NATIONAL SENIOR CERTIFICATE GRADE 11 - Mindset Learn

NATIONAL

SENIOR CERTIFICATE

GRADE 11

MATHEMATICAL LITERACY P1

EXEMPLAR 2013

MEMORANDUM

MARKS: 100

SYMBOL

M

MA

CA

A

C

S

RT /RG

F

SF

O

P

R

EXPLANATION

Method

Method with accuracy

Consistent accuracy

Accuracy

Conversion

Simplification

Reading from a table/Reading from a graph

Choosing the correct formula

Substitution in a formula

Opinion

Penalty, e.g. for no units, incorrect rounding off etc.

Rounding off/Reason

This memorandum consists of 7 pages.

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NSC ¨C Grade 11 Exemplar ¨C Memorandum

DBE/2013

QUESTION 1 [19 marks]

Ques

Solution

Explanation

Topic

1.1.1

2,35 ? = 2,35 ¡Á 1 000 m ?

= 2 350 m ? ?A

1A answer

M

L1

1.1.2

1.2.1

R 19,99

3 kg ?M

= R6,66 ?A

Cost per kg =

?SF

R 53,75 ? R 43,00

Percentage mark-up =

¡Á 100%

R 43,00

R10,75

=

¡Á 100% ?S

R 43

= 25% ?CA

(1)

1M dividing by 3 kg

F

L1

1A answer

(2)

F

L2

1SF substitution

1S simplification

1CA answer

(3)

1.2.2

1.2.3

1.2.4

R2 000 ?M

R54

= 37,04 ?A

Number of kilograms =

A = R76,00 + 30% of R76,00 ?M

= R76,00 + R22,80 ?A

= R98,80 ?CA

OR

?A

A = 1,3 ¡Á R76,00 ?M

= R98,80 ?CA

?M

?M

Cost = 1,2 kg ¡Á R53,75 + 0,5 kg ¡Á R85,00

= R64,50 + R42,50 ?S

= R107,00 ?CA

F

L1

1M dividing

1A answer

(2)

F

L2

1M adding 30%

1A mark up

1CA answer

(3)

F

L1(2)

L2(2)

2M using correct

selling prices

1S simplification

1CA answer

(4)

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NSC ¨C Grade 11 Exemplar ¨C Memorandum

Ques

Solution

1.2.5

?M

Cost price = 50 ¡Á R43,00

= R2 150 ?S

Selling price = 50 ¡Á R53,75

= R2 687,50 ?S

Profit

= R2 687,50 ¨C R2 150

= R537,50 ?CA

DBE/2013

Explanation

Topic

1M/A multiplying

1S simplifying

F

L1(2)

L2(1)

L3(2)

1S selling price

1CA answer

OR

Profit per kilogram = R53,75 ¨C R43,00 ?M

= R10,75 ?S

1M/A multiplying

1S simplifying

Profit on 50 kg = 50 ¡Á R10,75 ?M

1M/A multiplying

= R537,50 ?CA

1CA answer

(4)

[19]

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NSC ¨C Grade 11 Exemplar ¨C Memorandum

DBE/2013

QUESTION 2 [15 marks]

Ques

Solution

2.1

Deposit = 10% of R6 599,99

10

=

¡Á R6 599,99 ?M

100

= R659,999

¡Ö R660 ?CA

2.2

2.4

Topic

F

L1(2)

1M finding 10%

1CA answer

(2)

Amount to be financed (P) = R6 599,99 ¨C R660,00

= R5 939,99 ?A

?A

?A

Interest = R5 939,99 ¡Á 0,115 ¡Á 2

= R1 366,20 for two years

Total due = R5 939,99 + R1 366,20

= R7 306,19

2.3

Explanation

?CA

R 7 306,19

Amount =

?M

24

= R304,4245.....

¡Ö R304,42 ?A

?A

?M

Height = 12 ¡Á 2,54 cm = 30,48 cm

F

L1(2)

1A subtracting deposit L2(2)

1A value of i.

1A two years

1CA adding i to

outstanding amount

(4)

F

L2

1M dividing by 24

1A answer

(2)

M

L1

1M multiplying by

2,54

1A height

(2)

2.5.1

Amount = 12 ¡Á R300?M

= R3 600 ?A

2.5.2

?M

?M

Amount = R300 ¨C (R120 + R150)

= R30 ?A

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F

1M multiplying by 12 L1

1A answer

(2)

F

1M subtracting from

L1(2)

R300

L2(1)

1M adding costs

1A answer

(3)

[15]

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NSC ¨C Grade 11 Exemplar ¨C Memorandum

DBE/2013

QUESTION 3 [20 marks]

Ques

Solution

Explanation

Topic

3.1.1

Height of tin = 430 mm = 43 cm

1A length

1C conversion

M

L1

(2)

3.1.2

Distance = 4 570 mm ¨C 1 780 mm

= 2 790 mm

?A

3.1.3

Length = 4 260 ¨C 2 610

= 1 650 mm ?A

Area = 1 600 ¡Á 1 650 ?SF

= 2 640 000 mm2 ?CA

3.1.4

3.2.1

3.2.2

?M

?SF

?C

?SF

?SF

A = 4,830 m (2 ¡Á 9,75mm + 6,4 m) ¨C 6,4 m ¡Á 0,43 m

= 4,83 m (25,9 m) ¨C 2,752 m2

?S

= 122,345 m2 ?CA

122,345

? ?M

8

= 15,293.. ? ?CA

¡Ö 16 ? ?R

Amount of paint =

16 ?

?M

5?

= 3,2 = 4 ?CA

Number of 5 ? =

Cost = 4 ¡Á R215,85 ?M

= R863,40 ?CA

M

1M subtracting correct

L1

values

1A correct answer

(2)

M

1 A value of length

L1(1)

L2(2)

1SF substitution

1CA answer

(3)

1C conversion

M

1SF value of k

L1(2)

1SF value of b

L2(2)

1SF value of t and p.

L3(2)

1S simplification

1CA area

(6)

M

1M dividing by 8

L1(2)

L2(1)

1CA amount of paint

1R rounding up

(3)

M/F

L1(2)

1M dividing by 5 ?

L2(2)

1 CA number of

containers

1M multiplying by cost

1CA cost of paint

(4)

[20]

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