Mathematics Grade 12 Paper 1 Eastern Cape Mock 2019 Errata

Downloaded from hlayiso.com 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)

Downloaded from hlayiso.com 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

Downloaded from hlayiso.com 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

Downloaded from hlayiso.com 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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