 3 correct
for 2 correct)
for 1 correct)
dp)
for
13
7
and
on the first branch
20
20
(0.65 and 0.35)
12 7 13
6
, , and on the
for
19 19 19
19
second branch
(accept 2 dp or better 0.6315...,
0.3684..., 0.6842..., 0.3157...)
ft from (a) as long as probabilities
less than 1
for
132
17
16
4 y2 1
y2 1
e.g. n2 – n2 + 2n – 1 or n2 + 2n + 1 − n2
e.g. n2 – (n – 1)2 or (n + 1)2 – n2
e.g. 2n – 1 is always
odd
3
A1
M1
M1
A1
SCB1 for eg (2n)² − (2n – 1)² or
(2n + 1)² − (2n)² oe
Total 3 marks
dep on M2 for eg 2n – 1 or 2n + 1
or – (2n + 1) oe and a suitable
conclusion
Correct expansion of brackets and
correct signs or a correct result
for setting up a correct algebraic
expression (any letter can be used)
for expanding the bracket and
rearranging for x so that the terms
in x are on one side of the correct
equation
4 y2 1
4 y 2  1
for x  2
or x 
y 1
1 y2
(need to see x = somewhere)
Total 4 marks
M1
y 2 x  x  4 y 2  1 or 4 y 2  1  x  y 2 x or
x
for removing the fraction
M1
x( y 2  1)  4 y 2  1 or 4 y 2  1  x(1  y 2 )
for squaring
M1
4
x 1
x4
2
y ( x  4)  x  1 or y 2 x  4 y 2  x  1
y2 
133
18
(b)
or all correct values in bars oe not added
(1 × 7) + (2.5 × 6) + (5 × 4.8) = 7 + 15 + 24 (= 46)
or
no. of sml squares
(48 × 5) + (6 × 25) + (7 × 10) = 240 + 150 + 70 (=460)
46
120
no. of sml squares =
(10 × 7) + (5 × 34) + (9 × 10) + (6 × 25) + (15 × 48)
= 70 + 170 + 90 + 150 + 720 (= 1200)
(a) (0.7 × 10) + (3.4 × 5) + (1 × 9) + (2.5 × 6) + (4.8 × 15)
= 7 + 17 + 9 + 15 + 72 (= 120)
9
15
46
120
2
3
M1 for a correct method to work out the area between 17
minutes and 35 minutes
eg using frequency density or number of small squares
oe
A1
46
for
oe (allow 2 dp or better 0.3833... or 38% or
120
better)
M1
n
q
where n ˂ 15 or
where q < 720 or
for
15
720
r
where r < 72 or
72
9
432
where m > 9 or
where p > 432
m
p
43.2
where t > 43.2
t
A1 9
oe
15
Total 5 marks
M1 for a correct method to work out the total area eg total
frequency or number of small squares or other correct
method (allow one error in method) [count use of 25
for 24 as one error]
or all correct values in bars oe not added
134
19
(b)
(a)
2  1
1  2
3
or m 
or − or −0.6
2  3
3  2
5
5
mp 
3
5
4  (6)  c oe eg 4 = −10 + c (c = 14)
3
5
y  4  ( x  6)
3
m
4x + k (oe)
Eg
5x – 3y + 42 = 0
y=
M1
4
numerical value k ≠ 7
Could be written in another form e.g. 3y +
12x = 20
for using 𝑚
for y = 4x or y = 4x + k where k is any
M1ft dep on previous M1 for substituting
(−6, 4) into linear equation formula
5
4  x  c to find value of c or
3
5
y  x  14 or y  1.66...x  14
3
A1 for correct simplified equation where all
values are integers
10x – 6y + 84 = 0 or
3y = 5x + 42 oe
Total 5 marks
M1ft for using m1  m2  1
B1
1
135
21
20
(a)(i)
(iii)
(b)
18( 7  1)
7 1
eg x² + 3x + 10 oe or
3
9
eg ( x  ) 2   4  6 oe
2
4
eg y – 6 = x² + 3x + 4
eg (x – 4)² + 3(x – 4) + 4 oe or
3
9
eg ( x   4) 2   4 oe or
2
4
3 7 3
eg
18
7 1

7 1
7 1
y = (x – 4)² + 3(x – 4) + 10
or
3
9
y  ( x   4) 2   4  6
2
4
(0, 6)
(2, 6)
3 7 3
2
2
3
A1
B1
B1
M1
A1
M1
M1
Total 4 marks
5
31
or
oe eg y = ( x  )2 
2
4
y = x² − 5x + 14 oe
for applying one of the
transformations to the equation
Total 3 marks
18
7 1

7 1
7 1
Dep on M1 for a correct
numerator and multiplying out the
denominator to 7 – 1 or 6
Dep on M2
Allow 3( 7 -1)
for
136
x
2
2
1  1  (4  1 12)
or
2 1
2
2x² + 2x – 24 (=0) or x² + x −12 (=0)
or 2x² + 2x =24 or x² + x =12
(x + 4)(x − 3) (= 0) or
x2  ( x  2)2  2( x  2)  24
y = 5 and y = −2
(−4, −2) and (3, 5)
y
  3  ( 3) 2  (4  1 10)
or
2 1
2
2
3 3

 y       10  0
2 2

2y² − 6y – 20 (=0) or y² − 3y – 10 (=0)
2y² − 6y = 20 or y² − 3y = 10
( y  5)( y  2)  0 or
1 1

 x       12  0
2 2

x = −4 and x = 3
(−4, −2) and (3, 5)
Alternative mark scheme for 22
( y  2)2  y 2  2 y  24
22
(−4, −2) and (3, 5)
(−4, −2) and (3, 5)
5
5
A1
A1
M1ft
A1
M1
A1
A1
M1ft
A1
M1
which expanded give 2 out of 3 terms correct
for both y values dep on M1
for both solutions dep on M1
Total 5 marks
3  9  40
) or if factorising, allow brackets
2
for substituting linear equation into the
quadratic equation
for a correct equation in the form
ay2 + by + c = 0 or ay2 + by = −c
dep on M1 for solving their quadratic equation
using any correct method (allow one sign error
and some simplification – allow as far as
for substituting linear equation into the
quadratic equation
for a correct equation in the form
ax2 + bx + c = 0 or ax2 + bx = −c
dep on M1 for solving their quadratic equation
using any correct method (allow one sign error
and some simplification – allow as far as
1  1  48
) or if factorising, allow brackets
2
which expanded give 2 out of 3 terms correct)
for both x values dep on M1
for both solutions dep on M1
137
23
uuuur
3
3
1
1 
 1
PM =  a  b  4a  (2b  4a)   a  b 
2
4
2
4 
 2
uuuur
1
AM  4a  (2b  4a) ( 2a  b)
2
uuuur
1
AM  2b  (4a  2b) ( 2a  b)
2
uuur 1
MA  (2b  4a)  2b ( 2a  b)
2
uuur 1
MA  (4a  2b)  4a ( 2a  b)
2
3
3
1
1
( AP : PM ) a  b : a  b oe
2
4
2
4
3
3
( AP : AM ) a  b : 2a  b ( 3 : 4) oe
2
4
1
1
( AM : PM ) 2a  b : a  b (= 4 : 1) oe
2
4
3
3
1
1
AP  3PM oe eg a  b  3( a  b) oe
2
4
2
4
4
AM  AP oe
3
AM  4PM oe
3:1
3
Total 3 marks
For use of a correct ratio or fraction
linking
AP and PM or
AP and AM or
AM and PM
(in either order)
vectors must be in form pa + qb
M1
A1
uuuur uuuur
uuur
for finding PM or AM or MA
M1
138
24
3x  1
x  2 x  3 3  2 x 
e.g.
A1
2
or
isw for incorrect denominator
expansion
Total 4 marks
2
3x  1
oe
12 x  9 x  4 x 3 
3x  1
or
x 12 x  9  4 x 2 
x  2 x  3
1  3x
3x  1
or
x  2 x  3 3  2 x 
Writing 1st fraction as a fraction
over a common denominator (can
be 2 separate fractions)
Complete factorisation of
numerator or denominator of 2nd
fraction
may be partially simplified
M1
M1
4
 3x  1 2 x  5 
2x  3

 2 x  5  2 x  3 x  3  2 x  3  2 x 
8 x  12  6 x  15
oe
 2 x  5 2 x  3
M1
or
x  3  2 x  3  2 x  or  3x  1 2 x  5
 2 x  5  2 x  3
4  2 x  3  3  2 x  5 
139
25
50
 2  50   50  1  k  oe
2 
33125  25 100  49k  oe
1325 = 100 + 49k oe
1225 = 49k oe
33125 
n = 50
25
3
A1
B1
M1
Total 3 marks
(k may be written as d)
For correct equation, using
formula with a = 50 and n = 50
substituted
(for this mark, allow n = 49)
140
26

1600
or
  "7.817..." "26.193..."( 643.315...)
x
"49.1196..."  2    "26.193..."
360
or
x
"643.315..."    "26.193..."2 
360
or
2 × π × “7.817…” (= 49.1196…)
l  "7.817..."2  252 (  686.1154...  26.193...)
1   25
3
192
r
( 61.1(154..)  7.8176...)
eg r 
1
1600     r 2  25 oe
3
107°
6
M1 Dep on M2 correct method to find slant
height of cone (radius of sector)
M1 for using 𝐶 2𝜋𝑟 oe using figures from
correct method
or
for using A   rl using figures from
correct method
M1
x
for using arc length = 2 r 
360
or
for using area of sector =
x
 r2 
360
A1 for 107° - 108°
Total 6 marks
M1 dep for correct rearrangement of volume
formula for r
M1 for substituting into volume formula for
cone correctly and equating to 1600
141