GCSE Mathematics Mark Scheme for Higher Tier Calculator Paper M4 Exam in 2019, Slides of Mathematics

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General Certificate of Secondary Education 2019

11938.01 F

MARK

SCHEME

Mathematics

M

Calculator Paper

Higher Tier

[GMC41]

TUESDAY 21 MAY, 9.15am–11.15am

GCSE MATHEMATICS

Introduction

The mark scheme normally provides the most popular solution to each question. Other solutions given by candidates are evaluated and credit given as appropriate; these alternative methods are not usually illustrated in the published mark scheme.

The marks awarded for each question are shown in the right hand column and they are prefixed by the letters M , A and MA as appropriate. The key to the mark scheme is given below:

M indicates marks for correct method.

A indicates marks for accurate working, whether in calculation, reading from tables, graphs or answers. Accuracy marks may depend on preceding M (method) marks, hence M0 A1 cannot be awarded, i.e. where the method is not correct no marks can be given.

MA indicates marks for combined method and accurate working.

A later part of a question may require a candidate to use an answer obtained from an earlier part of the same question. A candidate who gets the wrong answer to the earlier part and goes on to the later part is naturally unaware that the wrong data is being used and is actually undertaking the solution of a parallel problem from the point at which the error occurred. If such a candidate continues to apply correct method, then the candidate’s individual working must be followed through from the error. If no further errors are made, then the candidate is penalised only for the initial error. Solutions containing two or more working or transcription errors are treated in the same way. This process is usually referred to as “follow-through marking” and allows a candidate to gain credit for that part of a solution which follows a working or transcription error.

Positive marking:

It is our intention to reward candidates for any demonstration of relevant knowledge, skills or understanding. For this reason we adopt a policy of following through their answers, that is, having penalised a candidate for an error, we mark the succeeding parts of the question using the candidate’s value or answers and award marks accordingly.

Some common examples of this occur in the following cases:

(a) a numerical error in one entry in a table of values might lead to several answers being incorrect, but these might not be essentially separate errors;

(b) readings taken from candidates’ inaccurate graphs may not agree with the answers expected but might be consistent with the graphs drawn.

When the candidate misreads a question in such a way as to make the question easier only a proportion of the marks will be available (based on the professional judgement of the examining team).

AVAILABLE

AVAILABLE

• 1 2.5 × 7 + 7.5 × 8 +12.5 × 5 + 17.5 × 5 + 22.5 × 4 + 27.5 × 1 (= 345) M1 A
  • 345/30 MA
  • = 11.5 A1
• 2 8 x – 12 – 2 x + 10 MA
  • = 6 x – 2 MA1
• 3 200 = 2 × 2 × 2 × 5 × 5 M1 A - 23 × 5^2 A1
• 4 2 x + 10 + 2 x + x + 20 = 180 M
  • 5 x = 150 MA
  • x = 30 MA
  • Smallest angle = 50 A1
• 5 Diagram correctly set up M
  • 42 + x^2 = 12^2 MA
  • x^2 = 128 MA
  • x = 11.3 MA
  • Perimeter = 4 + 12 + 11. 3 = 27.3 MA1
• 6 (a) 20% increase, e.g. 120% of 150 = 180 M1 A - 20% decrease, e.g. 80% of 180 = 144 – decrease MA
  • (b) % decrease = 4% M1 A
  • (c) (20% decrease) = 0.8 × 1.2 (20% increase) = same M1 A - (or similar explanation/calculation to justify answer)
• 7 SA of sphere = 4 × π × 62 = 452 MA
  • SA hemisphere = 226 MA
  • base = π × 62 = 113 MA
  • total =
  • Martha is correct MA1
• 8 distance travelled in 1 minute = 720m MA
  • sin 20 = x /720 MA
  • x = 720 sin 20 MA
  • x = 246.25m A1
• 9 ( x – 4)( x + 3) = 0 M1 A - x = 4 x = –3 A1
• 10 2( a – 1) + ( a + 1) = 12 MA
  • 2 a – 2 + a + 1 = 12 MA
  • 3 a = 13 MA
  • a = 4 31 MA1
• 11 y = 3 x + c ( c = any numerical value, c 5) M1 A1
• 12 107.5% = 29455 MA
  • 100% = 29455 ÷ 107.5 × 100 MA
  • = £27400 A
  • 100% = 29455 ÷ 1.075 M1 A Alternative Solution
  • = £27400 A1
• 13 (a) Cumulative frequency graph and scale M1 A
  • (b) 160 – (reading from 55 on their graph) A2
    • Allow A1 for reading at (correct reading approx. 160 – 72 = 88)
• 14 Year 8: 756126 × 50 = 8 or 9 M1 A1
• 15 (a) 5.75 – 3.25 = 2.5 MA
  • (b) 3 155 85.. = 1.857142 M1 A1
• 16 (a) perimeter of rectangle = 16 +4 x MA - Side of square = 4 + x A
  • (b) (i) (4 + x )^2 = 16 x + 4 MA - 16 + 8 x + x^2 = 16 x + 4 MA - x^2 – 8 x + 12 = 0 A
    • (ii) ( x – 2)( x – 6) = 0 MA
      • x = 2, 6 A1

22 (a) (2 a – b )( a + 4 b ) M1 A

(b) (^) ( )( )

x x

x x x x x

x 2 1 4

^ + h + # – MA

x x

x x x x x

x 2 1 4

] ]

]

g g

g MA

x

x x 4 x

] g MA

x^ or

x x

x 4

] g (^) – A1 6

Total 100

11938.01 F 77

AVAILABLE MARKS