PiXL AQA Style Paper 3H (March 2017) Mark Scheme

Q Answer Mark Comments

1 5 × 10 B1 −4

2 2n + 3 B1

3 10 B1

4 2x : 4x and 2x : x B1 Both must be indicated

5 (a) Point correctly plotted at ( 22 , 40 ) B1

5 (b)

“Mary sells fewer hot drinks on warmer days”, or “Mary sells more hot drinks on colder days”, or “The hotter the temperature, the fewer drinks Mary sells” or “The colder the temperature, the more drinks Mary sells”, oe

B1

Must include reference to both temperature and number of drinks sold, with valid connection between them. Unexplained repetition of “negative correlation” B0.

5 (c)

Line of best fit seen M1

Any value in range 60 to 80 A1 Additional guidance

If no line of best fit seen, cannot award method mark (even if lines are drawn parallel/perpendicular to either axis from 18°C). M0 A1

5 (d) “More people visiting the town” “Bank holiday”, “School holiday” “Coach party visited”, “Other coffee outlet nearby closed for day”.

B1 Any reason that would significantly increase the numbers of people visiting the cafeteria.

6

Reaches 4x = −14 or better M1

x = −3.5 or x = −312 A1 Must be simplified and not improper if given

using a fraction. Additional guidance

Do not allow unsimplified fractions (including −144 , − 7

2 , −324 M1 A0 ) or

unrounded decimals (for example −3.50).

Q Answer Mark Comments

7 (a)

Evidence of attempt to calculate numerator and denominator before undertaking division, and sight of two numbers divided.

M1 At least one must be correct 15.7 + 8.2 = 23.9 2.38 × 4.19 = 9.9722

2.3966627 A1 Tolerate some rounding of answer after correct numerator and denominator have been divided.

Additional guidance Do not tolerate any rounding of numerator or denominator before division, for example 23.9 ÷ 9.97. M1 A0

7 (b) 2.40 B1 2.4 is B0.

8 (a) a and b B1

8 (b)

Finds d or corresponding angle next to 125° = 55°. M1

Valid method to find remaining angle in triangle (= 85°) M1

95° A1

9 7 B1

10

Total value of tickets sold is £300 M1

Attempt to find total value of prizes (must see addition of 60 × 1.50 = 90 (or 60 × 150 = 9000) and 6 × 20 = 120; at least one of 90 (or 9000) or 120 must also be correct.

M1

£210 (or £90 profit) must be correct and “Yes” ticked B1

Additional guidance “Yes” ticked with no working M0 M0 B0

Q Answer Mark Comments 11 (a) 10 < x ≤ 20 B1

11 (b)

23 × 5 + 18 × 15 + 8 × 25 + (1 ×) 35 (= 620) M1 oe

“their 620” ÷ 50 (= 12.4) M1

12.4 A1 Allow rounding to 12

11 (c)

Earthworm must be in last class interval, so 35cm increases to 40cm B1 oe

Estimated total increases so “it increases my estimate” box ticked B1

Additional guidance Can award B1 mark if “it increases my estimate” box ticked without working or explanation. B0 B1

Q Answer Mark Comments

12

Alternative method 1 (distance converted to miles)

420km × 58 M1 (= 262.5) oe

23

10.5 − “their 7” − “their 2.1” (= 4.9)

× 10.5 (= 7) and 20% of 10.5 (= 2.1) M1

“their 4.9” × 56 (= 274.4) M1

262.5 < 274.4 (both must be correct) and “Yes, I have enough fuel” box ticked.

B1

Alternative method 2 (range converted to kilometres)

56 miles × 85 M1 (= 89.6) oe

23

10.5 − “their 7” − “their 2.1” (= 4.9)

× 10.5 (= 7) and 20% of 10.5 (= 2.1) M1

“their 4.9” × 89.6 (= 439.04) M1

439.04 > 420 (439.04 must be correct) and “Yes, I have enough fuel” box ticked.

B1

Alternative method 3 (range converted to miles; fuel fraction first)

420km × 58 M1 (= 262.5) oe

23 − 1

5 = 715 and M1 7

15 × 10.5 (= 4.9)

“their 4.9” × 56 (= 274.4) M1

262.5 < 274.4 (both must be correct) and “Yes, I have enough fuel” box ticked.

B1

Alternative method 4 (range converted to kilometres; fuel fraction first)

56 miles × 85 M1 (= 89.6) oe

23 − 1

5 = 715 and M1 7

15 × 10.5 (= 4.9)

“their 4.9” × 89.6 (= 439.04) M1

439.04 > 420 (439.04 must be correct) and “Yes, I have enough fuel” box ticked.

B1

Additional guidance “Yes, I have enough fuel” ticked with no working M0 M0 M0 B0

Q Answer Mark Comments

13 (a)

Either m = −13 B1 or c = 2 seen

y = −13 B1 x + 2

Additional guidance Allow follow through from incorrect gradient for final equation if method for finding gradient is clear and gradient given is negative and c is correct (eg gradient stated as −3 followed by y = −3x + 2).

B0 B1

Look out for working on diagram for gradient, but do not allow any attempt at accurate drawing or measurement from diagram provided.

13 (b)

m = 3, or −1original gradient ( )3

11

−−

, or ,

or “gradients multiply to −1” seen B1 May be implied.

R is at ( 0 , −18 ) B1

y = 3x − 18 B1

Additional guidance Allow follow through from incorrect gradient in 13 (a) for first B1 mark if

statement about −1original gradient

or “gradients multiply to −1” is seen

Allow follow through from incorrect gradient in 13 (a) for final equation if first B1 is awarded and gradient given is positive and c is correct (eg gradient stated as 4 followed by y = 4x − 18). Under certain circumstances, y = 4x − 18 or similar could receive B1 B1 B1.

Look out for working on diagram for gradient, but do not allow any attempt at accurate drawing or measurement from diagram provided.

Q Answer Mark Comments

14

Alternative method 1 (a first)

10a − 15b = 65, 12a + 15b = 12

M1

Correctly multiplies both equations and matches co-efficient of b. Attempts appropriate choice of addition or subtraction for their pair of equations.

a = 312 A1 or a = 3.5 If a fraction is used for a, must be simplified

and not improper.

b = −2 A1

Alternative method 2 (b first)

4a − 6b = 26, 4a + 5b = 4

M1

Correctly multiplies at least one equation and matches co-efficient of a Attempts appropriate choice of addition or subtraction for their pair of equations.

b = −2 A1

a = 312 A1 or a = 3.5 If a fraction is used for a, must be simplified

and not improper.

Alternative method 3 (a substitution)

1354432 =

−

−aa

or 10a − 12 + 12a = 65

or 43

13254 =

−

+aa

or 12a + 10a − 65 = 12

M1 Correctly rearranges to obtain equation for a and attempts to solve it.

a = 312 A1 or a = 3.5 If a fraction is used for a, must be simplified

and not improper.

b = −2 A1

Alternative method 4 (b substitution)

1334542 =−

− bb

or 8 − 10b − 12b = 52

or 452

3134 =+

+ bb

or 52 + 12b + 10b = 8

M1 Correctly rearranges to obtain equation for b and attempts to solve it.

b = −2 A1

a = 312 A1 or a = 3.5 If a fraction is used for a, must be simplified

and not improper. Additional guidance

No marks for correct answer based on trial and improvement. M0 A0 A0 No marks for correct answer with no working. M0 A0 A0

There are numerous other correct equations for M1 (generally multiples of those given) – allow these if correct, awarding marks as above.

Those given are the most likely to appear.

Q Answer Mark Comments

15 (a)

3 parts = (180 + 60) = 240, so total number of trees in B is 400 M1

A1 340 (must be in correct part of diagram)

A1 170 (must be in correct part of diagram)

15 (b)

Fraction with either numerator (= 180) or denominator (= 240) correct B1 oe

14 B1 or 0.25 or 25% Fraction must be simplified.

Additional guidance Do not accept ratio for second B1 mark B1 B0 or B0 B0

16 tany = 11

6 M1 oe

61.4 A1

17

States 65% = (£)104

or 104 ÷ 0.65, or 104 ÷ 65100

104 ÷ 65 × 100 (tolerate just 104 ÷ 65)

, or

or 104 × 10065

M1

oe

160 A1 Additional guidance

Do not allow any method based on finding any percentage of £104 M0 A0

18

−1… or …2 B1

−1, 0, 1, 2 B1 Check for 0 Additional guidance

Integers can be in any order For first B1 −1 must clearly be the start of the list or 2 clearly the end.

“−1 and 2” or “−1 to 2” for final answer. B1 B0

A B ξ

180 60 340 170

Q Answer Mark Comments

19

Any combination of cube root and minus 5 (only) seen B1

f−1 3 5−x(x) = B1

Additional guidance

Accept use of ( )31

5−x throughout for 3 5−x , etc.

Both cube root and minus 5 (and no other operations) must appear in some form for M1; see examples:

Combination of cube root and minus 5 seen, eg f−1 3 x(x) = − 5, etc M1 A0

One incorrect operation, eg f−1 3 x(x) = + 5, etc. M0 A0

Only one operation. eg f−1 3 x(x) = , etc. M0 A0

Combination of cube root and minus 5 seen, but with extra operations,

eg f−1

253 −x(x) = , etc

M0 A0

Special case; correct answer but with extra operations, eg f−1 3 5−x(x) = + 1

M1 A0

20

Angle COA (or DOA) is 64° because radius and tangent meet at 90°, so OAD is a right angled triangle.

M1 “radius and tangent meet at 90°”, oe, must be seen for this M1

CDA = 12

so CDA is “their 64°” ÷ 2

COA, because angle at

circumference = half angle at centre, or COA = 2 × CDA, because angle at centre = double angle at circumference

M1

Statement of “angle at centre = double angle at circumference” can be implied, but, if not, “their 64°” ÷ 2 must be present for this M1.

32° M1 21 (a) D B1 Accept any clear indication

21 (b)

y = kx² or x2 B1 y = k oe

k = 208 M1

y = 13 A1

Q Answer Mark Comments

22 (a)

12 M1 × 3x × (3x + 5) = 24

Likely to be obtained from formula, but allow equivalent from splitting trapezium into two shapes (eg rectangle and triangle).

Multiplies to Error! Bookmark not

defined.92 x2

so 9x + 15

2 x = 24 2

and so to result. + 15x − 48 = 0

A1

22 (b)

x = 6

1634255 −××−±− M1 Tolerate one sign slip-up inside square root, but not the initial −5.

Obtains x = 1.6218 A1 Ignore x = −3.2885; no need to see it or reason why it is discarded.

Attempts Pythagoras with 3 × “their 1.6218” and “their 1.6218” + 5 to find missing side.

M1

Reaches 22 6218.68655.4 + o M1

Missing side is 8.2171 A1

22.9 A1

23

5.225

25.2

2 ×+=x (= 2.25) M1 2.5 substituted to obtain first iterate.

25.225

225.2

3 ×+=x = 2.236111

236111.225

2236111.2

4 ×+=x

= 2.236068

236068.225

2236068.2

5 ×+=x

= 2.236068

M1

Obtains correct values for 3x and 4x and makes clear that, being equal to four decimal places, third decimal place will remain unchanged.

2.236 A1 Must be correctly rounded

Q Answer Mark Comments

24

Either error interval for thickness of a DVD case 1.45cm ≤ thickness ≤ 1.55cm, or 14.5mm ≤ thickness ≤ 15.5mm or error interval for length of shelf 39.5cm ≤ length ≤ 40.5cm, or 395mm ≤ length ≤ 405cm

M1 If division using correct bounds is seen this can be considered to be implied.

39.5cm ÷ 1.55cm or 395mm ÷ 15.5mm M1

Must have lower bound of length of shelf divided by upper bound of thickness of DVD case.

25 A1 Additional guidance

Units should be consistent for two chosen measurements at second M1 mark, but accept appropriate multiplication or division by 10 later.

25

( )RPQsin9

35sin7

=°

B1

sin(RPQ) = 9 × sin35°7 M1

RPQ = 47.515° A1

Attempts to find PQR (= 97.485°) from other angles in triangle. M1

12 M1 × 7 × 9 × sin(“their 97.485°”)

31.2 A1