# Mathematics Grade 12 Paper 1 Eastern Cape Mock 2019 Errata

> Grade 12 Mathematics Paper 1 errata from the Eastern Cape Mock Exam 2019, 5 pages. Read online or get the free PDF download.

- Page: https://www.hlayiso.com/library/2019-grade-12-maths-p1-errata-prep-exams/92e012a4-ed6f-4afd-be8c-0b32a8aaf531
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 12
- Document type: Errata
- Year: 2019
- Exam period: Eastern Cape Mock Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 5

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## About this document

Mathematics · Grade 12 · Eastern Cape Mock Exam · 2019. Errata, 5 pages. Read online or download the PDF.

Read it free online on the page above; download the PDF from there.

## Document text

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### Page 1

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EXAMINATIONS & ASSESSMENT
Steve Vukile Tshwete Complex, Private Bagx0032, Zwelitsha, 5605
REPUBLIC OF SOUTH AFRICA, Website: www.ecdoe.gov.za
E-mail: nomvuyo.mbeleki@ecdoe.gov.za
dRef. No.    13/P                                       Tel.:     (040) 604 7028
Enquiries:   Ms N. Mbeleki                              Fax:      (043) 604 7789

                                              ERRATA
TO:              PRINCIPALS OF GRADE 12 SCHOOLS
                 SUBJECT ADVISORS

FROM:            CES: INSTRUMENT DEVELOPMENT UNIT
                 MS N. MBELEKI

DATE:            29 AUGUST 2019

SUBJECT:         GRADE 12 MATHEMATICS PAPER 1 ERRATA

1. The Grade 12 Mathematics Paper 1 was written on Friday, 23 August 2019. The Directorate
   received reports from some schools and relevant stakeholders about some challenges
   experienced during the writing of Grade 12 Mathematics question paper.

2. In order to address these challenges, the following standardised approach to marking should
   be adopted across the Province:

The word “repetition” and the fact that the word contains 2 x “R’s” contributed to the ambiguity of
this question.

Below are solutions that may be considered for this examination ONLY.

QUESTION 11.2

 11.2   Consider the letters of the word: NUMERATOR.

        11.2.1    How many 9 letter word-arrangements can be formed, if repetition of
                  letters is allowed?                                                           (1)

        11.2.2    How many 9 letter word-arrangements can be formed, if all 4 vowels
                  are never together and repetition of letters is not allowed?                  (3)

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         11.2.3   An 8 letter word-arrangement is made from the word NUMERATOR. All
                  the vowels must be included in this word-arrangement and repetition of
                  letters is not allowed. What is the probability that all odd-number spaces
                  are occupied by vowels?                                                                           (4)

                                                                                                                  [15]

ORIGINAL MARKING GUIDELINE

11.2.1      9 9 or / of 387 420 489                                99                                                  (1)

11.2.2      If vowels are together/As die vokale saam is:
               6! 4!                                              6!     4!
           If vowels are not all together:
              As die vokale nie almal saam is nie :
               9!  (6! 4!)
                                                                   subtracting from

             345600                                                  aftrekking vanaf 9!

                                                                   answer / antwoord

                                                                                                                        (3)

11.2.3      Vowels in odd spaces :
             4  5  3  4  2  3 1 2
             (4  3  2  1)  (5  4  3  2)
             4!  120
             2880                                                        

                                           2880
            Probability 
                               (9  8  7  6  5  4  3  2)
                                                                  
                                2880                               Vowels in odd spaces       Vokale in onewe spasies
                                                                   (98765432)
                                                                                          /
                                                                                                (98765432)
                               362880
                                  1
                             
                                 126

                                                                   answer / antwoord

                                                                                                                        (4)

                                                                                                                    [15]

                             ERRATA: MATHEMATICS P1- PREPARATORY EXAMINATIONS

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CORRECTION TO MARKING GUIDELINE

11.2.1   Since there are only 8 distinct letters the answer
         should be:

         Omdat daar slegs 8 unieke letters is, is die
         antwoord:

         89 or / of 134 217 728
                                                                89

                                                                                           (1)

11.2.2   If vowels are together/As die vokale saam is:
           6!
               4!                                             
                                                                   6!
                                                                       4!
           2!                                                      2!
         If vowels are not all together:
           As die vokale nie almal saam is nie :
                                                                                      9!
            9!  6!                                            subtracting from
                 4!                                                               2!
            2!  2!   
                                                                                      9!
                                                                   aftrekking vanaf
          172800                                                                     2!

                                                                answer / antwoord

         OR/OF                                                                             (3)

         The ambiguity might lead to the understanding
         that there are only 8 distinct letters but the word
         must contain 9 letters.

         NO SOLUTION / GEEN OPLOSSING

11.2.3

         Since there are only 4 distinct consonants:

                        ERRATA: MATHEMATICS P1- PREPARATORY EXAMINATIONS

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          Omdat daar slegs 4 unieke konsonante is:

                                                                 4!  4!
          Vowels in odd spaces :
           4  4  3   2  2  1  1
           (4  3  2  1)  (4  3  2  1)                   
                                                                 Vowels in odd spaces       Vokale in onewe spasies
           4!  4!                                                 (87654321)
                                                                                        /
                                                                                              (87654321)

           576

                                          576
           Probability 
                             (8  7  6  5  4  3  2 1)
                                                                 answer / antwoord
                              576
                           
                             40320
                               1
                           
                               70                                                                                     (4)

                                                                                                                  [15]

After considering the impact of the ambiguity of the question, schools are informed not to
mark this sub-question 11.2 and mark out of 142.

Yours in Quality Education

____________________________________                                   ______________
CES: INSTRUMENT DEVELOPMENT UNIT                                       DATE
MS N. MBELEKI

                            ERRATA: MATHEMATICS P1- PREPARATORY EXAMINATIONS

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building blocks for growth.

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