Pt1420 Unit 1 Assignment 2 Maths

NATIONAL SENIOR CERTIFICATE

GRADE 12

MATHEMATICS P1 NOVEMBER 2011 MEMORANDUM

MARKS: 150

This memorandum consists of 28 pages.

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Mathematics/P1

2 NSC – Memorandum

DBE/November 2011

NOTE: • If a candidate answers a question TWICE, only mark the FIRST attempt. • If a candidate has crossed out an attempt of a question and not redone the question, mark the crossed out version. • Consistent Accuracy applies in all aspects of the marking memorandum. QUESTION 1 1.1.1 x( x + 1) = 6 x2 + x = 6 x2 + x − 6 = 0 (x + 3)(x − 2) = 0 x = −3 or 2 OR Note: Answers by inspection: award 3/3 marks Note: Answer only of x = 2 : award 1/3 marks Note: If candidate converts equation to linear: award 0/3
OR

Pn = 3.2 n −1 Which is a multiple of 3 Qn = 6 n − 3 = 3(2n − 1) Which is also a multiple of 3 Since Tn = Q2 k −1 or Tn = P2 k for all n ∈ N , Tn is always divisible by 3
OR

factors 3.2 n −1

factors 3(2n − 1) (2)

The odd terms are odd multiples of 3 and the even terms are 3 times a power of 2. This means that all the terms are multiples of 3 and are therefore divisible by 3.

odd multiples of 3 3 times a power of 2 (2) [9]

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12 NSC – Memorandum

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QUESTION
uses P = : max 2/4 marks 2

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20 NSC – Memorandum

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7.2

Radesh:
A = P(1 + in ) = 8 550 Bonus = 0,05 × 6 000 = 300 = 6 000(1 + 0,085 × 5)

A = 6 000 + 8,5% of 6000 × 5 OR = 6000 + 510 × 5 = 6000 + 2550 = 8 550 8 550

Received = 8 550 + 300 = R 8 850
Thandi: n A = P(1 + i )

R8 850

⎛ 0,08 ⎞ = 6 000⎜1 + ⎟ 4 ⎠ ⎝ = R 8 915,68

20

n = 20 0,08 i= 4 answer choice made (6)
0,15 1 or or 0,0125 12 80 n = 18 n = 18