OAQA - 2E Marking Scheme, Exams of Mathematics

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INTERNATIONAL GCSE

MATHEMATICS

9260/ 2 E

Paper 2E Extension Tier

Mark scheme

June 2022

Version: 1.0 Final

226Y9260/2E/MS

Key to mark types and abbreviations

M Method marks are awarded for a correct method which could lead to a correct answer.

A Accuracy marks are awarded when following on from a correct method. It is not necessary to always see the method. This can be implied.

B Marks awarded independent of method.

ft Follow through marks. Marks awarded for correct working following a mistake in an earlier step.

SC Special case. Marks awarded for a common misinterpretation which has some mathematical worth.

M dep A method mark dependent on a previous method mark being awarded.

B dep A mark that can only be awarded if a previous independent mark has been awarded.

oe Or equivalent. Accept answers that are equivalent.

eg accept 0.5 as well as 21

[a, b] Accept values between a and b inclusive.

[a, b) Accept values a ⩽ value < b

3.14 … Accept answers which begin 3.14 eg 3.14, 3.142, 3.

Use of brackets It is not necessary to see the bracketed work to award the marks.

Q Answer Mark Comments

1^2 B

Q Answer Mark Comments

2 5.000 06 B

Q Answer Mark Comments

nx 4

B

Q Answer Mark Comments

4 B

Q Answer Mark Comments

Alternative method 1

A pair of arcs, centre Y , equal radius

(± 2 mm), intersescting YX and YZ

or a single arc, centre Y , intersecting YX and YZ

M

oe eg a single arc, centre Y , radius YX intersecting YZ

Two intersecting arcs, equal radius

(± 2 mm), with centres at the

intersections on YX and YZ

M1dep

if a single arc, centre Y , radius YX intersecting YZ has been drawn the centres will be at X and the intersection on YZ

Fully correct method of construction of angle bisector and P marked 6 cm from Y in a position where YP bisects the angle XYZ

A

± 2 mm

Alternative method 2

A pair of arcs, centre Y , equal radius

(± 2 mm), intersecting YX and YZ

or a single arc, centre Y , intersecting YX and YZ

M

oe eg a single arc, centre Y , radius YX intersecting YZ

Two further arcs, centre Y , equal

radius (±^2 mm), intersecting^ YX^ and

YZ or a further single arc, centre Y , intersecting YX and YZ and two lines from the YX intersection of one arc to the YZ intersection of the other arc

M1dep

if a single arc, centre Y , radius YX , intersecting YZ has been drawn, one line will join X to the intersection on YZ

Fully correct method of construction of angle bisector and P marked 6 cm from Y in a position where YP bisects the angle XYZ

A

± 2 mm

Additional guidance is on the next page

Y^ Z

cont.

Additional Guidance

The angle bisector need not be drawn

Ignore incorrect arc(s)

Condone P not labelled if clearly indicated eg by a cross or dot

If P is not labelled or clearly indicated by a cross or dot, take the end of the angle bisector as the position of P

The angle bisector need not pass through the intersection of the second pair of arcs but must do so if it were extended

Accept an arc touching YX and/or YZ

No arc, centre Y , seen on YZ M

Method for alt 2

M

X

Q Answer Mark Comments

Alternative method 1

6 2 and 4.5 2 or 36 and 20. or 56.

M

oe

62 +4.5^2

or 36 +20 25.

or 56 25.

or 7.

M1dep

oe

4 5. 3

× 5 or

× 5 oe is M

11 + 4.5 + their 7.5 + x = 30

or

30 – 11 – 4.5 – their 7.

M1dep

oe any letter

7 A

Alternative method 2

(Angle BDC =) tan− 1 4 5.

or [36.86, 37] or

(angle BCD =)

tan^ − 1 6 4 5 or [53, 53.14]

M

sin their [36.86, 37]

or

sin their [53, 53.14]

or [7.477, 7.513]

M1dep

oe eg

cos their [36.86, 37]

cos their [53, 53.14]

11 + 4.5 + their [7.477, 7.513] + x =

or

30 – 11 – 4.5 – their [7.477, 7.513]

M1dep

oe any letter

7 A

Additional Guidance for question 8 is on the next page

cont.

Additional Guidance

The first two method marks may be awarded for correct work with no or incorrect answer, even if this is seen amongst multiple attempts

x^2 = 6 2 + 4.5^2 – 2 × 6 × 4.5 × cos 90 M

Angle BDC = 36.9 and x^2 = 11 2 + 62 – 2 × 11 × 6 × cos 36.9 M1M

Decimal answer rounded to 7 M1M1M1A

Q Answer Mark Comments

Alternative method 1

9 p^2 + 45 p M

6 p^2 + 24 p

or – 6 p^2 – 24 p

M

c = 3 and d = 28 A1ft^ ft M1M0 or M0M1 with at most 1 error

Alternative method 2

9 p^2 – 6 p^2 = cp^2 or 3 p^2 = cp^2

or 9 – 6 = c

M

equating p^2 terms or coefficients

implied by c = 3

45 p – 24 p + 7 p = dp or 28 p = dp

or 45 – 24 + 7 = d

M

equating p terms or coefficients

implied by d = 28

c = 3 and d = 28 A1ft^ ft M1M0 or M0M1 with at most 1 error

Additional Guidance

Allow recovery of missing brackets, eg

9 p^2 + 45 p – 6 p^2 + 24 p + 7 p = 3 p^2 + 28 p c = 3 and d = 28 M1M1A

3 p^2 + 28 p in working implies the method marks M1M1A

c = 3 p^2 and d = 28 p implies the method marks M1M1A

9 p^2 + 45 p – 6 p^2 – 24 p = 3 p^2 + 21 p c = 3 and d = 21 M1M1A

9 p^2 + 45 p – 6 p^2 + 24 p + 7 p = 3 p^2 + 76 p c = 3 and d = 76 (one error) M1M0A1ft

9 p^2 + 45 p – 5 p^2 – 24 p + 7 p = 4 p^2 + 28 p c = 4 and d = 28 (one error) M1M0A1ft

9 p^2 + 35 p – (6 p^2 + 24 p ) + 7 p = 3 p^2 + 18 p c = 3 and d = 18 (one error) M0M1A1ft

9 p^2 + 45 p – 6 p^2 + 24 p = 3 p^2 + 69 p c = 3 and d = 69 (two errors) M1M0A0ft

Q Answer Mark Comments

(5 – 2) × 180

or 5 × –

or 5 × 108 or 540

M

oe

103 + 89 + 150 + x + x + 50 = their

or

their 540 – 103 – 89 – 150 or 198

M

oe eg 2 x + 392 = 540

their 540 must be an integer > 392

their 540 – 103 – 89 – 150 – 50

or

their 198 – 50 or 148

M1dep

oe eg 2 x = 148

dep on 2nd M

74 A

Q Answer Mark Comments

Additional Guidance

• 1.4(0) × 6 – 1.4(0) × 6 × 0. Alternative method 1 – cost of 6 rolls each week
• or 8.4(0) – 1.
• or 1.4(0) × 6 × 0.
• or 7.
  • M
    • oe week
• (1.4(0) × 2 + 1.4(0) ÷ 2) ×
• or 3.5(0) ×
• or 1.4(0) × 4 + 1.4(0) × 2 ÷
• or 1.4(0) ×
• or
  • M
    • oe week
• (1.4(0) + 1.4(0) × 0.75) ×
• or 2.45 ×
• or 4.2(0) + 3.
• or 7.
  • M
    • oe week
• 7.14 and 7 and 7.35 and 2 A
• 1.4(0) × 6 × 0.15 or 1.26 M1 oe week Alternative method 2 – saving made each week
• 1.4(0) ÷ 2 × 2 or 1.4(0) M1 oe week
• 1.4(0) × 3 × 0.25 or 1.05 M1 oe week
• 1.26 and 1.4(0) and 1.05 and 2 A
• 6 × 0.85 or 5.1 M1 oe week Alternative method 3 – equivalent rolls being paid for
• 2 × 2.5 or 5 M1 oe week
• 3 × 1.75 or 5.25 M1 oe week
• 5.1 and 5 and 5.25 and 2 A

Q Answer Mark Comments

15 n^2 + n ⩾ 0 B

Q Answer Mark Comments

Alternative method 1

25 + c + 26 + d + 9 + 3 = 120

or 63 + c + d = 120

or c + d = 57

M

oe may be implied by a pair of non-negative

integers for c and d totalling 57

(0 × 25 or 0)

and 1 × c or c

and 2 × 26 or 52

and 3 × d or 3 d

and 4 × 9 or 36 and 5 × 3 or 15

M

oe allow 1 error

((0 × 25) + 1 × c + 2 × 26 + 3 × d + 4

× 9 + 5 × 3) ÷ 120 = 2

or

c + 3 d = 240 – 2 × 26 – 4 × 9 – 5 × 3

or c + 3 d = 137

M1dep

oe dep on second M

3 d – d = 137 – 57

or d = 40

or 3 c – c = 3 × 57 – 137

or c = 17

M

oe eliminates one unknown from their two linear equations

c = 17 and d = 40 A

Alternative method 2 and Additional Guidance is on the next page

Q Answer Mark Comments

cont.

Alternative method 2

25 + c + 26 + d + 9 + 3 = 120

or 63 + c + d = 120

or c + d = 57

M

M

M

oe

Test two values totalling 57 in correct mean calculation

M1 implies M

Second test of two values totalling 57 in correct mean calculation

M1dep

c = 17 and d = 40 A

Additional Guidance

Alt 2 scores up to M3 or M3A

c = 20 and d = 37 M

((0 +) c + 52 + 3 d + 36 + 15) ÷ 6 implies the second method mark (alt 1) M

(0 +) c + 52 + 3 d + 36 + 15 = 120 implies the second method mark (alt 1) M

25 + c + 26 + d + 9 + 3 2

= 120 M

25 + c + 26 + d + 9 + 3 6

= 2 M

Beware 120 – (2 × 26 + 4 × 9 + 5 × 3) = 17 M

Q Answer Mark Comments

52 × 2 or 104

or 30 × 0.75 or 22.

or 104 + 22.5 or 126.

M

oe

stage 1 or stage 2 distance or total

48 × (2 + 0.75 + 1)

or 48 × 3.

or 180

or

. .

× + × + x

M

oe

total distance

any letter

their 180 – their 126.

or

48 × 3.75 – (52 × 2 + 30 × 0.75)

M1dep

oe

dep on M

their 180 and their 126.5 must both be from fully correct methods

A

oe

SC3 39.1 or 48.

Additional Guidance

M1 may be awarded for correct work with no or incorrect answer, even if this is seen amongst multiple attempts

Q Answer Mark Comments

19 (15, 0) B

Q Answer Mark Comments

Alternative method 1

( AOC =) 360 – 10 x M

( ABC =) 5 x M

5 x + 2 x + 360 – 10 x + y = 360

M1dep

oe dep on M

y = 3 x A

Alternative method 2

( AOC =) 360 – 10 x M

( ABO =) 2 x and ( OBC =) y M

2 x + 2 x + y + y + 360 – 10 x = 360

M1dep

oe dep on M

y = 3 x A

Alternative method 3

( ABC =) 5 x M

( ABO =) 2 x and ( OBC ) = y M

2 x + y = 5 x

M1dep

oe dep on M

y = 3 x A

Additional Guidance

Working for M1M1 may be seen on diagram

Ignore reasons

The first two marks may be awarded even if not subsequently used

3 x on answer line with y = 3 x in working M1M1M1A