Mark Scheme

Mock Set 4

Pearson Edexcel GCSE (9 – 1)

In Mathematics (1MA1)
Higher (Calculator) Paper 2H
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Summer 2018
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© Pearson Education Ltd 2018
General marking guidance
These notes offer general guidance, but the specific notes for examiners appertaining to individual questions take precedence.

1 All candidates must receive the same treatment. Examiners must mark the last candidate in exactly the same way as they mark the
first.

Where some judgement is required, mark schemes will provide the principles by which marks will be awarded;
exemplification/indicative content will not be exhaustive. When examiners are in doubt regarding the application of the mark scheme to
a candidate’s response, the response should be sent to review.

2 All the marks on the mark scheme are designed to be awarded; mark schemes should be applied positively. Examiners should also be
prepared to award zero marks if the candidate’s response is not worthy of credit according to the mark scheme. If there is a wrong
answer (or no answer) indicated on the answer line always check the working in the body of the script (and on any diagrams), and
award any marks appropriate from the mark scheme.

Questions where working is not required: In general, the correct answer should be given full marks.
Questions that specifically require working: In general, candidates who do not show working on this type of question will get no
marks – full details will be given in the mark scheme for each individual question.

3 Crossed out work

This should be marked unless the candidate has replaced it with
an alternative response.

4 Choice of method
If there is a choice of methods shown, mark the method that leads to the answer given on the answer line.

If no answer appears on the answer line, mark both methods then award the lower number of marks.

5 Incorrect method
If it is clear from the working that the “correct” answer has been obtained from incorrect working, award 0 marks. Send the response
to review for your Team Leader to check.

6 Follow through marks

Follow through marks which involve a single stage calculation can be awarded without working as you can check the answer, but if
ambiguous do not award.
Follow through marks which involve more than one stage of calculation can only be awarded on sight of the relevant working, even if it
appears obvious that there is only one way you could get the answer given.
7 Ignoring subsequent work
It is appropriate to ignore subsequent work when the additional work does not change the answer in a way that is inappropriate for the
question or its context. (eg. an incorrectly cancelled fraction when the unsimplified fraction would gain full marks).
It is not appropriate to ignore subsequent work when the additional work essentially makes the answer incorrect (eg. incorrect
algebraic simplification).

8 Probability
Probability answers must be given as a fraction, percentage or decimal. If a candidate gives a decimal equivalent to a probability, this
should be written to at least 2 decimal places (unless tenths).
Incorrect notation should lose the accuracy marks, but be awarded any implied method marks.
If a probability fraction is given then cancelled incorrectly, ignore the incorrectly cancelled answer.

9 Linear equations
Unless indicated otherwise in the mark scheme, full marks can be gained if the solution alone is given on the answer line, or otherwise
unambiguously identified in working (without contradiction elsewhere). Where the correct solution only is shown substituted, but not
identified as the solution, the accuracy mark is lost but any method marks can be awarded (embedded answers).

10 Range of answers
Unless otherwise stated, when an answer is given as a range (e.g 3.5 – 4.2) then this is inclusive of the end points (e.g 3.5, 4.2) and
all numbers within the range.

11 Number in brackets after a calculation

Where there is a number in brackets after a calculation E.g. 2 × 6 (=12) then the mark can be awarded either for the correct method,
implied by the calculation or for the correct answer to the calculation.

12 Use of inverted commas

Some numbers in the mark scheme will appear inside inverted commas E.g. “12” × 50 ; the number in inverted commas cannot be any
number – it must come from a correct method or process but the candidate may make an arithmetic error in their working.

13 Word in square brackets

Where a word is used in square brackets E.g. [area] × 1.5 : the value used for [area] does not have to come from a correct method or
process but is the value that the candidate believes is the area. If there are any constraints on the value that can be used, details will
be given in the mark scheme.

14 Misread
If a candidate misreads a number from the question. Eg. uses 252 instead of 255; method or process marks may be awarded provided
the question has not been simplified. Examiners should send any instance of a suspected misread to review.
Guidance on the use of abbreviations within this mark scheme

M method mark awarded for a correct method or partial method

P process mark awarded for a correct process as part of a problem solving question

A accuracy mark (awarded after a correct method or process; if no method or

process is seen then full marks for the question are implied but see individual
mark schemes for more details)

C communication mark awarded for a fully correct statement(s)

with no contradiction or ambiguity

B unconditional accuracy mark (no method needed)

oe or equivalent

cao correct answer only

ft follow through (when appropriate as per mark scheme)

sc special case

dep dependent (on a previous mark)

indep independent

awrt answer which rounds to

isw ignore subsequent working

Paper: 1MA1/2H
Question Answer Mark Mark scheme Additional guidance
1 3360, 4200 M1 for a complete method to find one part of the ratio
eg 7560 ÷ 9 × 4 (=3360) or 7560 ÷ 9 × 5 (=4200)

A1 3360, 4200 If numbers are not in the correct order award

M1 only
2 (a)(i) 1, 3, 5, 15 B1 cao Accept numbers in any order
(ii) 3 B1 cao
(iii) 1, 3, 5, 9, 15, 21 B1 cao Accept numbers in any order

(b) 8 M1 for
#
where a > 8 or
%
where b < 11
$ &&
11
C1 # Accept the equivalents of 0.72, 0.727…, 72%
oe
&&
or 72.7(2727…)%
3 14.5 to 14.53 M1 for substituting into Pythagoras’ theorem
eg 16² = 6.7² + a² or 16² − 6.7²

M1 for a complete method to find the unknown length Method must show they understand to square;
eg 256 44.89 16 × 16 6.7 × 6.7 is sufficient

A1 Answers in the range 14.5 to 14.53

4 (a) 104 000 B1 cao

(b) 6 × 10-2 A1 cao

(c) 0.65 M1 (300.3 × 106) ÷ ( 4.62 × 108) (=0.65) oe

A1 cao Accept 6.5 ´ 10-1

Paper: 1MA1/2H
Question Answer Mark Mark scheme Additional guidance
5 15488 P1 for showing a process to depreciate by 20% Could be shown in several stages
eg 0.8 × 25000 (=20000) oe eg 25000 - (25000 ´20 ÷ 100)

P1 for showing a process to depreciate by 12% as a Or alternative methods

second stage eg “20000” × 0.88 (= 17600) eg “20000” - (20000 ´ (100 - 12) ÷ 100)

A1 cao

6 7.5 P1 for using p × 4² ÷ 2 (=25.13..)

P1 for finding the area of the trapezium “25.13” must come from a correct method
eg 64 −“25.13” (=38.86..) involving p.

P1 for finding the sum of AB and DC

eg “38.86..” ÷ 5 × 2 (=15.54..)

P1 for complete process to find the missing length

eg “15.54..” − 8 (=7.546..)

A1 7.5 to 7.6 Accept answers in the range 7.5 to 7.6

If an answer is given in the range but then
incorrectly rounded award full marks.
7 4.5 P1 for starting to use inverse proportionality
eg 3 × 4 × y = 4200
or 4200 ÷ (3 × 4) (=350)

P1 for a complete method eg 3150 ÷ (“350” × 2) (=4.5)

A1 4.5 oe 1
Accept 4 or 4 hours 30 minutes
2
Paper: 1MA1/2H
Question Answer Mark Mark scheme Additional guidance
3
8 12000 P1 for 12 180.9 = 1.005 × P
or 12180.9 ÷ 1.005 (=12 120.29..)

P1 for 12 180.9 ÷ “1.0053” (=12 000)

A1 cao Do not award this mark unless working is

shown leading to the correct answer.
9 20 P1 for showing a process to find the interior angle of the Award if 108 is shown as an interior angle on
pentagon eg (5 − 2) ÷ 5 × 180 (=108) the diagram (eg at SPT) but not contradicted
by other working or placed incorrectly
P1 for finding the interior angle of the unknown polygon Award if 162 is shown as an interior angle on
eg 360 − 90− “108” (=162) the diagram (eg at RPT) but not contradicted
by other working or placed incorrectly
P1 for a complete method to find n
eg 360 ÷ (180 − “162”) (=20)

A1 cao Do not award this mark unless working is

shown leading to the correct answer.
10 3 8 M1 for a method to eliminate one variable (condone one arithmetic error)
,
5 5
A1 3 8 Accept equivalent forms of these answers.
for either x = or y = oe
5 5

M1 (dep) for substituting found value in one of the (condone one arithmetic error)
equations
or appropriate method after starting again
A1 3 8 Accept equivalent forms of these answers.
for x = and y = oe
5 5
Paper: 1MA1/2H
Question Answer Mark Mark scheme Additional guidance
11 (a)(i) box plot B1 for a fully correct box plot

(ii) 20,170,200 B1 smallest value 20, lower quartile 170 and median 200 Shown in the table or otherwise associated
with the correct values.
(b) Statements C2 for two comments one about median and one about “in context” means some reference to number
IQR; one must be in context. of lorries

(C1 for one comment about median or about IQR)

12 25.5 P1 for process to find DH eg 15 ÷ DH = sin 64 all 3 elements of sin64, DH and 15 must be
present in a correct equation
P1 for process to find AD eg AD ÷ 15 = tan 28 all 3 elements of tan28, AD and 15 must be
present in a correct equation
P1 (dep on P1) for a full method to find angle AHD eg
tan-1 (“7.9756” ÷ “16.68902911”)

A1 Answer in the range 25.5 to 25.6 If an answer is given in the range but then
incorrectly rounded award full marks.
13 (a) −4 B1 cao

(b) 2r - 3t M1 for ./0 or .12

A1 2r - 3t oe Accept answer written as u2r-3t

3 6 6
(c) 16 2 or 8 8 M1 for 8 4 or 22.62... or n² = 8³ or 53 = 84

A1 16 2 or 8 8 Accept the answer written as 512

Paper: 1MA1/2H
Question Answer Mark Mark scheme Additional guidance
14 18 P1 for starting the process eg using
1
and
:
9 &&
77 or finding the lowest multiple

P1 1 :
for a complete process eg ×
9 &&

A1 &# Accept equivalent fractions, 0.23….. or

oe
99
23.(….)%
15 232 M1 14 × 8 (=112) or 10 × 12 (=120)

C1 14 × 8 + 10 × 12 oe

16 (a) Graph M1 Sketches ; = =² An approximate sketch of the graph is all that

is needed; no other information required.

M1 for a partially correct reflection of ; = =² in the line The line y = x does not need to be shown.
;== If only part of y=x2 has been drawn and
therefore only part of the reflection shown,
then award M0 M1.

A1 for ; = =² and ; / = = sketched. Curves must pass through the origin.

The label C is not needed if the intention is
clear.
(b) A,D,G,E B3 for all 4 correct

(B2 for 3 correct)

(B1 for 2 correct)

Paper: 1MA1/2H
Question Answer Mark Mark scheme Additional guidance
17 201468 or P1 for substitution eg 200720 = 1.04 (200000 - G)
201469 or 200720 ÷ 1.04 (=193000)

P1 (dep P1) for process to isolate G

eg G =200000 – (200720 ÷ 1.04) =7000

P1 (dep P2) for a full process to find P17 using “7000”

A1 201468 or 201469

18 & M1 for a first step to solve the quadratic e.g. factorisation:

−1 < x < 1
/
2= + 3 (= − 1)
B&± &4 BD×/×(B1)
or using the formula
/×/

A1 &
for −1 / and 1
Sketch drawn C2 1
Solution set drawn for 1 < x <1
2
(C1 for a correct solution set drawn for two values (not Any attempt must at least show a circle at two
the correct solutions) or an attempt to draw the values, and some attempt to add lines.
1
correct solution set for 1 < x < 1 with some errors.
2
19 /FG1H M1 for JK = b − a or KJ = a – b
I
or the correct use of the ratio

M1 1
for a complete method eg (b−a)+a
I

A1 /FG1%
oe
I
Paper: 1MA1/2H
Question Answer Mark Mark scheme Additional guidance
20 (a) Show M1 for a method to show ff(x) as an unsimplified fraction Note M marks can be awarded in either order
6LM
&B
6NM
eg ff(x) = 6LM
&G
6NM

M1 for a full method to write either the numerator or

&GOB(&BO)
denominator as a single fraction eg
&GOG(&BO)

C1 for a complete method with correct working

(b) &BO B1 &BO

f-1(x) = for f-1(x) =
&GO &GO

21 1 P1 for showing for showing Assuming x =P(R) and y =P(Y)

&1
3 x² + y² = oe x
2
y
2
13
1: + = oe
n n 36
P1 for setting up an equation for using x² + y² = 13
&
in x eg (1 − x)x = oe
D
P1 for a process to solve the for showing
equation in x y y 1
eg 1 = oe
n n 4
P1 for a full method to find P(Y)

A1 &
from correct working 1
1 An answer of alone is insufficient for any
3
marks; it must be supported by working
shown.
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