# Mathematics Grade 11 March Test 2020 Paper & Memo

> Grade 11 Mathematics question paper and memo from the March Test 2020, 14 pages. Read online or get the free PDF download.

- Page: https://www.hlayiso.com/library/maths-kzn-march-qp-memo-2020-grade-11/4929dfd4-fba7-4970-92a9-c16401a5a74d
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 11
- Document type: Question paper and memo
- Year: 2020
- Exam period: March Test
- Curriculum: CAPS
- Pages: 14

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## About this document

Mathematics · Grade 11 · March Test · 2020. Question paper and memorandum, 14 pages. Read online or download the PDF.

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## Document text

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Mathematics

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NSC

m 2020 Common Test

INSTRUCTIONS AND INFORMATION

Read the following instructions carefully before answering the questions.

This question paper consists of 5 questions.
Answer ALL the questions.

Number the answers correctly according to the numbering system used in this
question paper.

Clearly show ALL calculations, diagrams, graphs, etc. which you have used in
determining your answers.

Answers only will NOT necessarily be awarded full marks.

‘You may use an approved scientific calculator (non-programmable and
non-graphical), unless stated otherwise.

If necessary, round off answers correct to TWO decimal places, unless stated
otherwise.

Diagrams are NOT necessarily drawn to scale.

TWO DIAGRAM SHEETS for QUESTION 4.1, QUESTION 4.2, QUESTION 5.1
and QUESTION 5.2 are attached at the end of this question paper. Detach the
DIAGRAM SHEETS and hand in together with your ANSWER BOOK.

Write neatly and legibly.

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Mathematics March 2020 Common Test
NSC
QUESTION 1
ll Solve for x:
Lid 2s(-4)=0 (2)
2)
1.1.2 4x?+11x=2 (answer correct to TWO decimal places) (4)
1.1.30 x-y8-2x =0 (5)
1.1.4 x(x~7)+12<0 (4)
1.2. Solve simultaneously for x and y:
x=3y-5 and (2x-y-2)(x+y)=0 (4)
[19]
QUESTION 2
2) Simplify WITHOUT the use of a calculator: VI8 +2V2 + z
Write your answer with a rational denominator. (3)
2.2 Solve for x without the use of a calculator:
2.2.1 16 =128 (4)
geet 321
222 9 24 (4)
(1)
QUESTION 3
3.1 If cosO= -2 and 0eé [180° : 360°], determine the value of tan@ by using a diagram,
and without the use of a calculator. (3)
cos115°.cos214°
3.2 Evaluate, without using of a calculator; ——_———_ 5
valuate, without using of a calculator’ 70865° sin 236° (5)
3.3. Simplify to a single term, without the use of a calculator:
cos(90° + x).tan(540° + x).cos(180° — x) + sin(-90°) q)
[15)

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Mathematics rch 2020 Common Test

GIVE REASONS FOR YOUR STATEMENTS AND CALCULATIONS IN
QUESTIONS 4 AND 5.

QUESTION 4

4.1 In the diagram, AB is a chord of the circle having centre O. C is a point on AB such
that ACO = 90°. OC =5 cm and AO = 10cm.

‘

Calculate, with reasons, the length of AB. (4)

4.2 In the diagram, O is the centre of the larger circle, which passes through P, R and S.
R is a point on chord PV of the smaller circle PVSO. SV is produced to T. RVT = 96°.

4.2.1 Calculate, with reasons, the size of R, . (7)
4.2.2 Prove that ARVS is isosceles. (3)
[14]

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Mathematics March 2020 Common Test

QUESTION 5

5.1 In the figure below, JLH is a tangent to the circle having centre O at L. K andM
are points on the circle.

K M
o°
J
L H
Prove the theorem which states that MLH =K. (5)

5.2 AE is a diameter of the circle in the diagram below. AE is produced to D. DB is a
tangent to the circle at B. DC is perpendicular to AD. CBA is a straight line.

BE and EC are drawn. Let A=x.

E
D
A x 1A\3 1
2
: 3
1
B
1
Cc
5.2.1 Prove that BCDE is a cyclic quadrilateral. (3)
5.2.2. Prove that c =X. (4)
5.2.3 Prove that E, =£,. (2)
5.2.4 Is CE a tangent to the circle through A, B and E? Give a reason for your answer. (2)
[16]
TOTAL: 75

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Mathematics     Downloaded from hlayiso.com March 2020 Common Test
                                         NSC

      NAME & SURNAME:

      DIAGRAM SHEET 1

                                      QUESTION 4.1

                                 A

II.
II.

a:
<C
w                                    QUESTION 4.2
I-

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Mathematics Down loaded from h layiso . com 2020 Common Test
NSC

t NAME & SURNAME:

DIAGRAM SHEET 2

QUESTION 5.1

TEAR OFF

QUESTION 5.2

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                                     NATIONAL
                                SENIOR CERTIFICATE

                                          GRADE 11

                                       MATHEMATICS

                                MARKING GUIDELINE

                                       COMMON TEST

                                         MARCH 2020

MARKS: 75

                             This marking guideline consists of 7 pages.

                                   GEOMETRY  MEETKUNDE
                     A mark for a correct statement
                     (A statement mark is independent of a reason)
                S
                     'n Punt vir 'n korrekte bewering
                     ('n Punt vir 'n bewering is onafhanklik van die rede)
                     A mark for the correct reason
                     (A reason mark may only be awarded if the statement is correct)
               R
                     'n Punt vir 'n korrekte rede
                     ('n Punt word slegs vir die rede toegeken as die bewering korrek is)

                     Award a mark if statement AND reason are both correct
              S/R
                     Ken 'n punt toe as die bewering EN rede beide korrek is

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Mathematics-(Grade 11)                                 2                           March 2020 Common Test
                                             NSC – Marking Guideline

QUESTION 1

   1.1.1           1
            2 x x    0
                   2
                              1                                                               1
             x  0 or x                                                x0  x
                              2                                                               2
                                                                                                         (2)

   1.1.2     4 x 2  11x  2
             4 x 2  11x  2  0                                        standard form
                  b  b 2  4ac
             x
                       2a
                  11  (11) 2  4(4)(2)                               substitution into the
             x                                                          correct formula
                           2(4)
             x  0,17 or x  2,92                                      x-value  x-value
                                                                                                         (4)

   1.1.3    x  8  2x  0
            x  8  2x                                                  isolate     8  2x
            x 2  8  2x                                               squaring both sides
            x 2  2x  8  0
            ( x  4)( x  2)  0                                        factors
            x  4 or x  2                                             rejecting x  4
             x  2 only                                                correct solution
                                                                                                         (5)
   1.1.4    x( x  7)  12  0
            x 2  7 x  12  0                                         standard form
            ( x  3)( x  4)  0
                                                                        method
                   +                          + OR
                         3         4                    3          4
                                                                       correct solution
             3  x  4 OR x  (3 ; 4)
                                                                                                         (4)
   1.2      x  3 y  5 ……….(1)
            (2 x  y  2)( x  y)  0 ………….(2)
            Substitute (1) into (2):
            (2(3 y  5)  y  2)(3 y  5  y)  0                      correct substitution
            (6 y  10  y  2)(4 y  5)  0
            (5 y  12)(4 y  5)  0                                     factors
                 12            5
            y        or y                                            values of y
                  5            4
                 11              5
            x       or x                                            values of x
                  5              4                                                                       (4)
                                                                                                        [19]

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Mathematics-(Grade 11)                              3                         March 2020 Common Test
                                          NSC – Marking Guideline

QUESTION 2

  2.1                            1
                18  2 2 
                                 2
                                     1                               for 3 2
            = 3 22 2
                                      2
                         1           2
            = 5 2                                                 rationalizing denominator
                    2                2
                    2
            = 5 2
                   2
              11 2
                                                                   answer
                2                                                                                  (3)

  2.2.1     x
                16  128
                                                                       4
                4

            2 2x    7                                               2  27
                                                                       x

            4
              7                                                     equating exponents
            x
               4
            x                                                      answer
               7
                                                                                                   (4)

  2.2.2     32 x 1  32 x 1
                               24
                   3x
            32 x (31  31 )                                         factorising
                                24
                    3x
                 2
            3 x  2   24
                 3                                                simplification
            3 9
              x

            3 x  32                                                like bases
            x2                                                      answer
                                                                                                   (4)

                                                                                               [11]

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Mathematics-(Grade 11)                                       4                               March 2020 Common Test
                                                   NSC – Marking Guideline

QUESTION 3

3.1
          y2  r 2  x2

                5 2   2
                                    2
                                                                                 correct sketch in correct
                21                                         –2
                                                                                   quadrant
              y   21
                                                        y
                                                                 5
                        21
          tan                                                                   value of y
                        2
                           21                                                     answer
                                                                                                                    (3)
                           2
3.2       cos115. cos 214
           cos 65.sin 236
               cos 65.  cos 34                                                 cos 65
          
               cos 65.  sin 56                                                  cos 34 
               cos 34                                                             sin 56 
          
               sin 56 
               cos 34         cos 56 
                      OR                                                      cos 34  OR sin 56 
               cos 34         cos 56 
           1                                                                    1
                                                                                                                     (5)

3.3                                                     
               cos 90  x . tan 540  x . cos 180  x  sin  90        
                                             
           ( sin x). tan 180  x ( cos x)  sin 90                            sin x
                                                                                  tan x
           ( sin x)(tan x)( cos x)  1
                                                                                   cos x
                      sin x 
           ( sin x)       ( cos x)  1                                       1
                      cos x 
                                                                                     sin x
           sin 2 x  1                                                          
                                                                                     cos x
               
            1 sin 2 x                                                         sin 2 x  1
            cos 2 x                                                             answer
                                                                                                                     (7)
                                                                                                                    [15]

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Mathematics-(Grade 11)                                 5                              March 2020 Common Test
                                             NSC – Marking Guideline

QUESTION 4

4.1     AC2  AO2  OC2                 [Pythagoras]                      S/ R
                  10 2  5 2
                  75

         AC  75  8,66 cm                                                value of AC

         AB  2  AC                    [line from centre  to chord]    R

                  17,32 cm                                               answer
                                                                                                             (4)
4.2     TV̂R  Ô1                  [ext.  of cyclic quad.]             S R
                  96     

        Ô 2  360   Ô1          [  s around a point]

             360   96                                                 size of Ô 2

             264                                                       SR
                 1
        R̂ 1      Ô 2             [  at centre = 2   at circumf.]
                 2
                                                                          size of R̂ 1
            
                 1
                 2
                   264 
             132                                                        answer

        R̂ 2  180   R̂ 1         [  s on a straight line]

             180  132
             48
                                                                                                              (7)

4.3        Ŝ1  RV̂T  R̂ 2               [ext.  of RVS ]             S/R
                  96  48     

                  48

          Ŝ1  R̂ 2                  [both  48 ]                     S

        VS  VR                      [sides opp. to =  s]              S/R
       And RVS is isosceles.                                                                                (3)
                                                                                                             [14]

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 Mathematics-(Grade 11)                                    6                         March 2020 Common Test
                                                 NSC – Marking Guideline

 QUESTION 5

5.1     Construction: Draw diameter LOP and join P and M..                  construction
                                     P

                            K                               M

                                     O

                                         2
                                             1
               J                         L                        H

        Proof:
                                                                            S /R
             L̂1  L̂ 2  90    
                                    [tangent  radius]
                 PM̂L  90         [  in a semi-circle]
               P̂  L̂ 2  90        [sum of  s of PML]                  S/R
                    P̂  L̂1
        But:         P̂  K̂        [  s in the same segment]              S R
                    L̂1  K̂                                                                              (5)

5.2.1          B̂3  90                     [  in a semi-circle]         S R
            B̂3  CD̂E                 [both  90  ]
            BCDE is a cyclic quadrilateral                                R
            [converse: ext.  of a cyclic quadrilateral]                                                   (3)

5.2.2    B̂ 2  Â = x                   [tan-chord-theorem]               S R
         B̂ 2  Ĉ1 = x                  [  s in the same segment]        S R
         Ĉ1  Â  x                                                                                     (4)

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 Mathematics-(Grade 11)                                                7                              March 2020 Common Test
                                                             NSC – Marking Guideline

5.2.3                  
         Ê1  180   Â  B̂3                        [sum of  s of ABE ]
                   180  x  90
                            
                                               
                                                    
                   90  x
                        
                                                                                        Eˆ1  90   x
                        
         Ê 3  180   Ĉ1  CD̂E              
                       
             180  x  90
                                       
                                                       [sum of  s of CDE ]
            90  x                                                                    Eˆ 3  90   x
         Ê1  Ê 3                                                                                                        (2)

5.2.4                   
         Ê 2  180   Ê1  Ê 3                     [  s on a straight line]
                       
            180  90  x  90  x
                                                  
                                                        
            2x                                                                         Ê 2  2 x
         Ê 2  Â
                                                                                         Ê 2  Â
         CE is not a tangent to the circle through A, B and E.
        [converse of tan-chord-theorem does not apply]                                                                      (2)

                                                                                                                           [16]

                                                                                                             TOTAL: 75

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