# Mathematics Grade 10 March Test 2020 Paper & Memo

> Grade 10 Mathematics question paper and memo from the March Test 2020, 13 pages. Read online or get the free PDF download.

- Page: https://www.hlayiso.com/library/math-march-qp-memo-2020-gr10/3a696c0c-2807-4079-9ebc-a054a8fedc57
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 10
- Document type: Question paper and memo
- Year: 2020
- Exam period: March Test
- Curriculum: CAPS
- Pages: 13

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Mathematics · Grade 10 · March Test · 2020. Question paper and memorandum, 13 pages. Read online or download the PDF.

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Mathematics 2 Common Test March 2020

NSC

INSTRUCTIONS AND INFORMATION

Read the following instructions carefully before answering the questions.

-

.

This question paper consists of 6 questions.
Answer ALL the questions.

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.

Write neatly and legibly.

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Mathematics 3 Common Test March 2020
NSC

QUESTION 1

1.1 Factorise the following expressions fully:
LLL xy? +43x7y (1)
1.1.2 x°-7x-18 (2)
11.30 x y-164+4y—4x? (3)

1.2 Simplify the following expressions fully:
12.1 (2x-1)(x? -3x+1) (3)
12.2 ee (4)

(x+2)tx(x+2) 2x44

Deen ee
12.30 ————_ 4
gn gre! 9-3" ¢ )
{17]
QUESTION 2
21 Solve for x:
2.1.1 x(2x-5)=0 (2)
2.12 3x’-2x-8=0 GB)
213. 5221-0 (2)
2.14 x=ytxy (3)
8x° -1
2.1.5 =1 4
2x-1 ®
2.2 The following inequality is given: -11<-2x+1<-9; wherex ER.
2.2.1 Solve for x. (3)
2.2.2 Hence, and without the use of a calculator, show that x = 29 would
satisfy the above inequality. (2)
[19]

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Mathematics 4 Common Test March 2020
NSC

QUESTION 3

31 Solve for x and y simultaneously:

2x-y=3
[ 6)
3x+2y =8
3.2 Giventhat M=2°? and M’ =16, determine the value of b. QB)
[8]
QUESTION 4
Various options are provided as possible answers to the following questions.
Write down the question number (4.1 — 4.5) and choose the answer by writing the
letter (A—D) next to the question number (4.1 — 4.5) in your answer book, for
example: 4.6) D
4.1 Which description below does NOT guarantee that a quadrilateral is a square?
A. Quadrilateral is both a rectangle and a rhombus
B. Quadrilateral is a parallelogram with perpendicular diagonals
Cc Quadrilateral has all sides equal and all angles equal
D. Quadrilateral has all right angles and has all sides equal qd)
4.2. Which of the following statements is true?
A. All quadrilaterals are rectangles
B. All quadrilaterals are squares
Cc. All rectangles are quadrilaterals
Dz. All quadrilaterals are parallelograms (1)
43 In the diagram below rectangle KLMN has KM = 6x + 16 and LN = 49.
Find the value of x.
K L
N M
A. x=5,5
B. x= 33
Cc, ee
D x =6,5 Q)

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Mathematics 5 Common Test March 2020
NSC

44 A quadrilateral with only one pair of opposite sides parallel is called a:

A. Trapezium

B. Square

e Kite

D. Rhombus (1)

4.5 Inquadrilateral ABCD, ACD =2x+4 and ACB=5x-11.
For what value of x is ABCD a rhombus?

"

GVaAW>S>
He HOM OH
ll
ADO P

(1)
[5]

Give reasons for your statements in the answers to QUESTIONS 5 and 6.

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Mathematics 6 Common Test March 2020
NSC

QUESTION 5

5.1 In the diagram below, straight lines DEF and TMB are parallel to each other.
It is also given that EM = EB and BEF = 64°,

D

64°

Calculate the size of MEB. (4)

5.2 In the diagram below, PQRS is a parallelogram.
PT = 4x -2, TR=x + 28, ST=4y-7 and TQ=y+ 14.

Ss

Determine, with reasons, the values of x and y. (4)

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Mathematics i Common Test March 2020
NSC

33 In the diagram below, ABCD and BECD are parallelograms with common base DC.
BC LBD and DAB=40° .

A B E
40°

D C

Determine the size of BEC . (4) [12]

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Mathematics 8 Common Test March 2020
NSC

QUESTION 6

6.1 In the diagram below, JKLM is a rectangle. MLPR is a rhombus.

JMK = RMP ; JMK =55° and MRP =70°

R P
Using the diagram, and giving reasons, determine:
6.1.1 MPR (2)
6.1.2 KML ()
6.13 KEP 2)

6.2 In the diagram below rectangle ABCD is given with AP = DN .

B
N
E
P
D
6.2.1 Prove that AABP=ADCN . (4)
6.2.2 Prove that AE = DE. (4)
[14]
GRAND TOTAL: [75]

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                                  NATIONAL
                             SENIOR CERTIFICATE

                                     GRADE 10

                                  MATHEMATICS

                                  COMMON TEST

                                    MARCH 2020

                            MARKING GUIDELINES

     MARKS:            75

     TIME:             1½ hours

                       This marking guideline consists of 5 pages.

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Mathematics                                               2                   March 2020 Common Test
                                               NSC – Marking Guidelines

QUESTION 1

1.1.1    xy 2  3x 2 y
          xy ( y  3x)
                                                                              answer                        (1)
1.1.2    x  7 x  18
           2

                                                                              (x – 9)
          ( x  9)( x  2)                                                   (x + 2)                       (2)
1.1.3    x y  16  4 y  4 x
           2                        2

          x 2 y  4 y  4 x 2  16
                                                                              grouping
          y ( x 2  4)  4( x 2  4)
          ( x 2  4)( y  4)                                                 common bracket
                                                                              answer                        (3)
1.2.1    (2 x  1)( x  3x  1)
                        2

                                                                              correct expansion
          2 x3  6 x 2  2 x  x 2  3x  1                                  – 7x2
          2 x3  7 x 2  5 x  1                                             + 5x                          (3)
1.2.2             x2  1            x 1                                      factorising of
                                                                              D.O.T.S
         ( x  2)  x( x  2) 2 x  4
                                                                              factorising of
            ( x  1)( x  1) 2( x  2)                                         common bracket
                             
            ( x  2)(1  x)       x 1                                        changing 
         2                                                                   answer                        (4)
1.2.3         2 2 n .3 3n
         22 n .4n  1 .12 3n
                     22 n.33n                                               prime bases
         
               22 n.22 n2.(22.3) 3n
                                                                              raising powers
           22 n2 n2 n2.33n
         
                26 n.33n                                                    simplification
          26 n26 n.33n3n
          22
         4                                                                   answer                        (4)
                                                                                                            [17]

2.1.1    x(2 x  5)  0
                               5                                              x=0
          x  0 or x                                                            5
                               2                                              x
                                                                                  2
                                                                                                             (2)
2.1.2    3x  2 x  8  0
               2

         (3x  4)( x  2)  0
                                                                              correct factors
                4                                                                    4
          x   or x  2                                                     x
                3                                                                    3
                                                                              x2                           (3)

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Mathematics                                              3                   March 2020 Common Test
                                              NSC – Marking Guidelines

2.1.3    52 x  1  1  0
         52 x 1  50                                                        50
         2x 1  0
                1
         x                                                                 answer                        (2)
                2
2.1.4    x  y  xy
         x  xy  y
         x(1  y )  y                                                       isolate x terms
                                                                             factorise x
                y
         x
              1 y                                                           answer                        (3)
2.1.5    8x  1
            3
                                                                   1
                  1                          Restriction : x 
         2x 1                                                     2
         (2 x  1)(4 x 2  2 x  1)                                      
                                    1                                        factorising
                  2x 1
         4 x2  2 x  1  1                                                  simplification
         4x  2x  0
              2

         2 x(2 x  1)  0                                                    factors
                                 1
          x  0 or x  
                                 2                                           both x values                 (4)
2.2.1    11  2 x  1  9
         12  2 x  10
                                                                              12  2 x  10
         6   x  5                                                       values
         5  x  6                                                          inequality “flip”             (3)
2.2.2    25  29  36                                                        creating inequality
           25          29     36
                                                                              25      29     36
          5  29  6

         Therefore          29 satisfies the inequality in 2.2.1                                            (2)
                                                                                                           [19]

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Mathematics                                                  4                 March 2020 Common Test
                                                NSC – Marking Guidelines

QUESTION 3

3.1      2x  y  3                    Eq1
         3x  2 y  8                  Eq 2
                                                                               setting up
         Eq1 2 :              4x  2 y  6                                     equations
         Eq 2 :                3x  2 y  8
                                                                               elimination to
         Eq1  Eq 2 :          7x       14
                                                                                solve for x or y
                                      x 2                                     x=2
         Sub x  2 into Eq1: 2(2)  y  3                                      substitution
                                               y  1
                                                                                y 1
                                                y 1                                                        (5)

                                               OR

         2x  y  3                    Eq1
         3x  2 y  8                  Eq 2

                                                                               setting up
         Eq1  Eq3 :                   y  2x  3                               equation
         Sub Eq3 into Eq 2 :           3x  2(2 x  3)  8                     substitution
                                       3x  4 x  6  8
                                       7 x  14
                                                                               x=2
                                       x 2
         Sub x  2 into Eq3 :          y  2(2)  3                            substitution
                                       y 1                                     y 1
                                                                                                            (5)
3.2      M  16
              b

                                                                               16 = 24
         M  2   b     4

         but M  20,2
                     (20,2 )b  24
                                                                               (20,2 )b  24
                     0, 2b  4
                     b  20                                                   answer
                                                                                                            (3)
                                                                                                            [8]

QUESTION 4

4.1      B                                                                     answer                      (1)
4.2      C                                                                     answer                      (1)
4.3      A                                                                     answer                      (1)
4.4      A                                                                     answer                      (1)
4.5      B                                                                     answer                      (1)
                                                                                                            [5]

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Mathematics                                          5                 March 2020 Common Test
                                      NSC – Marking Guidelines

QUESTION 5

5.1       ˆ  64
         BEF                   ( given)
                                                                       S
           ˆ  64
         TBE                  (alt s, DF TB)                         R
            ˆ  64
          EMB                 (s opp  sides)                        S/R
             ˆ  52
          MEB                 ( sum s )                             S/R                          (4)
5.2      4 x  2  x  28      (diagonals of parm)                     S/R
         3x  30
         x  10                                                        x = 10
         4 y  7  y  14      (diagonals of parm)                     S/R
         3 y  21
         y3
                                                                       y=3                          (4)
5.3       ˆ  40
         EBC                   (corresponding s; AD BC )              S
                                                                       R
          ˆ  90
         BCE                   (alternate s; BD EC )                  S/R
            ˆ  50
          BEC                 ( sum s )
                                                                       S/R                          (4)
                                                                                                    [12]

QUESTION 6

6.1.1     ˆ  55
         MPR                   (s opp  sides)                        S
                                                                       R                            (2)
6.1.2     ˆ  35
         KML                   (adj comp s)                           S/R
                                                                                                     (1)
6.1.3     ˆ  90
         MLP                   ( prop of rect )                        S/R
                                                                       S/R
          ˆ  70
         MLP                   (opp s rhombus)
             ˆ  160
          KLP                                                         S                            (3)
6.2.1    In ABP and DCN
                                                                       S/R
         1.    Bˆ  Cˆ  90              ( prop of rect )
         2.          AP  DN              ( given)                     S/R
         3.          AB  DC              (opp sides of rect )         S/R
                   ABP and DCN         ( RHS )                      S/R
                                                                                                     (4)
6.2.2      ˆ  NDC
         BAP     ˆ                        (congruent s proved )       S/R
             ˆ  ADE
          DAE     ˆ                      (adj compl s)               S               R
          AE  DE                        ( sides opp  s )           S/R                          (4)
                                                                                                    [14]

                                                                       TOTAL MARKS:                 75

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