Notesale: Turn your study into money

Document Preview

Extracts from the notes are below, to see the PDF you'll receive please use the links above

---

MATHEMATICS

OCR 2024
GCSE Mathematics
J560/05 Paper 5 (Higher Tier)
With Marking Scheme Merged

Oxford Cambridge and RSA

Monday 3 June 2024 – Morning
GCSE (9–1) Mathematics
J560/05 Paper 5 (Higher Tier)
Time allowed: 1 hour 30 minutes
You must have:
• the Formulae Sheet for Higher Tier (inside
this document)

H

You can use:
• geometrical instruments
• tracing paper
Do not use:
• a calculator
*

J

5

6

0

0

5

*

Please write clearly in black ink
...


Centre number

Candidate number

First name(s)
Last name
INSTRUCTIONS
• Use black ink
...

• Write your answer to each question in the space provided
...
The question numbers must be clearly shown
...

• Where appropriate, your answer should be supported with working
...

INFORMATION
• The total mark for this paper is 100
...

• This document has 20 pages
...


2
1

Work out
...
2 ' 0
...
[2]

2

Kai has these four number cards
...

(a) Complete the table to show all of the possible differences
...


(b)
...

1

of the journey is on roads with a speed limit of 40 miles per hour
...


Ryan leaves home at 08 50 and does not exceed the speed limits on the journey
...

You must show your working
...
[6]
(b) Write down an assumption you have made when working out the answer to part (a)
...


...


60

50

40

Mathematics

30

20

10

00

10

20

30

40

50

60

Science
(a) Describe the type of correlation shown in the scatter diagram
...
[1]
(b) One pupil took the Science test but was then ill during the Mathematics test and had to leave
early
...


[1]

(c) By drawing a line of best fit, estimate the test score in Mathematics for a pupil who scored
28 in the Science test
...
[2]
(d) Explain why using the scatter diagram to estimate the test score in Mathematics for a pupil
who scored 60 in Science may be unreliable
...


...


(e)
...


A

C

B

Using a ruler and compasses only, construct and shade the region which is closer to AB than
AC
...
The density of bronze is 8
...


3

By rounding each value correct to one significant figure, work out an estimate for the mass of the
bronze ornament
...
g [3]

7

3

A bottle contains 1 4 litres of cordial
...

Cups that can hold

1
6 of a litre are completely filled with this orange squash
...
You must show your working
...
[6]
© OCR 2024

7
8

(a) y is directly proportional to x
...


(a)
...

Write down the percentage decrease in z when x is increased by 100%
...

9

% [1]

The following kinematics formulas may be used in this question
...

2
The particle accelerates uniformly at 3 m/s for 4 seconds
...



...


Not to scale
a cm

14 cm

(a + 5) cm

The equilateral triangle has the same perimeter as the rectangle
...


a =
...


2b

a

(a) Write vector a as a column vector
...


JK
(b)

NO

K

O

K

O

L

P

[1]

(c) On the grid below, draw the vector a – b
...

Sasha says
After one year, my investment will get £50 in interest and will be worth £1050
...

Is Sasha correct?
Give a reason for your answer
...
because
...
[1]

(b) Sasha buys a car
...


The graph shows some information about the value of the car
...


(b) a =
...
[4]
© OCR 2024

11
13 Solve the inequality
...



...

2
5

4
9

6
13

8
17

Find the nth term of the sequence
...
[3]
© OCR 2024

Turn over

12
15 Expand and simplify
...
[3]

16 Two prisms, A and B, are mathematically similar
...
The height of prism A is 6 cm
...



...


P

P = 1
...


Q is the midpoint of PR
...

Give your answer as a mixed number in its simplest form
...



...

A cone has radius R cm and height 2R cm
...

Write R in terms of x
...


The volume V of a cone with radius r and height h is V =

1 2
3 rr h
...
[4]

© OCR 2024

15
19 (a) Describe fully the single transformation that is equivalent to:
•
•

a rotation of 20° anticlockwise about the origin, followed by
a rotation of 70° clockwise about the origin
...


...

y
Not to scale
(1, 4)

(1, 1)

(5, 4)

(5, 1)

O

x

The rectangle is reflected in the line y = 4
...


(b)
...
Angle OAB = 30°
...

Give your answer in its simplest terms in the form a
You must show your working
...



...

You must show your working
...
) and (
...

Indicate any values where the graph crosses the axes
...

Indicate any values where the graph crosses the x-axis
...
The results are shown on the Venn diagram
...

Find the probability that this person watches only one other sport
...
[2]
(b) Two of the 50 people are chosen at random
...

[3]
35

END OF QUESTION PAPER
© OCR 2024

20
EXTRA ANSWER SPACE
If you need extra space use this lined page
...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...



...


...


Higher
GCSE
Mathematics - Paper 5
J560/05: Paper 5 (Higher tier)
General Certificate of Secondary Education

Mark Scheme for June 2024

Oxford Cambridge and RSA Examinations

OCR (Oxford Cambridge and RSA) is a leading UK awarding body, providing a wide range of
qualifications to meet the needs of candidates of all ages and abilities
...


It is also responsible for developing new specifications to meet national requirements and
the needs of students and teachers
...

This mark scheme is published as an aid to teachers and students, to indicate the requirements
of the examination
...
It does not
indicate the details of the discussions which took place at an examiners’ meeting before
marking commenced
...

Mark schemes should be read in conjunction with the published question papers and the
report on the examination
...


Make sure that you have accessed and completed the relevant training packages for on-screen marking: RM Assessor Online Training; OCR
Essential Guide to Marking
...


Make sure that you have read and understood the mark scheme and the question paper for this unit
...


3
...


MARKING
1
...


2
...


3
...
It is essential that you meet the RM Assessor 50% and 100% deadlines
...


4
...


5
...
Where no
alternative response has been provided, examiners should give candidates the benefit of the doubt and mark the crossed out response
where legible
...


When a candidate provides contradictory responses, then no mark should be awarded, even if one of the answers is correct
...


On each blank page the annotation BP must be inserted to confirm that the page has been checked
...


2

J560/05

Mark Scheme

June 2024

7
...

Team Leaders must confirm the correct use of the NR button with their markers before live marking commences and should check this
when reviewing scripts
...


The RM Assessor comments box is used by the Principal Examiner or your Team Leader to explain the marking of the practice responses
...
Do not use the comments box for any other reason
...

9
...
Please follow the direction of your Team Leader about which questions you should report on and how to submit your
report
...


10
...
These must be used whenever appropriate during your marking
...

For responses that are not awarded either 0 or full marks, you must make it clear how you have arrived at the mark you have awarded and
all responses must have enough annotation for a reviewer to decide if the mark awarded is correct without having to mark it independently
...


4

J560/05

Mark Scheme

June 2024

Subject-Specific Marking Instructions
11
...

A marks are for an accurate answer and depend on preceding M (method) marks
...

B marks are independent of M (method) marks and are for a correct final answer, a partially correct answer, or a correct intermediate
stage
...


12
...

- figs 237, for example, means any answer with only these digits
...
g
...
37, 2
...
00237 would be acceptable but 23070 or 2374 would not
...

- nfww means not from wrong working
...

- rot means rounded or truncated
...

- dep means that the marks are dependent on the marks indicated
...

- with correct working means that full marks must not be awarded without some working
...


13
...


14
...

Do not award the marks if the answer was obtained from an incorrect method, i
...
incorrect working is seen and the correct answer
clearly follows from it
...


Where follow through (FT) is indicated in the mark scheme, marks can be awarded where the candidate’s work follows correctly from a
previous answer whether or not it was correct
...
You may find it easier to mark these questions candidate by candidate rather than question by question
...
g
...
Answers to part questions which are being followed through are indicated
by e
...
FT 3 × their (a)
...


In questions with no final answer line, make no deductions for wrong work after an acceptable answer (i
...
isw) unless the mark
scheme says otherwise, indicated by the instruction ‘mark final answer’
...


In questions with a final answer line and incorrect answer given:
(i) If the correct answer is seen in the body of working and the answer given on the answer line is a clear transcription error allow full marks
unless the mark scheme says ‘mark final answer’
...


(ii) If the correct answer is seen in the body of working but the answer line is blank, allow full marks
...


✓

next to the correct

(iii) If the correct answer is seen in the body of working but a completely different answer is seen on the answer line, then accuracy marks
for the answer are lost
...
Use the M0, M1,
M2 annotations as appropriate and place the annotation next to the wrong answer
...


In questions with a final answer line:
(i) If one answer is provided on the answer line, mark the method that leads to that answer
...

(ii) If more than one answer is provided on the answer line and there is a single method provided, award method marks only
...


19
...

(ii) If more than one response is provided, award marks for the poorer response unless the candidate has clearly indicated which response
is to be marked
...


When the data of a question is consistently misread in such a way as not to alter the nature or difficulty of the question, please follow the
candidate’s work and allow follow through for A and B marks
...
M marks are not deducted for misreads
...


6

J560/05

Mark Scheme

June 2024

21
...
For example, an answer in the mark scheme is 15
...
The
candidate then rounds or truncates this to 15
...
Allow full marks for the 15
...


22
...


23
...
If in doubt, consult your
Team Leader
...


If in any case the mark scheme operates with considerable unfairness consult your Team Leader
...
g
...
g
...
5[h], 2 [h] or 30 [mins]

nfww
M1 implied by 20 [miles] nfww

10

AND
Method to find time in 2nd stage
M2 for 0
...
6 [h] isw,
8

5

5

96[mins] or 1[h] 36[min] nfww
or
M1 for 0
...
g
...
g
...
g he travels at a constant speed
Mark best response as long as not contradictory
or incorrect
See Appendix 1

4

(a)

Positive

1

Ignore embellishments

4

(b)

Indicates the point (39, 10)

1

Ignore circles around the points ≤ 30 for both
Science and Maths as this is working for part (e)

10

J560/05

4

Question
(c)

Mark Scheme
Answer
Ruled line of best fit

Mark
2

and
answer FT ± 0
...
g
...


June 2024

Part marks and guidance
B1 for ruled line of best fit
Use overlay for LOBF, ruled line needs to
or answer FT ± 0
...


11

7

20

oe or (k > 7)

J560/05
Question
5

Mark Scheme
Answer
acceptable bisector of angle A
with two pairs of supporting arcs

Region to the left of their bisector
shaded

6

1800 final answer
and 200 and 9 shown

Mark
M2

B1

3

June 2024
Part marks and guidance

M1 for acceptable bisector of angle A with
no or incorrect arcs

Tolerance ± 2°
Use overlay
Accept dashed or solid line for bisector
If additional incorrect bisectors are drawn
then this is choice and M0 unless the
shading indicates they have chosen the
correct bisector

Dep on ruled line from angle A reaching
BC

Accept any clear indication for shading

M2 for 200 × 9
or M1 for their 200 × their 9
or B1 for 200 and 9 seen

12

For M1 e
...
uses 198 and 8
...
g
...
5 nfww or
oe or 14 nfww
4
4
oe e
...
M3 for 1750 ÷ 1000 × 6 [or × 8] oe
1
7 6
If or 7 used as ratio then max mark is M3 for
oe
7
4 1
isw (leads to answer 73
...
g
...
75 ÷ 0
...
75 ÷ 0
...
g
...
160 for 0
...
75 ÷

1
6

or 1
...
25 seen

J560/05

Mark Scheme
Mark

Answer

Question

June 2024
Part marks and guidance

8

(a)

100

1

Not e
...
100x, 100k

8

(b)

50

1

Not e
...
–50, 50x, 50k

24 nfww

4

9

M1 for [v = ] [0+] 3 × 4
M2 for their (3 × 4)2 ÷ (2 × 3) or better
or M1 for their (3 × 4)2 = 2 × 3 × s

10

8

4

M2 for
a + a + a + 5 + a + 5 = 42 oe
or M1 for 3 × 14 oe
or a + a + a + 5 + a + 5 oe

nfww – not from 2 × 3 × 4
M1 implied by [v =] 12 but not if obtained from
0 = u + 3 × 4 , this gets M0
If eqn not used and d = st answer 12 then M0
Condone other variable used for s but not v,
u, a or t
See AG if other kinematics formulas used
Allow M2 if correct expression seen first and
then incorrectly simplified before in eqn = 42
M1 implied by 42 or 4a + 10 oe

AND
M1 for a =

42−10

oe
4

or FT their equation of the form
ka + c = d oe to a =
11

(a)

−3

2
B1 for answer

1
11

(b)

k

Correct response
e
...
a vector (

−3
or

) where p2

1FT
+ q2

−

oe

No fraction line in vector – but penalise 1
mark only in (a) and (b)

k

1

Allow correct or FT their (a)
Must be numeric and column vector

= 10
or FT their (a)

FT only from written equation
Where k, c and d are positive integers and
k>1
Allow M1 for one trial into 2(a + 5) + 2a
evaluated correctly

e
...


3

or

1
but not their

14

1

or
3
−3
from (a)

(√ 10) etc

1

0

J560/05
Question
11 (c)

Mark Scheme
Answer
Vector a – b correctly drawn with
correct direction arrow on the
resultant

June 2024
Part marks and guidance

Mark
3

B2 for a – b correctly drawn but with an
incorrect or no direction arrow
or
or B1 for their (−3) − (

0 ) or (−3)

1

a-b

Full marks may be within a triangle with
correct direction arrow
B2 may be within a triangle with no direction
arrow[s]

−2

3

0 ) oe on grid

or for drawing (−3) − (

1

−2

as two sides of a potential triangle
(condone missing/wrong arrows)
If 0 scored, SC1 for vector (−3) drawn
5

on grid with correct direction arrow
12

(a)

No oe
AND
correct valid reason or correct
supporting values
e
...

• The value of the interest
changes each year as the
amount grows
• It is exponential growth
• Compound interest means the
interest grows each year

1

e
...

Accept e
...

• There will 5% interest on the £50 as
well as an extra £50 oe
• It will increase by 5% of 1050
• Finds £1102
...
5[0] or 52
...
8

June 2024
Part marks and guidance

B1 for [a = ] 8000
AND
B3 for [b = ] 0
...
g
...
8

6400

M2 for 8000 oe or 80% oe

or

oe 0
...
g
...
g
...
g
...
g
...
g
...
g
...
g
...
g
...
g
...
g
...
g
...
4 or 0
...
6 - 1
...
1oe
or M1 for 1
...
4 oe
If 0 or 1 scored, instead award
1
SC2 for answer 2
9
Alt method
11
M2 for
oe fraction
9
or M1 for 1
...
2 oe seen
M2 for

15

+

15

9

or M1 for

− their

11

9

15

oe

9

− their

11

9

oe

9

If 0 or 1 scored, instead award
SC2 for answer 2

19

1
9

1

9 , correct working may

be shown on a diagram
For M1 accept 1
...
22[2…] or better
For M2, accept any clear indication that the
decimal figure 1 recurs e
...
2
...
111… For
M1 accept 1
...
44[4…] or better
Accept oe e
...
1
...
4 )

Allow oe for any pair allowing ‘recurrence’ to
be removed, accept e
...
1
...
22…

For M2 and M1 must be using a
common denominator for FT and could
work with mixed numbers
...
g
...
14, 3
...
g
...
g
...
g
...
12
...
cos 120
12sin120
2
2
or
, [2×]
oe
12
(12sin30)
sin30
where x is ½ AB
M1 for any correct implicit method for finding
AB or ½ AB

In this method and other methods used, B1 is
awarded for the correct trig value[s]
associated with their method for find [½]AB,
even if their method is incorrect, not just seen
in a table of trig values unless selected
Award maximum of 6 marks if answer incorrect

If 0, 1 or 2, scored instead award

SC3 for answer 12 3 + 16 π
If 0 or 1 scored, instead award
SC2 for 12 3 or 16 π in answer

21

SEE AG

J560/05
Question
21

Mark Scheme
Answer
5 25
−
, oe and (1, 3)
3 3
with correct working

Mark
6

June 2024
Part marks and guidance
Accept ( –1
...
666 to –1
...
33[3]…)
Correct working requires evidence of at least M2M2

M2 for 3x2 + 2x – 5 [= 0]

For M2 accept e
...
5 – 2x – 3x2 [= 0]

or M1 for 3x2 = 5 – 2x
M2 for (3x + 5)(x – 1) [= 0]
or M1 for (3x + a)(x + b) where ab = –5
or 3b + a = 2
or for correct partial factors
3x(x – 1) + 5(x – 1) or x(3x + 5) – [1] (3x + 5)

A1dep on M2M2 for either pair of coordinates
correct
or for both x values correct or both
y values correct
If 0, 1 or 2 scored, instead award
5 25
− , oe and
SC3 for answers
3 3
(1, 3)
If 0 or 1 scored, instead award
5 25
− ,
SC2 for answer
oe or for both x values
3
3
correct or both y values correct
If 0 scored, SC1 for one answer (1, 3)

22

Strict FT their 3-term quadratic equation or
expression
e
...
If M2 awarded for 5 – 2x – 3x2 [= 0] then factors
should be correct for this equation for M2 or M1
Accept correct use of quad formula or completing
the square, M2 if completely correct, M1 if one
error in substitution in formula or (x + a)2 correct if
completing the square
If A1 for correct x–values or y – values after correct
partial factorisation award M2 for factors
See AG for work with equations in terms of y

J560/05
Question
22 (a)

Mark Scheme
Answer
Correct sketch with y – intercept 1
indicated

Mark
2

22 (b)

Correct sketch of y = cos x with 90
and 270 indicated and starting at 1
on y-axis

90

23

(a)

Part marks and guidance
For 2 marks, condone curve touching but not
crossing x- axis
For 2 marks or B1 mark intention
For 2 marks no ruled sections
B1 for increasing exponential curve
with no ruled sections
or for any sketch with a single
y – intercept at 1

1
2

9
oe
16

2

See AG
For 2 marks mark intention

B1 for cos shape graph with period
360 starting at (0, k), k > 0 and
amplitude k
Or
for cos shape graph starting at (0, 1)
with amplitude 1 and with a period a
factor of 360
Or
for cos shape graph with period 360
with 90 and 270 indicated where
graph crosses x - axis

270

June 2024

B1 for answer

k
9
or and must be a
16
k

e
...
y = 2cosx
For B1 allow poor curvature, mark intention

e
...
y = cos 2x,

May be errors with amplitude
See AG
Do not accept ratio or words
isw conversion/cancelling

proper fraction
23

(b)

10 7 + 7 10 oe
50 49 50 49
2
leading to
with no errors in
35
processing seen

3

10
M2 for 2

10

7

50

49

oe

or M1 for
10
7
7
10
oe and
oe or
oe and
oe
50
49
50
49
seen

23

7

1
=

Award 3 marks for e
...


50

49

1
2

and

35

35

0
...
142 to 0
...
35 and 0
...
g
...
6 is incorrect, should be 16 Award B1M3

Example B

Example A

1:7

1
...
6
0
...
75 ÷ 0
...
The only error is 16 should be
14 which is arithmetic
...
75 ÷ 0
...


×134 = 1214

12 1

×6=73
...
5 = 87 cups

Example G

7
192

,

1

=

8

7
4

×

7

=

16 × 18 = 481 [

Award B1M2

32

192

Example H

There is an error in dividing 74 by 81 After this error the remaining steps
imply 16 × 18 and earn M2

7

× 6 is embedded within lines 1 and 2 and

4

Example F
Example E

3
4

192
4

192Additional÷4=48 guidance question 9

27

Non-standard approach using ratio after 16 × 18 = 481
and finding an equivalent fraction over 48 resulting in
84 cups

litres cordial for 1 cup]

7

7

4

4

[litres] = 48 + 36
84 cups

B1 for 8 in line 1
Works in ml and earns M4 for
a correct method 1750 ÷ 1000 × 8
× 6 oe , the only error is using 160
but this comes from 1000 ÷ 6 which
is correct
Award B1M4

=

84
48

Award 6 marks for convincing
alternate approach

J560/05

Mark Scheme

Additional guidance Question 9
Candidates
may choose to use Kinematics formulas other than those given
2
If s = ut + ½ at used instead of given formulae, allow M3 for [s = ] [0 × 4 +] 12 × 3 × 42

OR
After v = 12

Allow M2 for [s = ] 0 +212 × 4 or [s = ] 12 × 4 – 12 × 3 × 42

28

June 2024

J560/05

Mark Scheme

June 2024

Additional guidance question 20
The length AB or

1

108 ) can be obtained by a number of different methods
...
g

---